Lectures on Quantum Mechanics: With Problems, Exercises and their Solutions [2ed.] 3319434780, 9783319434780, 9783319434797

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Graduate Texts in Physics

Jean-Louis Basdevant

Lectures on Quantum Mechanics With Problems, Exercises and Their Solutions Second Edition

Graduate Texts in Physics Series editors Kurt H. Becker, Polytechnic School of Engineering, Brooklyn, USA Jean-Marc Di Meglio, Université Paris Diderot, Paris, France Sadri Hassani, Illinois State University, Normal, USA Bill Munro, NTT Basic Research Laboratories, Atsugi, Japan Richard Needs, University of Cambridge, Cambridge, UK William T. Rhodes, Florida Atlantic University, Boca Raton, USA Susan Scott, Australian National University, Acton, Australia H. Eugene Stanley, Boston University, Boston, USA Martin Stutzmann, TU München, Garching, Germany Andreas Wipf, Friedrich-Schiller-Universität Jena, Jena, Germany

Graduate Texts in Physics Graduate Texts in Physics publishes core learning/teaching material for graduate- and advanced-level undergraduate courses on topics of current and emerging fields within physics, both pure and applied. These textbooks serve students at the MS- or PhD-level and their instructors as comprehensive sources of principles, definitions, derivations, experiments and applications (as relevant) for their mastery and teaching, respectively. International in scope and relevance, the textbooks correspond to course syllabi sufficiently to serve as required reading. Their didactic style, comprehensiveness and coverage of fundamental material also make them suitable as introductions or references for scientists entering, or requiring timely knowledge of, a research field.

More information about this series at http://www.springer.com/series/8431

Jean-Louis Basdevant

Lectures on Quantum Mechanics With Problems, Exercises and Their Solutions Second Edition

123

Jean-Louis Basdevant Département de Physique Ecole Polytechnique Palaiseau Cedex France

ISSN 1868-4513 Graduate Texts in Physics ISBN 978-3-319-43478-0 DOI 10.1007/978-3-319-43479-7

ISSN 1868-4521

(electronic)

ISBN 978-3-319-43479-7

(eBook)

Library of Congress Control Number: 2016947037 © Springer International Publishing Switzerland 2007, 2016 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper This Springer imprint is published by Springer Nature The registered company is Springer International Publishing AG Switzerland

Preface

This book is a revised and extended version of my introductory lectures on quantum mechanics that I delivered for many years at the Ecole Polytechnique. It is not a textbook. I was dragged into writing it by friends, among whom are many former students of mine. I have published two books in the same Springer collection with my colleague and friend Jean Dalibard. One is a textbook: Quantum Mechanics, J.-L. Basdevant and Jean Dalibard, Heidelberg: Springer-Verlag, 2002 (revised in 2005). The other one is a collection of problems and their solutions: The Quantum Mechanics Solver, J.-L. Basdevant and Jean Dalibard, Heidelberg: Springer-Verlag, 2000 (completely revised in 2005). All of these problems concern contemporary experimental or theoretical developments, some of which had appeared in the specialized literature a year or so before we gave them as written examinations. Needless to say that if the second of these books is somewhat unusual, there are dozens of excellent textbooks on quantum mechanics, among which some masterpieces which I often consult and refer to. The remarks that eventually convinced me to write the present text are twofold. The textbooks I had written on the subject, both in French and in English, were terribly deprived of life, action, thoughts, and questioning which I always liked to put in the narrative account of the ideas and applications of the subject, during my lectures. The human aspect of the experimental investigations and of the ensuing discovery of basic principles made the lectures lively (besides the fact that the minds need to rest for few minutes after following a difficult argument). I always thought that teaching science is incomplete if it does not incorporate the human dimension, be it of the lecturer, of the audience and of the topic to which it is devoted. The second is that my original publication was totally deprived of any exercise or problem. Very many remarks were that the book helped a lot to understand quantum mechanics, but it did not help much to work out applications. Therefore it has been one of my major goals to fill that void. The present book contains ten

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problems, placed when the tools to solve them have been treated, and followed by their solutions and possible comments. Many of them refer to quite modern physics (there is no overlap with The Quantum Mechanics Solver). It also contains sixty or so shorter exercises whose solutions are given at the end of the book. I must say a few words about the content of this book. First, of course, my lectures evolved quite a lot in 25 years. Actually they were never the same from one year to the next. Minds evolve; student’s minds as well as mine. Science evolves: during that period there appeared numerous crucial experimental and technological steps forward. So each lecture itself must be considered as a superposition of texts and topics, which I could not have covered completely in about an hour and a half. I used to make selections according to my mood, to latest experimental results, to the evolution of the student’s minds in mathematics, in physics, and in regard to the world they were facing. The first lecture always consisted of a general description of contemporary physics and of the various courses that students were offered in their curriculum. I have reproduced an example in Chap. 1. Another point which is amusing and quite characteristic of the French higher educational system (devised more than two centuries ago) is that the students of the Ecole Polytechnique, who were all selected after a stiff entrance examination, and whose ambitions in life were diverse—in science, in industry, in business, and in high public office—all had to follow this introductory physics course. The official reason put forward was that whatever they were going to work on later, Quantum Mechanics and Fundamental physics would be indispensable in their occupation, as would Pure and Applied Mathematics. The famous mathematician Laurent Schwartz, the man I admired most, who was my colleague, liked to be asked the question: what’s the use of doing mathematics? “It’s very simple,” he said. Mathematicians study Lp spaces, negligible sets, and representable functors. One must certainly do mathematics. Because mathematics allows to do physics. Physics allows to make refrigerators. Refrigerators allow to keep Lobsters, and Lobsters are useful for mathematicians who can eat them and therefore be in a good mood to do Lp spaces, negligible sets, and representable functors. It’s obviously useful to do mathematics.

I also tried to attract students to physics. I must admit that when Laurent Schwartz retired, we discussed that point, and we came to the conclusion that obviously it was impossible to convince our students that what we had taught them would be indispensable, even if they were to manage a large company: “I do not manage to persuade them because they know very well that I am objectively wrong […] on that issue I failed completely,” he said. And I fully agreed with him. Nevertheless, quantum mechanics is an ideal subject because one can be interested in it for a variety of reasons such as the physics itself, the mathematical structure of the theory, its technological spinoffs, as well as its philosophical or cultural aspects. And the task was basically to think about the pedagogical aspects, in order to satisfy audiences that went up to 500 students during the last ten years. I do think it is a part of their indisputable personal culture. It seems difficult to grasp the concepts and the functioning of quantum mechanics past a certain age.

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I thank Jean Dalibard, who is now my successor, and Philippe Grangier for their constant help during the last 10–15 years. They are in particular responsible for part of the text on quantum entanglement and Bell’s inequalities, of which they are worldwide known specialists. I am deeply grateful to James Rich and to Alfred Vidal-Madjar. Both of them contributed immensely on all of this book. Discussing with them was an everlasting pleasure, and they taught me a lot of physics. I want to thank Jean-Michel Bony and to pay a tribute to the memory of Laurent Schwartz. Both had the patience to explain to me with an incredible profoundness and clarity the mathematical subtleties of quantum mechanics. This enabled me to eliminate most of the unnecessary mathematical complications at this stage, and still to be able to answer the questions of my more mathematically minded students. Indeed, if quantum mechanics has been a rich field of investigation for mathematicians, it is really the physics that is subtle in it. Paris, France

Jean-Louis Basdevant

Contents

1

The Appeal of Physics . . . . . . . . . . . . . . . . . . . . . . . . 1.1 The Interplay of the Eye and the Mind . . . . . . 1.2 Advanced Technologies . . . . . . . . . . . . . . . . . 1.3 The Pillars of Contemporary Physics . . . . . . . 1.3.1 Mysteries of Light . . . . . . . . . . . . . . 1.3.2 Fundamental Structure of Matter . . . 1.4 The Infinitely Complex . . . . . . . . . . . . . . . . . . 1.5 The Universe . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Physical Constants . . . . . . . . . . . . . . . . . . . . .

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A Quantum Phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Wave Behavior of Particles. . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Interferences . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Wave Behavior of Matter . . . . . . . . . . . . . . . . . 2.1.3 Analysis of the Phenomenon . . . . . . . . . . . . . . . 2.2 Probabilistic Nature of Quantum Phenomena . . . . . . . . . . 2.2.1 Random Behavior of Particles . . . . . . . . . . . . . . 2.2.2 A Nonclassical Probabilistic Phenomenon . . . . . 2.3 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Appendix: Notions on Probabilities . . . . . . . . . . . . . . . . . 2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Wave Function, Schrödinger Equation . . 3.1 Terminology and Methodology . . . . 3.2 Principles of Wave Mechanics . . . . 3.2.1 The Wave Function . . . . . 3.2.2 Schrödinger Equation. . . . 3.3 Superposition Principle . . . . . . . . . . 3.4 Wave Packets . . . . . . . . . . . . . . . . . 3.4.1 Free Wave Packets . . . . . 3.4.2 Fourier Transforms . . . . . 3.4.3 Shape of Wave Packets . .

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4

Physical Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Statement of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Physical Quantities . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Position and Momentum . . . . . . . . . . . . . . . . . . 4.2 Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Position Observable . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Momentum Observable . . . . . . . . . . . . . . . . . . . 4.2.3 Correspondence Principle . . . . . . . . . . . . . . . . . 4.2.4 Historical Landmarks . . . . . . . . . . . . . . . . . . . . . 4.3 A Counterexample of Einstein and Its Consequences . . . . 4.3.1 What Do We Know After a Measurement? . . . . 4.3.2 Eigenstates and Eigenvalues of an Observable. . 4.3.3 Wave Packet Reduction . . . . . . . . . . . . . . . . . . . 4.4 The Specific Role of Energy . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 The Schrödinger Equation, Time and Energy . . 4.4.3 Stationary States . . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 Motion: Interference of Stationary States . . . . . . 4.5 Schrödinger’s Cat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Energy Quantization . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Bound States and Scattering States . 5.1.2 One-Dimensional Problems . . . . . . . 5.2 The Harmonic Oscillator . . . . . . . . . . . . . . . . . 5.3 Square Well Potentials . . . . . . . . . . . . . . . . . . 5.4 Double Well, the Ammonia Molecule . . . . . . . 5.4.1 The Model . . . . . . . . . . . . . . . . . . . . 5.4.2 Stationary States, the Tunnel Effect . 5.4.3 Energy Levels . . . . . . . . . . . . . . . . . 5.4.4 Wave Functions . . . . . . . . . . . . . . . . 5.4.5 Inversion of the Molecule . . . . . . . .

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3.7 3.8 3.9 3.10 3.11 3.12

Historical Landmarks . . . . . . . . . . . . . . . . . . . Momentum Probability Law . . . . . . . . . . . . . . 3.6.1 Free Particle. . . . . . . . . . . . . . . . . . . 3.6.2 General Case . . . . . . . . . . . . . . . . . . Heisenberg Uncertainty Relations . . . . . . . . . . Controversies and Paradoxes. . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix: Dirac δ “Function”, Distributions. . Appendix: Fourier Transformation . . . . . . . . . 3.11.1 Uncertainty Relation . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

5.5 5.6 5.7 5.8

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Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . 6.1 Hilbert Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Dirac Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.3 Syntax Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.4 Projectors; Decomposition of the Identity . . . . . 6.3 Measurement Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Eigenvectors and Eigenvalues of an Observable 6.3.2 Results of the Measurement of a Physical Quantity . . . . . . . . . . . . . . . . . . . . 6.3.3 Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.4 The Riesz Spectral Theorem . . . . . . . . . . . . . . . 6.3.5 Physical Meaning of Various Representations . . 6.4 Principles of Quantum Mechanics . . . . . . . . . . . . . . . . . . . 6.5 Heisenberg’s Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 The Polarization of Light, Quantum “Logic” . . . . . . . . . . 6.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Two-State Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 The NH3 Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 “Two-State” System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Matrix Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 NH3 in an Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Uniform Constant Field . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 Weak and Strong Field Regimes . . . . . . . . . . . . . . . . 7.4.3 Other Two-State Systems. . . . . . . . . . . . . . . . . . . . . . 7.5 Motion of Ammonia Molecule in an Inhomogeneous Field . . . 7.5.1 Force on the Molecule in an Inhomogeneous Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.2 Population Inversion . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Reaction to an Oscillating Field, the Maser . . . . . . . . . . . . . . . 7.7 Principle and Applications of the Maser . . . . . . . . . . . . . . . . . . 7.7.1 Amplifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.2 Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.3 Atomic Clocks and the GPS . . . . . . . . . . . . . . . . . . . 7.7.4 Tests of Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . .

147 148 148 151 154 155 156 157 158

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Illustrations and Applications of the Tunnel Effect . Tunneling Microscopy, Nanotechnologies . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem. The Ramsauer Effect . . . . . . . . . . . . . . . . 5.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . Problem. Colored Centers in Ionic Cristals . . . . . . . 5.9.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . .

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7.8 7.9

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9

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem: Neutrino Oscillations . . . . . . . . . . . . . . . . . . . . . 7.9.1 Mechanism of the Oscillations; Reactor Neutrinos . . . . . . . . . . . . . . . . . . . . . . . 7.9.2 Oscillations of Three Species; Atmospheric Neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.9.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.9.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Approximation Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Definition of the Problem . . . . . . . . . . . . . . . . . . . . . 9.1.2 First Order Perturbation Theory . . . . . . . . . . . . . . . . . 9.1.3 Second Order Perturbation to the Energy Levels . . . . 9.2 The Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Algebra of Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Commutation of Observables . . . . . . . . . . . . . . . . . 8.1.1 Fundamental Commutation Relation . . . . 8.1.2 Other Commutation Relations . . . . . . . . . 8.1.3 Dirac in the Summer of 1925 . . . . . . . . . 8.2 Uncertainty Relations . . . . . . . . . . . . . . . . . . . . . . . 8.3 Evolution of Physical Quantities . . . . . . . . . . . . . . . 8.3.1 Evolution of an Expectation Value . . . . . 8.3.2 Particle in a Potential, Classical Limit . . . 8.3.3 Conservation Laws . . . . . . . . . . . . . . . . . 8.4 Algebraic Resolution of the Harmonic Oscillator . . 8.5 Commuting Observables . . . . . . . . . . . . . . . . . . . . . 8.5.1 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2 Example. . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Tensor Structure of Quantum Mechanics . 8.5.4 Complete Set of Commuting Observables (CSCO) . . . . . . . . . . . . . . . . 8.5.5 Completely Prepared Quantum State . . . . 8.6 Sunday September 20, 1925 . . . . . . . . . . . . . . . . . . 8.7 Exercices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 Problem. Quasi-Classical States of the Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 Problem. Benzene and C8 Molecules . . . . . . . . . . . 8.9.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 8.10 Problem. Conductibility of Crystals; Band Theory . 8.10.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . .

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10 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Fundamental Commutation Relation . . . . . . . . . . . . . . . . . . . . . 10.1.1 Classical Angular Momentum . . . . . . . . . . . . . . . . . . 10.1.2 Definition of an Angular Momentum Observable . . . 10.1.3 Results of the Quantization . . . . . . . . . . . . . . . . . . . . 10.2 Proof of the Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 Statement of the Problem. . . . . . . . . . . . . . . . . . . . . . 10.2.2 Vectors jj; m [ and Eigenvalues j and m . . . . . . . . . J x  i^ Jy . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Operators ^J  ¼ ^ 10.2.4 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Orbital Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.1 Formulae in Spherical Coordinates . . . . . . . . . . . . . . 10.3.2 Integer Values of m and ‘ . . . . . . . . . . . . . . . . . . . . . 10.3.3 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Rotation Energy of a Diatomic Molecule . . . . . . . . . . . . . . . . . 10.5 Interstellar Molecules, the Origin of Life . . . . . . . . . . . . . . . . . 10.6 Angular Momentum and Magnetic Moment . . . . . . . . . . . . . . . 10.6.1 Classical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.2 Quantum Transposition . . . . . . . . . . . . . . . . . . . . . . . 10.6.3 Experimental Consequences . . . . . . . . . . . . . . . . . . . . 10.6.4 Larmor Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.5 What About Half-Integer Values of j and m? . . . . . . 10.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

229 230 230 230 231 231 231 233 233 235 236 236 236 237 239 241 245 246 247 248 249 250 250

11 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Two-Body Problem; Relative Motion . . . . . . . . . . . 11.2 Motion in a Central Potential . . . . . . . . . . . . . . . . . 11.3 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . 11.3.1 Atomic Units; Fine Structure Constant . . 11.3.2 The Dimensionless Radial Equation . . . . 11.3.3 Spectrum of Hydrogen . . . . . . . . . . . . . . . 11.3.4 Stationary States of the Hydrogen Atom . 11.3.5 Dimensions and Orders of Magnitude . . . 11.3.6 Historical Landmarks . . . . . . . . . . . . . . . . 11.4 Muonic Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Problem. Decay of a Tritium Atom . . . . . . . . . . . . 11.6.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . .

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12 Spin 1/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Experimental Results . . . . . . . . . . . . . . . . . . . 12.2 Spin 1/2 Formalism . . . . . . . . . . . . . . . . . . . . 12.3 Complete Description of a Spin 1/2 Particle . . 12.3.1 Observables . . . . . . . . . . . . . . . . . . . 12.4 Physical Spin Effects . . . . . . . . . . . . . . . . . . .

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12.5 12.6 12.7

Spin Magnetic Moment. . . . . . . . . . . . . . . . . . . . . . . . . . . The Stern–Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . Principle of the Experiment . . . . . . . . . . . . . . . . . . . . . . . 12.7.1 Semi-classical Analysis . . . . . . . . . . . . . . . . . . . 12.7.2 Experimental Results . . . . . . . . . . . . . . . . . . . . . 12.7.3 Explanation of the Stern–Gerlach Experiment . . 12.7.4 Successive Stern–Gerlach Setups . . . . . . . . . . . . 12.7.5 Measurement Along an Arbitrary Axis . . . . . . . 12.8 The Discovery of Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.8.1 The Hidden Sides of the Stern–Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.8.2 Einstein and Ehrenfest’s Objections . . . . . . . . . . 12.8.3 Anomalous Zeeman Effect . . . . . . . . . . . . . . . . . 12.8.4 Bohr's Challenge to Pauli . . . . . . . . . . . . . . . . . 12.8.5 The Spin Hypothesis . . . . . . . . . . . . . . . . . . . . . 12.8.6 The Fine Structure of Atomic Lines . . . . . . . . . 12.9 Magnetism, Magnetic Resonance . . . . . . . . . . . . . . . . . . . 12.9.1 Spin Effects, Larmor Precession . . . . . . . . . . . . 12.9.2 Larmor Precession in a Fixed Magnetic Field . . 12.9.3 Rabi's Calculation and Experiment . . . . . . . . . . 12.9.4 Nuclear Magnetic Resonance . . . . . . . . . . . . . . . 12.9.5 Magnetic Moments of Elementary Particles . . . . 12.10 Entertainment: Rotation by 2π of a Spin 1/2 . . . . . . . . . . 12.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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294 296 297 298 298 299 300 301 301 302 306 307 308 310

13 Addition of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Addition of Angular Momenta . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.1 A Simple Case: The Addition of Two Spins 1/2 . . . . 13.1.2 Addition of Two Arbitrary Angular Momenta . . . . . . 13.2 One-Electron Atoms, Spectroscopic Notations . . . . . . . . . . . . . 13.2.1 Fine Structure of Monovalent Atoms . . . . . . . . . . . . . 13.3 Hyperfine Structure; The 21 cm Line of Hydrogen . . . . . . . . . . 13.4 Radioastronomy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 The 21-cm Line of Hydrogen . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6 The Intergalactic Medium; Star Wars . . . . . . . . . . . . . . . . . . . . 13.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

313 313 315 318 322 323 325 330 333 336 341

14 Identical Particles, the Pauli Principle . . . . . . . . . . . . . . . . . . . . 14.1 Indistinguishability of Two Identical Particles . . . . . . . . . 14.1.1 Identical Particles in Classical Physics . . . . . . . . 14.1.2 The Quantum Problem . . . . . . . . . . . . . . . . . . . . 14.1.3 Example of Ambiguities . . . . . . . . . . . . . . . . . . 14.2 Two-Particle System; The Exchange Operator . . . . . . . . . 14.2.1 The Hilbert Space for the Two-Particle System .

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14.2.2

14.3

14.4

14.5 14.6

The Exchange Operator Between Two Identical Particles . . . . . . . . . . . . . . . . . . . 14.2.3 Symmetry of the States . . . . . . . . . . . . . . . . . . . The Pauli Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.3.1 The Case of Two Particles . . . . . . . . . . . . . . . . . 14.3.2 Independent Fermions and Exclusion Principle . 14.3.3 The Case of N Identical Particles . . . . . . . . . . . Physical Consequences of the Pauli Principle . . . . . . . . . . 14.4.1 Exchange Force Between Two Fermions . . . . . . 14.4.2 The Ground State of N Identical Independent Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4.3 Behavior of Fermion and Boson Systems at Low Temperatures . . . . . . . . . . . . . . . . . . . . . 14.4.4 Stimulated Emission and the Laser Effect . . . . . 14.4.5 Uncertainty Relations for a System of N Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . Problem: Discovery of the Pauli Principle . . . . . . . . . . . . 14.5.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem. Heisenberg Relations for Fermions. The Way to Macroscopic Systems . . . . . . . . . . . . . . . . . . 14.6.1 Uncertainty Relations for N Fermions . . . . . . . . 14.6.2 White Dwarfs and the Chandrasekhar Mass . . . 14.6.3 Neutron stars . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6.4 Mini-boson Stars . . . . . . . . . . . . . . . . . . . . . . . . 14.6.5 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Lagrangian Formalism and the Least Action Principle . . . . . . . 15.2 Canonical Formalism of Hamilton and Jacobi. . . . . . . . . . . . . . 15.3 Analytical Mechanics and Quantum Mechanics . . . . . . . . . . . . 15.4 Classical Charged Particle in an Electromagnetic Field . . . . . . . 15.5 Lorentz Force in Quantum Mechanics . . . . . . . . . . . . . . . . . . . 15.5.1 Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.2 Gauge Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.3 The Hydrogen Atom Without Spin in a Uniform Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5.4 Spin 1/2 Particle in an Electromagnetic Field . . . . . . 15.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.7 Problem. Landau Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.7.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

385 386 389 391 393 394 394 394 396 397 397 399 400

16 The Evolution of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403 16.1 Time-Dependent Perturbation Theory . . . . . . . . . . . . . . . . . . . . 403 16.2 Interaction of an Atom with an Electromagnetic Wave . . . . . . . 407

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16.4

16.5 16.6

16.2.1 The Electric Dipole Approximation . . . . . . . . . . 16.2.2 Justification of the Electric Dipole Interaction . . 16.2.3 Absorption of Energy by an Atom . . . . . . . . . . 16.2.4 Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.5 Spontaneous Emission . . . . . . . . . . . . . . . . . . . . Decay of a System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.1 The Radioactivity of 57 Fe . . . . . . . . . . . . . . . . . 16.3.2 The Fermi Golden Rule . . . . . . . . . . . . . . . . . . . 16.3.3 Orders of Magnitude . . . . . . . . . . . . . . . . . . . . . 16.3.4 Behavior for Long Times . . . . . . . . . . . . . . . . . The Time-Energy Uncertainty Relation . . . . . . . . . . . . . . . 16.4.1 Isolated Systems and Intrinsic Interpretations . . 16.4.2 Interpretation of Landau and Peierls . . . . . . . . . 16.4.3 The Einstein–Bohr Controversy . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem. Molecular Lasers . . . . . . . . . . . . . . . . . . . . . . . . 16.6.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 Entangled States. The Way of Paradoxes . . . . . . . . . . . . 17.1 The EPR Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 The Version of David Bohm. . . . . . . . . . . . . . . . . . 17.2.1 Bell's Inequality . . . . . . . . . . . . . . . . . . . . 17.2.2 Experimental Tests . . . . . . . . . . . . . . . . . 17.3 The GHZ Experiment . . . . . . . . . . . . . . . . . . . . . . . 17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment . . . . . . . . . . . . . . . . . . . . . . . 17.5 The Quantum Computer . . . . . . . . . . . . . . . . . . . . . 17.6 Quantum Teleportation . . . . . . . . . . . . . . . . . . . . . . 17.7 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . .

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18 Solutions to the Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495

Chapter 1

The Appeal of Physics

1.1 The Interplay of the Eye and the Mind Physics is a fascinating adventure between the eye and the mind, between the world of phenomena and the world of ideas. Physicists look at Nature and ask questions to which they try and imagine answers. And the interplay, often the quarrel, between what we observe and what we construct “logically” is always a source of amazement. A discovery is often the seed of creation of new and unthought ideas to explore. For instance: why do stars shine? It’s important. The sun is an ordinary star, similar to 80 % of the 200 billion stars of our galaxy. But it is unique and incomparable, because it is our star. In mass, the sun is made of 75 % hydrogen and 25 % helium (actually a plasma of electrons and nuclei). Its parameters are radius R = 700,000 km, mass M = 2 1030 kg, power (luminosity) L = 4 1023 kW, surface temperature T = 6000 K. One mustn’t overestimate the power of the sun. We are much more efficient. If you calculate the power-to-mass ratio, the sun has a score of 0.2 mW/kg which is very small. We consume on the average 2000 kilo-calories per day; that is, 100 W, 25 % of which is used by the brain. Our brain has a power of 25 W! It is consequently 10,000 times more powerful than the sun for a given mass! One always tells kids they are brilliant, but without explaining why, and where that can lead them. So, why does the sun shine? One usually thinks that it shines because of the powerful thermonuclear reactions that take place inside it. But that is not really true! I want to show you that, contrary to common prejudices, it is gravitation that makes stars shine and that thermonuclear reactions cool them permanently. • Stars shine because they are hot and any hot body radiates energy. • They are hot because of gravity. Stars are huge masses of gas, mainly hydrogen, which are strongly compressed by the inward pressure of their own weight. This © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_1

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brings them to high temperatures. Stars are self-gravitating systems in equilibrium under their own weight. • OK, but you may object that a hot compressed gas loses energy by radiating. If it loses energy, then it contracts and it cools down. • Well, the amazing thing is that, on the contrary, a self-gravitating gas does contract when it loses energy, but its temperature increases! This is understandable. If the size of a self-gravitating system decreases, the gravitational inward pressure increases and, in order to maintain equilibrium, the thermal outward pressure increases, (the mean energy of the constituents of the gas must increase). Because temperature reflects motion (i.e., kinetic energy of the constituents of a gas), if the particles move faster, the temperature increases. Therefore, if a selfgravitating system contracts, its temperature increases. This is quite easy to formulate. A star such as the sun can be represented by an ideal gas of N ∼ 1057 particles, at some average temperature T . The temperature itself varies from 15 million degrees in the center to 6000 degrees on the surface. • The gravitational potential energy of a sphere of mass M and radius R is proportional to Newton’s constant, to the square of the mass, and to the inverse of the radius E G = −γG M 2 /R, where γ is a dimensionless constant of order 1 (γ = 3/5 if the mass distribution is uniform in the sphere). The potential energy is negative because one must give energy to the system in order to dissociate it. • In a self-gravitating system, the total kinetic energy E kin of the orbiting particles, that is, the internal energy U of the gas, is equal to half of the absolute value of the potential energy: U = E kin = 1/2|E G |. This is obvious for circular orbits around a massive center, and it can be generalized with no great difficulty. • Therefore, the total energy of the star is E = E kin + E G = −(1/2)γG M 2 /R . Again, it is negative because the system is bound. One must bring energy in order to dissociate it. • On the other hand, the average temperature T  of the star is related to the mean kinetic energy of the constituents by Boltzmann’s relation (3/2)N kT  = E kin = (1/2)γG M 2 /R, with N ∼ 1057 . • When the star radiates, it loses energy. Its energy decreases and becomes more negative, therefore its radius decreases, it is compressed, and its temperature increases. When the star loses energy it radiates more and more strongly. Therefore, stars shine because of gravitation. Since it was formed in a molecular cloud, the sun, whose present mass and radius we know, has lost a gravitational energy of ΔE  1041 J; its average temperature is of the order of T   3 million degrees, which is quite acceptable. Now, we must think! Paleontologists teach us the following. • The “blue-green algae” or cyanobacteria, who are responsible for the birth of life on earth because they manufactured the oxygen in the atmosphere, existed 3.5 billion years ago.

1.1 The Interplay of the Eye and the Mind

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• Our cousin, the Kenyapithecus, lived 15 million years ago, and our ancestors Lucy, an Ethiopian Australopithecus afarensis, as well as her cousin Abel in Chad, lived 3.5 million years ago. Orrorin (the ancestor of the millennium) lived 6 million years ago in Kenya, and the present champion, Toumaï lived in Chad 7 million years ago. • Dinosaurs lived 200 million years ago; they consumed a lot of greenery. Therefore, the sun must have been stable during all that time. It must have had approximately the same power (the weather stayed roughly the same) and the same external temperature (the sun radiates in the visible part of the spectrum where photosynthesis takes place, allowing vegetables to grow). Now, if the sun was stable, we can evaluate roughly when it started to shine. If it had the same power L for a long time, we can evaluate the time that it took to get rid of the energy ΔE ∼ 1041 J, that is, a time t = ΔE/L ∼ 10 million years. The sun started shining 10 million years ago. Therefore, the sun is only 10 million years old! Consequently, we have just proven scientifically that dinosaurs never existed; they were simply invented to make Jurassic Park. The Kenyapithecus was just invented to give us a superiority complex. Because the sun did not shine at that time! Something is wrong in our above reasoning. Of course it may be that, in order to save energy, the Creator turned on the sun once in a while, just when it was necessary for archaeologic discoveries. Since it is not my purpose here to inspire religious vocations, we must find something else. Actually, the answer is right in front of us. Suppose there is a source of energy in the sun, and that, at some temperature, something ignites in the gas. The combustion releases energy. It increases the energy of the gas, which becomes less negative. Therefore the gas expands, which is understandable. But if its radius increases, then its temperature decreases! If a self-gravitating gas loses energy, its temperature increases; if it gains energy, its temperature decreases. It has a negative specific heat. And that’s great. A combustion stabilizes the star’s temperature. An excess of combustion cools the gas and slows down combustion. Conversely, an insufficient combustion rate heats the gas and revives combustion. The combustion energy is peacefully evacuated at constant temperature. The system is self-regulated. The energy we receive is indeed due to thermonuclear reactions, but as long as the combustion lasts, the star evacuates that energy in stable conditions. That is exactly what we need for blue-green algae, dinosaurs, Lucy, Orrorin and their companions! So, our star is stable, but for how long? As long as the fuel is not exhausted. With the mass and power of the sun, one can check that if the combustion were chemical, for instance, 2H → H2 + 4, 5 eV, that is, 2 eV per proton, the available energy would be 1038 J; the lifetime of the sun would be at most 30,000 years, which is much too short. On the contrary, nuclear fusion reactions such as 4 p →4 H e + 27 MeV, that is,  7,000,000 eV per proton,

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1 The Appeal of Physics

is a million times more energetic, which leads us to roughly ten billion years. And we have made in three pages a theory of the sun which is not bad at all in first approximation! The conclusion is that stars shine because of gravitation, which compresses them and heats them. Nuclear reactions, which should make them explode, simply allow them to react against gravitational collapse. They cool down the stars permanently and give them a long lifetime. The sun has been shining for 4.5 billion years and will continue to do so for another 5 billion years. That is an example of the confrontation of a physicist’s ideas with the observed world. And that’s what is interesting in physics. If the ideas we have do not correspond to what we see, we must find other ideas. One cannot change Nature with speeches. In physics, one can make mistakes but one cannot cheat. One can do lots of things with speeches. In 1894, Edwin J. Goodwin, a country doctor in Indiana, found that the number π was too complicated. So he decided that from then on, π should be equal to 3 (π version 3.0),1 which is much simpler for everyone. Well, that doesn’t work so well! One can observe that if π were equal to 3, four inches of tires would be missing on bicycles, which would be uncomfortable, and five inches of stripes would be missing on a French colonel’s hat, which would be inelegant. However, Goodwin convinced a state representative who introduced a bill “The House bill No. 246, Indiana State Legislature 1897” which decided that from then on the number π should be equal to 3. The full House passed the bill by a vote of 67 to 0. At the Indiana Senate, the bill was nearly passed, but one senator observed that the General Assembly lacked the power to define mathematical truth. He added that he thought consideration of such a proposition was not dignified or worthy of the Senate. He moved the indefinite postponement of the bill, and the motion carried. Politicians learn to make speeches and scientists learn to use their intelligence. It is a radically different way of thinking. The two methods happen to be useful in practice: there exist scientists who can explain their findings, and there exist intelligent politicians.

1.2 Advanced Technologies There are many other reasons to learn physics, of course. Our world is filled with advanced technologies such as the Internet, GPS, optoelectronics, nanotechnologies, and so on. Many of these new technologies come from the results of fundamental research obtained in the last 10 or 20 years, sometimes in very recent years. Very many of them are tightly linked with quantum mechanical techniques, and we will cross a few of them in the course of this book. 1 Actually

the figure he proposed was 3.2, but it’s less fun.

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5

Fig. 1.1 Left Forest of microlasers, each of which is a pile of pancakes of alternating slices of GaAs and GaAl semiconductors. The diameter of each element is 0.5 µm, the height is 7 µm. (Courtesy Emmanuel Rosencher) Right Scanning electron microscope (SEM) image of quantum dots fabricated through electron beam lithography on a bidimensional GaAl layer. These structures are used to study the behavior of electrons, which are confined into tiny spaces approximately 10 electrons per dot. The diameter of each quantum dot is 200 nm (Image: C.P. Garcia, V. Pellegrini, NEST (INFM), Pisa.)

Figure 1.1 shows two examples of nanotechnology components. On the left, the details of a sample of microelectronics. This was made in the 1990s. It consists of a forest of microlasers each of which is a pile of 10 nm pancakes of alternating slices of gallium–arsenide and gallium–aluminum semiconductors. We come back to such devices. These components have numerous applications in infrared technologies. Infrared sensors are used as temperature sensors for night vision, on automobiles to see pedestrians at night, in rescuing operations in the ocean, to measure the temperature of the earth and of the ocean from satellites, in telecommunications with fiber optics, and so on. On the right, there is a much more recent arrangement of quantum dots fabricated in 2010 through electron beam lithography and subsequent dry-chemical etching on a quasi bidimensional layer (GaAl heterostructure) by C.P. Garcia, V. Pellegrini in Pisa. These structures are used to study the behavior of electrons, which are confined into tiny spaces approximately 10 electrons per dot. The diameter of each quantum dot is 200 nm (which means that a billion of these structures easily fit on the tip of your finger). What is really amazing is the size. The size of each individual elements, of the order of 10 nm, is that of a virus, the smallest living object. In order to imagine the order of magnitude, if instead of making lasers or dots one made letters (which is quite possible, even though it may seem ridiculous) one could write and read on 1 mm2 of silicon, the complete works of Sigmund Freud, Carl von Clausewitz, Karl Marx, Shakespeare, Snoopy, Charlie Brown, Anaïs Nin and Leo Tolstoy etc. (which may be useful during a boring lecture). These components are called quantum components because in order to conceive, to manufacture, and to use such components, one cannot bypass quantum physics.

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1 The Appeal of Physics

Micro- and nanotechnologies are undergoing tremendous progress at present. In electronics, one of the present world records consists of a transistor that is 18 nm long, a hundred times smaller than the smallest present transistors. One could put three billion such transistors on a dime. It is the physical lower limit due to the Heisenberg inequalities. One builds automated microsystems that possess the three functions of being sensors, of processing information, and of activating a reaction or a response. Such systems are found in all sectors of technology, from electronic equipment of cars up to medicine, including telecommunications, computers, and space technologies. Medical researchers are developing customized nanoparticles the size of blood cells that can deliver drugs directly to diseased cells in the body. In the early 2000s usage for cancer detection and treatment has become a systematic and incredibly successful direction. Nanotechnology increases the capabilities of electronics devices while they reduce their weight and power consumption. Nanotechnological solar cells can be manufactured at significantly lower cost than conventional solar cells. Nanotechnologies may make space-flight more practical. They are used as chemical sensors for a variety of applications, to improve the quality of air and water.2 One can multiply the number of such examples. Here again, whatever one’s own perspectives are, one must be familiar with such developments, be it only to position oneself in front of them. One must be capable of inventing and acting.

1.3 The Pillars of Contemporary Physics In order to understand contemporary physics, three fundamental links are necessary: quantum mechanics, statistical mechanics, and relativity. Quantum mechanics, which is dealt with in this book, is the complete and fundamental theory of structures and processes at the microscopic scale, that is, atomic, molecular, or nuclear scales. It is the fundamental and inescapable field. All physics is quantum physics. The first success of quantum mechanics is to explain the structure of matter, atoms and molecules. But it is in the interaction of atoms and molecules with radiation that one finds the greatest progress, both fundamental and technological, in recent years.

1.3.1 Mysteries of Light Light has always been considered as a physical phenomenon of its own, the great mystery. It is our first tool to explore the world, to probe the cosmos as well as the infinitely small. In physics, light delivered simultaneously the two great discoveries of the 20th century: relativity with Michelson, Einstein, Lorentz, and Poincaré in 1905,

2 See

for instance http://www.understandingnano.com/nanotech-applications.html.

1.3 The Pillars of Contemporary Physics

7

and quantum physics with the black-body theory of Planck in 1900 and Einstein’s photon in 1905. The nature of light has always been a mysterious and fundamental question. The first theory of light originated from the importance given to light rays. Just look at Fig. 1.2. This drawing seems quite ordinary, not at all scientific. Fifty percent of children draw the sun that way. But 50 % is extraordinary, because you have never seen those light rays attached to the sun. This child cannot explain why she drew them, but for everybody their presence is perfectly natural. In nature, one can see light rays only under special circumstances, when light is partially screened by clouds or trees. And the fact that light rays are straight, and that they materialize the perfect straight lines of geometry was always considered as fundamental. For thousands of years, a sacred character was attributed to light rays, as one can see in Fig. 1.3. In Egyptian as well as in Christian culture, light rays are a medium through which the beyond becomes accessible to humans. In the 18th century, Newton decided that light was made of corpuscles, because only particles can travel along straight lines. However, since the end of the 17th century, interference and diffraction phenomena were known and the 19th century saw the triumph of wave optics. Nobody could imagine the incredible answer of quantum theory. Einstein understood in 1905 that light was both wavelike and corpusclelike. Quantum optics, that is, the quantum description of electromagnetic radiation, also plays a decisive role in modern science and technology. The interaction between radiation and matter has produced laser physics. Lasers beams are the modern legendary light rays. The manipulation of cold atoms with laser beams is one of the highlights of present fundamental research. There are numerous practical applications: inertial controlled fusion, optoelectronics, gyrolasers, and others. Intensive work is carried out on optical computers.

Fig. 1.2 Child’s drawing

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1 The Appeal of Physics

Fig. 1.3 Left Stele of Taperet (around 900–800 B.C.) Taperet worships the sun god Horakhty whose rays are materialized by lily flowers of all colors (Le Louvre Museum, Paris.) Right Il Sodoma, Saint Sebastian (1526) (Galleria Pitti, Florence.)

1.3.2 Fundamental Structure of Matter Elementary particle physics started a bit more than one century ago with the discovery of the electron by J.J. Thomson in 1897. It tries to answer two questions: • What is the world made of? • How does the world work? In one century, one has found a nearly complete answer. At present, we possess a simple theory of the Universe, called the Standard model, in which a small number of elementary constituents of matter, quarks and leptons, interact through a simple set of forces. And that theory explains all natural phenomena! In October 1989, a measurement, done in the CERN LEP collider in Geneva, allowed us to count the number of different constituents of matter. There are 24 of them. The validity of the Standard model is constantly verified experimentally more and more accurately. The next to last elements, the top quark and the τ neutrino, were observed respectively in 1995 and 2001. The discovery of the Higgs boson at the CERN Large Hadron Collider facility in June 2012 was a celebrated event. Many physicists consider the Standard model to be very close to the end of the story in the infinitely small structure of matter, and, for the moment, there is no experimental evidence against that. It is a problem of esthetics and a semi-metaphysical problem, namely the whereabouts of the big bang. Matter is made of atoms. In 1910, Rutherford discovered that atoms are made of tiny but heavy nuclei bound to electrons by electromagnetic forces. In the 1930s, people showed that nuclei also have an internal structure. They are systems of nucleons (protons and neutrons), bound by nuclear forces of small range and large intensity. Then, in the 1960s, people understood that nucleons are not elementary either. They

1.3 The Pillars of Contemporary Physics

9

have an internal structure: they are systems of three quarks. There are two sorts of quarks, the u (up) quark of charge +2/3 and the d (down) quark of charge −1/3. The proton is a (uud) system, and the neutron a (udd) system. Quarks are imprisoned against each other by “gluons”. What is amazing in the Standard model is that apparently quarks and leptons (electrons, neutrinos, etc.) are experimentally pointlike. “After” them, there is nothing else. Electrons and quarks are elementary down to 10−18 m. They are the true elements of matter. Actually, this end of the story is a problem. The model works too well! Pointlike objects are not consistent with what we know from quantum field theory or from general relativity. At very short distances, it seems that the notion of particles must be replaced by some other concept: superstrings, which are extended objects. This is one of the major problems of fundamental physics. This problem is related to something we have not yet mastered, unifying general relativity, which is primarily a geometrical theory, with quantum mechanics which is basically nongeometrical. In this problem, we might find the answer to fascinating questions such as: why is the dimensionality of space equal to three? The answer is probably that actually there are more dimensions, ten altogether, but that the additional ones cannot be seen with the naked eye. Like a bug on a straw, it seems that the bug moves up and down on a one-dimensional space, the straw, but the bug itself knows that it can also turn around along the surface of the straw, and its world is two-dimensional. Nuclear physics (i.e., the physics of atomic nuclei) is a beautiful and complex fundamental field of research, but it is also an engineering science that plays a considerable role in our societies. It has many aspects. In medicine, nuclear magnetic resonance imaging, as well as the various applications of radioactivity, and proton and heavy ion therapy, are revolutions in medical diagnosis and therapy. It is needless to emphasize the problems of energy in the world. It is a fact that in order to dismantle a nuclear plant, it takes 50 years, and in order to launch a new nuclear option (in fusion or in fission) it will take 30 or 40 years. In any case, we are concerned with that question for many reasons, in particular because of safety and the disposal of nuclear waste.

1.4 The Infinitely Complex Now, it is very nice to know the laws of physics at the microscopic scale, but we must some day turn back to the physical world at our scale, namely macroscopic physics. When we eat a pound of strawberry pie, we don’t think we’re eating half a pound of protons, half a pound of neutrons, and a little overweight of electrons. It’s perfectly true, but it’s silly, it’s perverse, and it’s disgusting. Statistical physics studies the global and collective behavior of large numbers of particles or systems whose individual properties are known. It is a great discovery of the last decades that one cannot reconstruct everything from the very beginning, that is, microscopic laws. As soon as one deals with large numbers of constituents,

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1 The Appeal of Physics

there appear new phenomena, new singularities or regularities that are specifically macroscopic. These are related to the number of constituents rather than to their specific nature. Examples are: • Collective effects, phase transitions • Shapes, ordered structures • Irreversibility, life and death This kind of problem (i.e., physics of the infinitely complex world) is one of the most fascinating fields of physics at present. To understand it, to dominate it, will have a considerable impact not only in physics, but in biology where reproducible ordered structures are fundamental, to some extent in economics, and maybe some day in sociology. The most fascinating system is the brain itself. At this point, there appears a much simpler and more relevant answer to the question of what is the use of doing physics. Physics is fun; it’s amusing. Take a simple example. The fact that water freezes at 0 ◦ Celsius is a very ancient scientific observation. Everyone knows that. At school, that property is used to define water: “Water is a colorless tasteless liquid, it is used to wash, some people even drink it, and it freezes at zero degrees Celsius!” But, one day, we learn physics. We learn that water is a liquid made of H2 O molecules that wander around at random. Ice is a crystal where the same molecules H2 O are well organized in a periodic structure. That’s really an amazing phenomenon! Why on earth do those molecules decide at 0◦ to settle down in an ordered structure? It is a mystery! We all know how difficult it is, after a break, to put in an ordered state a number of children at a playground or scientists whose natural tendency is to be dispersed. Therefore, because we have learned some physics, we discover a very deep aspect in a very familiar fact: the freezing of water. And that’s when we make progress. But, in order to do that, one must learn to observe and ask oneself questions about reality. Creativity is much more important than knowledge or equations, and it is fundamental to develop it and to preserve it. Physics, and in particular experimental physics, is an excellent field for that operation. Materials Physics of condensed matter, as opposed to corpuscular physics, is a broad domain common to physics, to mechanics, to chemistry, and to biology. Materials have perhaps the most important role in the evolution of science and technology, including semiconductors, steels, concretes, composite materials, glasses, polymers, paints, and so on. Practically all the important breakthroughs of the progress of mankind are associated with the discovery and the use of new materials: think of stones, flint, bronze 10,000 years ago, iron, more recently aluminum and aeronautics, silicon, electronics and computer science, carbon nanotubes and their derivatives. Up to the 1970s, it was customary to differentiate between the mechanical properties of solids, that is metallurgy, and electrical properties. But thanks to quantum

1.4 The Infinitely Complex

11

physics and statistical physics, materials science has become a unified theory, because we can understand it from its microscopic aspect. Solids are aggregates of atoms or molecules that are bound by the electrons of crystalline bonds. These electrons form a more or less hard cement that determines the mechanical properties, resistance, hardness, and plasticity. And it is, in turn, the physics of these electrons that determines the electrical and thermal properties. All these properties are intimately connected. At first, it is difficult to appreciate the importance and the depth of such a global synthetic understanding. Metallurgy was for a long time purely empirical. By manipulating such and such a mixture, one used to obtain such and such a result; knowledge was transmitted by word of mouth. Sometimes it was great, such as in Syria in the 13th century. There was a problem in the weapons industry for making swords. Iron is a resistant material, but it is soft and iron swords got bent easily. On the other hand, carbide is hard, but it breaks easily. Damascus steel consisted of alternating sheets of iron and carbide. This allowed them to make swords that were both hard and resistant (sometimes physics isn’t that funny; it would have been much more fun if the result had been soft and fragile). It was a revolution in weaponry, and it is very clever from the modern point of view; in fact this is an example of composite materials. The best composite materials that people try to imitate are biological composites such as bones or shells. These associate the hardness of limestone apatite, which is fragile but hard, with the resistance of biological collagen. For modern purposes, one must conceive a material directly in view of the function it should have, namely the desired mechanical, electrical, chemical, and optical properties. And this is done more and more systematically. We have already mentioned nanotechnologies. Let’s add that in recent years there has been a technological breakthrough with what one calls as smart materials, for instance, materials with shape memory. A piece of material can have some shape (think of a metal wire) that we can change. The surprise is that a smart material recovers its initial shape if it is heated. This does not occur with just any material. The alloys with shape memory are, for instance, metal alloys (such as nickel–titanium) that undergo a phase transition between two crystalline structures, martensite and austenite, called a martensitic transition at some temperature. One can give a material the shape one wants above the transition temperature. It holds that shape below the transition point, but one can change this shape by a plastic deformation. If after that change, one heats the material, it recovers its original shape because there are domains that “remember” the initial shape and convey the structure to the entire material. The industrial issues are huge. The applications of such materials are found in many different domains such as opening up satellite antennas, bone or tooth prosthesis, and heart and blood vessel surgery. One can crumple pieces of smart material at usual temperature (20 to 25 ◦ C) and insert them in a blood vessel. Once they have reached their destination they open up and take their functional shape at the temperature of the human body, 37 ◦ C.

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There exist in addition hysteresis phenomena. One can “educate” such materials and construct artificial “muscles” that can transform heat into work. Again, industrial issues are huge.

1.5 The Universe To end this brief panorama of physics, one should mention astrophysics. The three basic fields—quantum mechanics, statistical physics, and relativity—are deeply connected in astrophysics, and in cosmology—the history of the Universe—together with General Relativity. This is a fascinating subject, one of the most exciting perhaps at present. New observations appear constantly. On September 14, 2015, LIGO, for the first time, physically sensed distortions in spacetime caused by passing gravitational waves generated by two colliding black holes nearly 1.3 billion light years away! This discovery is undoubtedly one of the greatest scientific achievements of our times.3 The reader will understand that this subject would take us too far in this introduction, together with questions such as: are we alone in the Universe? More and more extrasolar planets are being discovered, around other stars, so are there other beings who think like we do in the Universe?

1.6 Physical Constants Units: Angström 1 Å = 10−10 m (∼size of an atom) Fermi 1 fm = 10−15 m (∼size of a nucleus) Electron-volt 1 eV = 1.60218 10−19 J. Fundamental Constants: Planck’s constant h = 6.6261 10−34 J s,  = h/2π = 1.05457 10−34 J s = 6.5821 10−22 MeV s Velocity of light

c = 299 792 458 m s−1 c = 197.327 MeV fm  1973 eV Å

Vacuum permeability μ0 = 4π10−7 H m−1 ,

3 See

0 μ0 c2 = 1

for instance https://www.ligo.caltech.edu/page/what-are-gw.

1.6 Physical Constants

13

Boltzmann’s constant k B = 1.38066 10−23 J K−1 = 8.6174 10−5 eV K−1 Avogadro’s number

N A = 6.0221 1023

Electron charge Electron mass

qe = −q = −1.60218 10−19 C and e2 = q 2 /(4π0 ) m e = 9.1094 10−31 kg, m e c2 = 0.51100 MeV

Proton mass

m p = 1.67262 10−27 kg, m p /m e = 1836.15

Neutron mass

m n = 1.67493 10−27 kg,

m p c2 = 938.27 MeV, m n c2 = 939.57 MeV

Fine structure constant (dimensionless) α = e /(c) = 1/137.036 Classical radius of the electron re = e2 /(m e c2 ) = 2.818 10−15 m Compton wavelength of the electron λc = h/(m e c) = 2.426 10−12 m Bohr radius a1 = 2 /(m e e2 ) = 0.52918 10−10 m Ionisation energy of Hydrogen E I = m e e4 /(22 ) = α2 m e c2 /2 = 13.6057 eV Rydberg’s constant R∞ = E I /(hc) = 1.09737 107 m−1 Bohr magneton μ B = qe /(2m e ) = −9.2740 10−24 J T−1 = −5.7884 10−5 eV T−1 Nuclear magneton μ N = q/(2m p ) = 5.0508 10−27 J T−1 = 3.1525 10−8 eV T−1 . 2

Updated values can be found at http://wulff.mit.edu/constants.html

Chapter 2

A Quantum Phenomenon

If, at a party, you ask someone to state a physics formula, the odds are that the answer will be E = mc2 . Nevertheless, the formula E = hν, which was written in the same year 1905 by the same Albert Einstein concerns their daily life considerably more. In fact, among the three great scientific events of the beginning of the 20th century, 1905 with the special relativity of Einstein, Lorentz, and Poincaré, 1915 with Einstein’s general relativity, an extraordinary reflection on gravitation, space, and time, and 1925 with the elaboration of quantum mechanics, it is certainly the last that has had the most profound impact on science and technology. The first Nobel prize for relativity was awarded in 1993 to Taylor and Hulse for the double pulsar. Nobel prizes for quantum mechanics can hardly be counted (of the order of 120) including Einstein’s for the photon in 1921. That reflects discoveries which have had important consequences. About 30 % of the gross internal product of the United States comes from byproducts of quantum mechanics. Quantum mechanics is inescapable. All physics is quantum physics, from elementary particles to the big bang, semiconductors, and solar energy cells. It is undoubtedly one of the greatest intellectual achievements of the history of mankind, probably the greatest of those that will remain from the 20th century, before psychoanalysis, computer science, or genome decoding. This theory exists. It is expressed in a simple set of axioms that we discuss in Chap. 6. Above all, this theory works. For a physicist, it even works too well, in some sense. One cannot determine its limits, except that during 10−43 s just after the big bang, we don’t know what replaced it. But afterwards, that is, nowadays, it seems unbeatable. However, this theory is subtle. One can only express it in mathematical language, which is quite frustrating for philosophers. Knowing mathematics is the entrance fee to the group of the happy few who can understand it, even though, as we show, the core of these mathematics is quite simple. It is the physics that is subtle. More important perhaps, we show how and why quantum mechanics is still a subject of debate as to its interpretation and its intellectual content. In some sense, mankind © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_2

15

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has made a beautiful and successful intellectual construction that escapes human understanding to some extent. As Richard Feynman put it: “I think I can safely say that nobody understands quantum mechanics”.1 The discovery of quantum mechanics could have happened by analyzing a variety of physical facts at the end of the 19th century. The notion of quanta was proposed in 1900 by Max Planck. Planck had found semi-empirically a remarkable formula to explain a problem that fascinated people, the spectrum of black-body radiation. The frequency distribution of radiation inside an oven at temperature T depends only on the temperature, not on the nature or shape of the oven. It is a universal law. Planck obtained the good result ν3 8πh , (2.1) u(ν) = 3 (hν/kT ) c e −1 where ν is the frequency, T the temperature, and k Boltzmann’s constant, by assuming that radiation of frequency ν can exchange energy with the inner surface of the oven only by discrete quantities that are integer multiples of an elementary energy quantum hν, ΔE = nhν. (2.2) Planck understood that the constant h in the above formula, which now bears his name and whose value is h ≈ 6, 62 10−34 j.s, is a fundamental constant of nature, as the velocity of light c in relativity and Newton’s constant G in gravitation. For technical simplicity, we mainly use the reduced Planck constant h ≈ 1, 05 10−34 j.s. ≡ 2π Planck’s formula works remarkably well. The direct verification would require us to be inside an oven. We have the great luck to live inside the cosmic background radiation of the big bang, which cooled down as the Universe expanded. The temperature of that radiation is at present 3 K. Its observation and its more and more precise measurement (Fig. 2.1) is perhaps the best observational evidence in favor of the big bang theory, as well as of Planck’s formula. Planck’s quanta were somewhat mysterious, and it was Einstein who made a decisive step forward in 1905, the same year as he did for Brownian motion theory and for special relativity. By performing a critique of Planck’s ideas, and for reasons due to equilibrium considerations (i.e., entropy) Einstein understood that the quantized aspect is not limited to the energy exchanges between radiation and matter, but that it must be present in the electromagnetic field itself. Light, which was known to be a wave propagation phenomenon since the beginning of the 19th century, must also

1 The

Character of Physical Law, MIT Press, Cambridge, MA 1965.

2 A Quantum Phenomenon

17

Fig. 2.1 Wave-number distribution of the cosmic background radiation measured in 1992 by the COBE satellite. The agreement between Planck’s formula at a temperature T = 2.728 K lies within the line (Photo credit: Mather et al., Astrophys. J., 420, 439, (1994). http://lambda.gsfc. nasa.gov/product/cobe/firas_ image.cfm.)

exhibit a particlelike behavior. Light of frequency ν is carried by particles, photons as the chemist Gilbert called them in 1926, of energy E = hν,

(2.3)

and momentum p = k, where k is the wave vector k = 2π/λ, as was proven experimentally by Compton in 1921. In that respect, Einstein understood an essential feature of quantum theory, the so-called “dual” manifestations of the properties of light, which appear to be both wavelike and particlelike. In the course of his work, Einstein found the explanation of the photoelectric effect, which was one of the first experimental confirmations of his ideas. Such ideas were considered revolutionary or even iconoclastic because they seemed to contradict Maxwell’s equations which were a great triumph of the 19th century. At the same time, atomic spectroscopy was one of the great enigmas of physics. The third breakthrough, which derives in some respect from Einstein’s ideas, came in 1913 from Niels Bohr. There are three parts in Bohr’s ideas and results. • He postulated that matter is also quantized and that there exist discrete energy levels for atoms, which was verified experimentally by Franck and Hertz in 1914. • He postulated that spectral lines which had been abundantly observed during the 19th century, came from transitions between these energy levels. When atoms absorb or emit radiation, the positions of spectral lines are given by the difference νnm =

|E n − E m | . h

(2.4)

• Finally, Bohr constructed an empirical model of the hydrogen atom that works remarkably well and gives the energy levels E n of this atom as

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2 A Quantum Phenomenon

En = −

mqe 4 , 2(4πε0 )2 2 n 2

(2.5)

where n is a positive integer. With that formula, where all physical constants are known from different experiments, the wavelengths λ = c/νnm of spectral lines coincide with experiment to one part in a thousand. Bohr’s formula (2.5) expresses the famous “Rydberg constant” of spectroscopists in terms of fundamental constants, which impressed people, in particular, Einstein.2 So we are facing three similar formulae, E = hν. The first (2.2) is an assumption about the interaction of radiation and matter, the second (2.3) has to do with radiation itself, and the third (2.4) is a property of atoms, namely matter. Bohr’s success was fantastic, but it was too easy. Actually one realized later on that it was a piece of luck due to the fact that the hydrogen atom is a simple twobody system. This easy result generated an obscure prequantum era, where people accumulated recipes for more complicated atoms, with fluctuating results deprived of any global coherence.

2.1 Wave Behavior of Particles The synthetic and coherent formulation of quantum mechanics was performed around 1925. It is due to an incredible collective work of talented people such as Louis de Broglie, Schrödinger, Heisenberg, Max Born, Dirac, Pauli, and Hilbert, among others. Never before, in physics, had one seen such a collective effort to find ideas capable of explaining physical phenomena. We are now going to discover some of the main features on a simple concrete experiment that shows the wavelike behavior of particles. This is symmetric in some respect to the particlelike behavior of light. We show that the behavior of matter at atomic scales does not follow what we expect from daily “common sense.” It is impossible to explain it with our immediate conceptions. In order to understand quantum mechanics, one must get rid of prejudices and ideas that seem obvious, and one must adopt a critical intellectual attitude in front of experimental facts.

2.1.1 Interferences Let us recall interference phenomena in wave physics, optics, or acoustics, in the simple case of Young slit fringes. 2 The

1/n 2 behavior was known since 1886 and Balmer’s empirical discovery.

2.1 Wave Behavior of Particles

19

Fig. 2.2 Sketch of a Young two-slit interference experiment

One sends a light beam on a screen pierced with two slits, and one observes the variation of the intensity of light on another screen as a function of the distance x to the center (Fig. 2.2). The two slits act as secondary sources in phase, and the amplitude of the wave at a point C of the screen is the algebraic sum of the amplitudes issued from each of them. If the two waves are in phase, the amplitude is twice as large. If they are out of phase by π the amplitude vanishes; there is no luminous energy at that point. And there exist all intermediate cases. In other words, the amplitude at some point is the sum of amplitudes reaching that point, Amplitude at C : AC = A1 + A2 , Intensity : I (x) = | AC |2 .

(2.6)

The amplitudes emitted by the two slits add up, the intensity is the square of that sum and it presents a periodic variation, the distance of fringes being x0 = λD/a.

2.1.2 Wave Behavior of Matter We turn to the wave behavior of matter. In 1923, Louis de Broglie made the bold but remarkable assumption that any particle of mass m and of velocity v possesses an “associated” wave of wavelength h (2.7) λ= ; p where p = mv is the momentum of the particle and p its norm.

20

2 A Quantum Phenomenon

Fig. 2.3 Double slit Young interference experiment performed with neon atoms cooled down to a milliKelvin (left part). Each point of the figure (right part) corresponds to the impact of an atom on the detector. Interference fringes are clearly visible

Louis de Broglie had many reasons to propose this. In particular he had in mind that the discrete energy levels of Bohr might come from a stationary wave phenomenon. This aspect struck the minds of people, in particular that of Einstein, who was enthusiastic. In order to verify such an assumption, it is natural to perform interference and diffraction experiments. The first experimental confirmation is due to Davisson and Germer in 1927. It is a diffraction experiment of an electron beam on a nickel crystal. It is more difficult to perform a Young double-slit interference experiment with electrons. However, a group of Japanese physicists from Nippon Electronics (NEC) performed in 1994 a beautiful interference experiment of cold atoms in Young slits. Neon atoms are initially trapped in stationary laser waves (so-called optical molasses). They are then released and undergo free fall across a two-slit device. The slits are 2 µm large, they are 6 µm apart. The scale in Fig. 2.3 is distorted. What do we observe in Fig. 2.3? The distribution of impacts of atoms on the detecting plate is similar to the optical intensity in the same device. The fringes are at the same positions provided Louis de Broglie’s relation is satisfied λ = h/ p. (Of course, one must take care of the uniform acceleration in this particular setup.) The same phenomenon can be observed with other particles: neutrons, helium atoms, hydrogen molecules, the same relation holds between the wavelength and the momentum. The present record is to perform interferences with large molecules such as fullerenes, that is, C60 molecules.3 Therefore matter particles exhibit a wave behavior with a wavelength given by de Broglie’s formula.

3 O. Nairz, M. Arndt, A. Zeilinger, American Journal of Physics, Vol. 71, 319 (2003), and references

therein.

2.1 Wave Behavior of Particles

21

2.1.3 Analysis of the Phenomenon Now, a number of questions are in order. What is this wave? And why is this result so extraordinary? It is extraordinary because atoms are known to be particles. An atom has a size of the order of an angström (0.1 nm) and it is pointlike at the scales of interest (µm or mm). With a counter, one can measure whether an atom has arrived at some point with an accuracy as fine as one wishes. When an atom is detected, it has a well-defined position; it does not break up into pieces; it is point-like. But a wave fills all space. A wave, on the surface of water, is the whole set of deformations on all points of that surface. So, what is a particle? Is it a pointlike object or is it spread out in the entire space? A simple glance at Fig. 2.4 shows that we are facing a conceptual contradiction. How can we escape this contradiction? Actually, the phenomenon is much richer than a simple wave phenomenon; we must observe experimental facts and use our critical minds.

Fig. 2.4 Top two source interferences on the surface of water; the radial lines are nodes of interferences. Bottom tracks of particles in the Aleph detector of LEP at CERN

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2 A Quantum Phenomenon

Since atoms are particles, we can send them individually, one at a time, and all in the same way. This proposition is perfectly decidable; it is feasible experimentally. We can trigger the device so that it releases atoms one after the other and that they are all released in the same way.

2.2 Probabilistic Nature of Quantum Phenomena 2.2.1 Random Behavior of Particles What do we observe? Actually, we can guess that from Fig. 2.3. • Each atom has a well-defined impact; an atom does not break into pieces. • But the positions of the impacts are distributed at random. In other words, to the same initial conditions, there correspond different impacts. In other words, atoms, or particles in general, have a random behavior. Each atom arrives where it wants, but the whole lot is distributed with a probability law similar to the intensity observed in optics or acoustics: P(z) ∝ I(optical) (λ = h/ p). Therefore, there is a second difference with classical physics: to identical initial conditions, there correspond different final conditions. The impact of a single particle is unpredictable, the whole set of impacts has a well defined probability distribution. But, one can object that random, or probabilistic, phenomena exist in classical physics, such as playing dice, or heads and tails, and so on. True, but the big problem is that this is by no means a classical probabilistic phenomenon, as in usual probability theory. Why is that?

2.2.2 A Nonclassical Probabilistic Phenomenon If we block one of the slits, the atoms will pass through the other one and their distribution on the detector shows no sign of any interference. If we block the other slit, the distribution is approximately the same, up to a small global shift (1 µm/1 mm) Now let’s make a logical reasoning and perform the critique of what we say. 1. We send the atoms one by one. These are independent phenomena; atoms don’t bother each other; they do not act on each other’s trajectory. 2. Each atom has certainly gone through one of the slits. 3. We can measure which slit each atom went through. There exist techniques for this; send light on the slits, put counters, and so on. It is feasible.

2.2 Probabilistic Nature of Quantum Phenomena

23

4. If we perform this measurement, we can separate the outgoing atoms in two samples, those that have passed through the first slit, and those that have passed through the second one. And we know where each atom arrived. 5. For those that passed through the first slit, everything is as if the second slit were blocked, and vice versa. Each sample shows no interference. Now, we have two independent samples, and we can bring them together. Classically, the result we would obtain by opening the two slits should be the sum, the superposition of the two distributions such as (2.5). But not at all! It’s even worse! Opening a second slit (i.e., giving an extra possibility for the atoms to reach the detector) has prevented the atoms from arriving at certain points. That’s really incredible to be able to stop some people from entering your house by opening another door! We must admit that the usual logical ideas of probability theory do not apply. We cannot explain the phenomenon in classical terms. It is a non-classical probabilistic phenomenon.

2.3 Conclusions At this point, it seems we are at a logical dead end. How can we find our way? Our argument, however logical it may seem, leads to wrong conclusions. There is something we haven’t thought about. Because physics is consistent. The answer is experimental. What actually happens is the following. 1. If we measure by which slit each atom passed, we can indeed make the separation and indeed we observe the sum of two distributions such as in Fig. 2.5. Therefore we no longer observe interferences; they disappear. It is another experiment!

Fig. 2.5 Same experiment as in Fig. 2.3 but opening only one slit. The interference fringes disappear and one observes a diffraction pattern (this figure is not experimental)

24

2 A Quantum Phenomenon

2. Conversely, if we do observe interferences, it is not possible to know through which slit each atom passed. We can talk about it, but we can’t do anything with it. Knowing by which slit an atom has passed in an interference experiment is a proposition that has no physical meaning; it is undecidable. It is perfectly correct to say that an atom passed through both holes at the same time, which seems paradoxical or absurd classically. What was wrong was to assume implicitly that, at the same time, we could measure by which slit each atom passed and observe interferences. We assumed that without checking it. We can draw the following conclusions. • First, a measurement perturbs the system. If we do not measure by which slit they pass, the atoms are capable of interfering. After we perform this measurement, they are in another state where they are no longer capable of interfering. They have been perturbed by the measurement. • Secondly and consequently, there is no trajectory in the classical sense. If we observe an atom in an interference experiment, we know when and where it was emitted and where and when it was detected, but we cannot say where it was in the meantime. However, these two ideas seemed obvious in classical physics. The fact that we can make a measurement as accurate as we wish without affecting the system is an old belief of physics. Physicists used to say that they just needed to improve the measuring apparatus. Quantum physics tells us that there is a absolute lower bound to the perturbation that a measurement produces. The notion of a trajectory, namely that there exists a set of points by which we can claim and verify that a particle has passed at each moment, is as old as mankind. Cavemen knew that intuitively when they went hunting. It took centuries to construct a theory of trajectories, to predict a trajectory in terms of initial conditions. Newton’s classical mechanics, celestial mechanics, ballistics, rests entirely on that notion, but its starting point is beaten up by the simple quantum phenomenon we just examined. Classically, we understand the motion of a particle by assuming that, at each moment one can measure the position of a projectile, that the collection of the results consists of a trajectory, and that we can draw a reproducible conclusion independent of the fact that we measure the positions at any moments. We learn these ideas as if they were obvious, but they are wrong. More precisely: in order to penetrate the quantum world, one must get rid of such ideas. Figure 2.6, or analogous ones, is completely wrong in quantum mechanics. Of course, one mustn’t go too far. These are very good approximations in the classical world. If a policeman stops you on a freeway saying you were driving at 80 miles an hour, the good attitude is to claim, “Not at all! I was driving peacefully at 35 mph on the little road under the bridge, and your radar perturbed me!” Unfortunately, he won’t believe you even if he knows some physics. Because it is Planck’s constant  that governs such effects. However, in quantum driving one must change the rules. Changing the rules consists of constructing the theory of all that.

2.3 Conclusions

25

Fig. 2.6 Stroboscopic picture of the free fall of an apple which then bounces on the floor. This is a good example of the a priori representation of an intuitive phenomenon that cannot show up in quantum mechanics (William McLaughlin, “The resolution of Zeno’s paradoxes,” Sci. Amer., 1994)

Phenomenological Description The interference phenomenon would be very complicated to explain if we did not have the luck that it so closely resembles usual interference, with, in addition, a simple formula for the wavelength λ = h/ p. So, let’s try and use the analogy with wave physics in order to formalize Louis de Broglie’s idea. Here, we should be able to explain the interference experiment in the following way. • The behavior of an atom of velocity v and momentum p = mv in the incoming beam corresponds to that of a monochromatic plane wave ψincident = e−i(ωt−p·r/) , k = p/, λ = 2π/k = h/ p, which has the good wave vector k = p/ and the good wavelength.

(2.8)

26

2 A Quantum Phenomenon

• After the two slits, the behavior is that of the sum of two waves each of which has been diffracted by a slit (2.9) ψoutgoing (x) = ψ1 + ψ2 , which would describe, respectively, the behavior of the atom if it passed through one of the slits, the other one being blocked. We can calculate the phase shift of these waves at any point because we know the wavelength. • Finally, the probability for an atom to reach some point C of the detector is simply the modulus squared of that sum P(C) = |ψC |2 .

(2.10)

We just follow the same argument as for usual interferences. We now have an answer to one of our questions above; what is the physical meaning of these waves? In usual wave physics, one manipulates electromagnetic or acoustic wave amplitudes which add up and whose modulus squared gives intensities, that is, energy densities. Our quantum waves are probability amplitudes. They add up and the modulus squared of the sum gives us probabilities, or probability densities. One does not work directly with probabilities but with these intermediate tools, these probability amplitudes that add up. The interference experiment gives us the wavelength, but not the frequency ω of the waves. Louis de Broglie made a good choice by assuming that this frequency is related to the energy of the particles in the same way as for Einstein’s photons ω = E/, that is, ν = E/ h,

(2.11)

where E = p 2 /2m is the kinetic energy of the atoms. This leads to the complete structure of de Broglie waves: ψincident = e−(i/)(Et−p·r) , where E = p 2 /2m ,

(2.12)

which is the probability amplitude for the presence of a particle at point r and time t of a particle of momentum p = mv. Remark Notice that because the kinetic energy and the momentum are related by E = p2 /2m, one can find with this expression a wave equation, which is satisfied whatever the value of the momentum p. Indeed, if we take the time derivative on one hand, and the Laplacian on the other, we obtain iE p2 ∂ ψincident = − ψincident , and Δψincident = − 2 ψincident , ∂t  

2.3 Conclusions

27

therefore, because E = p 2 /2m, we have the wave equation i

∂ψ 2 =− Δψ, ∂t 2m

(2.13)

which is nothing but the Schrödinger equation for a free particle.4 Of course, we are not completely finished. For instance, atoms have a particlelike behavior that is obscure in all that. But we’re getting closer.

2.4 Appendix: Notions on Probabilities Probabilistic Phenomena Consider a set of phenomena of the same nature on which we repeatedly make the same observation or measurement. For instance, play dice, measure a temperature or an economic parameter etc. Each observation belongs to some set Ω of outcomes. This set can be discrete, continuous, or a more complicated object such as a set of functions. The set Ω is the set of a priori possible outcomes of the experiment. One also speaks of events: “the roulette number is even”, “the observed temperature is between T0 and T1 ”, etc. Suppose we repeat an experiment a large number of times N , Ω being the set of possible outcomes. Consider a specific event α and Nα the number of times, among N , where α occurs. The observed number Nα depends on the specific sequence of experiments. One calls the empirical frequency of the event α in this sequence of experiments, the ratio: f α (N ) = Nα /N . The fundamental empirical observation is that when N becomes large, if the successive repetitions of the experiment are done independently (the result of an experiment has no a priori influence on the conditions in which the other experiments are done), the frequencies f α (N ) tend, for each event α, to a well defined limit. To each event α there corresponds a number P(α) called the probability of event α, related to the empirical frequency by the relation: P(α) = lim f α (N ). N →∞

4 It

is surprising that de Broglie didn’t think of writing this equation, or its relativistic equivalent— since he used the relativistic energy-momentum relation E 2 = ( p 2 c2 + m 2 c4 ).

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2 A Quantum Phenomenon

Clearly, one has P(α) ≥ 0, P(Ω) = 1, P(∅) = 0, and if (Ai )i∈I is a finite family of disjoint events:   P(Ai ). P( Ai ) = i∈I

i∈I

Examples of Probability Laws Discrete Laws The Simple Alternative In this case, there are only two possible outcomes, α = 1 or 2 (example: heads or tails). We note p the probability of the outcome 1 and q that of the outcome 2. We obviously have p + q = 1. The Generalized Alternative There are n outcomes α = 1, 2 . . . n. For instance one can place in an urn m 1 balls marked with the sign 1, m 2 balls marked 2, . . .. If the draw does not distinguish the balls, the probability law consists in the set of numbers p1 , p2 , . . ., pn such that: mα p α = n β=1

with

mβ

n 

pα = 1.

α=1

Probability Laws on R or Rn n A probability law P on R (resp.  +∞R ) is said to be of density p,np being a positive integrable function such that −∞ p(x) d x = 1 (resp. R n p(x) d x = 1), if, for any interval (resp. any volume) I :

 P(I ) =

p(x) d x. I

It is useful to treat the discrete and continuous cases in the same formalism by working with the distribution function: F(t) = P(] − ∞, t]). Examples 1. Exponential law:

 p(x) =

which yields:

 F(t) =

t

−∞

λe−λx if x ≥ 0 (λ > 0) , 0 if x < 0. 

p(x)d x =

0 if t < 0 . 1 − e−λt if t > 0

2.4 Appendix: Notions on Probabilities

29

Fig. 2.7 The Gaussian probability law for μ = 0 and σ = 1

2. Gauss’s law of parameters μ, σ (Fig. 2.7): p(x) =

1 (x − μ)2 with μ ∈ R, σ ∈ R∗ . √ exp − 2σ 2 σ 2π

(2.14)

Random Variables Definition Consider the example of the game with n outcomes α1 , . . ., αn of respective probabilities p1 , . . ., pn . If, in this game, we win some amount of money xα when the outcome is α, the number xα which is a function of the (random) outcome of the experiment is called a random variable. In the above example, the set of the {xα } is discrete. One calls a discrete random variable x a set of numbers xα (positive, negative, complex) each of which is associated to an outcome of a discrete random event. The couples {xα , pα } define the probability law of the random variable x. In the same way, one can consider continuous random variables. Let x be a random variable which takes its values in an interval [a, b]. The probability density p(x) (positive or zero) defines the probability law of this random variable if the probability to find, in an experiment, a value between x and x + d x is p(x) d x. We obviously b have a p(x) d x = 1. The generalization to Rn is straightforward. Conditional Probabilities Consider two types of events [ A] and [B]. We are led to defining the conditional probability of the event B knowing A, noted P(B/A) by P(B/A) =

P(B ∩ A) as long as P(A) > 0. P(A)

If X is a discrete random variable, one can define the conditional probability P(B|X = x) of the event B when X = x, i.e. knowing the event {X = x}.

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2 A Quantum Phenomenon

Example: The Exponential Decay Law When a radioactive particle exists at time t, its probability to decay in the time interval ]t, t + Δt] is independent of its past history. Therefore the conditional probability that the time X at which the particle decays is between t and t + Δt, knowing that {X > t}, is independent of t and equal to P{0 < X ≤ Δt}: P{0 < X ≤ Δt} =

P{t < X ≤ t + Δt} . P{X > t}

If we call F the distribution function of X we obtain the functional relation: F(Δt) =

F(t + Δt) − F(t) . 1 − F(t)

The function F therefore satisfies the differential equation: F  (t) = F  (0) (1 − F(t)). Therefore, setting λ = p(0) = F  (0) (λ is a decay rate), we get F(t) = 1 − e−λt . The density of the law of X is then:  p(x) =

λe−λx for x ≥ 0 . 0 for x < 0

Since λ has the dimension of the inverse of a time, we can note it 1/τ where τ is the lifetime (or mean life). This exponential law is met in many practical applications (physics, pharmacology, reliability, etc.). Independent Random Variables Consider two discrete random variables X and Y with values in E 1 and E 2 respectively. One says that X and Y are two independent variables if the observation of X does not give any information on Y , and vice-versa. In other words, the conditional probability to find x if one knows y is independent of y (and vice versa). This can be expressed in a symmetric form in x and y by: P({X = x, Y = y}) = P({X = x}) P({Y = y}).

The variables X and Y are independent if and only if the law of the couple (X, Y ) is the product of the laws of X and of Y . Binomial Law and the Gaussian Approximation Consider an experiment consisting in repeating N consecutive times and independently an experiment with two outcomes (for instance heads or tails). The first outcome, noted 1, has a probability p to happen, and the second, noted 0, has the prob-

2.4 Appendix: Notions on Probabilities

31

ability q = 1 − p to happen. Such a sequence of experiments is called a Bernoulli sequence. Since the successive partial experiments are assumed to be independent, the probability for a given sequence (x1 , . . . , x N ) is given by: P(x1 , . . . , xn ) = P[X 1 = x1 ] . . . P[X N = x N ] = p k q N −k , where k is the number of 1 in the sequence (x1 . . . x N ). We now consider the random variable X = X 1 + · · · + X N representing the number of times 1 appears in the N successive draws:  N P[X = k] = p k q N −k ≡ b(k; N , p). k This law b(k; N , p) is called the binomial law of parameters N and p. Normal Approximation of the Binomial Law √ Using Stirling’s formula: n! ∼ 2πn n n e−n , we obtain for n  1:

b(k; n, p) ∼

1 (k − np)2 exp − , 2πnpq 2npq

i.e. a Gaussian law for k, with μ = np and σ =

√ npq.

Moments of a Probability Distribution Mean Value or Expectation Value Consider a function ϕ(x) of the random variable x (ϕ(x) is a new random variable). We define its mean value or equivalently expectation value ϕ as ϕ =

 ϕ(xα ) pα discrete case  bα ϕ(x) p(x) d x continuous case a

(a < x < b)

We note x the mean value of the variable x itself:  x = x p(x) d x. This quantity is equivalently called the mathematical expectation, or expectation value: if we gain the amount x α when the result is α, then we expect to gain on the average x. Expectation Values of Usual Laws 1. Variable of the simple alternative: X  = p. 2. Binomial law b(k; n, p): k = np.

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2 A Quantum Phenomenon

3. Geometric law P{X = k} = (1 − p) p k (k ≥ 0): X  = p/(1 − p). 4. Poisson law P{X = k} = e−λ λk /k! (k ≥ 0): X  = λ. Example In the exponential decay above, the mean time that the particle spends before it decays, or the expectation value of its lifetime, is:  t = 0

∞

t −t/τ dt = τ . e τ

Variance and Mean Square Deviation Consider a real random variable x whose expectation value is x = m. The mean square deviation of x, noted σ or Δx, is defined by: (Δx)2 = σ 2 = (x − x)2 , σ 2 is also called the variance of the probability law. One readily checks, by expanding the square term, that: σ 2 = x 2  − 2xx + x2 = x 2  − x2 . The smaller σ is, the more probable it is to find a value of x close to the mean value. The quantity σ measures the deviation from the mean value. Variance of Usual Laws 1. The simple alternative: σ 2 = p(1 − p). 2. Binomial law: σ 2 = np(1 − p). Note that the relative dispersion σ/X  tends to zero as n −1/2 when n → ∞. 3. Gaussian law: the variance coincides with the parameter σ 2 of (2.14). 4. Geometric law: σ 2 = p/(1 − p)2 . 5. Poisson law of parameter λ : σ 2 = λ. Bienaymé–Tchebycheff Inequality Note m the mean value and σ 2 the variance of the discrete real variable X . One can show that: (2.15) P({|X − m| ≥ τ σ}) ≤ 1/τ 2 , which proves that for a small variance, there is a small probability to find X far from its expectation value. Error Function In the particular case of the Gaussian law (m, σ), one calls error function Φ(τ ) the quantity P({|X − m| ≤ τ σ}). One has:

2.4 Appendix: Notions on Probabilities

 Φ(τ ) =

33 +τ σ τσ

1 2 2 √ e−x /2σ d x. σ 2π

Some values of Φ(τ ) are the following τ 1 2 3 Φ(τ ) 0.68 0.95 0.99

2.5 Exercises 1. Distribution of impacts We observe the impacts on a target in the x y plane. The observable is assumed to obey a probability law of density p(x, y) = (2πσ 2 )−1 exp(−ρ2 /(2σ 2 )) where ρ = (x 2 + y 2 )1/2 is the distance from the origin of the impact point. What is the probability law of ρ? 2. Is this a fair game? Suppose that one offers you the following game: Bet one euro and throw three dice. If number 6 (or any number you choose in advance) does not show up, you lose your bet; you get paid 2 euros if it shows up on one dice, 3 euros if it shows up on two, and 6 euros if it shows up on the three of them. Calculate the expectation value of what you gain (which is negative if you lose) and see if it is reasonable to play. 3. Spatial distribution of the molecules in a gas Consider in a volume V (22.4 l for instance) N molecules (6 × 1023 for instance). Consider an enclosed volume v (10−3 cm3 ). How many molecules are there on the average in v? What are the fluctuations of this number?

Chapter 3

Wave Function, Schrödinger Equation

In the first chapter, we described an interference experiment of atoms which, as we have understood, is both a wave and a probabilistic phenomenon. We now want to construct the theory of this experiment. More generally, we want to find the quantum theory of the simplest problem of classical mechanics; the nonrelativistic motion of a particle of mass m in a field of force. This is called wave mechanics. It is due to de Broglie and to Schrödinger. We generalize it later on. We do not want to say what the nature of an atom or an electron is; we simply want to determine their behavior in a field of force. In celestial mechanics, one does not worry about the nature of planets. They are considered as points whose motion we can calculate.

3.1 Terminology and Methodology Terminology Before we start, we must agree on the meaning of words and on the methodology. We cannot avoid using ordinary language. Words are necessary. But words can also be traps when discussing phenomena that are so new and unusual. We constantly use the following words: physical system, state, physical quantities. The foundation of physics is experimental observation and the measurement process that consists of characterizing aspects of reality that we observe, by numbers. These aspects of reality are elaborated into concepts of physical quantities (for instance, velocity, energy, electric intensity, etc.). In given circumstances, we say that a physical system (i.e., an object pertaining to reality) is in a certain state. The state of the system is “the way the object is” (i.e., the particular form in which its reality can manifest itself). That is what we are interested © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_3

35

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3 Wave Function, Schrödinger Equation

in. We want to know the state of an atom in space, not its internal structure, which we study later. We possess some more or less complete knowledge of a state if we perform some set of measurements of physical quantities on the system. We give a name to the system. We can call it a particle, an atom, or an electron. But a particle is simply for the moment an object with a well-defined mass m and electric charge q, and which preserves any internal structure it may have in the experiments of interest. We are not interested for the moment in the possible internal degrees of freedom of the particle. Methodology The construction of our theory has the following elements that we describe in the classical case of Newtonian mechanics of a massive particle in a field of force that we assume to derive from a potential energy V (r). 1. We must first describe the state of the system. This means associating with this state a mathematical representation that defines it from an operational point of view. In Newton’s theory, the state of a massive particle in space is described at time t by six numbers; its position r and velocity v, or its momentum p = mv. 2. Then, we must know the law that governs the time evolution of the state of the system when it is placed in given conditions; that is, we must be able to predict the state at time t, given the state at time t = 0. In Newton’s theory, it is the fundamental law of dynamics d p/dt = f , that allows us to calculate the trajectory. 3. Next, we must know the laws that enable us to calculate the results of measurements of physical quantities, laws that transform the mathematical representation of the state of a system into measurable numbers. In Newton’s theory, physical quantities are functions of the state variables r and p. 4. Finally, we must address a question that is absent in Newtonian theory. In what does the measurement process result? What do we know after a measurement? In this chapter, we study the first two questions. We examine the two others in the next chapter. In quantum mechanics, there is no direct intuitive link between physical concepts and their mathematical representation as we can find in classical physics. For instance, any child knows that instantaneous velocity and acceleration exist because in a car there is an object, the speedometer, whose indications correspond to the various physical feelings of the child. When later on, the child learns the mathematical notion of derivatives, it is quite natural to associate that notion with the physical concept that has become familiar. So, what we do is to place ourselves in the position of de Broglie, Schrödinger, Einstein, and Born, after they have investigated all alternative possibilities. • We give the principles of the theory. • We check that they are consistent, and that they account for observed phenomena; we will then understand how the theory works.

3.1 Terminology and Methodology

37

• Finally, we say a few words about how such ideas came to the minds of people. This is useful for understanding the theoretical scheme and approaching it in a human way. Principles were not dictated by any superior being. It is human scientists who fought with reality in order to elaborate them. Later on, we shall be capable of understanding that the mathematical structure which is perhaps the most important in quantum mechanics is the first of the four elementary operations, addition. But it takes some time to fully appreciate that.

3.2 Principles of Wave Mechanics The Interference Experiment In the previous chapter, we have described the experimental aspects of quantum interferences of particles on the specific case of atoms. A beam of particles of given momentum p is sent on a diffracting setup. This gives rise to an interference pattern similar to light interferences. The impact point of a given atom can be anywhere with a probability law proportional to the intensity of light fringes. We attempted to make an empirical description using de Broglie waves, which are probability amplitudes that add and whose modulus squared gives a probability ψincident = e−(i/)(Et−p·r) , where E = p 2 /2m.

(3.1)

These waves obey the free particle Schrödinger equation i

∂ψ 2 =− Δψ. ∂t 2m

(3.2)

We now go back to a deductive presentation.

3.2.1 The Wave Function • Description Schrödinger and Born understood in 1926 that the complete description of a particle in space at time t is performed with a complex wave function ψ(r, t), whose physical interpretation is that the probability d P(r) to find the particle in a vicinity d 3 r of the point r is given by d P(r) = |ψ(r, t)|2 d 3r.

(3.3)

38

3 Wave Function, Schrödinger Equation

This is indeed a probabilistic description, but it is a nonclassical one. One cannot describe a quantum system with probabilities only. The wave function is a probability amplitude (it is necessarily complex). Its modulus squared gives the probability density of finding the particle at point r at time t and, quite naturally, the integral of this quantity over all space is equal to one:  |ψ(r, t)|2 d 3r = 1,

(3.4)

which is obvious but fundamental (the particle must surely be somewhere). • Probabilistic interpretation The meaning of this description is the following. We prepare N atoms independently, in the same state, so that, when each of them is measured, they are described by strictly the same wave function. Then the result of a position measurement is for each of them as accurate as we wish (limited by the accuracy of the measuring apparatus) but is not the same for all. The set of impacts is distributed in space with the probability density |ψ(r, t)|2 . One can plot the histogram of the distribution. The set of N measurements is characterized by an expectation value x and a root mean square dispersion (or simply dispersion, for short) Δx,  (3.5) x = x |ψ(r, t)|2 d 3 r ; similarly, the square of the dispersion (Δx)2 is by definition (Δx)2 = x 2  − (x)2 = (x − x)2 .

(3.6)

It is a theorem of probability theory that the probability of finding a result within a few times Δx of the value x is close to one. If the accuracy δx of the measuring apparatus is not as fine as the dispersion Δx we can say the particle has a welldefined position. A probabilistic description can perfectly well accommodate the description of a “pointlike” object (Fig. 3.1).

3.2.2 Schrödinger Equation In 1926, Schrödinger discovered that when the particle is placed in a field of force that derives from a potential V (r), the time evolution of the wave function, therefore of the state of the particle, is given by the partial differential equation

3.2 Principles of Wave Mechanics

39

Fig. 3.1 Simple example of a histogram of the distribution of a position measurement for particles prepared all in the same state ψ

i

∂ 2 ψ(r, t) = − Δψ(r, t) + V (r) ψ(r, t). ∂t 2m

(3.7)

It is the same as previously (3.2) for a free particle. The incorporation of the forces lies in the second term on the right hand side. 1. The Schrödinger equation is the wave equation. It replaces the fundamental law of dynamics (mγ = f ). 2. It is a partial differential equation of first order in time. Therefore, if the state ψ(r, 0) is known at some initial time t = 0, the equation determines the state ψ(r, t) at any further time t. 3. One cannot prove this equation (any more than one can prove mγ = f ). Its justification is that it works: when one calculates, it gives the correct results. When he tried to apply de Broglie’s idea to the hydrogen atom, Schrödinger wrote several other equations. These were more clever, in his mind, because they had a relativistic structure. But, if they gave the correct 1/n 2 term, they did not give the proper relativistic corrections, which were small but known. His wife said that for three weeks, he had been in a terrible mood. Then, he surrendered and abandoned relativity, saying that “God decided that things were so.” Actually, when he did this, he did not know about the electron spin, which contributes in an important way to these corrections. That is a piece of luck. Most of quantum physics is nonrelativistic, therefore comparatively simple. 4. Nearly all quantum mechanical problems consist of solving that equation. We come back to it constantly. At this point, our real problem is to become familiar with this new concept of a wave function. We want to understand its structure, its properties. We want to understand why we need a whole function in order to describe the state of a particle, whereas for Newton six numbers were sufficient.

40

3 Wave Function, Schrödinger Equation

3.3 Superposition Principle Consider our empirical analysis of atom interferences and de Broglie waves. Everywhere the atom is free, and the potential vanishes (except on the screen which atoms cannot cross). Starting from Louis de Broglie’s choice for the frequency ω, we had deduced the Schrödinger equation in the absence of forces, with V = 0. Conversely, the de Broglie waves are particular solutions of the Schrödinger equation for V = 0. These two principles are the mathematical formulation of de Broglie’s idea. Schrödinger’s real contribution lies in the incorporation of forces. The previous analysis seems correct; it will account for experimental observation provided one condition is satisfied. This condition is that after the slits the wave function at a point C is indeed the sum ψC = ψ A + ψ B

(3.8)

of the wave function which would describe the atom if it went through A (B being blocked) and of the wave function which would describe the atom if it went through B ( A being blocked), each of which would give diffraction distributions showing no interference fringes. Here, we have in front of us the most fundamental thing of all this chapter. Wave functions have the right to add up. That is the fundamental property of wave functions. We promised addition; there it is! More generally, let ψ1 and ψ2 be two wave functions, then the combination Ψ = αψ1 + βψ2 ,

(3.9)

where α and β are complex numbers (one can produce a phase shift or an attenuation of a wave) belong to the family of wave functions. This sum is a possible wave function, and this first principle is called the superposition principle. The superposition property is completely connected with the interference phenomenon. It is much more important than the formula λ = h/ p; any other formula would not change the fundamentals of the experiment. In other words, two states have the right to add up in order to construct a third one. That is the fundamental notion. What is amazing with mathematicians is that when they see simple things such as that, they understand immediately underlying structures. They tell us that the set of wave functions {ψ(r,  t)} is a complex vector space. If one imposes the normalization condition (3.4), |ψ|2 = 1, it is what one calls a Hilbert space, the space of square integrable functions. This property, which expresses interferences, is much more important than the concept of a wave function itself. Does this theory account for Young slit interferences?

3.3 Superposition Principle

41

Yes, because the Schrödinger equation is linear and any linear combination of solutions is a solution. If one formulates the problem of sending a plane wave from the left of the slits, one can prove that in the vicinity of the axis and, at large enough distances, the usual interference formulae apply. It is a complicated mathematical problem, but it has a well-defined solution. (Note that the problem of interferences on the surface of water presented in the previous chapter, Fig. 2.4, is much simpler mathematically, because the two sources are independent. Here, one must take into account that it is the same plane wave that is incident on the slits.)

3.4 Wave Packets However, de Broglie waves are not wave functions. In fact they are not normalizable. This is not a very difficult problem. It is frequently encountered in wave physics: a plane wave does not exist in practice. It is a useful idealization that makes calculations easy, but physically it would fill all space at all times. A physical wave is always localized in space at a given moment, and localized in time at a given place. A wave is never exactly monochromatic; there is always some dispersion in frequency and in wavelength.

3.4.1 Free Wave Packets The representation of a realistic physical situation is a linear superposition of monochromatic plane waves, each of which is a particular solution of the Schrödinger equation, of the form  ψ(r, t) =

ϕ( p) e−(i/)(Et−p·r)

d3 p p2 , where E = , (2π)3/2 2m

(3.10)

such that all these waves interfere destructively outside some region of space. This is called a wave packet. In this formula, we have introduced the constant (2π)3/2 for convenience, and the complex function ϕ( p) is arbitrary (it determines ψ) except that expression exists and we want it to be properly normalized (we want that  the |ψ|2 = 1). This expression satisfies the Schrödinger for a free particle. One proves in mathematics that it is the general solution of the free Schrödinger equation.

3.4.2 Fourier Transforms Note: some further mathematical considerations are treated in Sect. 3.11. However, we want the resulting wave function to be normalizable. How can we do this without having to check each time?

42

3 Wave Function, Schrödinger Equation

The tool that answers this question is the Fourier transformation. The Fourier transformation is one of the most important mathematical structures discovered in the 19th century. It has numerous applications in mathematics, in electronics, in physics, in chemistry, and so on. For our purpose, we only need to have in mind one definition and three consequences. One morning in 1812, Fourier found out that he could solve many problems by transforming a function g of a real variable k into another function f of a real variable x with the formula  1 (3.11) eikx g(k) dk. f (x) = √ 2π • The inverse transformation, which allows us to get back g knowing f , is given by 1 g(k) = √ 2π



e−ikx f (x) d x.

(3.12)

The similarity between the two expressions (3.11) and (3.12) is such that we can say that f and g are Fourier transforms of each other. • The second property is that the Fourier transformation is what is called an isometry. If f 1 (x) and f 2 (x) are, respectively, Fourier transforms of g1 (k) and g2 (k), then we have the Parseval–Plancherel theorem:   f 1∗ (x) f 2 (x) d x = g1∗ (k) g2 (k) dk. (3.13) Of course in all this, we assume that all expressions exist and behave properly. • The third property is that the more the support of |g(k)|2 is concentrated (around 2 (and vice versa). If we normalize some value k0 ), the larger  is 2the support of| f (x)| 2 f and g to one, so that | f | d x = 1 and |g| dk = 1, we can consider |g(k)|2 and | f (x)|2 as probability laws for the variables k and x, respectively, and if we consider the resulting expectation values and dispersions  k =

k |g(k)|2 dk ; (Δk)2 = k 2  − k2 ,

(3.14)

and similarly for x and Δx in terms of f , the product of the dispersions Δx and Δk is constrained by the inequality Δx Δk ≥ 1/2.

(3.15)

3.4 Wave Packets

43

3.4.3 Shape of Wave Packets Coming back to wave packets (3.10), the constant  is present in order to have dimensionless quantities because p.r has the dimension of an action. We see that ψ is the Fourier transform of ϕ( p) e−i Et/ which implies that 

 |ψ(r, t)|2 d 3 r =

|ϕ( p)|2 d 3 p.

(3.16)

  Therefore, |ψ|2 = 1 if and only if |ϕ|2 = 1. In the choice of ϕ above, we only need to have |ϕ|2 = 1; ϕ is otherwise arbitrary. We see that: 1. If ϕ is very concentrated around some value p0 the wave function will be close to a monochromatic plane wave in a large region of space. It is therefore a realistic wave function for a beam of atoms. A simple case is represented in Fig. 3.2. If we look at the wave function closely, it resembles a plane wave; if we look at it from far away, it may seem to be a concentrated distribution. 2. Conversely, it suffices to exchange the two functions in order to obtain the wave function of an atom whose position is well localized in space; that is, ψ is concentrated near r0 , as can be seen in the symmetric Fig. 3.3.

Fig. 3.2 Wave packet obtained with a square form of φ( p) localized in the vicinity of a value p0 (top). The two figures at the bottom show the probability distribution (actually the square of the real part) with two scales of the variable which differ by a factor of 10

44

3 Wave Function, Schrödinger Equation

Fig. 3.3 Real part of the wave function ψ(x) corresponding to a localized function φ( p) in the vicinity of p0 and vice versa, in two cases

3.5 Historical Landmarks Now, how and when did these ideas emerge? In 1923, Louis de Broglie, who initially studied history, proposed his wave hypothesis. His thesis is a beautiful text with five chapters. The Science Faculty of the Sorbonne was a little bit embarrassed. Was that brilliant or foolish? • Paul Langevin gave de Broglie’s thesis to Einstein in April 1924, and Einstein was very enthusiastic for several reasons. • First of all, the wave hypothesis could lead to energy quantization as a stationary wave problem, provided one used proper mathematics. This quantization could be achieved without abandoning the continuity of physical laws. Bohr kept on saying that discontinuity was a fundamental property of matter, Einstein was shocked by such an idea. • In addition, Einstein managed to explain problems of statistical physics with the wave assumption. He advertised the idea, saying that it was much more than an analogy with photons; it was very deep. • At the beginning, Schrödinger found that the idea was elegant, but he was skeptical because of relativity. It was Einstein who urged Schrödinger to work on the subject. Actually, Schrödinger called that the wave theory of Einstein and de Broglie. He acknowledged that without Einstein he would not have done anything. • It is Schrödinger who introduced the Greek letter ψ for the wave function, which has become a tradition. However, he made a mistake in its interpretation. • It is Max Born (a mathematician, who had the chair of theoretical physics in Göttingen) who gave the probabilistic interpretation at the end of 1926. What happened is that, using the Schrödinger equation, Born calculated the scattering of electrons

3.5 Historical Landmarks

45

on nuclei. He noticed that the square of the wave function |ψ|2 gave the value of the measured intensity. At the same time, the development of Geiger–Müller counters provided an alternative measurement technique. Instead of measuring electric current intensities, one could count numbers of electrons; in other words, one did statistics. Hence the interpretation of |ψ|2 as a probability density. This probabilistic interpretation achieved the synthesis of two “complementary” aspects of the behavior of particles (electrons or atoms). In the same experiment, an atom behaves as a wave when it interacts with the Young slits, and as a particle when it is detected (a detector and a diffracting system do not act in the same way). That is the real problem. An atom is not schizophrenic; it is neither a wave nor a particle in the classical sense. It is a well-defined quantum object. In given experimental situations, this quantum object may behave as a Newtonian particle or as a wave, which seems contradictory to those who possess only classical concepts.

3.6 Momentum Probability Law So, everything seems to work well up to now. But one can go much further.

3.6.1 Free Particle We said that the wave function ψ describes completely the state of the particle at time t. Therefore, it must contain the information on the particle’s velocity or momentum. But we do not see that information at the moment. So, let’s invent. The Fourier transformation is fantastic because it enables us to invent. Now, we use mathematics just as in the case of instantaneous velocity and derivatives. In fact, the properties of the Fourier transformation suggest that the probability law of momentum is given by |ϕ( p)|2 ; that is, d P( p) = |ϕ( p)|2 d 3 p.

(3.17)

This can be proven1 ; we won’t do this here. We have an idea and we wish to see if it is plausible and consistent, and if it works.

1 The

proof can be found in R.P. Feynman’s Thesis Ph.D. Princeton University, 1942; See R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals McGraw-Hill, New York, p. 96–100; There is a direct derivation in J.-L. Basdevant and J. Dalibard, Quantum Mechanics, Chap. 2, Sect. 6, 2005.

46

3 Wave Function, Schrödinger Equation

1. |ϕ|2 is nonnegative, and, because |ψ|2 is normalized to one, then |ϕ|2 is also normalized to one, its integral is equal to one. Therefore it has all the properties of a probability law. 2. There is a one-to-one correspondence between ϕ and ψ, which satisfies our requirement that the information on the momentum is contained in the wave function. If our idea is true, the Fourier transformation extracts the information on the momentum of the particle from the wave function. 3. De Broglie’s assumption says that the wave function of a free particle of welldefined momentum p0 should be a monochromatic plane wave. In order to get close to that, one must find a function ϕ that is large in the vicinity of p0 and small elsewhere as in Fig. 3.2. Now, we see that we can use a mathematical notion in order to represent a physical idea. The Fourier transform of the wave function is the probability amplitude for the momentum of the particle (again, this can be proven). That’s a fantastic physical result! If the particle is localized in momentum p, then it is wavelike in position x. If it is localized in position, then it is wavelike in momentum, as one can see in Fig. 3.3. The space of positions and the space of momenta are reciprocal spaces, conjugate of each other. In order to describe the state of the particle, we can use either of them, in a symmetrical way.

3.6.2 General Case In fact, not only is this true, but it is general. For a particle placed in an arbitrary potential, if we consider the Fourier transform of the wave function,  ϕ( p, t) =

ψ(r, t) e−(i/)(p·r)

d 3r , (2π)3/2

(3.18)

then ϕ( p, t) is the probability amplitude for the momentum, d P( p) = |ϕ( p, t)|2 d 3 p.

(3.19)

The free particle case is a particular case for the behavior in time. Now, we can calculate various expectation values of the momentum or of functions of the momentum.

3.7 Heisenberg Uncertainty Relations Here, we face an extraordinary discovery, a central result, and, at the same time, a tragedy. Indeed, it is a consequence of Fourier analysis and of Eq. (3.15) that, whatever the wave function is, whatever the state of the system, the dispersions on measurements of positions and momenta along the same axis always satisfy the inequalities

3.7 Heisenberg Uncertainty Relations

Δx Δpx ≥

   , Δy Δp y ≥ , Δz Δpz ≥ , 2 2 2

47

(3.20)

which are called the Heisenberg uncertainty relations. Here is a true physical constraint on a wave packet. If we compress it in one variable, it expands in the other! If it is compressed in position (i.e., localized) then it must be spread out in momentum. If it is compressed in momentum, it is spread out in space. What is the physical meaning of Heisenberg’s inequalities? • Once more, suppose we prepare N systems in the same state. For half of them, we measure their positions x; for the other half, we measure their momenta px . Whatever way we prepare the state of these systems, the dispersions obey these inequalities. • These are intrinsic properties of the quantum description of the state of a particle. • Heisenberg uncertainty relations have nothing to do with the accuracy of measurements. Each measurement is done with as great an accuracy as one wishes. They have nothing to do with the perturbation that a measurement causes to a system, inasmuch as each particle is measured only once. • In other words, the position and momentum of a particle are defined numerically only within limits that obey these inequalities. There exists some “fuzziness” in the numerical definition of these two physical quantities. If we prepare particles all at the same point, they will have very different velocities. If we prepare particles with a well-defined velocity, then they will be spread out in a large region of space. • Newton’s starting point must be abandoned. One cannot speak simultaneously of x and p. The starting point of classical mechanics is destroyed. Some comments are in order. A plane wave corresponds to the limit Δp = 0. Then Δx is infinite. In an interference experiment, the beam, which is well defined in momentum, is spread out in position. The atoms pass through both slits at the same time. We cannot “aim” at one of the slits and observe interferences. The classical limit (i.e. how does this relate to classical physics) can be seen in a variety of ways that are more or less equivalent. One possibility is that the orders of magnitude of x and p are so large that /2 is not a realistic constraint. This is the case for macroscopic systems. Another possibility is that the accuracy of the measuring devices is such that one cannot detect the quantum dispersions Δx and Δp. We show later on how one recovers quantitatively the laws of classical mechanics themselves. Size and Energy of a Quantum System Our third comment is that now we can do a lot of physics. In fact, uncertainty relations enable us to estimate orders of magnitude of various effects without solving complicated equations. In the lowest energy state of a quantum system, its “ground state,” that is, when it is no longer excited, the product Δx Δp is of the order of ,

48

3 Wave Function, Schrödinger Equation

Δx Δpx ∼ .

(3.21)

This allows us to estimate quickly orders of magnitude. Of course we will always miss a numerical factor of order 1. This factor can only be obtained by solving equations, but we will know the orders of magnitude. For instance, there is a relation between the size and the energy of a quantum system. Consider a particle that is bound to a fixed center, for instance an electron around a proton in an atom, or nucleons (neutrons and protons) in a nucleus. We assume the center is fixed at position r0 ; that is, we neglect quantum effects of the center which is very massive. The particle has some wave function. In the center of mass system, we can choose the origin of coordinates so that we have by assumption r = 0,  p = 0. The dispersion Δx is of the order of r0 , the size of the system. The square of the momentum dispersion Δp is equal to 2m times the kinetic energy Δp 2 =  p 2  = 2m E kin . There is therefore a relation between the kinetic energy and the size of the system E kin  

2 . 2mr02

The smaller the system is, the larger its kinetic energy. For an external electron in an atom, the size is of the order of an Angström, and we obtain a kinetic energy of a few eV. In a nucleus, we lose a factor of 2000 because of the mass, but the size, one fermi, gives a factor of 1010 larger; we therefore obtain energies of the order of tens of MeV. We do obtain the correct orders of magnitude, because the kinetic energy, binding energy, and potential energy are of the same order of magnitude unless there exists a pathology. This also explains why, in order to probe matter at short distances, one must use large energies. It is necessary to use powerful particle accelerators. Stability of Matter More important is that uncertainty relations allow us to prove the stability of matter, which is one of the greatest contradictions of classical physics. In the world of Newton and Maxwell, matter should be unstable and the world should collapse. This is an inevitable consequence of the theories of Newton and Maxwell. Consider the very simple case of a hydrogen atom; one electron orbiting around a proton in the Coulomb potential V (r ) = −qe2 /4π 0 r . Suppose the orbit is circular, for simplicity, of radius r . Mechanical equilibrium implies mv 2 /r = qe2 /4π 0 r 2 , and the energy of the electron is therefore:

3.7 Heisenberg Uncertainty Relations

49

Fig. 3.4 Effective interaction potential due to the interplay between Coulomb attraction and quantum repulsion owing to uncertainty relations in an atom

E=

p2 1 q2 + V (r ) = − . 2m e 2 4π 0 r

This energy is not bounded from below; the more the radius decreases the more the energy decreases. Now, in its circular motion, the electron is accelerated. The consequence of Maxwell’s equations is that it must then radiate and lose energy. Therefore, from the classical point of view, matter is unstable. The electron should radiate continuously and it should collapse on the nucleus (r → 0) by radiating an infinite amount of energy. It is perhaps the most serious problem of classical physics, although people did not realize that because the electron and the nucleus were found rather late. Uncertainty relations preserve us from this catastrophic fate and suppress this inconsistency. Let r  be the average distance of the electron and the proton, which we consider fixed (the recoil is negligible). The Coulomb energy is of the order of qe2 /4π 0 r . If we use the order of magnitude (3.21), the kinetic energy is E k ≥ 2 /2m e r 2 . Therefore, the total energy is of the order of E≥

2 q2 − . 2m e r 2 4π 0 r 

(3.22)

This quantity is bounded from below. Its minimum is attained for r  = 4π 0 2 / (m e q 2 ) ∼ 0.53 10−10 m, which is bounded from below, and leads to E min

me ∼ − 2 2



q2 4π 0

2 = −13.6 eV.

This is a fundamental result. The uncertainty relations put a lower bound on the average distance of the electron and the proton, as well as on their potential energy, and their total energy. This explains the stability of matter (Fig. 3.4).

50

3 Wave Function, Schrödinger Equation

The above argument is not rigorous; it can be made rigorous because one can prove other forms of the uncertainty relations, in particular that for any system we have  p 2  ≥ 2 1/r 2 . If we apply this to E =  p 2 /2m − (qe2 /4π 0 )1/r  the argument becomes rigorous: 1/r  is bounded from above. The uncertainty relations create a “Heisenberg pressure,” which reacts against the fact that the electron and proton come too close to each other. Consequently, this results in an equilibrium situation and the quantum impossibility that matter collapses. This term has the behavior of a centrifugal force, but it does not come from angular momentum. We show later on that, in its quantum ground state, the electron of the hydrogen atom has zero angular momentum and does not radiate.

3.8 Controversies and Paradoxes Heisenberg was a young assistant of Max Born and, as early as 1924, he had elaborated his own version of quantum mechanics which seems at first very different from what we are doing (we come back to this point). It was only in 1927 that he stated his uncertainty principle (the proof was given later). It is one of the most fundamental elements of quantum mechanics. It is compulsory for the consistency of quantum mechanics. If an experiment contradicts Heisenberg’s relations, all of quantum mechanics is destroyed. The 1927 Solvay Congress The uncertainty principle set off terrible debates. Some people were astonished; some others were enthusiastic. And there was the tragedy we mentioned above. The year 1927 was a breaking off for one of the founders of quantum mechanics, and not the least, Albert Einstein, one of the greatest physicists in history. In 1905 he had originated the theory of Brownian motion, he had invented special relativity, and he had invented the photon, for which he was awarded the Nobel prize in 1921. In 1915, he had constructed general relativity, a most beautiful theory. He had made tremendous contributions to statistical physics and to quantum physics. In 1917, he understood the existence of stimulated emission, a keypoint for laser physics. At the fifth Solvay Congress, in Brussels in October 1927, he rose up and revolted against the probabilistic aspect of quantum mechanics, and against uncertainty relations which he disliked profoundly. Einstein did not understand what was going on. Einstein’s revolt concerned two points. One is the notion of a complete description of reality. He thought that a complete description is possible in principle, but that the probabilistic description is simply quicker to handle and more convenient. The other reason is the notion of determinism: same causes produce same effects. Einstein said the famous: “God does not throw dice!” Actually, Einstein’s words are in a letter he wrote to Max Born on December 4th, 1926,2 “The theory produces 2 “Die Theorie liefert viel, aber dem Geheimnis des Alten bringt sie uns kaum näder. Jedenfalls bin ich überzeugt, dass der nicht würfelt.”

3.8 Controversies and Paradoxes

51

a good deal but hardly brings us closer to the secret of the Old One. I am at all events convinced that He does not play dice.” Because the theory works and gives good results, it must be some intermediate step: there must exist a more complete underlying theory, involving, for instance, “hidden variables”3 to which we do not yet have any access and over which we average things. And, at the Solvay meeting, Einstein refused and reacted abruptly. This was due partly to the dogmatic attitude of Niels Bohr. So Einstein attacked; in particular he attacked uncertainty relations. Uncertainty relations are, in that respect, similar to the Carnot principle and perpetual motion in thermodynamics. Einstein accumulated counterexamples. But, of course, his counterexamples were far from being obvious. One had to work hard to disprove them. The EPR Paradox In 1935, Einstein proposed a famous paradox: the EPR—Einstein, Podolsky, and Rosen—paradox which we shall describe in Chap. 17. In that “gedanken experiment,” by considering a couple of correlated particles, he showed how because of momentum conservation, one could in principle know as accurately as one wished both the position and the momentum of one of the particles, which “beats” uncertainty relations. Hidden Variables, Bell’s Inequalities In another version, proposed 20 years later by David Bohm, with spin, it is a genuine puzzle. This goes so far that physicists addressed the question as to whether there exists a more complete theory involving “hidden variables” of which we would have some ignorance. After all, when we play cards, we are blind to the identity of each card that is dealt. But if one used sophisticated enough devices, one could certainly tell the difference between the ace of spades and the seven of hearts. In order to play end enjoy it, we forbid ourselves to have a more complete knowledge which exists in principle. The amazing thing is that John Bell, in 1965, was able to show that this assumption leads to quantitative measurable consequences that are in opposition to the predictions of quantum mechanics.4 We come back to this in Chap. 17. The Experimental Test This attracted the attention of the general public and the press. There are philosophical considerations attached to that problem. In 1979, at the Cordoba Colloquium, there were many discussions on parapsychology, levitation, oriental philosophy, and other bold considerations. This was reported by the French newspaper Le Monde of October 24, 1979 (considered as very serious intellectually). All of that was based on the Einstein–Podolsky–Rosen paradox. But exactly three years later, on December 15, 3 Actually 4 J.S.

Einstein never used that word. Bell, Physics 1, 195 (1964).

52

3 Wave Function, Schrödinger Equation

1982, in the same newspaper, the phraseology changed. One can see the words: “Experiment might ruin Einstein’s hopes,” “God probably plays dice,” and so on. Why did that change of phraseology occur? Because, since the late 1970s, a series of experiments have been performed, among which are those of Clauser and Freedman5 and of Alain Aspect and his group6 in Orsay, to try to see whether one could falsify quantum mechanics. In fact the results were negative! Quantum mechanics holds, even though it hurts some “common sense.” We must face experimental results and revise our way of thinking. These are extraordinary experiments which we discuss in Chap. 17. We see what kind of thoughts Feynman had when he said in 1965 “I think I can safely say that nobody understands quantum mechanics”.7

3.9 Exercises 1. Spreading of the wave packet of a free particle a. Consider a free particle moving along the axis x. Show that the time derivative of x 2 t can be written: dx 2 t = A(t) dt

i A(t) = m

with



 x

∂ψ ∂ψ ∗ ψ − ψ∗ ∂x ∂x

 d x.

b. Calculate the time derivative of A(t) and show that: dA = B(t) dt

with

c. Show that B(t) is constant. d. By setting: v12 =

2 m2

B(t) =



22 m2



∂ψ ∂ψ ∗ d x. ∂x ∂x

∂ψ ∂ψ ∗ dx ∂x ∂x

and ξ0 = A(0), show that x 2 t = x 2 0 + ξ0 t + v12 t 2 . e. Show that (Δxt )2 = (Δx0 )2 + ξ1 t + (Δv)2 t 2 holds, with: i ξ1 = m



 x

∂ψ0∗ ∂ψ0 ψ0 − ψ0∗ ∂x ∂x

 d x − 2x0 v0 ,

5 S.J. Freedman, J.F. Clauser, Experimental test of local hidden-variable theories, Phys. Rev. Lett. 28.938, 1972. 6 Experimental Realization of Einstein–Podolsky–Rosen–Bohm Gedankenexperiment: A New Violation of Bell’s Inequalities, A. Aspect, P. Grangier, and G. Roger, Physical Review Letters, 49, 91 (1982); Experimental Test of Bell’s Inequalities Using Time-Varying Analyzers, A. Aspect, J. Dalibard and G. Roger, Physical Review Letters, Vol. 49, 1804 (1982). 7 The Character of Physical Law, MIT Press, Cambridge, MA 1965.

3.9 Exercises

53

where ψ0 ≡ ψ(r, 0). The coefficient ξ1 can be interpreted physically using the results of the next chapter, as the correlation at time 0 between position and velocity: ξ1 /2 = xv0 − x0 v0 . One can verify that the constraint on ξ1 resulting from the fact that (Δxt )2 > 0 is equivalent to the condition that Δxt Δpt ≥ /2 at each time t. 2. The Gaussian wave packet Consider the wave packet given by:   ( p − p 0 )2 ϕ( p) = (πσ 2 2 )−1/4 exp − 2σ 2 2

(3.23)

a. For t = 0 show that Δx Δp = /2. b. Show that the spatial width of the wave packet at time t is given by: Δx 2 (t) =

1 2



t 2 σ 2 2 1 + σ2 m2

 .

(3.24)

3. Characteristic size and energy in a linear or quadratic potential Using an argument similar to that of the stability of matter, evaluate the characteristic size and energy of a particle with mass m moving in (i) a one-dimensional harmonic potential V (x) = mω 2 x 2 /2; (ii) a one-dimensional linear potential V (x) = α|x|.

3.10 Appendix: Dirac δ “Function”, Distributions Definition of δ(x) We often refer to point-like objects in physics. The mass density ρ(r) (or the charge density) of such an object is not a function in the usual sense, since it is everywhere zero except at point r 0 , but its “integral” is finite:  ρ(r) d 3r = m. The δ “function”, introduced by Paul Dirac, can describe such a density. Its mathematical definition was elaborated by the mathematician Laurent Schwartz in the framework of distribution theory, which we shall briefly describe in the next section. In this section we present the (mathematically improper) names and formalism used by physicists. For a real variable x, the “function” δ(x) has the following properties:  δ(x) = 0 for x  = 0

+∞

and −∞

δ(x) d x = 1.

(3.25)

54

3 Wave Function, Schrödinger Equation

For any function F(x) regular at x = 0, we have by definition:  F(x) δ(x) d x = F(0).

(3.26)

By a change of variables, one can define the function δ(x − x 0 ) for which:  F(x) δ(x − x 0 ) d x = F(x0 ).

(3.27)

The generalization to several dimensions is straightforward. Consider for instance r = (x, y, z), we will have: δ(r − r 0 ) = δ(x − x0 ) δ(y − y0 ) δ(z − z 0 ), that is to say:

(3.28)

 F(r) δ(r − r 0 ) d 3r = F(r 0 ).

Examples of Functions Which Tend to δ(x) One can construct distributions which are nearly point-like, using functions concentrated in the vicinity of a point x0 (Fig. 3.5). In order to do so, we consider sequences of functions depending on a parameter which determines their width (yε (x), gσ (x) in the first two following examples). Although these functions have no limit in the usual sense when their width goes to zero, the integral of their product with any function F regular at x = x0 remains well defined and tends to the limit F(x0 ).

Fig. 3.5 Examples of functions concentrated in the vicinity of a point, whose limit, in the sense of distributions, is equal to δ(x)

3.10 Appendix: Dirac δ “Function”, Distributions

55

1. Consider the sequence of functions yε (x) (Fig. 3.5a) defined by:  yε = Then  +∞ −∞

F(x) yε (x) d x =

1 ε



1/ε for |x| ≤ ε/2 . 0 |x| > ε/2

ε/2 −ε/2

(3.29)

F(x) d x = F(θε/2) with

− 1 ≤ θ ≤ 1.

In the limit ε → 0, yε (x) “tends” to δ(x).

1 exp(−x 2 /2σ 2 ). 2. Gaussian function (Fig. 3.5b): gσ (x) = √ 2πσ By the change of variables y = x/σ, 

+∞

−∞

1 F(x) gσ (x) d x = √ 2π



+∞

e−y

2

/2

F(σ y) dy.

−∞

In the limit σ → 0, the above integral remains well defined and gives the result F(0). As σ → 0, gσ (x) “tends” to δ(x). 3. Square of a “sine cardinal” (Fig. 3.5c): sin2 (xY )/(πx 2 Y ) with Y → ∞. Since  +∞  +∞ sin2 x sin2 xY d x = π, we have d x = 1. 2 x πx 2 Y −∞ −∞ 4. “Sine  cardinal” (Fig. 3.5d): sin(xY )/(πx) with  Y+∞→ ∞. +∞ sin x sin xY d x = π, we have for all Y : d x = 1. Since x πx −∞ −∞ The last case differs from the previous examples in the sense that for x  = 0, the function sin(xY )/(πx) does not tend to zero in the sense of functions when Y → ∞. On the contrary, it oscillates more and more rapidly. It is only “on the average” that it vanishes. Properties of δ(x) 1. δ(x) is an even function: δ(x − x0 ) = δ(x0 − x) (just make the change of variables in (3.27)). 1 2. We have δ(ax) = δ(x) (a real). Indeed we find for a > 0: |a| 

+∞

−∞

 F(x) δ(ax) d x =

+∞ −∞

F(u/a) δ(u)

du 1 = F(0). a a

For a < 0, we use the fact that δ is even. Distributions The previous notions can be formalized rigorously using distribution theory. We sketch this theory in order to extract some useful results.

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3 Wave Function, Schrödinger Equation

The Space S We shall consider the vector space S whose elements are complex valued functions ϕ(x) of one (or several) real variable x and which satisfy the following conditions: the functions ϕ(x) are indefinitely differentiable and, as x tends to infinity, they tend to zero, as well as all their derivatives, more rapidly than any power of 1/|x|. For 2 2 example, the functions e−x , x n e−x are elements of S. Linear Functionals A continuous linear functional f on space S is a mapping of S onto the complex numbers ( f : S → C), such that to each ϕ in S, there corresponds the complex number noted ( f, ϕ). This mapping satisfies the following properties: 1. Linearity: ( f, α1 ϕ1 + α2 ϕ2 ) = α1 ( f, ϕ1 ) + α2 ( f, ϕ2 ),

(3.30)

whatever the complex numbers α1 and α2 and the functions ϕ1 and ϕ2 belonging to S. 2. Continuity: if the sequence of functions ϕ1 , ϕ2 , . . ., ϕn tends to zero in S, the sequence of numbers ( f, ϕ1 ), ( f, ϕ2 ), . . ., ( f, ϕn ) tends to zero.  N.B. We say that the sequence ϕn tends to zero if x k (d/d x)k ϕn tends to zero uniformly in x whatever the integers k and k  positive or zero. These functionals are called tempered distributions and their set is named S  . Example 1. Let f (x) be a locally integrable function which remains bounded by a power of |x| as |x| → ∞. One can associate to it a functional, also noted f , with the formula:  ( f, ϕ) = f (x) ϕ(x) d x for any ϕ in S. (3.31) 2. Dirac δ distribution. It is the functional which to any function ϕ(x) in S associates the number ϕ(0). This is written as: (δ, ϕ) = ϕ(0).

(3.32)

It is convenient for physicists (but improper) to write:  δ(x) ϕ(x) d x = ϕ(0). In the examples we showed in the previous section, the statement yε or gσ tend to δ is incorrect. However, the statement: lim (gσ , ϕ) = (δ, ϕ)

σ→0

for any ϕ in S

(3.33)

3.10 Appendix: Dirac δ “Function”, Distributions

57

is perfectly correct. One says that gσ (or yε , . . .) tends to δ in the sense of distributions. Derivation of a Distribution When the distribution is associated to a differentiable function f (x), as in (3.31), one can write, all operations being legitimate: ( f  , ϕ) =



+∞

−∞

d f (x) ϕ(x) d x = − dx



+∞ −∞

f (x)

dϕ d x = −( f, ϕ ). dx

One defines the derivative d f /d x or f  of an arbitrary linear functional f in the set of tempered distributions by the relation: 

   dϕ df , ϕ = − f, . dx dx

(3.34)

Example 1. Derivative δ  of δ:

(δ  , ϕ) = −(δ, ϕ ) = −ϕ (0),

(3.35)

 which the physicists will write as δ  (x) ϕ(x) d x = −ϕ (0). 2. Consider the step function (Heaviside’s function) defined by:  Θ(x) =

0 1

x =

ψ ∗ (r, t) x ψ(r, t)d 3 r , < V > =



ψ ∗ (r, t)V (r)ψ(r, t)d 3r, (4.7)

which indeed have good form. The position and potential energy observables are the multiplication of the wave function by x, y, and z, or by V (r).

4.2.2 Momentum Observable But for momentum it is less obvious, except that the form  < px >=

ϕ∗ (r, t) px ϕ(r, t)d 3 p

(4.8)

has good structure. In fact, if we remember the definition of the Fourier transform, its inversion, and the Plancherel Theorem (3.13), we see that we simply have to find a function whose Fourier transform is px ϕ. But that is very simple; we only need to take the derivative ∂ ψ(r, t) = ∂x or



 ∂ ψ(r, t) = i ∂x



d3 p i px , ϕ( p, t) e(i/)p·r  (2π)3/2

(4.9)

d3 p . (2π)3/2

(4.10)

px ϕ( p, t) e(i/)p·r

Therefore (/i)∂ψ/∂x is the Fourier transform of px ϕ  ∂ ψ ←→ px ϕ, i ∂x and thanks to Plancherel, this is becoming really interesting and nontrivial.

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We obtain  < px >=

ϕ∗ (r, t) px ϕ(r, t)d 3 p =



ψ ∗ (r, t)





 ∂ ψ(r, t) i ∂x

d 3 r. (4.11)

In other words, in our principle, the observable pˆ x associated with the x component of the momentum is  ∂ . (4.12) pˆ x = i ∂x This holds for other components, and the vector observable pˆ is simply pˆ =

 ∇. i

(4.13)

And there we win! We don’t have to calculate the Fourier transform. We know how to express  px  or the expectation value of any function of the momentum directly in terms of the wave function. For instance, the expectation value of the kinetic energy, E k = p 2 /2m in Eq. (4.3) is  −2 (4.14) )Δψ(r, t) d 3 r. E k  = ψ ∗ (r, t) ( 2m

4.2.3 Correspondence Principle Now, we are nearly home. We just have to fix the form of other observables. In order to do this, we simply remember classical physics and use a principle that ensures, as we show later on, that in the classical limit one recovers the classical equations. This is called the correspondence principle. In classical mechanics, physical quantities are functions A(r, p) of the position and momentum variables. The correspondence principle consists of choosing in quantum mechanics the same functions of the position and momentum observables. To the quantity A(r, p) there corresponds the observable Aˆ = A(ˆr , pˆ ) . For instance −2  ∂  ∂ Δ , Lˆ = r × ∇ , that is, Lˆ z = (x − y ). Eˆ k = 2m i i ∂y ∂x

4.2 Observables

69

4.2.4 Historical Landmarks Why is this principle of physical quantities so difficult to accept for us, whereas it caused no difficulty for the people who built quantum mechanics? Because we are missing a link (or some mathematics) and because history is not a logical sequence but an arborescence. In 1926, there were actually two versions of quantum mechanics. There was wave mechanics (in 1923 de Broglie’s hypothesis, in 1926 Schrödinger’s equation). But there was also matrix mechanics, which had been constructed since 1924 in a most inspired way by Heisenberg, and had been developed by Heisenberg, Born, and Jordan. Furthermore, both versions gave good results. So people quarreled. Are electrons or atoms waves or matrices? That was a terrible question. Physicists discussed and argued about fundamental questions, but not this question of observables, which is basically technical. Born had declared Schrödinger’s works, namely finding energy levels as stationary wave problems, as being of “unsurpassed greatness in theoretical physics.” In 1926, Schrödinger had given a series of talks in Copenhagen. There, he, who was a man of great distinction and class, had been literally attacked by Niels Bohr, rigid and dogmatic, to such an extent that he said, “If one must stick to that idea of quantum jumps, I really regret getting involved in all this business!” Niels Bohr had answered him: “But, Herr Schrödinger, you shouldn’t, given all the publicity that your works have given to our ideas”. One can imagine the atmosphere. But Schrödinger and Dirac, the little Mozart of quantum mechanics, were wise people and they brought back peace, as we show in Chap. 6, because they knew some mathematics. Independently, at the end of 1926, they showed that the two approaches were equivalent, and they saw this principle emerge in a natural way. There was no problem. They had both the physical concepts and the mathematical tools. Wave mechanics was interested primarily in the state of a system; matrix mechanics concentrated primarily on physical quantities. For our information, the importance of the Fourier transform, which has been very useful for us, was pointed out quite late, in 1927, by C.G. Darwin.

4.3 A Counterexample of Einstein and Its Consequences At this stage, in order to progress, it is useful to analyze one of Einstein’s counterexamples. This will allow us to address the question, “OK, God plays dice, but is it a full time job?” In other words, are there situations where indeterminism disappears and God ceases to throw dice? Einstein did not like the probabilistic aspect of the theory. So he attacked the root of the problem. Indeterminism is inscribed in the uncertainty relations, so why not break them to pieces, once and for all.

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Fig. 4.1 Gedanken experiment of Einstein in order to beat Heisenberg’s uncertainty relations

Fig. 4.2 Velocity distribution of the particles after they pass through the second diaphragm, owing to the uncertainty relations Fig. 4.3 Diffraction of a wave by a small hole

The first counterexample, “gedanken experiment,” of Einstein is the following. Between two diaphragms, one puts a device made of two cogwheels that let pass particles of a well-defined velocity (as in Fizeau’s measurement of the velocity of light). The second diaphragm has a dimension δz as small as one wishes (Figs. 4.1, 4.2 and 4.3). The cogwheel device ensures that particles which pass through the second diaphragm have a well-defined longitudinal velocity vlong . But, in order for the particles to pass through the diaphragm, it is necessary for them to have a transverse velocity vt less than vt ≤ vlong δz/L.

4.3 A Counterexample of Einstein and Its Consequences

71

We can make the size of the diaphragm δz as small as we wish. Therefore, immediately after the particles have passed the diaphragm, we are sure of their positions along z, up to δz, and we know their velocity vt up to vlong δz/L. We therefore beat Heisenberg! • “Nein, nein, nein! Not at all!” said Heisenberg to Einstein. • “Warum nicht? Why not?” asked Einstein. • “Look at the result of the experiment. If you measure the velocity vz after particles have passed through the diaphragm, for a large number of particles you will not find only one value up to vlong δz/L, but a whole lot of possible values.” • “Why?” asked Einstein. • “Because of the uncertainty relations! These particles are emitted by a small hole and have a small spreading δz in position along z. Therefore, they have a large spreading in momentum Δpz ≥ /2δz (Fig. 4.2).” • “You’re making fun of me,” said Einstein. “If you use uncertainty relations to prove uncertainty relations, something’s going wrong in your mind!” • “OK, you prefer de Broglie waves,” said Heisenberg, “but it’s exactly the same result. The wave diffracted by a small hole is more and more dispersed in the wave vector as the hole gets smaller (Fig. 4.3). Uncertainty relations also exist in usual wave physics. They express globally one aspect of diffraction. It is only because we apply them to particles on which we have some a priori knowledge that we are shocked.” The only thing we know with certainty is that after passing through the second diaphragm, the particles have passed through a hole of size δz, therefore there is a dispersion in momentum Δpz which is larger than /2δz in the statistical sense (i.e., if we measure a large number of particles). This is what Heisenberg explained to Einstein: the longitudinal and transverse velocities belong to the past; talking about them is history or philosophy. One cannot do anything with that. And Heisenberg added, “It is a matter of personal belief, without any operational content, to decide whether the past history of a particle has a physical reality. It is something one cannot measure.” Heisenberg came close to convincing Einstein just before the 1927 Solvay meeting.

4.3.1 What Do We Know After a Measurement? But we have learned a lot with this counterexample. We know something for sure after the particle exits. It has passed through the hole. Therefore, its position is well-defined (up to δz). But, placing a hole and checking that a particle has passed through it is simply one of the many ways to measure the position of that particle. We could use any measuring apparatus and measure any physical quantity; the situation would be the same. 1. Therefore, we know something for sure after a measurement on a single system. We have found some well-defined measured value (up to δz in this particular case).

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Therefore, if we redo the measurement immediately afterwards (immediately means that the wave function hasn’t had time to evolve appreciably) we will find the same result z 0 up to δz, and only that result. Therefore, in performing that measurement, we have obtained information on the state of the particle after the measurement process. A measurement on a single system provides us with some information on the state of the system after the measurement. This can also be considered as preparing the state of the system. 2. However, can we check that Einstein was partially right and that the velocity was well-defined by his device? Yes, of course; we must put a speedometer before the diaphragm. For N particles, the velocity is well-defined, but then a position measurement will show that the particles are delocalized; that is, the position dispersion Δz is very large before the diaphragm. And we can also move the hole around in all possible positions, which will give us the position probability before particles pass through the hole. After that, by applying the Fourier transformation (which in full rigor means making an assumption on the phase) we will also get a measurement of the velocity. It is by a measurement on N identical systems that we can acquire information on the state before the measurement. 3. We can imagine the origin of many philosophical questions at the time, and still now: what is physical information? What is reality? Does reality exist independently from the fact that one observes and measures? Does a tree falling in a forest make noise even if no one is there to listen to it?

4.3.2 Eigenstates and Eigenvalues of an Observable Now, obviously, this analysis implies much more. Let’s pursue this argument. 1. Just after a position measurement, there is no longer any determinism on the position. A further position measurement will give us a known value; the value of the position is sure and well-defined. The state is such that God has stopped playing dice! 2. The completely general consequence is that, for any physical quantity A, there must exist particular states such that the result of a measurement is sure and well-defined. 3. What are the corresponding particular wave functions? This is formalized using ˆ the notions of eigenfunctions and eigenvalues of the observable A. A function ψα is an eigenfunction of Aˆ if the application of Aˆ on this function gives the same function multiplied by a number aα which is the corresponding eigenvalue ˆ α (r, t) = aα ψα (r, t). Aψ With this definition, the theorem is very simple.

(4.15)

4.3 A Counterexample of Einstein and Its Consequences

73

ˆ then the result of a measurement of A is Theorem 1 If ψ is an eigenfunction of A, certain and equal to the corresponding eigenvalue. Proof. The expectation value of A is aα , and the expectation value of A2 is aα2 , therefore the dispersion vanishes, and there is no uncertainty. Indeed, we have  a =

ˆ α d 3 r = aα ψα∗ Aψ 

but a  = 2

ψα∗ Aˆ 2 ψα

 |ψα |2 d 3r = aα ,

(4.16)

 d r= 3

aα2

|ψα |2 d 3r = aα2 ,

therefore (Δa)2 = a 2  − a2 = 0 QED. Later on, we show that the converse is true. Therefore the result of a measurement ˆ We can is well-defined if and only if the wave function ψ is an eigenfunction of A. also say that “the value of A is well-defined.” Careful: all of this is easier to formulate and to understand with discrete probabilities than with continuous probabilities. Hence the answer to the question, “What are the possible results of a measurement?” Obviously these are the eigenvalues of the corresponding observable. If we are sure that in a further measurement, we will find the same value ai that we have already found, that means that the number ai found previously is one of the eigenˆ values of A.

4.3.3 Wave Packet Reduction Now, the last consequence and not the least. The measurement, which we can decide to perform at some instant, is an irreversible act that changes the wave function completely. Before the measurement of a given quantity, the particle had some wave function that led to a variety of possible results in a measurement of that quantity. After the measurement, it has another wave function for which the result is unique. This is completely irreversible! Once we have made the measurement, we cannot decide that we haven’t made it and come back to the previous wave function. In other words, a measurement, which we are free to perform or not, modifies the wave function instantaneously in all space. Because of the very intuitive example presented above, this is called the reduction of the wave packet. Before, it was spread out in space; just after the measurement it is very concentrated. And this wave packet reduction postulate is essential in the formalization of quantum mechanics. It is understandable that there lies the source of serious problems which still last in the interpretation of quantum mechanics. That seems to be an instantaneous phenomenon at a distance. Isn’t that in contradiction to relativity?

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4 Physical Quantities

In fact this is the point on which Einstein constructed his EPR paradox a few years later. No signal can travel faster than light; it is not possible to change the wave function instantaneously at a distance. However, this is what experiments on Bell’s inequalities have shown to occur. Something does happen instantaneously at a distance. Experiment says that the predictions of quantum mechanics are right. But no information can travel faster than light because, in order to observe the phenomenon (the correlation of the values of two quantities at a distance), one must phone other people at other places to know what happened there. We now have all the elements of the theory. Notice that the language is getting closer and closer to that of linear algebra: we have spoken of linear mappings, we used the words Hermitian, eigenvalues, and indeed, in a while, quantum mechanics becomes more and more similar to matrix calculus.

4.4 The Specific Role of Energy In the following, we study various systems, and various physical quantities. But one of these quantities has a special role and we encounter it constantly because it has considerable importance. That is energy.

4.4.1 The Hamiltonian How did Schrödinger test his equations? • He was looking for wave equations. The problem was to find energy levels as the result of a stationary wave problem. Therefore, Schrödinger looked for stationary solutions of his equations (he actually wrote several before he ended up with the good one), solutions of the form ψ(r, t) = φ(r)e−iωt .

(4.17)

Inserting this into the Schrödinger equation, the term e−iωt factorizes and we obtain a time-independent equation −2 Δφ(r) + V (r)φ(r) = ωφ(r). 2m

(4.18)

• Schrödinger knew a lot of mathematics. He knew that the physically acceptable solutions (i.e., for us square integrable functions) of such a problem form a discrete set {φn (r), ωn }. Therefore, the frequencies and the energy levels are quantized E n = ωn .

4.4 The Specific Role of Energy

75

• But all of this must be consistent. If we are seeking energy levels of atoms, we are looking for particular solutions such that the energy is well-defined. Now, let’s consider a particle in a potential V (r ). Classically its total energy is E = p 2 /2m + V , and the corresponding observable is therefore pˆ 2 2 Hˆ = +V =− Δ + V. 2m 2m

(4.19)

The equation −

2 Δψ(r, t) + V ψ(r, t) = Eψ(r, t) , that is, Hˆ ψ(r, t) = Eψ(r, t), 2m

(4.20)

is the eigenvalue equation for the energy. It is the same as (4.18), since the time variable does not play any role. We can suppress it as in (4.18) which is called the time-independent Schrödinger equation. The energy observable Hˆ (4.19) is called the Hamiltonian of the system. How is it that Hamilton, who lived from 1805 to 1865, is involved in quantum mechanics, although he existed one century before its discovery? Well, he deserves it! The story is fascinating. Although he did not know quantum mechanics, Hamilton had understood that the structure of classical mechanics and quantum mechanics is the same. He had first shown that geometrical optics was a limit of wave optics for small wavelengths. He was fascinated by variational principles, in particular by the similarity between Fermat’s principle in optics and the least action principle of Maupertuis. In 1828, Hamilton wrote the following extraordinary observation, “Newtonian mechanics is only one of the possible mechanics. It corresponds to the same limit as geometrical optics compared to wave optics, whereas geometrical optics is only an approximation.” Nobody noticed, even Hamilton himself. The great mathematician Felix Klein pointed that out with some regrets in 1891. Of course there was no experiment at that time which showed that this idea could have any application. Planck’s constant appeared nowhere.

4.4.2 The Schrödinger Equation, Time and Energy We observe a most remarkable property. The Schrödinger can be written in the form i

∂ ψ = Hˆ ψ. ∂t

(4.21)

This is the true Schrödinger equation. It is “simpler” than the other one. Naturally it boils down to the same thing in the present context. But it is more general, as we show later. It is valid for any system and, when we generalize quantum mechanics,

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the wave function ψ is replaced by another object, a vector in Hilbert space, the Hamiltonian Hˆ is an operator in that space, and the Schrödinger equation is simply (4.21). If we look at it from that point of view, it tells us some thing remarkable. What determines the time evolution of the state of a system is the energy observable! Caution! The information on the energy of a system is contained in the wave function, and Hˆ is the tool that enables us to extract it. Nevertheless, there exists a relation between two fundamental but equally mysterious physical concepts: energy and time. We must be careful. Quite often, people think that scientists are partially idiotic and that they know only complicated things such as mathematics. In order to make you feel comfortable at parties, they will ask you about Fermat’s theorem, black holes, or what was there before the big bang. But take care, if they ask you what are time or energy, it’s far better to shift the conversation to football, the global temperature increase, or tornadoes. The word energy is everywhere. Time is one of the most difficult physical concepts to define. I’ve looked it up in the Oxford Concise Dictionary, in order to see simple definitions. • Energy is “a body’s power of doing work by virtue of stresses resulting from its reaction to other bodies.” • Time is “The progress of continued existence viewed as affecting persons or things.” That’s superb, but we haven’t really made much progress. We don’t know what time is. Does time simply exist? • The past no longer exists. • The future does not yet exist. • The present instant is of measure zero in the mathematical sense; as soon as it’s arrived, it’s already gone. In the Confessions, Saint Augustine wrote: “What is time? If no one asks me, I know. As soon as someone asks the question, and I want to explain it, I no longer know.” It’s true that we don’t know what time is, no more than what energy is (as opposed to other concepts such as velocity, flux, wavelength, and so on). What is extraordinary is to see that, if we don’t know what time and energy are, there is a well-defined relation between the two concepts. It is before us in quantum mechanics. It also exists in Hamilton’s analytical mechanics.

4.4.3 Stationary States In order to see that in a more concrete way, we use the notions introduced above. Consider an isolated system, that is, a system whose potential energy does not depend on time.

4.4 The Specific Role of Energy

77

We have just seen that states with well-defined energies, eigenstates of energy (energy levels of atoms as for Schrödinger) for which ΔE = 0, have no uncertainty in the energy. They have a particularly simple time dependence ψn (r, t) = φn (r)e−i En t/ .

(4.22)

It is then remarkable that in such cases |ψ|2 does not depend on time. The position probability law, the expectation of the position, does not depend on time. The system does not move! No physical quantity changes! Such states are called stationary states. The functions φn (r ) satisfy the eigenvalue equation (4.18), that is, Hˆ φn (r) = E n φn (r), in the space variables only, the time-independent Schrödinger equation. Therefore, if a system has a well-defined energy, it cannot evolve. For that system, time does not exist. No evolution, no motion can occur. If energy is well-defined, there are no oscillations for a pendulum, no Kepler motion along an orbit, and so on. The planetary model of atoms is wrong. Electrons don’t orbit around the nucleus. If the energy is well-defined, the system is frozen in time. That’s a rather frightening observation.

4.4.4 Motion: Interference of Stationary States What causes motion? (This seems to be a question for Greek philosophers in antiquity.) In order for motion to appear, the wave function must be a linear superposition of stationary states in interference. For instance, one can easily check that a superposition of two stationary states of different energies E 1 and E 2 ψ = λϕ1 e−i E1 t/ + μϕ2 e−i E2 t/

(4.23)

has a position probability |ψ|2 in which the crossed term depends on time. However, for such a system ΔE = 0, the energy is not well-defined. Motion occurs, and time exists, only if energy is not well-defined, and if God plays dice with energy. Then, there are given nonvanishing probabilities of finding E 1 , E 2 , and so on. In order for motion to appear, the system must be an interference between states of different energies. That’s a rather fascinating relation between time and energy. Actually, we have just written a fundamental technical result, useful for all that follows. Suppose that at t = 0 the wave function is a given superposition of states with well-defined energies

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4 Physical Quantities

ψ(r, t = 0) =



cn φn (r ),

(4.24)

n

where the φn (r ) are the energy eigenfunctions. Then its time evolution can be written directly, without solving any equation: ψ(r, t) =



cn φn (r )e−i En t/ .

(4.25)

n

One can readily check that this expression satisfies the Schrödinger equation (4.21) with the boundary condition (4.24). The evolution is known; motion is known. We show later on that it is a fundamental theorem of Hilbert space analysis that any wave function can be written in that way. Therefore the solution of the evolution problem in quantum mechanics, for an isolated system, always involves finding its energy eigenfunctions and eigenvalues, that is, solving the time-independent Schrödinger equation or, equivalently, the eigenvalue problem of the Hamiltonian. The time evolution follows immediately.

4.5 Schrödinger’s Cat To end this Chapter, we show one of the most famous paradoxes on the problem of measurement in quantum mechanics, in particular on the superposition of states and wave packet reduction. It’s necessary for our culture; people also like to talk about it at parties. It’s not that important if one does not understand everything at the beginning. That’s something that always makes me sad. Schrödinger was really an extraordinary man. The photograph (Fig. 4.4) shows that he was somewhat fancy, but a very pleasant and friendly man. But he had an enormous defect: he didn’t like cats. I like cats very much. The Dreadful Idea In 1935, Schrödinger had the following monstrous idea. Suppose one puts a cat in a steel chamber (Fig. 4.5) with a diabolic device consisting of a single atom of radioactive chlorine 39 which decays with a mean life of 60 min into argon 39. This means that the probability for the atom to decay within an hour is 50 %. If it decays, it emits an electron that is detected by a Geiger–Müller counter, which itself operates an apparatus, a hammer that falls on a capsule of cyanide, which kills the cat instantly. It’s simply revolting. Now, one observes the cat after an hour; the horrible questions are the following: 1. What killed the cat? 2. When did the cat die? (Of course, if the cat is still alive, we redo the experiment an hour later, which is horrendous.)

4.5 Schrödinger’s Cat

79

Fig. 4.4 Erwin Schrödinger in 1956. (All rights reserved)

Fig. 4.5 Diabolic device imagined by Schrödinger

The atom, more exactly its nucleus, is a quantum object. Its state is not described by a wave function of the type we have seen, but by another mathematical object with the same name ψ. The atom has a 50 % probability of having decayed after one hour, therefore its quantum state after an hour is 1 ψ(atom) = √ (ψ(nondecayed atom) + ψ(decayed atom)). 2

(4.26)

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4 Physical Quantities

Fig. 4.6 Siné: Schrödinger’s cat (© Siné)

It is an equal weight quantum superposition. As long as one has not measured whether it has decayed, the atom is in a superposition of the two states, nondecayed atom and decayed atom. But, because of the diabolic device, the entire system, including the cat, has become a quantum system. And, the cat is alive if the atom has not decayed; it is dead if the atom has decayed. To describe a cat requires a very complicated wave function, with some 1027 variables. But that’s inessential. The state of the cat can be inferred directly from the state of the atom. It is a linear superposition of the states {live cat} and {dead cat}, both at the same time, in interference 1 ψ(cat) = √ (ψ(live cat) + ψ(dead cat)). 2

(4.27)

The cat is both alive and dead. It is an abominable state, quite inconceivable and very difficult to represent, as one can see in Fig. 4.6! So, coming back to our outrageous questions: what killed the cat? When did the cat die? • Cyanide? No, it’s simplistic; it was in a capsule that must be broken. Similarly, the hammer must be released. • The decay? Maybe, but the atom is a quantum system. Its state is a superposition; the superposition must cease; that is, there must be a wave packet reduction in order to know whether the atom has decayed. Then it will be possible to accuse the atom. • Therefore, we must have a measuring instrument in order to know whether the atom has decayed. • The horror is that the porthole and the cat form an instrument that enables us to see whether the atom has decayed. • Therefore, one must observe the cat in order to reduce the wave packet of the atom! • It is by observing the cat that we reduce the wave packet of the cat. • It is when we observe the cat that we commit the atrocity of killing it, destroying any hope for the poor animal. • Before we looked at it, the cat was in a much more profitable state (4.27)!

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81

Fig. 4.7 Consequence of the observation of the cat

Wigner had suggested that observation is a transcendental act of conscience destroys the superposition. He thus addressed the question of the role of the observer in quantum physics. But one can reply that one can put a camera with shape recognition that types whether the cat is dead or alive and puts the answer in an envelope. One reads the letter a year after. If Wigner was right, the transcendental act of conscience would go backwards in time, because we exert it a year later (Fig. 4.7). Actually, in this disgusting example, what is shocking is not so much the wave packet reduction but the superposition. The quantum superposition of states, which seems natural when we apply it to objects deprived of souls, such as atoms or electrons, becomes very disconcerting if we apply it to familiar objects, as in Fig. 4.8. The Classical World In order to settle the matter, I gave that as a written examination to my students. Other students are lucky to escape such torture. The answer is that it is quite conceivable to manufacture paradoxical states such as the superposition {live and dead cat}, that is, paradoxical macroscopic states. But such states are extremely vulnerable and fragile. They imply a coherence, or a conspiracy of the 1027 particles, which gets destroyed in an incredibly short time because of the (thermal) interaction with the environment. This is called “decoherence” theory. The Nobel Prize in physics 2012 was awarded to jointly to Serge Haroche and David J. Wineland “for ground-breaking experimental methods that enable measuring and manipulation of individual quantum systems”. In 1996,

Fig. 4.8 The superposition principle applied to familiar objects. (Courtesy of Siné.)

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4 Physical Quantities

S. Haroche and his collaborators1 managed to study the behavior of Schrödinger “kittens”, i.e. systems involving a small number of atoms, and to verify their predicted decoherence. They were able to make a movie of the time evolution of a cat-state to a classical state by decoherence.2 Consequently, one must point out that the so-called “macroscopic world”, the world of large objects, is not identical to the world of objects that follow classical physical laws. The classical world is the world of large objects or systems that are furthermore stable under quantum fluctuations in their interactions with the external world. I tried to explain that to my own cat, but either he didn’t understand or he just wasn’t interested.

4.6 Exercises 1. Expectation values and variances

√ Consider the one-dimensional wave function ψ(x) = 2/a sin(πx/a) if 0 ≤ x ≤ a, and ψ(x) = 0 otherwise. Calculate x, Δx,  p, Δp and the product Δx Δp. 2. The mean kinetic energy is positive Verify that for any wave function ψ(x), the expectation value  p 2  is positive. 3. Real wave functions Consider a real one-dimensional wave function ψ(x). Show that  p = 0. 4. Translation in momentum space Consider a one-dimensional wave function ψ(x) such that  p = q and Δp = σ. What are the values of  p and Δp for the wave function ψ(x)ei p0 x/ ? 5. The first Hermite function Show that the wave function ψ(x) = e−x (x 2 − ∂ 2 /∂x 2 ) with eigenvalue 1.

2

/2

is an eigenfunction of the operator

1 M. Brune, E. Hagley, J. Dreyer, X. Matre, A. Maali, C. Wunderlich, J. M. Raimond and S. Haroche, Phys. Rev. Lett. 77, 4887 (1996). 2 S. Delglise, I. Dotsenko, C. Sayrin, J. Bernu, M. Brune, J.M. Raimond and S. Haroche, Nature 455, 510 (2008).

Chapter 5

Energy Quantization

We now solve quantum mechanical problems in order to see how the theory works. We we consider the motion of particles in simple potentials and the quantization of energy. We are going to do three things. • First explain the position of the problem. • Next, see in two simple cases the origin of quantization of energy levels. • And finally, we study an example that is basically as simple as these two but much more subtle in its consequences. This is a model of the ammonia molecule NH3 , on which true quantum mechanics effects appear. We will discover the tunnel effect, one of the most important quantum effects. By generalizing our results, this leads us to modern applications and nanotechnologies.

5.1 Methodology The solution of such a problem consists of solving the Schrödinger equation. The Hamiltonian of a particle in a potential V is  Hˆ = − Δ + V (r). 2m 2

(5.1)

We study solutions of given energy (energy is conserved; it is a constant of the motion). These are stationary states which have a very simple time-dependence ψn (r, t) = φn (r) e−i En t/ .

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_5

(5.2)

83

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5 Energy Quantization

The functions φn are the eigenfunctions of the Hamiltonian Hˆ and E n are the corresponding eigenvalues Hˆ φn (r) = E n φn (r). (5.3) One must impose boundary conditions on the wave function.

5.1.1 Bound States and Scattering States In classical physics, there is a difference between two regimes in the motion of a particle in a potential according to the value of the energy. If it is greater than the value of the potential at infinity, E ≥ V (∞), the system is in a scattering state. The particle goes to infinity as t tends to infinity, and extracts itself from the field of force. If the energy is smaller than the value of the potential at infinity, E < V (∞), it is a bound state. The particle is on an orbit. At any time its position remains confined in a finite region of space. The same distinction exists in quantum mechanics, but going from one regime to the other is not as simple as in classical physics, both physically and technically. In classical physics, we are always interested in the trajectory. In quantum mechanics, the physical quantities of interest are not the same in the two cases. The general problem of three-dimensional scattering, which is of great importance particularly in atomic, nuclear and particle physics, is technically complicated. Here, we are only interested in bound states. A bound state of well-defined energy is defined by the fact that its wave function satisfies the eigenvalue equation of the Hamiltonian (5.3) and that it is square integrable. In other words it is a “good” wave function  |ψn (r)|2 d 3r = 1.

(5.4)

This condition is essential and corresponds to the fact that classically the particle is confined in a finite region of space. Therefore, we are interested in finding the set {φn , E n } of eigenfunctions and corresponding eigenvalues of the Hamiltonian, with this boundary condition (5.4). In mathematics, one can prove that it is a discrete set. This is the origin of energy quantization: the E n are the energy levels of the system. The most general bound state is a linear superposition of stationary bound states. It evolves in a nontrivial way: a wave packet moves around in the potential. In the limit of large energy values, this motion can become the classical motion. If we relax the normalization condition (5.4), there exist solutions of the equation Hˆ ψ(r) = Eψ(r),

(5.5)

5.1 Methodology

85

for a continuous set of values of the energy E. Such solutions must be properly interpreted, and they correspond to scattering states. Their linear superpositions are wave packets that move in space.

5.1.2 One-Dimensional Problems Here, we concentrate on simple one-dimensional problems. For instance, a marble on a rail. This is simpler technically, we treat three-dimensional problems later. The eigenvalue equation is then an ordinary second-order differential equation −

2  ψ (x) + V (x)ψn (x) = E n ψn (x). 2m n

(5.6)

We wish to determine the functions {ψn (x)} and the numbers {E n } with the normalization condition  |ψn (x)|2 d x = 1. (5.7) In addition, we consider simple potentials for which analytical solutions exist, in order to become familiar with the physics.

5.2 The Harmonic Oscillator A first example is the harmonic potential V (x) =

1 mω 2 x 2 . 2

(5.8)

This corresponds classically to the sinusoidal motion, of frequency ω, of a particle elastically bound to a center (here x0 = 0). It is called a harmonic oscillator. It only has bound states (V (∞) = ∞). The eigenvalue equation is −

2 d 2 ψ(x) 1 + mω 2 x 2 ψ(x) = E ψ(x). 2m d x 2 2

(5.9)

The harmonic oscillator has numerous applications. For motions of small amplitude around its equilibrium position, any system has harmonic oscillations. Eigenvalues, Eigenfunctions This equation is a classic of 19th century mathematics. We turn to dimensionless quantities

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5 Energy Quantization

ε=

 E x , y = √ , with a = , ω mω a

(5.10)

and the eigenvalue equation becomes 1 2

 y2 −

d2 dy 2

 φ(y) = ε φ(y).

(5.11)

√ √ The square integrable solutions ψ(x) = a φ(x/ a), were found by Charles Hermite: d n  −y 2  2 2 φn (y) = cn e y /2 (5.12) e = cn e−y /2 Hn (y), dy n where Hn (y) is a polynomial of degree n, called a Hermite polynomial. It contains only even (resp., odd) powers of y if n is even (resp., odd), and the normalization constant is cn = π −1/4 2−n/2 (n!)−1/2 . The corresponding eigenvalues are 1 εn = n + , n nonnegative integer. 2

(5.13)

The differential equation (5.11) has solutions for all positive values of ε. However, 2 in general, these solutions increase at infinity as  e+y /2 . It is only for the set of values (5.13) of ε that the solutions are square integrable. Energy Levels, Eigenfunctions The energy levels of the one-dimension harmonic oscillator are therefore equalspaced 1 1 (5.14) E n = (n + ) ω = (n + ) hν. 2 2 Planck had correctly guessed the term nhν. Energy exchanges of the oscillator in transitions between levels occur for integer multiples of ω = hν, ΔE nn  = (n − n  )hν. But Planck had no means (and no reason) to guess the presence of the constant ω/2, which is called the zero point energy, and which can be measured. This constant is essential in order to satisfy uncertainty relations. In fact the classical equilibrium state of the oscillator consists of being at rest at the minimum of the potential. In other words, its velocity is zero and its position is at the minimum of the potential. This is contrary to Heisenberg’s uncertainty relations. The zero-point energy is inevitable and measurable.1 This result, which can be extended easily to three dimensions, has numerous applications. It allows us to understand the vibration spectra of molecules, and the specific 1 This energy may manifest itself in what is called the “vacuum energy” of the Universe in cosmology.

5.2 The Harmonic Oscillator

87

Fig. 5.1 The first four Hermite functions (abscissa: xa 1/2 ); ψ0 (x) is a Gaussian; ψ1 (x) is this Gaussian multiplied by x(2a)1/2 , and so on.

heats of solids (molecules in a crystal vibrate around their equilibrium position). It is a basic tool in field quantization and in relativistic quantum physics. The eigenfunctions are real and orthogonal 

ψn∗ (x) ψn  (x) d x = δn,n  .

(5.15)

From the definition of Hermite functions (5.12), one obtains the action of the operators x and d/d x on ψn (x): 

√ √ 2 x ψn (x) = n + 1 ψn+1 (x) + n ψn−1 (x) a √ √ √ d 2a ψn (x) = n ψn−1 (x) − n + 1 ψn+1 (x). dx

(5.16) (5.17)

The first four Hermite polynomials are (Fig. 5.1): H0 (x) = 1,

H1 (x) = 2x,

H2 (x) = 4x 2 − 1,

H3 (x) = 8x 3 − 12x.

5.3 Square Well Potentials Square Potentials Even simpler models consist of what are known as “square potentials”, which are piecewise constant. The solutions are, piecewise, exponential or sinusoidal according to the sign of V − E (E is a number we want to determine). 1. In regions where V = V0 and E − V0 > 0, the wave functions have a sinusoidal behavior ψ(x) ∝ e±ikx with k = 2m(E − V0 )/2 . ±K x 2. When E −V 0 < 0, the wave functions have an exponential behavior ψ(x) ∝ e 2 with K = 2m(V0 − E)/ .

88

5 Energy Quantization

Fig. 5.2 Sandwich of AlGaAs–GaAs–AlGaAs. The central slice of GaAs has a width of 6 nm. On the vertical axis, the aluminum concentration is indicated. The shape corresponds to the variation of the potential as “seen” by a conduction electron (i.e., the electrostatic potential averaged over one period of the crystal lattice) (Photograph due to Abbas Ourmazd, ATT Bell Labs.)

In order to obtain the bound states of such systems, we require as usual that the wave functions be square integrable. But in addition, the wave functions must be continuous as well as their first derivative at the points of discontinuity of the potential. This prescription is a consequence of the Schrödinger equation. It can be proven easily. One can understand it by considering a step in the potential as the limit of a continuous function (for instance, V ∼ limλ→0 tanh(x/λ)) and by reductio ad absurdum. This procedure allows us to determine the values of the energy E, as we will see. In modern microelectronic technologies, such potentials have many applications. In “sandwiches” of alternating thin layers of semiconductors (Ga As and Ga Al As), shown in Fig. 5.2, one can manufacture 2–5 nm wide quantum wells (we showed an example of this in the first chapter). The quantum confinement of electrons in such domains has opened a new era in electronics, computer components, energy, and medicine.2 These are used in optoelectronics, because the corresponding transitions between electron levels (ΔE ∼ 50–200 meV) are in the infrared part of the spectrum. At the end of this chapter, we show how one can obtain such a picture with quantum tunneling microscopes. What is important in such models is to understand the physics, that is, the qualitative results. Exact calculations can then be given safely to computers. Symmetric Square Well Energy quantization is a theorem, but let’s see how it occurs. We consider the energy levels of a particle in a symmetric potential well of depth V0 and width 2a, centered at x = 0. 2 Tagged: quantum dots, MIT Technology Review Cambridge Mass. Nov.2015; M.A. Reed, Quantum Dots, Sci. Amer., Jan. 93 ; L.L. Chanz and L. Esaki, Semiconductor Quantum Heterostructures, Phys. Today, 45, p. 36 (1992).

5.3 Square Well Potentials

89

We choose the origin of energies at the bottom of the potential well, so that the energy E, which we want to determine, is the kinetic energy of the particle inside the well. We are only interested in bound states, states for which 0 ≤ E < V0 , which in classical physics correspond to a particle that is confined inside the well. We assume the energy is not sufficient for the particle to jump out of the well. We set  k=

2m E , and K = 2

and we have k2 + K 2 =



2m(V0 − E) , 2

2mV0 . 2

(5.18)

(5.19)

It is straightforward to solve the Schrödinger equation. The wave functions are exponentials on the left and on the right (regions I  and I ) and sinusoids in the middle (region I I ). An important simplification comes from the symmetry of the problem V (x) = V (−x). One can indeed classify the solutions in two categories: symmetric (or even) solutions, and antisymmetric (or odd) solutions. In fact, because the Hamiltonian is symmetric Hˆ (x) = Hˆ (−x), if we change x into −x, we obtain for any solution ψ(x) Hˆ (x)ψ(x) = Eψ(x), and Hˆ (x)ψ(−x) = Eψ(−x). In other words, if ψ(x) is a solution of the Schrödinger equation, then ψ(−x) is also a solution for the same eigenvalue of the energy. Therefore, ψ(x) ± ψ(−x) is either a solution for the same value of the energy, or it is identically zero. This is a particular case of an important feature in quantum mechanics. To the invariance laws of the Hamiltonian, there correspond symmetry properties of the solutions. (This can also be obtained directly here by performing the calculation; we could have made the same remark on the harmonic oscillator.) The symmetric (ψ S ) and antisymmetric (ψ A ) solutions have the forms ψ S : (I  ) ψ(x) = B e K x , (I I ) ψ(x) = A cos kx, (I ) ψ(x) = B e−K x , ψ A : (I  ) ψ(x) = −D e K x , (I I ) ψ(x) = C sin kx , (I ) ψ(x) = D e−K x , (5.20) where the constants A, B, C, and D are determined by the continuity of ψ and ψ  at x = ±a. Notice that in (5.20) we have omitted terms that increase exponentially at infinity because of the normalizability condition. We must join these expressions at x = ±a. When V is continuous, the equation handles the problem by itself. If not, as is the case here, we must impose that ψ and ψ  are continuous at x = ±a.

90

5 Energy Quantization

We obtain ψ S : A cos ka = Be−K a , and k A sin ka = K Ce−K a ψ A : C sin ka = De−K a , and kC cos ka = −K De−K a ,

(5.21)

and, if we take the ratios, k tan ka = K , for (ψ S ), and − k cot ka = K , for (ψ A ).

(5.22)

These two relations between k and K must be completed by the definition (5.19) which can be written as 2mV0 a 2 . (5.23) k2a2 + K 2a2 = 2 In the (ka, K a) plane, this is the equation of a circle. We must therefore find the intersections of this circle with the curves K a = ka tan ka and K a = −ka cot ka as represented in Fig. 5.3b. Suppose the width 2a of the well is given, these intersections kn form a finite set, and they correspond alternatively to even and odd solutions. The number of solutions, or bound states, increases with V0 (deeper well). There is only one bound state if V0 is less than √ π 2 2 a 2mV0 π . < or V0 <  2 8ma 2

(5.24)

The energy levels E n = 2 kn2 /2m are quantized. This quantization is not a consequence of the continuity conditions, which simply enable us to calculate the eigenvalues E n , but of the normalizability that eliminates exponentially increasing terms in (5.20).

Fig. 5.3 Square well potential: a shape of the potential; b graphical solution giving the energy levels; c limit of an infinitely deep well.

5.3 Square Well Potentials

91

Fig. 5.4 Temperature distribution in a breath (artificial colors). The background is black, the breath comes out of the mouth at body temperature and it expands at room temperature (Courtesy of Emmanuel Rosencher.)

The solutions can be classified by increasing values of the energy E, according to the number of nodes of the wave functions (Sturm–Liouville theorem). The lowest energy state is called the ground state. Notice an essential difference from classical mechanics: the particle has a nonvanishing probability to be in the classically forbidden regions, (I ) and (I  ), where its kinetic energy would be negative and is therefore energetically illegal. However, it does not propagate in those regions; it can penetrate them, but it bounces off the edges x = ±a. The wave function decreases exponentially with a mean penetration distance of 1/K . This is analogous to the skin effect in electromagnetism. We can see how the “classical limit” appears: 1/K → 0, if we make  → 0 or if the mass is large m → ∞. As we said, quantum wells such as the one represented in Fig. 5.2 have allowed decisive improvements in advanced technologies, in particular infrared technologies such as shown in Fig. 5.4. Infinite Well, Particle in a Box Another simple, but interesting, limiting case is that of an infinitely deep well V0 = ∞. The particle is confined inside a region that we place for convenience between x = 0 and x = L as in Fig. 5.3c. In this limit of the previous case, the wave function vanishes, ψ(x) = 0, outside the interval [0, L]. The continuity conditions are then different: only the wave function is continuous (this can be seen by taking the limit V0 → ∞ on the above solutions). The normalized eigenfunctions of the Hamiltonian are then:  ψn (x) =

2 nπx sin( ) n integer > 0. L L

(5.25)

92

5 Energy Quantization

The corresponding energy levels are En = n2

π 2 2 n integer > 0. 2m L 2

(5.26)

Quantization appears here as a simple stationary wave phenomenon. This calculation can easily be generalized to three dimensions, that is, the case of a particle in a box. If we consider a rectangular box of sides (a, b, c), the solutions can be factorized: √ 8 n2π y n 3 πz n 1 πx ) sin( ) sin( ), (5.27) sin( Ψn 1 ,n 2 ,n 3 (r) = √ a b c abc   2 π 2 n 21 n 22 n 23 + + . (5.28) E = E n 1 ,n 2 ,n 3 = 2m a 2 b2 c2 This very simple result has numerous applications. Many systems can be approximated by infinite wells: molecules in a gas, neutrons inside a neutron star, conduction electrons in a metal, and others.

5.4 Double Well, the Ammonia Molecule Consider now a similar problem, where we really do quantum mechanics and discover unexpected results. This is the case of a symmetric double potential well. At the beginning it is similar to the infinite potential well, but with a potential barrier in the middle: in other words, an infinite well containing two wells of width a centered respectively at ±b, and separated by a barrier of height V0 and width Δ = 2b − a.

5.4.1 The Model Consider a concrete example, the ammonia molecule NH3 . In its lowest energy states, this molecule has the shape of a pyramid with the nitrogen atom at the top and the three hydrogen atoms at the base, on an equilateral triangle. It is a complex object made of 14 particles (10 electrons and 4 nuclei) and there are many possible motions of this system. However, the lowest energy motions correspond to the global displacement of the triangle of hydrogen atoms, which we call collectively a “particle of mass m”, with respect to the nitrogen atom along the symmetry axis of the molecule. When the abscissa x of this plane varies along x > 0, the potential energy of the system has a minimum that corresponds to a classical equilibrium configuration at x = b. However, the molecule can invert itself, as an umbrella, and there exists another symmetric stable configuration for a negative value of x = −b. The potential energy is symmetric with two minima and a maximum in between. This maximum corresponds to an unstable configuration where the four atoms are in the same plane.

5.4 Double Well, the Ammonia Molecule

93

Fig. 5.5 The ammonia molecule: a the two classical configurations; b the actual molecular potential energy (full line) and the simplified model (dashed line) that describes the reversal of the molecule

These two configurations of classical equilibrium are not physically equivalent because the molecule possesses an intrinsic angular momentum, and one can define the “right” and the “left”, as in an umbrella. This model can be solved numerically, but it suffices to study the square well model V (x) drawn in a dashed line in Fig. 5.5b. This potential consists of two wells of width a centered at b and −b, respectively, and separated by a barrier of height V0 and width Δ = 2b − a. We ask the following question. For E > V0 there are periodic oscillations from left to right. But when E < V0 , for a given value of the energy E, there are always two possible classical configurations of same energy, one in the left-hand side well, the other in the right-hand side well. In particular, there are two positions of equilibrium, one on the left, the other on the right; both have the same energy. What is the quantum situation?

5.4.2 Stationary States, the Tunnel Effect From the calculational point of view, the problem is strictly analogous to what we just did. We consider the energy levels of a particle of mass m such that classically the particle cannot cross the potential barrier: E < V0 . As previously, we define  k=

2m E and K = 2



2m(V0 − E) . 2

(5.29)

94

5 Energy Quantization

Fig. 5.6 Symmetric solution (a), and antisymmetric solution (b) of lowest energy in the double square well potential which is a model of the ammonia molecule

The problem has the symmetry (x ↔ −x). We can classify the solutions according to their parity or symmetry. The solutions are sinusoidal in the regions L and R and exponential in the middle region M. The wave functions must vanish for x = ±(b + a/2), and the eigenfunctions of the Hamiltonian are of the form ψ(x) = ±λ sin k(b + a/2 + x) L region = λ sin k(b + a/2 − x) R region ψ(x) = μ cosh K x Symmetric solution M region ψ(x) = μ sinh K x Antisymmetric solution

(5.30)

The two lowest energy solutions are represented in Fig. 5.6. The ground state is symmetric; the first excited state is antisymmetric. We observe that the wave functions exist in the classically illegal middle region E < V0 , which is not surprising. That comes from the exponential of the simple symmetric well which is cut off at a finite distance. This results in a symmetrization or antisymmetrization of the wave functions. Therefore, in all stationary states, the particle has the same probability to be on the right and on the left. So it is a two-well problem, but, because the particle has a non-vanishing probability to be in the classically forbidden middle region, these two wells are coupled by the tunnel effect. Things resemble a classical situation where one would have drilled a narrow quantum tunnel in the intermediate potential barrier, in order to allow the particle to communicate between the two wells.

5.4 Double Well, the Ammonia Molecule

95

5.4.3 Energy Levels The continuity of the function and of its derivative at points x = ±(b − a/2) yields the conditions: k coth K (b − a/2) for a symmetric solution ψ S , K k tan ka = − tanh K (b − a/2) for an antisymmetric solution ψ A . K

tan ka = −

This, together with the condition k 2 + K 2 = 2mV0 /2 gives transcendental equations that can be solved numerically. However, in order to understand the physics of the problem in a simple  manner, we assume the orders of magnitude are such that: V0 E, that is, K  2mV0 /2 = constant and K Δ 1, which is quite reasonable in the specific case of the NH3 molecule. Under such conditions, we end up with tan ka  −

k 1 ± 2e−K Δ , K

(5.31)

where the + sign corresponds to ψ S , and the − sign to ψ A . With this equation, we can calculate the quantized values of ka. These values appear on the graph in Fig. 5.7 as the positions of the intersections of the successive branches of y = tan ka with the two straight lines y = −ε A ka and y = −ε S ka. These intersections are located in the vicinity of ka ∼ π. The two constants ε A and ε S are: εA =



1 1 1 − 2e−K Δ , ε S = 1 + 2e−K Δ . Ka Ka

(5.32)

They are close to each other, and such that ε A < ε S  1, because K a ka ∼ π.

Fig. 5.7 a Graphical determination of the energy levels in the double well; b the two first levels are lower than the ground state of a single well centered at L or R (E 0 → E 0 ), and there is a splitting by tunneling between these two levels (E 0 → E A and E S )

96

5 Energy Quantization

For K infinite, V0 infinite, we recover two independent wells. The particle is in one of them. The energies are E n = n 2 π 2 2 /2ma 2 . The order of magnitude of E S and E A is E 0 = π 2 2 /2ma 2 . For finite K (V0 finite), the two levels are shifted downwards to E 0 , which is intuitively understandable. The two wells communicate by quantum tunneling and the particle “feels” an effective well broader than a. This lowering is accompanied by a splitting in two sublevels as can be seen in Fig. 5.7. The symmetric state is more tightly bound. In our approximation (K k, K a 1), we obtain kS ∼

π π , kA ∼ , a(1 + ε S ) a(1 + ε A )

(5.33)

with ε S and ε A  1. Using (5.31) and (5.33), we obtain a mean energy E 0 = (E A + E S )/2   2 π 2 2  E0  . (5.34) 1− 2ma 2 Ka The splitting E A − E S of the two levels by the tunnel effect is E A − ES ≡ 2 A 

2 π 2 2ma 2

or A≡



1 1 2 π 2 8e−K Δ − ,  2 2 (1 + ε A ) (1 + ε S ) 2ma 2 K a

E A − ES 2 π 2 4 −K Δ .  e 2 2ma 2 K a

(5.35)

(5.36)

√ Because K  2mV0 / in this approximation, the splitting tends to zero exponentially when the width Δ, or the height V0 , of the potential barrier increase. In fact, in all its consequences, namely the splitting E A − E S and the value of the wave function in the middle region, the tunnel effect is proportional to e−K Δ where  Δ = 2b − a and K ∼ 2m(V0 )/2 (or the square of this for probabilities). We also notice that A → 0 extremely rapidly in the limit  → 0. This is in contrast with the polynomial level spacing, in , of the harmonic oscillator or potential well levels. Here, the splitting is exponential in −1/. We therefore observe a first difference with the classical case. There are indeed two lowest lying energy states, but classically they have the same energy (E 0 ), whereas here they are split. This qualitative feature is general, for any symmetric double well of arbitrary shape.

5.4.4 Wave Functions The two corresponding wave functions, which are shown in Fig. 5.6, are such that the probability of finding the particle on the left or on the right is the same. If the particle

5.4 Double Well, the Ammonia Molecule

97

is in a well-defined energy state, it has the same probability of being on either side. It is both on the right and on the left. That result is really contrary to classical observations. Classically, in its two lowest energy states, the particle is either on the right or on the left. Here it is both on the right and on the left at the same time. Is that a real difference? That’s not really convincing for the moment, because of the statistical interpretation. If, classically, we fix the condition that the energy be minimum, then it is natural to find half of the particles on the right and the other half on the left. We must find something else. Something else means doing true quantum physics, taking into account the time parameter.

5.4.5 Inversion of the Molecule How can we put the particle on the right- or left-hand side in quantum mechanics? We must prepare it in a state where its wave function is concentrated on the right or on the left. But we know how to do that! We just need to look at the wave functions in Fig. 5.6. We just have to take the sum and the difference of these two eigenfunctions, according to the superposition principle, √ ψ R = (ψ S + ψ A )/ 2 ,

√ and ψ L = (ψ S − ψ A )/ 2.

(5.37)

The resulting wave functions describe states where almost all the probability is concentrated on one side only, on the left for ψ L and on the right for ψ R . (Actually the residual probability is of the order of ≈ e−2K Δ , which is very small.) These two states correspond to the “classical” configurations, molecule on the right and molecule on the left (Fig. 5.8). This observation is interesting. We can phrase it in the following way. ψ R is a linear superposition of states with a well-defined energy that interferes destructively

Fig. 5.8 Classical configurations of the ammonia molecule

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5 Energy Quantization

on the left. Similarly, ψ L is a linear superposition of states with a well-defined energy that interferes destructively on the right. In this way, interferences are really simple; there is nothing more from the mathematical point of view. We can invert these relations and say that states of well-defined energies, which are on both sides, are interfering superpositions of “contradictory” classical states. Classically, one is either on the right or on the left. Nobody is both on the right and on the left: well nearly nobody, some people once in a while, but very few. Here, it seems perfectly natural. But that is what seemed shocking with Schrödinger’s cat! What was inconceivable for a cat seems perfectly natural here. Notice that, at √that time, we had forgotten a possible state ψ− (cat) = (ψ(live cat) − ψ(dead cat)) 2. Frankly, the second elementary operation, subtraction, is just as noble as the first one. But mind you: the probabilities to be on either side (or dead or alive) are the same although the two states are definitely not the same! Puzzling! Now we can apply what we know on time evolution. The state ψ R is not a stationary state; the system does not have a well-defined energy: p(Es) = p(Ea) = 1/2. Therefore it must evolve in a non-trivial way with time. Its time evolution is:

1 ψ(x, t) = √ ψ S (x) e−i E S t/ + ψ A (x) e−i E A t/ 2 −i E S t/

e = √ ψ S (x) + ψ A (x)e−iωt , 2

(5.38)

where we have introduced the Bohr frequency ω = E A − E S = 2 A of the system. Now, we face something truly astonishing and unpredictable. At time T = π/ω the particle is on the left! Its wave function is ψ L , up to a phase factor, ψ(x, T ) =

e−i E S T / √ (ψ S (x) − ψ A (x)) . 2

This phenomenon is really contrary to classical mechanics where, if at t = 0 the particle is at rest in one of the wells, it stays there! In quantum mechanics, there is a permanent oscillation between the two wells at the frequency ω = 2 A/. The wave function and the probability density “flow” periodically from one well to the other. The particle keeps on shifting from one side to the other. The cat oscillates permanently between life and death. That is truly non-classical. In the specific case of NH3 , this phenomenon is called the “inversion of the ammonia molecule.” If, at t = 0, we prepare it in a classical configuration, it reverses itself periodically at the Bohr frequency. The inversion frequency can be measured very accurately. In the lowest energy state, the splitting is 2 A ≈ 10−4 eV, hence a frequency ν = 24 GHz, a wavelength λ =1.25 cm, and a period τ = 4.2 10−11 s. In fact, NH3 possesses an electric dipole moment. The center of gravity of positive charges is different from the center of gravity of negative charges (the nitrogen atom

5.4 Double Well, the Ammonia Molecule

99

attracts electrons more strongly). This electric dipole moment D is inverted when the molecule reverts. This produces the emission or absorption of a radio wave at that frequency. (Actually, at the microscopic level, a molecule can absorb or emit a photon of energy hν ≈ 10−4 eV in a transition between the two states. It is only on a macroscopic sample that the electric dipole argument is valid.) That is a characteristic line of NH3 , a fingerprint of that molecule which is used in radioastronomy to detect ammonia in the interstellar medium. We come back later to an important application, the ammonia maser. It was totally impossible to predict the order of magnitude of that frequency qualitatively, by dimensional analysis, before we understood the mechanism of the tunnel effect and before we found the exponential (5.36).

5.5 Illustrations and Applications of the Tunnel Effect This brings us to a series of remarks on the tunnel effect, of great importance in quantum mechanics. Sensitivity to the Parameters The exponential dependence of the splitting is fundamental. An exponential varies very rapidly. This same effect, this same mechanism explains phenomena whose orders of magnitude are incredibly different. Consider for instance NH3 and similar molecules ND3 , PH3 , AsH3 . This was treated in detail by Townes and Schawlow3 who gave the form of realistic potentials in this kind of physics. It is instructive to go from NH3 to AsH3 : NH3 : V0 = 0, 25 eV , ND3 : V0 = 0, 25 m d PH3 : V0 = 0, 75 eV , AsH3 : V0 = 1, 50 eV ,

b = 0, 4 Å : ν0 = 2, 4 1010 Hz , = 2m p , b = 0, 4 Å : ν0 = 1 600 Hz , b=1Å : b=2Å :

ν0 = 2, 4 1010 Hz , ν0 = 1, 6 10−8 Hz .

A change by a factor of 6 in V0 and 5 in the size b produces a spectacular decrease of 18 orders of magnitude in the inversion frequency between NH3 and AsH3 . For AsH3 , the frequency is one inversion in two years, which is not measurable. One can only calculate it theoretically. In other words, AsH3 , which seems quite similar to NH3 from a chemical point of view, behaves as a classical structure from the point of view studied here, simply because the arsenic atom is 5 times larger than nitrogen. There are a variety of similar situations where, by quantum tunneling, a process occurs through a nonclassical transition across a potential barrier. The first physical phenomenon that was explained in this way was the alpha decay of nuclei. This was 3 C.H.

Townes and A.L. Schawlow, Microwave Spectroscopy, Chap. 12. New York: McGraw-Hill, (1955).

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Fig. 5.9 Double potential well as seen by an electron when two atomsare separated by a distance Δ

understood by Gamow in 1928. Many other phenomena have the same origin, such as catalysis, the formation of interstellar molecules on interstellar dust, nuclear fusion and fission, and so on. Molecular Structure Valence Electrons Similar examples are provided by electrons in molecules. Consider, for instance, the case of two identical atoms located at a distance Δ of each other. An external electron “sees” a double well as shown in Fig. 5.9 (for simplicity, we assume the atoms are at a given distance that we can vary). We choose the origin of energies such that V → 0 for x → ∞. If Δ is sufficiently large, one can safely consider that V ∼ 0 halfway between the atoms. In order to get from one atom to the other, an electron in an energy level E 0 < 0 must cross a potential barrier of height −E 0 and width Δ. We can calculate the typical time T it takes to get from one atom to the other. Suppose the kinetic energy E k of the electron in one well is of the order of owing to the virial the binding energy |E 0 | (for the hydrogen atom, this is exact, √ theorem). In the exponential of the tunnel effect, we have K = 2m|E 0 |/. For an electron bound in an atom, one has in good approximation K a ∼ 1. The exponential dependence of the oscillation frequency in terms of the parameter K Δ remains true. We can use (5.36) in the form A ∼ E k e−K Δ ∼ |E 0 |e−K Δ , up to a numerical factor of order one. In a molecule or in a crystal, the distances of atoms are of the order of 0.1 nm. In a gas at usual temperatures and pressures, they are roughly ten times larger (∼3 nm). The binding energies of valence electrons in an atom are of the order of a few eV. One then finds: Molecule: Δ = 0.2 nm |E 0 | = 4 eV A = 1 eV T = 10−15 s , Gas: Δ = 3.0 nm |E 0 | = 4 eV A = 10−12 eV T = 10−3 s . The time to get from one atom to the other is very small for valence electrons in a molecule or in a solid. These electrons are completely delocalized in the molecular

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101

structure. Conversely, this phenomenon is completely negligible in gases. In fact, owing to thermal motion, two molecules in a gas remain at a distance of, say, 3 nm for a length of time smaller than 10−10 s. Quantum tunneling oscillations have a period of 10−3 s and they cannot occur appreciably on such a small time scale. Therefore, the idea that in a gas at room temprature each electron belongs to a given molecule is quite acceptable. Molecular Binding Similarly, in this result, one can find a starting idea for the explanation of molecular binding, which cannot be understood classically. Consider the simplest molecule, the H+ 2 ion, made of two protons and one electron. The decrease E 0 → E 0 explains that it is more favorable energetically for the electron to have an equal probability to be on both protons rather that being bound to one of them and let the other live its own life. Classically, the moon belongs to the earth. It could belong to the planet Mars, but it has chosen the earth and it sticks to that choice. However, quantum mechanically, the fact that the electron belongs to both protons at the same time stabilizes it. This effect increases as the distance between protons decreases. However, a compromise must be found, owing to the Coulomb repulsion of protons. An equilibrium situation results. After doing the calculations, one can prove that the H+ 2 ion is bound. If, in addition, one takes into account spin and the Pauli principle, the splitting and the fact that the symmetric state is more tightly bound E S < E A accounts for the chemical covalent bond. In other words, quantum tunneling is responsible for our existence! Potential Barriers One can find exercises of the following type. A particle of energy E hits a potential barrier of width Δ and height V0 > E. Classically, the particle should bounce back. What is the tunneling probability? For an electron and with atomic orders of magnitude E = 1 eV, V0 = 2 eV, Δ ≈ 0,1 nm, we obtain a probability of p = 80% that the particle crosses the barrier which is completely anticlassical. In the same conditions, a proton m p ∼ 2000m e has a probability of crossing the barrier of p ≈ 10−19 because of the mass effect. In other words, saying that nuclei, protons and neutrons, have well-defined positions in an atom or a molecule makes sense. But at nuclear scales V0 = 2E ≈ 10 MeV and Δ ≈ 1 fm, the probability is 80 %, and a proton is delocalized in a nucleus.

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5.6 Tunneling Microscopy, Nanotechnologies An important practical application is the construction by Binnig and Rohrer in the 1980 s of the scanning tunneling microscope (STM, 1986 Nobel prize). A conducting tip is moved along a surface at a very short distance Δ ≈ 10Å. A potential difference is applied, and the electrons pass from the surface to the tip by the tunnel effect. The current is extremely sensitive to the distance (actually to the electrostatic potential V0 ). That way, one can detect incredibly fine details ≈0.01 nm, and one can make a map of the surface. An example is shown in Fig. 5.10. Nanotechnologies With the tunnel effect one gets near to science fiction by inventing nanotechnologies, that is, creating operation techniques at the nanometer scale, at distances comparable to the size of the smallest living systems, viruses.4 On September 29, 1989, D. Eigler, research engineer at IBM, was able to manipulate individual atoms on a metal surface. He picked them with a tip and put them on another site of the surface (the device is similar to that shown in Fig. 5.10). Eigler first managed to write the letters IBM with 35 xenon atoms on a nickel substrate. One year later, he was able to construct an electronic switch whose moving part was made of a single atom (5 nm high)! In Fig. 5.11, one can see the progressive construction of a “coral reef” of 48 iron atoms on a nickel substrate by atom manipulation. Figure 5.12 shows a side view of

Fig. 5.10 a Principle of a tunneling microscope. A thin tip is moved in the vicinity of a solid surface with piezoelectric transductors. One adjusts the distance of the tip to the surface in such a way that electric current due to tunneling between the surface and the tip is constant. This provides a mapmaking of the electron density distribution (actually the electrostatic potential) at the surface of the crystal. An example is shown in b where one can see a surface of In Sb. The Sb atoms appear to be raised. The actual size of the sample in the figure is ∼3 nm (After Y. Liang et al., J. Vac. Sci. Technol. B9, 730, (1991).) 4 See for instance Zooming into the nanoworld http://www.nano.geo.uni-muenchen.de/SW/images/

zoom.html..

5.6 Tunneling Microscopy, Nanotechnologies

103

Fig. 5.11 Progressive construction of a “coral reef” made of 48 Fe atoms on an nickel crystalline surface by atom manipulation. M.F. Crommie, C.P. Lutz, D.M. Eigler, E.J. Heller. Waves on a metal surface and quantum corrals. Surface Review and Letters 2 (1), 127–137 (1995) Fig. 5.12 Side view of the electron “lake” inside the coral reef of the previous picture. The waves are stationary de Broglie waves. STM Image Gallery-IBM

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the corral reef. The “waves” correspond to the surface density of electrons trapped inside the structure. In other words we see the ground state of de Broglie waves in a circular two-dimensional well directly (this is a consequence of the Pauli principle; we see the probability densities of higher excitations because the lower ones are already occupied). We could have considered this problem in Sect. 5.3. Classical Limit These developments are truly impressive. However, what is even more striking perhaps is what one cannot do with the tunnel effect. Indeed, quantum mechanics opens the possibility of fantastic dreams. One can get to another galaxy by quantum tunneling and see extraterrestrials. One can win a bike race by crossing mountains by quantum tunneling without any fear of being dope-tested, or, more seriously, see a piece of dust cross a tulle curtain by quantum tunneling. In principle, quantum mechanics gives us the possibility to do all that. Unfortunately, the probability is very small. One can, as an exercise, calculate the probabilities and one ends up with the most extravagantly small numbers one can imagine p ∼ exp(−10(30 to 65) ). Such numbers are impossible to write in the binary system! In order to win the Nobel prize, it is undoubtedly much more promising to buy a cat. The probability that a cat will type the unified theory of the Universe, or Shakespeare’s works by walking at random on the keyboard of a computer without making any mistake is enormously larger: p ∼ exp(−10(4 to 6) ). All of this is meaningless for simple reasons. There exist other terms in the Hamiltonian of a piece of dust that allow it to cross a curtain: someone can do the cleaning; there can be a small hole in the curtain, and so on. But it is fascinating to see that an effect which is so important for our existence cannot manifest itself openly at our scale.

5.7 Exercises 1. Uncertainty relation for the harmonic oscillator Using the recursion relations satisfied by the Hermite functions (5.16), show that, in a state of energy E n given by (5.13) one has x =  p = 0. Calculate x 2  and  p2  and show that the zero point energy is essential in order to preserve the uncertainty relations. 2. Time evolution of a one-dimensional harmonic oscillator Consider a harmonic oscillator of Hamiltonian Hˆ = pˆ 2 /2m + mω 2 xˆ 2 /2 and its first two normalized eigenfunctions φ0 (x) and φ1 (x). Consider a system which at time t = 0 has the wave function: ψ(x, t = 0) = cos θ φ0 (x) + sin θ φ1 (x) with 0 ≤ θ < π .

5.7 Exercises

105

a. What is the wave function ψ(x, t) at time t? b. Calculate the expectation values E, E 2  and ΔE 2 = E 2  − E2 . Explain their time-dependence. c. Calculate the time evolution of x, x 2  and Δx. 3. Three-dimensional harmonic oscillator Consider in three dimensions a particle of mass m and the Hamiltonian Hˆ = pˆ 2 /2m+ mω 2 rˆ 2 /2 where rˆ2 = xˆ 2 + yˆ 2 + zˆ 2 . a. What are the energy levels and their degeneracies? b. How do these results change in the case of an anisotropic potential:

V = m ω12 x 2 + ω22 y 2 + ω32 z 2 /2 ? 4. One-dimensional infinite potential well Consider an infinite potential well of width a: V (x) = 0 for 0 < x < a and V = ∞ otherwise. a. Show that in the energy eigenstate ψn (x), one has x = a/2 and Δx 2 = a 2 (1 − 6/n 2 π 2 )/12. b. Consider the wave-function ψ(x) = Ax(a − x). (i) What is the probability pn to find the particle in the n-th excited state? (ii) From this set of probabilities, calculate the expectation values E and E 2  for that wave function.

−2n = π 2 /8 for n = 1, π 4 /96 for n = 2, and π 6 /960 We recall that ∞ k=0 (2k + 1) for n = 3. c. Check that if one applies blindly the correspondence principle, i.e. if one uses Hˆ 2 = (2 /2m)2 d 4 /d x 4 in the definition of E 2 , one obtains the absurd result ΔE 2 < 0. What is the reason for this? 5. Isotropic states of the hydrogen atom The energy levels of spherically symmetric states of the hydrogen atom can be obtained in the following one-dimensional calculation. Consider an electron of mass m in the potential V (x) such that V = ∞ if x ≤ 0 and V = −A/x if x > 0, where A = q 2 /4πε0 , and q is the elementary charge. We set α = q 2 /(4πε0 c)  1/137 (dimensionless constant) where c is the velocity of light. a. Show that the wave function ψ(x) = C x e−x/a for x ≥ 0 and ψ(x) = 0 for x < 0, is an eigenfunction of the Hamiltonian with energy E for a given value of a. Express E and a in terms of m, α,  and c. b. Calculate the numerical values of E and a. One can use mc2 = 5.11 105 eV and c = 197 eV nm.

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c. Determine the normalization constant C in terms of a. d. Calculate the expectation value of 1/x in the state |ψ and deduce from that the expectation value of the kinetic energy. What is the relation, valid also in classical mechanics, between these two quantities? 6. δ-function potentials a. Consider a particle of mass m in the one-dimensional potential V (x) = αδ(x), α < 0. We are interested in bound states (E < 0). (i) Assuming that the wave function ψ(x) is continuous at x = 0 (which can be proven), find the relation between the discontinuity of its derivative and ψ(0) by integrating the Schrödinger equation between x = −ε and x = +ε. (ii) How many bound states are there? With what energies? We set K = √ −2m E/ and λ0 = −2 /mα. b. Consider the double δ-function potential: V (x) = α (δ(x + d/2) + δ(x − d/2)) . (i) Write the general form of bound state wave functions. What is the quantization condition? (ii) Discuss the number of bound states as a function of the distance d between the two wells.

5.8 Problem. The Ramsauer Effect In 1921, Ramsauer noticed that for some particular values of the incident energy, rare gases such as helium, argon or neon were transparent to low-energy electron beams. This can be explained in the following one-dimensional model. Consider a stationary solution of the Schrödinger equation of positive energy E, for a particle of mass m in the following one-dimensional potential (V0 > 0): V (x) = 0 for |x| > a

,

V (x) = −V0 for |x| ≤ a.

We set q 2 = 2m(V0 + E)/2 , k 2 = 2m E/2 and we are interested in a solution of the form ψ(x) = eikx + A e−ikx ψ(x) = B e

iq x

ψ(x) = D e

ikx

+Ce

−iq x

x ≤ −a, − a < x ≤ a, x >a.

5.8 Problem. The Ramsauer Effect

107

1. Write the continuity relations at x = −a and x = a. 2. Setting Δ = (q + k)2 − e4iqa (q − k)2 , calculate the transmission probability T = |D|2 . Calculate the reflection probability R = | A|2 . Check that R + T = 1. 3. Show that T = 1 for some values of the energy. Interpret this result and the Ramsauer effect. 4. In Helium, the lowest energy at which the phenomenon occurs is E = 0.7 eV. Assuming that the radius of the atom is a = 0.1 nm, calculate the depth V0 of the potential well inside the atom in this model. 5. How does the reflection coefficient behave as the energy E tends to zero? When one sends very slow hydrogen atoms on a liquid helium surface, these atoms bounce back elastically instead of being adsorbed. Explain this phenomenon qualitatively.

5.8.1 Solution 1. The continuity equations are at x = −a: e−ika + Ae+ika = Be−iqa + Ceiqa ik(e−ika − Aeika ) = iq(Be−iqa − Ceiqa ), and at x = +a: Beiqa + Ce−iqa = Deika and iq(Beiqa − Ce−iqa ) = ik Deika . 2. Setting Δ = (q + k)2 − e4iqa (q − k)2 , one obtains: D=

4kq −2i(k−q)a e Δ

A=

(k 2 − q 2 ) −2ika (1 − e4iqa ) . e Δ

We have |Δ|2 = 16k 2 q 2 + 4(k 2 − q 2 )2 sin2 2qa and: R = | A|2 =

4(k 2 − q 2 )2 sin2 2qa |Δ|2

T = |D|2 =

16k 2 q 2 |Δ|2

where R + T = 1. 3. For all values of q such that sin 2qa = 0, i.e. qa = nπ/2, the transmission probability is equal to 1, and there is no reflection, T = 1 , R = 0. This happens when the size of the well 2a is a multiple of λ/2, where λ = 2π/q is the de Broglie wavelength of the particle inside the potential well. All the reflected waves interfere destructively and the well becomes transparent to the incident wave (more precisely, the wave reflected in x = −a, which does not enter the

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well, interferes destructively in the backward direction with the sum of all the waves undergoing multiple reflections in the well). 4. The corresponding energies are: En =

n 2 π 2 2 − V0 , 8ma 2

Choosing n = 1 and E = 0.7 eV, we obtain V0 = 9.4 − 0.7 = 8.7 eV. 5. When E tends to 0, k also tends to 0 and the transmission probability vanishes. The incident particle is reflected by the potential well. The sticking of the hydrogen atoms on the liquid helium surface occurs when the hydrogen atoms enters the potential well at the vicinity of the surface. In this well, the hydrogen atom may loose energy via the emission of a wave propagating at the surface of the liquid (ripplon): after such a process, the energy of the hydrogen atom is too low to exit the well, and the atom is trapped at the surface of the liquid. At very low incident energy, incoming hydrogen atoms have a vanishing probability to enter the well, and hence the absorption probability tends to zero.

5.9 Problem. Colored Centers in Ionic Cristals Consider the diatomic crystal Na Cl. It is called an ionic crystal because, when the crystal forms, the outer electron of a sodium atom is transfered to a chlorine atom. Hence, in the crystal, the electronic configuration is (Na+ , Cl− ), and the electrostatic interaction between the Na+ and Cl− ions is responsible for the binding of the structure. The crystal is face centered cubic for both ions. Schematically, the cristal, as seen parallel to one face of the cube can be represented as in Fig. 5.13. Such a

Fig. 5.13 Structure of the ionic Na+ Cl−

5.9 Problem. Colored Centers in Ionic Cristals

109

Fig. 5.14 Structure of an F center in a NaCl crystal

structure, called the Na Cl structure, is very frequently encountered. It is in particular the structure of all alkali halides. These crystals are transparent if they are sufficiently pure. However, if they are irradiated by energetic photons (X or γ rays) alkali halides become coloured. The reason for this is the following. A photon can eject an anion from its site, creating an unoccupied site called a vacancy. This anion vacancy, surrounded by positively charged ions, can trap an electron and restore the local electrical neutrality of the crystal. The trapped electron has a series of energy levels. It can absorb light and jump from the ground state to an excited state. This process is responsible for the colour of the crystal. The electron traped in the vacancy is called a coloured centre, or F center (from the german Farbezentrum). The structure of an Fcenter is shown on Fig. 5.14. The Mollwo-Ivey Law Let a be the lattice spacing, i.e. the distance between two neighbouring ions Na+ and Cl− . Measurements of the wavelengths λ or energies ε of absorption lines on various alkali halides have been performed by Mollwo and Ivey. The results are displayed on Fig. 5.15. They show that the variation of the absorption energy with the lattice spacing a follows a simple law. 1. Express the empirical law that emerges from these measurements as ε = K an

(5.39)

where ε is in eV and a is in Å. This is called the Mollwo-Ivey law. Since, in good approximation, the absorption energy ε depends only on the lattice spacing a and not on its particular nature, one may assume that the shapes of Fcenters are the same for all of these crystals and that they only differ by their sizes. The simplest model one can build consists in assuming that the Z positive ions nearest neighbours to the Fcenter form a cubic square well potential inside which

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Fig. 5.15 Energy of the absorption peaks of various alkali halides, versus the lattice spacing a

the electron is trapped. In first approximation, we shall assume it is an infinitely deep potential well: V = 0 for 0 < x < a, 0 < y < a, and 0 < z < a V = ∞ for x < 0, y < 0, z < 0, or x > a, y > a, z > a. 2. What is the number Z of positive ions nearest neighbours to an F center? 3. Give the energy levels E 1 and E 2 of the ground state and of the first excited state in the potential well, and the corresponding wave functions. What is the degeneracy of each level? 4. Assuming that the absorption of light is due to the transition of the electrons from E 1 to E 2 , express the absorption energy in terms of the lattice spacing a. Since this model leads to an expression of the form (5.39), compare the experimental and the theoretical values of the exponent n and of the constant K . 5. Clearly, the previous simple model accounts quite successfully for the exponent n but not for the constant K . In order to cure this defect, we remark that the size of the square well is rather arbitrary. By introducing an effective size a0 = αa, choose α in order to fit the experimental date. Give a brief physical comment on the effective size a0 of the well. Plot the theoretical curve on Fig. 5.15. The Jahn–Teller Effect When a state of a non linear molecule is degenerate, one can show that a distortion of the molecule lifts the degeneracy and stabilizes the molecule. This general effect is called the Jahn–Teller effect.5 Here, the Fcenter and the surrounding ions can be considered as a pseudomolecule which can undergo a Jahn–Teller distortion, as we shall now see. 5 H.A.Jahn

and E.Teller, Proc.Royal Soc. A161, 220, 1937.

5.9 Problem. Colored Centers in Ionic Cristals

111

Fig. 5.16 Distortion of an F center

1. Let us distort the potential well of the vacancy into a parallelepiped as shown on Fig. 5.16. The lengths along the x and y axes are equal, we note them c, and the length along the z axis is b. It is reasonable to assume that, owing to the rigidity of the crystal, this distortion occurs at constant volume, i.e. a03 = bc2 . The distortion is characterized by the parameter η = b/c. Show that this distortion lifts the degeneracy of the excited level E 2 . Calculate the dependence of the excited levels on the parameter η. Show that, for one of the excited states (specify which one), the energy has a minimum E 20 for a certain value η0 of the distortion. Is the Fcenter stretched along the z axis or flattened against the (x, y) plane? 2. Calculate the variation with respect to η of the ground state energy E 1 . Calculate the value E 10 = E 1 (η0 ). 3. Plot the variations of E 1 and E 2 as a function of η. The Stokes Shift We can now give a simple account of the absorption and emission of light by an F center. In section 1, we have described the absorption of light by an F center. After a time of the order of 10−6 s, the electron makes a transition to the ground state and emits radiation, called “luminescence”. Experiment shows that the emission lines are systematically shifted towards longer wavelengths—or equivalently smaller energies—than the corresponding absorption lines. This shift, an example of which is shown on Fig. 5.17 is called the Stokes shift. 1. Let us first assume that most lines are shifted to the infrared part of the spectrum, which is not visible. Under this assumption, by what simple mechanism do the F centers colour a crystal when the crystal is placed in visible light? 2. What are, respectively, the colours of the crystals KI, KCl, and NaCl after they have been exposed to X-rays ? We recall that the colours of the spectrum of visible light are, for increasing values of the energy, red (from 1.65 to 2.0 eV), orange (from 2.0 to 2.1 eV), yellow (from 2.1 to 2.3 eV), green (from 2.3 to 2.55 eV), blue (from 2.55 to 2.65 eV) and violet (from 2.65 to 3.1 eV).

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Fig. 5.17 Absorption and emission spectra of an F center in KBr at low temperature. The maxima of the two lines are respectively at 2.06 and 0.92 eV. Data are taken from W. Gebbart and A. Kuhnert, Physica Status Solidi, Vol. 14, p. 157 (1966)

We also recall that “complementary colours” are colours which, when associated, give back white light. The main couples of complementary colours are yellowviolet, red-green and blue-orange. Hence when blue is absorbed by a substance in natural white light, the substance appears to be yellow. 3. We shall now attempt to give a simple description of the Stokes shift. We shall assume that the elecronic excitation or de-excitation times are negligible compared to typical times for local distortions of the crystal, these being, in turn, much shorter than the lifetimes of the excited states (of the order of 10−6 s). Under these assumptions, give a simple description of the absorption and emission of light by an F center, using the results of Sect. 2. 4. More quantitatively, show that the results of section 2 give a good account of the experimental result shown on Fig. 5.17. 5. Justify the assumption made in question 3.1 by showing that for most crystals of Fig. 5.15, the emission line is in the infrared part of the spectrum. Specify for which crystals this occurs.

5.9.1 Solution Section 1 1. The experimental points lie on a straight line in a log-log plot. The experimental law is of the form ε = K a n with K  68 and n  −1.85. 2. There are Z = 6 positive ions at a distance a/2 of the Fcenter. 3. Choosing the origin at a vertex of the cube, (a) the ground state, with energy E 1 = 32 π 2 /(2ma 2 ), is not degenerate; its wave function is ψ = (2/a)3/2 sin(πx/a) sin(π y/a) sin(πz/a);

5.9 Problem. Colored Centers in Ionic Cristals

113

Fig. 5.18 Absorption lines of F centers in various alkali halides; comparison of the data and the model developed in question 1.5

(b) the first excited state has a three-fold degeneracy ψ2x , ψ2y , ψ2z with, for instance, ψ2z = (2/a)3/2 sin(πx/a) sin(π y/a) sin(2πz/a) corresponding to an energy E 2 = 62 π 2 /(2ma 2 ). 4. The transition E 1 → E 2 corresponds to the absorption of an energy ε = E 2 − E 1 = 32 π 2 /(2ma 2 ) where a is the lattice spacing. This expression is of the type (5.39) with K = 112 and n = −2. The value of n is close to what is experimentally observed (−1.85). The constant K is quite overestimated. 5. If the effective extension of the potential is a0 = αa, the theoretical formule becomes ε = 32 π 2 /(2ma 2 α2 ). Using the value α = 1.13 corresponding to K = 87 (and n = −2), one obtains a good fit to the data as shown on Fig. 5.18. The effective size of the cube is 13 % greater than the lattice size. This is not surprising since the six neighbouring positive ions each attract the electron of the F center. In a more realistic potential model of the F center, the probability for the electron to be outside the vacancy should be non-zero. Section 2 1. Consider the state ψ2z = (2/a0 )3/2 sin(πx/a0 ) sin(π y/a0 ) sin(2πz/a0 ) . Under the distortion, it becomes:  ψ2z = (2/c)(2/b)1/2 sin(πx/c) sin(π y/c) sin(2πz/b) ,

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and the corresponding energy E 2 = 6π 2 2 /(2ma02 ) becomes  = E 2z

2 π 2 2 4 ( 2 + 2) . 2m c b

Setting η = b/c, and imposing that the distortion occurs at constant volume, a03 = c2 b, one has c = a0 η −1/3 and b = a0 η 2/3 , hence  = E 2z

2 π 2 (2η 2/3 + 4η −4/3 ) . 2ma02

Similarly, one finds that   E 2x = E 2y =

2 π 2 (5η 2/3 + η −4/3 ) . 2ma02

   Clearly, E 2x = E 2y on one hand, and E 2z on the other are different from E 2 , and different from one another. The distortion lifts partially the degeneracy.   and E 2x with respect to η, we find that both If we study the variation of E 2z energies have minimum values:  • E 2z is minimum for η = 2, where it reaches the value  (2) = 4.76 E 2z  is minimum for η = • E 2x

2 π 2 ; 2ma02

√ (2/5)  0.63, where it reaches the value

 2 π 2  ( (2/5)) = 5.52 . E 2x 2ma02 Both values are smaller than E 2 . The first is the absolute minimum. Hence the energy of the first excited state has the minimum value E 20 = 4.76 (2 π 2 )/ (2ma02 ), for a value η = 2 of the distortion parameter. Since η > 1 this corresponds to an F center stretched along the z axis. 2. When the F center is distorted, the ground state energy is E 1 =

2 π 2 2 1 2 π 2 (2η 2/3 + η −4/3 ) . ( 2 + 2) = 2m c b 2ma02

This function is minimum for η = 1, i.e. an undistorted center. Any distortion will increase the energy of the ground state. We have, in particular at η = 2 where the excited state energy is minimum, E 1 (η0 ) = E 1 (2) = 3.57 2 π 2 /(2ma02 ) . 3. The variations of the energy levels with the distortion are shown on Fig. 5.19.

5.9 Problem. Colored Centers in Ionic Cristals

115

Fig. 5.19 Variation of the energy levels, in units 2 π 2 /2ma02 , with the distortion parameter η

Section 3 1. If the emission of light is in the infrared part of the spectrum, it will not produce a colouring of the crystal. The colour is only due to absorption. The observed colour is the complementary colour to that of the absorbed radiation (of energy ε). 2. Among the crystals mentioned, NaCl absorbs violet light (ε  2.75 eV), its colour is therefore yellow. Similarly, KI is green, and KCl is violet. The first time this problem was given to students, it was actully accompanied with three plastic bags containing respectively yellow, green and light violet crystals. The question was to determine what type of alkali halide was contained in each of them. The crystals had been irradiated overnight in a Van de Graaf accelerator. 3. If an F center distorts itself after absorbing energy, i.e. when it is in the excited state, its energy will decrease down to E 20 . If it de-excites, it will emit a photon of smaller energy than the energy of the absorbed photon. Hence the Stokes shift. Using the Franck–Condon principle, one may represent the successive steps of the absorption-emission process as follows (Fig. 5.20). • 0: F center in its ground state; • 1: Absorption of a photon of energy ε = E 2 − E 1 , instantaneous transition to the degenerate state ψ2 ; • 2: Distortion of the F center. The electronic energy decreases down to E 20 . The corresponding energy difference E 2 − E 20 is transfered to the thermal vibrations (phonons) of the crystal.

Fig. 5.20 Schematic time description of the absorption-emission process by an F center

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• 3: De-excitation. This process is instantaneous, and occurs on a distorted center. The emitted photon has energy E 20 − E 1 (η0 ). • 4: The F center recovers its original symmetry. The corresponding energy E 1 (η0 ) − E 1 , is again released in the crystal thermal vibrations. 4. From the previous considerations, the energy of the emission line is, within our model, 2 π 2 2 π 2 (4.76 − 3.57) = 1.19 . ε = E 20 − E 1 (η0 ) = 2 2ma0 2ma02 This emission energy is smaller than the absorption energy; the ratio is ε /ε ∼ 0.4. The experimental result for KBr is ∼0.44 (see Fig. 5.17). The agreement of the model with experiment is quite acceptable. 5. The ratio ε /ε calculated above does not depend on a0 , and therefore should not depend on the nature of the cristal. For an absorption energy near the upper part of the visible spectrum, i.e. ∼3.1 eV, the calculated emission energy is of the order of 3.1 × 0.4 = 1.14 eV which lies in the infrared region. We therefore conclude that if the absorbed light is in the visible part of the spectrum, (crystals ranging from RbI to KF on Fig. 5.15) then the emission lines lie outside of the visible spectrum. We assumed that in Sect. 3.1. Further Comments on F centers 1. The mechanism by which the F centers form is yet unclear. There are several proposals (the most plausible beeing due to Pooley) which are based on the assumption that the X-ray photons can ionize the anions once ( A− → A) or twice ( A → A+ ). The resulting species, either electrically neutral or positively charged, is then in a very unstable situation in the middle of all the positive ions. It is then ejected from its site, leaving behind a vacancy (F center) and reaching an interstitial position. The colour can also be obtained by adding impurities (such as a few Ca++ ions in NaCl) to the crystal. This the reason why many minerals with a marked ionic character are found coloured in nature, while they are transparent if they are pure (like quartz). They were contaminated by other ions when they crystallized. 2. The model of Sect. 1 accounts for the Mollwo-Ivey law quite reasonably. It is, of course, very simplistic. The actual potential is by no means infinitely deep. By electron spin resonance experiments, one can show that the wave function extends up to the eighth ionic shell surrounding the F center, i.e. much further than a0 /2. 3. The F centers can move around. A nearby anion can jump in the F center, which therefore moves in the reverse direction. This process involves the crossing of a potential barrier, and is favored by an increase in temperature. The mobility of an F center increases with the temperature.

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Owing to this mobility, the F centers tend to disappear, for instance when they reach the surface of the crystal. One can see the colour disappear if the crystals are heated. The colour can also disappear progressively if the crystals are exposed to natural light. In fact, the F centers can then be ionized by ultra-violet photons, which can eject the electron from its vacancy.

Chapter 6

Principles of Quantum Mechanics

During the years 1925–1927, quantum mechanics took shape and accumulated successes. But three persons, and not the least, addressed the question of its structure. In Zürich there was Erwin Schrödinger, in Göttingen David Hilbert, and in Cambridge Paul Adrien Maurice Dirac (Fig. 6.1). Hilbert was 65; he was considered the greatest living mathematician since the death of Poincaré. Schrödinger was 40. Dirac was a young 23-year-old student in Cambridge, simply brilliant. Their thoughts, which we can follow, generated the basic principles of quantum mechanics. Here, we do the following. • First we write a more concise and general formulation of what we have done by using the formalism of Hilbert space and the notations invented by Dirac.

Fig. 6.1 Schrödinger, Hilbert, and Dirac at the end of the 1920s (All rights reserved)

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_6

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• Then, we prove in a simple manner some results we have already guessed on observables, in order to be able to state the general principles of quantum mechanics, valid for any system. • This allows us to discover at last Heisenberg’s matrices and the “matrix mechanics” that had been elaborated by Heisenberg, Born, Jordan, and Pauli between 1924 and 1925. • In the course of this, we understand quite simply how Schrödinger and Dirac showed, in 1926, the equivalence of the two approaches: wave mechanics and matrix mechanics. • Finally, we illustrate these principles on a quantum phenomenon that is directly visible and familiar. It is the only one: the polarization of light. From the point of view of Hilbert, Schrödinger, and Dirac, what were the problems? First there were two versions of quantum mechanics and one had to impose some order. But even without that, our theory is nice and appealing, but it is not very pleasant esthetically. First, it is restricted to the motion of a particle in space, and it must be generalized. Also, it is somewhat ambiguous. In wave mechanics, one describes the state of a particle in space at time t by a wave function ψ(r, t). However, that the description is not unique. The Fourier transform of the wave function ϕ( p, t) is a completely equivalent description of the state of the particle. We can perfectly well state our principles using ϕ( p, t). For instance, the position observable x is then xˆ = i

∂ . ∂ px

To get a clear insight of this consists of doing theoretical physics. We now speak about mathematics. We do not want to do mathematics in the sense of taking care of rigor and convergence questions, but we want to use mathematics and to see how this allows to understand the physics better. Actually, the only difficulty is the language. After one has become familiar with the language, things become much simpler; life is easier. But learning a new language is always difficult at the beginning.

6.1 Hilbert Space Consider two wave functions and their Fourier transforms. Because of Plancherel’s theorem, the two following integrals are equal, 

ψ1∗ (r, t)ψ2 (r, t) d 3 r =



ϕ∗1 ( p, t)ϕ2 ( p, t) d 3 p.

(6.1)

When mathematicians see such properties, they understand the underlying structures. Indeed such integrals can be viewed as scalar products. The extraordinary idea of people such as Banach, Hilbert, and Fréchet was to consider functions as vectors or

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121

points in vector spaces, and to use a geometric language in order to solve problems of analysis. What happens here is quite similar to what happens in ordinary geometry. A given vector can be represented by different sets of coordinates in different reference systems. But the lengths, the angles, namely scalar products, are independent of the specific reference system. We show that there are, in fact, many representations of the state of a system. Each one has its own advantages. Two-Dimensional Space Before entering the core of the subject, we can recall some simple notions about Hermitian spaces, that is, complex vector spaces of finite dimension. These notions are useful in what follows. In two dimensions, for simplicity, one can represent a vector u by the column matrix of its components. The conjugate vector is the line matrix u¯ where we transpose and take the complex conjugate of each coordinate. We denote v|u as the Hermitian scalar product of u and v:  u=

u1 u2



  , u¯ = u ∗1 , u ∗2 , v|u = v1∗ u 1 + v2∗ u 2 .

(6.2)

This scalar product is positive definite and the norm u of a vector is defined by u2 = u|u.

(6.3)

The Hermitian conjugate M † of a matrix M is obtained by transposing and taking complex conjugates of numbers Mi†j = (M ji )∗ . A matrix is said to be Hermitian if it is equal to its Hermitian conjugate M = M † . The eigenvalues of a Hermitian matrix are real. The corresponding eigenvectors, normalized to one, form an orthonormal basis, or a Hermitian basis of the space. Square Integrable Functions Let us now consider square integrable functions that are of interest in quantum mechanics. A complex function f (x) of a real variable x is said to be square integrable if it satisfies  ∞ | f (x)|2 d x < ∞. (6.4) −∞

In mathematics, one calls this set of functions L2 (R), so we write f ∈ L2 (R). Square integrable functions form a complex vector space (any linear combination of square integrable functions is square integrable). The extension to three dimensions (or three variables) g(x, y, z) is straightforward; the corresponding space is denoted L2 (R3 ). At this point, we consider the results of Charles Hermite in 1860. In fact, considering complex functions, Hermite defined an Hermitian scalar product of two square

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integrable functions f and g by 

g ∗ (x) f (x) d x.

g| f  =

(6.5)

This is linear in f and antilinear in g, and it possesses the Hermitian symmetry g| f  =  f |g∗ .

(6.6)

This allows us to define the norm  f  of the function f by  f = 2

| f (x)|2 d x.

(6.7)

Algebraically, it is exactly the same as in finite-dimensional spaces considered above. It is convergence, that is, topological properties, that is different. Now, Hermite made a very remarkable discovery. Without being aware of that, he studied the quantum harmonic oscillator eigenvalue problem (x 2 − or

d2 )ϕn (x) = εn ϕn (x) dx2

2 ˆ n (x) = εn ϕn (x) with hˆ = (x 2 − d ), hϕ dx2

(6.8)

and he found all the square integrable solutions {ϕn (x), εn }, ϕn (x) = γn e x

2

/2

d n  −x 2  e , εn = 2n + 1, n integer ≥ 0. dxn

(6.9)

These functions are normalized to one (ϕn  = 1) if γn = π −1/4 2−n/2 (n!)−1/2 .

(6.10)

The Hermite functions ϕn (x) are orthonormal (i.e., orthogonal and normalized to one) as one can check; they form a free orthonormal set. But Hermite found a most remarkable property. All square integrable functions can be expanded on the set of Hermite functions, ∀ f ∈ L2 (R) ,

f (x) =

∞

Cn ϕn (x),

(6.11)

n=0

where the components Cn of f are Cn = ϕn | f ,

(6.12)

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123

because the ϕn (x) are orthogonal and normalized. In other words the Hermite functions are a complete set, called a Hilbert basis of L2 (R). That is the great discovery! Square integrable functions form a Hilbert space which has the three properties of being a complex vector space, where a Hermitian positive definite scalar product is defined, and that contains Hilbert bases (the ϕn (x) are an example of such a basis for the space of square integrable functions in one variable L2 (R)). This space is infinite-dimensional. For mathematicians it is more interesting to study than a two-dimensional space, but psychologically, for us it is essentially similar. We are not concerned with topological properties (even though we mention some of them, and they play an important role in more elaborate quantum mechanical problems). The algebraic rules are the same as in finite-dimensional spaces. Therefore, in a Hilbert basis, f is entirely determined by the set of its components f (x) ⇐⇒ {Cn }. Actually, one can “forget” about the elements of the basis ϕn (x) which are simply kept in a catalogue, and work directly with the components {Cn } that define the vector f ∈ L2 (R). For instance, consider a function g, whose expansion is g(x) =

∞

Bn ϕn (x).

(6.13)

n=0

The scalar product g| f  is expressed simply in terms of the components of f and g, Cn = ϕn | f  and Bn = ϕn |g, as g| f  =



Bn∗ Cn ,

(6.14)

|Cn |2 ,

(6.15)

and the norm of f is given by  f 2 =



which is simply the extension of the Pythagorean theorem. (In infinite dimensions, mathematicians call that the Bessel–Parseval theorem.) If f is normalized to 1, we have Cn 2 = 1. One thing that one does not fully appreciate at first is that the geometrical properties of a Hilbert space are very similar to those of a Euclidian space: the Pythagorean theorem and the triangle inequality hold in both. The result of the application of hˆ (Eq. (6.8)) to f is

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hˆ f (x) =



Cn εn ϕn (x).

(6.16)

Hence the scalar product of hˆ f and f (or, more generally, any function g) is  f |hˆ f  =



εn |Cn |2 .

(6.17)

We can therefore use a geometrical language in order to speak about problems of analysis.

6.2 Dirac Formalism Coming back to quantum mechanics, Hilbert and Dirac understood in 1927 that the wave function ψ and its Fourier transform ϕ are simply two representations of the same unique mathematical object, a vector in Hilbert space. And they were able to give a clear formulation of the theory with these ideas. Dirac invented notations that have been adopted by mathematicians. From now on, we say that the state of a system is described a any time t by a state vector (6.18) |Ψ (t), which belongs to a Hilbert space E H , and which Dirac calls a “ket”. The functions ψ and ϕ are simply particular representations of this vector. For a particle in three-dimensional space R 3 , the Hilbert space is the space of square integrable functions in three variables L2 (R 3 ), but this can be generalized to any system.

6.2.1 Notations Vectors 1. The vectors are denoted by the symbol: |name. In (6.18) we have indicated the presence of the time variable. Of course, there exist fixed vectors, such as the elements of a basis that can be denoted |ϕn  or |n. 2. The Hermitian scalar product of ψ1 and ψ2 is denoted ψ2 |ψ1  = ψ1 |ψ2 ∗ .

(6.19)

It is not commutative, it has Hermitian symmetry, and it is linear on the right and antilinear on the left. 3. This scalar product is positive definite which allows us to define the norm ψ of a vector by

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125

ψ2 = ψ|ψ.

(6.20)

The norm of a vector ψ vanishes if and only if ψ is the null vector. The norm of a state vector is always equal to one. 4. At the beginning, it is useful to have a little dictionary, where one can see that Dirac’s notations are obviously quicker to use (Fig. 6.2). 5. The elements of the dual space are denoted ψ(t)| ∈ E H∗

(6.21)

and called bras (this is justified owing to a theorem of F. Riesz). This notation comes from the scalar product. A rule in Dirac’s formalism is the contraction of products. When a bra u| is on the left of a ket |v the expression contracts in a “bracket”, that is, the number λ = u|v. In other words, the bra “eats” the ket and gives this number. We encounter other examples of this contraction rule, which comes from the tensor structure of quantum mechanics. 6. Hilbert basis A Hilbert basis is a free, orthonormal and complete set of vectors {|n} :

n|m = δnm .

Fig. 6.2 Dirac versus wave functions dictionary

Dictionary Dirac formalism

Wave functions

|ϕ

ϕ(r)

|ψ(t)

ψ(r, t)

ψ2 |ψ1 





ψ2 = ψ|ψ ˆ 1 ψ2 |A|ψ ˆ a = ψ|A|ψ

ψ2∗ (r)ψ1 (r) d3 r

 

|ψ(r)|2 d3 r

ˆ 1 (r) d3 r ψ2∗ (r)Aψ ˆ ψ ∗ (r)Aψ(r) d3 r

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Any vector |ψ can be expanded on this basis as |ψ =



Cn |n, with Cn = n|ψ.

(6.22)

n

Cn is the component of |ψ along |n. All Hilbert spaces possess Hilbert bases.

6.2.2 Operators Consider now linear operators, that is, linear mappings of the space onto itself. We will keep the same notation Aˆ as before: ˆ |χ = A|ψ , (ψ , χ) ∈ E H .

(6.23)

1. We are interested in the following numbers which are the scalar products of (6.23) with some other vector: ˆ 1 ). ψ2 |χ1  = ψ2 |( A|ψ

(6.24)

One can show that this product is associative, that is, one can define the action of Aˆ in the dual space ˆ 1 ) = (ψ2 | A)|ψ ˆ 1  = ψ2 | A|ψ ˆ 1 . ψ2 |χ1  = ψ2 |( A|ψ

(6.25)

We call this scalar product the matrix element of Aˆ between ψ1 and ψ2 . This is the same as in finite-dimensional spaces. In ordinary matrix calculus, this expression would be of the type v¯ Mu, where M is a matrix. ˆ 1 , In order to be rigorous, mathematicians have another way of writing: ψ2 | Aψ which does not show the useful associativity. Of course, they are right to do so in their work. One cannot bypass rigor. In infinite-dimensional spaces there exist operators whose “domain” (i.e., the set of functions on which they act safely and produce square integrable results) is not the entire space. For instance, if we multiply a square integrable function by x, the result is not always square integrable. Here, we do not worry about such a question. Again, the basic theme in all this is that we act as if we were in finite-dimensional spaces. To a large extent the topological questions can be put aside for what concerns us at the present level. 2. Adjoint operators. The adjoint operator Aˆ † of an operator Aˆ can be defined by the relation ˆ 1  = ψ1 | A|ψ ˆ 2 ∗ , ∀ψ1 , ψ2 ∈ E H . (6.26) ψ2 | A|ψ

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127

It is the same definition as for matrices: we transpose and take the complex conjugate. 3. Self-adjoint operators. An operator Aˆ is said to be self-adjoint or Hermitian if Aˆ = Aˆ † , or, equivalently ˆ ψ| A|ψ is real for all vectors ψ of E H .

(6.27)

But we already know this expression. If we look at that above dictionary, (6.27) is the expectation value of the quantity A. Now we remark that the results of measurements, in particular expectation values, are real numbers. Therefore, observables are Hermitian operators Aˆ = Aˆ † as we announced in (4.6). Theorem 2 Observables are Hermitian operators. 4. Commutation of observables. We can see easily with the correspondence principle that observables do not commute in general. The product of Aˆ and Bˆ is not the same as the product of Bˆ ˆ One defines the commutator [ A, ˆ B] ˆ of two operators as and A. ˆ B] ˆ ≡ Aˆ Bˆ − Bˆ A. ˆ [ A,

(6.28)

One can check the fundamental commutation relation [x, ˆ pˆ x ] = i Iˆ,

(6.29)

where Iˆ is the identity operator. We show later on that commutation relations between observables play a fundamental role. They allow us, in particular, to derive uncertainty relations for any couple of physical quantities.

6.2.3 Syntax Rules Before we come back to observables and measurement results, we make two observations on syntax rules in Dirac’s formalism. 1. Contraction of products. In this formalism, expressions are products of terms. Such products can be contracted. We have seen this for the scalar products; it is true for any expression (as we have said, this comes from the tensor structure of the theory). With this rule, we discover a special kind of operators of the form |uv|.

(6.30)

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This object is a linear operator because if we apply it to a ket |ψ, the bra v| eats the ket |ψ, which gives the number λ = v|ψ, and one ends up with the vector λ|u. 2. Hermitian conjugate of an expression. Just as in finite-dimensional spaces, one must transpose and take the conjugate of each term. In other words, we reverse the order of the factors, and we change the kets into bras and vice versa, the operators in their adjoints and the numbers in their complex conjugates. The Hermitian ˆ conjugate of λ|ϕψ| Aˆ † Bˆ is λ∗ Bˆ † A|ψϕ|.

6.2.4 Projectors; Decomposition of the Identity Projector. Consider a Hilbert basis {|n}. The operator Pˆn = |nn|

(6.31)

is the projector on the basis vector |n (this remark applies to any vector that is normalized to one). Indeed we have Pˆn2 = Pˆn and Pˆn |ψ = Cn |ψ, where Cn is the component (6.22) of |ψ on the vector |n. One can define a projector Pˆν on a subspace n ∈ {ν} by Pˆν =



|nn|.

(6.32)

n∈{ν}

If we extend this to the entire space, we obtain the important closure relation, also called the decomposition of the identity

|nn| = Iˆ.

(6.33)

n

6.3 Measurement Results We can now understand that, in a measurement of a quantity A, ˆ • The possible results of the measurement are the eigenvalues an of the observable A, • And the probability of finding the result an is the modulus square of the scalar product of the state vector with the corresponding eigenvector: p(an ) = |ϕn |ψ(t)|2 .

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129

6.3.1 Eigenvectors and Eigenvalues of an Observable We denote |ϕn  the eigenvectors of Aˆ and an the corresponding eigenvalues, that is, by definition ˆ n  = an |ϕn . A|ϕ (6.34) We constantly use the following theorems, well known in finite-dimensional spaces. Theorem 3 The eigenvalues an of an Hermitian operator are real. The proof is straightforward. If we multiply (6.34) on the left by ϕn |, we obtain ˆ n  = an ϕn |ϕn . ϕn | A|ϕ The left-hand side is real because Aˆ is Hermitian, and ϕn |ϕn  is real and positive. Therefore an is a real number. Theorem 4 The eigenvectors corresponding to different eigenvalues are orthogonal. We multiply (6.34) on the left by another eigenvector ϕm |; we obtain ˆ n  = an ϕm |ϕn  = am ϕm |ϕn . ϕm | A|ϕ In the last expression, Aˆ acts on the left on ϕm |. Therefore, we obtain (an − am )ϕm |ϕn  = 0,

(6.35)

so that either an = am , or, if an = am , ϕm |ϕn  = 0. We can therefore choose a set of eigenvectors {ϕn } that are orthonormal. Let us mention a minor and inessential technical difficulty in this context. It can happen that to the same eigenvalue an there correspond several independent eigenvectors |ϕn,k  ˆ n,r  = an |ϕn,r  , r = 1, . . . , k. A|ϕ (6.36) In such a case, one says that the eigenvalue an is degenerate with a degeneracy of order k. Notice that the projector on the eigensubspace of an , of dimension k, is Pˆn =

k

|ϕn,r ϕn,r |,

r=1

where we assume we have chosen the |ϕn,k  to be orthonormal ϕn  ,k  |ϕn,k  = δnn  δkk  .

(6.37)

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6.3.2 Results of the Measurement of a Physical Quantity Theorem 5 The result of the measurement of a quantity A on a system is certain (i.e., with probability one) if and only if the system is in an eigenstate of the observable ˆ A. We have already seen in Chap. 4 that if ψ is an eigenvector of Aˆ then the result of the measurement is certain since the dispersion vanishes, Δa = 0. The converse is easy to prove. Consider the norm of the vector |χ = ( Aˆ − a Iˆ)|ψ, where Iˆ is the identity. We obtain ( Aˆ − a Iˆ)|ψ2 = ψ|( Aˆ − a Iˆ)2 |ψ = ψ| Aˆ 2 |ψ − a2 = Δa 2 .

(6.38)

If the dispersion is zero Δa 2 = 0, the norm of the vector ( Aˆ − a Iˆ)|ψ vanishes. ˆ This vector is therefore the null vector and A|ψ = a |ψ. Therefore, if Δa = 0, |ψ is necessarily an eigenvector of Aˆ with eigenvalue a. Theorem 6 The result of a measurement of a quantity A on a single system is one ˆ of the eigenvalues an of the observable A. In Chap. 4, we made the following remarks. • For a given system and a given quantity, nothing restricts the accuracy of the measurement. One finds some value with the accuracy allowed by the measuring instruments. • By consistency, if, immediately after the measurement, we perform another measurement of the same quantity on the system that has already been measured, we will find the same answer with probability one. (This is a postulate on the consistency of physics; no experiment has ever contradicted this.) • Therefore, the measurement on a system is actually a means to prepare this system in a new state for which we know the value of A exactly (Δa = 0). ˆ and • Owing to Theorem 5, the system is then necessarily in an eigenstate of A, ˆ therefore the value found previously is an eigenvalue of A. • Obviously, after the measurement, the state vector of the system is in the eigenspace corresponding to the eigenvalue that has been obtained.

6.3.3 Probabilities What is the probability of finding the result an by measuring A on a system whose state is ψ? There, the answer comes from geometry. In fact, the question is to know “how much” of the eigenvector |ψn  does the state vector contain. By the superposition principle, we can understand that if it is 100 % |ψn  the probability is equal to 1, and if it does not have any component along |ψn , if it doesn’t contain it at all, the probability is zero; one will never find the value an .

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131

How can we evaluate this probability? The answer is that the probability is the modulus squared of the component of the state vector |ψ along the normalized eigenvector |ϕn , or, equivalently, of the scalar product of |ψ and |ϕn , p(an ) = |ϕn |ψ(t)|2 .

(6.39)

The scalar product ϕn |ψ(t) is the probability amplitude α(an ) to find an . Do we really find that by some magic inspiration? No; it is practically written above in (6.17) in a particular case. Let’s go back to Sect. 6.1, to the Eqs. (6.15) and (6.17), and to the case of the ˆ n (x) = E n ϕn (x). harmonic oscillator. The {ϕn } are the eigenstates of the energy: hϕ The components Cn = ϕn |ψ of the state vector on this basis are these scalar products. The normalization of the state vector (i.e., the probabilistic interpretation) and the Pythagorean theorem tell us that ∞

|Cn |2 = 1,

(6.40)

n=0

as should be satisfied by a probability law. The expectation value of the energy (6.17) is indeed given in terms of the possible issues E n by E =

∞

E n |Cn |2 = 1.

(6.41)

n=0

This shows that in this case the |Cn |2 = |ϕn |ψ|2 are the probabilities of finding E n . Indeed, it is a theorem, which is not difficult to prove for a finite set, that if we know the outcomes of a probability law and its moments (i.e., E k  for all values of the integer k) we know the probabilities (obviously we can calculate the expectation value of any power of hˆ and obtain the value of E k ). And that is completely general! Above, we never referred to the specific form of ˆ of ϕn , or of the values E n . We simply used the fact that the {ϕn } form a Hilbert h, ˆ basis, and that they are the eigenvectors of h.

6.3.4 The Riesz Spectral Theorem This property relies on the spectral theorem of Frederic Riesz, which is a fundamental theorem of Hilbert space analysis. Theorem 7 Spectral theorem. The set {|ϕn } of eigenvectors of an Hermitian operator Aˆ forms a Hilbert basis of the space. This is well known in finite-dimensional spaces and matrix calculus.

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We have deliberately stated it in a mathematically incorrect way for an infinite dimensional space. The true statement is: To any self-adjoint operator, there corresponds a decomposition of the identity and a spectral decomposition. In other words, if we forget about possible degeneracies for simplicity, ˆ 1. Any vector |ψ can be decomposed on the basis{|ϕn } of the eigenvectors of A, ∀ |ψ, |ψ =



Cn |ϕn  , with Cn = ϕn |ψ.

(6.42)

n

If |ψ is a normalized state vector, then ψ|ψ =



|Cn |2 = 1.

(6.43)

n

2. The decomposition of the identity, or closure relation, is Iˆ =



|ϕn ϕn |.

(6.44)

n

3. The operator Aˆ has a spectral decomposition; that is, Aˆ =



an |ϕn ϕn |.

(6.45)

n

4. Therefore

ˆ A|ψ =

n

Cn an |ϕn  , and a =



an |Cn |2 .

(6.46)

n

We can read in (6.43) and (6.46) that the numbers {|Cn |2 } are the probabilities of finding the results an , as previously. As stated in Theorem 7, this “theorem” isn’t quite true; why? It is true in spirit, but not in form; it lacks rigor. In fact, there exist pathologies. For instance, there exist operators which, when they are applied to some vectors, “push them out” of the Hilbert space, such as x or d/d x. Their eigenvectors do not belong to the Hilbert space, but to another space called the space of eigendistributions. The eigenfunctions of pˆ x are not square integrable, because they are plane waves ∝ ei p0 x/ . But the amazing fact is that one can nevertheless expand any square integrable function on this set, this is simply the Fourier transformation f (x) = (2π)−1/2 g( p)ei px/ dp. One can expand a square integrable function on a continuous set of functions that does not belong to the Hilbert space, a continuous basis that belongs to another space. Of course, it is quite feasible to write quantum mechanics in a rigorous way, but it is tedious and too complicated at our stage, and it does not bring anything new physically. It suffices to be aware of that.

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133

6.3.5 Physical Meaning of Various Representations At this point, things are becoming physically interesting. In a given Hilbert basis, it is obvious that the state vector |ψ is completely determined by the set of its components {Cn }, |ψ ↔ {Cn = ϕn |ψ} which we can write as a column vector, the corresponding bra being the conjugate line vector. This representation of the state vector is completely equivalent to the wave function ψ(r, t). Therefore, there are not only two, but an infinite number of equivalent representations of the state of the system. What is their physical meaning? In the basis of the eigenstates of the Hamiltonian, the interpretation of this representation is simple and crystal clear: the Cn ’s are the amplitudes to find E n in an energy measurement. Therefore, • The representation ψ(r, t) is more convenient if we are interested in the properties of the particle in space, • Its Fourier transform ϕ( p, t) is more convenient if we are interested in its momentum properties, • And the components {Cn } in the basis of energy eigenstates are more convenient if we are interested in the energy of the particle. But, owing to Riesz’s theorem, this can be done with any physical quantity, for instance, the angular momentum, which we examine later on and which also has discrete eigenvalues. This can be thought of as a “generalization” of the properties of the Fourier transform.

6.4 Principles of Quantum Mechanics We are now able to state the general principles of quantum mechanics. Up to a technical detail, which we discuss below, these are the following three principles. The Principles I. Superposition principle With each physical system is associated a Hilbert space E H . The state of the system is defined at any instant by a vector |ψ(t) of E H normalized to one.

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 Comment. This means that any linear superposition of state vectors |ψ = Ci |ψi , with Ci complex such that |ψ is normalized, is a possible state vector. Notice that the convention ψ = 1 leaves an indetermination. A state vector is defined up to an arbitrary phase factor eiδ . However, it is an overall phase factor: the relative phases of different states of the system are essential. If |ψ1  = eiδ1 |ψ1  and |ψ2  = eiδ2 |ψ2 , the superposition of states C1 |ψ1  + C2 |ψ2  is different from the superposition C1 |ψ1  + C2 |ψ2 . II. Physical quantities 1. To each physical quantity A there corresponds a linear Hermitian operator Aˆ in E H : Aˆ is the observable corresponding to the quantity A. 2. Let |ψ be the state of the system when a measurement of A is performed. Whatever the state |ψ is, the only possible results of the measurement are the eigenˆ values an of the observable A. 3. Denoting by Pˆn the projector of the subspace associated with the eigenvalue an , the probability of finding the value an in a measurement of A is: P(an ) = ψn 2 where |ψn  = Pˆn |ψ.

(6.47)

This reduces to (6.39) in the absence of degeneracies, but takes into account all cases in a geometrical way. 4. Immediately after a measurement of A that has given the value an , the system is in a new state |ψ  : |ψn  |ψ   = . (6.48) ψn  Comments. Relation (6.47) can be written in the equivalent forms: P(an ) = ψ| Pˆn |ψ = |ψ|ψn |2 .

(6.49)

Principle II.2 is called the principle of quantization and II.3 is the principle of spectral decomposition. Principle II.4 is the principle of wave packet reduction. It is the quantitative form of the fact that a measurement perturbs the system. III. Evolution in time Let |ψ(t) be the state of a system at time t. As long as no measurement is performed on the system, its evolution in time is given by the Schrödinger equation: i

d |ψ(t) = Hˆ |ψ(t), dt

(6.50)

where Hˆ is the observable energy, or the Hamiltonian of the system. The state vector ψ(t) depends on time and evolves in Hilbert space according to this first-order ordinary differential equation.

6.4 Principles of Quantum Mechanics

135

The Case of a Continuous Spectrum This is the case for the position and momentum variables. In such a case, the only prediction that makes sense is to find a result inside some range [a, a + da[. The discrete probability law (6.47) is replaced by a continuous law. In the case of the position observable x, the law is P(x) d x = |ψ(x)|2 d x,

(6.51)

where ψ(x) is the wave function of Chap. 3, and similarly for the momentum variable p. Interest of This Synthetic Formulation This formulation has many advantages. It is general. It is precise mathematically speaking. It exhibits the important features of the theory. It is instructive to compare what we just did with the history of Maxwell’s equations. On October 27, 1864, Maxwell presented his memoir on the unification of electricity and magnetism to the Royal Society. Maxwell used 283 symbols to write his equations. Here is a sample: ⎧ dγ ⎪ ⎨ dy − Magnetic Force (α, β, γ) dα − dz ⎪ ⎩ dβ − dx

dβ dz dγ dx dα dy

= 4π p  = 4πq  . = 4πr 

“In these equations for the electromagnetic field, we have introduced twenty variable quantities,” said Maxwell, who added “Between these twenty quantities, we have found twenty equations. Therefore these equations are sufficient to determine all the quantities involved provided we know the conditions of the problem.” Of course, we can write the equations in a simpler way by using vectors and vector analysis which exhibits the rotation invariance of the equations. ∇ · B = 0, ∇ · E =

∂B j ∂E ρ , ∇×E=− + , c2 ∇ × B = . ε0 ∂t ε0 ∂t

This way of writing uses only 59 symbols and it leads easily to results such as the propagation equation in vacuum 

 ∂2 − Δ E = 0. c2 ∂t 2

However, relativistic invariance is the fundamental property that underlies Maxwell’s equations and when a theorist writes them, they appear as ∂μ Fμν = j ν , where relativistic invariance is explicit, and one uses only 8 symbols.

(6.52)

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Of course, when it comes to constructing the antenna of a satellite, we must come back to more concrete quantities and recall that in (6.52) there exist a number of implicit conventions (ε0 = c = 1, Aμ = (ϕ, A), F μν = ∂ μ Aν − ∂ ν Aμ , jμ = (ρ, j ), and so on). But the proof of intermediate results is much simpler.

6.5 Heisenberg’s Matrices We can now understand what Heisenberg’s matrices are. Matrix Representation of Operators A vector of the Hilbert space can be represented in a Hilbert basis {|ϕn } by a column vector made of its components Cn . It is therefore natural to expect that any linear operator Aˆ acting on this space has a matrix representation in the same basis {|ϕn }. In fact, consider a vector |ψ and its expansion |ψ =



Cn |ϕn .

(6.53)

n

If we apply a linear operator Aˆ on this, we obtain another vector |χ with components {Bn } in the basis {|ϕn } ˆ Bn |ϕn . (6.54) A|ψ = |χ = n

If we multiply (6.54) on the left by the basis vector |ϕn  we obtain ˆ Bn = ϕn | A|ψ =

ˆ m Cm , ϕn | A|ϕ

(6.55)

m

where we have inserted the expansion (6.53) of |ψ. This is simply the matrix relation between the coefficients {Bn } and {Cn }, Bn =



An,m Cm ,

(6.56)

m

where the matrix elements An,m are given by ˆ m . An,m = ϕn | A|ϕ

(6.57)

This is the expected result: an observable Aˆ is represented in this basis by the matrix (An,m ), which acts on the line or column vectors above.

6.5 Heisenberg’s Matrices

137

Matrices X and P It is interesting to write the matrix representations of the operators x and px in the basis of the harmonic oscillator eigenstates. In order to do this, we can use the recursion relations (5.16) of the Hermite functions (Chap. 5). We obtain: √ ⎞ ⎛ 1 √0 0 . . . √0 ⎜ 1 0  2 √0 . . . ⎟ ⎟ √  ⎜ ⎜ 0 2 √0 3 ...⎟ (6.58) xˆ ⇒ ⎟ , ⎜ 2 mω ⎜ 0 0 3 0 ...⎟ ⎠ ⎝ .. .. .. .. . . . . √ ⎞ 1 √0 0 ... 0 √ ⎜− 1 0 2 √0 . . . ⎟ ⎟ ⎜ √ ⎜ 0 − 2 0 3 ...⎟ ⎟. ⎜ √ ⎜ 0 0 − 3 0 ...⎟ ⎠ ⎝ .. .. .. .. . . . . ⎛  pˆ ⇒ −i

mω 2

(6.59)

In this basis, the matrix of the Hamiltonian is of course diagonal. These are examples of Heisenberg’s matrices ! But how did he end up there? Heisenberg’s Thoughts In 1924–1925, Heisenberg was a young assistant of Max Born in Göttingen (he was 23). He did things that his contemporaries found difficult to comprehend. Heisenberg was an amazing person. It is difficult to understand his way of thinking. He was both very simple and infused with Northern philosophy. He admired Niels Bohr who was a propagandist of positivism and of Kierkegaard’s principle, “Any new experimental field can only be analyzed with concepts of its own. It is not possible to use concepts and principles used previously in other contexts.” And Göttingen was a fabulous place. The philosopher Husserl had been there. He was the founder of phenomenology, “One must come back from discourses and opinions to facts,” that is, describe what one observes in the simplest possible language before trying to interpret it. Husserl had left his student Wittgenstein, a great specialist of language, who said that “It is the language which must adapt to facts and not the reverse. If one attempts to adapt the interpretation of a phenomenon with a language which is already formed and filled with a priori’s, one is bound to draw wrong conclusions on the nature of things.” So, Heisenberg said to himself, “I cannot talk about the position x and velocity v of an electron in an atom. I can only talk about what I am able to see, the positions and the intensities of spectral lines.” In the classical theory of radiation, a fundamental consequence of Maxwell’s equations and of relativity is that if a charged particle is accelerated, it radiates energy. The simplest example is a charged dipole with a sinusoidal motion x = a cos(ωt). It

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radiates and loses energy. At large distances, the field decreases not as 1/r 2 as for a fixed charge, but as 1/r . Therefore, the energy radiated through a sphere of radius R is independent of R; its flux is conserved. The radiated amplitude is proportional to the acceleration E ∝ x  ∝ aω 2 , and the total radiated power is therefore proportional to P ∼ a 2 ω 4 . An arbitrary periodic motion of period T = 2π/ω can be expanded in a Fourier series: an einωt , x= n

and the resulting radiation occurs at frequencies νn = nω/2π with intensities proportional to In ∝ n 4 ω 4 an2 . Now, said Heisenberg, in atoms there are two indices. The atomic frequencies follow the rule νnm = (E n − E m )/ h. So, Heisenberg introduced “quantum amplitudes” or “quantum quantities” with two indices, and a standard time behavior A =⇒ Anm e−i(En −Em )/. (There is a technical difference between the Schrödinger picture, where the state vector evolves and the observables are fixed, and Heisenberg’s where the state vector is fixed, but the observables evolve in time.) The positions of spectral lines are νnm = (E n − E m )/ h and their intensities are proportional to | Anm |2 . Using such ideas, Heisenberg developed a theory that worked quite well. However, he realized that in order for the product of two quantum quantities to have a proper time-dependence and to be expressed in terms of the two quantities, he needed to invent a “symbolic multiplication,” for instance, A2 →





Ank e−i(En −Ek )/ Akm e−i(Ek −Em )/ = (

k

Ank Akm ) e−i(En −Em )/

k

in other words, (A2 )nm =



Ank Akm .

k

Max Born was intrigued. He was a mathematician, and that reminded him of his youth. Those are just the rules of matrix calculus! Born was very excited. The matrices of Heisenberg are there in front of us in (6.58). What seems incredible is that in 1925 mathematicians knew of the existence of matrices, as examples of noncommutative algebras, but they considered them as very formal objects, and they weren’t used to working with matrices in practice. So, Born was very excited: “We must absolutely go further in Heisenberg’s ideas! In his works, there is an underlying structure to be explored!” He publicized Heisenberg’s work widely.

6.5 Heisenberg’s Matrices

139

On July 17, 1925, Born took the train from Göttingen to Hanover and, in the same compartment, he met Pauli. He was excited and said to Pauli, “You know, Heisenberg’s quantities are matrices! Do you want to collaborate?” Pauli, who was very young but very famous because of his monumental treatise on relativity, which he had written in 1922 when he was 22, was a tough character. Most of the time, he found other people’s ideas either stupid or obvious. So he replied to Born, “You’re going to spoil Heisenberg’s beautiful physical ideas with your futile mathematics.” They quarreled. In the same compartment, there was a young and shy mathematics assistant, Pascual Jordan, who said to Born, once the train had arrived and Pauli had left, “I have worked on matrices. Maybe I can help you.” On the evening of the next day, Born and Jordan established the fundamental commutation relation between the matrices (X ) and (P) h i I, (6.60) (X )(P) − (P)(X ) = 2π where I is the unit matrix, that Born called the fundamental equation of the quanten mechanik,1 that is, the mechanics specific to quanta. This led to three fundamental articles of Born, Heisenberg, and Jordan. One must say that in the meantime, Pauli, who was pragmatic, had reconsidered his opinion on matrices, and he had calculated the energy levels of the hydrogen atom with matrices in 1925 before Schrödinger’s calculation.2 Pauli managed to calculate all eigenvalues of an infinite matrix, and some effects that were not measured until the 1980s! Now, Born, Heisenberg, and Jordan went to see Hilbert, who was the great mathematics professor in Göttingen, and questioned him about matrices. “It’s very formal,” said Hilbert. “The only case when I observed it was useful in practice is in eigenvalue problems related to differential equations with boundary conditions.” Born, Heisenberg and Jordan retired politely, thinking that poor Hilbert did not understand the issue. Six months later, Hilbert had fun saying, “If these arrogant youngsters had listened to me, they would have found the Schrödinger equation six months before him.” Indeed the Schrödinger equation was greeted as the great step forward because it enabled the calculation of energy levels as a stationary wave problem, in opposition to quantum restrictions on classical trajectories, as advocated by Bohr. We come back to Dirac in Chap. 8. He appeared in the same summer of 1925; he was just 23. We have done, backwards, the work of unification performed by Schrödinger at the end of 1926 and independently by Dirac at the beginning of 1927. Schrödinger knew a lot of mathematics; he had worked on eigenvalue problems. Dirac was simply a young genius. This brought to an end the quarrels between the pros of wave mechanics and the pros of matrix mechanics. On the other hand, this opened a new field of research to

formula pq − qp = i h/(2π) is carved on the gravestone of Born and his wife in Göttingen. order to do this, Pauli used the SU (2) × SU (2) symmetry of the hydrogen atom.

1 The 2 In

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mathematicians. In 1927 Hilbert and von Neumann laid the mathematical foundations of quantum mechanics where one can find the first steps of the theory of distributions developed by Laurent Schwartz in 1946.

6.6 The Polarization of Light, Quantum “Logic” In order to illustrate all this and to show once more that the most important mathematical structure of quantum mechanics is addition, we examine a quantum phenomenon that is the only one directly visible. Light waves are transverse and they possess a polarization that describes the behavior of the electric field in the plane transverse to the electric field. There are several types of polarizations. The light coming out of a projector is nonpolarized; it is in a statistical mixture of polarization states. A polarizer, for instance a polaroid, filters linear polarization along its optical axis (it can be an anisotropic medium that absorbs light whose polarization is perpendicular to the axis). In Fig. 6.3, left, this axis is assumed to be horizontal. If we place another polarizer, also called an analyzer, at an angle θ with the first one, the transmitted intensity is proportional to I ∝ cos2 θ (Fig. 6.3, middle). At θ = 45◦ the intensity is half of that which came out of the first polarizer. If the axis of the analyzer is vertical, at θ = 90◦ , no light comes out (Fig. 6.3, right). Classically, this is how Fresnel understood and explained the phenomenon. It seems to be elementary geometry. Notice that it happens in the same manner independently of the wavelength. But actually, this is purely a quantum phenomenon; it can and it must be described exactly by this quantum state formalism. (This can be done in classical optics as Stokes understood in the 19th century.) In fact: 1. Light is composed of photons. 2. Photons are elementary particles; they cannot be broken into pieces.

Fig. 6.3 Outgoing light from a horizontal polarizer (left). Intensity across an analyzer at an angle θ (middle); extinction if the analyzer is at 90◦ of the first one (right)

6.6 The Polarization of Light, Quantum “Logic”

141

Fig. 6.4 Reappearance of light issuing from two crossed polarizers if a third polarizer at some angle is inserted between them

So, let’s discuss the matter in terms of photons.3 When a photon impinges on a polaroid, either it goes through it or it doesn’t; there is no other choice for it. Of course, with a macroscopic source, a lot of photons are produced. A light beam of 1 watt carries ∼1018 photons per second. At an angle θ, it is a fraction cos2 θ of these photons that comes out of the analyzer. In other words, each photon has a probability of cos2 θ to get through. This is even clearer if we cross the polarizer and analyzer at a right angle θ = 90◦ . Nothing gets through. States of orthogonal polarizations are incompatible; there is a zero probability that a photon in the horizontal polarization state can be found in the vertical polarization state. Now we can observe an amazing phenomenon. If we insert a third polarizer at some non-zero angle, say 45◦ , between the crossed polarizers, the light reappears (Fig. 6.4) although we have inserted an absorbing object that can only reject all photons polarized perpendicular to its axis! (Actually, is it really the only thing it can do? No! It’s not a triviality to say that it is also able to let photons pass if their polarization is parallel to its axis.) We know the solution. We must describe polarization states of photons in a twodimensional Hilbert space. In this space, we can choose as basis states the states of horizontal and vertical linear polarization, which we denote | → and | ↑.

(6.61)

If the photon is in the state | → it passes through the horizontal polarizer with probability 1. If it is in the state | ↑, it is absorbed by this polarizer, but it passes through a vertical polarizer with probability one. By definition, these states are orthogonal ↑ | → = 0 (Fig. 6.3). We denote |θ as the state of a photon polarized linearly along a direction at an angle θ with the horizontal axis (0 ≤ θ < π). This state is a linear superposition of

3 Polarization

of light comes from the fact that the photon is a “spin one” particle. It is a pointlike massless particle that carries an intrinsic angular momentum whose projection on the direction of propagation is either + or −. The reader will understand that such odd properties are outside the scope of this book.

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the basis states (6.61), as is the orthogonal state |θ + π/2: |θ = cos θ| → + sin θ| ↑, |θ + π/2 = − sin θ| → + cos θ| ↑.

(6.62)

For the particular value θ = π/4, we have (this does not restrict the generality of our argument) 1 1 |  = √ (| → + | ↑), |  = √ (−| → + | ↑); 2 2

(6.63)

this relation can be inverted: 1 1 | → = √ (|  + | ), | ↑ = √ (−|  + | ). 2 2

(6.64)

The explanation of the observations (Fig. 6.3) is that the probability for a horizontally polarized photon to get through a polarizer at an angle θ is p(→, θ) = |θ| →|2 = cos2 θ

(6.65)

as announced before; that is, p(→, 45◦ ) = 1/2. We now come to the observation of Fig. 6.4. After crossing the polaroid successfully at 45◦ , by the principle of wave packet reduction, the photon is necessarily in the state | , which can be decomposed according to (6.63). In this new state, it is in both states | → and | ↑. It is therefore natural to observe that it can cross the vertical polarizer with probability 1/2, whereas this was forbidden in the absence of the intermediate polarizer. If this latter polarizer is at an angle θ, the probability of finding that a photon crosses the entire setup is p(→, θ, ↑) = cos2 θ sin2 θ.

(6.66)

Notice that, although all we have done here is ordinary Euclidian geometry in two dimensions, it is necessary for the space to be complex, i.e. Hermitian, in order to describe all the pure polarization states. There exist states with complex components such as 1 (6.67) |Ψ L ,R  = √ (| → ± i| ↑) . 2 One can check that such states keep the same form under an arbitrary rotation of the linear polarization basis states. These states correspond to left and right circular polarized states (more generally, elliptic polarization states). Quantum “Logic” This allows us to understand better the difference between classical and quantum logics, namely the difference between “or"and “and.”

6.6 The Polarization of Light, Quantum “Logic”

143

The polaroids are filters that let photons pass if their polarization is along their axis and eject them otherwise. Let’s use a metaphor (after all, at the end of such a Chapter, we deserve it!). Let’s use other words. Instead of horizontal and vertical, let’s say ladies and gentlemen. A horizontal polaroid is a filter that only allows ladies to pass. Similarly, a vertical one only lets gentlemen pass. Of course, if one places the two filters one after the other, no one passes. And, if we want to know how many gentlemen are in a population, it suffices to put a vertical polarizer and to measure the outgoing intensity. Let’s say that a polaroid at 45◦ allows smokers to get through, and a polaroid at 135◦ does so for the nonsmokers. No smoker is a nonsmoker. Now assume we are quantum kids trying to understand set theory. We first make a selection of ladies with a horizontal polarizer. Then, in this new sample, we select those who smoke with a polarizer at 45◦ , as in Fig. 6.5 middle. In classical logic, the succession of the two filters allows the passage of people who are both ladies and smokers, that is, the intersection of the sets {ladies} and {smokers}. We can in fact check that none of them is a nonsmoker. But see what happens if we look at who they are! We observe with great horror that, by placing a vertically oriented polarizer, half of the people we had selected initially get through (Fig. 6.5 right). In other words half of these ladies are gentlemen! In quantum mechanics there is no way out but to conclude that half of the ladies who smoke are in fact gentlemen! Therefore there appears to be a very different way of conceiving logic for quantum children when they play. If they play with cubes and spheres each of which can be either blue or red, it’s a hard task, if not an impossible one, to try to find the intersections of sets. Whatever you do in quantum mechanics, smokers are always both ladies and gentlemen, ladies are always both smokers and nonsmokers. All that is simply the mechanism of the superposition principle and the reduction of the wave packet.

Fig. 6.5 Difference between the quantum and and the classical or: outgoing light after crossing a series of polarizers. Left, only one polarizer, middle, a second polarizer at 45 ◦ , right a third polarizer on top of the second, perpendicular to the first one

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6.7 Exercises 1. Translation and Rotation Operators a. Consider a one-dimensional problem and a wave function ψ(x) which can be ˆ , where x0 expanded in a Taylor series. Show that the operator Tˆ (x 0 ) = e−i x0 p/ is a length and pˆ is the momentum operator, is such that: Tˆ (x0 )ψ(x) = ψ(x − x0 ).  N.B. The expansion ei uˆ = ∞ ˆ n /n! is mathematically legitimate. n=0 (i u) b. We now consider a two-dimensional problem in an x y plane and we define the z component of the angular momentum operator by:   ∂ ∂ ∂ Lˆ z = xˆ pˆ y − yˆ pˆ x = −i x −y = −i , ∂y ∂x ∂θ 1/2  and θ = where the polar coordinates r, θ are defined by r = x 2 + y 2 ˆ ˆ arctan(y/x). Show that the operator R(ϕ) = e−iϕ L z / , where ϕ is dimensionless, is such that : ˆ R(ϕ) ψ(r, θ) = ψ(r, θ − ϕ). 2. The Evolution Operator Consider a system whose Hamiltonian does not depend on time (isolated system). Show that the state vector at time t, denoted |ψ(t), can be deduced from the state vector |ψ(t0 ) at initial time using: |ψ(t) = Uˆ (t − t0 ) |ψ(t0 )

with

ˆ

U (τ ) = e−i H τ / .

(6.68)

Show that Uˆ (τ ) is unitary, i.e., Uˆ † = Uˆ −1 . 3. Heisenberg Representation Consider an isolated system whose Hamiltonian is Hˆ . We denote |ψ(0) the state vector of the system at time t = 0. We want to calculate the expectation value a(t) of the results of the measurement of an observable Aˆ at time t. a. Express a(t) in terms of |ψ(0), Aˆ and the evolution operator Uˆ (t) defined in the previous exercise. ˆ for the b. Show that a(t) can be written as the expectation value of an operator A(t) ˆ is determined from: state |ψ(0). Show that A(t) i

ˆ d A(t) ˆ = [ A(t), Hˆ ] dt

and

ˆ ˆ A(0) = A.

(6.69)

6.7 Exercises

145

This approach is called Heisenberg representation (or Heisenberg picture): the state vector is time independent, and the operators obey the Heisenberg equation (6.69). 4. Dirac Formalism with a Two-State Problem Consider two normalized eigenstates |ψ1  and |ψ2  of a Hamiltonian Hˆ corresponding to different eigenvalues E 1 and E 2 (one can set E 1 − E 2 = ω) a. Show that |ψ1  and |ψ2  are orthogonal. √ b. Consider the state |ψ−  = {|ψ1  − |ψ2 }/ 2, calculate the expectation value E of the energy and the dispersion ΔE in this state. c. Assume that at t = 0 the system is in the state |ψ(t = 0) = |ψ− . What is the state of the system |ψ(t) at time t? ˆ 2  = |ψ1 . What are ˆ 1  = |ψ2 , and A|ψ d. Consider an observable Aˆ defined by A|ψ the eigenvalues a of Aˆ in the subspace generated by |ψ1  and |ψ2 ? e. Construct the corresponding combinations of |ψ1  and |ψ2 , which are eigenvecˆ tors of A. f. Assume that at t = 0 the system is in the state |ψ−  corresponding to the eigenvalue a = −1. What is the probability to find a = −1 in a measurement of A at a later time t?

Chapter 7

Two-State Systems

In the previous chapter, we saw how Heisenberg’s matrix mechanics arose in 1924–1925. What we want to do here is to come back to the problem of the NH3 molecule, seen in Chap. 5 and to do matrix mechanics on this particular case, in a similar way to what we did in wave mechanics by considering the simple problem of the motion of a particle in space. This will allow us to become familiar with Dirac’s formalism and to treat problems in a much simpler way, mathematically speaking, than with wave functions. We show that a great deal of quantum mechanics can be performed with very little mathematics, namely two-dimensional matrices. We have seen a first example of this with the polarization of the photon in the previous chapter. In other cases, the link with physics might have been obscure if we had started quantum mechanics in this way. We might have had problems in relating simple calculations with actual physical phenomena if we hadn’t already treated the problem of the NH3 molecule as in Chap. 5. This leads us to a series of applications. Here, we consider the behavior of the ammonia molecule in an electric field. This enables us to understand the mechanism of masers and to see various applications such as atomic clocks, the GPS system, and the tests of the predictions of relativity on time. The problem we present at the end of this chapter is devoted to a remarkable phenomenon, which is impossible to imagine with classical concepts: neutrino oscillations. It is only after having understood quantum oscillations in two-state systems that one can understand this amazing discovery where an elementary particle in vacuum can change into another one periodically. The experimental results on this question have been awarded many awards, among which the 2002 and 2015 Nobel Prizes for Physics.

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_7

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7 Two-State Systems

7.1 The NH3 Molecule We recall what we found on NH3 . We had made a model of the inversion motion of this molecule by a symmetric double well whose minima correspond to the two classical equilibrium configurations. We saw that the lowest energy level is actually split in two sublevels by the tunnel effect: • E S = E 0 − A0 , corresponds to a symmetric wave function ψ S , and • E A = E 0 + A0 , corresponds to an antisymmetric wave function ψ A . We constructed states corresponding to the “left and right” classical configurations as linear superpositions of these stationary states. And we understood the inversion motion of the molecule between these states. In the transitions between the two energy levels, the molecule emits or absorbs radiation at the Bohr frequency ν = 24 GHz, related to the splitting. This emission can easily be detected and measured since the molecule has an electric dipole moment. Therefore one can measure directly the splitting E S − E A . It is even measured so accurately that it is nonsense to try to calculate it theoretically from first principles.

7.2 “Two-State” System This is where we follow Heisenberg. We study physical processes that involve states of the NH3 molecule which are superpositions of only the two lowest energy states |ψ = a|ψ S  + b|ψ A . In other words, we are interested in physical situations where the state vector |ψ of the molecule remains in a two-dimensional subspace of the Hilbert space, which is infinite-dimensional. This type of situation is called a two-state system or a two-level system, that is, d(E H ) = 2. There are an infinite number of states, but all are linear combinations of two of them. We are going to do quantum mechanics in the case of a two-dimensional Hilbert space where the mathematics are simple. Of course, this is conceivable mathematically. • Can it be achieved physically? • Does it have any physical interest? Yes! It is indeed Heisenberg’s starting point! In N H3 , one knows everything in principle, but it’s an awfully complicated problem. It is a system of 14 particles, 10 electrons, and 4 nuclei, with pairwise Coulomb interactions. The Hamiltonian is

7.2 “Two-State” System

Hˆ =

149

10 4 10 4     pi2 pn2 qe qn + − 2M 2m 4πε |r n − r i | n e 0 n=1 n=1 i=1 i=1

+

4 10 10 4 1 qn qm 1  qe2   + , 2 i=1 j=1 4πε0 |r i − r j | 2 n=1 m=1 4πε0 |r n − r m |

(7.1)

where (r i , pi ) are the positions and momenta of electrons, and (qn , Mn ), (r n , pn ) are the charges, masses, positions, and momenta of the nuclei. Our potential model allowed us to understand qualitatively the tunnel effect, without which the radiowave of ν = 24 GHz would be a mystery, because the energy splitting is comparatively very small. But the general problem of finding these wave functions in 42 variables is much too complicated. And, above all, it is completely uninteresting. The description of the positions of these 14 particles is of no interest. However, we know for sure, owing to the spectral theorem, 1. That there exists a set of energy levels E n and corresponding eigenfunctions ψn . (It is an imaginary catalogue. We cannot write these functions, but we only need to know they exist; we can talk about them even though we do not know them.) 2. And we know the {E n } experimentally. For instance, there exists a whole part of the spectrum due to the inversion motion discussed in Chap. 5. It is a series of levels whose energies are higher and higher and where the splitting increases with the energy until it fades away, as depicted in Fig. 7.1. The following orders of magnitude are of interest. The lowest energy splitting is 2 A0 ≈ 10−4 eV, the splitting between this doublet and the following one is E 1 − E 0 ≈ 0.12 eV, and the splitting of the latter doublet (E 1 ) is 2 A1 ≈ 4 × 10−3 eV. If we denote this imaginary catalogue {|ψn }, we can represent the most general state of the N H3 molecule as

Fig. 7.1 Energy levels and corresponding wave functions in a double well. The abscissa axis is taken, for a given wave function, at the value of the corresponding energy level (Courtesy of Denis Gratias)

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7 Two-State Systems

|ψ =



an |ψn ,

(7.2)

n

or by an infinite column vector whose components an are the probability amplitudes of finding the system in each energy level ⎞ a1 ⎜ a2 ⎟ ⎜ ⎟ ⎜ a3 ⎟ ⎜ ⎟ 2 . ⎟ |ψ : ⎜ ⎜ .. ⎟ with probabilities P(E n ) = |an | ⎜ ⎟ ⎜ an ⎟ ⎝ ⎠ .. . ⎛

by definition of the {an }. This is what Heisenberg does, and it’s more realistic than a description in space by a wave function of the 14 particles because 1. One can measures the values E n , 2. And one can control much more easily the energy of such a complex system than all these positions. It is not deeper than a wave function representation, but it is not less deep for the moment. Now, is it possible to construct physically states that are only superpositions of the two lowest energy states? Yes, of course, because we control the energy. For instance, in a gas at temperature T , we know that the ratio of populations of molecules in energy states E 1 and E 2 is given by Boltzmann’s factor n(E2)/n(E1) = e−(E2−E1)/kT . Therefore, inserting the above values, • At 100 K: N A ≈ N S , n(E1)/n(E0) ≈ 6 × 10−7 which is a small probability, one makes the approximation to consider it to be zero; • If that’s not enough, we can go down to 50 K, p ≈ 3×10−13 , or to 25 K, p ≈ 10−25 (not a single molecule in 22.4 L). In order for the first excited level E 1 to play a role and for p(E1) to be nonnegligible, one must reach temperatures of T ≈ 1300 K. Temperature gives us a cutoff on the finite number of significant components of the state vector. We know this type of situation. Gravity holds us on the ground. If we navigate, we live in a effective two-dimensional world in first approximation. If we climb mountains or fly in a spacecraft, then we must worry about the third dimension.

7.3 Matrix Quantum Mechanics

151

7.3 Matrix Quantum Mechanics Vectors Consider a two-dimensional subspace of E H generated by the set {|ψ S , |ψ A } which forms a basis of the subspace. We can represent an arbitrary state |ψ, by a twocomponent vector, |ψ = a|ψ S  + b|ψ A , with P(E S ) = |a|2 , P(E A ) = |b2 |.

(7.3)

In a matrix representation, this is written as





1 0 a , |ψ A  : , |ψ : , |ψ S  : 0 1 b 1 for instance, |ψ R/L  : √ 2



1 . ±1

Hamiltonian What is the expression of the Hamiltonian? Of course, not the full Hamiltonian of the molecule, but its restriction to the subspace of interest. The basis states are by definition energy eigenstates of Hˆ which is a diagonal matrix, and, in the subspace, it is the 2 × 2 matrix:

0 E0 − A . (7.4) Hˆ = 0 E0 + A The Schrödinger equation is d i |ψ(t) = Hˆ |ψ(t), with |ψ(t) = dt



α(t) . β(t)

(7.5)

This gives us two uncoupled equations whose solutions are |ψ(t) = e−i(E0 t/)



a eiωt/2 b e−iωt/2

,

(7.6)

where we have introduced the Bohr frequency ω = 2 A/. Observables In fact, in this subspace, any linear operator, any observable, is a 2 × 2 matrix. The “restriction” of an observable to the subspace is the first 2 × 2 block of the infinite matrix in the basis of our imaginary catalogue. Now, how are we going to do calculations? In wave mechanics, we have a differential equation and a potential. What should we do here?

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7 Two-State Systems

Simply make models of observables (i.e., 2×2 matrices) instead of making models of potentials as in wave mechanics. Eigenvalues and Eigenvectors of a Hermitian 2 × 2 Matrix Before we continue, it is useful to recall a few formulae on 2 × 2 matrices. Any Hermitian 2 × 2 matrix can be written as

a c eiφ ˆ A= , (7.7) c e−iφ b where a, b, c are real and φ is a phase. The eigenvalues of Aˆ are λ± =

1 a + b ± (a − b)2 + 4c2 , 2

(7.8)

corresponding to eigenvectors |ψ+  =

cos θ sin θ e−iφ



, |ψ−  =

− sin θ cos θ e−iφ

, with tan 2θ =

2c . (7.9) (a − b)

Position Observable The following observable plays a central role here. We can define a “position” observable by the matrix

01 , (7.10) Xˆ = x0 10 where x0 is a length parameter, roughly speaking equal to the positions of the minima of the double well. The observable Xˆ has eigenvalues ±x0 and eigenvectors 1 |ψ±  = √ 2



1 ±1

= |ψ R/L ,

(7.11)

that is, the “classical” configurations of Chap. 5. This is indeed the restriction of the position observable xˆ in the previous sense. It has the same structure as the first 2 × 2 block of Eq. (6.58) in the harmonic oscillator basis. Strictly speaking, it is not the position observable but rather the disposition with respect to the center. Our assumptions forbid us to talk about the position with greater accuracy than the half-width of each well. This is the consequence of the Heisenberg relations. If we prepare a wave function whose dispersion in x is much smaller than Δx = a/2 (i.e., the half width of one of the wells) then one cannot remain in the twodimensional subspace. One must attain the levels E 1 . We need more eigenfunctions in order to localize the particle better as one can understand in Fig. 7.1.

7.3 Matrix Quantum Mechanics

153

Examples We can do a series of simple exercises. Consider the state |ψ(t) =

a eiωt/2 b e−iωt/2

, with ω = 2 A.

(7.12)

The expectation value of X in this state is X  = ψ(t)| Xˆ |ψ(t) = 2x0 Re(a ∗ b e−iωt );

(7.13)

√ that is, for |ψ(0) = |ψ R  with a = b = 1/ 2, X  = |ψ R | Xˆ |ψ(t)|2 = x0 cos ωt,

(7.14)

and a probability Pt (X = +x0 ) = |ψ R |ψ(t)|2 = cos2 (

ωt ), 2

(7.15)

which shows simply the inversion of the NH3 molecule. For the moment, there is nothing really new, but calculations are very simple. We have got rid of the potential and of wave functions. All the interesting physics is in the value of the parameter A which is given by experiment, and in the parameter x0 which defines the size of the molecule. This is a model. In order to improve it, we must increase the size of the Hilbert subspace, namely take more terms into account in the expansion (7.2). Basis of Classical Configurations It is interesting to make a change of basis and in particular to express vectors and operators in the alternative basis of classical equilibrium configurations {|ψ R , |ψ L }. We then express quantum effects in terms of classical situations. The Hamiltonian is not diagonal in this basis, as opposed to X : Hˆ =



E 0 −A −A E 0

,

Xˆ = x0



10 . 01

(7.16)

The off-diagonal terms are transition terms that allow transitions L ↔ R, the inversion of the molecule. If A = 0, the two classical configurations have the same energy. The energy levels are the eigenvalues of H . The calculation is straightforward: we indeed obtain the eigenvalues and eigenvectors we know √ E − = E 0 − A, |ψ S  = (|ψ R  + |ψ L )/ √2, E + = E 0 + A, |ψ A  = (|ψ R  − |ψ L )/ 2.

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7 Two-State Systems

Interference and Measurement At this point, it is instructive to apply the principles. Suppose we start with an energy eigenstate, say |ψ S , 1 |ψ S  = √ (|ψ R  + |ψ L ). 2 √ If we measure X , we can find ±x0 with probabilities 1/2 = (1/ 2)2 . Suppose the measurement has given the result +x0 ; the state after the measurement is 1 (7.17) |ψ R  = √ (|ψ S  + |ψ A ). 2 If we measure X again immediately afterwards, before the oscillation is appreciable, we find +x0 with probability 1; the state after the measurement is |ψ R . Now, suppose that, on this new state |ψ R , we measure not X but the energy E which we are sure was E = E S when we started. One can read in (7.17) that we do not always find E S but the two possibilities E S and E A , each with a probability of 1/2. We see in this case how the measurement has perturbed the system. At the beginning, the state was |ψ S ; at the end it is a mixture of |ψ S  and |ψ A  in interference, for which E = (E S + E A )/2. All of this results from the superposition principle on one hand and the filtering of which a measurement consists. We remark that, consequently, a position measurement implies a minimum energy exchange with the system. Here, on the average, the exchange of energy is equal to A. Needless to say that all of this is quite similar to what we saw with the photon polarization at the end of Chap. 6.

7.4 NH3 in an Electric Field Let us come back to NH3 . In order to make those energy or position measurements, we must control the energy in a more refined way than thermal motion. The temperature has allowed us to be in a subspace; now we want to manufacture individual states of a given energy. We have seen that the NH3 molecule has an electric dipole moment (Fig. 7.2). We can make this electric dipole interact with an electric field. How can we describe this problem? Classically, if a system with an electric dipole moment D is placed in an electric field E it acquires a potential energy W = −E · D. If we assume that E and D are parallel, this becomes W = −E D. (7.18)

7.4 NH3 in an Electric Field

155

Fig. 7.2 The two classical configurations of the molecule NH3 and the associated electric dipole

If we place the NH3 molecule in an electric field E, what is the quantum potential ˆ according to the correspondence energy Wˆ , which is related to the observable D, principle, by ˆ Wˆ = −E D? (7.19) The answer is simple. In the particular configurations of interest here, Dˆ is simply proportional to Xˆ , that is, Dˆ = q0 Xˆ =



0 d0 d0 0



, Wˆ =



0 −d0 E −d0 E 0

(7.20)

where q0 is an effective charge and d0 is an electric dipole moment that is measured experimentally: d0 ≈ 3 × 10−11 (eV)/(V/m) = 5 × 10−30 C.m. In other words, if we measure X and we find ±x0 with some probabilities, a measurement of D will give ±d0 with the same probabilities. And Wˆ is the product of Dˆ times the number E, the value of the electric field. ˆ The only difficulty, here, is to accept that it is a good model for the observable D, ˆ within our assumptions, to be proportional to X . The potential energy observable Wˆ is simply the product of the observable Dˆ by the numerical value of the applied electric field. The only real justification for this choice is that it works very well. From then on, things are quite simple. The Hamiltonian of the molecule in a field is the sum of the free Hamiltonian (7.4) and the potential energy (7.20).

7.4.1 Uniform Constant Field In a uniform constant field, if we set η = d0 E, the Hamiltonian is Hˆ =



−η E0 − A −η E0 + A

.

(7.21)

Finding the energy levels and the corresponding eigenstates amounts to diagonalizing this matrix.

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7 Two-State Systems

Fig. 7.3 Energy levels of a NH3 molecule in an electric field

Turning to (7.8) and (7.9), the eigenvalues and eigenvectors of Hˆ are E− = E0 − E+ = E0 +



A2

+

η2

|ψ−  =

A2 + η 2 |ψ+  =

cos θ sin θ



− sin θ cos θ

,

(7.22) with tan 2θ = η/A.

(7.23)

The variation of these energy levels with the applied electric field E is represented in Fig. 7.3. The validity of the result rests on the condition that E must not be large enough to reach the levels E 1 . Otherwise one would have to take care of higher levels (4-, 6-, etc. level systems). The value of d0 leaves a lot of space.

7.4.2 Weak and Strong Field Regimes It is interesting to consider two limits: the weak field and the strong field limits. The borderline between the two domains, weak and strong fields, is grosso modo E ∼ A/d0 ≈ 1.7 × 106 V/m for NH3 . The quantity d02 /A is called the polarizability of the molecule. It is large for NH3 in its ground state because the splitting A0 is small. We notice that if we consider excited levels E 1 , A1 ≈ 40 A0 , the polarizability is much smaller. The borderline is then around 7 × 107 V/m. For usual fields of 106 V/m, the polarization of the states E 1 is completely frozen, these states do not participate in energy exchanges through an electric field; our starting assumption is again consistent. Weak Field In the weak field regime, such that E d0 /A or θ 1, the levels and eigenstates are to lowest order in E:

7.4 NH3 in an Electric Field

157



d02 E 2 E∓  E0 ∓ A + , 2A d0 E d0 E |ψ A , |ψ+   |ψ A  − |ψ S . |ψ−   |ψ S  + 2A 2A

(7.24) (7.25)

This is understandable. In the absence of a field, the molecule has a symmetric probability and D = 0. The effect of the field is to polarize the molecule, which acquires a mean electric dipole moment proportional to the field D ≈ ±d02 E/A, hence an energy which is quadratic in E. Strong Field For strong fields, E d0 /A or θ  π/4, the effect of the field dominates over the tunnel effect; the molecule is completely polarized. The eigenstates are close to the classical configurations |ψ R/L  with D = ±d0 and the energies are E ± d0 E; the response to the field is linear as for a classical dipole. There is a competition between two effects: • The tunnel effect tends to symmetrize the molecule, which results in a vanishing dipole moment ⇒ D = 0. • The field pulls the molecule toward the classical configurations |ψ R/L , where it has a dipole moment D = ±d0 .

7.4.3 Other Two-State Systems We have made progress on the ammonia molecule. All this could have been done with wave functions, but it would soon have been complicated. Here, the calculation of energy levels amounts to diagonalizing a matrix. We come back below to the ammonia maser. We insist that this is simply an example, because the same mathematics, finite dimensional matrix calculus, applies to many effects. Some of them are exactly finite-dimensional such as • The spin 1/2 of particles such as electrons, protons, quarks, and so on • The polarization of the photon • The physics of “strange” neutral mesons K 0  K¯ 0 , and “beautiful” mesons B0  B¯ 0 , • The universality of weak interactions and the mixing matrix of d, s, b quarks • The quantum oscillations of neutrinos, which we show below. It can also be an approximate model as for lasers, the chemical bond, and nuclear magnetic resonance.

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7 Two-State Systems

7.5 Motion of Ammonia Molecule in an Inhomogeneous Field We come back to the case of the ammonia molecule. With the tools we have developed, we can understand the principle of the maser, which has been a revolution in the physics of radiowaves, in telecommunications, and in astrophysics. We do not prove technical results here. They are intuitive and we come back to them in the similar, but simpler, case of the electron spin in Chap. 12.

7.5.1 Force on the Molecule in an Inhomogeneous Field How can we separate energy eigenstates of the molecule? How can we prepare a sample of NH3 in only one energy state? Consider for definiteness the weak field case. The eigenstates are close to |ψ S  and |ψ A  with energies E∓ = E0 ∓



A2 + d02 E 2  E 0 ∓ A ∓

d02 E 2 . 2A

(7.26)

The last term is simply the potential energy V∓ of the molecule in the field. We see that it is different according to the internal quantum state of the molecule. We prepare a molecular beam. Suppose, first that all molecules are in the state |ψ S . If these molecules which are “big” classical objects cross a region where there is an inhomogeneous field, i.e. ∇E 2  = 0, their energy depends on the point where they are, therefore a force will act on them F− = −∇ V− = +

d02 ∇E 2 . 2A

(7.27)

Similarly, if they are all in the state |ψ A  the force acting on them is F+ = −∇ V+ = −

d02 ∇E 2 . 2A

(7.28)

The force is different according to the internal state of the molecule. The two forces actually have opposite signs. Therefore, the molecular beam will follow a different path according to whether the internal state of the molecules is |ψ S  or |ψ A . What is quantum mechanical is the internal state of the molecule, not the motion of its center of gravity. What happens if the molecule is in a superposition of the states |ψ S  and |ψ A ? This problem is slightly more complicated. The state of a molecule must be represented by a vector such as we have done above, but the components of this vector depend on the position R of the center of gravity of the molecule,

7.5 Motion of Ammonia Molecule in an Inhomogeneous Field

159

Fig. 7.4 Stabilization of the beam |ψ+  and divergence of the beam |ψ−  in an electric quadrupole field (E 2 ∝ y 2 + z 2 )

|ψ =

Ψ1 (R, t) . Ψ2 (R, t)

The probabilistic interpretation is that |Ψ1 |2 is the probability density of being at point R in the internal state |ψ S , and similarly |Ψ2 |2 is the probability density of being at point R in the state |ψ A . This can be proven. A similar problem arises in the simpler, case of the electron with its spin in the Stern–Gerlach experiment, Chap. 12. What is quite interesting is that the respective expectation values of the positions of the wave packets Ψ1 (R, t) and Ψ2 (R, t) each evolve according to Newton’s laws with, respectively, the potentials (7.27) and (7.28). All happens as if different forces acted on these components. Therefore an inhomogeneous field allows to perform the following operations (Fig. 7.4). 1. One observes a spatial separation of the molecules according to their internal state. This device is a filter that selects the states |ψ S  and |ψ A . Notice also that in a well chosen field one can select any real linear superposition of |ψ S  and |ψ A . 2. Here we face an incredible phenomenon. There are only two quantum trajectories whereas classically, if the electric dipole moments were oriented at random there should be a continuous set of impacts on a screen. The first time physicists saw such a behavior was in the experiment of Stern and Gerlach (Chap. 12). They thought it was an experimental proof of the quantization of trajectories advocated by Bohr. 3. But if we can select the states |ψ S  and |ψ A  in space, that means we measure their energies. If a molecule arrives above, it has an internal quantum energy E S ; if a molecule arrives below, it has energy E A . The respective numbers of them give the probabilities that in the incoming beam the molecules are in the states |ψ S  and |ψ A ; we obtain information on the initial state. 4. This apparatus is a concrete example of a quantum mechanical measuring apparatus. It transfers internal quantum degrees of freedom into classical space–time properties.

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7 Two-State Systems

5. It is also a device to prepare the molecules in the states |ψ S  or |ψ A , or in linear superpositions of them (it is sufficient to vary the value of the field E in the device). Here we can at last speak of the measurement question in quantum mechanics. We see that in order to make the separation, one must have, for a given velocity, a minimal length, therefore a minimal time. A measurement is never instantaneous or pointlike. A measurement always has a finite extension in space and time. A little calculation (again with spin 1/2) shows that the condition for the measurement to be possible is universal. One must have the inequality ΔE Δt > ,

(7.29)

where ΔE is the transverse kinetic energy acquired by the molecules and Δt the time they spend in the inhomogeneous field zone. This is called the “time-energy uncertainty relation” by analogy with the previous ones. Actually, there is a big difference because time t is not an intrinsic observable of a system (even though it can be measured).

7.5.2 Population Inversion Actually, in the case of NH3 , if one uses a molecular beam and an electric quadrupole field, the beam |ψ A  is stable and can be focused, whereas the beam |ψ S  is unstable and gets dispersed. What is interesting is that by this technique (a molecular beam, a diaphragm, an inhomogeneous field), one can perform what is called a population inversion. One can select all molecules in the state Ψ A . One breaks the (A) − (S) thermal equilibrium there was in the initial beam. This is only one example of a technique of population inversion. There are many others. We notice that in order for this population inversion to occur, it is not necessary to align the axis of a molecule along the field. Whatever the value of E, the ψ A beam is focused and ψ S is dispersed.

7.6 Reaction to an Oscillating Field, the Maser In order to make a maser, we will force the molecules in the state |ψ A  to give back their energy 2 A by making a transition to the state |ψ S . We must force them because spontaneously, these molecules do fall back in the state |ψ S , but they do it very slowly. The mean time to do this transition is of the order of one month, which is much too long for our purpose. In quantum physics, a system can absorb a photon of energy hν and reach an excited state; it can fall back in the initial level by emitting a photon spontaneously.

7.6 Reaction to an Oscillating Field, the Maser

161

However, there exists a third mechanism which was understood by Einstein as soon as 1917, called stimulated emission. If the excited system is placed in an electromagnetic field properly tuned to the Bohr frequency, it can undergo a transition to the lower state very rapidly. This is done by a resonance mechanism. Technically, this exercise is not much more difficult than the previous one, but it is new. We describe the position of the problem and the result. Again the calculation itself is done later on with spin 1/2 and magnetic resonance. We place the molecule in an oscillating field E = E0 cos ωt, and we set η = d0 E0 . The Hamiltonian is Hˆ =



E 0 − A −η cos ωt −η cos ωt E 0 + A

.

(7.30)

We see the difference with previous problems. The Hamiltonian H now depends explicitly on time. The system is not isolated, one cannot speak of stationary states. We must solve the Schrödinger equation. If we write |ψ(t) = a(t)|ψ S  + b(t)|ψ A , we must solve the equation i(d/dt)|ψ(t) = Hˆ |ψ(t) in order to determine the evolution of the system. If we write the state vector of a molecule as

a(t) |ψ(t) = , (7.31) b(t) the Schrödinger equation is a first-order differential system: ia˙ = (E 0 − A)a − ηb cos ωt ib˙ = (E 0 + A)b − ηa cos ωt.

(7.32) (7.33)

Setting a(t) = e−i(E0 −A)t/ α(t) and b(t) = e−i(E0 +A)t/ β(t), we obtain:   2i α˙ = − ω1 β ei(ω−ω0 )t + e−i(ω+ω0 )t ,   2i β˙ = − ω1 α e−i(ω−ω0 )t + ei(ω+ω0 )t .

(7.34) (7.35)

This set of coupled equations involves three frequencies: ω, ω0 =

2A η d0 E0 , and ω1 = = .   

(7.36)

Starting with state |ψ A  at t = 0, a(0) = 0 and b(0) = 1, we want to calculate the probability |a(t)|2 of finding the system in state |ψ S  at time t.

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7 Two-State Systems

Physically, this differential system corresponds to forced oscillations with a resonance phenomenon at ω = ω0 .1 There is no analytic solution, however, one obtains a good approximation in the vicinity of the resonance ω ∼ ω0 , if we neglect terms that oscillate rapidly in time e±i(ω+ω0 )t . This leads to an exactly soluble problem, for which we can give the solution. Near the resonance, the transition probability PA→S (t) that at time t the molecules undergo a transition to the state |ψ S  under the influence of the oscillating field, and that they release their energy 2 A = E A − E S is given by ω12 sin2 PA→S (t)  (ω − ω0 )2 + ω12



(ω − ω0

)2

+

ω12

t . 2

(7.37)

This formula is due to Rabi. As can be seen in Figure 7.5a, the probability PA→S (t) oscillates in time between 0 and a maximum value Pmax given by Pmax =

ω12 . (ω − ω0 )2 + ω12

When the frequency ω of the applied field is varied (Fig. 7.5b), the maximum probability Pmax has a characteristic resonant behavior, with a maximum equal to 1 at the resonance, that is, for ω = ω0 . The width at half maximum of the resonance curve is ω1 . If the frequency of the external field is tuned in the vicinity of the resonance, |ω − ω0 | ω1 , practically all the molecules will release their energy 2 A at the time T = π/ω1 . This energy emission occurs in the form of an electromagnetic radiation of frequency ν = ω0 /2π = 24 GHz. This is called stimulated emission. The smaller ω1 , the narrower the resonance curve of Fig. 7.5b is, and the greater the time to obtain that emission.

Fig. 7.5 Rabi oscillations: a Probability of finding the molecule in the state |ψ S  as a function of time. b Resonance curve showing the maximum transition probability as a function of the external field frequency ω

there are two resonances at ω = ω0 and ω = −ω0 , but the two values are equivalent for what concerns us here.

1 Actually,

7.6 Reaction to an Oscillating Field, the Maser

163

Therefore, if spontaneously the molecules decay in ≈1 month, with this phenomenon one can force them to do it quickly, and to beat the natural time constant of the process. This happens whatever the intensity of the exciting field, provided it is properly tuned.

7.7 Principle and Applications of the Maser In practice, a NH3 maser (microwave amplification by stimulated emission of radiation), which was invented by Townes in 1951, works in the following way, represented in Fig. 7.6. One starts with a molecular beam of velocity v coming from an oven at a temperature of 100 K. One then separates the molecules in the state |ψ A  by an electric quadrupole field. The beam then enters a high-frequency cavity where there is a field E0 cos ωt and whose length L is adjusted2 so that L/v = T = (2n +1)π/ω1 . The outgoing molecules are in the state |ψ S  and they have released their energy 2 A in the cavity in the form of an electromagnetic radiation of frequency ω0 . In the setup made by Townes in 1951, a beam of 1014 molecules s−1 gave a power of 10−9 W on a single frequency 24 GHz with a width of 3000 Hz, a quality factor of Q = 107 . This maser effect, which is based on two elements, population inversion (which breaks the thermodynamic equilibrium) and stimulated emission which generates the quick and coherent transition, was discovered by Townes in 1951 (1964 Nobel prize). It has numerous applications. Lasers (light) work in a different frequency range, with a more sophisticated technique of population inversion, optical pumping. They are treated as three- or four-state systems. There is no basic difference from what we have seen otherwise. There are essentially three types of applications of such devices.

Fig. 7.6 Sketch of a NH3 maser device

2 It

is not necessary that L be exactly adjusted to the correct value; the transition probability is appreciable if one doesn’t have the bad luck of falling on the unfavorable values T = 2nπ/ω1 . In practice a feedback device adjusts the length of the cavity so that the signal is maximum.

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7 Two-State Systems

7.7.1 Amplifiers One can amplify in a selective way, and without any background noise, a very weak signal. Hence there are very important applications in radioastronomy in order to study the interstellar medium, as we show in Chap. 13. Initially, Townes used a molecular beam of 1014 molecules per second. Nowadays one uses solid-state masers, such as a crystal of ruby (a crystal of Al2 O3 with Cr+3 ions of a concentration of ∼0.05 %). This allows gains of 36 dB. Ruby masers were used in 1965 by A. Penzias and R.W. Wilson when they discovered the cosmic background radiation at 3 K, one of the major observational proofs in favor of the big bang. The mechanism of the amplification is simple if one visualizes it in terms of photons. The population inversion is achieved on a macroscopic number of atoms, such as in a crystal of ruby. As shown in Fig. 7.7, a photon coming from the interstellar medium will induce the transition of a first atom by stimulated emission. This results in two tuned photons both of which can in turn generate stimulated emission on other atoms, leading to four photons, and so on. The chain reaction yields the amplification. The device presented in Fig. 7.6 is not an amplifier but an emitter or an oscillator, because the outgoing intensity is independent of the incoming intensity. In order to see the system act as an amplifier, one must calculate its response to an incoherent signal spread out in frequency. Our calculation only concerns a monochromatic coherent field.

7.7.2 Oscillators In Fig. 7.6, we actually see an oscillator. A field as small as one wishes can be selfproduced in the cavity; one evacuates the electromagnetic wave of frequency 24 GHz, which results in a very monochromatic output wave.

Fig. 7.7 Chain reaction of a tuned photon on a set of excited atoms

7.7 Principle and Applications of the Maser

165

7.7.3 Atomic Clocks and the GPS As oscillators, masers have allowed the construction of atomic clocks, which are the present standard time-keeping devices. Such devices use, for instance, a jet of cesium atoms (isotope 133 Cs). The ground state of cesium is split by the hyperfine interaction, which we describe later on. The physical origin of this effect is magnetic instead of being electric. It results from the magnetic interaction of the nuclear spin and the spin of the valence electron, and a splitting in two levels |g1  and |g2 , of energies E 1 and E 2 . Since 1967, the definition of the hertz is given by the fact that the 133 Cs hyperfine frequency ν12 = (E 2 − E 1 )/ h is equal to 9,192,631,770 Hz. It is a considerable improvement compared to astronomical definitions. In order to transform the above device into an atomic clock, we prepare a jet of cesium atoms in the state |g1 . These atoms cross a cavity inside which one injects an electromagnetic wave of frequency ν, and one adjusts ν in order to maximize the number of outgoing atoms in the state |g2 . The frequency ν is thus locked at ν12 . One can measure a time interval by simply counting the number of oscillations during that time interval. The principle is shown in Fig. 7.8. Atomic clocks were first developed in the 1950s by Zacharias at the National Institute of Standards and Technology (NIST). The clock NIST-7 together with ten others in the world thus serves as a “time keeper.” Present clocks have a relative accuracy of 10−15 , so that the time standard is the most accurate of all standards. This explains that, in 1983, the definition of the meter was changed by fixing the value of the velocity of light to an integer value: c = 299,792,458 ms−1 . Such accuracy is essential both in applied physics and in a variety of technical devices, such as positioning and navigation, be it terrestrial with the GPS system (Fig. 7.9), or on satellites. They are commonly used to guide ships, they

Fig. 7.8 Sketch of an atomic clock (Courtesy of Patrick Bouchareine)

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7 Two-State Systems

Fig. 7.9 Satellite of the GPS Navstar system. There are 24 of them orbiting around the Earth

equip airplanes. In 1990, they appeared in automobiles where they are commonly used by now: almost all common automobiles have a GPS guiding system for daily purposes. They exist in cell phones, in mountain climbing safety devices, in elaborate wrist watches etc. Perhaps more amazing is their use in space navigation. The recent mission of the Cassini spacecraft, launched in 2000, saw, on January 14 2005, the Huygens probe land on the planet Titan. Titan is a satellite of Saturn that is similar to the Earth because of its atmosphere, though it is much colder (in some sense it is an Earth in a refrigerator). This has yielded an impressive number of results and will continue to do so. In the final phase, the probe had to guide itself by its own means and land properly (the time of propagation of a signal to the earth is one hour). (See http:// www.esa.int/SPECIALS/Cassini-Huygens/index.html.)

7.7.4 Tests of Relativity As soon as atomic clocks appeared, people thought of directly verifying the predictions of relativity on time, in particular the twin paradox. In this “paradox” the twin who has traveled comes back younger. His proper time is shorter than the time of the other twin who stayed on Earth. His clock must therefore be late compared to the clocks that remained on Earth, t=

t0 1−

v2 c2

, Δt = t − t0  t0

v2 , 2c2

(7.38)

7.7 Principle and Applications of the Maser

167

where t is the time measured by the twin who stayed on Earth, and t0 the time of the twin who traveled. In 1975, Alley3 sent 4 atomic cesium clocks for a 15-h flight on an airplane and compared them with 11 clocks that stayed on Earth. Those clocks should have been late. Not at all; they they had gained 47 × 10−9 s! In fact, there is an effect of the opposite sign due to gravitation and predicted by general relativity:

Δt t

GR

ΔΦ = 2 , c



Δt t

=− SR

v2 , 2c2

where Φ is the gravitational potential. In the conditions of flight of Alley’s clocks, the result is the sum of



Δt Δt −9 = 53 × 10 s, = −6 × 10−9 s. t GR t SR The predictions of relativity are verified here with an accuracy of 1 %. These effects were measured in 1979 by R.F.C. Vessot and collaborators.4 A hydrogen maser was sent to an altitude of 10,000 km by a Scout rocket, and the variation in time of its frequency was made as the gravitational potential increased (algebraically). There are many corrections, in particular due to the Doppler effect of the spacecraft and to the Earth’s rotation. It was possible to test the predictions of general relativity on the variation of the pace of a clock as a function of the gravitational field with a relative accuracy of 7 × 10−5 . This was done by comparison with atomic clocks, or masers, on Earth. Up to now, it has been one of the best verifications of general relativity. The recording of the beats between the embarked maser and a test maser on Earth is shown in Fig. 7.10. (These are actually beats between signals that are first recorded and then treated in order to take into account all physical corrections.) A simple calculation shows that with this accuracy, one must specify the exact altitude of a clock up to one meter or so, in order to know one’s time. Time is longer at the top of a conference hall than at the bottom. Students know this very well.

3 C.

Alley, “Proper Time Experiments in Gravitational Fields with Atomic Clocks, Aircraft, and Laser Light Pulses,” in Quantum Optics, Experimental Gravity, and Measurement Theory, eds. Pierre Meystre and Marlan O. Scully, Proceedings Conf. Bad Windsheim 1981, 1983 Plenum Press New York, ISBN 0-306-41354-X, pg 363427. 4 R.F.C. Vessot, M.W. Levine, E.M. Mattison, E.L. Blomberg, T.E. Hoffman, G.U. Nystrom, B.F. Farrel, R. Decher, P.B. Eby, C.R. Baugher, J.W. Watts, D.L. Teuber, and F.D. Wills, “Test of Relativistic Gravitation with a Space-Borne Hydrogen Maser”, Phys. Rev. Lett. 45, 2081, (1980).

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7 Two-State Systems

(a)

(b)

(c)

(d)

(e)

Fig. 7.10 Beats between a maser onboard the spacecraft launched by a Scout rocket and a maser on Earth at various instants in GMT. a Signal of the dipole antenna; the pointer shows the delicate moment when the spacecraft separated from the rocket (it was important that the maser onboard had not been damaged by vibrations during takeoff). During this first phase, the special relativity effect due to the velocity is dominant. b Time interval of “zero beat” during ascent when the velocity effect and the gravitational effect, of opposite signs, cancel each other. c Beat at the apogee, entirely due to the gravitational effect of general relativity. Its frequency is 0.9 Hz. d Zero beat at descent. e End of the experiment. The spacecraft enters the atmosphere and the maser onboard ceases to work (Courtesy of R.F.C. Vessot)

7.8 Exercises

169

7.8 Exercises 1. Linear three-atom molecule We consider the states of an electron in a linear three-atom molecule (such as N3 or C3 ) with equally spaced atoms L , C, R at a distance d from one another. ˆ corresponding to Let |ψ L , |ψC  and |ψ R  be the eigenstates of an observable B, an electron localized respectively in the vicinity of the atoms L, C and R: ˆ L  = −d|ψ L ; B|ψ

ˆ C  = 0; B|ψ

ˆ R  = +d|ψ R . B|ψ

In the basis {|ψ L , |ψC , |ψ R }, the Hamiltonian of the system is represented by the matrix: ⎛ ⎞ E 0 −a 0 Hˆ = ⎝ −a E 0 −a ⎠ a > 0. 0 −a E 0 a. Calculate the energy levels and eigenstates of Hˆ . b. Consider the ground state; what are the probabilities to find the electron in the vicinity of L, C and R? c. Suppose the electron is in the state |ψ L , and we measure its energy. What values can we find, with what probabilities? Calculate E and ΔE in this state. 2. Crystallized Violet and Malachite Green The active principle of the dye 42555 (called “Crystallized Violet”) is the organic monovalent cation C[C6 H4 N(CH3 )2 ]+ 3 . The skeleton of this ion is made of three identical branches (Fig. 7.11). The electronic deficit responsible for the positive charge can be taken from either of these three branches. One can treat the electronic state of this ion as a three-state system. The Hamiltonian Hˆ is not diagonal in the basis {|1, |2, |3} (which we assume orthonormal) because of tunnelling between these classical configurations.

Fig. 7.11 The three possible configurations of the molecule

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7 Two-State Systems

a. We work in the basis {|1, |2, |3} corresponding to “classical configurations”. We choose the origin of energies in order to have 1| Hˆ |1 = 2| Hˆ |2 = 3| Hˆ |3 = 0. We set 1| Hˆ |2 = 2| Hˆ |3 = 3| Hˆ |1 = −A where A is real positive (A > 0). Write the matrix Hˆ in this basis. Comparing with the case of the ammonia molecule N H3 , justify briefly the choice of this matrix. √ √ b. Consider the states |φ1  = (|1 + |2 + |3)/ 3 and |φ2  = (|2 − |3)/ 2. Calculate the expectation value E and the dispersion ΔE in each of these states. Interpret the result. c. Determine the energy levels of the system. Give a corresponding orthonormal eigenbasis. Is this basis unique? d. The value of A is A ≈ 0.75 eV. Why is this ion violet? We recall that the colors of the spectrum of natural light are, for increasing energies (E = hc/λ), red (from ≈1.65 to 2.0 eV); orange (from ≈ 2.0 to 2.1 eV); yellow (from ≈2.1 to 2.3 eV); green (from ≈2.3 to 2.55 eV); blue (from ≈2.55 to 2.65 eV); violet (from ≈ 2.65 to 3.1 eV). The main couples of “complementary colors” which produce white light when they are associated, are yellow-violet, red-green and blue-orange. e. One replaces the N(CH3 )2 group of the upper branch by a hydrogen atom. We assume that the sole effect of this substitution is to increase 1| Hˆ |1 by an amount Δ > 0, and that it leaves the other matrix elements of Hˆ unchanged. (i) Show that A is still an eigenvalue of the Hamiltonian. What are the other energy levels of this new system? (ii) How do they behave in the limits Δ A and Δ A? f. This modified ion (coloring 42,000 “malachite green”) absorbs light of two wavelengths: 620 and 450 nm. Calculate Δ and comment on the agreement between theory and experiment. One can use hc ≈ 1240 eV.nm.

7.9 Problem: Neutrino Oscillations We consider here a quantum oscillation effect that is completely contrary to common intuition. It is an oscillation between two or more types of pointlike elementary particles. The effect consists of the fact that if we prepare a particle in the vacuum, it can transform spontaneously and periodically into one or several types of different elementary particles. In other words, we face a phenomenon of “successive periodic hermaphroditism” between elementary particles.5

5 Successive

hermaphroditism exists in zoology, for instance, in several fish species.

7.9 Problem: Neutrino Oscillations

171

Lepton Families In β decay or, more generally, in Weak interactions, the electron is always associated with a neutral particle, the neutrino νe . There exists in nature another particle, the μ lepton, or muon, whose physical properties seem completely analogous to those of the electron, except for its mass m μ  200 m e . The muon has the same Weak interactions as the electron, but it is associated to a different neutrino, the νμ . A neutrino beam produced in an accelerator can interact with a neutron (n) in a nucleus and give rise to the reactions νe + n → p + e

and

νμ + n → p + μ,

(7.39)

whereas the reactions νe + n → p + μ or νμ + n → p + e are never observed. The reactions (7.39) are used in practice in order to detect neutrinos. Similarly, a π − meson can decay via the modes π − → μ + ν¯μ (dominant mode) and π − → e + ν¯e ,

(7.40)

whereas π − → μ + ν¯e or π − → e + ν¯μ are never observed. This is how one can produce neutrinos abundantly (it is easy to produce π mesons). In (7.40) we have introduced the antiparticles ν¯μ et ν¯e . There is a (quasi) strict symmetry between particles and their antiparticles, so that, in the same way as the electron is associated with the neutrino νe , the antielectron, or positron, e+ is associated with the antineutrino ν¯e . One observes the “charge-conjugate” reactions of (7.39) and (7.40) ν¯e + p → n + e+ ,

ν¯μ + p → n + μ+

and π + → μ+ + νμ .

(7.41)

In all what follows, what we will say about neutrinos holds symmetrically for antineutrinos. The experimental discovery of the electron antineutrino is due to Cowan and Reines in 1956. They operated near the Savannah River nuclear reactor, and they observed the reaction ν¯e + p → n + e+ , and not ν¯e + n → p + e.

(7.42)

The ν¯e antineutrinos came from the many beta decays of the type n → p + e + ν¯ in the reactor. (The second reaction in (7.42) does not occur because the electron is associated with its neutrino and not with the antineutrino ν¯e which is the partner of the antielectron e+ .) In 1975, a third lepton, the τ , was discovered. It is much more massive, m μ  3500 m e , it is associated with its own neutrino ντ , and it obeys the same physical laws as the two lighter leptons, except for mass effects. Since the 1990s, the experimental measurements at the LEP colliding ring in CERN have shown that these three neutrinos νe , νμ , ντ (and their antiparticles) are the only ones of their kinds (at least for masses less that 100 GeV/c2 ).

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7 Two-State Systems

For a long time, physicists believed that neutrinos were zero-mass particles, as is the photon. In any case, their masses (multiplied by c2 ) are considerably smaller than the energies involved in experiments where they are observed. Therefore, many experimental limits on these masses are consistent with zero. However, both theoretical and cosmological arguments suggested that this might not be the case. The proof that neutrino masses are not all zero is a great discovery of the 1990s. For this discovery, the 2002 Nobel prize was awarded to Raymond Davis Jr. a pioneer of this physics, and to Masatoshi Koshiba who led experiments performed in Japan with detectors particularly well adapted to that kind of physics. The Japanese experiments have gathered an impressive amount of results. The 2015 Nobel prize for physics was awarded to Takaaki Kajita (SuperKamiokande) and Arthur B. McDonald (SNO) “for the discovery of neutrino oscillations, which shows that neutrinos have mass”. In the present study, we show how the mass differences of neutrinos can be measured by a quantum oscillation effect. The idea is that the “flavor” neutrinos νe , νμ and ντ , which are produced or detected experimentally are not eigenstates of the mass, but rather linear combinations of mass eigenstates ν1 , ν2 , ν3 , with masses m1, m2, m3. The neutrinos observed on earth have various origins. They can be produced in accelerators, in nuclear reactors, and also in the atmosphere by cosmic rays, or in thermonuclear reaction inside stars, in particular the core of the sun, and in supernovae explosions.

7.9.1 Mechanism of the Oscillations; Reactor Neutrinos In this first part, we consider oscillations between two types of neutrinos, the νe and the νμ . This simple case will allow us to understand the underlying physics of the general case. We will analyze the data obtained with nuclear reactors. The average energy of the (anti-)neutrinos produced in reactors is E = 4 MeV, with a dispersion of the same order. In all what follows, we will assume that if m is the neutrino mass and p and E its momentum and energy, the mass is so small that the energy of a neutrino of mass m and momentum p is E=



p 2 c2 + m 2 c4  pc +

m 2 c4 , 2 pc

(7.43)

and that the neutrino propagates to very good approximation at the velocity of light c. Let Hˆ be the Hamiltonian of a free neutrino of momentum p, which we assume to be well defined. We note |ν1  and |ν2  the two eigenstates of Hˆ :

7.9 Problem: Neutrino Oscillations

Hˆ |ν j  = E j |ν j ,

173

E j = pc +

m 2j c4 2 pc

,

j = 1, 2.

m 1 and m 2 are the respective masses of the states |ν1  and |ν2 , and we assume m 1 = m 2 . The oscillations of freely propagating neutrinos come from the following quantum effect. If the physical states of the neutrinos which are produced (reactions (7.40)) or detected (reactions (7.39)) are not |ν1  and |ν2 , but linear combinations of these: |νe  = |ν1  cos θ + |ν2  sin θ,

|νμ  = −|ν1  sin θ + |ν2  cos θ

(7.44)

where θ is a mixing angle to be determined, these linear combination of energy eigenstates oscillate in time and this leads to measurable effects. 7.9.1 At time t = 0, one produces a neutrino of momentum p in the state |νe . Calculate the state |ν(t) at time t in terms of |ν1  and |ν2 . 7.9.2 What is the probability Pe for this neutrino to be detected in the state |νe  at time t? The result will be expressed in terms of the mixing angle θ and of the oscillation length L L=

4π p , |Δm 2 | c2

Δm 2 = m 21 − m 22 .

(7.45)

7.9.3 Calculate the oscillation length L for an energy E  pc = 4 MeV and a mass difference Δm 2 c4 = 10−4 eV2 . 7.9.4 One measures the neutrino fluxes with a detector located at a distance  from the production area. Express the probability Pe as a function of the distance  = ct. 7.9.5 The mass of the muon satisfies m μ c2 = 106 MeV. Conclude that in such an experiment one cannot detect muon neutrinos νμ with the reaction (7.39). We recall that m p c2 = 938.27 MeV and m n c2 = 939.57 MeV. 7.9.6 The detectors measure neutrino fluxes with an accuracy of ∼10%. (a) Assuming Δm 2 c4 = 10−4 eV2 , determine the minimal distance min where to put a detector in order to detect an oscillation effect. For this calculation, assume the mixing in (7.44) is maximum, i.e. θ = π/4. (b) How does min change if the mixing is not maximum? 7.9.7 Several experiments on neutrinos produced by nuclear energy plants have been performed in Chooz and in Bugey in France. The most recent data comes from the KamLAND collaboration, in Japan. The results are given on Fig. 7.12. (a) Explain the results of Fig. 7.12, except that of KamLAND.

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7 Two-State Systems

Fig. 7.12 Ratio between the numbers of observed electron neutrinos and those expected in the absence of oscillations as a function of the distance  to the reactor

(b) The KamLAND experiment, which started operating in 2002,6 consisted in measuring the neutrinos coming from all the (numerous) reactors in Japan and neighboring countries, which amounts to taking an average distance of  = 180 km. Putting together that data and the results of numerous experiments performed on solar neutrinos, the physicists of Kamland come to the following results: |Δm 2 | c4 = 7.1 (± 0.4) × 10−5 eV2 ,

tan2 θ = 0.45 (± 0.02).

(7.46)

Show that these values are consistent with the result Pe = 0.61 (± 0.10) of Fig. 7.12.

7.9.2 Oscillations of Three Species; Atmospheric Neutrinos We now consider the general formalism with three neutrino species. We denote |να , α = e, μ, τ the “flavor” neutrinos and |νi , i = 1, 2, 3 the mass eigenstates. These two bases are related to one another by the Maki–Nagawaka–Sakata (MNS) matrix Uˆ , ⎛ ⎞ 3 Ue1 Ue2 Ue3  (7.47) Uαi |νi , Uˆ = ⎝ Uμ1 Uμ2 Uμ3 ⎠ |να  = Uτ 1 Uτ 2 Uτ 3 i=1  ∗ This matrix is unitary ( i Uβi Uαi = δαβ ) and it can be written as:

6 See

for instance Atsuto Suzuki, Proc. of Nobel Symposium 129, Physica Scripta T121 (2005).

7.9 Problem: Neutrino Oscillations

175

⎛

⎞⎛ ⎞⎛ ⎞ 1 0 0 c13 0 s13 e−iδ c12 s12 0 1 0 ⎠ ⎝ −s12 c12 0 ⎠ Uˆ = ⎝ 0 c23 s23 ⎠ ⎝ 0 iδ 0 −s23 c23 0 0 1 −s13 e 0 c13 where ci j = cos θi j and si j = sin θi j . The complete experimental solution of the problem would consist in measuring the three mixing angles θ12 , θ23 , θ13 , the phase δ, and the three masses m 1 , m 2 , m 3 . We consider situations such that (7.43) is still valid. 7.9.1 At time t = 0 a neutrino is produced with momentum p in the state |ν(0) = |να . Express, in terms of the matrix elements Uαi , its state at a later time t. Write the probability Pα→β (t) to observe a neutrino of flavor β at time t. 7.9.2 We define the oscillation lengths at an energy E  pc by: Li j =

4π p , |Δm i2j |c2

Δm i2j = m i2 − m 2j .

(7.48)

Notice that there are only two independent oscillation lengths since Δm 212 + Δm 223 + Δm 231 = 0. For neutrinos of energy E = 4 GeV, calculate the oscillation lengths L 12 and L 23 . We will choose for |Δm 212 | the result given in (7.46), and we will choose |Δm 223 | c4 = 2.5 × 10−3 eV2 , a value which will be justified later on. 7.9.3 The neutrino counters have an accuracy of the order of 10 % and the energy is E = 4 GeV. Above which distances 12 and 23 of the production point of the neutrinos can one hope to detect oscillations coming from the superpositions 1 ↔ 2 and 2 ↔ 3? 7.9.4 The Super-Kamiokande experiment, performed in 1998, consists in detecting “atmospheric” neutrinos. Such neutrinos are produced in the collision of high energy cosmic rays with nuclei in the atmosphere at high altitudes. In a series of reactions, π ± mesons are produced abundantly, and they decay through the chain: π − → μ− + ν¯μ

followed by

μ− → e− + ν¯e + νμ ,

(7.49)

and an analogous chain for π + mesons. The neutrino fluxes are detected in an underground detector by the reactions (7.39) and (7.41). To simplify things, we assume that all muons decay before reaching the surface of the Earth. Deduce that, in the absence of neutrino oscillations, the expected ratio between electron and muon neutrinos Rμ/e = would be equal to 2.

N (νμ ) + N (ν¯μ ) N (νe ) + N (ν¯e )

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7 Two-State Systems

7.9.5 The corrections to the ratio Rμ/e due to the fact that part of the muons reach the ground can be calculated accurately. Once this correction is made, one finds, by comparing the measured and calculated values for Rμ/e (Rμ/e )measured = 0.64 (± 0.05). (Rμ/e )calculated In order to explain this relative decrease of the number of νμ ’s, one can think of oscillations of the types νμ  νe and νμ  ντ . The Super-Kamiokande experiment consists in varying the time of flight of the neutrinos by measuring selectively the direction where they come from, as indicated on Fig. 7.13. The neutrinos coming from above (cos α ∼ 1) have traveled a distance equal to the atmospheric height plus the depth of the detector, while those coming from the bottom (cos α ∼ −1) have crossed the diameter of the Earth (13,400 km). Given the weakness of the interaction of neutrinos with matter, one can consider that the neutrinos propagate freely on a measurable distance between a few tens of km and 13,400 km. The neutrino energies are typically 4 GeV in this experiment. Can one observe a νe  νμ oscillation of the type studied in the first part? 7.9.6 The angular distributions of the νe and the νμ are represented on Fig. 7.13, together with the distributions one would observe in the absence of oscillations. Explain why this data is compatible with the fact that one observes a νμ  ντ oscillation, no νe  ντ oscillation, and no νe  νμ oscillation. 7.9.7 In view of the above results, we assume that there is only a two-neutrino oscillation phenomenon: νμ  ντ in such an observation. We therefore use the same formalism as in the first part, except that we change the names of particles. By comparing the muon neutrino flux coming from above and from below, give an estimate of the mixing angle θ23 . In order to take into account the large energy dispersion of cosmic rays, and therefore of atmospheric neutrinos, we replace the oscillating factor sin2 (π/L 23 ) by its mean value 1/2 if  L 23 . The complete results published by the Super-Kamiokande experiment are |Δm 223 | c4 = 2.5 × 10−3 eV2 , θ23 = π/4, θ13 = 0. Do they agree with the above considerations?

7.9.3 Solution Mechanism of the Oscillations: Reactor Neutrinos 7.3.1 Initially, the neutrino state is |ν(0) = |νe  = |ν1  cos θ+|ν2  sin θ. Therefore, we have at time t

7.9 Problem: Neutrino Oscillations

177

Fig. 7.13 Left production of atmospheric neutrinos in collisions of cosmic rays with terrestrial atmospheric nuclei. The underground detector measures the flux of electron and muon neutrinos as a function of the zenithal angle α. Right number of atmospheric neutrinos detected in the SuperKamiokande experiment as a function of the zenithal angle (this picture is drawn after K. Tanyaka, XXII Physics in Collisions Conference, Stanford 2002)

|ν(t) = |ν1  cos θ e−iE1 t/ + |ν2  sin θ e−iE2 t/ . 7.3.2 The probability to find this neutrino in the state |νe  at time t is 2  Pe (t) = |νe |ν(t)|2 = cos2 θ e−iE1 t/ + sin2 θ e−iE2 t/  , which gives, after a simple calculation: Pe (t) = 1 − sin (2θ) sin 2

2

(E 1 − E 2 )t 2

.

We have E 1 − E 2 = (m 21 − m 22 )c4 /(2 pc). Defining the oscillation length by L = 4π p/(|Δm 2 | c2 ), we obtain Pe (t) = 1 − sin (2θ) sin 2

2

πct L

.

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7 Two-State Systems

7.3.3 For an energy E = pc = 4 MeV and a mass difference Δm 2 c4 = 10−4 eV2 , we obtain an oscillation length L = 100 km. 7.3.4 The time of flight is t = /c. The probability Pe () is therefore

π . Pe () = 1 − sin2 (2θ) sin2 L

(7.50)

7.3.5 A νμ energy of only 4 MeV is below the threshold of the reaction νμ + n → p + μ. Therefore this reaction does not occur with reactor neutrinos, and one cannot measure the νμ flux. 7.3.6 In order to detect a significant decrease in the neutrino flux νe , we must have sin2 (2θ) sin2

π L

> 0.1.

(a) For the maximum mixing θ = π/4, i.e. sin2 (2θ) = 1, this implies π/L > 0.32 or  > L/10. For E = 4 MeV and Δm 2 c4 = 10−4 eV2 , one finds  > 10 km. The typical distances necessary to observe this phenomenon are of the order of a fraction of the oscillation length. (b) If the mixing is not maximum, one must operate at distances  greater than L/10. Note that is the mixing angle is too small, (sin2 (2θ) < 0.1 i.e. θ < π/10), the oscillation amplitude is too weak to be detected, whatever the distance . In that case, one must improve the detection efficiency to obtain a positive conclusion. 7.3.7 (a) In all experiments except KamLAND, the distance is smaller than 1 km. Therefore, in all of these experiments |1 − Pe | ≤ 10−3 . The oscillation effect is not detectable if the estimate |Δm 2 | c4 ∼ 10−4 eV2 is correct. (b) For |Δm 2 | c4 = 7.1 × 10−5 eV2 , tan2 θ = 0.45 and  = 180 km, we obtain Pe = 0.50 which agrees with the measurement. The theoretical prediction taking into account the effects due to the dispersion in energy is drawn on Fig. 7.14. We see incidentally how important it is to control error bars in such an experiment. Oscillations of Three Species: Atmospheric Neutrinos 7.3.1 At time t = 0, we have: |ν(0) = |να  =



Uα j |ν j ,

j

and therefore at time t: |ν(t) = e−i pct/

 j

Uα j e−im j c

2 3

t/(2 p)

|ν j  .

7.9 Problem: Neutrino Oscillations

179

Fig. 7.14 Experimental points of Fig. 7.12 and the theoretical prediction of (7.50) (sinusoidal function damped by energy dispersion affects). This curve is a best fit of solar neutrino data. We notice that the KamLAND data point corresponds to the second oscillation of the curve

We conclude that the probability Pα→β to observe a neutrino of flavor β at time t is  2    2  ∗  2 3 −im j c t/(2 p)     Pα→β (t) = νβ |ν(t) =  Uβ j Uα j e  .  j  7.3.2 We have L i j = 4πE/(|Δm i2j | c3 ). The oscillation lengths are proportional to the energy. We can use the result of question 1.3, with a conversion factor of 1000 to go from 4 MeV to 4 GeV. • For |Δm 212 | c4 = 7.1 × 10−5 eV2 , we find L 12 = 140,000 km. • For |Δm 223 | c4 = 2.5 × 10−3 eV2 , we find L 23 = 4000 km. 7.3.3 We want to know the minimal distance necessary in order to observe oscillations. We assume that both mixing angles θ12 and θ23 are equal to π/4, which corresponds to maximum mixing. We saw in the first part that if this mixing is not maximum, the visibility of the oscillations is reduced and that the distance which is necessary to observe the oscillation phenomenon is increased. By resuming the argument of the first part, we find that the modification of the neutrino flux of a given species is detectable beyond a distance i j such that sin2 (πi j /L i j ) ≥ 0.1 i.e. i j ≥ L i j /10. This corresponds to 12 ≥ 14,000 km for the oscillation resulting from the superposition 1 ↔ 2, and 23 ≥ 400 km for the oscillation resulting from the superposition 2 ↔ 3. 7.3.4 The factor of 2 between the expected muon and electron neutrino fluxes comes from a simple counting. Each particle π − (resp. π + ) gives rise to a νμ , a ν¯μ and a ν¯e (resp. a νμ , a ν¯μ and a νe ). In practice, part of the muons reach the ground before

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decaying, which modifies this ratio. Naturally, this effect is taken into account in an accurate treatment of the data. 7.3.5 For an energy of 4 GeV, we have found that the minimum distance to observe the oscillation resulting from the 1 ↔ 2 superposition is 14,000 km. We therefore remark that the oscillations νe  νμ , corresponding to the mixing 1 ↔ 2 which we studied in the first part cannot be observed at terrestrial distances. At such energies (4 GeV) and for evolution times corresponding at most to the diameter of the Earth (0.04 s), the energy difference E 1 − E 2 and the oscillations that it induces can be neglected. However, if the estimate |Δm 223 | c4 > 10−3 eV2 is correct, the terrestrial distance scales allow in principle to observe oscillations resulting from 2 ↔ 3 and 1 ↔ 3 superpositions, which correspond to νμ  ντ or νe  ντ . 7.3.6 The angular distribution (therefore the distribution in ) observed for the νe ’s does not show any deviation from the prediction made without any oscillation. However, there is a clear indication for νμ oscillations: there is a deficit of muon neutrinos coming from below, i.e. those which have had a long time to evolve. The deficit in muon neutrinos is not due to the oscillation νe  νμ of the first part. Indeed, we have seen in the previous question that this oscillation is negligible at time scales of interest. The experimental data of Fig. 7.13 confirm this observation. The deficit in muon neutrinos coming from below is not accompanied with an increase of electron neutrinos. The effect can only be due to a νμ  ντ oscillation.7 No oscillation νe  ντ appears in the data. In the framework of the present model, this is interpreted as the signature of a very small (if not zero) θ13 mixing angle. 7.3.7 Going back to the probability (7.50) written in question 1.4, the probability for an atmospheric muon neutrino νμ to be detected as a νμ is: P() = 1 − sin (2θ23 ) sin 2

2

π , L 23

(7.51)

where the averaging is performed on the energy distribution of the neutrino. If we measure the neutrino flux coming from the top, we have  L 23 , which gives Ptop = 1. If the neutrino comes from the bottom, the term sin2 (π/L 23 ) averages to 1/2 and we find: 1 Pbottom = 1 − sin2 (2θ23 ). 2 The experimental data indicate that for −1 ≤ cos α ≤ −0.5, Pbottom = 1/2. The distribution is very flat at a value of 100 events, i.e. half of the top value (200 events). We deduce that sin2 (2θ23 ) = 1, i.e. θ23 = π/4 and a maximum mixing angle for νμ  ντ . The results published by Super-Kamiokande fully agree with this analysis. 7 For completeness, physicists have also examined the possibility of a “sterile” neutrino oscillation,

i.e. an oscillation with a neutrino which would have no detectable interaction with matter.

7.9 Problem: Neutrino Oscillations

181

7.9.4 Comments The difficulty of such experiments comes from the smallness of the neutrino interaction cross sections with matter. The detectors are enormous water tanks, where about ten events per day are observed (for instance ν¯e + p → e+ + n). The “accuracy” of a detector comes mainly from the statistics, i.e. the total number of events observed. An amazing feature of neutrino oscillations is that they are genuine quantum effects whose wavelengths are macroscopic, owing to the smallness of the neutrino masses or, equivalently, energies, and therefore take place over distances of the size of the earth. The unexpected phenomenon in this physics is that quantum oscillations are not restricted to “geometrical” structures such as the NH3 molecule. Oscillations appear between elementary pointlike particles of different species! We notice that in the case of neutrinos, quantum oscillation phenomena can only be observed at macroscopic distances because of the smallness of mass differences. It is amusing that phenomena that are so far from our usual intuition can only be observed at very large distance scales. In 1998, the first undoubted observation of the oscillation ντ  νμ was announced in Japan by the Super-Kamiokande experiment (Y. Fukuda et al., Phys. Rev. Lett. 81, 1562 (1998)). This experiment uses a detector containing 50,000 tons of water, inside which 11,500 photomutipliers detect the Cherenkov light of the electrons or muons produced. About 60 ντ ’s were also detected, but this figure is too small to give further information. An accelerator experiment confirmed the results afterwards (K2K collaboration, Phys. Rev. Lett. 90, 041801 (2003)). The KamLAND experiment is a collaboration between Japanese, American and Chinese physicists. The detector is a 1000 m3 volume filled with liquid scintillator (an organic liquid with global formula C-H). The name means KAMioka Liquid scintillator Anti-Neutrino Detector. Reference: KamLAND Collaboration, Phys. Rev. Lett. 90, 021802 (2003); see also http://kamland.lbl.gov/. Very many experimental results come from solar neutrinos, which we have not dealt with here. This problem is extremely important, but somewhat too complex for our purpose. The pioneering work is due to Davis in his celebrated paper of 1964 (R. Davis Jr., Phys. Rev Lett. 13, 303 (1964)). Davis operated on a 37 Cl perchlorethylene detector and counted the number of 37 Ar atoms produced. In 25 years, his overall statistics has been 2200 events, i.e. one atom every 3 days! This was much smaller than what was expected from solarmodel calculations. At first, it seemed that some neutrinos were being lost during their travel between the Sun and the Earth. In 1991, the SAGE experiment done with Gallium confirmed the deficit (A. I. Abasov et al., Phys. Rev Lett. 67, 3332 (1991) and J. N. Abdurashitov et al., Phys. Rev Lett. 83, 4686 (1999)). In 1992, the GALLEX experiment, using a Gallium target in the Gran Sasso, also confirmed the solar neutrino deficit (P. Anselmann et al., Phys. Lett. B285, 376 (1992)). In 2001 the Sudbury Neutrino Observatory (SNO), whose detector uses 1000 tonnes of heavy water, and which is built 2 Km deep in a mine near Sudbury, Ontario

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in Canada, gave decisive experimental results on solar neutrinos (Q.R. Ahmad et al., Phys. Rev. Lett. 87, 071307 (2001) and 89, 011301 (2002); see also M.B. Smy, Mod. Phys. Lett. A 17, 2163 (2002)). The 2002 Nobel prize for physics was awarded to Raymond Davis Jr., the pioneer of this chapter of neutrino physics, and to Masatoshi Koshiba the leader of the reactor neutrino KamLAND collaboration. The 2015 Nobel prize for physics was awarded to Takaaki Kajita (SuperKamiokande) and Arthur B. McDonald (SNO).

Chapter 8

Algebra of Observables

Our axioms may be very elegant and profound, but they seem to be deprived of substance. Within this framework, how does one treat a particular case? What are the Hamiltonian and the observables for a given system? In wave mechanics, we have operators pˆ x = −i(∂/∂x), Hˆ = p 2 /2m + V and we solve differential equations, but what should we do here? It is true that for the ammonia molecule in the previous chapter, we have guessed quite easily how to proceed, but what is the general method? The answer lies in the first major discovery of Paul Adrien Maurice Dirac, who was an unusual person and who arrived on the scene of quantum mechanics during the summer of 1925. Dirac’s answer seemed surprising: in the general formalism of Hilbert space, the structures that play a key role and allow us to perform calculations are not the special form of such or such an observable, but rather the algebraic relations between the observables, in particular their commutation relations. These algebraic relations have the same form in all representations, and, as we shall see, they enable us to perform actual calculations.

8.1 Commutation of Observables We know that in general two observables Aˆ and Bˆ do not commute. Their commutator ˆ B] ˆ is defined by [ A, ˆ B] ˆ = Aˆ Bˆ − Bˆ A. ˆ [ A,

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_8

(8.1)

183

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8.1.1 Fundamental Commutation Relation We have already seen that between position and momentum observables, we have, in wave mechanics (8.2) [x, ˆ pˆ x ] = i Iˆ, where Iˆ is the identity operator. This relation is independent of the representation we use. In Heisenberg’s matrix mechanics it was found by Born and Jordan in July 1925, Eq. (6.60). This fundamental commutation relation is actually the definition of position and momentum observables along the same axis. In three-dimensional space, if we note xˆ j ( j = 1, 2, 3) the components of a position vector and similarly for momentum pˆ j , the set [xˆ j , pˆ k ] = iδ j,k Iˆ, [xˆ j , xˆk ] = [ pˆ j , pˆ k ] = 0 (8.3) defines a couple of vector observables of position xˆ and momentum pˆ . And that’s what we use in order to do calculations.

8.1.2 Other Commutation Relations For observables that have a classical analogue, we still use the correspondence principle. For instance, for the angular momentum Lˆ = rˆ × pˆ one can easily obtain, using the fundamental relation (8.2), the commutation relations [ Lˆ x , Lˆ y ] = i Lˆ z ,

(8.4)

and two others related by cyclic permutations. The three relations can be put together in the form: ˆ Lˆ × Lˆ = i L. (8.5) We show in the next chapter that what defines in general an angular momentum observable Jˆ is not the particular form Lˆ = rˆ × pˆ (valid for a particle in space), but the algebraic relations ˆ Jˆ × Jˆ = i J. (8.6) We show how this algebra allows us to calculate numbers, and furthermore that it is more general than Lˆ = rˆ × pˆ and leads to the existence of purely quantum angular momenta, which have no classical analogues, such as the spin 1/2 of the electron.

8.1 Commutation of Observables

185

When there is no classical analogue, there is no substitute to guessing or postulating the corresponding algebra of observables, by keeping in mind the symmetries of the problem. Here, we illustrate this on some simple and important results. First, we give the general proof of uncertainty relations for any couple of variables. Then we consider the time evolution of the expectation value of a physical quantity and the Ehrenfest theorem, which enables us to see how the classical limit appears in quantum mechanics. Next, we give an algebraic solution of the harmonic oscillator problem, due to Dirac, which plays a crucial role in quantum field theory. Next, we say a few words about observables that commute, in order to have the concept of a complete set of commuting observables, which is essential in order to treat problems in several variables, in particular, three-dimensional problems. Finally we say a few words about Dirac’s first major discovery in quantum mechanics.

8.1.3 Dirac in the Summer of 1925 Before we start, let us say a few words about how Dirac entered quantum mechanics in 1925. Dirac was a student in Cambridge. He was born on August 8, 1902 so he was barely 23. One can talk for hours about him. He was completely unusual. Very discreet, very polite, and careful, he thought a lot. He had a great culture, in particular in mathematics. He knew about noncommutative algebras, but let’s not anticipate. At that time, Cambridge had an impressive intellectual richness and was similar to Göttingen in that respect. The professors were Keynes in economics, J.J. Thomson, E. Rutherford, A.S. Eddington, and Fowler in physics, as well as others. And there, by accident, everything got started. On July 28, 1925, Heisenberg was invited to give a talk in Cambridge at the very fashionable Kapitza club. That talk had no influence on Dirac (who kept thinking instead of listening); neither of them remembered seeing the other on that day. But Heisenberg gave his article to Fowler who gave it to Dirac. He read the paper with some difficulty: it was full of philosophical ideas, but the mathematical formalism was clumsy. Dirac had nothing against philosophy, but he didn’t particularly care for it in that context; what he wanted was good mathematics. Very rapidly he said, “This will lead us nowhere.” Two weeks later, he walked into Fowler’s office and said, “It’s remarkable; it contains the key of quantum mechanics!” What had happened is that Heisenberg had cracked up! Since 1924, he had elaborated a system of calculational rules and a theory that worked well. Then Born had gotten involved in the whole business, and he had found that there were matrices. They had seen Hilbert. And Heisenberg, a positivist, had learned something tragic about his theory: some physical quantities did not commute! The product of a and b was not the same as the product of b and a! And that was contrary to any physical sense. The product of two quantities had never depended on the order. So Heisenberg was terrified, “My theory is not beautiful!” He was so

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terrified that he had put all the dust under the carpet and he wrote his papers in such a way to hide this noncommutativity as well as he could. Dirac, very meticulously, had redone the calculations step by step, and he realized that everything boiled down to this noncommutativity. In particular, he proved independently from Born and Jordan the fundamental relation [x, ˆ pˆ x ] = i Iˆ in August 1925 without mentioning matrices. He knew the existence of noncommutative algebras. So he made an attempt to modify the classical equations in order to take into account this noncommutativity. After all nothing dictates that physical quantities should commute. Some time after, Dirac said, “You know, Heisenberg was scared. He was scared by these foreign mathematics. I can understand that he was scared: all his theory was at stake. I had an enormous advantage over him: I had no risk.” So, Dirac constructed his own version of quantum mechanics, based on noncommutative algebras, that is, on commutators.

8.2 Uncertainty Relations In 1927 Heisenberg stated his uncertainty principle. When Dirac saw it, he simply said, “Oh yes, indeed, I proved that in 1925.” The general proof of uncertainty relations for any couple of physical quantities is the following. ˆ Let Consider two quantities A and B, and the corresponding observables Aˆ and B. |ψ be the state of the system. The measurement of A and of B gives the expectation values a and b, and the dispersions Δa and Δb. We want to relate Δa and Δb ˆ for two given observables Aˆ and B. A simple calculation based on the Schwarz inequality and the manipulation of Dirac’s formalism shows that, if |ψ is the state of the system, then Δa Δb ≥

1 ˆ B]|ψ|. ˆ |ψ|[ A, 2

(8.7)

Therefore ˆ B] ˆ  = 0, then, in general the two dispersions on 1. If Aˆ and Bˆ do not commute, [ A, A and B cannot be made simultaneously as small as possible except in special ˆ B]|ψ ˆ cases where the state |ψ is such that ψ|[ A, = 0. 2. For x and px , using (8.2), we obtain Heisenberg’s Δ x Δ px ≥ /2. We no longer refer to the Fourier transformation, we can forget about it. 3. The uncertainty relations are therefore generalized to any couple of observables. They come from the noncommutativity of the corresponding observables. Proof We first center the variables, that is, we set Aˆ  = Aˆ − a and Bˆ  = Bˆ − b, so that  Aˆ   =  Bˆ   = 0. We then have

8.2 Uncertainty Relations

187

(Δa)2 = ψ| Aˆ 2 |ψ, and (Δb)2 = ψ|B 2 |ψ. Consider any state |ψ and the vector ( Aˆ  + iλ Bˆ  )|ψ, with λ real. The square of the norm of this vector is: ( Aˆ  + iλ Bˆ  )|ψ2 = ψ|( Aˆ  − iλ Bˆ  )( Aˆ  + iλ Bˆ  )|ψ = ψ| Aˆ 2 |ψ + λ2 ψ| Bˆ 2 |ψ + iλψ|[ Aˆ  , Bˆ  ]|ψ = Δa 2 + λ2 Δb2 + iλψ|[ Aˆ  , Bˆ  ]|ψ. Because Aˆ  and Bˆ  are Hermitian, the operator i[ Aˆ  , Bˆ  ] is also Hermitian and the last term is real. The above expression is the square of the norm of a vector, thus it must be positive whatever the value of λ. Because it is positive for λ → ∞ it must not change sign. Therefore the discriminant of the trinomial in λ must be negative, ˆ B], ˆ and, since [ Aˆ  , Bˆ  ] = [ A, Δa Δb ≥

1 ˆ B]|ψ| ˆ |ψ|[ A, 2

Q.E.D.

(8.8)

8.3 Evolution of Physical Quantities We now address the following problem: what is the time variation of the expectation value of a physical quantity?

8.3.1 Evolution of an Expectation Value Consider the expectation value a of a physical quantity ˆ a = ψ| A|ψ. We take the time derivative:       d d ∂ ˆ d ˆ ˆ a = ψ| A|ψ + ψ| A |ψ + ψ| A |ψ . dt dt ∂t dt Using the Schrödinger equation and its Hermitian conjugate expression: i

d|ψ = Hˆ |ψ, and dt

− i

dψ| = ψ| Hˆ ; dt

188

we obtain:

8 Algebra of Observables

ˆ 1 d ˆ Hˆ ]|ψ + ψ| ∂ A |ψ. a = ψ|[ A, dt i ∂t

(8.9)

This formula is due to Ehrenfest (1927) (but Dirac had found it in 1925). We remark that, in a similar way to the Schrödinger equation, the Hamiltonian governs the time evolution of a physical expectation value. Here it does so through its commutator with the observable. If the operator Aˆ does not explicitly depend on time, we have 1 d ˆ Hˆ ]|ψ. a = ψ|[ A, dt i

(8.10)

8.3.2 Particle in a Potential, Classical Limit We call qi (i = 1, 2, 3) the three position variables x, y, z and pi (i = 1, 2, 3) the coordinates of the momentum px , p y , pz . The operators qˆi and pˆ i obey the canonical commutation relations: [qˆi , qˆ j ] = 0, [ pˆ i , pˆ j ] = 0, [qˆ j , pˆ k ] = iδ j,k .

(8.11)

From these relations one can prove the commutation relations [qˆ j , pˆ mj ] = m(i) pˆ m−1 , [ pˆ j , qˆ nj ] = −n(i)qˆ n−1 , j j

(8.12)

which we can generalize to any function Fˆ = F(qˆi , pˆ i ) of the operators qˆi and pˆ i which can be expanded in a power series ˆ ˆ ˆ = i ∂ F , and [ pˆ j , F] ˆ = −i ∂ F . [qˆ j , F] ∂ pˆ j ∂ qˆ j

(8.13)

If the Hamiltonian does not depend explicitly on time, choosing Fˆ = Hˆ , we obtain the evolution equations:   ∂ Hˆ d qi  = , dt ∂ pˆ j

  ∂ Hˆ d pj = − . dt ∂ qˆ j

(8.14)

These forms have a great similarity to the equations of Hamilton’s analytical mechanics, as we shall see in Chap. 15.

8.3 Evolution of Physical Quantities

189

The Hamiltonian of a particle in a potential is pˆ 2 + V (r). Hˆ = 2m

(8.15)

Substituting in (8.14), we obtain  p dr = dt m d p = −∇V (r). dt

(8.16) (8.17)

Equation (8.16) is the correct definition of the group velocity of a wave packet. It is the same for expectation values as the classical relation between the position and the velocity. Actually, Eq. (8.17) differs slightly from the classical equation, which would be for expectation values  d p  , = −∇V (r) r=r dt because, in general, f (r)  =  f (r). However, if the distribution in r is peaked around some value r 0 , then r ∼ r 0 and f (r) ∼  f (r), and in this case Eqs. (8.16) and (8.17) are, for the expectation values, essentially the same as the classical equations of the motion. This observation is the Ehrenfest theorem (1927). One-Dimensional Classical Limit In one dimension we have   dV d V  d .  p = − = −  dt dx d x x=x We expand the function f (x) = −∂V /∂x in the vicinity of x = x and we obtain 1 f (x) = f (x) + (x − x) f  (x) + (x − x)2 f  (x) + · · · 2 that is, taking the expectation value,  f  = f (x) +

Δx 2  f (x) + · · · 2

where Δx 2 = (x − x)2 . The nonclassical term in the evolution of the expectation value will be negligible if |Δx 2 f  (x)/ f (x)| 1,

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or, if we come back to the potential V ,      2 ∂3 V    Δx   ∂V  ,    3 ∂x ∂x  that is, if the potential varies slowly on the extension of the wave packet (or on an interval of the order of the de Broglie wavelength). That is the actual content of the Ehrenfest theorem. We recall that a macroscopic system (i.e., a system whose orders of magnitude are such that  seems negligible) is not necessarily a classical system. We can perfectly well consider a wave packet having two peaks located at a large macroscopic distance from each other. As we said in Chap. 4 such a state, macroscopic but quantum mechanical, is very unstable. One can check that for a harmonic oscillator, V = mω02 x 2 /2, or, more generally if the potential is a second degree polynomial, the Ehrenfest theorem gives identically the classical equation of motion d 2 x/dt 2 = −ω02 x.

8.3.3 Conservation Laws There are two cases where da/dt = 0. ˆ Hˆ ] = 0. The quantity • Either the observable commutes with the Hamiltonian, [ A, A is always conserved, it is a constant of the motion. ˆ Hˆ ]|ψ = 0. This is in particular the case for • Or the state |ψ is such that ψ|[ A, stationary states, that is eigenstates of Hˆ for which no quantity evolves in time. Here are some examples of applications of these results. Conservation of the Norm Consider the identity operator Aˆ = Iˆ; this gives the conservation of the norm d ψ|ψ = 0, dt that is to say, conservation of probability. Conservation of Energy for an Isolated System Consider a time-independent problem; the choice Aˆ = Hˆ gives d E = 0. dt

8.3 Evolution of Physical Quantities

191

Conservation of Momentum Consider the motion of a free particle of Hamiltonian Hˆ = pˆ 2 /2m. The observables pˆ x , pˆ y , pˆ z commute with Hˆ and we obtain conservation of the momentum: d  pi  = 0, i = x, y, z. dt This can be generalized to an N particle system whose Hamiltonian is translation invariant (pairwise interactions V (xi − x j )). The total momentum  P =  pi  is conserved.

8.4 Algebraic Resolution of the Harmonic Oscillator In order to see how the principles work, and to see the importance of the commutation relations, we show how the one-dimensional harmonic oscillator problem can be solved using the algebra of observables. This calculation is also due to Dirac. In order to simplify the proof, we make use of some qualitative results we already know. Consider the Hamiltonian 1 pˆ 2 + mω 2 xˆ 2 . Hˆ = 2m 2 With the change of observables Xˆ = xˆ we obtain

with



mω , 

pˆ Pˆ = √ , mω

(8.18)

ˆ Hˆ = ω H,

(8.19)

1  ˆ2 X + Pˆ 2 . Hˆ = 2

(8.20)

We must solve the eigenvalue problem ˆ H|ν = εν |ν, where we have assumed that the eigenvalues εν are not degenerate. We know this result from Chap. 5, but this can be proven in the present context. The commutation relation of Xˆ and Pˆ can be deduced from (8.2), ˆ = i. [ Xˆ , P]

(8.21)

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Operators aˆ , aˆ † , and Nˆ In order to solve the eigenvalue problem we introduce the following operators 1 1 ˆ ˆ aˆ = √ ( Xˆ + i P), aˆ † = √ ( Xˆ − i P) 2 2

(8.22)

whose commutator is [a, ˆ aˆ † ] = 1.

(8.23)

1 Nˆ = aˆ † aˆ = ( Xˆ 2 + Pˆ 2 − 1) 2

(8.24)

Consider now the operator

that satisfies the commutation relations [ Nˆ , a] ˆ = −a, ˆ [ Nˆ , aˆ † ] = aˆ † . We have

(8.25)

1 Hˆ = Nˆ + Iˆ, 2

therefore Hˆ and Nˆ have the same eigenvectors. Let ν be the eigenvalues of Nˆ and |ν its eigenvectors; we have, coming back to the initial Hamiltonian   1 Hˆ |ν = ν + ω|ν. 2

(8.26)

Determination of the Eigenvalues The determination of the eigenvalues comes from the following lemmas. Lemma 1 The eigenvalues ν of the operator Nˆ are positive or zero. In fact, consider a vector |ν and the norm of the vector a|ν: ˆ 2 = ν|aˆ † a|ν ˆ = ν| Nˆ |ν = νν|ν = ν|ν2 . a|ν ˆ

Therefore ν ≥ 0 and

a|ν ˆ = 0, if and only if ν = 0.

(8.27)

(8.28)

Lemma 2 The vector a|ν ˆ is either an eigenvector of Nˆ , corresponding to the eigenvalue ν − 1, or the null vector. In fact, consider the vector Nˆ a|ν. ˆ If we use the commutation relation of Nˆ and a, ˆ we obtain

8.4 Algebraic Resolution of the Harmonic Oscillator

193

Nˆ (a|ν) ˆ = aˆ Nˆ |ν − a|ν ˆ = ν a|ν ˆ − a|ν ˆ = (ν − 1)(a|ν). ˆ Therefore • Either ν − 1 is an eigenvalue of Nˆ and a|ν ˆ is an associated eigenvector; • Or ν − 1 is not an eigenvalue of Nˆ and a|ν ˆ is the null vector. By an analogous argument one can show that: Lemma 3 If ν + 1 is an eigenvalue of Nˆ , aˆ † |ν is an eigenvector associated with the eigenvalue ν + 1 and aˆ † |ν2 = (ν + 1)|ν2 .

(8.29)

It is now simple to prove that: Theorem The eigenvalues of Nˆ are the nonnegative integers. In fact, because the eigenvalues are nonnegative, one of them must be smaller than the others. Let us call it νmin . Since νmin is the smallest eigenvalue, νmin − 1 is not an eigenvalue. Therefore, a|ν ˆ min  is the null vector and its norm is zero. Inasmuch as 2 = ν|ν2 for all values of ν, this means that νmin = 0. we have shown that a|ν ˆ Starting with this eigenvalue and the corresponding eigenvector |νmin = 0, we can generate all other eigenvalues and corresponding eigenvectors by repeatedly applying the operator aˆ † , at each step we add 1: (. . . ν → ν + 1 → ν + 2 . . .). We recover the energy levels of the harmonic oscillator. Eigenstates Ground State The ground state |0 satisfies (8.28) a|0 ˆ = 0, or



Xˆ + i Pˆ |0 = 0.

(8.30)

In the language of wave functions, this amounts to 

mω d x+  dx

 ϕ0 (x) = 0,

(8.31)

where ϕ0 (x) is the ground state wave function. The solution is ϕ0 (x) = C0 e−(mω/)x

2

/2

,

where C0 is a normalization constant. We recover the result of Chap. 5.

(8.32)

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Excited States We assume the eigenstates are normalized: n|n = 1. Owing to the above Lemmas 1 and 2 and to Eqs. (8.27) and (8.29), a|n ˆ =

√

n|n − 1, aˆ † |n =

√ n + 1|n + 1.

(8.33)

Hence the name of annihilation operator (for a) ˆ and of creation operator (for aˆ † ) because they allow us to get from an energy state (n + 1/2)ω to energy states (n + 1/2 ∓ 1)ω, and they respectively annihilate or create an energy quantum ω. Similarly, the operator Nˆ corresponds to the number of quanta in the state |n. The sequence of states |n is generated from the ground state |0 by applying repeatedly the operator aˆ † , 1 |n = √ (aˆ † )n |0. (8.34) n! This allows us to find the wave function ϕn (x) of the energy state (n + 1/2)ω in terms of the ground state wave function: 1 1 ϕn (x) = √ √ n! 2n

n



mω  d ϕ0 (x). − x  mω d x

(8.35)

This is an explicit and compact formula for the Hermite functions. Similarly, one can show with (8.33) and the definition of aˆ and aˆ † , that we have

√  √ n + 1|n + 1 + n|n − 1 2mω

√ mω √ n + 1|n + 1 − n|n − 1 . p|n ˆ =i 2

x|n ˆ =

(8.36) (8.37)

We see, in this example, the elegance and the power of Dirac’s algebraic method. This treatment of the harmonic oscillator, and the operators a, ˆ aˆ † , and Nˆ , are fundamental tools in many branches of physics such as quantum field theory, statistical mechanics, and the many-body problem.

8.5 Commuting Observables When two observables commute, there is no constraint such as the uncertainty relations. This case is, however, important in practice.

8.5 Commuting Observables

195

8.5.1 Theorem We know that if two matrices commute, one can diagonalize them simultaneously. This remains true in the infinite-dimensional case. If two observables Aˆ and Bˆ commute, there exists a common eigenbasis of these two observables. ˆ B, ˆ This theorem is generalized immediately to the case of several observables A, ˆ which all commute. C, ˆ where the index rα means that an eigenProof Let {|α, rα } be the eigenvectors of A, vector associated with an eigenvalue aα belongs to an eigensubspace of dimension dα ≥ 1, ˆ rα  = aα |α, rα , r = 1, . . . , dα . A|α, (8.38) ˆ B] ˆ = 0, i.e. By assumption, we have [ A, ˆ rα  = aα B|α, ˆ ˆ Aˆ B|α, rα  = Bˆ A|α, rα , r = 1, . . . , dα .

(8.39)

ˆ Therefore, the vector B|α, rα  is an eigenvector of Aˆ with the eigenvalues aα . It therefore belongs to the corresponding eigensubspace. We call this vector |α, β, kαβ ; the index kαβ means that again this vector may be nonunique. Therefore, this vector is a linear combination of the vectors {|α, rα }, that is, ˆ B|α, rα  =



brα |α, rα ,

rα

which can be diagonalized with no difficulty. In other words, if Aˆ and Bˆ commute, they possess a common eigenbasis. The reciprocal is simple. The Riesz theorem says that the orthonormal eigenvectors of an observable form a Hilbert basis. Suppose Aˆ and Bˆ have in common the basis {|ψn } with eigenvalues an and bn : ˆ n  = an |ψn  and B|ψ ˆ n  = bn |ψn . A|ψ

(8.40)

If we apply Bˆ to the first expression and Aˆ to the second, and subtract, we obtain ˆ n  = (an bn − bn an )|ψn  = 0. ( Aˆ Bˆ − Bˆ A)|ψ Because {|ψn } is a Hilbert basis, we therefore have ˆ B]|ψ ˆ [ A, = 0, whatever |ψ; ˆ B] ˆ = 0. which means that [ A,

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8 Algebra of Observables

8.5.2 Example Actually, we have not yet seen examples of this because we have considered only one-dimensional problems. But the statement is nothing very complicated. Consider, for instance, an isotropic two-dimensional harmonic oscillator. The eigenvalue problem of the Hamiltonian is a priori a difficult problem because it seems to be a partial differential equation in two variables. But the Hamiltonian can be written as the sum of two independent Hamiltonians acting on different variables: 1 2 ∂ 2 1 2 ∂ 2 2 2 + x − + mω 2 y 2 = Hˆ x + Hˆ y . mω Hˆ = − 2m ∂x 2 2 2m ∂ y 2 2

(8.41)

The two operators Hˆ x and Hˆ y , which are both operators in one variable and which act on different variables commute obviously. One can solve the eigenvalue problems of Hˆ x and Hˆ y separately: Hˆ x ϕn (x) = E n ϕn (x);

Hˆ y ϕn (y) = E n ϕn (y).

The eigenvalues of Hˆ are the sums of eigenvalues of Hˆ x and Hˆ y with eigenfunctions that are the products of corresponding eigenfunctions: E n = E n 1 + E n 2 = (n 1 + n 2 + 1)ω, ψn (x, y) = ϕn 1 (x)ϕn 2 (y). In other words, a sum of Hamiltonians that commute has for eigenvalues the sum of eigenvalues of each of them, and for eigenfunctions the product of corresponding eigenfunctions.

8.5.3 Tensor Structure of Quantum Mechanics This example has another interest. It shows that in the case of systems with several degrees of freedom, the Hilbert space is the tensor product of the Hilbert spaces in which each individual degree of freedom is described. In the above example, it is the fashionable way of saying that the products of eigenfunctions of the one-dimensional harmonic oscillator ψn,m (x, y) = ϕn (x)ϕm (y)

(8.42)

form a basis of square integrable functions in two variables (x, y), or that the space of square integrable functions in two variables L2 (R2 ) is the tensor product of two spaces L2 (R) of square integrable functions in one variable.

8.5 Commuting Observables

197

The tensor structure of quantum mechanics is important and useful in complex systems. One can find it at various degrees of sophistication in the literature.1 In Dirac’s notations the elements of the basis ϕn 1 (x)ϕm 2 (y) are written as |ψn,m  = |1 : ϕn  ⊗ |2 : ϕm ,

(8.43)

where (1) and (2) stand for the two degrees of freedom (or subsystems) and n and m are the corresponding eigenstates. The symbol ⊗ stands for “tensor product,” which is just an ordinary product in Eq. (8.42). Any state of the global system |ψ can be written as  cn,m |1 : ϕn  ⊗ |2 : ϕm . (8.44) |ψ = n,m

An important property is that the Hermitian scalar product of two factorizable vectors |u ⊗ |v and |u   ⊗ |v   factorizes as the product (u  | ⊗ v  |)(|u ⊗ |v = u  |uv  |v.

(8.45)

In Chap. 12 we show, with spin 1/2, an example just as simple but not as trivial of a tensor product of Hilbert spaces.

8.5.4 Complete Set of Commuting Observables (CSCO) This brings us to a notion that is useful both conceptually and technically. ˆ B, ˆ C, ˆ . . ., is said to form a complete set of commuting A set of operators A, observables (CSCO) if the common eigenbasis of this set of operators is unique. In other words, to each set of eigenvalues aα , bβ , cγ , . . . there corresponds a single eigenvector |αβγ . . . (up to a phase factor). If an operator Oˆ commutes with all of the operators of the CSCO, then it is a function of these operators. For a given system, there exists an infinite number of CSCOs. We show in the following chapters that one chooses a CSCO according to criteria of convenience. Neither the nature nor the number of observables that form a CSCO are fixed a priori. For a one-dimensional harmonic oscillator, the Hamiltonian pˆ 2 1 Hˆ x = x + mω 2 xˆ 2 2m 2 is a CSCO by itself. There is only one eigenbasis of Hˆ x formed by the Hermite functions ψn (x). For the two-dimensional isotropic oscillator above (8.41), this is not the case. A possible basis is formed by the set of functions {ψn 1 (x)ψn 2 (y)}, where ψn 1 (x) and 1 See

J.-L. Basdevant and J. Dalibard, Quantum Mechanics, Chap. 5, Sect. 6.

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8 Algebra of Observables

ψn 2 (y) are, respectively, the eigenfunctions of Hˆ x and Hˆ y . The eigenvalue corresponding to ψn 1 (x)ψn 2 (y) is E n 1 ,n 2 = ω(n 1 + n 2 + 1), which is degenerate, except for n 1 = n 2 = 0. This implies that there are actually several eigenbases of Hˆ (actually an infinite number). For instance, in the subspace associated with 2ω, the possible base elements are: {cos θψ1 (x)ψ2 (y) + sin θψ2 (x)ψ1 (y), − sin θψ1 (x)ψ2 (y) + cos θψ2 (x)ψ1 (y)}. Therefore, Hˆ is not a CSCO by itself. A possible CSCO is the set of two Hamiltonians { Hˆ x , Hˆ y }. In fact, the two eigenvalues {E n x = (n x + 1/2)ω, E n y = (n y + 1/2)ω} uniquely specify an eigenvector. This does not imply that in this problem a CSCO is necessarily composed of two observables. A rigorous theoretician may object that the operator Hˆ π = Hˆ x + π Hˆ y forms a CSCO by itself. Indeed, its eigenvalues are n π = (n x + 1/2) + π(n y + 1/2), and since π is transcendental there is a unique couple of integers (n x , n y ) corresponding to a given eigenvalue. An experimentalist will reply that it is simpler to measure two numbers (n x , n y ) rather than investing in a measuring apparatus that directly gives the answer n π .

8.5.5 Completely Prepared Quantum State Why is the notion of a completely prepared quantum state important physically? If we want to specify as accurately as possible the initial conditions of an experiment, we must know whether we start from a specific quantum state or with some ill-defined situation. In the above case of the isotropic harmonic oscillator, if we know the initial energy nω, we only know that the initial state belongs to a subspace of dimension n, generated by the n functions ψn 1 (x)ψn 2 (y) with n 1 + n 2 + 1 = n. A measurement of the energy is not sufficient to specify unambiguously the initial state. If we measure the vibrational energies along both the x and y axes, we know the state. One says that one deals with a completely prepared quantum state. More generally, consider two observables Aˆ and Bˆ that commute. There exists a ˆ where the index γ indicates that Aˆ and common eigenbasis {|α, β, γ} of Aˆ and B, ˆ B are not necessarily a CSCO by themselves. Consider a state |ψ, and suppose that by measuring A on |ψ, we find the value a0 . After the measurement, the state of the system has changed. It is ⎛ ⎞  |α0 ; β, γ  α0 ; β, γ  |⎠ |ψ, |ψ → |ψ0  = λ ⎝ β,γ 

where λ is a normalization factor such that ψ0 |ψ0  = 1.

8.5 Commuting Observables

199

On this new state |ψ0 , we measure B. Suppose the result is b1 . Similarly, we obtain another state |ψ1  after the measurement: 

 |ψ0  → |ψ1  = λ





|α, β1 ; γα, β1 ; γ|

|ψ0 .

αγ

If we insert the expression of |ψ0  in this formula, we obtain |ψ1  = λλ



|α, β1 ; γα, β1 ; γ|α0 , β; γ  α0 , β; γ  |ψ.

αβγ 

But by assumption, α, β; γ|α , β  ; γ   = δαα δββ  δγγ  . Therefore, the expansion of |ψ1  reduces to |ψ1  = λλ



|α0 , β1 ; γα0 , β1 , γ|ψ.

γ

In other words, |ψ1  is still an eigenvector of Aˆ with the same eigenvalue a0 . This is an important result. If two observables Aˆ and Bˆ commute, if we successively measure A, with a result a0 , and then B with a result b1 , this second measurement does not affect the value found previously for A. If we redo a measurement of A, we find the same result a0 with probability one. This result can be extended to any number of commuting observables. If one measures all the physical quantities of a CSCO, any (“immediate”) new measurement of one of these quantities will always give the same result. The state vector defined by this series of measurements is defined uniquely. One says that with this series of measurements one obtains a state that is completely prepared or completely known.

8.6 Sunday September 20, 1925 Let us get back to Dirac, in the summer of 1925. His problem was to find how to incorporate noncommutativity in classical mechanics. Dirac knew that in analytical mechanics, which had been developed one century before, Hamilton had found a quite fruitful formulation.2 In this version of mechanics, the state of a particle is described at any time by its position x and its momentum p. The system is characterized by a Hamilton function or Hamiltonian H which, for a particle of mass m in a potential V (x) is

2 See, for instance, J.-L. Basdevant, Variational Principles in Physics, Chap. 4. New York: Springer

(2006).

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8 Algebra of Observables

H=

p2 + V (x). 2m

(8.46)

The evolution equations of the state variables x and p are then given by the canonical equations ∂H dp ∂H dx = , and =− . (8.47) dt ∂p dt ∂x Dirac wanted to construct his quantum mechanics with noncommutative algebras, that is, with commutators. On Sunday, September 20, 1925, Dirac committed a crime! He was very well organized and he had the habit of working hard all week and of relaxing on Sundays by talking a walk. But on that Sunday he committed a crime. In fact, during his walk he started thinking. He had given himself the right to walk when he was thinking, but never to think on his Sunday walk! He thought about commutators, and, suddenly, he thought about something he had been told of, and about which he had read. He thought about the works of a former student of Polytechnique, in Paris, Siméon-Denis Poisson! And Dirac rushed to the library, but there was some suspense because “On Sundays it’s closed.” He had to wait. On Monday morning, he rushed in and read about what Carl Gustav Jacob Jacobi had called the greatest discovery of Poisson: the Poisson brackets. Consider two physical quantities f and g, which are functions of the state variables (x, p); the Poisson bracket of f and g is the quantity  { f, g} =

∂ f ∂g ∂ f ∂g − ∂x ∂ p ∂ p ∂x

 .

(8.48)

For the state variables (x, p) we find the relation {x, p} = 1, and {x, f } =

∂f ∂f , { p, f } = − . ∂p ∂x

(8.49)

(8.50)

In three dimensions, (xi , pi ) (i = 1, 2, 3), (8.49) generalizes as {xi , x j } = 0, { pi , p j } = 0, {xi , p j } = δi j .

(8.51)

The time evolution of a physical quantity f (x, p) is ∂f ∂f df =( x˙ + p). ˙ f˙ = dt ∂x ∂p

(8.52)

8.6 Sunday September 20, 1925

201

But, using Hamilton’s equations (8.47), one obtains f˙ = { f, H }.

(8.53)

In particular, the equations (8.47) can be written in the symmetric way x˙ = {x, H },

p˙ = { p, H }.

(8.54)

Dirac was fascinated. The quantum commutators, divided by i, play a completely similar role as the Poisson brackets do in analytical mechanics. Just compare the fundamental commutation relation (8.2) and the relation (8.49), and, similarly, the Ehrenfest theorem (8.10) and Eq. (15.20). That is how Dirac understood the actual form of the correspondence principle. One must, in the classical equations, replace the Poisson brackets by the quantum commutators divided by i. Dirac called the quantum physical quantities “q-numbers” which are noncommutative contrary to the classical commutative “c-numbers”. He finished his work on November 7, 1925, and he published his article “The Fundamental Equations of Quantum Mechanics” in December 1925. In November, he wrote to Heisenberg who replied that his work was “most beautiful and remarkable.” He was apparently insensitive to the publication of the works of Born and Jordan in November 1925 and of Born, Heisenberg, and Jordan in January 1926, where one can find a number of his results (proven independently).

8.7 Exercices 1. Commutator algebra Show the following equalities: ˆ Bˆ C] ˆ = [ A, ˆ B] ˆ Cˆ + B[ ˆ A, ˆ C] ˆ , [ A, ˆ Bˆ n ] = [ A,

n−1 

ˆ B] ˆ Bˆ n−s−1 Bˆ s [ A,

s=0

ˆ [ B, ˆ C]] ˆ + [ B, ˆ [C, ˆ A]] ˆ + [C, ˆ [ A, ˆ B]] ˆ = 0 (Jacobi identity) . [ A, 2. Classical equations of motion for the harmonic oscillator Show that for a harmonic oscillator V (x) = mω 2 x 2 /2, the Ehrenfest theorem gives identically the classical equation of motion: d 2 x = −ω 2 x. dt 2

202

8 Algebra of Observables

3. Conservation law Consider a system of two particles interacting through a potential V (r 1 − r 2 ). Check that the total momentum P = p1 + p2 is conserved. Show that this property can be extended to a system of n interacting particles. 4. Hermite functions Prove from (8.33) and the definition of aˆ and aˆ † , the recursion relations for the Hermite functions:

√  √ n + 1|n + 1 + n|n − 1 , (8.55) x|n ˆ = 2mω

√ mω √ n + 1|n + 1 − n|n − 1 . (8.56) p|n ˆ =i 2 5. Generalized uncertainty relations  a. Consider, in 3 dimensions, the radial variable r = x 2 + y 2 + z 2 and a real function f (r ) of this variable. Show that the commutator of pˆ x with f (ˆr ) is: xˆ [ pˆ x , fˆ] = −i f  (ˆr ), r where f  (r ) is the derivative of f . b. Consider the operator Aˆ x = pˆ x − iλxˆ f (ˆr ) where λ is a real number. • Calculate the square of the norm of Aˆ x |ψ for an arbitrary vector |ψ. • Add the analogous relations for Aˆ y and Aˆ z , and derive an inequality relating  p 2 , r 2 f 2 ,  f  and r f  , which holds for any function f and any state |ψ. c. Considering the cases f = 1, f = 1/r , and f = 1/r 2 show that one has the following relations in three dimensions:  p 2 r 2  ≥

9 2 1 2 1  ,  p 2  ≥ 2  2 ,  p 2  ≥  2 . 4 r 4 r

d. Harmonic oscillator. The Hamiltonian of a three-dimensional harmonic oscillator is Hˆ = pˆ 2 /2m + mω 2 rˆ 2 /2. • Using the first inequality, find a lower bound for the ground state energy of this oscillator, and explain why this bound is equal to the ground state energy. • Write the differential equation satisfied by the corresponding ground state wave function and calculate this wave function. e. Hydrogen atom. The Hamiltonian of the hydrogen atom is, considering the proton mass as very large compared to the electron mass,

8.7 Exercices

203

e2 pˆ 2 − Hˆ = 2m e r where, for simplicity, we set e2 = q 2 /4π0 . • Using the second inequality, find a lower bound for the ground state energy of the hydrogen atom, and explain why this bound is equal to the ground state energy. • Write the differential equation satisfied by the corresponding ground state wave function φ(r ) and calculate this wave function. 7. Time-energy uncertainty relation Consider a state |ψ of a system whose energy dispersion is ΔE, and an observable Aˆ whose expectation value and dispersion are a and Δa. Using the commutation relations, show that following inequality holds:  Δa ΔE ≥ 2

   da     dt  .

Deduce from this that if the typical evolution time-scale τ of the system is defined by τ = |Δa/(da/dt)|, one has the inequality τ ΔE ≥ /2. 8. Virial Theorem ˆ where Consider a one-dimensional system of Hamiltonian Hˆ = pˆ 2 /2m + V (x) V (x) = λx n . a. Calculate the commutator [ Hˆ , xˆ p]. ˆ b. By taking the expectation value of this commutator, show that in any eigenstate of Hˆ , one has the relation: 2T  = nV  , where Tˆ = pˆ 2 /2m is the kinetic energy operator. Check this relation on the harmonic oscillator. c. Generalize this result to three dimensions by calculating [ Hˆ , rˆ · pˆ ] and considering a potential V (r) which is a homogeneous function of the variables x, y, z, of degree n. A homogeneous function of degree n satisfies V (αx, αy, αz) = αn V (x, y, z) and r · ∇V = nV . d. Show that for an arbitrary potential V (r ) one has the general relation: 2T  = r

∂V . ∂r

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8 Algebra of Observables

8.8 Problem. Quasi-Classical States of the Harmonic Oscillator We consider a one-dimensional harmonic oscillator of frequency ω and we study the eigenstates |α of the annihilation operator: a|α ˆ = α|α where α is a complex number. We expand |α on the basis {|n}: |α =



Cn |n.

n

1. Determination of α: (a) (b) (c) (d) (e)

Write the recursion relation between the coefficients Cn . Express the Cn ’s in terms of the first coefficient C0 . Calculate the coefficients Cn by normalizing |α, i.e. α|α = 1. What are the allowed values for the number α ? In an energy measurement on the state |α, what is the probability to find the value E n = (n + 1/2)ω?

2. Consider a state |α. Starting from the expression of the Hamiltonian and the definition of this state: (a) Calculate the expectation value E, (b) Calculate the expectation value of the square of the energy E 2  (use the commutator of aˆ and aˆ † ), (c) Deduce the value of the dispersion ΔE in that state. (d) In what sense can one say that the energy is defined more and more accurately if |α|  1? 3. Calculate x, Δx,  p, Δp in a state |α. In that state, what is the value of the product Δx Δp? 4. We assume that at t = 0, the oscillator is in the state |α. (a) Write the state |ψ(t) of the system at time t. (b) Show that the state |ψ(t) is also an eigenstate of the operator aˆ and give the corresponding eigenvalue. (c) We set α = α0 eiφ with α0 real positive. What are, at time t, the values of x,  p and Δx Δp? 5. We now determine the wave functions corresponding to |α. (a) Check that the change of variables from x and p to X and P leads to the expression of the operator Pˆ acting on wave functions: ψ(X, t)

8.8 Problem. Quasi-Classical States of the Harmonic Oscillator

205

∂ Pˆ = −i . ∂X Give the corresponding expression for the operator Xˆ acting on functions ϕ(P, t). (b) Calculate the wave function ψα (X ) of the state |α. (c) Calculate the Fourier transform ϕα (P) of this wave function. (d) Starting from the time dependence of |ψα (X, t)|2 and |ϕα (P, t)|2 , explain the results obtained previously.

8.8.1 Solution 1. By definition, we have: a|α ˆ =

∞ 

∞  √ Cn n |n − 1 = α Cn |n.

n=1

n=0

√ The ensuing recursion relation Cn n = αCn−1 allows to calculate the coefficients Cn in terms of C0 and α, whatever the value of the complex number α: αn Cn = √ C0 . n! For any α, we therefore obtain: |α = C0

∞  αn √ |n n! n=0

and, by normalizing the result: α|α = |C0 |2

∞  |α|2n n=0

n!

= e|α| |C0 |2 ⇒ C0 = e−|α| 2

up to an arbitrary phase factor. The probability p(E n ) to find E n is: p(E n ) = |n|α|2 = e−|α| |α|2n /n! 2

which is a Poisson distribution.

2

/2

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8 Algebra of Observables

2. The expectation value of the energy is obtained by using: E = α| Hˆ |α = ωα|(aˆ † aˆ + 1/2)|α = (|α|2 + 1/2)ω ˆ 2 + aˆ † aˆ + 1/4]|α. E 2  = 2 ω 2 α|[(aˆ † a) We have [a, ˆ aˆ † ] = 1 and therefore: (aˆ † a) ˆ 2 = aˆ † aˆ aˆ † aˆ = aˆ † aˆ † aˆ aˆ + aˆ † a, ˆ hence: E 2  = 2 ω 2 (|α|4 + 2|α|2 + 1/4) and the variance: ΔE 2 = E 2  − E2 = 2 ω 2 |α|2 , ΔE = ω|α| , or, equivalently ΔE/E = |α|/(|α|2 + 1/2) ∼ 1/|α| for |α|  1. The relative dispersion of the energy ΔE/E goes to zero as |α| increases. 3. The calculation of these expectation values yields: √ √ X  = α|(aˆ + aˆ † )/ 2|α = (α + α∗ )/ 2 √ √ P = α|(aˆ − aˆ † )/i 2|α = i(α∗ − α)/ 2 ˆ X 2  = α|(aˆ 2 + aˆ +2 + aˆ aˆ † + aˆ † a)|α/2  2  ∗2 2 = α + α + 2|α| + 1 ΔX 2 = X 2  − X 2 = 1/2 therefore: Δx =



 /2mω and x = (α + α∗ ) /2mω.

Similarly, we obtain: Δp = and for all α:



 mω/2 and  p = i(α∗ − α) mω/2 , Δx Δp = /2 .

Since the lower bound of the Heisenberg inequality is attained whatever the value of α, the X or P representation of |α is a Gaussian function of X or P. We will check it explicitly in the following. We remark that x and  p can be as large as one wants if we increase |α| whereas Δx and Δp remain constant (and of course compatible with the uncertainty relations). As for the energy, the position and momentum become well defined in relative value as |α| becomes large.

8.8 Problem. Quasi-Classical States of the Harmonic Oscillator

207

4. Time evolution: we start with |ψ(0) = |α, hence: |ψ(t) = e−|α|

2

/2

∞  αn −i(n+1/2)ωt |α = e−iωt/2 |αe−iωt . e n ! n=0

Therefore |ψ(t) is an eigenstate of aˆ with the eigenvalue β = αe−iωt . From the result of question (3), we obtain: √ X t = (αe−iωt + α∗ eiωt )/ 2

√ Pt = i(α∗ eiωt − αe−iωt )/ 2 .

Setting α = α0 eiϕ with α0 > 0, we obtain:

2 cos(ωt − ϕ) = x0 cos(ωt − ϕ) mω √  pt = −α0 2mω sin(ωt − ϕ) = − p0 sin(ωt − ϕ) xt = α0

with, naturally, Δxt Δpt = /2. The time evolution of xt and  pt is the same as for a classical oscillator (cf. Exercise 3). 5. We have

 ∂ ∂ p ˆ = −i Pˆ = √ = −i mω ∂x ∂X mω and similarly Xˆ =



√ mω ∂ ∂ xˆ = i mω =i .  ∂p ∂P

In terms of the variable X , we find:   ∂ 1 X+ ψα (X ) = αψα (X ) √ ∂X 2  √ whose solution is ψα (X ) = C exp −(X − α 2)2 /2 . In terms of the variable P, we have:   i ∂ P+ ϕα (P) = αϕα (P) √ ∂P 2  √ and the solution is ϕα (P) = C  exp −(P + iα 2)2 /2 . The wave function is a real Gaussian centered at X  multiplied by a plane wave of wave vector P. This wave function is called a minimal wave packet because the probability distribution is the same √ as for the ground state of the oscillator, except that it is shifted by X 0 = (α 2). In particular the Heisenberg inequality is saturated at any time. The time evolution consists in replacing α by α exp(−iωt).

208

8 Algebra of Observables

The oscillation of the center of the wave function is the same as that of a classical oscillator. The states α are genuine quantum states in the sense that they satisfy all conditions of quantum mechanics. However, the physical properties of an oscillator prepared in the state |α, with |α|  1, are very similar to those of a classical oscillator. Traditionally, these states are called “quasi-classical” or “coherent” states of the harmonic oscillator and they play a key role in the quantum theory of radiation.

8.9 Problem. Benzene and C8 Molecules Consider the states of an electron in a hexagonal molecule C6 formed of 6 equalspaced atoms. The distance between two neighboring atoms is denoted d. We note |ξn , n = 1, . . . , 6 the states localized respectively in the vicinity of the atoms n = 1, . . . , 6. We assume that ξn |ξm  = δn,m . The Hamiltonian Hˆ of this system is defined in the basis {|ξn }, by Hˆ = E 0 Iˆ + Wˆ with: Wˆ |ξn  = −A(|ξn+1  + |ξn−1 ) and A > 0. We use here the cyclic conditions |ξ7  ≡ |ξ1  and |ξ0  ≡ |ξ6 . We note |ψn  and E n , n = 1, . . . , 6 the eigenstates of Wˆ and the corresponding eigenvalues. For simplicity, we chose the origin of energies such that E 0 = 0. ˆ n  = |ξn+1 . We define the rotation operator Rˆ by R|ξ ˆ 1. What are the eigenvalues λk k = 1, . . . , 6 of R? 2. The eigenvector corresponding to λk is noted |φk  = 6p=1 ck, p |ξ p . Write the recursion relation between the coefficients ck, p and determine these coefficients by normalizing |φk . 3. Check that the vectors |φk  form an orthonormal basis of the six-dimensional space under consideration. 4. Check that the same vectors |φk  are eigenvectors of the operator Rˆ −1 = Rˆ † ˆ n  = |ξn−1  and calculate the corresponding eigenvalues. defined by R|ξ ˆ 5. Show that W and Rˆ commute. What conclusions can we draw from that? 6. Express Wˆ in terms of Rˆ and Rˆ −1 . Deduce the eigenstates of Wˆ and the corresponding eigenvalues. Discuss the degeneracies of the energy levels. 7. Consider now a regular 8-center closed chain of atoms (called the cyclooctatetraene molecule). (a) Using a method similar to the preceding one, deduce the energy levels for an electron moving on this chain. Discuss the degeneracies of these levels. (b) At time t = 0 the electron is assumed to be localized on the site n = 1, |ψ(t = 0) = |ξ1 . Calculate the probability p1 (t) to find the electron again on the site n = 1 at a later time t; we set ω = A/.

8.9 Problem. Benzene and C8 Molecules

209

(c) Does there exist a time t  = 0 for which p1 (t) = 1? Explain why. Is the propagation of an electron on the chain periodic? 8. Consider now an electron on a closed chain of N sites, located regularly on a circle with a distance d between two adjacent sites. The states localized in the vicinity of each center n = 1, . . . , N are denoted |ξn . The Hamiltonian is defined as above by Hˆ = E 0 Iˆ + Wˆ with Wˆ |ξn  = −A(|ξn+1  + |ξn−1 ) and A > 0. By extending the argument above, calculate the energy levels and the corresponding eigenstates. What happens in the limit of a chain of infinite length?

8.9.1 Solution 1. One has obviously Rˆ 6 = Iˆ. Therefore, λ6k = 1 and λk = e2ikπ/6 k = 1, . . . , 6. ˆ k  = e2ikπ/6 |φk  gives the recursion relation 2. The definition R|φ e2ikπ/6 ck, p = ck, p−1 .

3. 4. 5.

6.

√ Therefore, we have up to an phase factor ck, p = e−2ikpπ/6 / 6. arbitrary  One has φk |φk   = (1/6) p e−2i(k −k) pπ/6 = δk,k  . A direct calculation gives Rˆ −1 |φk  = e−2ikπ/6 |φk . The eigenvalues of Rˆ −1 = Rˆ † ∗ are λ−1 k = λk . ˆ k  = |ξk  + |ξk+2  = Rˆ Wˆ |ξk . Therefore Wˆ and A direct calculation gives Wˆ R|ξ ˆ which are respectively Hermitian and unitary operators, commute and possess R, a common eigenbasis. We have Wˆ = −A( Rˆ + Rˆ −1 ). The eigenvectors of Wˆ are therefore: 1  −2ikpπ/6 |φk  = √ e |ξ p  6 p=1 6

k = 1, . . . , 6 ,

with eigenvalues E k = −2 A cos(2kπ/6). The ground state E 6 = −2 A is nondegenerate, the levels E 5 = E 1 = −A and E 4 = E 2 = A are twice degenerate, the level E 3 = 2 A is non-degenerate. 7. (a) Using a method similar to the preceding one, we obtain the eight following energy levels: • E 8 = −2 A is the√ground state (non degenerate). • E 7 = E 1 = −A 2 (two-fold degenerate), (two-fold degenerate), • E6 = E2 = 0 √ • E 5 = E 3 = A 2 (two-fold degenerate), • E 4 = 2 A (non degenerate).

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8 Algebra of Observables

√ (b) Using φk |ξ1  = eikπ/4 / 8, we write: 1  ikπ/4 |ψ(t = 0) = |ξ1  = √ e |φk  8 k=1 8

and therefore:

1  ikπ/4 −i Ek t/ |ψ(t) = √ e e |φk  . 8 k=1 8

The probability to find the electron again on the site n = 1 is p1 (t) = |ξ1 |ψ(t)|2 , i.e. p1 (t) = |(1/8) k e−i Ek t/ |2 . This yields, putting ω = A/:  8 2  1   2 √ 1   −i E k t/   p1 (t) =  e  =  1 + 2 cos(ωt 2) + cos(2ωt)  . 8  4 k=1 (c) We find of course p1 (0) = 1. √ To get p1 (t) = 1 at a later time, one should 2) = 1 and cos(2ωt) = 1. This would mean find t  = 0 such that cos(ωt √ Taking the ratio that ωt 2 = 2N π and 2ωt = 2N  π with N and N  integers. √ √ of these two quantities, we see that one should have 2 = N  /N , i.e. 2 rational! Consequently, the particle never reaches again its initial state on the site 1, and the evolution of the state of the system is not periodic. However one can show that the system comes back as close to the initial state as one wishes, if one waits for a long enough time, which is called ergodicity. This type of time evolution is said to be quasi-periodic, it has similarities with quasicristalline spatial structures. Notice that with 2 centers (NH3 molecule), 4 centers, or with Benzene, with 6 centers, this does not happen because the energy levels have rational ratios and the evolution is periodic. Above n = 6 centers, one always observes the same ergodic phenomenon; the values of cos(2K π/n) have always irrational ratios for integer values of k. 8. These results can readily be extended to N centers, as we shall examine in the next problem. The levels are E n = −2 A cos(2nπ/N ) which are all twice degenerate except the ground state n = N and the highest level n = N /2 if N is even. These levels populate an energy band of fixed width 4 A, which becomes a√continuum in the limit N → ∞. The eigenstates are of the form |φn  = (1/ N ) Np=1 e−2inpπ/N |ξ p .

8.10 Problem. Conductibility of Crystals; Band Theory The previous results above on cyclic molecules give an explanation for the electric conductibility of metals. At first sight, this phenomenon is paradoxical. Metals are compact cristalline structures, with atomic distances of the order of an angström,

8.10 Problem. Conductibility of Crystals; Band Theory

211

and, from a mechanical point of view, one would imagine that the mean free path of an electron in a medium of such a large density is extremely small (a few angströms). The quantum result that we want to elicit here is that, contrary to first intuition, an electron in such a periodic structure can propagate freely provided its energy is located inside an energy band that generalizes, in the case of a periodic potential, the notion of energy levels in simple potentials. That free propagation entails the electric conductibility of materials. Electrons in a Periodic Potential Instead of a circular structure as for benzene and octene, we consider the behavior of an electron in a long linear periodic chain of potential wells. We consider a onedimensional problem. We assume there are on the x axis an infinite set of equal-spaced attractive atomic centers. A given center is at a distance xn = n d from the origin (Fig. 8.1). In that set of potentials, we place an electron of mass m. The detailed structure of the potential is of no importance here. Let V (x) be the potential of the center n = 0. In the absence of other centers the hamiltonian of the electron in this potential is hˆ = p 2 /2m + V (x). We assume V (x) has only a single bound state of energy E 0 and wave function ϕ(x) ˆ hϕ(x) = E 0 ϕ(x). This wave function has an extension of the order of λ0 , i.e. it decreases as e−|x|/λ0 (or faster) when |x| → ∞. We want to find the eigenfunctions and eigenvalues of the hamiltonian Hˆ = p 2 /2m +

+∞ 

V (x − xn ).

n=−∞

The method used for the ammonia molecule in Chap. 7 is going to simplify greatly the problem. We will make a matrix model for the hamiltonian similar to what we

Fig. 8.1 Periodic potential seen by an electron in a crystal

212

8 Algebra of Observables

Fig. 8.2 One-dimensional square wells. Wave function φ(x) localized on one of the wells

have seen for the ammonia molecule NH3 . In fact we can approximate the row of atoms to the limit as N → ∞ of a set of N square well potentials of width a and depth V0 (Fig. 8.2). We assume that the distance d between the centers is much larger than the extension λ0 of the wave function ϕ(x) in a single center, d  λ0 . In other words, we place ourselves in conditions such that the tunneling between two neighboring centers is small but non-zero, but that the tunneling between more distant centers is negligible. Basis of Localized States (i) The eigenfunctions of the hamiltonian Hˆ are equally distributed on the different potential wells. In the energy eigenstates, the electron has the same probability to be on all sites. (ii) Appropriate linear combinations of the eigenfunctions localize the electron in a given potential well; we note by ψn the orthonormal combination that localizes the electron in the vicinity of the potential well n. The state |n of an electron localized on the potential well n corresponds to the wave function ψn (x), with n|n   = δn,n  . The periodicity of the problem (invariance of the hamiltonian in the transformation x → x + m d, m integer) implies that ψn (x) = ψ0 (x − xn ) where xn = n d. (iii) Since, by assumption, the tunneling is weak, the functions ψn (x) are close to the atomic wave functions ϕ(x − xn ) that the electron would have on the site n in the absence of the other sites: ψn (x) ∼ ϕ(x − xn ). course, the set {ϕ(x − xn )} is not orthogonal, but, within our assumptions,  Of ϕ∗ (x − xn )ϕ(x − xm )d x is small for m  = n. (iv) The set of states {|n} is an orthonormal basis of the space of the electron’s states, which we call: the basis of localized states (this can be considered as a definition).

8.10 Problem. Conductibility of Crystals; Band Theory

213

It is in that basis that we make a model for the hamiltonian in a very similar manner as what we did in Eq. (7.16). Model Hamiltonian The model hamiltonian in the basis {|n} of localized states is as follows The diagonal elements are simply equal to E 0 , the energy of an electron in a center in the absence of coupling of the centers by quantum tunneling: n| Hˆ |n = E 0 We set equal to −A, where A is a (small) real constant, the elements of the two first parallels to the main diagonal n| Hˆ |n + 1 = n − 1| Hˆ |n = −A. This energy A characterizes the possibility for the electron to jump from site n to the immediate neighboring sites: n + 1 and n − 1. Finally, we assume that all other matrix elements of Hˆ vanish, since the tunnel effect decreases exponentially with the distance. n| Hˆ |n + m = 0, |m| ≥ 2 In the basis {|n} of localized states, the hamiltonian is represented by infinite matrix:    Row Column n − 2 n − 1 n n + 1 n + 2     .... : ... ... .. ... ... ......    n-1  . . . . . . 0 −A E 0 −A 0 0 0 . . . . . .   0 0 . . . . . .  −A E 0 −A n H =  . . . . . . 0 0 (8.57)  n+1 . . . . . . 0 0 −A 0 . . . . . . 0 −A E 0   . . . . . . 0 0 0 0 −A E 0 −A . . . . . . n+2   .... ... ... .. ... ... ......  : Stationary States, Energy Bands 1. We write the stationary states as |Ψk (t) = e

−iE k t 

|Ψk ,

where |Ψk  =

 n

and E k is the corresponding eigenvalue.

Cnk |n,

214

8 Algebra of Observables

Write the eigenvalue problem which determines the eigenvalues E k and the corresponding complex coefficients {Cnk }. 2. Show that these equations are all satisfied if one sets Cnk = eikxn

(xn = n d)

where k is an arbitrary wave number (or a wave vector). 3. What is the eigenvalue E k of the energy corresponding to a given value of k? 4. Why can one restrict the value of k to the interval −π/d ≤ k ≤ π/d ? 5. What is the degeneracy of a given energy level E k ? 6. Show that in a stationary state, the probability |Cnk |2 to find the electron on a given site is the same on all sites. (Since the linear chain is of infinite length, the wave functions Ψk (x) are not normalizable.) 7. Show that the energy E of the electron can take any value in an interval between E 0 − 2 A and E 0 + 2 A, called an energy band of width 4 A. 8. Write the wave function Ψk (x) of an eigenstate, called a Bloch function, in terms of the localized wave functions ψn (x). Make use of the translational invariance to express that in terms of ψ0 (x − xn ). 9. Replacing the variable k by a more physical quantity q = k and writing E(q) instead of E(k), rewrite the form of a stationary wave function and its approximation in terms of the atomic wave functions ϕn (x). Show that the coefficients of the atomic wave functions are the values at x = xn of a monochromatic plane wave Φ(x, t). 10. Interpret the result as the free propagation of a particle of momentum q and energy E(q). 11. Build a wave packet of such free waves. 12. Assuming the energy E(q) is close to the minimum of the energy band, kd = qd/ 1 show that this corresponds to the propagation of a particle of effective mass m eff which is different from the electron mass: m eff = 2 /2 A d 2 .

8.10 Problem. Conductibility of Crystals; Band Theory

215

8.10.1 Solution 1. The stationary states |ψ(t) = e

−iEt 



Cn |n

n

are obtained by solving the eigenvalue problem Hˆ |ψ = E|ψ or



Hnm Cm = E Cn

m

which determines the eigenvalues E and the corresponding complex coefficients {Cn }. This is an infinite set of linear equations, part of which is ⎧ ⎪ = E Cn−1 ⎨−A Cn−2 + E 0 Cn−1 − A Cn −A Cn−1 + E 0 Cn − A Cn+1 = E Cn ⎪ ⎩ −A Cn + E 0 Cn+1 − A Cn+2 = E Cn+1 These equations are all satisfied if one sets Cn = eikxn

(xn = n d)

where k is an arbitrary wave number (or a wave vector) 2. This corresponds to the energy eigenvalue E(k) = E 0 − 2 A cos(k d) 3. The solutions are invariant under the transformation kd → kd + 2mπ (m integer), so one can restrict kd to the interval [−π, +π] or −π/d ≤ k ≤ π/d. 4. E(k) is invariant under k → −k; there is a twofold degeneracy. 5. For a givent value of k, we have |Ψ (t) = e−iE(k)t/ |Ψk 

with |Ψk  =



eikxn |n.

n

In a stationary state, the probability |Cn |2 to find the electron in the vicinity of each center is the same.

216

8 Algebra of Observables

Fig. 8.3 Energy band in an infinite one-dimensional periodic structure

The state |Ψk  is therefore not normalizable, because we have assumed the crystal is infinite. 6. The energy E of the particle can take any value between E 0 − 2 A and E 0 + 2 A. This is called an allowed energy band, of width 4 A (Fig. 8.3). 7. The stationary state |Ψk  is Ψk (x) =



eikxn ψn (x)

n

and since ψn (x) = ψ0 (x − xn ), this can be rewritten as Ψk (x) = eikx



e−ik(x−xn ) ψ0 (x − xn ),

n

called a Bloch function. In this form, the stationary states appear as the product of eikx by a periodic function of x, of period d, invariant under the transformation x → x + md, m integer. This reflects the invariance property of the problem. 8. Setting q = k and writing E(q) instead of E(k), we obtain ψq (x, t) =

 n

e−i(E(q)t−q xn )/ ψn (x) ∼



e−i(E(q)t−q xn )/ ϕn (x)

n

where ϕ(x) is the atomic wave function (in the absence of coupling) and we have set ϕn (x) = ϕ(x − xn ). 9. In this form the stationary wave function appears as a superposition of atomic wave functions of each site, whose coefficients are the values at x = xn , i.e. at the site, of the function

8.10 Problem. Conductibility of Crystals; Band Theory

217

Fig. 8.4 Propagation of a plane wave inside an energy band

Φ(x, t) = e−i(E(q)t−q x)/ . This function Φ is a monochromatic plane wave propagating alongs +O x (q > 0) or −O x (q < 0), which, by itself would represent the free motion in space of a particle of momentum q and energy E(q). The function ψ is a sum of the local, atomic, functions ψn modulated by the enveloping function Φ(x, t) that propagates freely. The real parts of ψq (x, t) and of Φ(x, t) at t = 0 are represented (Fig. 8.4). 10. In order to observe a propagation of the electron we must build a wave packet of the type ψ(x, t) = n γn (t)ψn (x) where  γn (t) =

g(q)e−i(E(q)t−q xn )/ dq/(2π)1/2

where γn (t) is the value at x = xn of the function  Γ (x, t) = g(q)e−i(E(q)t−q x)/ dq/(2π)1/2 . which is a wave packet where the relation between E(q) and q is more complicated than for a free particle in space. 11. Assuming that the wave packet consists in a superposition of states near the minimum of the energy band, i.e. kd = qd/ 1, only the small values of q will contribute and one can expand E(q) as E(q)  E 0 − 2 A + A

d 2q 2 . 2

where the relation between the energy and the momentum is the same as for a free particle of effective mass different from the electron mass

218

8 Algebra of Observables

m eff = 2 /2 A d 2 . In general, for electrons in a good metal conductor, the effective mass is of the same order as the electron mass; m eff /m e is of the order of 1, 6 for Be, 1, 2 for Na, 1, 0 for Cu. That is not at all the case in a semiconductor, for instance m eff /m e = 0, 067 for Gallium Ga As. Note that if we had enriched our model by adding terms −B on the next parallels to the diagonal, the result would be similar E (k) = E 0 − 2 A cos kb − 2B cos 2kb.

Chapter 9

Approximation Methods

In quantum mechanics the number of problems for which there exist analytical solutions is rather restricted. In general one must resort to approximation methods. In this chapter, we present two of these methods: perturbation theory and the variational method. We are mainly interested in applications.

9.1 Perturbation Theory 9.1.1 Definition of the Problem Consider the eigenvalue problem: Hˆ |ψ = W |ψ

(9.1)

where the Hamiltonian Hˆ is the sum of a dominant term Hˆ 0 , whose eigenvalues and eigenstates are known, and a perturbation which we write as λ Hˆ 1 , where λ is a real parameter: Hˆ = Hˆ 0 + λ Hˆ 1 . (9.2) The solution of the eigenvalue problem of Hˆ 0 is: Hˆ 0 |n, r  = E n |n, r ,

r = 1, 2, . . . , pn

(9.3)

where the degeneracy of the eigenvalue E n is pn , and where the pn orthonormal eigenstates |n, r  with r = 1, 2, . . . , pn span the eigensubspace E n . We assume that the term λ Hˆ 1 is sufficiently weak to bring only small perturbations to the spectrum of Hˆ 0 . © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_9

219

220

9 Approximation Methods

Power Expansion of Energies and Eigenstates We assume that the energy levels W of Hˆ vary analytically in λ. Therefore, if λ is small, these levels and the corresponding states will be close to those of the nonperturbed Hamiltonian Hˆ 0 . Perturbation theory consists in expanding |ψ and W in powers of λ: |ψ = |ψ 0  + λ |ψ 1  + λ2 |ψ 2  + . . . W = W (0) + λ W (1) + λ2 W (2) + . . .

(9.4) (9.5)

and in calculating the coefficients of the expansion. The method consists of inserting these expansions in the eigenvalue equation (9.1): 

Hˆ 0 + λ Hˆ 1

 

 |ψ 0  + λ |ψ 1  + . . . =  (0)   W + λ W (1) + . . . |ψ 0  + λ |ψ 1  + . . .

(9.6)

and in identifying each order in powers of λ: Hˆ 0 |ψ 0  = W (0) |ψ 0  Hˆ 0 |ψ  + Hˆ 1 |ψ 0  = W (0) |ψ 1  + W (1) |ψ 0  Hˆ 0 |ψ 2  + Hˆ 1 |ψ 1  = W (0) |ψ 2  + W (1) |ψ 1  + W (2) |ψ 0  ... = ... 1

(9.7) (9.8) (9.9)

We have to take into account the normalization condition:   1 = ψ|ψ = ψ 0 |ψ 0  + λ ψ 0 |ψ 1  + ψ 1 |ψ 0  + . . .

(9.10)

which yields: ψ 0 |ψ 0  = 1 Reψ 0 |ψ 1  = 0 ... = 0

(9.11) (9.12)

Since the description of the perturbed state is done in the same Hilbert space as for the unperturbed state, each term |ψ i  can be expanded in the original eigenbasis of H0 : pn  i |ψ i  = γn,r |n, r  (9.13) n

r=1

9.1 Perturbation Theory

221

The series of Eqs. (9.7)–(9.11) provides recursion relations for calculating all terms |ψ i  and W (i) . At a given order, we get the corresponding approximation to the exact solution. We note that Eq. (9.7) implies that |ψ 0  is an eigenvector of H0 , and that W (0) is an eigenvalue of H0 . Therefore, to lowest order: W (0) = E n

(9.14)

and |ψ 0  is a vector of the corresponding eigensubspace En .

9.1.2 First Order Perturbation Theory First Order Perturbation in the Non Degenerate Case If the level E n is not degenerate, we simply note |n the corresponding eigenvector. The solution to first order is particularly simple. Let |ψn  = |ψn0  + λ|ψn1  + . . . be the perturbed state and Wn = Wn(0) + λ Wn(1) + . . . the corresponding energy level. Equation (9.7) then implies: Wn(0) = E n ,

|ψn0  = |n

(9.15)

i.e. the perturbed state and energy level are close to the unperturbed ones. We take the scalar product of Eq. (9.8) with the vector n|. Taking into account (9.15) and the fact that n| Hˆ 0 = E n n|, we obtain, by setting ΔE n(1) = λ Wn(1) , ΔE n(1) = n|λ Hˆ 1 |n.

(9.16)

To first order, the energy shift ΔE n of the level E n is equal to the expectation value of the perturbing Hamiltonian in the unperturbed state |n. First Order Perturbation in the Degenerate Case Suppose that the level E n of Hˆ 0 has a pn -fold degeneracy. We note |n, r , r = 1, . . . , pn an orthonormal basis of the corresponding eigensubspace. In general the perturbation λ Hˆ 1 will lift the degeneracy and the level E n will be split in pn sub(1) , q = 1, . . . , pn . We denote |ψn,q  the corresponding eigenstates levels E n + λWn,q 0 and |ψn,q  the zeroth order in λ of each of these eigenstates. 0 0 Each |ψn,q  belongs to the eigensubspace En . There is no reason why |ψn,q  should coincide with one of the basis vectors |n, r , since these can been chosen arbitrarily. It is rather a linear combination of them, and we have in general: 0  |ψn,q

=

pn 

Cq,r |n, r 

r=1

where we want to determine the coefficients Cq,r .

(9.17)

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9 Approximation Methods

We multiply (9.8) on the left by n, r  | and we obtain: pn  (1) n, r  |λ Hˆ 1 |n, r  Cq,r = λ Wn,q Cq,r  .

(9.18)

r=1

For each value of q, this is nothing but the eigenvalue problem for the pn × pn (1) (1) matrix n, r  |λ Hˆ 1 |n, r . The pn shifts ΔE n,q = λWn,q of the level E n are given by 1 the solutions of the so-called secular equation :    n, 1|λ Hˆ 1 |n, 1 − ΔE ... n, 1|λ Hˆ 1 |n, pn     .. ..  = 0.  . n, r |λ Hˆ 1 |n, r  − ΔE .    n, p |λ Hˆ |n, 1 ... n, pn |λ Hˆ 1 |n, pn  − ΔE  n 1 We also obtain the Cq,r and therefore the eigenstates to zeroth order in λ corresponding to these eigenvalues. Summary

In all cases, degenerate or not, the first order energy shift to a level E n is obtained by diagonalizing the restriction of the perturbing Hamiltonian to the corresponding subspace. First Order Perturbation to the Eigenstates Consider the non degenerate case. Using (9.8) and taking the scalar product with the eigenstate |k for k  = n, we obtain: (E n − E k ) k|ψn1  = k| Hˆ 1 |n. Therefore we can write |ψn1  as: |ψn1  = |n n|ψn1  +

 k| Hˆ 1 |n k =n

En − Ek

|k.

(9.19)

  of phase |ψn  → Equation (9.11) implies Re n|ψn1  = 0. By making the change  eiλα |ψn  in (9.4) we can choose α such that Im n|ψn1  = 0 without loss of

1 Perturbation theory was first used in celestial mechanics by Laplace and Lagrange. The initial purpose was to calculate the long term (secular) perturbations of the motions of planets around the sun (dominant term), due to the gravitational field of the other planets (perturbation). Poisson and Cauchy showed that the problem was basically an eigenvalue problem (6 × 6 matrices for Saturn, up to 8 × 8 for Neptune).

9.1 Perturbation Theory

223

generality. The first order perturbation |ψn1  to the state vector is then completely determined as: |ψn1  =

 k| Hˆ 1 |n k =n

En − Ek

|k.

(9.20)

9.1.3 Second Order Perturbation to the Energy Levels We consider the non-degenerate case for simplicity. Using the above result for the first order perturbation to the state vector, and taking the scalar product of Eq. (9.9) with the eigenstate |n, one obtains the second order correction to the energy of the eigenstates:  |k| Hˆ 1 |n|2 . (9.21) ΔE n(2) = λ2 Wn(2) = λ2 En − Ek k =n Example: Anharmonic Potential Consider a harmonic potential which is perturbed by a quartic potential: pˆ 2 1 + mω 2 xˆ 2 , Hˆ 0 = 2m 2

λ Hˆ 1 = λ

m 2 ω3 4 xˆ . 

(9.22)

where λ is a dimensionless real parameter. Using the expression of xˆ in terms of annihilation and creation operators, we find the shift of the energy level E n = (n + 1/2) ω at first order in λ: ΔE n(1) = λ

m 2 ω3 3λ n|xˆ 4 |n = ω (2n 2 + 2n + 1).  4

(9.23)

Remarks on the Convergence of Perturbation Theory In using the expansions (9.4) and (9.5), we implicitly assumed that the solution could be expanded in a power series in λ, therefore that it is analytic in the vicinity of λ = 0 and that the series converges for λ sufficiently small. The case of the anharmonic potential is somewhat pathological in the sense that one can prove that the power series expansion in λ never converges: the series has a vanishing radius of convergence! Nevertheless, the result (9.23) is a good approximation as long as the correction to the unperturbed term (n + 1/2)ω is small. For a fixed value of λ (small compared to unity) this will only occur for values of n smaller than some value n max (λ) since the correction increases as n 2 . This can be understood physically since:

224

9 Approximation Methods

• The term proportional to x 4 is only small if the extension of the wave function is not too large; it becomes dominant as soon as x 2  is large. • For λ ≥ 0 the potential mω 2 xˆ 2 /2 + λ Hˆ 1 has bound states, while for λ < 0 (even arbitrarily small) the force becomes repulsive for x sufficiently large. The Hamiltonian is no longer bounded from below and there are no bound states. Therefore when one crosses the value λ = 0, the physical nature of the problem changes dramatically. This is reflected in the mathematical properties of the solution; there is a singularity at λ = 0 and the power series expansion around the origin has a vanishing radius of convergence. A well-known example of a series which does not converge, but whose first terms give an excellent approximation of the exact answer is Stirling’s formula, used to approximate Euler’s Gamma function:  Γ (x) =

2π  x x 1 1 + . . . . + 1+ x e 12 x 288 x 2

This is called an asymptotic series, which is safely used in computers, although it is not convergent.

9.2 The Variational Method The variational method, which is very convenient for estimating the approximate value of energy levels (mostly the ground state) and which is frequently used in quantum chemistry. The Ground State The first use of the variational method is to derive an upper bound on the ground state energy of a quantum system. It is based on the following theorem: Let |ψ be any normalized state; the expectation value of a Hamiltonian Hˆ in this state is always larger than or equal to the ground state energy E 0 of this Hamiltonian: ψ| Hˆ |ψ ≥ E 0

f or any |ψ.

(9.24)

To prove this result, we expand |ψ on an eigenbasis of Hˆ : |ψ =

 n

Cn |n ,



Cn Cn∗ = 1

n

with Hˆ |n = E n |n and by definition E 0 ≤ E n . Calculating ψ| Hˆ |ψ − E 0 , we obtain:

9.2 The Variational Method

ψ| Hˆ |ψ − E 0 =



225

E n Cn Cn∗ − E 0

n



Cn Cn∗ =

 (E n − E 0 )|Cn |2 ≥ 0

n

n

which proves (9.24). Alternatively, one may simply observe that if the spectrum of an operator is bounded from below, the expectation value of this operator is necessarily greater than or equal to the lower bound of the spectrum. In practice this result is used in the following way. We choose a state |ψ which depends on some parameters and we calculate E in this state. The minimum value we find by varying the parameters gives an approximation for the ground state energy, which is furthermore an upper bound for this energy level. Example Consider the harmonic oscillator Hˆ = pˆ 2 /2m + mω 2 xˆ 2 /2 and the normalized test function:  2a 3 1 . ψa (x) = π x 2 + a2 In this case there is a single variational parameter a and we obtain: E(a) = ψa | Hˆ |ψa  =



 ψa (x)

2 d 2 1 − + m ω2 x 2 2m d x 2 2

ψa (x) d x.

We can compute E(a) by using: 

+∞ −∞

dx π = x 2 + a2 a

and its derivatives with respect to a. We obtain: 2 1 + mω 2 a 2 4ma 2 2 √ which is minimum for a 2 = /(mω 2), hence: E(a) =

ω E min = √ . 2 This gives an upper bound to the exact result ω/2. The difference between the exact result and the value derived from the variational method could be further reduced by choosing more elaborate test functions with several variational parameters. Had we chosen Gaussian functions as the set of test functions, we would have obtained the exact result of course, since the true ground state of Hˆ would have been an element of this set.

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9 Approximation Methods

Relation with Perturbation Theory The following result is useful.

The first order of perturbation theory is an upper bound for the ground state energy. Indeed, to first order perturbation theory, the ground state energy is: W0 = ψ0 | ( Hˆ 0 + λ Hˆ 1 ) |ψ0  where |ψ0  is the ground state wave function of H0 . Because of theorem (9.24), W0 is an upper bound to the ground state energy of H0 + λH1 . Other Levels One can generalize the variational method to other states by using the following theorem:

The function |ψ −→ E ψ =

ψ| Hˆ |ψ ψ|ψ

is stationary in |ψ if and only if |ψ is an eigenstate of Hˆ . To prove this result we consider a variation |δψ of |ψ, i.e. |ψ → |ψ + |δψ. Expanding the above formula to first order, we find: ψ|ψ δ E ψ = δψ|( Hˆ − E ψ )|ψ + ψ|( Hˆ − E ψ )|δψ. If |ψ is an eigenstate of Hˆ with eigenvalue E, then E ψ = E and ( Hˆ − E ψ )|ψ = 0. Consequently δ E ψ = 0 whatever the infinitesimal variation |δψ. Conversely, if δ E ψ = 0 whatever the variation |δψ, we must have: δψ|( Hˆ − E ψ )| ψ + ψ|( Hˆ − E ψ )|δψ = 0. This must happen in particular if we make the choice: |δψ = η ( Hˆ − E ψ )|ψ, where η is an infinitesimal number. Inserting this in the above formula, we obtain: ψ|( Hˆ − E ψ )2 |ψ = 0. The norm of the vector ( Hˆ − E ψ )|ψ vanishes, therefore: ( Hˆ − E ψ )|ψ = 0. This means that |ψ is an eigenvector of Hˆ with the eigenvalue E ψ .

9.2 The Variational Method

227

In practice, we can use this result in the following way. We choose a set of wave functions (or state vectors) which depend on a set of parameters which we call α collectively. We calculate the expectation value of the energy E(α) for these wave functions. All the extrema of E(α) with respect to the variations of α will be approximations to the energy levels. Of course, these extrema will not be in general exact solutions, since the choice of test wave functions does not cover the entire Hilbert space.

9.3 Exercises 1. Calculations of energy levels Consider a particle of mass m placed in the isotropic 3D potential V (r ) ∝ r β . We choose the normalized Gaussian test function: ψa (r) = (a/π)3/4 exp(−ar 2 /2).

(9.25)

We find in this state:  p2  =

3 2 a 2

r β  = a −β/2

Γ (3/2 + β/2) Γ (3/2)

This gives an upper bound for the ground state of: • The harmonic potential (β = 2) for which we recover the exact result. • The Coulomb potential V (r ) = −e2 /r ; one finds: E0 = −

4 me4 1 me4 to be compared with the exact result: − . 2 3π  2 2

• The linear potential V (r ) = gr ; one finds: E0 =

81 2π

1/3

2 g 2 2m



1/3

2.345

2 g 2 2m

1/3

to be compared with the coefficient 2.338 of the exact result. 2. Uncertainty relations Using the inequality (9.24) for systems whose ground state is known, we can derive uncertainty relations between  p 2  and r α , where α is a given exponent. a. The r 2   p 2  uncertainty relation. Consider a one-dimensional harmonic oscillator, whose ground state is ω/2. Show that the inequality  p2 x 2  ≥ 2 /4 holds whatever the state |ψ. Extend the result to three dimensions.

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9 Approximation Methods

b. The 1/r   p 2  uncertainty relation. Consider the hydrogen atom Hamiltonian H = pˆ 2 /2m − e2 /ˆr and its ground state energy which is E 0 = −me4 /(22 ) (see Chap. 11). Show that  p 2  ≥ 2  r1 2 for all |ψ. 3. Comparison of the ground states of two potentials Consider two potentials V1 (r) and V2 (r) such that V1 (r) < V2 (r) in any point r. Show that the energy of the ground state of a particle moving in the potential V1 is always lower than the energy of the ground state for the particle moving in V2 . 4. Existence of a bound state in a potential well Consider a particle moving at one dimension in a potential V (x) which tends to zero in ±∞ and which is such that V (x) ≤ 0 for all x. Show that there is always at least one bound state for this motion. Is this result still valid in three dimensions? 5. Generalized Heisenberg inequalities Consider the Hamiltonian Hˆ = p 2 /2m + gr α where g and α have the same sign and where α > −2. The energy levels E n of Hˆ can be derived from the eigenvalues εn of the operator (−Δρ + ηρα ) (where ρ is a dimensionless variable and where η = |α|/α) by the scaling law: E n = εn |g|

2/(α+2)

2 2m

α/(α+2) ,

1/(α+2)  . as one can check directly by making the scaling r = ρ 2 /(2m|g|) Show, using the variational method, that the following general relation holds: α 2/α

 p  r  2

≥ κ

2

with

κ = |α| 2

2/α

|ε0 | α+2

(α+2)/α

where ε0 is the smallest eigenvalue of the operator −Δρ + ηρα .

,

Chapter 10

Angular Momentum

The rotations of systems play a fundamental role in physics, be it in atomic spectra, in magnetic resonance imaging as well as in the stabilization of space crafts. The conservation laws of angular momentum play a central role, as important as energy conservation. We make use of this fact when we study the hydrogen atom. The quantization of angular momentum displays a special feature in that it is universal. Contrary to energy levels, which depend on the system under consideration, the possible discrete values of the angular momentum must be chosen in a given universal catalogue which we establish. They depend on the system under consideration, but they are “ready to wear” and not “made-to-measure.” The reason is that Planck’s constant has the dimension of an angular momentum, and that  is the natural unit of angular momenta. Hence the quantization involves dimensionless factors. The field of applications of our results is very broad. • We use these results in order to study the hydrogen atom or the spin of the electron and other particles. • The rotation spectra of molecules play an important role, both in chemistry and in astrophysics. • The angular distributions of final particles in collisions or decays in nuclear and elementary particle physics allow us to determine the structure of particles and the nature of fundamental interactions. • Last, and not least, all of magnetism, on which we base the experimental analysis of our results, comes from rotating charges. We will discover that there exist angular momenta that do not have any classical analogues, in particular spin 1/2 to which we devote Chap. 12 and which is one of the most revolutionary discoveries of the 1920s.

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_10

229

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10 Angular Momentum

10.1 Fundamental Commutation Relation 10.1.1 Classical Angular Momentum Our starting point is the correspondence principle. The angular momentum L of a particle of momentum p and position r with respect to the origin is classically the vector product L = r × p. Quantum mechanically we therefore postulate that the vector observable Lˆ (i.e., the set of three observables) corresponding to this angular momentum is Lˆ = rˆ × pˆ , or, in the wave function formalism,  ˆ Lˆ = rˆ × ∇. i However, as said previously, more than this particular representation, it is the commutation relations of these three observables Lˆ x , Lˆ y , and Lˆ z , that are of interest because they are independent of the representation. These algebraic relations are, as pointed out in (8.2): [Lˆ x , Lˆ y ] = iLˆ z ,

(10.1)

and two other relations obtained by cyclic permutations. These can be put together in the compact form ˆ Lˆ × Lˆ = iL. (10.2)

10.1.2 Definition of an Angular Momentum Observable However we shall discover that there exist in nature angular momenta which do not have a classical analog. Therefore, we take as a general definition of any angular momentum observable Jˆ the algebraic relation between its coordinates ˆ or [Jˆx , Jˆy ] = iJˆz . Jˆ × Jˆ = iJ,

(10.3)

By definition, any triplet of observables corresponding to the components of a vector quantity that satisfy this relation is an angular momentum observable. We keep the letter L in the specific case of orbital angular momenta. We first show how one can obtain numbers (i.e., eigenvalues) by manipulating this algebra. This was done by Heisenberg and Jordan in 1925 (but they missed the most interesting part of their results). This method is quite similar to Dirac’s method for the harmonic oscillator (Chap. 8 Sect. 8.4). The results are obtained in a much simpler way than with wave functions.

10.1 Fundamental Commutation Relation

231

10.1.3 Results of the Quantization The results we obtain are as follows. Consider a vector observable Jˆ that has the algebraic properties (10.3) and the corresponding physical quantity. It is an experimental result that 1. Whatever the system under consideration, a measurement of the square of the angular momentum (10.4) Jˆ 2 = Jˆx2 + Jˆy2 + Jˆz2 gives one of the values j(j + 1)2 , where 2j is an integer. 2. After finding this value for the square of the angular momentum, the measurement of one of the components of J along any axis gives one of the (2j + 1) values m, where m ∈ {j, j − 1, . . . , −j}. 3. In the case of orbital angular momenta (which have a classical analogue) the numbers j and m are integers. Actually, the algebra (10.3) is very important in mathematics. It is called the Lie algebra of the group of rotations in three dimensions. And these results were proven as soon as 1913 by Elie Cartan in his classification of Lie groups. Why the physicists of the 1920 s didn’t know it is a mystery! Elie Cartan did not forget to point out their ignorance in his book Lectures on the Theory of Spinors in 1935. How did Cartan proceed, inasmuch as he did not know the Planck constant ? Because  plays no role in this game.  is the natural unit of angular momenta. In (10.3), if we divide both sides by 2 we obtain an algebraic relation between ˆ dimensionless operators. Let Kˆ = J/, we obtain [Kˆ x , Kˆ y ] = iKˆ z

ˆ or Kˆ × Kˆ = iK.

10.2 Proof of the Quantization 10.2.1 Statement of the Problem Elie Cartan’s proof is crystal clear. The problem is to find all matrices that satisfy the relation (10.3) and to calculate their eigenvalues and eigenvectors. 1. Jˆx , Jˆy , and Jˆz do not commute. We can only diagonalize one of them. However, they obviously have the same eigenvalues.

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10 Angular Momentum

2. Looking at the physics of the problem enables us to understand a methodology that the mathematicians Sophus Lie and Elie Cartan knew at the end of the 19th century. The square of the angular momentum Jˆ 2 commutes with the three components [Jˆ 2 , Jˆx ] = [Jˆ 2 , Jˆy ] = [Jˆ 2 , Jˆz ] = 0. This is a simple and direct calculation. It is easy to understand because J 2 does not depend on the system of axes. It is rotation invariant and it makes no discrimination among the components. Therefore, we can diagonalize simultaneously Jˆ 2 and any one of the components ˆ of J, for instance, Jˆz . Note that Jˆ 2 and one of the components is the maximal set of operators that are functions of only Jˆx , Jˆy , and Jˆz , which can be diagonalized simultaneously. Jˆ 2 and Jˆz form a CSCO if one considers only angular momentum. Our problem is to find the eigenvectors common to Jˆ 2 and Jˆz and the corresponding eigenvalues. Analytic solution At this point, we could use wave functions. Because L is invariant under dilatation, it acts only on angular variables. The calculation of eigenfunctions and eigenvalues had been done in the 19th century by Legendre and Fourier. If we work in spherical variables with Oz as the polar axis, r is the radial coordinate, θ the colatitude, and φ the azimuth. The expression of Lˆ 2 and Lˆ z is not too complicated  ∂ , Lˆ z = i ∂ϕ Lˆ 2 = −2



∂ 1 ∂2 1 ∂ sin θ + sin θ ∂θ ∂θ sin2 θ ∂ϕ2

(10.5)  .

(10.6)

This is what Legendre and Fourier did. We look for eigenfunctions Ym (θ, ϕ) common to Lˆ 2 and Lˆ z . These are called the spherical harmonics: Lˆ 2 Ym = λ2 2 Ym , and Lˆ z Ym = μYm , where λ2 and μ are the eigenvalues. However, this resolution of differential equations is tedious and, more important, we would miss half of the results. The Ym will be given below. Cartan’s algebraic relation is much more elegant.

10.2 Proof of the Quantization

233

10.2.2 Vectors |j, m > and Eigenvalues j and m We look for the eigenvectors common to Jˆ 2 and Jˆz , and the corresponding eigenvalues. We define the vectors |j, m > and the numbers j and m as Jˆ 2 |j, m = j(j + 1)2 |j, m Jˆz |j, m = m|j, m

(10.7) (10.8)

with j ≥ 0 (the eigenvalues of Jˆ 2 are positive because ψ|Jˆ 2 |ψ ≥ 0). For the moment, no other constraint exists on the possible values of j and m. We assume these vectors are orthonormal j, m|j , m  = δj,j δm,m . This eigenbasis is unique, because Jˆ 2 and Jˆz form a CSCO. There are two indices because two operators are diagonalized simultaneously. The method consists of first finding the eigenvalues of Jˆ 2 , then for a given value of Jˆ 2 , we search the eigenvalues of Jˆz in the corresponding eigensubspace. This is geometrically intuitive. We fix the norm |j| of a vector, and we seek the values of its components. Classically, all values between j and −j are allowed; quantum mechanically, there is only a finite discrete set of values for the projection.

10.2.3 Operators Jˆ ± = Jˆ x ± iJˆ y We now follow Elie Cartan. The technique is similar to Dirac’s creation and annihilation operators. i. Consider the two operators Jˆ+ and Jˆ− : Jˆ+ = Jˆx + iJˆy , and Jˆ− = Jˆx − iJˆy .

(10.9)

Jˆ+ and Jˆ− are Hermitian conjugates of each other Jˆ+† = Jˆ− , Jˆ−† = Jˆ+ . We show that these operators enable us to move from one vector to another in the eigensubspace of Jˆ 2 by increasing by one unit of  the eigenvalues of Jˆz . ii. Commutation relation of Jˆ± with Jˆ 2 and Jˆz Because Jˆ± are linear combinations of Jˆx and Jˆy , which commute with Jˆ 2 , they commute with Jˆ 2 [Jˆ 2 , Jˆ± ] = 0. (10.10) However, Jˆ± do not commute with Jˆz . In fact, using the relations (10.3): [Jˆz , Jˆ± ] = [Jˆz , Jˆx ] ± i[Jˆz , Jˆy ] = iJˆy ± i(−iJˆx )

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10 Angular Momentum

therefore

[Jˆz , Jˆ± ] = ±Jˆ± .

(10.11)

iii. The states Jˆ± |j, m and their normalization We apply these commutation relations to a vector |j, m >. We see that Jˆ 2 Jˆ± |j, m = Jˆ± Jˆ 2 |j, m = j(j + 1)2 Jˆ± |j, m Jˆz Jˆ± |j, m = (Jˆ± Jˆz ± Jˆ± )|j, m = (m ± 1)Jˆ± |j, m.

(10.12) (10.13)

Therefore the vectors Jˆ± |j, m are either eigenvectors of Jˆ 2 and Jˆz with eigenvalues j(j + 1)2 and (m ± 1), or equal to the null vector (Jˆ± |j, m = 0) if j(j + 1)2 and (m ± 1) are not eigenvalues of Jˆ 2 and Jˆz . Thus, the operators Jˆ± can be repeatedly applied to any eigenvector and this increases or decreases the value of m by any integer in the eigensubspace of Jˆ 2 . They allow us to move around in this subspace. However, as our physical intuition tells us, to a given value of j, the projection m is bounded on both sides: −j ≤ m ≤ j. In fact, the square of the norm of the vector Jˆ± |j, m is Jˆ± |j, m2 = j, m|Jˆ±† Jˆ± |j, m = j, m|Jˆ∓ Jˆ± |j, m. However Jˆ∓ Jˆ± = (Jˆx ∓ iJˆy )(Jˆx ± iJˆy ) = Jˆx2 + Jˆy2 ± i[Jˆx , Jˆy ] = Jˆ 2 − Jˆz2 ∓ Jˆz . Therefore, owing to (10.7), we have: Jˆ± |j, m2 = j(j + 1)2 − m2 2 ∓ m2 = (j(j + 1) − m(m ± 1))2 ≥ 0. (10.14) We therefore conclude that: 1. Because the square of the norm of a vector is positive, we necessarily have the inequalities: −j ≤ m ≤ j. (10.15) 2. If m + 1 ≤ j (or m − 1 ≥ −j), the vector Jˆ± |j, m is nonzero and it is proportional to the vector |j, m ± 1. The proportionality coefficient is deduced from the norm calculated above: Jˆ± |j, m =



j(j + 1) − m(m ± 1)  |j, m ± 1

(we make an implicit choice for the phase).

(10.16)

10.2 Proof of the Quantization

235

3. If the eigenvalues m = j and m = −j exist, we have Jˆ+ |j, j = 0, Jˆ− |j, −j = 0.

10.2.4 Quantization Consider the maximum value of m, mmax . This means that mmax + 1 is not an eigenvalue. Therefore the vector Jˆ+ |j, mmax  is the null vector, and its norm is zero. Therefore, according to (10.14), mmax = j and the vector |j, j exists. If we apply repeatedly the operator Jˆ− to this vector |j, j, we generate a whole series of eigenvectors of Jˆz corresponding to the eigenvalues (j − 1), (j − 2), and so on. However, there exists a minimum value mmin such that mmin − 1 is not an eigenvalue and the vector Jˆ− |j, mmin  is the null vector. Therefore, because of (10.14), mmin = −j. Consequently, in the repeated application of Jˆ− to the vector |j, j, there exists an integer N such that j − N = −j. In other words, the eigenvalues of the square of the angular momentum (10.7) are such that 2j is an integer j = N/2. (10.17) For a given value of j = N/2, the corresponding eigensubspace is of dimension 2j + 1 = N + 1. The eigenvalues of Jz corresponding to the set of the N + 1 values m ∈ {−j, −j + 1, . . . , j − 1, j}

(10.18)

are either integers or half integers according to the value of j. There we are! We have found the eigenvalues, that is, the catalogue for which we ˆ the eigenvalues of the were looking. If Jˆ is an observable such that Jˆ × Jˆ = iJ, 2 2 2 2 2 ˆ ˆ ˆ ˆ observable J = Jx + Jy + Jz are of the form j(j + 1) , where j is either an integer or a half integer, positive or zero. The eigenvalues of the observable Jˆz are of the form m, where m is an integer or a half-integer. For a system in an eigenstate of Jˆ 2 corresponding to the value j, the only possible values of m are the 2j + 1 numbers {−j, −j + 1, . . . , j − 1, j}.

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10 Angular Momentum

Construction of the States |j, m This algebraic analysis allows us to construct the eigenstates |j, m for a given j, starting from (10.16). At this stage, we are on a formal level. Concrete applications come later. The state |j, j satisfies Jˆ+ |j, j = 0, which defines it, as we show. The states |j, m, m = j − n, are obtained by using (10.16) and by repeatedly applying the operator Jˆ− : |j, j − n = γn (Jˆ− )n |j, j, where the γn are calculated with (10.16) and the normalization of the |j, m.

10.3 Orbital Angular Momenta Consider now the orbital angular momentum of a particle with respect to the origin Lˆ = rˆ × pˆ , which we mentioned in the beginning. After the results of the previous section, if ( + 1)2 are the eigenvalues of Lˆ 2 ( ≥ 0) and m the eigenvalues of Lˆ z , then 2 and 2m are integers. However, in this case  and m are integers.

10.3.1 Formulae in Spherical Coordinates In addition to the formulae (10.5) and (10.6) above, a useful formula for our purpose in spherical coordinates is the following. The operators Lˆ ± have the form:   ˆL± = Lˆ x ± iLˆ y = e±iϕ ± ∂ + i cot θ ∂ . ∂θ ∂ϕ

(10.19)

10.3.2 Integer Values of m and  The state of a particle in space can be described by a wave function ψ(r) = ψ(x, y, z). In spherical coordinates, this wave function becomes a function (r, θ, ϕ). The operator Lˆ z has the very simple form  ∂ . Lˆ z = i ∂ϕ

(10.20)

10.3 Orbital Angular Momenta

237

Consider an eigenstate of the projection on the z-axis of the angular momentum of the particle with the eigenvalues m. The corresponding wave function ψm (r) satisfies Lˆ z ψm (r) = mψm (r). The form (10.20) of Lˆ z gives us the simple ϕ dependence of the wave function ψm (r) = Φm (r, θ)eimϕ , where Φm (r, θ) is arbitrary at this point. The function ψm (r) is a particular case of a wave function ψ(x, y, z). In the change ϕ → ϕ + 2π, x, y and z do not change and the function ψm is unchanged. It must therefore be a periodic function of ϕ with period 2π. Therefore eimϕ = eim(ϕ+2π) ⇒ ei2πm = 1. Therefore, in the case of an orbital angular momentum, m must be an integer. In the above analysis, we have seen that m and j differ by an integer. Therefore, for an orbital angular momentum the value of  is a nonnegative integer.

10.3.3 Spherical Harmonics The eigenfunctions common to Lˆ 2 and Lˆ z , denoted Ym (θ, ϕ), are called the spherical harmonics. The spherical harmonics associated with the eigenvalues ( + 1)2 and m satisfy: Lˆ 2 Ym (θ, ϕ) = ( + 1)2 Ym (θ, ϕ) Lˆ z Ym (θ, ϕ) = m Ym (θ, ϕ).

(10.21) (10.22)

We have just seen that their ϕ dependence is simply (eimϕ ), so that they factorize as Ym (θ, ϕ) = F,m (θ) eimϕ . The spherical harmonics form a Hilbert basis of square integrable functions on the sphere of radius one. They are completely defined in the following way. 1. They are normalized:  

 m ∗  Y (θ, ϕ) Ym (θ, ϕ) sin θ dθ dϕ = δ, δm,m .

2. Their phases are such that the recursion relation (10.16) which we repeat below, is satisfied and that Y0 (0, 0) is real and positive (this is a convention):

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10 Angular Momentum

Lˆ ± Ym (θ, ϕ) = 3. Starting from the relation

we obtain with (10.19)



( + 1) − m(m ± 1)  Ym±1 (θ, ϕ). Lˆ + Y (θ, ϕ) = 0

(10.23)

(10.24)

Y (θ, ϕ) = C(sin θ) eiϕ ,

where the normalization constant C is determined by the above constraints. Examples of Spherical Harmonics The spherical harmonics play an important role in atomic and molecular physics. They form with their linear combinations the atomic orbitals of one-external electron atoms, in particular atomic hydrogen which we describe in the next Chapter. The first are: =0 =1

Fig. 10.1 Graph as a function of the polar angle θ of |Ym (θ, ϕ)|2 = |F,m (θ)|2 for  = 0, 1, 2 and |m| ≤ l

1 Y00 (θ, ϕ) = √ 4π  3 Y11 (θ, ϕ) = − sin θ eiϕ 8π  3 0 cos θ Y1 (θ, ϕ) = 4π  3 Y1−1 (θ, ϕ) = sin θ e−iϕ . 8π

(10.25) (10.26) (10.27) (10.28)

10.3 Orbital Angular Momenta

239

In Fig. 10.1, the squares |Ym (θ, ϕ)|2 = |F,m (θ)|2 of spherical harmonics corresponding to the first values of  and m are represented in polar coordinates in terms of θ. Wave Function of a Particle in an Eigenstate of the Orbital Angular Momentum The wave functions ψ,m (r) of particles in an eigenstate of the orbital angular momentum are therefore of the form ψ,m (r) = R,m (r)Ym (θ, ϕ). The radial dependence of these functions is contained in the radial wave function R,m (r), which can have any form a priori.

10.4 Rotation Energy of a Diatomic Molecule It seems that the first one who understood empirically the quantization of angular momentum was Ehrenfest (in an article published on June 15, 1913) just before Bohr’s article on the hydrogen atom (published in July 1913). Ehrenfest noticed that  had the dimension of an angular momentum, and he postulated the quantization, without giving any proof, in order to explain the variation with temperature of the specific heats of diatomic molecular gases (he found an improvement on the theory of Einstein and Stern). Diatomic Molecule A simple illustration of the quantization of the values of Lˆ 2 is obtained through the rotational energy spectrum of a molecule. Such a spectrum is presented in Fig. 10.2 for the diatomic cesium molecule Cs2 . It has been obtained1 by measuring the frequency of the photons needed to ionize the Cs2 molecules that are formed in a very cold atomic vapor of cesium atoms (temperature ∼100 µK). The data of Fig. 10.2, which represent only a small fraction of the total spectrum, exhibit a series of peaks characteristic of a quantized rotational energy. One can visualize a diatomic molecule formed by two atoms of mass M separated by a distance R as a two-body system bound by a harmonic potential. Classically, if the interatomic distance R is in its equilibrium position, the molecule has a rotational energy: L2 , (10.29) Erot = 2I where I = MR2 /2 is the moment of inertia of the system and L is its angular momentum with respect to its center of gravity. In quantum mechanics, this result transposes into: 1 These data, corresponding to the seventeenth excited vibrational state, are extracted from A. Fioretti

et al., Eur. Phys. J. D 5, 389 (1999).

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10 Angular Momentum

Erot () =

2 ( + 1) , 2I

(10.30)

where the rotational energy is quantized. Formula 10.30 gives a very good account for the series of peaks of Fig. 10.2. The distance between two consecutive peaks increases linearly with the peak index, as expected from: Erot () − Erot ( − 1) =

2 . I

The moment of inertia deduced from this spectrum corresponds to a distance R = 1.3 nm between the two cesium atoms. This distance, much larger than the usual interatomic spacing in diatomic molecules, indicates that the Cs2 dimer is actually prepared in a long-range molecular state. If one investigates the absorption spectrum of the cold molecular gas on a much wider range, one finds several series of lines such as the one of Fig. 10.2. Each series corresponds to a given vibrational state of the molecule. The moments of inertia associated with these series differ slightly from one another: this is a consequence of the variation of the average distance between the two atoms in the various vibrational states of the molecule. The study of rotational excitations of molecules is an important field of research in physics, chemistry, and astrophysics.

Fig. 10.2 Rotational spectrum of cold Cs2 molecules, showing the quantization of Lˆ 2 . This spectrum is obtained by measuring the number of molecular ions produced by a laser beam crossing the assembly of cold molecules, as a function of the laser frequency ν. The height of each peak is proportional to the population of the corresponding rotational level . (Courtesy of Pierre Pillet)

10.5 Interstellar Molecules, the Origin of Life

241

10.5 Interstellar Molecules, the Origin of Life The observation of molecular clouds is of considerable interest because it has led in the last decades to the discovery of more and more interstellar molecules. At first, these were simply unusual astrophysical observations, but quite rapidly they raised the problem of the origin of life. Rotation Spectra of Molecules We have mentioned above the rotation spectra of molecules. Considering a molecule as a rigid rotator with principal axes x, y and z, and corresponding moments of inertia Ix , Iy , and Iz , the energy spectrum comes from the Hamiltonian: Lˆ y2 Lˆ 2 Lˆ 2 + z. Hˆ R = x + 2Ix 2Iy 2Iz If two moments of inertia are equal, for instance, Ix = Iy ≡ I, and the third one is very much smaller, as is the case for a diatomic molecule, the spectrum is particularly simple. The energy levels are  El,m = 2

m2 l(l + 1) − m2 + 2I 2Iz

 .

The energy difference of consecutive levels increases linearly with the angular momentum , 2 (10.31) Erot () − Erot ( − 1) = . I The CO Molecule The CO molecule plays an important role in astrophysics. It emits in the millimeter range. The symmetric molecules such as H2 or O2 have the defect that they do not have an electric dipole moment (such as the molecule NH3 ) and therefore they do not have an electric dipole emission. The nonsymmetric molecule CO does possess a permanent electric dipole moment and it emits intensively in these transitions. This molecule has a size R = 0.1128 nm and the masses MC = 12 amu, MO = 15.99492 amu, which leads to a frequency of the transition  = 2 →  = 1 of 230.54 GHz, or equivalently a series of transitions of frequencies ν = 115.27  with  = 1, 2, . . .. Carbon and oxygen are comparatively abundant in the interstellar medium because these elements are synthesized in most stars. The relative abundance of the CO molecule is 10−5 compared the hydrogen atom, the most abundant element in the universe. In general, its distribution is more dense near the center of galaxies than on the edge. This molecule radiates strongly because it has a permanent electric dipole moment, and it is easy to observe. It is a very useful radioastronomical indicator.

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Carbon monoxide CO has a length R = 0.1128 nm and masses MC = 12 amu, MO = 15.99492 amu, which corresponds to a transition frequency  = 2 →  = 1 of 230.54 GHz. A famous example, which is very rich and was one of the first to be analyzed is the group of three galaxies M81, M82, NGC 3077. This group is located in the constellation Ursa Major at 2.5 Mpc. These three galaxies (and other smaller ones) seem to orbit around each other quietly. M82 has an anomalous shape with a central prominence, perpendicular to its plane. For a long time the galaxy M82 was considered as irregular, and that its nucleus was undergoing an eruption or an explosion. In fact, the visible picture, on the lefthand side of Fig. 10.3, shows the unusual shape of a jellyfish. The radioastronomical observation of carbon monoxide CO, shown on the right, reveals that actually everything is perfectly normal. The central deformation of M82 results from important tidal effects produced on M82 by the presence of the nearby much more massive M81 galaxy. Interstellar Molecules Many interstellar molecules have been identified by now in molecular clouds, which, in turn, generate stars. Figure 10.4 shows one of these natural sources of millimetric emission, a molecular interstellar cloud in the Orion nebula, the closest region of star formation, 1500 light-years from the sun, in the same spiral arm. At present, more than 200 molecules have been detected. Some of them are not observed in laboratories because they are too unstable at room temperature, such as acetylenic nitriles HC3 N, HC5 N, and so on, up to HC11 N. These are linear molecules whose moments of inertia can be calculated quite simply, therefore their spectra can be predicted with (10.31). (These molecules are identified by the quantum theory of angular momentum!). A well-known example is the case of fullerenes, spherical C60 molecules that were identified in 1985 thanks to their spectrum, which is calculable. They were also found in meteorites, and they were synthesized in laboratories. Such molecules generated a major breakthrough in nanotechnologies. The 1996 Nobel prize in chemistry was awarded to Harold Kroto, Robert Curl, and Richard Smalley for the discovery of this new chapter of the chemistry of carbon.

Fig. 10.3 M82 galaxy. Left “Anomalous” aspect in the optical range with the shape of a jellyfish. Right Radio emission at 230.54 GHz of carbon monoxide which shows a perfectly normal distribution in a ring around the galactic nucleus. (Photo credit: John Stawn Ward, Thesis www.jsward. com/publications/thesis.pdf.)

10.5 Interstellar Molecules, the Origin of Life

243

Fig. 10.4 Left Orion nebula. (Photo credit: NASA, C. R. O’Dell and S. K. Wong, http://hubblesite. org/newscenter/newsdesk/archive/releases/1995/45/.) Right Level curves of formaldehyde HCHO in the Trifida nebula, where stars also form. (Photo credit: James E. Brau, University of Oregon http://physics.uoregon.edu/~jimbrau/BrauImNew/Chap18/FG18_21.jpg.)

Fig. 10.5 Molecular spectrum from the Orion nebula in the frequency range 215–235 GHz. (Credit: Craig Kulesa, http://loke.as.arizona.edu/~ckulesa/research/overview.html.)

Coming back to the Orion nebula, the great discovery lies in a large family of organic molecules. Figure 10.5 shows this diversity in the (small) frequency range 213–233 GHz. One observes: • The strong intensity of the 12 CO peak and its isotope 13 CO • Numerous organic molecules such as HCHO, CH3 OH, C2 H5 OH, and glycine, the simplest aminoacid (outside the spectral range of the Figure) • Molecules that are unknown in laboratories such as acetylenic nitriles mentioned above

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10 Angular Momentum

The Origin of Life The observation of such molecules is interesting in many respects. It raises the question of the origin of life. In the spectrum of Orion, in Fig. 10.5, two types of molecules are particularly interesting: HCN and HCHO, because these are very reactive molecules with which one can construct quite easily aminoacids, therefore biological molecules. These organic molecules and the mechanism of their formation obviously raises the question of the place from where life originates. The presence of an aminoacid, glycine, whose formula is NH2 CH2 COOH is quite significant in this respect. It is quite possible that not only amino acids, but DNA can form in extrastellar regions. However, this idea must be compared with the amazing observation made in San Diego, in 1953, by Stanley Miller, who was at that time a young 19-year-old chemistry student. He set up the initial conditions of the terrestrial atmosphere. In a mixture of NH3 , CH4 , H2 O, and H2 , he provoked artificial thunderstorms with electric discharges of 60,000 V. Within a week, he obtained 10 of the 20 amino acids that form living matter! This is a strong argument in favor of the terrestrial beginning of prebiotic chemistry. This latter result has to be put together with measurements made on objects that belong to the solar system, in particular, comets and meteorites. Comets are difficult to observe. They were formed 4.5 billion years ago, at the same time as the sun, in the external regions of the original nebula. Since then, they have spent most of their time in the outer and cold regions of the solar system. Comet nuclei have evolved very little since they were formed. Their chemical composition gives an access to the chemical composition of the solar nebula when it was formed, 4.5 billion years ago. In 1997, the Hale–Bopp comet, which came close to the sun, allowed a considerable improvement of the catalogue of comet molecules. Observations at the Millimetric Radioastronomy Institute (IRAM), and at the Caltech Submillimeter Observatory (CSO) displayed the existence of seven new molecules: sulphur monoxide SO and dioxide SO2 , formic acid HCOOH, formamide NH2 CHO, cyanoacetylene HC3 N, methyl formiate HCOOCH3 , and ethanal CH3 CHO. These observations confirmed the presence of HNCO and OCS, which had been identified one year before in the comet Hyakutake. Two dozen interesting organic molecules have now been identified in comets. The chemical composition of meteorites is easier to analyze. In 1969, 10 amino acids, out of the 20 which form DNA and RNA, were discovered in a meteorite that landed in Murchison, Australia. Most of them were the same as those found by Miller. They were all present in the racemic form (equal amounts of enantiomers) which excludes any terrestrial contamination. It is difficult to avoid relating this observation with the richness of interstellar molecules. Therefore, the hypothesis that life can have an extraterrestrial or an interstellar origin cannot be dismissed. It may be that a conjunction of both processes is even more favorable. For instance, life could originate in the encounter of extraterrestrial amino acids and terrestrial nucleic acids. This field of research, quite close to fundamental research, opens fascinating perspectives.

10.5 Interstellar Molecules, the Origin of Life

245

Observations of the Rosetta Probe on Churyumov Gerasimenko In October 2015, ESA s Rosetta spacecraft has made the first in situ detection of oxygen molecules outgassing from a comet, a surprising observation that suggests they were incorporated into the comet during its formation, more than 4 billion years ago. Rosetta has been studying Comet 67P for over a year and has detected an abundance of different gases pouring from its nucleus. Water vapour, carbon monoxide and carbon dioxide are the most prolific, with a rich array of other nitrogen-, sulphurand carbon-bearing species, and even noble gases also recorded. Oxygen is the third most abundant element in the Universe, but the simplest molecular version of the gas, O2 , has proven surprisingly hard to track down, even in star-forming clouds, because it is highly reactive and readily breaks apart to bind with other atoms and molecules. Despite its detection on the icy moons of Jupiter and Saturn, O2 had been missing in the inventory of volatile species associated with comets until now. It was a great surprise to find an unexpectedly high abundance of O2 from 1–10% relative to H2 O, at least an order of magnitude higher than expected from molecular clouds. This observation means in particular that one must revise seriously our models of Solar System evolution and the formation of the earth.

10.6 Angular Momentum and Magnetic Moment How can we directly compare our results with experiment? Molecular spectra are interesting, but they do not give us access to nonclassical angular momenta. In fact, we have discovered possible half-integer values of the angular momentum, and we know that angular momenta with classical analogues correspond to integer values of (j, m). Are such half-integer values mathematical artifacts or do they exist in nature? It is quite possible to measure angular momenta directly, but at our level the description of the corresponding experiments would be complicated. We rely on a phenomenon that is intuitively related to angular momentum, that is, magnetism, which has numerous applications from nuclear magnetic resonance to superconductivity. The experimental evidence for the quantization of angular momenta relies to a large extent on the fact that when charged particles rotate, they possess magnetic moments.

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10.6.1 Classical Model Classically, a rotating charge distribution has a magnetic moment proportional to its angular momentum. We can make a classical model of an atom (let’s say hydrogen for simplicity) by considering a particle with mass me and charge −q (the electron), rotating with a uniform velocity v along a circle of radius r centered on a charge +q. This positive fixed charge represents the nucleus and it is supposed to be much heavier than the electron. The angular momentum of this system is: L = r × p = me rv u,

(10.32)

where u is the unit vector orthogonal to the orbital plane of the electron. The magnetic moment of this elementary current loop is: µ = IS u,

(10.33)

where I = −qv/(2πr) is the intensity in the loop, and S = πr 2 is the loop area. We then find a remarkably simple relation between the angular momentum and the magnetic moment of this classical system: µ = γ0 L, with γ0 =

−q . 2me

(10.34)

Note that the proportionality coefficient, called the gyromagnetic ratio, does not depend on the radius r of the trajectory of the electron, nor on its velocity v. Strictly speaking, the presence of an external magnetic field perturbs the electronic motion and modifies this very simple relation, but one can show that this perturbation is very weak for realistic fields, and we neglect it here. If we place this magnetic moment in a magnetic field B the system has a magnetic energy (10.35) WM = −µ · B, and a torque Γ =µ×B

(10.36)

is exerted on the magnetic moment. From (10.35) one could naively expect that the magnetic moment of the atom would get aligned with the local magnetic field, as does the needle of a compass. However, the proportionality between the magnetic moment and the angular momentum gives rise to a radically different phenomenon, analogous to the gyroscopic effect. The evolution equation of the angular momentum is dL/dt = Γ . The proportionality between L and µ then implies:

10.6 Angular Momentum and Magnetic Moment

247

Fig. 10.6 Time evolution of the components of a magnetic moment placed in a field B(r) along the z-axis.

dµ = −γ0 B × µ. dt

(10.37)

Consequently, for an atom at r, the magnetic moment does not align with the axis of the local magnetic field B(r), but it precesses around this axis with the angular frequency: (10.38) ω0 = −γ0 B(r), as shown in Fig. 10.6. The quantity ω0 is called the Larmor frequency. This precession phenomenon is very important in practice. It is a particular case of a general theorem of electrodynamics2 proven by Larmor in 1897. This problem was considered independently the same year by H. A. Lorentz.

10.6.2 Quantum Transposition The quantum transposition of this result consists of assuming that the same proportionality relation remains true in quantum mechanics. Any system in an eigenstate ˆ that is of the square of the angular momentum Jˆ 2 possesses a magnetic moment µ proportional to Jˆ ˆ µ ˆ = γ J. (10.39) This is an hypothesis. It is verified experimentally.

2 See,

for example, J. D. Jackson, Classical Electrodynamics. New York: Wiley, (1975).

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10 Angular Momentum

For an orbital electron, the gyromagnetic factor is the same as in classical physics γ0 =

−q . 2me

(10.40)

In general the gyromagnetic factor γ of (10.39) depends on the value of j in a complex system, such as a nucleus.

10.6.3 Experimental Consequences In a magnetic field B, the system has a magnetic potential energy Hˆ M which we deduce from (10.35), ˆ · B. (10.41) Hˆ M = −µ The quantization of angular momenta therefore leads to a perfectly analogous quantization of magnetic moments, up to a coefficient. But measuring magnetic moments is easier and more intuitive than measuring angular momenta. Orbital Angular Momenta ˆ we deduce the following results, anticipating the concluFrom the properties of L, sions of the next Chapter. 1. Consider an electron moving in a central potential. We suppose that the electron is in a given energy level En and in an eigenstate of the orbital angular momentum with eigenvalue ( + 1)2 . As a consequence of rotation invariance, in the absence of an external magnetic field, the 2 + 1 states corresponding to m = −, . . . , + have the same energy En . Let us call these states |n, , m; they are eigenstates of Lˆ z with eigenvalues m. Using the assumption (10.41), the state |n, , m is also an eigenstate of μˆ z . The corresponding eigenvalue is μz = γ0 m. The negative quantity μB = γ0  =

−q ∼ −9.27 10−24 J T−1 2me

(10.42)

ˆ we is called the Bohr magneton. From the properties of orbital angular momenta L, can deduce the following. 2. If we place the system in a magnetic field B parallel to z, the degeneracy is lifted. The state |n, , m is an eigenstate of the observable Hˆ M , with eigenvalue: Wm = −mμB B.

10.6 Angular Momentum and Magnetic Moment

249

We therefore expect to observe a splitting of the atomic energy level En into 2 + 1 sublevels, equally spaced by the interval E = −μB B. This is called the Zeeman effect. It can be observed in a transition En → En . In the absence of a magnetic field, this transition occurs at a single frequency (En − En )/2π. If we apply a field B, several lines appear. The number of such lines is directly related to the angular momenta  and  of the initial and final levels. Notice that if all angular momenta were orbital, then 2j + 1 would always be odd. This simple qualitative prediction existed in classical physics. It was made by Lorentz and by Larmor in 1897. (They both had in mind a three-dimensional harmonic motion of electrons.)

10.6.4 Larmor Precession Another consequence of this proportionality relation between Jˆ and µ ˆ is the Larmor precession phenomenon, which takes place at the quantum level for the expectation values µ. The Ehrenfest theorem yields: 1 d µ = [µ, ˆ Hˆ M ], dt i where we assume that only the term Hˆ M in the Hamiltonian does not commute with µ. ˆ Indeed the other terms of the Hamiltonian are supposed to be rotation invariant. Hence they commute with Jˆ and with µ. ˆ Owing to (10.3), the commutation relations of µ ˆ are: µ ˆ ×µ ˆ = iγ µ. ˆ Therefore, a simple calculation yields: d µ = −γB × µ. dt The expectation value µ satisfies the same equations of motion as those found above for the classical quantity (10.37). This comes from the fact that the Hamiltonian is linear in µ. ˆ The measurement of the Larmor precession frequency provides a direct determination of the gyromagnetic ratio γ and a consistency test of the results. We therefore have an experimental means to measure angular momenta via the measurements of magnetic moments.

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10 Angular Momentum

10.6.5 What About Half-Integer Values of j and m? To conclude this Chapter, we come back to the half-integer values of j and m that we found in the general derivation of the eigenvalues of angular momenta. Concerning an orbital angular momentum, we did not accept such values. Nevertheless, one may wonder whether these values appear in Nature, or whether they are simply a mathematical artifact. If all angular momenta are orbital angular momenta, j is an integer and 2j+1 is odd. Initially, Zeeman performed his experiments with cadmium and zinc (atoms with two external electrons), and he saw, as expected, an odd number of lines. But then there was a drama. In fact, Zeeman continued his experiments on alkali atoms, sodium, potassium, and he found an even number of lines. This was called the anomalous Zeeman effect. It is astonishing that, for 25 years, nobody was able to explain those even numbers. The anomalous Zeeman effect seemed to be the greatest challenge given to the physics community. Of course we come back to this when we study spin 1/2.

10.7 Exercises 1. Rotation invariant operator Show that if an operator Aˆ commutes with two components of the angular momentum (e.g. Jˆx and Jˆy ), it also commutes with the third component (e.g. Jˆz ). 2. Commutation relations for rˆ and pˆ Prove the following commutation relations: [Lˆ j , xˆ k ] = iεjk xˆ 

[Lˆ j , pˆ k ] = iεjk pˆ  ,

where εijk = 1 (resp. −1) if (i, j, k) is an even (resp. odd) permutation of (x, y, z), and εijk = 0 otherwise. Deduce the following identity: ˆ pˆ 2 ] = [L, ˆ rˆ 2 ] = 0. [L, 3. Rotation invariant potential Consider a particle in a potential V (r). What is the condition on V (r) in order for L to be a constant of the motion? 4. Unit angular momentum Consider a system in an eigenstate of Lˆ 2 with eigenvalue 22 , i.e.  = 1. a. Starting from the action of the operators Lˆ + and Lˆ − on the basis states {|, m} common to Lˆ 2 and Lˆ z , find the matrices which represent Lˆ x , Lˆ y and Lˆ z . b. Give in terms of the angles θ and ϕ the probability density for a system in the eigenstate of Lˆ 2 and Lˆ x corresponding to the eigenvalues  = 1 and mx = 1.

10.7 Exercises

251

5. Commutation relations for Jˆ x2 , Jˆ y2 and Jˆ z2 a. Show that [Jˆx2 , Jˆy2 ] = [Jˆy2 , Jˆz2 ] = [Jˆz2 , Jˆx2 ]. b. Show that these three commutators vanish in j = 0 or j = 1/2 states; for example: j, m1 |[Jˆz2 , Jˆx2 ]|j, m2  = 0 for any relevant pair m1 , m2 in the case j = 0 or j = 1/2. c. Show that they also vanish in j = 1 states. Find the common eigenbasis to Jˆx2 , Jˆy2 and Jˆz2 in this case.

Chapter 11

The Hydrogen Atom

The explanation of spectroscopic data was one of the first great victories of quantum theory. In modern science and technology, the mastery of atomic physics is responsible for decisive progress ranging from laser technology to the exploration of the cosmos. The particular case of the hydrogen atom is perhaps the most striking. Its particularly simple spectrum delivered the first clues of quantum laws. It has been used as a testbed for the development of quantum theory. Its hyperfine structure is responsible both for the hydrogen maser and for a revolution in astrophysics, because the hydrogen atom is the most abundant element in the universe and its 21-cm line has been extensively studied in radioastronomy to probe the structure of the interstellar and intergalactic media. Furthermore, the hydrogen atom is probably the physical system that is known with the greatest accuracy. It can be calculated “completely” in the sense that the accuracy of present experimental results is the same as the accuracy of theoretical computer calculations, of the order of 10−12 to 10−13 relative accuracy (the only competitor is celestial mechanics). Here, we first consider technical points, that is, how a two-body problem, where the potential depends only on the distance of the particles, reduces to a one-particle problem. In the case of a central potential, we use the invariance properties of the problem in order to choose the CSCO made up with the Hamiltonian and the angular momentum, Hˆ , Lˆ 2 and Lˆ z , and we show how the traditional quantum numbers {n, l, m} used in atomic physics appear. We study the Coulomb potential and we calculate the bound state energies of hydrogen in the nonrelativistic approximation and recover the E n = −E I /n 2 formula obtained by Bohr in 1913. We end with some considerations on similar atoms, muonic atoms, where the electron is replaced by a heavier sibling, the muon whose mass is 207 times larger. Taking into account relativistic kinematics and spin effects requires a formalism not covered in this book: the Dirac equation. These corrections are small compared to the leading terms. Up to that point, the problem can be solved analytically. Other fine structure effects, such as the Lamb shift, require the more elaborate formalism of quantum field theory. The theoretical treatment of complex atoms (i.e., atoms with more than one electron) involves serious computational problems, even at the nonrelativistic stage. The helium atom with its two electrons can only be calculated numerically. Actually © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_11

253

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11 The Hydrogen Atom

this calculation was considered as the first true test of quantum mechanics, because the much simpler case of hydrogen could be treated successfully by several other approaches issued from the “old” Bohr–Sommerfeld quantum theory. Owing to the accuracy of present numerical calculations, the helium atom is considered to be known exactly.1

11.1 Two-Body Problem; Relative Motion Consider a system of two particles, of masses M1 and M2 , and of positions r 1 and r 2 , whose mutual interaction is given by a potential V (r 1 − r 2 ). The potential depends only on the relative position of the particles. The Hamiltonian is: 2

2

pˆ pˆ Hˆ = 1 + 2 + V ( rˆ 1 − rˆ 2 ), 2M1 2M2

(11.1)

and the system is described by wave functions Ψ (r 1 , r 2 ). We can separate the global motion of the center of mass of the system and the relative motion of the two particles. We introduce the position and momentum operators of the center of mass ˆ = M1 rˆ 1 + M2 rˆ 2 , R M1 + M2

Pˆ = pˆ 1 + pˆ 2 ,

(11.2)

and the relative position and momentum operators: rˆ = rˆ 1 − rˆ 2 ,

pˆ =

M2 pˆ 1 − M1 pˆ 2 . M1 + M2

(11.3)

We can rewrite the Hamiltonian as Hˆ = Hˆ c.m. + Hˆ rel , with:

2

Pˆ Hˆ c.m. = , 2M

(11.4)

2

pˆ Hˆ rel = + V ( rˆ ). 2μ

(11.5)

We have introduced the total mass M and the reduced mass μ: M = M1 + M2 , μ =

1 T.

M1 M2 . M1 + M2

Kinoshita, “Ground state of the helium atom”, Phys. Rev., 105, 1490 (1957).

(11.6)

11.1 Two-Body Problem; Relative Motion

255

Just as in classical mechanics, the Hamiltonian Hˆ separates in the sum of (i) the Hamiltonian Hˆ c.m. describing the free motion of the center of mass (momentum P, total mass M) and (ii) the Hamiltonian Hˆ rel which describes the relative motion of the two particles in the potential V (r) (momentum p, reduced mass μ). ˆ and {xˆi } and { pˆ i } those of rˆ and Let { Xˆ i } and { Pˆi } be the coordinates of Rˆ and P, pˆ . The commutation relations are:

and

[ Xˆ j , Pˆk ] = iδ jk , [xˆ j , pˆ k ] = iδ jk

(11.7)

[ Xˆ j , pˆ k ] = 0, [xˆ j , Pˆk ] = 0.

(11.8)

In other words, the position and momentum operators of the center-of-mass and of the relative variables obey the canonical commutation relations (11.7), and any variable associated with the center-of-mass motion commutes with any other variable associated with the relative motion (11.8). These commutation relations imply: ˆ Hˆ ] = 0, [ Hˆ , Hˆ rel ] = 0. ˆ Hˆ rel ] = 0, [ P, [ P,

(11.9)

Consequently there exists a basis of eigenfunctions of Hˆ that are simultaneously eigenfunctions of Pˆ and Hˆ rel . The eigenfunctions of Pˆ are the plane waves ei K ·R , where K is an arbitrary wavevector. Consequently the desired basis of eigenfunctions of Hˆ has the form: Ψ (R, r) = ei K ·R ψ(r), where ψ(r) is an eigenfunction of Hˆ rel : Hˆ rel ψ(r) = E ψ(r).

(11.10)

The eigenvalues E tot of Hˆ are E tot =

2 K 2 + E, 2M

(11.11)

the sum of the kinetic energy of the global system ( Hˆ c.m. ) and the internal energy ( Hˆ rel ). One consequence of this is the conservation of the total momentum d P/dt = 0. This is due to the fact that the potential depends only on the relative variable r = r 1 − r 2 ; in other words, the Hamiltonian of the system is translation invariant. Because rˆ and pˆ have canonical commutation relations, the problem reduces to the quantum motion of a particle of mass μ in the potential V (r). For an atomic system made of an electron (M1 = m e ) and the rest of the atom (M2 ), we have

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11 The Hydrogen Atom

M2  m e . Therefore we can neglect the small difference between the reduced mass μ and the electron mass m e , remembering that it is easy to correct for reduced mass effects if necessary. We are interested in the eigenvalue problem of Hrel .

11.2 Motion in a Central Potential The Coulomb potential is central, that is, it only depends on the distance r = |r| of the two particles. The problem is therefore rotation invariant. Spherical Coordinates, CSCO Owing to the symmetry of the problem, it is appropriate to work in spherical coordinates. The Laplacian Δ has the following expression (which is easily obtained by writing the square of the angular momentum Lˆ = −ir × ∇ and expanding) Δ=

1 1 ∂2 r − 2 2 Lˆ 2 . 2 r ∂r r 

(11.12)

Equation (11.10) is then written as 



Lˆ 2 2 1 ∂ 2 r + + V (r ) − 2m e r ∂r 2 2m e r 2

ψ(r) = E ψ(r).

(11.13)

The Hamiltonian Hˆ rel commutes with the three angular momentum operators Lˆ i , i = x, y, z. Each Lˆ i commutes with Lˆ 2 . In addition Lˆ i only acts on the variables θ and ϕ, and it commutes with r, ∂/∂r, V (r ). In other words, the Hamiltonian Hˆ rel , which from now on is denoted Hˆ for simplicity, commutes with the angular momentum: ˆ = 0. [ Hˆ , L] Consequently Hˆ , Lˆ 2 and a given component of Lˆ (e.g., Lˆ z ) form a set of commuting observables. We verify a posteriori that this set is complete by checking that the basis corresponding to the common eigenfunction is unique. ˆ = 0 implies the conservation of the angular momentum: The relation [ Hˆ , L] dL/dt = 0. This is due to the fact that the Hamiltonian of the system is rotation invariant. ˆ Lˆ 2 , and Lˆ z Eigenfunctions Common to H, Separation of the Angular Variables Part of the eigenvalue problem (11.13) is already solved because we know the form of the eigenfunctions common to Lˆ 2 and Lˆ z . These are the spherical harmonics. We separate the variables in the following way

11.2 Motion in a Central Potential

257

ψ,m (r) = R (r ) Y,m (θ, ϕ),

(11.14)

L ψ,m (r) = ( + 1) ψ,m (r), Lˆ z ψ,m (r) = m ψ,m (r),

(11.15)

ˆ2

2

(11.16)

where  and m are integers, with |m| ≤ . Substituting in (11.13), the eigenvalue equation becomes: 

 ( + 1)2 2 1 d 2 r+ + V (r ) R (r ) = E R (r ). − 2m e r dr 2 2m e r 2

(11.17)

This equation is independent of the quantum number m. This is why we have not put an index m for the unknown function R (r ) in Eq. (11.14). This differential equation is the radial equation and R (r ) is called the radial wave function. The normalization of the wave function, which we must impose on finding bound  states, is |ψ(r)|2 d 3r = 1; in spherical coordinates, 



∞

d 2Ω

dr r 2 |ψ(r, θ, ϕ)|2 = 1.

0

Here Ω is the solid angle with d 2 Ω = sin θ dθ dϕ. Because the spherical harmonics are normalized, we obtain, for the radial wave function R (r ), the condition 

∞

dr r 2 |R (r )|2 = 1.

(11.18)

0

Introducing the reduced wave function u  (r ) = r R (r ), the Schrödinger equation becomes:   ( + 1)2 2 d 2 + + V (r ) u  (r ) = E u  (r ), (11.19) − 2m e dr 2 2m e r 2 ∞ with 0 |u  (r )|2 dr = 1. One can prove that any normalizable solution R (r ) is bounded at the origin, and therefore u  (0) = 0. This equation has the structure of the Schrödinger equation describing the one dimensional motion of a particle of mass m e in the potential: Veff (r ) = V (r ) +

( + 1)2 . 2m e r 2

(11.20)

This effective potential is the superposition of the interaction potential between the two particles 1 and 2, and a centrifugal barrier term which is repulsive and increases as the angular momentum increases, Fig. 11.1.

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11 The Hydrogen Atom

Fig. 11.1 Effective potential that enters the one-dimensional Schrödinger equation for the reduced radial wave function u  (r ). For  = 0 (left) the motion occurs in the “bare” potential V (r ); for   = 0 (right) the effective potential is the superposition of V (r ) and of the centrifugal barrier ( + 1)2 /(2m e r 2 ). The Figure is drawn for a Coulomb potential V (r ) ∝ 1/r

Quantum Numbers The energy levels depend on the parameter , but they do not depend on the projection m. For each value of , corresponding to a given value of the square of the angular momentum, each level has a degeneracy of degree (2 + 1). For a given , we deal with a one-dimensional problem of the same type as studied in Chap. 5. The bound state energy levels (E < 0) correspond to solutions R (r ) that satisfy (11.18). The Radial Quantum Number n For a given , we can arrange the possible values of the bound state energies in an increasing sequence, which we label by an integer n (n = 0, 1, 2 . . .), the state n = 0 being the most strongly bound. Depending on the potential, this sequence may be finite (as for a square well potential) or infinite (as for the Coulomb potential). The general mathematical properties of the differential equation (11.17), together with the conditions that R (0) is finite and that R (r ) can be normalized (11.18), show that this number n corresponds to the number of nodes of the radial wave function; the number of times it vanishes between r = 0 and r = ∞. This is independent of the form of the potential V (r ) (provided it is not too pathological). The quantum number n is called the radial quantum number. A radial wave function, defined by the two quantum numbers  and n and normalized to unity, is unique (up to a phase factor). The eigenvalues of the Hamiltonian are therefore labeled in general by the two quantum numbers n and . They do not depend on the quantum number m as a consequence of the rotation invariance of the system. This means that the 2 + 1 states corresponding to given values of n and  and to different values of m, have the same energy and are degenerate. These general considerations apply to any two-body system with a central potential: the hydrogen atom and also to a certain extent alkali atoms, diatomic molecules, the deuteron, and quark systems.

11.2 Motion in a Central Potential

259

The Principal Quantum Number n In the following section, we solve Eq. (11.19) in the case of a Coulomb potential V (r ) = −q 2 /4πε0 r . In this particular case, the energy levels only depend on the quantity n +  + 1. It is therefore customary to label atomic levels with the three quantum numbers , m, and the positive integer n, called principal quantum number, defined by the relation: n = n +  + 1. The energy eigenstates are then classified by increasing values of n, (n = 1, 2, 3, . . .). The classification of atomic states by the three integers (n, , m) is just a redefinition of a catalogue with respect to the classification in terms of (n , , m). For a given value of n, there are only n possible values of :  = 0, 1, . . . , n − 1. For each value of , there are 2 + 1 possible values of m. The wave function of an energy eigenstate is labeled with the three corresponding quantum numbers (ψn,,m (r)) and the corresponding energy is denoted E n, . Spectroscopic notation (states s, p, d, f , . . .) The measurement of the energy levels of an atom often comes from the observation of the wavelengths of its spectral lines. We show in Fig. 11.2 the energies E n, of the valence electron of sodium and some of the observed transitions. Each horizontal line represents a state; the number on the left is the value of the principal quantum number n. Each column corresponds to a given value of . The energy of the state is given on the vertical axis (for instance, E 3,0 = −5.13 eV). On the right, we give the energy levels E n of hydrogen, which, as we show, only depend on n. The quantum theory of the emission of a photon by an excited atom imposes selection rules. In the transition from a state (n, ) to a state (n 0 , 0 ) by emission of a photon of energy ω = E n, − E n 0 ,0 , all transitions are not allowed. Only the transitions for which  = 0 ± 1 are intense (see Chap. 16, Sect. 16.2.4). Experimental observations in the 19th century showed that one can group the lines in series which were given names according to their aspect. In the case of sodium, after the theory had been understood, it turned out that these series correspond to the following transitions, the sharp series ω = E n,=0 − E 3,1 the principal series ω = E n,=1 − E 3,0 the diffuse series ω = E n,=2 − E 3,1 the fundamental series ω = E n,=3 − E 3,2 . Each of these four series corresponds to transitions from a state of given  (and various values of n) to a well-defined state. Consequently, the tradition consists of attributing to a given value of  the initial of the corresponding series (spectroscopic notation): Symbolic letter: s p d f g h Corresponding value of : 0 1 2 3 4 5.

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11 The Hydrogen Atom

Fig. 11.2 Energy levels of the external electron of sodium (left) and energy levels of hydrogen (right)

A state of well-defined energy is then denoted by a number (the value of n) followed by a letter (corresponding to the value of ): n = 1,  = 0 : state 1s ; n = 3,  = 2 : state 3d.

11.3 The Hydrogen Atom We now consider the specific case of the hydrogen atom. Here, we consider the problem in its first approximation, where we neglect spin effects. We consider the problem of a particle of mass m e in the Coulomb field of the proton, which is considered infinitely massive (the reduced mass correction is straightforward): V (r ) = −

q2 e2 =− . 4πε0 r r

11.3 The Hydrogen Atom

261

q is the elementary charge and we set e2 = q 2 /4πε0 . The radial equation is: 

( + 1)2 e2 2 1 d 2 r + − − 2m e r dr 2 2m e r 2 r

 R (r ) = E R (r ).

(11.21)

11.3.1 Atomic Units; Fine Structure Constant The above equation involves three constants:  (action), m e (mass), and e2 (product of an energy by a length). Using these three constants, it is useful to form a length unit and an energy unit relevant for our problem. Fine Structure Constant We notice that e2 / has the dimension of a velocity. Unless the differential equation had pathologies (which is not the case), it must represent the typical velocity v of the electron in the lowest energy levels of the hydrogen atom. This velocity must be compared with the velocity of light c, which is the absolute velocity standard in physics. The ratio between these two velocities is a dimensionless constant α, which is a combination of the fundamental constants q, , and c: α=

q2 1 e2 = ∼ . c 4πε0 c 137

The smallness of this constant α guarantees that the nonrelativistic approximation is acceptable up to effects of the order of v 2 /c2 ∼ 10−4 . The constant α is called, for (unfortunate) historical reasons, the fine structure constant. A more appropriate terminology would have been: fundamental constant of electromagnetic interactions. Any charge Q is an integer multiple of the elementary charge Q = Zq (or an integer multiple of q/3 if one incorporates quarks). Therefore the fundamental form of Coulomb’s law between two charges Q = Zq and Q = Z q is V (r ) = αZ Z (c/r ), with Z and Z integers, which only involves mechanical quantities. The introduction of electric units and of ε0 is only a convenient manner to describe macroscopic cases where Z and Z are very large. The experimental determination of the fundamental constant α is a key point in physics: 1/α = 137.035 999 139 (031). The fact that this is a dimensionless number stirred minds at the beginning. One cannot change the value of α by changing units. For a long time, after this discovery (i.e., the discovery of the universality of Planck’s constant) people have tried to obtain it starting from transcendental num2 bers e, π, the Euler constant γ, and so on. For instance e(−π /2)  1/139, or better 2 (1/2)π (−e /2)  137.3. The truth came with the great revolution of quantum field theory introduced by Ken Wilson in the 1970s. What is called the renormalization group enabled us to understand that the value of the fine structure constant, as well as other dimensionless

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11 The Hydrogen Atom

constants in elementary interactions, depends on the energy, or on the distance. In fact, experiments in the LEP collider at CERN have shown that around 100 GeV center of mass energy, the value of the fine structure constant is larger, α  1/129. There is some hope to calculate it in unified field or superstring theories. Atomic Units The length unit of the problem is the Bohr radius: a1 =

2 1  = ∼ 0.53Å, 2 mee α mec

where /m e c is the Compton wavelength of the electron. The Bohr radius is the typical size of an atom. The energy unit relevant for the hydrogen atom is: EI =

m e e4 1 = m e c2 α2 ∼ 13.6 eV 2 2 2

which, as we show, is the ionization energy of the atom. The electron-volt is a typical energy for external atomic electrons. The atomic time scale is 2π3 /(m e e4 ) ∼ 1.5 10−16 s. It represents the period of the classical circular motion of the electron around the proton for the energy −E I .

11.3.2 The Dimensionless Radial Equation Having identified the relevant length and energy scales of the problem, we introduce the dimensionless quantities ρ = r/a1 and ε = −E/E I . We define ε with a minus sign so that this quantity is positive if we are dealing with a bound state, whose energy is negative. We obtain the following dimensionless equation, 

 ( + 1) 2 1 d2 ρ− + − ε R (ρ) = 0. ρ dρ2 ρ2 ρ

(11.22)

This equation is well known to mathematicians. Laguerre gave the solutions in 1860. The following properties can be proven. 1. For each value of , we obtain a infinite set of normalizable solutions labeled by an integer n = 0, 1, . . .: √

R(ρ) = e−

ερ

ρ Q n , (ρ),

(11.23)

where Q n , (ρ) = C0 + C1 ρ + · · · + Cn ρn is called a Laguerre polynomial of degree n . It has n real zeros between ρ = 0 and ρ = +∞.

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263

Table 11.1 Radial wave functions Rn, (ρ) for the Coulomb problem, for n = 1, 2, 3

n=1

=0

n=2

=0 =1

n=3

=0 =1 =2

2 e−ρ 1  ρ −ρ/2 1− e √ 2 2 1 √ ρ e−ρ/2 2 6   2 2 2 2 ρ + ρ 1 − e−ρ/3 33/2 3 27 25/2  ρ −ρ/3 e ρ 1− 7/2 3 6 23/2 √ ρ2 e−ρ/3 9/2 5 3

2. These normalizable solutions correspond to the eigenvalues ε=

1 . (n +  + 1)2

(11.24)

As already mentioned above, the integer n corresponds to the number of nodes of the radial wave function and is called the radial quantum number. The principal quantum number is the integer n = n +  + 1. The first radial wave functions Rn, (ρ) are given in Table 11.1. We remark that, owing to (11.24), εn = 1/n 2 is an eigenvalue of all radial equations corresponding to values of  smaller than n :  = 0, 1, . . . , n − 1. We do not give a rigorous proof here that the normalizable solutions of (11.22) can indeed be cast in the form (11.23), but we can check that these solutions make sense both around the origin and at infinity. Around ρ = 0 The Coulomb term 1/ρ and the constant term are negligible compared with the centrifugal term ( + 1)/ρ2 (for   = 0). Assuming a power law dependence Rn, (ρ) ∝ ρs around ρ = 0, one finds that the only possible exponent s compatible with a normalizable solution is s =  (s = − − 1 is not square-integrable for  ≥ 1). This corresponds to the expansion of (11.23) around ρ = 0. Notice that in an s-wave ( = 0), a function behaving as 1/r is square-integrable; however, it does not satisfy the Schrödinger equation because Δ(1/r ) = −4πδ(r) where δ(r) is the Dirac distribution which is not a square integrable function. At Infinity Keeping only the leading terms in the expansion in Rn, , we have: √

Rn, (ρ) ∼ e−

ερ

 Cn ρn−1 + Cn −1 ρn−2 + · · ·

If one injects this expansion √into the differential equation (11.22), it is immediate to check that the term in e− ερ ρn−1 always cancels out of the equation, and the

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11 The Hydrogen Atom

√ √ coefficient of the next term e− ερ ρn−2 is proportional to Cn (1 − n ε). Therefore this term also cancels out for the particular choice ε = 1/n 2 . The subsequent terms of the expansion, which depend on the centrifugal barrier, allow the determination of the coefficients Cn , Cn −1 , . . . , C0 . Coming back to the initial variables for length and energy, we can summarize the above results. Each solution of the Schrödinger equation (11.13) corresponding to a bound state for the Coulomb problem is labeled by three integers (or quantum numbers):

n = 0, 1, 2, . . . ,  = 0, 1, . . . , n − 1, m = −, . . . , .

The energy of a solution depends only on the principal quantum number n: En = −

EI , n2

with

EI =

m e e4 ∼ 13.6 eV. 22

To each energy level, there correspond several possible values of the angular momentum. The total degeneracy (in  and m) of a level with given n is n−1 (2 + 1) = n 2 . =0

The wave function corresponding to a given set n, , m is unique (up to a phase factor) and it reads: ψn,,m (r) = Y,m (θ, ϕ) e 

−r/(n a1 )



r a1



r × C0 + C1 + · · · + Cn−−1 a1



r a1

n−−1  .

(11.25)

where the Ck ’s (k = 0, . . . , n −  − 1) are the coefficients of the Laguerre polynomials and where a1 = 2 /(m e e2 ) ∼ 0.53 Å.

11.3.3 Spectrum of Hydrogen In Fig. 11.3 we represent the energies E n of the hydrogen atom. Each line represents an energy level, the number on the right is the value of n, the column corresponds to a given value of , and we give the value of the energy on the vertical axis. The selection rule given for the observable spectral lines of the sodium atom  = 0 ± 1 still holds. The most famous series is the Balmer series. It corresponds to transitions from states ns to the state 2 p:

11.3 The Hydrogen Atom

265

Fig. 11.3 Energy levels of hydrogen

ω = E n − E 2 = 13.6

n2 − 4 eV. 4n 2

The first lines of the Balmer series are in the visible part of the spectrum (ω ∼ 2 to 3 eV; λ ∼ 0.5 µm). The Lyman series, corresponding to transitions to the ground state, lies in the ultraviolet (λ ≤ 121.5 nm).

11.3.4 Stationary States of the Hydrogen Atom The Ground State (1s) The ground state corresponds to n = 1, therefore  = 0 and m = 0 (1s state in the spectroscopic √ language). Because the spherical harmonic Y0,0 (θ, ϕ) is a constant equal to 1/ 4π, the normalized wave function of this state is: e−r/a1 . ψ1,0,0 (r) = πa13 The probability of finding the electron in a spherical shell of thickness dr , represented in Fig. 11.4, is:

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11 The Hydrogen Atom

Fig. 11.4 Radial probability density P(r ), giving the probability of finding the electron between r and r + dr in a hydrogen atom prepared in its ground state

P(r )dr = |ψ1,0,0 (r)|2 4πr 2 dr. The probability density per unit volume is proportional to the exponential function e−2r/a1 , and it is maximum for r = 0. The most probable distance between the electron and the proton is the Bohr radius a1 = 0.53 Å. Other States Figure 11.5 represents the radial probability density Pn, (r ) = r 2 |Rn, (r )|2 for various states n, . We note the reduction of the number of nodes of the radial wave function as  increases for a given n. For a level n, l, the function Pn, (r ) has n = n −  − 1 zeroes, where n is the degree of the corresponding Laguerre polynomial. In particular, for  = n − 1, we remark that P(r ) has a single maximum, located at a distance r = n 2 a1 (Eq. 11.25). Figure 11.6 represents some spatial probability densities |ψn,,m (r)|2 in a plane y = 0 (these are axial symmetric functions around the z-axis). For large quantum numbers n  1, we notice that one gets closer to “classical” situations, corresponding to a localized particle.

Fig. 11.5 Radial probability density r 2 |Rn, (r )|2 of the states n = 2, 3, 4 of hydrogen

11.3 The Hydrogen Atom

267

Fig. 11.6 Probability density |ψn,,m (r)|2 in the y = 0 plane for n = 6,  = 5 (mesh size: 60 a1 × 60 a1 ). For m = 0, the particle is localized in the vicinity of the z axis. For large |m| (in particular m = ±5), the particle is localized in the plane z = 0, in the vicinity of a circle centered at the origin, of radius r = 30 a1 (circular state). The vertical scale of the surface m = 0 has been reduced by a factor of 2 with respect to the five other surfaces in order to improve the visibility

11.3.5 Dimensions and Orders of Magnitude Consider a hydrogen atom prepared in a stationary state |n, , m. Using the virial theorem, one can show that the classical relation between the kinetic energy and the potential energy still holds for the expectation values of these quantities: p2 EI  = −E n = 2 , 2m e n 2 −e 2E I =  = 2 En = − 2 . r n

E n(kin) = 

(11.26)

E n(pot)

(11.27)

Using the properties of Laguerre polynomials, one finds that the mean radius has the following variation with n and ,

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11 The Hydrogen Atom

r  =

 a1 2 3n − ( + 1) , 2

(11.28)

as well as  1 1 2 n 2 a12 2 1   = 2 ,  2 = 3 , r 2  = 5n + 1 − 3( + 1) . 2 r n a1 r 2 n (2 + 1) a1 Also, setting ρ = r/a1 , one gets for p > −2 − 1:  p

p+1 p ρ  − (2 p + 1)ρ p−1  + (2 + 1)2 − p 2 ρ p−2  = 0. 2 n 4 Remark Because of the n 2 variation of the mean atomic radius, the maximal probability density for a given radial wave function decreases like 1/n 4 . This is why a readjustment of scales is necessary in order to visualize Fig. 11.5 properly.

11.3.6 Historical Landmarks The importance of this calculation done by Schrödinger in 1926, is not so much in the result, which was known, but in the fact that it was obtained by a systematic method, which could be generalized to other cases. Spectroscopy The law involving integers was a crowning achievement for spectroscopy. After Newton’s discovery of the decomposition of white light in 1680, Wollaston, in 1802, and mainly Fraunhofer, after 1814, had discovered dark and shining lines in the solar spectrum and in stars. People had realized that spectral lines observed in the light emitted by a body were characteristic of the elements in this body. The observation of regularities allowed the attribution of a series of lines to a given element (there was a catalogue of 5000 lines in 1890). Very soon, people had the idea that spectral lines were analogous to harmonics in acoustics and that there should exist some simple relations between them. But that didn’t lead very far. Fundamental physicists considered spectroscopy to be too complicated and outside the scope of physics, such as the harmonics of a piano, which depend on the shape of the instrument as a whole. Why? Because the spectra were too complicated. The Spectrum of Hydrogen One had to start with the spectrum of the simplest atom, hydrogen. But people didn’t know that, because the spectrum of atomic hydrogen was found late. It was difficult to obtain the spectrum of atomic hydrogen. One had to operate with a discharge tube,

11.3 The Hydrogen Atom

269

and disentangle the spectra of atomic and molecular hydrogen, and of other species such as nitrogen and water. The first line of hydrogen was discovered in 1853 by the Swedish physicist Anders Jonas Ångström. Then in 20 years, he found three others that were called, including the first, α, β, γ, δ. In 1881 Huggins discovered in the star α of the Lyre, ten other lines “whose sequence seemed to be following the visible lines of hydrogen.” Hydrogen is abundant in stars, but one had to guess it at that time! Balmer Integers played an important role in science in the 19th century. Examples can be found in chemical reactions, atomic theory, classification and evolution of species in zoology and in botany, and so on. It is by chance that, in 1885, Balmer, who was a high school teacher in Basel, and who was fascinated by numerology, learned the positions of the first four lines of hydrogen. He realized that the wavelengths of the lines could be represented with an accuracy of 10−3 by a formula involving integers λ= A

n2 , with A = 0.3646 µm. n2 − 4

And this law was good to one part in a thousand when applied to the ten lines of Huggins! Balmer predicted the next ones and the limit. Although he was not a physicist, Balmer found the simplicity of the formula quite striking. He called that constant A the “fundamental number” of hydrogen. In his 1885 paper, he wrote, “It appears to me that hydrogen . . . more than any other substance is destined to open new paths to the knowledge of the structure of matter and its properties.” In 1912, Niels Bohr, who was 27, was working with Rutherford on an atomic model. He was not aware of Balmer’s formula and of analogous results obtained by Rydberg for alkali atoms. One day, by chance, he learned the existence of Balmer’s formula; it only took him a few weeks to construct his celebrated model of the hydrogen atom, which is one of the turning points of quantum physics.

11.4 Muonic Atoms To end this chapter, we describe an application of what we have done in nuclear and elementary particle physics. The μ lepton, or muon, discovered in 1937, has physical properties of a heavy electron. As is the electron, it is elementary or pointlike; it has the same electric charge, the same spin, but it is 200 times more massive: m μ = 206.8 m e . It is unstable and it decays into an electron and two neutrinos: μ → e + ν e + νμ with a lifetime of 2 × 10−6 s. Therefore, physically it is a true heavy electron except that it is unstable.

270

11 The Hydrogen Atom

In particle accelerators, one can produce muons, slow them down, and have them captured by atoms where they form hydrogenlike systems. In a complex atom, the muon is not constrained with respect to electrons by the Pauli principle. The muon expels electrons, cascades from one level to another, and eventually falls in the vicinity of the nucleus at a distance aμ = 2 /Z m μ e2 , which is 200 times smaller than the distance of the internal electrons. Therefore, it forms a hydrogenlike atom, because it does not feel the screened electron electric field. The lifetime of the muon is much longer than the total time of the cascades (∼10−14 s). It is also much longer than the typical atomic time 3 /m μ e4 ∼ 10−19 s. The muon can therefore be considered as stable on these time scales. It then forms a muonic atom. The Bohr radius of a muonic atom is of the same order as a nuclear radius. The effect of external electrons is screened. Consider lead (Z = 82), whose nuclear radius is R ≈ 8.5 fm. One finds aμ ≈ 3.1 fm, which means that the μ penetrates the nucleus noticeably. In fact, in the ground state, it has a 90 % probability of being inside the nucleus. The description of the nucleus as a point particle is inadequate. Consequently, spectra of muonic atoms provide information on the structure of nuclei, in particular concerning their charge distributions. For a spherical nucleus, the potential is harmonic inside the nucleus (assuming a constant charge density), and Coulomblike outside the nucleus. If the nucleus is deformed, flattened, or cigar-shaped, spherical symmetry is broken, and the levels will no longer be degenerate in the magnetic quantum number m. This results in a splitting of the spectral lines, which enables us to determine the charge distribution, that is the proton distribution inside the nucleus. Figure 11.7, obtained at CERN, shows the spectra of muonic atoms in the cases of gold (Z = 79), which is a spherical nucleus, and of uranium (Z = 92), which is a deformed nucleus. We notice the more complicated structure of the higher energy

Fig. 11.7 Transition line from the 2 p level (actually split into two sublevels 2 p1/2 and 2 p3/2 , to the 2s level in muonic atoms of gold (Z = 79, A = 197) and uranium (Z = 92, A = 238) (scale in keV). Gold is spherical, and its spectrum has a simple shape; uranium is deformed, and the upper peak is split in four lines (CERN document)

11.4 Muonic Atoms

271

line for uranium. This is a very accurate method to determine the deformations of nuclei. The existence of the muon has been a mystery for more than 40 years. When it was discovered, Rabi said, “Who ordered that?” Why a heavy electron? All matter we know about around us can be built starting with protons, neutrons, electrons, and neutrinos, or, in terms of fundamental constituents, with the family of quarks and leptons {u, d, e, ν}, using the Schrödinger equation. (Of course, to construct a zebra or a humming bird, there are simpler methods, and we are interested in simple features such as 2 ears, 4 feet, 1 tail for a rabbit, 2 humps for a camel. But with a powerful enough computer, it’s in principle feasible). So, why a heavy electron, with which one can imagine a Gulliver universe: atoms, molecules, a chemistry, a biology, 200 times smaller, but 200 times more energetic than the matter we know? Muons are useful. They have many applications, such as probing nuclei, probing crystals, and probing pyramids, but why do they exist? What role is the muon supposed to play? In 1974, with the discovery of a new quark, the charm quark c, it was realized that the muon forms, with its neutrino, the charm quark, and the strange quark s, discovered in the 1940s, a new family of quarks and leptons (c, s, μ, νμ ). This family generates at higher scales, a new atomic and nuclear physics, but its members are unstable. In 1975–1976, with the discovery of a new lepton τ , another new quark was discovered, the b quark (beautiful or bottom), and, in 1995, the top quark t, hence a third family (t, b, τ , ντ ).2 By July 2000, the direct evidence for the τ neutrino by the DONUT collaboration in Fermilab, filled completely the family of quarks and leptons of the standard model. In 1989 at the LEP electron collider, it was proven that the (light) fundamental constituents of matter consist of these three families only. Present ideas are that a more fundamental theory could not exist, and that the big bang could not have happened so nicely, if these two extra (useless) families of quarks and leptons did not exist. They are of importance in order to create the world. But, after that, they are merely toys for physicists. However, we still do not understand the masses of these quarks and leptons (therefore their stability). The origin of mass is one of the great problems of modern physics. The discovery of the Higgs boson at CERN in June 2012 is a celebrated first step in that direction. To end this chapter, we note that in January 2014, the ASACUSA experiment3 at CERN succeeded in producing a beam of antihydrogen atoms, i.e., positrons bound to antiprotons. The comparisons of hydrogen and antihydrogen atoms constitute one of the best ways to perform highly precise tests of matter/antimatter symmetry.

2 M.

Perl, “The leptons after 100 years,” Phys. Today, October (1997), p. 34. N. et al. A source of antihydrogen for in-flight hyperfine spectroscopy. Nat. Commun. 5:3089 doi:10.1038/ncomms4089 (2014); and further references therein.

3 Kuroda,

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11 The Hydrogen Atom

11.5 Exercises 1. Expectation value of r for the Coulomb problem Consider the dimensionless radial equation for the hydrogen atom: 

 ( + 1) 2 d2 − + u n, (ρ) = ε u n, (ρ), dρ2 ρ2 ρ

(11.29)

u n, (ρ) = ρRn, (ρ) is the reduced wave function and satisfies the conditions where ∞ 2 |u n, (ρ)| dρ = 1 and u n, (0) = 0. 0 a. By multiplying this equation by ρ u n, (ρ) and by integrating over ρ, show that: ρ ( + 1) + −2= n2 n2



+∞ 0

ρ u n, (ρ) u n, (ρ) dr.

One can use the result 1/ρ = 1/n 2 deduced from the virial theorem (11.27). b. By multiplying the Schrödinger equation by ρ2 u n, (ρ) and by integrating over ρ, show that:  +∞ ρ − 1 = − ρ u n, (ρ) u n, (ρ) dr. n2 0 c. Deduce from the above results that ρ = (3n 2 − ( + 1))/2. 2. Three-dimensional harmonic oscillator in spherical coordinates We treat the three-dimensional harmonic oscillator problem as a central potential problem. Consider the Hamiltonian: 1 pˆ 2 + mω 2 rˆ 2 Hˆ = 2m 2 with rˆ 2 = xˆ 2 + yˆ 2 + zˆ 2 .

√ a. Introduce the dimensionless quantities ρ = r mω/ and = E/ω. Show that the radial equation (11.17) becomes: 



( + 1) 1 d2 + + ρ2 − 2 − ρ dρ2 ρ2

R (ρ) = 0.

(11.30)

b. One can prove that the normalisable solutions of (11.30) are labelled by an integer n : 2 Rn , (ρ) = ρ Pn , (ρ) e−ρ /2 , where Pn , (ρ) is a polynomial of degree n . These solutions correspond to particular values of :

11.5 Exercises

273

= 2n +  + 3/2. In the following we set n = 2n + . Show that one recovers the levels E = ω(n 1 + n 2 + n 3 + 3/2) (n i integer ≥ 0) obtained in Cartesian coordinates in Chap. 4 (Exercise 3), and associated to the eigenstates |n 1 ; n 2 ; n 3 . To what values of the angular momentum do the energy levels E n correspond? c. Give the explicit correspondence between the states |n 1 ; n 2 ; n 3  and |n, , m for n = n 1 + n 2 + n 3 = 1. 3. Relation between the Coulomb problem and the harmonic oscillator Consider the three-dimensional harmonic oscillator problem treated in the previous exercise, but writing the potential as V (r ) = K 2 mω 2 r 2 /2 where K is dimensionless, in order to keep track of the parameters. Using the dimensionless variable ρ of the previous exercise, the radial equation for the reduced wave function u = ρR(ρ) is: 

 ( + 1) E d2 2 2 − −K ρ +2 u(ρ) = 0. dρ2 ρ2 ω

(11.31)

Similarly, consider the Coulomb problem with a potential V (r ) = −Z e2 /r , where Z is dimensionless. The radial equation for the variable ρ = r/a1 is: 

( + 1) 2Z E d2 − + + 2 2 dρ ρ ρ EI

 u(ρ) = 0.

(11.32)

√ a. Show that under the transformation u(ρ) = x α f (x) where x = ρ, and with an appropriate choice of α, one can cast the hydrogen problem (11.32) in the same form as the harmonic oscillator problem (11.31). b. Discuss the correspondence between the parameters of the two problems. c. Recalling the results of the previous exercise, find the energy levels of the hydrogen atom. 4. Confirm or invalidate the following assertions ˆ = 0, the energy levels do not depend on m (i.e., on the eigenvalues of a. If [ Hˆ , L] ˆ the projection of one of the components of the angular momentum L). b. If [ Hˆ , Lˆ 2 ] = 0, the energy levels do not depend on . 5. Centrifugal barrier effects Consider a central potential. We note E  the lowest energy-level for a given . Show that E  increases with . 6. Algebraic method for the hydrogen atom We consider the radial dimension less equation for the Coulomb problem (11.29) and we introduce the operators:

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11 The Hydrogen Atom

A−  =

d +1 1 + − dρ ρ +1

A+  =

d +1 1 − + . dρ ρ +1

+ a. Calculate A−  A . Show that (11.29) can be written:



  + A− u A =

−   

b. Show that: − − + A+  A = A+1 A+1 −

1 ( + 1)2

 u.

(11.33)

1 1 + . 2 ( + 2) ( + 1)2

+ By multiplying (11.33) by A+  , show that A u  (ρ) satisfies the radial equation with the same eigenvalue ε but for an angular momentum  =  + 1. c. Similarly, show that A− −1 u  (ρ) satisfies the radial equation with the same eigenvalue ε but for an angular momentum  =  − 1. + d. Calculate the expectation value of A−  A with the radial function u  (ρ), and show 2 that ε ≤ 1/( + 1) . e. Show that, for a given value of ε, there exists a maximum value max of the angular momentum such that ε = 1/n 2 , where we have set n = max + 1. Show that the corresponding radial wave function u max (ρ) satisfies the differential equation:



 n 1 d − + u max (ρ) = 0. dρ ρ n

f. Deduce from these results the energy levels and the corresponding wave functions of the hydrogen atom. 7. Molecular potential Consider a central potential of the form: V (r ) = A/r 2 − B/r

(A, B > 0).

We want to calculate the energy levels of a particle of mass m e in this potential. a. Write the radial equation. b. By a change of notations, reduce this equation to an eigenvalue problem which is formally identical to the Kepler problem. Check that one can solve this equation using the same arguments as for the hydrogen atom. c. Give the explicit values of the energy levels in terms of A and B.

11.6 Problem. Decay of a Tritium Atom

275

11.6 Problem. Decay of a Tritium Atom In all this problem, we consider nuclei as infinitely massive compared to the electron, of mass m. We note a1 the Bohr radius a1 = 2 /me2 and E I = mc2 α2 /2 ∼ 13.6 eV the ionisation energy of the hydrogen atom, where α is the fine structure constant. We note e the unit charge, and we work with units such that 4π 0 = 1. The nucleus of the tritium atom is the isotope 3 H , of charge Z = 1. In the ground state |ψ0  of this atom, the wave function of the electron (n = 1, l = 0, m = 0) is the same as in the usual hydrogen atom: 1 e−r/a1 . ψ0 (r) =  (πa1 3 )

(11.34)

The tritium nucleus is radioactive and transforms into helium 3 through β decay: 3

H → 3 H e + e− + ν¯

(11.35)

(ν¯ is an antineutrino), where the emitted electron has an energy of the order of 15 keV and the the helium nucleus 3 H e has charge Z = 2. The decay is an instantaneous process, the β electron is emitted with a large velocity and leaves the atomic system very rapidly. Consequently, an ionized 3 H e+ atom is formed, for which, at the time t0 of the decay, the wave function of the electron is practically the same as in tritium, and we shall assume it is still given by Eq. (11.35). We note |n, l, m the states of the ionized helium atom which is a hydrogen-like system, i.e., one electron placed in the Coulomb field of a nucleus of charge 2. The Energy Crisis in Tritium Decay (1.1) Write the hamiltonian Hˆ 1 of the atomic electron before the decay and the hamiltonian Hˆ 2 of this electron after the decay (when the potential term has suddenly changed.) (1.2) What are, in terms of E I , the energy levels of the 3 H e+ atom? Give its Bohr radius and its ground state wave function ϕ110 (r). (1.3) Calculate the expectation value E of the energy of the electron after the decay. One can, for instance make use of the fact that ψ0 |(1/r )|ψ0  = 1/a1 and that Hˆ 2 = Hˆ 1 − e2 /r . Give the value of E in eV. (1.4) Express in terms of |ψ0  and |n, l, m the probability amplitude c(n, l, m) and the probability p(n, l, m) to find the electron in the state |n, l, m of 3 H e+ after the decay. Show that only the probabilities pn = p (n, 0, 0) do not vanish. (1.5) Calculate the probability p1 to find the electron in the ground state of 3 H e+ . What is the corresponding contribution to E? (1.6) A numerical calculation gives the following values:

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11 The Hydrogen Atom

p2 =

1 , 4

∞

pn = 0.02137,

n=3

∞ pn = 0.00177. n2 n=3

 Calculate the probability ∞ n=1 pn to find the atomic electron in a bound state of 3 H e+ and the corresponding contribution to E. What do you think of the result? (1.7) Experimentally, in the β decay of the tritium atom one observes that, in about 3 % of the events, there are two outgoing electrons, one with a mean kinetic energy E c  ∼ 15 keV, the other with E c  ∼ 34.3 eV, thus leaving a completely ionized 3 H e2+ nucleus, as if the β decay electron “ejected” the atomic electron. Can you explain this phenomenon?

11.6.1 Solution (2.1) The two hamiltonians are Hˆ 1 = p 2 /2m − e2 /r , and Hˆ 2 = p 2 /2m − 2e2 /r. (2.2) The levels of a hydrogen-like atom of nuclear charge Z are E n = −Z 2 E I /n 2 2 and, in this case, E n = −4E I /n . The new Bohr radius is a2 = a1 /2, and the wave −r/a2 / (πa2 3 ). function ϕ100 (r) = e (2.3) The expectation value of the electron energy in the new nuclear configuration is: E = ψ0 |H2 |ψ0  = ψ0 |H1 |ψ0  − e2 ψ0 |(1/r )|ψ0 , which amounts to E = −E I − e2 /a1 = −3E I ∼ −40.8 eV. (2.4) By definition, the probability amplitude is c(n, l, m) = n, l, m|ψ0 , and the probability p(n, l, m) = |n, l, m|ψ0 |2. The analytic form is c(n, l, m) = Rnl (r ) Ylm∗ (θ, φ) ψ0 (r)d 3 r, where Rnl (r ) are the radial wave functions of the 3 H e+ hydrogen-like atom. Since ψ0 is of the form ψ0 (r) = χ(r )Y00 (θ, φ), the orthogonality of spherical harmonics implies p(n, l, m) = 0 if l, m  = 0. (2.5) The probability amplitude in the lowest energy state is  ( p1 )1/2 = 4π

√ e−r/a2 e−r/a1 2

r dr = 16 2/27. πa23 πa13

Hence the probability p1 = 0.70233 and the contribution to the energy p1 E 1 = −38.2 eV. (2.6) With the numerical values given in the text, one has p2 E 2 = −E I /4= −3.4 eV, ∞ and p = ∞ 1 pn = 0.9737. The contribution to E is E B  = 1 pn E n = −3.0664 E I = −41.7 eV. The total probability is smaller than 1; there exists a non-zero probability (1 − p) = 0.026 that the atomic electron is not bound in the final state.

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277

The contribution of bound states E B  = −41.7 is smaller than the total expectation value of the energy E. The probability (1 − p) corresponds therefore to a positive electron energy, that is to say an ionization of 3 H e+ into 3 H e2+ with emission of the atomic electron. (2.7) There is necessarily a probability 1 − p = 0.026 for the atomic electron not to be bound in helium, therefore that the atom be ionized in the decay. If the mean kinetic energy of the expelled electron is E c ∼ 34.3 eV, this represents a contribution of the order of (1 − p)E c ∼ +0.89 eV to the mean energy which compensates the apparent energy deficit noted above. Comment: This type of reaction has been intensively studied in order to determine the neutrino mass. If M1 and M2 are the masses of the two nuclei, E β the energy of the β electron, E the energy of the atomic electron, and E ν¯ the neutrino energy, energy conservation is, for each event: M1 c2 − E I = M2 c2 + E β + E ν¯ + E. For a given value of E, the determination of the maximum energy of the β electron (which covers all the spectrum up to 19 keV in the tritium atom case) provides a method for determining the minimum value m ν¯ c2 of E ν¯ through this energy balance. An important theoretical problem is that current experiments are performed on molecular tritium (HT or TT molecules) and that molecular wave functions are not known explicitly, contrary to the atomic case considered here.

Chapter 12

Spin 1/2

Spin 1/2 is the first truly revolutionary discovery of quantum mechanics. The properties of this physical quantity in itself, the importance of its existence, and the universality of its physical effects were totally unexpected. The physical phenomenon is the following. In order to describe completely the physics of an electron, one cannot use only its degrees of freedom corresponding to translations in space. One must take into account the existence of an internal degree of freedom that corresponds to an intrinsic angular momentum. In other words, the electron, which is a pointlike particle, “spins” on itself. We use quotation marks for the word “spins”. One must be cautious with words, because this intrinsic angular momentum is purely a quantum phenomenon. It has no classical analogue, except that it is an angular momentum. One can use analogies and imagine that the electron is a sort of quantum top. But we must keep in mind the word “quantum”. The electron is a pointlike object down to distances of 10−18 m. One must admit that a pointlike object can possess an intrinsic angular momentum. (As we have already pointed out, in this respect, the photon, which is pointlike, has an intrinsic angular momentum, and is a zero mass particle, is from this point of view even more strange.)

12.1 Experimental Results Experimentally, this intrinsic angular momentum, called spin, has the following manifestations (we do not enter in any technical detail).

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_12

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12 Spin 1/2

1. If we measure the projection of the spin along any axis, whatever the state of the electron, we find either of two possibilities +/2, or

− /2.

There are two and only two possible results for this measurement. 2. Consequently, if one measures the square of any component of the spin, the result is 2 /4 with probability one. 3. Therefore, a measurement of the square of the spin S 2 = Sx2 + Sy2 + Sz2 gives the result 32 . S2 = 4 4. A system that has a classical analogue, such as a rotating molecule, can rotate more or less rapidly on itself. Its intrinsic angular momentum can take various values. However, for the electron, as well as for many other particles, it is an amazing fact that the square of its spin S 2 is always the same. It is fixed: all electrons in the universe have the same values of the square of their spins S 2 = 32 /4. The electron “spins” on itself, but it is not possible to make it spin faster. One can imagine that people did not come to that conclusion immediately. The discovery of the spin 1/2 of the electron is perhaps the most breathtaking story of quantum mechanics. The elaboration of the concept of spin was certainly the most difficult step of all quantum theory during the first quarter of the 20th century. It is a real suspense that could be called the various appearances of the number 2 in physics. There are many numbers in physics; it is difficult to find a simpler one than that. And that number 2 appeared in a variety of phenomena and enigmas that seemed to have nothing to do a priori with one another, or to have a common explanation. The explanation was simple, but it was revolutionary. For the first time people were facing a purely quantum effect, with no classical analogue. Nearly all the physical world depends on this quantity, the spin 1/2. The challenge existed for a quarter of a century (since 1897). Perhaps, there was never such a long collective effort to understand a physical structure. It is almost impossible to say who discovered spin 1/2, even though one personality dominates, Pauli, who put all his energy into finding the solution. We show that in order to manipulate spin 1/2, and understand technicalities we essentially know everything already. We have done it more or less on two-state systems. But for anybody, it is a complicated matter to have a really intuitive representation of spin 1/2. It is really a personal affair, as can be seen in Fig. 12.1.These gentlemen are simply discussing spin effects.

12.2 Spin 1/2 Formalism

281

Fig. 12.1 Three physicists discussing the optimal way to measure spin effects in proton collisions at the Argonne ZGS accelerator (CERN document)

12.2 Spin 1/2 Formalism The measurement results fit perfectly with the general framework of the theory of angular momenta. The electron spin is a half-integer angular momentum that corresponds to the quantum numbers j = 1/2, m = 1/2. The corresponding Hilbert space is two-dimensional. Any spin state is a linear superposition of two basis states and the degree of freedom corresponding to spin is described in a two-dimensional Hilbert space: Espin . Representation in a Particular Basis We choose a basis of states in which both Sˆ 2 and Sˆ z are diagonal, which we denote {|+, |−}:   32 |±. Sˆ z |+ = |+, Sˆ z |− = − |−, Sˆ 2 |± = 2 2 4

(12.1)

Using the notation of Chap. 10, the states |± would be |j = 1/2, m = ±1/2. The action of Sˆ x and Sˆ y on the elements of the basis is written as (see Eq. (10.16)) Sˆ x |+ = /2|−, Sˆ y |+ = i/2|−,

Sˆ x |− = /2|+ Sˆ y |− = −i/2|+.

(12.2) (12.3)

An arbitrary spin state |Σ can be written as: |Σ = α |+ + β |−, |α|2 + |β|2 = 1.

(12.4)

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The probabilities of finding +/2 and −/2 in a measurement of Sz on this state are P(+/2) = |α|2 , P(−/2) = |β|2 . We will often call |+ and |− respectively the “spin up” and “spin down” states with respect to the z axis, but we keep in mind that it has very little geometrical meaning. Matrix Representation It is convenient to use matrix representations for the states and the operators: |+ =

      1 0 α , |− = , |Σ = . 0 1 β

(12.5)

We can use the Pauli matrices σˆ ≡ {σˆ x , σˆ y , σˆ z }  σˆ x =

     01 0 −i 1 0 , σˆ y = , σˆ z = 10 i 0 0 −1

(12.6)

which satisfy the commutation relations σˆ × σˆ = 2i σ. ˆ

(12.7)

The spin observables are represented as  Sˆ = σ. ˆ 2

(12.8)

In this basis, the eigenstates |±x of Sˆ x and |±y of Sˆ y are: 1 |±x = √ 2



   1 1 1 , |±y = √ . ±1 ±i 2

(12.9)

The major complication, compared to the ammonia molecule of Chap. 7 is to deal with a vector degree of freedom, and a vector observable. But by studying angular momenta, we understood how to deal with that situation. We remark that, as we already know in different systems, the state |+ for which we are sure that Sz = /2 can also be considered as a linear superposition with equal weights of Sx = ±/2 or Sy = ±/2. If the spin “points” upwards, it points both on the left and on the right with equal probabilities. This is again the superposition principle. Notice that we just said that the spin points upwards, and we have spoken classically of a purely quantum property. That is not too severe, because we have a mathematical dictionary to translate it properly. But remember that Galileo just escaped being burned to death, and Giordano Bruno was burned to death, because they said that it is the earth that rotates and that the sun is fixed. Nevertheless, this does not

12.2 Spin 1/2 Formalism

283

prevent people nowadays from saying that the sun rises and sets everyday. Once things have been understood, it’s much simpler, in everyday life, to come back to traditional ways of visualizing things.

12.3 Complete Description of a Spin 1/2 Particle We now turn to the complete description of the state of an electron taking into account both the space degrees of freedom and the spin. Here, we follow a more direct path than the traditional way of showing a nontrivial tensor product of Hilbert spaces. The result is quite intuitive. We deal with two random variables, position and spin along the z-axis. We need the probability law of this couple of variables. The state of a particle in space is described by a square-integrable function that belongs to what we call the “external” Hilbert space Eexternal . At any point in space, the spin state is described by a two-dimensional vector of Espin , the “internal” degree of freedom. We therefore need two wave functions to describe the state of the electron in space with its spin. We must double the number of wave functions (or the “dimension” of the Hilbert space). A tensor product of spaces is as simple as that in this particular case. There are several possible representations of the states, together with corresponding representations of observables. Choosing a particular representation is a matter of convenience. Mixed Representation The state is represented by a vector of Espin whose components are square-integrable functions: (12.10) ψ+ (r, t) |+ + ψ− (r, t) |−. The physical interpretation of this representation is as follows. |ψ+ (r, t)|2 d 3 r (resp., |ψ− (r, t)|2 d 3 r) is the probability of finding the particle in a volume d 3 r around the point r, with a spin component +/2 (resp., −/2) along the z-axis. Two-Component Wave Function The state vector is represented in the form: 

 ψ+ (r, t) , ψ− (r, t)

(12.11)

The physical interpretation of ψ+ and ψ− as probability amplitudes for the couple of random variables (r, Sz ) is the same as above.

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12.3.1 Observables The space observables (x, id/dx) only act on wave functions ψ± (r, t). They do not make any difference between spin states. In the matrix representation the space observables are diagonal 2 × 2 matrices whose elements are operators and that do not act on spin variables. Spin observables are 2 × 2 matrices with numerical coefficients seen above. They do not act on space variables. In the general case, there exist observables that can act on both sets of variables. These are simply the products of the 2 × 2 matrices by the operators; for instance, rˆ · Sˆ = x Sˆ x + ySˆ y + zSˆ z . More generally, the products of operators Aˆ ext Bˆ sp follow the rule     Aˆ ext Bˆ sp (ψσ (r, t)|σ) = Aˆ ext ψσ (r, t) Bˆ sp |σ , σ = ±.



(12.12)

Functions of a Two-Valued Variable. In some problems, it is useful to use a single wave function depending on four variables (plus time) ψ(r, σ; t), where the fourth variable σ can take the values ±1. One has obviously ψ(r, σ; t) = ψσ (r, t), σ = ±1.

Atomic States In many problems of atomic physics, it is useful to use the quantum numbers n, , m to classify the states |n, , m that form a basis of Eexternal . The introduction of spin is done in the space spanned by the family {|n, , m |σ} where the spin quantum number can take the two values ±1. It is convenient to use the compact notation |n, , m, σ,

(12.13)

where the states of an electron are described by four quantum numbers. The action of space operators on the states |n, , m is known (see Chap. 11), therefore the action of general operators on the states |n, , m, σ can readily be inferred from the considerations developed above.

12.4 Physical Spin Effects

285

12.4 Physical Spin Effects Physical spin effects belong to the following main categories. 1. There are angular momentum effects, in particular in nuclear and particle physics. It is compulsory to take into account spin in order to observe conservation of the total angular momentum. The structure of fundamental interactions relies on the spin 1/2 of the electron, the quarks, and the neutrinos. This is too difficult to examine here. 2. There are surprising and revolutionary effects compared to classical physics that are due to the Pauli principle. Such effects are fundamental in order to understand the structure of matter, atoms, molecules, solids, and liquids. We show some of them in Chap. 14. The Pauli principle plays a central role in statistical physics and in chemistry. 3. There are magnetic effects that actually are at the origin of the discovery of spin 1/2. We mainly use those effects in order to develop our experimental discussion.

12.5 Spin Magnetic Moment Indeed, all what has been said above is for the moment simple matrix calculus. But it will acquire a much greater physical flavor because to the spin of the electron (or any spin 1/2 particle) there corresponds an intrinsic magnetic moment, a spin magnetic moment, that is proportional to the spin. ˆ μ ˆ = γ Sˆ = μ0 σ.

(12.14)

The gyromagnetic ratio γ, the value of which we come back to, is a characteristic of the particle under consideration. ˆ on the state vectors is obvious. The action of μ, ˆ which is proportional to S, It is through measurements of this magnetic moment that we can find the easiest access to spin measurements. For instance, if we place this spin magnetic moment in a field B, there is an interaction Hamiltonian Wˆ = −μ ˆ · B, (12.15) which leads to a number of experimental observations. Hamiltonian of a One-Electron Atom In what follows we consider an atom with one external electron in its ground state. Globally, the atom is electrically neutral. However, it carries the spin 1/2 of the external electron, and the corresponding magnetic moment.

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Suppose the atom is moving in space and that it is placed in a potential V (r), and that furthermore it is placed in a magnetic field B. The magnetic potential energy is given by (12.15). The Hamiltonian is the sum of two terms Hˆ = Hˆ ext + Wˆ , where

(12.16)

pˆ 2 Hˆ ext = + V (ˆr) 2m

is of the type seen in previous Chapters. In particular, Hˆ ext does not act on the spin variables. The operator Wˆ is given by (12.15). It acts in Espin through the three operators μˆ x , μˆ y , μˆ z . If the field is inhomogeneous, it also acts on space variables through the three functions Bx (ˆr), By (ˆr), Bz (ˆr). The Schrödinger equation is i

d ˆ |ψ = H|ψ. dt

(12.17)

If we choose the representation of states (12.11) and if we decompose on the orthonormal basis {|+, |−}, we obtain the coupled differential system   2 ∂ ψ+ (r, t) = − Δ + V (r ) ψ+ (r, t) ∂t 2m ++|Wˆ |+ ψ+ (r, t) + +|Wˆ |− ψ− (r, t),   2 ∂ i ψ− (r, t) = − Δ + V (r ) ψ− (r, t) ∂t 2m +−|Wˆ |+ ψ+ (r, t) + −|Wˆ |− ψ− (r, t). i

The matrix elements of Wˆ in the basis {|+, |−} are functions of the external space variables. They add to the usual potential energy terms, which are diagonal, and, in general, they couple the evolution equations of the components ψ+ and ψ− . We now have all the tools to be able to do physics and to discover the suspense of the discovery of spin 1/2.

12.6 The Stern–Gerlach Experiment The first measurement of the intrinsic magnetic moment of the electron, and the first undeniable appearance of the number 2 came in 1921–1922 with the experiment of Stern and Gerlach.

12.7 Principle of the Experiment

287

12.7 Principle of the Experiment A collimated beam of silver atoms is sent in a region where an inhomogeneous magnetic field is applied along the z-direction, perpendicular to the initial velocity of the atoms (Fig. 12.2a). The possible deflection of the beam by the field gradient is then measured by observing the impacts of the atoms on a detection plate perpendicular to the initial direction of the beam. The silver atom has one external electron in an orbital angular momentum state  = 0. Therefore, the atom’s magnetic moment is equal to the intrinsic magnetic moment of the valence electron.

12.7.1 Semi-classical Analysis We first analyze this experiment within classical mechanics. The atoms are neutral and they are not subject to a magnetic Lorentz force. However, because they have a nonvanishing magnetic moment μ, the force Fz = μz

∂Bz , ∂z

(12.18)

parallel to the z-direction, acts on them and deflects their trajectory.

Fig. 12.2 a The Stern–Gerlach experiment: atoms from a collimated beam are deflected as they cross a region where an inhomogeneous magnetic field is applied. This experiment can be interpreted as a measurement of the component of the atomic magnetic moment along the direction of the field (z in the Figure). b Magnetic gradient between the polar pieces of the magnet

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We recall that when a magnetic moment μ is placed in a magnetic field B the magnetic interaction energy is W = −μ · B, (12.19) and the torque Γ =μ×B

(12.20)

is exerted on the magnetic moment. In addition, if the magnetic field is inhomogeneous, the force (12.18), or more generally F = ∇(μ · B) =



μi (t) ∇Bi ,

(12.21)

i=x,y,z

acts on the dipole. As we have seen in (10.37), the magnetic moment of an atom at r does not align with the axis of the local magnetic field B(r), but it precesses along this axis with the Larmor frequency (12.22) ω0 = −γ0 B(r). We assume that the classical trajectory of the atoms lies in the plane of symmetry x = 0 of the magnet (see Fig. 12.2b). Along this trajectory the magnetic field is always parallel to the z-axis, so that the Larmor precession takes place around z. Also, owing to the symmetry of the device, the quantities ∂Bz /∂x and ∂Bz /∂y vanish along the atomic beam trajectory (we neglect possible edge effects). If the displacement of the magnetic moment during a single precession period 2π/ω0 is small compared with the typical variation scale of the magnetic field, we can average the force (12.21) over the Larmor period. The contributions of μx and μy to (12.21) then vanish, and one is left only with the z-component of the force Fz = μz (t) ∂Bz /∂z. In addition we deduce from (10.37) that μz stays constant as the atom moves in the magnetic field gradient, which is helpful to justify the result (12.18).

12.7.2 Experimental Results In the absence of a magnetic field gradient one observes a single spot on the detecting plate, in the vicinity of x = z = 0 (Fig. 12.3a). The magnetic field gradient provides a way to measure the z-component of the magnetic moment of the atoms as they enter the field zone. Let us assume that all the atoms carry the same magnetic moment of norm μ0 , and that this moment is oriented at random when an atom enters the field zone. Classically, this should produce some continuous distribution of μz between the two extreme values −μ0 and +μ0 , so that one would expect that the impacts of the atoms on the screen form an extended line parallel to z (Fig. 12.3b). The endpoints of the line correspond to atoms whose magnetic moments are oriented respectively such as μz = +μ0 and μz = −μ0 .

12.7 Principle of the Experiment

289

Fig. 12.3 Possible results of a Stern–Gerlach experiment. a In the absence of a magnetic gradient no deflection of the atomic trajectories occurs, and the atoms form a single spot around the point x = z = 0; each dot represents the impact of an atom on the detection screen. b Simulation of the result expected from classical mechanics, assuming that all atoms carry the same magnetic moment μ0 with a random orientation; the distribution of the z-component of the magnetic moment is uniform between −μ0 and +μ0 . c Simulation of the result found experimentally with silver atoms: the experiment, which can be considered as a measurement of the z-component of the magnetic moment, yields only the two results +μ0 and −μ0

The experiment is difficult. But the observation differs radically from this classical prediction. The set of impacts never forms a continuous line on the screen. For atoms such as silver, the impacts are grouped in two spots corresponding to μz = +μ0 and μz = −μ0 , with μ0 = 9.27 10−24 J T−1 (Fig. 12.3c). The result μ0 of the Stern–Gerlach experiment is consistent with: μ0 =  |γ0 | =

q , 2me

(12.23)

which amounts to taking L =  in (10.34). The quantity (12.23) is the absolute value of the Bohr magneton.

12.7.3 Explanation of the Stern–Gerlach Experiment Our theory explains the observed spatial separation of the states |±z . We consider an incident atomic beam propagating along y; each atom possesses a magnetic moment. In a region of length L, a magnetic field B parallel to z is applied with a gradient along z: (12.24) B(r) = Bz (r)uz , with Bz (r) = B0 + b z. In full rigor, the Eq. (12.24) is incorrect because the field B(r) does not satisfy ∇.B = 0. A more realistic calculation can be done with a field B = B0 uz + b (zuz − xux ) which satisfies Maxwell equations. If the dominant part of the field B0 uz is much larger than the transverse field −b xux on the transverse extension Δx of the

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atomic wave packet (i.e., B0  b Δx), the eigenstates of −μ ˆ · B remain practically equal to |±z and the present approach is valid. Under these conditions, the Schrödinger equation (12.17) can be decoupled into two equations:  2  pˆ ∂ − μ0 B ψ+ (r, t), i ψ+ (r, t) = ∂t 2m  2  ∂ pˆ i ψ− (r, t) = + μ0 B ψ− (r, t). ∂t 2m

(12.25) (12.26)

These two equations are both of the same type as the Schrödinger equation seen in Chap. 5, but the potential is not the same for ψ+ and ψ− . In order to proceed further we set:  (12.27) π± = |ψ± (r, t)|2 d 3 r, π+ + π− = 1, where π+ and π− are the probabilities of finding μz = +μ0 and μz = −μ0 . We deduce from (12.25) and (12.26) that dπ− dπ+ = = 0. dt dt

(12.28)

√ φ± (r, t) = ψ± (r, t)/ π± ,

(12.29)

We define the functions:

which are the conditional probability amplitudes of particles for which μz = ±μ0 . These normalized functions also satisfy the Schrödinger-type equations (12.25) and (12.26). We now define:  (12.30) r±  = r |φ± (r, t)|2 d 3 r,   p±  = φ∗± (r, t) ∇φ± (r, t) d 3 r, (12.31) i where r+  (resp., r− ) is the average position of particles for which μz = +μ0 (resp., μz = −μ0 ), and p±  are their average momenta. A simple application of the Ehrenfest theorem gives: (d/dt)r±  = p± /m, (d/dt)px±  = (d/dt)py±  = 0, (d/dt)pz±  = ±μ0 b .

(12.32) (12.33) (12.34)

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291

At t = 0 we assume that r±  = 0, px±  = pz±  = 0, py±  = mv. We obtain at time t: x±  = 0, y±  = vt, z±  = ±μ0 b t 2 /2m.

(12.35)

Therefore there is a spatial separation along z of the initial beam into two beams. One beam corresponds to μz = +μ0 , and the other to μz = −μ0 . When the beams leave the magnet of length L, their separation is δz = z+  − z−  =

μ0 b L 2 . m v2

(12.36)

If the field gradient is sufficiently strong so that δz > Δz (separation larger than the spatial extension of each wave packet), we obtain two beams: one in the internal state |+, the other in the state |−. Therefore, the formalism we have developed in this section explains completely the Stern–Gerlach experiment and its results. As was announced in the case of the population inversion for the ammonia molecule in Chap. 7, this experiment elicits two fundamental aspects of a measurement process in quantum mechanics: • A measurement requires a finite spatial extension (δz = 0 if L = 0). • A measurement is never instantaneous (δz = 0 if T = L/v = 0). These two aspects were absent in the formulation of the principles of quantum mechanics presented in Chapter 6. Finally a simple inspection of the evolution of the expectation value of the separation of the two spots, and of their dispersion at the exit of the magnet, leads to the following result. Let T = L/v be the time the atoms spend in the inhomogeneous magnetic field, and let us note E⊥ = p2z /2m the transverse energy communicated to the atom by the field gradient. In order to observe the splitting, the following condition must be satisfied, T E⊥ ≥ /2. This condition, where the value of the field gradient has disappeared, is one important aspect of the so-called time-energy uncertainty relation that appears in any quantum measurement.

12.7.4 Successive Stern–Gerlach Setups Consider now the experimental situation shown in Fig. 12.4. We place two consecutive magnets. The first has a field gradient directed along z and it splits the incident

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Fig. 12.4 A beam of silver atoms crosses two magnetic field zones. The first creates a field gradient along z, the second a field gradient along x. After the first magnet, a shutter only lets the atoms in the internal state |+z pursue. The second magnet allows us to perform a measurement of the xcomponent of the magnetic moment. One finds the two results +μ0 and −μ0 with equal probabilities

beam into two beams corresponding to the two internal states |+z and |−z . When the beams leave the field zone, we stop the beam corresponding to the state |−z and we keep only the beam in the state |+z . This latter beam is then sent into another Stern–Gerlach device whose axis is along the x-axis, orthogonal to z. We therefore perform a measurement of the x-component of the atomic magnetic moment, whose corresponding observable is μˆ x . The result observed experimentally is that the beam is again split in two beams of equal intensities corresponding to values of the magnetic moment along x equal to +μ0 and −μ0 , respectively. This, naturally, is the manifestation of the superposition principle. The eigenstates of μˆ x and of μˆ y are 1 1 |±x = √ (|+z ± |−z ) , |±y = √ (|+z ± i|−z ) . 2 2 There is nothing surprising.

(12.37)

12.7 Principle of the Experiment

293

12.7.5 Measurement Along an Arbitrary Axis Suppose we are interested in measuring the component of the magnetic moment along an arbitrary axis. This is shown in Fig. 12.5. We place a Stern–Gerlach apparatus along an arbitrary direction defined by the unit vector uθ such that: uθ = ux sin θ + uz cos θ.

(12.38)

This corresponds to a measurement of the component μθ of the magnetic moment along uθ (i.e., μθ = μx sin θ + μz cos θ). By the correspondence principle, the corresponding observable is:  μˆ θ = μˆ x sin θ + μˆ z cos θ = μ0

cos θ sin θ sin θ − cos θ

 .

(12.39)

This choice guarantees that the expectation values μx , μy , and μz  of the components of the magnetic moment transform as the components of a usual three-vector under rotations, which is essential. Just as μˆ x , μˆ y , μˆ z , the operator μˆ θ has the eigenvalues +μ0 and −μ0 . Its eigenvectors are:

Fig. 12.5 A beam of silver atoms is prepared in the state |+z . It then crosses a field gradient directed along uθ . In this measurement of the component of the magnetic moment along uθ , the two possible results are +μ0 and −μ0 with respective probabilities cos2 θ/2 and sin2 θ/2. The graph in the lower right corner shows a typical result for θ = π/4

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 cos(θ/2) , sin(θ/2)   − sin(θ/2) . |−θ = −|+z sin (θ/2) + |−z cos (θ/2) = cos(θ/2)

|+θ =

|+z cos (θ/2) + |−z sin (θ/2) =

(12.40) (12.41)

The experimental observations are the following. If a beam prepared in the state |+z is sent in the field gradient directed along uθ , one finds that this beam is split into two beams corresponding to a magnetic moment along uθ equal to +μ0 and −μ0 , with relative intensities I+ (θ) = I+ (0) cos2 (θ/2) and I− (θ) = I+ (0) sin2 (θ/2). In order to account for this result, we apply the principles of Chap. 6. A measurement of μˆ θ can give two possible values, the eigenvalues +μ0 and −μ0 ; if the initial system is in the state |+z , the respective probabilities for these two issues are: p+ = |θ +|+z |2 = cos2 (θ/2) p− = |θ −|+z |2 = sin2 (θ/2).

(12.42) (12.43)

Therefore, Eq. (12.39) explains why the experimental measurement, which involves a large number of atoms, gives two spots with relative intensities cos2 (θ/2) and sin2 (θ/2). The measurement only gives a result with probability 1 when θ is 0 or π, that is, when the preparation axis uz and the measurement axis uθ are parallel or antiparallel.

12.8 The Discovery of Spin 12.8.1 The Hidden Sides of the Stern–Gerlach Experiment Coming back to Stern and Gerlach, one naturally thinks that they have discovered spin 1/2 ! They have found the quantization of the magnetic moment, the superposition principle, and the explanation of the electron’s magnetic moment if not its spin. It is with that type of reasoning and observations that Fresnel had founded the wave theory of light, in particular, the laws of polarization, which was a similar and difficult problem in the 19th century. Absolutely not! The result was considered 1. As perfectly natural 2. And as a brilliant confirmation of the old quantum theory of Bohr and Sommerfeld (remember this happened in 1922) If people had thought about it, they might have found the answer. But at that time, physicists were concerned with quite a different problem: they wanted to prove the old quantum theory and the quantization of trajectories that Bohr had used in his model of the hydrogen atom.

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Actually the Stern–Gerlach experiment had been induced by theorists, Born, Bohr, Sommerfeld, and Pauli, for months if not for years, and they had predicted the result. In physics, there is some sort of credibility principle for an experimental result. One believes in a result if and only if a theory has predicted it. An example is the discovery of the 3 K cosmic background radiation by Penzias and Wilson. It was considered as a background noise until they were told about the prediction of Gamow of the existence of background radiation in the big bang theory. The quantization of Lz had been guessed very early. Sommerfeld, in the old quantum theory, had predicted the spatial quantization of trajectories and the directional quantization in a magnetic field B. He knew that μ = γL, and that the orbital gyromagnetic ratio is q/2m. Sommerfeld understood the principle of the experiment as soon as 1918. And he expected a lot from it because it would have been the first proof of quantization in a nonradiative process. But one could argue that there should be three spots and not two. Imagine an electron in a circular uniform motion around a proton. The quantization of angular momentum is an integer multiple of . In a magnetic field, the plane of the trajectory could have three directions corresponding, respectively, to an angular momentum parallel, antiparallel, or perpendicular to the field B with Lz = , Lz = −, or Lz = 0. Not at all! As soon as 1918, Bohr proved that the trajectory Lz = 0 was unstable. One should therefore observe only two spots Lz = ±1.1 Max Born insisted in 1920 (he was 30), “This experiment must absolutely be done.” At that time, Born was a professor in Frankfurt, where there was an artist of atomic and molecular beams, Otto Stern (32 at that time), but Stern wasn’t interested. So, Born, who was a mathematician, decided to do experiments. And he managed to do so thanks to a talented assistant Fraulein Elizabeth Bormann. This new activity of Born was a surprise to all physicists. (One day, Rutherford asked him if he had a relative doing experiments. Born answered, “No, but I have a good assistant.”) But Born had to face the facts; he suffered from the Pauli effect: the better you are as a theorist, the more you are a disastrous experimentalist. Whenever Pauli entered a laboratory, everything went wrong. One day, in Göttingen, an experimental setup of Franck exploded. Everyone looked for Pauli, but there was no trace of him. Some time later, someone learned that at the precise time of the explosion, Pauli was on a train, which had stopped in Göttingen, on the way from Munich to Hanover. The Pauli effect acted at a distance! Born eventually convinced Stern. Actually, Stern did not know what to think. At first he proposed the experiment, but some time later he was skeptical, “Quantum restrictions on trajectories are simply calculational rules. I’m going to show once for all that what theorists say is nonsense.” However, Stern suffered somewhat from the Pauli effect. All his experimental setups were constructed by his technician. He knew remarkably how to conceive them, but he wasn’t very skillful.

1 It

is one of the few times that Bohr predicted correctly the result of an experiment. He did it with a wrong argument.

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And that time, it was too difficult. Neither the technician nor Fraulein Bormann succeeded. Fortunately, Gerlach, who was a very talented 21-year-old experimentalist, had just arrived in Frankfurt, after graduating in Tübingen. Born said “Thank God, now we have at last someone who knows how to do experiments!” Gerlach took care of everything–the technician, Fraulein Bormann, and Stern’s ideas–and he did the experiment. He was successful, and found the two spots. It seemed to be a triumph for Sommerfeld. Pauli (22 at that time) congratulated Gerlach and said to: “Let us hope now that the old unbeliever Stern will now be convinced of directional quantization!” The triumph was even greater because, by measuring μ0 they found to a few percent q μ0 = |γ0orb | = , (12.44) 2me exactly the prediction of Bohr and Sommerfeld! At that time, nobody could suspect that Nature had played a bad trick. Equation (12.44) must be read as μspin = (

q  q )( ), and not ( ) ! me 2 2me

In other words, the spin gyromagnetic ratio is twice the orbital gyromagnetic ratio, and the angular momentum is /2. Dirac proved that in 1927, for any charged pointlike spin 1/2 particle, in his theory of a relativistic electron. Einstein used to say that the Lord is not mean, but he is subtle. On that point, the Lord had really been nasty!

12.8.2 Einstein and Ehrenfest’s Objections However, there were some skeptical people who thought about the physics. Because the experiment is amazing, it is completely opposed to classical conceptions. Einstein and Ehrenfest performed a critical analysis of the experiment in semi-classical terms as in Sommerfeld’s theory. They calculated the time it would take a loop of current to orient itself in the magnetic field, and they found a value of t  109 s, that is, 30 years! Yet the atoms stay in the inhomogeneous field for 10−4 s. Einstein and Ehrenfest concluded: “We must make a complete revisal of our classical ideas …this experimental shows a conceptual dead end …we must find a new quantum idea.” Indeed, for us, in quantum mechanics, this is not a problem. In the initial beam there is a probability amplitude for the spin to point upwards or downwards, and the orientation of the magnetic moment is “already there,” to speak in classical terms.

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Now, one can think that with Einstein getting involved in that business, and with that conclusion of his, people started wondering and found spin 1/2. The discovery of the number 2 must have occurred at that time. Not at all; nobody paid any attention.

12.8.3 Anomalous Zeeman Effect At that time, physicists were all concerned with a true intellectual challenge that seems to have nothing to do with all that, the anomalous Zeeman effect. We place an atom, prepared in a state of energy E and angular momentum j, in a magnetic field B parallel to z. The magnetic energy is: Wˆ = −μ ˆ · B.

(12.45)

The corresponding level is split in 2j + 1 sublevels of respective energies: E − γB0 m, m = −j, . . . , j. A corresponding splitting of each line is observed in the spectrum. If all angular momenta are orbital angular momenta (i.e., they have a classical interpretation), j must be an integer. In that case 2j + 1 is odd and we expect a splitting in an odd number of levels. The splitting of spectral lines in a magnetic field, first observed by Zeeman in the period 1896–1903, showed that in many cases, in particular for alkali atoms, this is not true. There is a splitting into an even number of levels. Faraday had been convinced as early as 1845 that there was a deep connection between optical and magnetic phenomena. In one of the last experiments of his life, in 1862, he attempted to find the influence of magnetic fields on radiation. Many technical problems prevented him from obtaining a positive answer. It was only in 1896 that these experiments were redone successfully by Zeeman. Already at that time, theorists, in particular H.A. Lorentz, predicted one should observe a splitting in an odd number of lines (1—no splitting—or 3). Zeeman first confirmed this result on the spectra of cadmium and zinc. The discovery, in the particular case of sodium, of what was to be called the “anomalous Zeeman effect”, namely an even number of lines, remained for more than 25 years a real challenge for the scientific community, which was totally confused by this phenomenon. It was only after many struggles, in the years 1925–1926, with the ideas of Pauli, and of Uhlenbeck and Goudsmit, that the introduction of the notion of spin completely clarified the problem. The “anomalous Zeeman effect” then appeared as a perfectly normal phenomenon. It was a real antique tragedy. One must realize that if we have the theory in front of us (i.e. that there exist half-integer angular momenta for which 2j + 1 is even) physicists did not have it at that time, despite the works of Elie Cartan in 1913.

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12.8.4 Bohr’s Challenge to Pauli The anomalous Zeeman effect seemed to be an impenetrable wall. Bohr thought there was only one person capable of solving that problem: Pauli, whom Bohr invited to Copenhagen in 1922. Pauli was 22, he had been Sommerfeld’s student in Munich, and he was known for his remarkable book on relativity. He had a difficult temper, as we said; for him most things were either obvious or stupid. One day, Ehrenfest had said to him, “You know, Herr Pauli, I find your paper on relativity much nicer than you are.” Pauli replied, “That’s funny! For me it’s the opposite, I find you much nicer than all the rubbish you write.” Of course, Pauli’s colleagues were terrified. Franck told him; “Please behave when you get to Copenhagen.” “I’ll have to learn Danish”, said Pauli. Niels Bohr asked Pauli two questions. • Why are there shells in complex atoms, containing 2, 8, 18, and so on; that is 2n2 electrons, n = 1, 2, 3 . . .? This remark had been made by Rydberg in 1903. Why aren’t the electrons all in the ground state? • What is the cause of the Zeeman effect? And Pauli started thinking. He thought very hard. So hard that he said, in his Nobel prize Chapter in 1945, that one day, as he was strolling in the beautiful streets of Copenhagen, a colleague had said to him in a friendly manner, “You look very unhappy;” whereupon Pauli replied fiercely, “How can one look happy when he is thinking about the anomalous Zeeman effect.” Pauli understood that both the anomalous Zeeman effect and the effect of the inner core of electrons are a conceptual dead end in classical terms. And, at the beginning of 1924, W. Pauli, Jr. wrote “The electron has a new, specifically quantum property, which corresponds to a two-valued physical quantity, which cannot be described classically. In an atom, an electron is characterized by four quantum numbers, (n, l, m, σ) where σ = ±1.” And Pauli stated the principle of exclusion of equivalent electrons, called by Heisenberg the “Pauli verbot,” which is very deep and also revolutionary. At this point, one is tempted to say, “Fine, Pauli found the answer.” Not at all.

12.8.5 The Spin Hypothesis The hypothesis of the electron spin, such as presented above, that the electron has an intrinsic angular momentum S whose measurement along an axis gives one of the two values ±/2, and whose associated magnetic moment μS = γS S is such that γS = q/m = 2γorbital , was proposed by two young Dutch physicists in 1925, Uhlenbeck and Goudsmit. Uhlenbeck hesitated between a career in physics and in history, and Goudsmit did not have his degree yet. As soon as they understood that their hypothesis explained many experimental facts which were obscure up to then,

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they discussed it with their professor, P. Ehrenfest, who encouraged them to publish their work. Their idea was received with various reactions in the physics community. Bohr was very enthusiastic, whereas Pauli and Lorentz found serious objections. The objection of Lorentz comes from relativity. If one imagines that the electron is a sphere whose electrostatic energy is equal to its mass me c2 , one finds a radius of the order of e2 /(me c2 ) and the equatorial velocity of the sphere is much larger than the velocity of light, if the angular momentum is /2 (one obtains veq ∼ c/α = 137c). We know that this objection is not valid, because spin is a purely quantum concept.

12.8.6 The Fine Structure of Atomic Lines In 1925, Pauli did not believe in spin. In January 1926 he called it a heresy, “Irrlehre.” Pauli’s criticism concerned what is known as fine structure effects. It is one of these effects of fine splitting of atomic lines that constitutes a world record in the agreement between experiment and theory. This splitting comes from the interaction of the electron’s spin magnetic moment with its orbital angular momentum. Consider the hydrogen atom and let us imagine we sit on the electron. It sees the proton circling around it with an angular momentum L and producing a magnetic field that interacts with the spin magnetic moment. The resulting interaction is W = 2α2 EI (a0 /r)3 (L · S)/2 . And, once all calculations are performed, this gives a splitting that is twice too large! Fortunately everything was saved by an English physicist. In March 1926, L. H. Thomas remarked that the rest frame of the electron is not an inertial frame, and that a correct relativistic calculation introduces a factor of 1/2 in the formula, called the Thomas precession.2 (Remember Pauli was a great specialist of relativity!) On March 12, 1926, Pauli wrote to Bohr: “I must surrender!” So there is the incredible story of the number two in physics. It appeared everywhere, but with traps. For a long time, nobody thought that all these numbers two had a common origin. There are lots of numbers in physics, but that one is quite simple. Spin was a two-valued quantum number. Its gyromagnetic ratio was twice the orbital gyromagnetic ratio. The “normal” Zeeman effect appeared in two-electron atoms. The anomalous Zeeman effect was a splitting in an even number of spectral lines. Complete electron shells in atoms contained 2n2 electrons. The exclusion principle said that two electrons could not be in the same state. The fine structure of atoms consisted of a splitting in two lines. The Thomas precession introduced a factor of 1/2. Maybe the Creator thought those physicists were going too far and he was annoyed to see them discover the structure of the world He had created. So He introduced this number two with all possible traps, thinking that they would stop teasing him. But He had forgotten that He had created Pauli. One cannot think of every detail, even when one is a Creator. 2 See,

for example, J. D. Jackson, Classical Electrodynamics, Sect. 11.8. New York: Wiley (1975).

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12.9 Magnetism, Magnetic Resonance In his Nobel Lecture, on December 12, 1946, Otto Stern spoke about his experiment, but he insisted much less on its revolutionary aspects than on the measurement of the parameter μ0 which he called the elementary magneton, the elementary quantum of magnetism. The discovery of magnetism goes very far back in time. It is probable that the first physicists of mankind knew this extraordinary phenomenon, which appeared as a force at a distance at a human scale. One can find examples of compasses in China in 2600 B.C., and with the Vikings in the 12th century. The treatise “De Magnete” of William Gilbert, in 1600, classifies magnetic phenomena into three categories. • The strong and permanent magnetism of substances such as iron, cobalt, and nickel • The weaker paramagnetism, induced by strong fields in crystals and in fluids • The diamagnetism, weak and repulsive, that appears in all substance and that appears as the square of the applied field All these forms of magnetism come from electrons. Paramagnetism is the response of a magnetic moment, be it orbital or spin, to an external field. Ferromagnetism does not come from the orbital motion of electrons, but from a subtle collective consequence of Pauli’s principle. In the transition metals, such as Fe, Co, Ni, the wave functions of external electrons are such that the electron spins point spontaneously in the same direction. This results in a permanent macroscopic magnetization. Stern knew he had discovered the elementary quantum of magnetism proposed by Pierre Weiss in 1911. For a long time, physicists had understood that ferromagnetism cannot be explained by currents. Weiss introduced the idea of an “elementary magneton,” analogous to the elementary charge, as being the “Greatest common divisor of molecular magnetic moments.”3 But, in the same type of experiment, Stern went much further. In 1933, he managed to measure the magnetic moment of the proton. This was much more difficult, because it is 1000 times smaller because of the mass factor (q/m) and one must suppress electronic effects. Stern used H2 or HD molecules, where the effect of paired electrons cancel. Stern found a gyromagnetic ratio roughly 2.5 times what one expects from a pointlike particle (i.e., the nuclear magneton μN = q/2mp = 3.1525 10−8 eV T−1 ). This observation was the starting point of the discovery of a fourth type of magnetism, nuclear magnetism, which is a major discovery of the 20th century. There exists a nuclear ferromagnetism, a nuclear paramagnetism. They are very weak, but their consequences are huge. This is when Rabi appears. He was born in 1899, he had worked on a thesis on magnetism at Cornell; then he had gone to Columbia where he learned some quantum mechanics. In the course of his work, he made a little calculation which had enormous 3 The

Weiss magneton was actually one fifth of the value found by Stern.

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consequences. In particular, between 1933 and 1939, he gained a factor of 1000 in accuracy compared to Stern’s measurements.

12.9.1 Spin Effects, Larmor Precession Uncorrelated Space and Spin Variables In most physical situations, such as the Stern–Gerlach experiment, the space and the spin variables are correlated. For instance, we only studied in Chap. 11 a first approximation to the hydrogen atom where we neglected spin effects. If we include the spin degree of freedom in this approximation, we find that the states |n, , m, + and |n, , m, − are degenerate. Actually, there exist corrections to this approximation, such as the fine structure of the hydrogen atom, that we mentioned above, which is due to the interaction between the spin magnetic moment and the electromagnetic field created by the proton. The degeneracy is then partially lifted and the new states are combinations of the initial states |n, , m, σ. In other words, in a given eigenstate of the total Hamiltonian, the spatial wave function of an electron depends on its spin state, and the two random variables r and Sz are correlated. There are many cases where this correlation is extremely weak. In such cases, the two variables r and Sz can be considered as independent and their probability law is factorized. Such a physical situation is represented by a factorized state vector:  α+ (t) . Φ(r, t) α− (t) 

(12.46)

If one performs spin measurements in this case, the results are independent of the position of the particle. The only relevant observables are 2 × 2 Hermitian matrices with numerical coefficient (which can depend on time). Such cases happen in practice, in particular in magnetic resonance experiments, where one can deal with electrons but also with other spin 1/2 particles, in particular protons. We then use terms such as spin state of the proton instead of state of the proton because the position of the proton in space does not play any role in the experiment under consideration.

12.9.2 Larmor Precession in a Fixed Magnetic Field We choose the z-axis parallel to a magnetic field B0 . Ignoring space variables, the Hamiltonian is: (12.47) Hˆ = −μ ˆ · B0 = −μ0 B0 σˆ z .

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We set − μ0 B0 / = ω0 /2, that is, ω0 = −γB0 .

(12.48)

The eigenstates of Hˆ are the eigenstates |+ and |− of σˆ z . Consider an arbitrary state |ψ(t) such that |ψ(0) = α|+ + β|− with |α|2 + 2 |β| = 1. Its time evolution is: |ψ(t) = α e−iω0 t/2 |+ + β eiω0 t/2 |−.

(12.49)

The expectation value μ reads:  μx  = 2μ0 Re α∗ β eiω0 t = C cos(ω0 t + ϕ),  μy  = 2μ0 Im α∗ β eiω0 t = C sin(ω0 t + ϕ),  μz  = μ0 |α|2 − |β|2 ,

(12.50) (12.51) (12.52)

where C and ϕ are, respectively, the modulus and phase of the complex number α∗ β. We recover the Larmor precession that we derived for an arbitrary angular momentum in Chap. 10, Sect. 10.6.4. The projection μz  of the magnetic moment along the field is time-independent, and the component of μ perpendicular to B rotates with the angular velocity ω0 . The fact that μz is a constant of motion is a ˆ μˆ z ] = 0 and of the Ehrenfest theorem. consequence of the commutation relation [H, This provides a simple method to measure the angular frequency ω0 . We place a coil in a plane parallel to B0 and we prepare a macroscopic quantity of spins all in the same spin state |ψ(0). The precession of μ at the frequency ω0 causes a periodic variation of the magnetic flux in the coil, and this induces an electric current at the same frequency. This method is, however, not as accurate as the resonance experiment we present below.

12.9.3 Rabi’s Calculation and Experiment Superposition of a Fixed Field and a Rotating Field A technique invented by Rabi in the 1930s, allows us to perform a very accurate measurement of ω0 by a resonance phenomenon. With this technique, he was able to gain a factor of 1000 in the accuracy of nuclear magnetic moments. We place the magnetic moment in a known field B0 , on which we superimpose a weak field B1 which rotates at a variable angular velocity ω in the xy plane. Such a field can be obtained with two coils along the x- and y-axes, with a current at a frequency ω and phase-shifted by π/2 (one works with radio frequencies). At the resonance, for ω = ω0 , the spin flips between the two possible states |±. Notice that this calculation, characteristic of a driven two-state system, is similar to that performed for the ammonia maser in Chap. 7.

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The form of the Hamiltonian is: Hˆ = −μ ˆ · B = −μ0 B0 σˆ z − μ0 B1 cos ωt σˆ x − μ0 B1 sin ωt σˆ y .

(12.53)

We set: |ψ(t) = a+ (t)|+ + a− (t)|−.

(12.54)

The Schrödinger equation yields the differential system: ω0 ω1 a+ + e−iωt a− , 2 2 ω1 iωt ω0 i a˙ − = e a+ − a− , 2 2 i a˙ + =

(12.55) (12.56)

where we have defined μ0 B0 / = −ω0 /2, μ0 B1 / = −ω1 /2. The change of functions b± (t) = exp(±iωt/2)a± (t) leads to: ω − ω0 ω1 b+ + b− . i b˙ + = − 2 2 ω1 ω − ω0 i b˙ − = b+ + b− . 2 2

(12.57) (12.58)

The above transformation is the quantum form of a change of reference frame. It transforms from the laboratory frame to the frame rotating with the magnetic field at an angular velocity ω around the z-axis. With this change of reference frame, the basis of Hilbert space is time-dependent, whereas the Hamiltonian is time-independent:  Hˆ˜ = 2



ω0 − ω ω1 ω 1 ω − ω0



  = − (ω − ω0 )σˆ z + ω1 σˆ x . 2 2

In this rotating reference system, the problem becomes time-independent! Equations (12.57), and (12.58) imply b¨ ± + (Ω/2)2 b± = 0 with: Ω 2 = (ω − ω0 )2 + ω12 .

(12.59)

Suppose the spin is initially in the state |+; that is, b− (0) = 0. One finds:   iω1 Ωt sin b− (t) = − Ω 2     Ωt ω − ω0 Ωt +i sin . b+ (t) = cos 2 Ω 2

(12.60) (12.61)

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Fig. 12.6 Rabi oscillations a slightly off resonance ω − ω0 = 3ω1 ; b at resonance ω = ω0

The probability that a measurement of Sz at time t gives the result −/2 is: P+→− (t) = |−|ψ(t)|2 = |a− (t)|2 = |b− (t)|2    ω 2 Ωt 1 = sin2 . Ω 2

(12.62)

This formula, which is due to Rabi, exhibits the resonance phenomenon: • If the frequency ω of the rotating field is noticeably different from the frequency ω0 we want to measure, more precisely, if |ω − ω0 |  ω1 , the probability that the spin flips, that is, that we measure Sz = −/2, is very small for all t. • If we choose ω = ω0 , then the probability for a spin flip is equal to one at times tn = (2n + 1)π/ω1 (n integer) even if the amplitude of the rotating field B1 is very small. • For |ω − ω0 | ∼ ω1 , the probability amplitude oscillates with an appreciable amplitude, smaller than one. In Fig. 12.6, we have drawn the time oscillation of the probability P+→− off resonance and at resonance. For a typical magnetic field of 1 Tesla, the resonance frequency is ωe /2π ∼ 28 Ghz for an electron, and 2.79 ωN /2π ∼ 43 MHz for a proton. These frequencies correspond to decametric waves in the nuclear case, and centimetric waves in the electronic case. At the resonance, the system periodically absorbs and emits the energy 2µ0 B by absorption and stimulated emission. Notice that, here, we have closed the loop, concerning the starting point in Chap. 2. We have proven in a particular case that absorption and stimulated emission of radiation do occur at the Bohr frequency of the system ν = ΔE/h. Rabi’s Experiment The resonance effect described above was understood by Rabi in 1939. It provides a very accurate measurement of a magnetic moment. The device used by Rabi consists

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Fig. 12.7 Apparatus developed by Rabi for observing the magnetic resonance effect. In the absence of magnetic resonance, all particles emitted in the state |+ reach the detector. If resonance occurs, the spins of the particles flip between the two magnets, and the signal drops

of a source, two Stern–Gerlach deflectors with magnetic fields in opposite directions, and a detector (Fig. 12.7). Between the two Stern–Gerlach magnets, one places a zone with a superposition of a uniform field B0 and a rotating field B1 , as described above. Consider first the effect of the two Stern–Gerlach magnets in the absence of the fields B0 and B1 . A particle emitted by the source in the spin state |+ undergoes two successive deflections in two opposite directions and it reaches the detector. When the fields B0 and B1 are present, this is not true anymore. If the frequency ω of the rotating field is close to the Larmor frequency ω0 , the resonance phenomenon will change the component μz of the particle. When such a spin flip occurs between the two Stern–Gerlach magnets, the two deflections have the same direction (upward in the case of Fig. 12.7), and the particle misses the detector. The signal registered on the detector as a function of the frequency of the rotating field B1 undergoes a sharp drop for ω = ω0 (Fig. 12.8). This leads to a measurement of the ratio |μ|/j = ω0 /B0 for a particle of angular momentum j. Actually, this measurement is so precise that the main source of error comes from the determination of B0 . In practice, as shown in Fig. 12.8, the frequency ω stays fixed and one varies the magnitude of the field B0 , or equivalently the frequency ω0 .

Fig. 12.8 Signal (obtained by Rabi) recorded on the detector of Fig. 12.7 with a beam of HD molecules, as a function of the field B0 (B1 = 10−4 T, ω/2π = 4 MHz)

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12 Spin 1/2

In 1933 Stern, with his apparatus, had measured the proton magnetic moment with 10 % accuracy. That was a very difficult experiment, because nuclear magnetic moments are 1000 times smaller than electronic ones. In 1939, with his resonance apparatus, Rabi gained a factor of 1000 in accuracy. The resonance is very selective in frequency and the presence of other magnetic moments causes no problem. Rabi’s result struck the minds of people. It was greeted as a great achievement. Stern remarked that Rabi had attained the theoretical accuracy of the measurement, which is fixed by the uncertainty relations. When Hulthén announced that Rabi was awarded the Nobel prize on December 10, 1944, on the Stockholm radio, he said that “By this method Rabi has literally established radio relations with the most subtle particles of matter, with the world of the electrons and of the atomic nucleus.” The Nobel prize for 1943 was awarded to Stern, that of 1944 to Rabi, and that of 1945 to Pauli. Rabi did not go to the reception. The world had undergone a terribly hard period.

12.9.4 Nuclear Magnetic Resonance Stern and Rabi agreed in 1943 on the fact that there was an intrinsic limitation with atomic and molecular beams, both in time and in intensity. They knew that eventually one would have to operate on condensed matter. The great breakthrough of the applications of nuclear magnetic resonance (N.M.R.) came with the works of Felix Bloch at Stanford and of Edward Purcell at MIT, in 1945. Owing to the development of radiowave technologies, Bloch and Purcell were able to operate on condensed matter, and not on molecular beams. One uses macroscopic numbers of spins, thereby obtaining much more intense signals and more manageable experiments. The resonance is observed, for instance, by measuring the absorption of the wave generating the rotating field B1 . The imbalance between the populations of the two states |+ and |−, which is necessary in order to get a signal, results from the conditions of thermal equilibrium. In a field B0 = 1 T, the magnetic energy for a proton is 2.79 µN B ∼ 10−7 eV and the relative population difference between the two spin states due to the Boltzmann factor at room temperature is π+ − π− ∼ 4 10−6 . This relative difference is small, but quite sufficient to observe a significant signal because one deals with samples containing a macroscopic number of spins (typically 1023 ). Bloch and Purcell discovered nuclear paramagnetism, a fourth type of magnetism. The applications of magnetic resonance are numerous in domains ranging from solid-state physics and low temperatures, chemistry, biology, or medicine. By its magnetic effects, spin can play the role of a local probe inside matter. NMR has transformed chemical analysis and the determination of the structure of molecules (see, for instance, Fig. 12.9). It has become an invaluable tool in molecular biology. Since 1980, NMR has also caused a revolution in medical diagnosis and physiology. Under the less frightening name of M.R.I. (Magnetic Resonance Imaging, since there is no dangerous nuclear reaction taking place) it allows us to measure and

12.9 Magnetism, Magnetic Resonance

307

Fig. 12.9 One of the first examples of nuclear magnetic resonance applied to chemistry: the resonance signal obtained with the protons of the ethanol molecule CH3 CH2 OH consists of a three-peak structure. These peaks are associated respectively with the three protons of the CH3 group, with the two protons of the CH2 group, and with the unique proton of the OH group. The magnetic field B0 is ∼0.8 T, and the total trace is 7.5 µT wide

visualize in three dimensions and with a spatial precision better than a millimeter the concentration in water of “soft matter” (muscles, brain, and so on), which, in contrast with bones, is difficult to observe with X-rays. One studies in this way the structure and the metabolism of living tissues; one can detect internal injuries, tumors, and so on. The nuclear spin, which was a curiosity for some visionary physicists of the 1940 and 1950s, has become one of the great hopes of modern medicine. One can, by now, visualize the activity of the brain in real-time. It is possible to localize and register the response of the visual cortex to some stimulation. The following step, after submitting a volunteer to a sequence of such excitations, is to ask her to think about the signal. The NMR response of the brain is the same as that obtained by an external stimulation. This may be considered a direct experimental proof that we think, which is somewhat comforting for the mind.

12.9.5 Magnetic Moments of Elementary Particles The electron, the proton, and the neutron are spin 1/2 particles. The corresponding ˆ Experiments spin magnetic moment is related to the spin S by the relation μ ˆ = γ S. give the following values of the gyromagnetic ratios electron γ  2γ0 = −q/me , proton γ  +2.79 q/mp , neutron γ  −1.91 q/mp .

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12 Spin 1/2

The possible results of the measurement of the component of these magnetic moments along a given axis are therefore: electron μz = ±μB = ∓q/2me , proton μz = ±2.79 q/2mp , neutron μz = ∓1.91 q/2mp .

The quantity μB = −9.274 10−24 J T−1 is called the Bohr magneton. The quantity μN = q/2mp = 5.051 10−27 J T−1 is called the nuclear magneton. Dirac’s relativistic theory of the electron explains the value of the electron magnetic moment:   q ˆ with ge = 2. S, μ ˆ = ge 2me The value measured experimentally for the gyromagnetic factor ge nearly coincides with this prediction. One can account for the slight difference between the experimental result and Dirac’s prediction by taking into account the coupling of the electron with the quantized electromagnetic field (quantum electrodynamics). This constitutes one of the most spectacular successes of fundamental physics. The experimental and theoretical values of the quantity ge coincide within the accuracy limits of experiments and of computer calculations. At present, one has for the electron, setting ge = 2(1 + a), atheo. = 0.001 159 652 18178 (77), aexp. = 0.001 159 652 18073 (28) ;

(12.63) (12.64)

the errors in parentheses bear on the two last digits. To lowest order in the fine structure constant, quantum electrodynamics gives the result a = α/2π = 0.00116. The coefficients +2.79 and −1.91 for the proton and for the neutron are due to the internal structure of these particles. They can be measured with great accuracy by magnetic resonance experiments: μp /μN = 2.792 847 386 (63) and μn /μN = −1.913 042 75 (45); they can be calculated with 10 % accuracy in the quark model.

12.10 Entertainment: Rotation by 2π of a Spin 1/2 It seems obvious and of common sense geometrically that the rotation of 2π of a system around a fixed axis is equivalent to the identity. However, strictly speaking this is not true for a spin 1/2 particle.

12.10 Entertainment: Rotation by 2π of a Spin 1/2

309

Let us return to the calculation of Sect. 12.9.2 and suppose that, at time t = 0 the state of the system is | + x: 1 |ψ(t = 0) = √ (|+ + |−) . 2 The mean value of the magnetic moment, given by (12.50), through (12.52), is μ = μ0 ux . Equation (12.49) gives the evolution of this state. After a time t = 2π/ω0 , classically, the system has precessed by an angle 2π around B. In quantum mechanics one verifies that the expectation value is back to its initial value μ = μ0 ux . What can we say, however, about the state vector? We can check that |ψ(t) is still an eigenvector of Sˆ x (or of μˆ x ) with an eigenvalue +/2. However, we notice that, quite surprisingly, the state vector has changed sign: 1 |ψ(t = 2π/ω0 ) = − √ (|+ + |−) = −|ψ(0). 2 A 2π rotation is therefore not equivalent to the identity for a spin 1/2. Only rotations of 4nπ give back the initial state identically. This property can also be guessed from the dependence eimϕ for orbital angular momenta: using the same formula for m = 1/2 and ϕ = 2π would give eiπ = −1. This peculiarity was understood as soon as spin 1/2 was discovered in 1926. It remained a controversial point for more than 50 years. Does the phase of the state vector after a rotation of 2π have a physical meaning? The positive experimental answer was only given in the 1980s in a series of remarkable experiments.4 The spin 1/2 particles are sent in a two-channel interferometer. In one of them a magnetic field rotates the spins by multiples of 2π. The change in sign of the wave function is observed by a displacement of the interference fringes and the experimental signal confirms that a rotation of 4π is needed to recover the fringe pattern that is measured in the absence of rotation. This property reflects an important mathematical structure that relates the two Lie groups SO(3) and SU(2). Rotations in Euclidian space R3 form the well-known group SO(3). Mathematically, one says that there is a local isomorphism between the Lie algebras of the two groups SO(3) and SU(2), but that these two groups are not globally isomorphic. This formalism, which was called spinor theory, was developed in the early 20th century by the mathematician Elie Cartan. The minus sign is completely equivalent to the fact that on a Moebius strip one must make an even number of rotations in order to get back to the starting point.

4 A.W. Overhauser, A.R. Collela and S.A. Werner, Phys. Rev. Lett. 33, 1237 (1974); 34, 1472 (1975);

35, 1053 (1975).

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12 Spin 1/2

12.11 Exercises

1. Determination of the magnetic state of a silver atom Consider a silver atom in an arbitrary state of its magnetic moment: α|+z + β|−z

with

|α|2 + |β|2 = 1.

(12.65)

a. Show that this is an eigenstate of u · μ ˆ with the eigenvalue +μ0 , where u is a unit vector whose direction will be determined. b. Alice sends to Bob one silver atom in the unknown state (12.65). Can Bob determine this state using Stern et Gerlach measurements? c. Alice sends to Bob N ( 1) silver atoms, all prepared in the same unknown state (12.65). Give a possible strategy for Bob to determine this state (within statistical errors). 2. Repeated measurement; quantum Zeno paradox The magnetic moment μ of a neutron can be described in the same way as the magnetic moment of a silver atom in the Stern-Gerlach experiment. If a neutron is placed in a uniform magnetic field B parallel to the z axis, it can be considered a two-state system for magnetic moment measurements (disregarding space variables). We note |+ and |− the eigenstates of the observable μˆ z . These eigenstates correspond to the two eigenvalues +μ0 and −μ0 . The Hamiltonian of the system in the field B is Hˆ = −Bμˆ z . We set ω = −2μ0 B/. a. Give the energy levels of the system. √ b. At time t = 0 the neutron is prepared in the state: |ψ(0) = (|+ + |−)/ 2. What results can be obtained by measuring μx on this state, with which probabilities? c. Write the state |ψ(T ) of the magnetic moment at a later time T . d. We measure μx at time T . What is the probability to find +μ0 ? e. We now perform on the same system a sequence of N successive measurements at times tp = pT /N p = 1, 2, . . . , N. What is the probability that all these measurements give the result μx = +μ0 ? f. What does this probability become if N → ∞? Interpret the result; do you think it makes sense physically? 3. Products of Pauli matrices Show that σˆ j σˆ k = δj,k + ij,k, σˆ l

(12.66)

where εj,k, = 1 (resp. −1) if (j, k, ) is an even (resp. odd) permutation of (x, y, z), and εj,k, = 0 otherwise.

12.11 Exercises

311

4. Algebra with Pauli matrices Consider the Pauli matrices σ, and two vectors A and B. Show that: (σ · A)(σ · B) = A · B + iσ · (A ∧ B).

5. Spin and orbital angular momentum Consider a spin 1/2 particle whose state is |ψ = ψ+ (r, t)|+ + ψ(r, t)|−. Let Sˆ be the spin observable and Lˆ the orbital angular momentum. We assume that: 



1 ψ+ (r) = R(r) Y0,0 (θ, ϕ) + √ Y1,0 (θ, ϕ) 3 R(r)  ψ− (r) = √ Y1,1 (θ, ϕ) − Y1,0 (θ, ϕ) . 3

a. What is the normalization condition on R(r)? b. What are the probabilities to find ±/2 in measurements of Sz or Sx ? c. What are the possible results of a measurement of Lz ? Give the corresponding probabilities?

Chapter 13

Addition of Angular Momenta

When atomic spectral lines are observed with sufficient resolution, they appear to have in general a complex structure, each line being in fact a group of nearby components. The (numerous) fine and hyperfine splittings of atomic levels are of particular importance because of what they revealed on atomic structure and in terms of applications. The origin of such structures lies in the magnetic interactions of the electron inside the atom, due to interactions of orbital and/or spin magnetic moments. Technically, this boils down to the notion of the addition of angular momenta in quantum mechanics, and the total angular momentum of a system. This notion is useful in many physical problems and we will give its basic elements in Sect. 13.1. In Sect. 13.2 we give a qualitative description of the spin—orbit interaction of the electron spin magnetic moment with the magnetic field originating from its orbital motion around the nucleus. A known example of the corresponding splitting is the yellow line (D-line) of sodium. In Sect. 13.3 we will describe more thoroughly the hyperfine interaction between the spin magnetic moments of the electron and of the proton in the ground state of hydrogen. This interaction produces a splitting which is responsible for the 21 cm line of hydrogen, of considerable interest in astrophysics. Its many analogs in alcali atoms have important practical applications, for instance in atomic clocks.

13.1 Addition of Angular Momenta The Total Angular Momentum Operator Consider two angular momentum observables Jˆ 1 and Jˆ 2 which, by definition, act in two different Hilbert spaces E1 and E2 . This may concern, for instance, a system of two particles: E1 (resp. E2 ) is then the space L2 (R 3 ) of square-integrable functions © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_13

313

314

13 Addition of Angular Momenta

in r 1 (resp. r 2 ). It can also concern a particle moving in space (E1 = L2 (R 3 )), which possesses an intrinsic angular momentum (E2 = Espin ). The Hilbert space of the total system is the (tensor) product: E = E1 ⊗ E2 . By definition, the total angular momentum observable of the system is: Jˆ = Jˆ 1 ⊗ Iˆ2 + Iˆ1 ⊗ Jˆ 2 ≡ Jˆ 1 + Jˆ 2 ,

(13.1)

where Iˆ1 (resp. Iˆ2 ) is the identity operator in E1 (resp. E2 ). This observable acts in E and it is an angular momentum observable. Indeed it satisfies the commutation relations: Jˆ × Jˆ = i Jˆ (13.2) since Jˆ 1 and Jˆ 2 commute. Therefore we know that we can diagonalize simultaneously Jˆ2 and Jˆz . We also know the set of their possible eigenvalues: 2 j ( j + 1) with 2 j integer for Jˆ2 , and m with m = − j, . . . , j − 1, j for Jˆz for a given j. The four angular momentum observables: Jˆ12 , Jˆ22 , Jˆ2 , Jˆz commute. Moreover this set forms a CSCO in the sense of Chap. 10, Sect. 10.2. Their common eigenbasis is therefore unique. We write their common eigenvectors as | j1 , j2 ; j, m. One has by definition: Jˆ12 | j1 , j2 ; j, m = Jˆ22 | j1 , j2 ; j, m = Jˆ2 | j1 , j2 ; j, m = Jˆz | j1 , j2 ; j, m =

j1 ( j1 + 1)2 | j1 , j2 ; j, m

(13.3)

2

j2 ( j2 + 1) | j1 , j2 ; j, m

(13.4)

j ( j + 1) | j1 , j2 ; j, m m | j1 , j2 ; j, m.

(13.5) (13.6)

2

As in Chap. 10, we omit the presence of other possible quantum numbers which would be cumbersome to write and are irrelevant in the present discussion. Uncoupled and Coupled Bases The Hilbert space E corresponding to the degrees of freedom associated with the angular momenta Jˆ 1 and Jˆ 2 is generated by the family of factorized states: {| j1 , m 1  ⊗ | j2 , m 2 } ≡ {| j1 , m 1 ; j2 , m 2 }.

13.1 Addition of Angular Momenta

315

In this basis, the observables Jˆ12 , Jˆ1z , Jˆ22 , Jˆ2z are diagonal. Consider the eigensubspace of the two observables Jˆ12 and Jˆ22 corresponding to given values j1 and j2 . The dimension of this subspace is (2 j1 + 1)(2 j2 + 1) and we ask the following question:

What are, in this subspace, the eigenvectors of Jˆ2 and Jˆz , and the corresponding eigenvalues j ( j + 1)2 and m? In other words, we want to perform in each eigen-subspace of Jˆ12 and Jˆ22 a change of basis to go from the uncoupled factorized eigenbasis common to { Jˆ12 , J1z , Jˆ22 , Jˆ2z } to the coupled eigenbasis common to { Jˆ12 , Jˆ22 , Jˆ2 , Jˆz }. The eigenvalues of Jˆ2 and Jˆz will be expressed as functions of j1 , j2 , m 1 and m 2 . Once the determination of the values of j is performed, we shall express the eigenstates | j1 , j2 ; j, m in terms of the states | j1 , m 1 ; j2 , m 2 : | j1 , j2 ; j, m =



j,m

C j1 ,m 1 ; j2 ,m 2 | j1 , m 1 ; j2 , m 2 

(13.7)

m1m2 j,m

C j1 ,m 1 ; j2 ,m 2 =  j1 , m 1 ; j2 , m 2 | j1 , j2 ; j, m.

(13.8)

j,m

The coefficients C j1 ,m 1 ; j2 ,m 2 of the change of basis (13.7) are called Clebsch–Gordan coefficients.

13.1.1 A Simple Case: The Addition of Two Spins 1/2 The (simplest) case of two spin 1/2 particles will be of particular interest in the following (21 cm line of hydrogen, Pauli principle, etc.). We will first treat it in an elementary way, before considering the general problem of the coupling of two arbitrary angular momenta. The Hilbert Space of the Problem Consider a system of two spin 1/2 particles, for instance the electron and the proton in a hydrogen atom, or the two electrons of a helium atom. We note the particles 1 and 2. The Hilbert space of the two spins Es is the (tensor) product of the two spin spaces (we omit here the external position variables) 1 2 ⊗ Espin . Es = Espin

(13.9)

Es is a four-dimensional space generated by the family {|σ1 ⊗|σ2 }, σ1 = ±, σ2 = ±, which we note in the simpler form: {|+ ; +, |+ ; −, |− ; +, |− ; −}.

(13.10)

316

13 Addition of Angular Momenta

The total spin operator is

Sˆ = Sˆ 1 + Sˆ 2 .

If we need to include external variables, he most general state (space + spin) |ψ of this two spin 1/2 particle system can be written as: |ψ = ψ++ (r 1 , r 2 )|+ ; + + ψ+− (r 1 , r 2 )|+ ; − + ψ−+ (r 1 , r 2 )|− ; + + ψ−− (r 1 , r 2 )|− ; −.

(13.11)

Matrix Representation We can use a matrix representation of the spin states and spin operators for this system. In the basis (13.10) a state is represented by a four-component vector. The observables Sˆ 1 and Sˆ 2 (extended to the tensor product space) can easily be written, using the Pauli matrices, and a block 2 × 2 notation for the 4 × 4 matrices: ⎛ Sˆ1x

0 ⎜ . .. = ⎜ 2⎝ Iˆ

⎞ .. . Iˆ ⎟ ... ...⎟ ⎠ .. . 0

⎛ Sˆ2x

⎛

Sˆ1y

Iˆ ⎜ . .. Sˆ1z = ⎜ 2⎝ 0

⎞ .. . 0 ⎟ ... ... ⎟ ⎠ .. ˆ . −I

⎞ .. . 0 ⎟ ... ...⎟ ⎠ .. . σˆ x

⎛

⎞ .. ˆ 0 . −i I ⎟ ⎜ . .. ... ... ⎟ = ⎜ ⎠ 2⎝ . i Iˆ .. 0 ⎛

σˆ ⎜ x . .. = ⎜ 2⎝ 0

Sˆ2y

⎞ .. σˆ . 0 ⎟ ⎜ y . .. ... ...⎟ = ⎜ ⎠ 2⎝ . 0 .. σˆ y ⎛

σˆ ⎜ z . .. Sˆ2z = ⎜ 2⎝ 0

⎞ .. . 0 ⎟ ... ...⎟ ⎠ .. . σˆ z

where Iˆ stands for the 2 × 2 identity matrix. Total Spin States We consider Es and we note |S, M the eigenstates of Sˆ 2 and Sˆ z with respective eigenvalues S(S + 1)2 and M. Since Sˆ z = Sˆ1z + Sˆ2z , the largest possible value of M is 21 + 21 = 1. The corresponding state is unique; it is the state |+ ; +. Similarly, the smallest possible value of M is − 21 − 21 = −1, and the corresponding eigenstate is |− ; −.

13.1 Addition of Angular Momenta

317

Let us calculate the action of the square of the total spin on these two vectors:

Sˆ 2 |+ ; + = Sˆ12 + Sˆ22 + 2 Sˆ 1 · Sˆ 2 |+ ; +  

3 2 3 2 2 =  +  + σˆ 1x σˆ 2x + σˆ 1y σˆ 2y + σˆ 1z σˆ 2z |+ ; + 4 4 2 = 22 |+ ; +. Similarly:

Sˆ 2 |− ; − = 22 |− ; −.

The two states |+ ; + and |− ; − are therefore eigenstates of Sˆ 2 with the eigenvalue 22 , which corresponds to an angular momentum equal to 1. With the notations of the first paragraph, we have therefore: |s1 = 21 , m 1 = 21 ; s2 = 21 , m 2 = 21  = |s1 = 21 , s2 = 21 ; S = 1, M = 1 and: |s1 = 21 , m 1 = − 21 ; s2 = 21 , m 2 = − 21  = |s1 = 21 , s2 = 21 ; S = 1, M = −1. Since we have recognized two states |S = 1, M = ±1 of angular momentum 1, we now look for the third one |S = 1, M = 0. In order to do this, we use the general relation found in Chap. 10: Sˆ− | j, m ∝ | j, m − 1 and we obtain:

Sˆ− |S = 1, M = 1 = Sˆ1− + Sˆ2− |+ ; +

∝

|− ; + + |+ ; −.

After normalization, we obtain the state: 1 |S = 1, M = 0 = √ (|+ ; − + |− ; +) . 2 One can check that this state is indeed an eigenstate of Sˆ 2 and Sˆ z with respective eigenvalues 22 and 0. We have identified a 3-dimensional subspace in Es corresponding to a total angular momentum equal to 1. The orthogonal subspace, of dimension 1, is generated by the vector: 1 √ (|+ ; − − |− ; +) . 2 One can readily verify that this vector is an eigenvector of Sˆ 2 and of Sˆ z with both eigenvalues equal to zero. To summarize, the total spin in the particular case j1 = j2 = 1/2 corresponds to: S = 1 or S = 0

318

13 Addition of Angular Momenta

and the four corresponding eigenstates, which form a basis of E H , are: ⎧ ⎨ |1, 1 = |+ ; + √ |1, M : |1, 0 = (|+ ; − + |− ; +)/ 2 ⎩ |1, − 1 = |− ; − √ |0, 0 : |0, 0 = (|+ ; − − |− ; +)/ 2.

(13.12) (13.13)

In this particular case of two spin 1/2 particles, we have solved the problem of Sect. 1.2 by decomposing the 4 = 2 × 2 dimensional space Es (tensor product of the two 2-dimensional spaces) into a direct sum of a one-dimensional space (S = 0) and a 3-dimensional space (S = 1), i.e. 2 × 2 = 1 + 3. One can check that any rotation of the (x, y, z) axes which transforms the m 1 = ±1 into m 1 = ±1 and, similarly, m 2 = ±1 into m 2 = ±1 will end up with a similar |1, M  = m 1 + m 2 , |0, 0 quadruplet. Notice that the state |0, 0 is rotation invariant. Symmetry Properties The following symmetry properties will be important when we consider identical particles and the Pauli principle. The three states |1, M are called collectively the triplet state of the two-spin system. They are symmetric with respect to the interchange of the z projections of the spins of the two particles, σ1 and σ2 . The state |0, 0 is called the singlet state and it is antisymmetric in the same exchange. In mathematical terms, if we define a s in Es by the relation: permutation operator Pˆ12 s Pˆ12 |σ1 ; σ2  = |σ2 ; σ1 ,

(13.14)

the triplet and singlet states are eigenvectors of this operator: s |1, M = |1, M Pˆ12

s Pˆ12 |0, 0 = −|0, 0.

(13.15)

13.1.2 Addition of Two Arbitrary Angular Momenta We now want to establish the following results: 1. Consider two angular momentum observables Jˆ 1 and Jˆ 2 . In the subspace corresponding to given values j1 and j2 , the possibles values for the quantum number j associated with the total angular momentum Jˆ are, assuming for definiteness that j1 ≥ j2 : j ∈ { j1 + j2 , j1 + j2 − 1, j1 + j2 − 2, . . . , j1 − j2 + 1 , j1 − j2 }.

13.1 Addition of Angular Momenta

319

2. For a given value of j, the quantum number m takes the values m ∈ { j, j − 1, j − 2, . . . , − j}. 3. The ensuing states | j, m ≡ | j1 , j2 ; j, m are linear combinations of the factorized states | j1 , m 1 ; j2 , m 2 . Assuming again, for definiteness, that j1 ≥ j2 , their total number k=− j2  S= (2( j1 + k) + 1) = (2 j1 + 1)(2 j2 + 1) (13.16) k= j2

is equal to the initial number of factorized states, as expected. Construction of the States Such that j = j1 + j2 Any vector | j1 , m 1 ; j2 , m 2  is an eigenvector of Jˆz = Jˆ1z + Jˆ2z with the eigenvalue m and m = m 1 + m 2 . One therefore deduces the following result:

The vector | j1 , j2 ; j, m corresponding to j = j1 + j2 and m = j1 + j2 exists and it is unique. Indeed the maximum values of m 1 and m 2 are j1 and j2 , therefore the maximum value of m is m max = j1 + j2 . We deduce that the maximal value of j is also jmax = j1 + j2 since the index m can take all the values m = − j, − j + 1 . . . , j in a given eigen-subspace of Jˆ2 . The only normalized vector in the Hilbert space which fulfills the condition m = m max (up to a phase factor) is: | j1 , m 1 = j1 ; j2 , m 2 = j2 . This vector is also an eigenstate of Jˆ2 with the eigenvalue j ( j +1)2 and j = j1 + j2 , as can be checked directly using the following expression: Jˆ2 = Jˆ12 + Jˆ22 + Jˆ1+ Jˆ2− + Jˆ1− Jˆ2+ + 2 Jˆ1z Jˆ2z . Consequently we can write: | j = j1 + j2 , m = j1 + j2  = |m 1 = j1 ; m 2 = j2 .

(13.17)

Remark Here and in what follows, we omit the indices j1 and j2 in the left and right-hand sides of (13.17); these are implicit in the form: | j, m ≡ | j1 , j2 ; j, m and |m 1 ; m 2  ≡ | j1 , m 1 ; j2 , m 2 . We now define as in Chap. 10 the raising and lowering operators: Jˆ+ = Jˆ1+ + Jˆ2+

Jˆ− = Jˆ1− + Jˆ2− .

320

13 Addition of Angular Momenta

We have:

Jˆ± | j, m ∝ | j, m ± 1, Jˆ1± |m 1 ; m 2  ∝ |m 1 ± 1; m 2 

Jˆ2± |m 1 ; m 2  ∝ |m 1 ; m 2 ± 1.

where the proportionality coefficients are given in (10.16). Starting with | j = j1 + j2 , m = j1 + j2  we can generate a series of 2( j1 + j2 ) + 1 states | j = j1 + j2 , m   for m  ∈ j1 + j2 − 1, . . . , −( j1 + j2 ) by applying repeatedly the operator Jˆ− . For instance, using the normalization coefficients given in (10.16) we find: (13.18) |ψa  = | j = j1 + j2 , m = j1 + j2 − 1   ∝ j1 |m 1 = j1 − 1; m 2 = j2  + j2 |m 1 = j1 ; m 2 = j2 − 1. Eigensubspaces of Jˆ z A graphical representation of the uncoupled basis states |m 1 ; m 2  is given in Fig. 13.1 in the particular case j1 = 3/2 and j2 = 1. In the m 1 , m 2 plane, each dot represents a basis state. A fixed m = m 1 + m 2 , corresponding to an eigensubspace E(m) of Jˆz , is represented by a straight dashed line. The point in the upper right corner corresponds to the state (13.17). As we already noted the dimension of this particular eigensubspace E( j1 + j2 ) is 1. The next dashed line corresponds to m = j1 + j2 − 1, and the corresponding eigensubspace E( j1 + j2 − 1) has dimension 2, with the possible basis: |m 1 = j1 − 1; m 2 = j2 

|m 1 = j1 ; m 2 = j2 − 1

(13.19)

In general the eigenvalue m of Jˆz has some degeneracy, except for m = ±( j1 + j2 ). By construction, each eigensubspace E(m) of Jˆz is invariant under the action of the Hermitian operators Jˆ+ Jˆ− and Jˆ− Jˆ+ . Indeed Jˆ+ (resp. Jˆ− ) globally increases (resp. decreases) the value of m 1 + m 2 by 1. Using the expression:



2 1 ˆ ˆ Jˆ2 = J+ J− + Jˆ− Jˆ+ + Jˆ1z + Jˆ2z , 2 it follows that E(m) is also globally invariant under the action of Jˆ2 . Construction of All States of the Coupled Basis The total dimension of the Hilbert space is (2 j1 + 1)(2 j2 + 1). Inside this space we have already identified the 2 j + 1 vectors of the coupled basis with j = j1 + j2 . We now give the principle of the determination of all remaining states of the coupled basis. Consider the subspace E( j1 + j2 −1) of Jˆz , whose possible basis is given in (13.19). Inside this subspace we have already identified the vector |ψa  given in (13.18), which is by construction an eigenvector of Jˆz and Jˆ2 with eigenvalues ( j1 + j2 − 1) and ( j1 + j2 )( j1 + j2 + 1)2 . Consider the vector of E( j1 + j2 − 1) orthogonal to |ψa :

13.1 Addition of Angular Momenta

321

Fig. 13.1 Representation of the uncoupled basis states |m 1 ; m 2 . The dashed lines represent the eigensubspaces E (m) of Jˆz . These subspaces are globally invariant under the action of Jˆ2 . The present figure corresponds to j1 = 3/2 and j2 = 1

|ψb  =



j2 |m 1 = j1 − 1; m 2 = j2  −



j1 |m 1 = j1 ; m 2 = j2 − 1.

Since E( j1 + j2 − 1) is globally invariant under the action of Jˆ2 , we can diagonalize this operator inside E( j1 + j2 − 1), and the corresponding eigenbasis is orthogonal. We know that |ψa  is an eigenvector of Jˆ2 . Therefore |ψb , which is orthogonal to |ψa , is also an eigenvector of Jˆ2 : Jˆ2 |ψb  = j ( j + 1)2 |ψb ,

(13.20)

and we want to determine the value of j. On one hand we have Jˆz |ψb  = m|ψb  with m = j1 + j2 − 1; therefore j ≥ j1 + j2 − 1 since one always has j ≥ m. On the other hand, we cannot have j = j1 + j2 , since this would mean there exists two independent vectors (|ψa  and |ψb ) corresponding to the same values of j and m (i.e. j1 + j2 and j1 + j2 − 1 respectively). This cannot be true since (i) there is only one vector corresponding to j = m = j1 + j2 and (ii) there is a one-to-one correspondence, through the action of Jˆ± , between states associated with ( j, m) and states associated with ( j, m ± 1). Consequently we must have j = j1 + j2 − 1 in (13.20): |ψb  ∝ | j = j1 + j2 − 1, m = j1 + j2 − 1. By applying repeatedly the operator Jˆ− to |ψb , we generate a new series of states, which are labelled | j = j1 + j2 − 1, m  . We have now identified all vectors in the two subspaces E( j1 + j2 ) and E( j1 + j2 − 1). We can repeat the same operation for the subspace E( j1 + j2 − 2) (whose

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13 Addition of Angular Momenta

dimension is 2( j1 + j2 ) − 3, with two vectors already identified), etc., until all the eigenstates |m 1 ; m 2  are used. This occurs when we reach a quantum number m such that the dimension of E(m) is smaller than or equal to the dimension of E(m + 1) (m = −1/2 for the values chosen in Fig. 13.1). Altogether, we find 2 jmin + 1 series of states, where jmin = min( j1 , j2 ). The possible values for j are therefore: j = j1 + j2 ,

j = j1 + j2 − 1, . . . ,

j = j1 + j2 − 2 jmin = | j1 − j2 |

The total number of states of the coupled and uncoupled bases coincide as seen in (13.16). (2 j1 + 1)(2 j2 + 1) = 2( j1 + j2 ) + 1 + 2( j1 + j2 ) − 1 + . . . +

2| j1 − j2 | + 1.

Mathematically we have decomposed the (2 j1 + 1) × (2 j2 + 1) dimensional space (tensor product of a 2 j1 + 1 and a 2 j2 + 1 dimensional space) into the direct sum of a 2( j1 + j2 ) + 1, a 2( j1 + j2 ) − 1, . . ., etc. dimensional space. Using this general procedure, one can determine the coefficients which relate the vectors of the uncoupled basis and those of the coupled one (the Clebsch–Gordan coefficients, defined in (13.7)). The general expression of a Clebsch–Gordan coefficient is quite involved and can be found elsewhere.1

13.2 One-Electron Atoms, Spectroscopic Notations In Chap. 11, we neglected spin effects in the hydrogen atom. If one takes spin into account, the classification of atomic states requires four quantum numbers: |n, , m, σ, σ = ±. The states σ = ± are degenerate in energy in the Coulomb approximation. The spin-orbit interaction which we will discuss later on lifts this degeneracy, and we will see that the energy eigenstates are the eigenstates |n, , j, m j  of the total angular momentum J = L+ S. Their energies do not depend on the quantum number m j giving the projection on z of J, since the Hamiltonian is rotation invariant. In the case of an electron of orbital angular momentum  and spin 1/2, the values of j are therefore: j =  ± 1/2, except for  = 0 in which case j = 1/2. One classifies the states according to the above quantum numbers. The spectroscopic notation consists in adding, on the right of the symbol (n) of Chap. 11, the value of j, for instance:

1 See

e.g. exercise 4 and A.R. Edmonds, Angular Momentum in Quantum Mechanics (Princeton University Press, 1950).

13.2 One-Electron Atoms, Spectroscopic Notations

2 p3/2

⇔

3d3/2

⇔

3 =+ 2 3 n = 3,  = 2, j = =  − 2 n = 2,  = 1, j =

323

1 , 2 1 . 2

13.2.1 Fine Structure of Monovalent Atoms The resonance line of monovalent atoms appears to be split into two components. One example is the yellow line of sodium, corresponding to the transition 3 p → 3s, which is used in highly efficient sodium vapor street lights. It is split into two lines called respectively D1 and D2 , of wavelengths λ1 589.6 nm and λ2 589.0 nm. The same effect is observed in the hydrogen atom: the Lyman α line, corresponding to the transition 2 p → 1s, is also split into two components. This splitting is due to the so-called spin-orbit interaction: the first excited level, which has an orbital angular momentum  = 1 ( p state), is cleaved into two sub-levels because of this interaction. One level corresponds to j = 3/2, the other to j = 1/2. The splitting is weak compared to the main effect, i.e. the energy difference between the initial levels (1s and 2 p for hydrogen). The 2 p3/2 − 2 p1/2 energy difference for hydrogen is roughly 4.5 × 10−5 eV; the 3 p3/2 − 3 p1/2 splitting in sodium is ∼2 × 10−3 eV. The physical origin of the spin-orbit coupling can be understood with a classical argument. Suppose we model the hydrogen atom as an electron orbiting with a velocity v around a proton. The proton is much heavier than the electron and it is assumed to be at rest in the laboratory frame. It creates an electrostatic field acting on the electron: q r. (13.21) E= 4πε0 r 3 In the rest frame of the electron, the proton moves at velocity −v and this gives rise, in addition to the electric field (13.21), to a magnetic field: B = −v × E/c2 =

q L. 4πε0 m e c2 r 3

(13.22)

Here L = m e r ×v stands for the angular momentum of the electron in the laboratory frame. In order to derive (13.22), we assume that |v|  c and we consider only the dominant terms in v/c. The spin magnetic moment of the electron μ ˆ s = −(q/m e ) Sˆ interacts with this magnetic field and this gives rise to a magnetic hamiltonian. The quantum Hamiltonian can be rewritten in the laboratory frame using the natural atomic units, i.e. the Bohr radius a1 and the ionization energy E I , together with the fine structure constant α2 :

2

For all details, see e.g. J.D. Jackson, Classical Electrodynamics, sect. 11.8 (Wiley, 1975).

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13 Addition of Angular Momenta

a 3 Lˆ · Sˆ 1 Wˆ s.o. = α2 E I . rˆ 2

(13.23)

Remarks 1. The spin-orbit coupling is a relativistic effect. We notice on (13.23) that, since ˆ 2 are of the order of 1, the spin-orbit coupling is of order a1 /r and Lˆ · S/ 2 2 α (1/137) compared to the main effect. This is indeed of order v 2 /c2 , since v/c ∼ α. 2. For the s states ( = 0), the term (13.23) vanishes. However one can show that there exists a relativistic shift of the levels, called the Darwin term, whose value is: WD =

π 2 e 2 2 |ψ(0)|2 . 2m 2e c2

This term vanishes for   = 0, since ψ(0) = 0 in that case. It only affects s waves. All the above terms can be obtained directly and exactly in the framework of the relativistic Dirac equation. When one solves this equation for the Coulomb potential, one finds that the states 2s1/2 and 2 p1/2 of hydrogen are degenerate, and that the state 2 p3/2 lies 10 GHz above these two states. Experimentally, one observes a splitting between the two states 2s1/2 and 2 p1/2 , called the Lamb shift in the name of its discoverer. This splitting, of the order of 1 GHz, is due to the coupling of the electron with the quantized electromagnetic field. The calculation of the Lamb shift by Bethe in 1947 was the first spectacular success of quantum electrodynamics. The spin-orbit splitting is small and we can calculate it by perturbation theory (Chap. 9). For instance, for the level n = 2 of hydrogen, we have to diagonalize the restriction of WS.O. (13.23) to the subspace generated by the 6 states |n = 2,  = 1, m, σ. This coupling Wˆ S.O. involves the scalar product Lˆ · Sˆ which is diagonal in the basis |n, , j, m j  of the eigenstates of the total angular momentum. We have the equality:



1 ˆ ˆ 2 − Lˆ 2 − Sˆ 2 = 1 Jˆ2 − Lˆ 2 − Sˆ 2 , ( L + S) Lˆ · Sˆ = 2 2 with eigenvalues ( j ( j + 1) − ( + 1) − 3/4)2 /2. Using (13.23) one therefore obtains the splitting of the j =  + 1/2 and j =  − 1/2 states (for instance 2 p3/2 and 2 p1/2 ): ΔE(n, ) ≡ E( j =  + 1/2) − E( j =  − 1/2) = ( + 1/2)An, , 

with: An, = α2 E I

|ψn,,m (r)|2

a 3 1

r

d 3r.

One can check easily that this quantity is independent of m and that its numerical value coincides with the experimental result for the 2 p1/2 –2 p3/2 splitting of hydrogen. For

13.2 One-Electron Atoms, Spectroscopic Notations

325

other atoms, one may observe more complicated effects. For instance, in sodium, there is inversion of the spin-orbit effect: E(3d3/2 ) > E(3d5/2 ). This comes from an effect of the core of internal electrons. Finally, one can understand the origin of the name fine structure constant for α which governs the order of magnitude of fine structure effects. The name had been introduced in 1920 by Sommerfeld who had calculated the fine structure of hydrogen in the framework of the old quantum theory, by considering the relativistic effect due to the eccentricity of the orbits, and not to the (yet to be discovered) spin effects. Sommerfeld’s calculation gave the good result, but this was simply another awkward coincidence due to the particular symmetries of the hydrogen problem and to the ensuing degeneracies in .

13.3 Hyperfine Structure; The 21 cm Line of Hydrogen An even smaller effect (splitting of the order of 6 × 10−6 eV) has very important practical applications. This effect comes from the magnetic interaction between the two spin magnetic moments of the electron and the proton: μ ˆ e = γe Sˆ e μ ˆ p = γ p Sˆ p

γe = −q/m e ,

(13.24)

γ p 2.79 q/m p .

(13.25)

This interaction is called the spin-spin, or hyperfine interaction. We will only consider its effect in the ground state of hydrogen n = 1,  = 0. Interaction Energy We neglect here effects due to the internal structure of the proton and we treat it as a point-like particle. The calculation of the magnetic field created at a point r by a magnetic dipole μ p located at the origin, is a well known problem in magnetostatics.3 The result can be written as:   3(μ p · r) r μ0 2μ0 (13.26) μ p δ(r). B(r) = − μp − + 3 2 4πr r 3 The interaction Hamiltonian between the magnetic moment μe of the electron and this magnetic field reads: ˆ Wˆ = −μ ˆ e · B. For r  = 0, Wˆ reduces to the usual dipole-dipole interaction (be it electric or magnetic): r = 0

3 See

μ0 Wˆ dip = 4πrˆ 3



 ˆ p · rˆ ) 3(μ ˆ e · rˆ )(μ ˆp− μ ˆe ·μ . rˆ 2

e.g. J.D. Jackson, Classical Electrodynamics, sect. 5.6 (Wiley, 1975).

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13 Addition of Angular Momenta

This interaction will not contribute to our calculation because of the following mathematical property. For any function g(r ), r = |r| regular at r = 0, the angular integration yields:  g(r ) Wdip (r) d 3r = 0.

(13.27)

The field (13.26) is singular at r = 0, owing to the term proportional to δ(r). This leads to a contact interaction: 2μ0 ˆ p δ(ˆr ). μ ˆ ·μ Wˆ cont = − 3 e The origin of this singularity at r = 0 lies in the assumption that the proton is point-like in our analysis. This entails that all magnetic field lines converge to the same point. A calculation taking into account the internal structure of the proton, its finite size, and the corresponding modification of the field is leads to a similar result, because the size of the proton is very small compared with the size of the probability distribution of the electron in the 1s state. This further calculation sheds light on the internal structure of the proton, and other nuclei such as deuterium or tritium. The point-like model is strictly valid for positronium, which is an atom made of an electron and a positron, both being point-like, and for muonium made of an electron and a positive muon both being point-like. The theoretical interest of Quantum Electrodynamics calculations of such transition frequencies, and the comparison with high accuracy experimental data can be found in S. G. Karshenboim and V.D. Ivanov.4 Perturbation Theory The observable Wˆ acts on space and on spin variables. We consider the orbital ground state of the hydrogen atom which, owing to spin variables, is a four state system. A state of this 4-dimensional subspace can be written as: |ψ = ψ100 (r) |Σ,

(13.28)

where  ψ100 (r) is the ground state wave function found in Chap. 11: ψ100 (r) = −r/a1 e / πa13 . We first take the expectation value over space variables, which results in an operator acting only on spin variables, Hˆ 1 =



∗ ψ100 (r) Wˆ ψ100 (r) d 3r.

then we diagonalize this latter operator. The contact term is readily evaluated as: 4 S.G.

Karshenboim and V.D. Ivanov, Hyperfine structure in hydrogen and helium ion, Phys. Lett. B 524, p. 259–264, 2002.

13.3 Hyperfine Structure; The 21 cm Line of Hydrogen

327

2μ0 ˆ p |ψ100 (0)|2 . μ ˆ ·μ Hˆ 1 = − 3 e

(13.29)

Hˆ 1 is an operator which acts only on spin states. It can be cast in the form: A Hˆ 1 = 2 Sˆ e · Sˆ p , 

(13.30)

where the constant A can be inferred from the values of γe , γ p , and ψ100 (0): A=−

16 2 μ0 4 me 2 γe γ p 2 = α EI . × 2.79 3 3 4π a1 3 mp

One obtains: A 5.87 × 10−6 eV

ν=

A

1417 MHz h

λ=

c ∼ 21 cm. ν

(13.31)

ˆ1 Diagonalization of H The diagonalization of Hˆ 1 in the Hilbert space of spin states is straightforward. Considering the total spin Sˆ = Sˆ e + Sˆ p , one has: 1 ˆ2 ˆ2 ˆ2

S − Se − S p , Sˆ e · Sˆ p = 2 which is diagonal in the basis of the eigenstates |S, M of the total spin, with eigenvalues: 2 with S = 0 or S = 1. (S(S + 1) − 3/2) 2 The ground state E 0 = −E I of the hydrogen atom is therefore split by the hyperfine interaction in two sublevels corresponding to the triplet |1, M and singlet |0, 0 states: triplet state |1, M, E + = E 0 + A/4 (13.32) singlet state |0, 0. E − = E 0 − 3A/4 The difference of these two energies is equal to A, i.e. 5.87 × 10−6 eV; it corresponds to the characteristic line of hydrogen at a wavelength λ ∼ 21 cm. Remarks (i) In its ground state, the hydrogen atom constitutes a four level system with two energy levels. By a method whose principle is similar to what we have discussed in Chap. 6, it is possible (but technically more complicated) to devise a hydrogen

328

13 Addition of Angular Momenta

maser.5 Among other things, this allows to measure the constant A, or equivalently, the frequency ν = A/ h with an impressive accuracy: 2 0  40 5 ν =  1 4   75 1 . 7  68 4 ±0. 0 0 1 7 Hz. A

B

C

D

E

• In this result, we have underlined several groups of digits. The first two (A) were obtained by Fermi in 1930; they correspond to the contact term considered above. The two following ones (B) are calculated using the Dirac equation, and the experimental value for the anomalous magnetic moment of the electron (deviation of the order of 10−3 ). Other corrections account for the two following decimals (C): relativistic vacuum polarization corrections, finite size of the nucleus, polarization of the nucleus, etc. The set (D,E) is out of range for theorists at present. Such an accuracy has, in particular, provided a means to test the predictions of general relativity.6 A hydrogen maser was sent in a rocket at an altitude of 10,000 km, and the variation of its frequency as the gravitational field and the velocity vary was measured. Despite numerous difficulties, it was possible to check the predictions of relativity with an accuracy of 7 × 10−5 , which is still one of the most accurate verifications of the theory (actually of the equivalence principle). (ii) The hyperfine splitting of alkali atoms has the same origin as that of hydrogen, although it is more difficult to calculate theoretically. It is measured with impressive accuracy. One observes the following frequencies: 7

Li Na 39 K 85 Rb 87 Rb 133 Cs 23

0.83 GHz 1.77 GHz 0.46 GHz 3.04 GHz 6.83 GHz 9.19 GHz

2s 3s 4s 5s

state state state state ” 6s state

This leads to the achievement of masers and atomic clocks.7 One of the many applications is the definition of the time standard based on the hyperfine effect of the isotope 133 of cesium in its ground state (ΔE ∼ 3.8 × 10−5 eV). One second is defined as being equal to 9 192 631 770 periods of the corresponding line. The relative accuracy of the practical realization of this definition is 10−15 , which we have mentioned in Sect. 7.7.3. Such an impressive precision has been made possible with the use of laser cooled atoms. However, this field is undergoing an impressive breakthrough with the use of optical wavelengths which at present have already achieved a 10−18 level of stability

5 H.M.

Goldenberg, D. Kleppner, and N.F. Ramsey, Phys. Rev. Lett. 8, 361 (1960). Vessot et al., Phys. Rev. Let. 45, 2081 (1980). 7 For the recent developments of atomic clocks, see e.g. W. Itano and N. Ramsey, Accurate measurement of time, Scientific American, p. 46, July 1993. 6 R.

13.3 Hyperfine Structure; The 21 cm Line of Hydrogen

329

and precision.8 It is quite probable that a new international time standard will be defined within a few years from now. The effect of an external magnetic field If we place the hydrogen atom in an external magnetic field, the magnetic Hamiltonian becomes: A ˆ e · B0 − μ ˆ p · B0. (13.33) Hˆ M = 2 Sˆ e · Sˆ p − μ  Here we do not take into account space variables, assuming that the Zeeman splitting is small enough so that first order perturbation theory is valid. The nuclear magneton μ N is much smaller than the Bohr magneton μ B . Therefore, we can neglect the last term in the expression (13.33). In this approximation, the diagonalization of HM is simple. We set η = qB0 /(2m e ) and tan 2θ = 2η/A, and we obtain the following splitting:

−(A/4) + −(A/4) −

 

(A/4) + η → |1, 1 (A/4) − η → |1, −1 A2 /4 + η 2 → cos θ |1, 0 + sin θ |0, 0 A2 /4 + η 2 → − sin θ |1, 0 + cos θ |0, 0

The levels are represented on Fig. 13.2.

Fig. 13.2 Zeeman splitting of the 21 cm line 8 B.J. Bloom, T.L. Nicholson, J.R. Williams, S.L. Campbell, M. Bishof, X. Zhang, W. Zhang, S.L. Bromley, J. Ye, An optical lattice clock with accuracy and stability at the 10−18 level, Nature 506, 71-75, 2014, doi:1038/nature12041.

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13 Addition of Angular Momenta

One observes, as for NH3 , a competition between the hyperfine coupling and the presence of the field. For weak fields, the states |1, 0 and |0, 0 are unaffected, whereas the energies of the states |1, 1 and |1, −1 vary linearly with B. There is a splitting of the 21 cm line into three components. For strong fields, the eigenstates are the factorized states |σe ; σ p . The transition region (η ∼ A) is around B ∼ 0.1 T.

13.4 Radioastronomy The main stages of the progress of astronomy are primarily due to technological progress in the means of observation: Galileo’s and Newton’s telescopes, photography, space telescopes, X-ray and γ-ray astronomy, space probes, cosmic neutrino detectors, and so on. This happened in parallel with progress in fundamental research that enables a more and more refined analysis and interpretation of the data.

Fig. 13.3 The Milky Way. Top separate views of the northern and southern hemispheres. Bottom the pictures have been put together in order to see the angular position of stars with respect to the plane of the galaxy, which is at a large angle from the plane of the solar system (this is what provokes the apparent asymmetries in the upper figure). One can see on the bottom right the two Magellanic clouds which are only visible in the southern hemisphere. On both pictures, the contrast has been strongly amplified (Photo credit Axel Mellinger; http://home.arcor-online.de/axel.mellinger/allsky. html.)

13.4 Radioastronomy

331

It seems easy to observe the cosmos. Our own galaxy, the Milky Way which is so beautiful to look at in the summer, is paradoxically one of the most difficult galaxies to explore. There are so many objects that one only sees a few of them: 6000 with the naked eye, 100 million with telescopes, among the 200 billion stars which it contains. The nearby stars are very bright but they screen everything that lies behind them. Since the 1950s, the development of radioastronomy has brought a fascinating development of our knowledge of the Universe. Radiowaves are more penetrating than light waves and they carry a lot of complementary information. In galaxies, matter exists in two main forms (let aside dark matter which amounts to 20 % of the mass of the universe). The first, which is directly visible, is condensed matter: stars at various stages of their evolution and planets which are now being discovered in other solar systems than ours. However there also exists a diffuse interstellar medium, composed mainly of atomic hydrogen, whose total mass is quite important (from 10 to 50 % of the total visible galactic mass). The temperature of these interstellar clouds is typically 100 K. Since the corresponding thermal energy kT ∼ 10−2 eV is much smaller than E I , hydrogen atoms cannot be appreciably excited by thermal collisions from the 1s ground state to the other states of the Lyman series. However the transitions between the two hyperfine states S = 1 and S = 0 occur easily. The emission of 21 cm radiation corresponds to the spontaneous transition from the S = 1 state to the S = 0 state. This emission is very weak, because the lifetime of the triplet S = 1 state is extremely long: τ ∼ 3.5 × 1014 s ∼ 107 years.9 Nevertheless the amount of atomic hydrogen in the interstellar medium is so large that an appreciable signal is observed. The observation of this line of hydrogen has deeply modified our understanding of the interstellar medium. The intensity of the line in a given direction gives the mass distribution of the amount of hydrogen. The Doppler shift allows to measure the velocities of the hydrogen clouds. The splitting of the line and its polarization provide a measurement of the magnetic field inside the interstellar medium. By analyzing the structure of our galaxy, the Milky Way (which is difficult to observe because we are in its plane), it has been possible to show that it is a spiral galaxy, of radius 50,000 light-years, and that we are at 30,000 light-years from the center (see Fig. 13.4). One can also measure the density of the interstellar medium (0.3 atoms.cm−3 on the average), its temperature (20–100 K), its structure (roughly one interstellar cloud every 1000 light-years along a line of sight), and its extension outside the plane of the galaxy (roughly 1000 light-years). The interstellar medium is a diffuse medium in the galaxies where the physical and chemical life of galaxies takes place. This medium has an overall large mass (10–50 % of the total mass of a galaxy). It is very dilute (1–20 atoms per cm3 ) and cold (50–100 K).

9 This very long lifetime is due to the combination of two facts; the energy difference is very small, and the emission proceeds through a magnetic dipole transition (while the atomic resonance lines correspond to electric dipole transitions).

332

13 Addition of Angular Momenta

Fig. 13.4 Left Spiral structure of the Milky Way as deduced from radioastronomical observations at a 21 cm wavelength. (Courtesy Frédéric Zantonio.) Right Barred spiral structure of the Milky Way

This medium consists primarily of atoms: 90 % atomic hydrogen and 10 % atomic helium (i.e., respectively, 75 and 25 % in mass) that were formed just after the big bang, during the first three minutes of the Universe. There are other atoms, such as carbon, calcium, potassium, and others, but in smaller amounts. They have been formed in stellar nucleosynthesis and have been ejected in the interstellar medium, for instance in supernovae explosions. The rest of the interstellar medium (from 1 to 50 %) consists of molecules such as H2 O, CO2 , C2 H5 OH which we have mentioned in Chap. 10, and interstellar dust. The interstellar medium causes spectacular phenomena in astronomical observations. Figure 13.5 shows veils of matter in the Cygnus nebula and between the Pleiade stars, which are young stars of our galaxy.

Fig. 13.5 Left veil of interstellar matter in the Cygnus (or Veil) nebula (Photo credit: HST WFPC2 http://casa.colorado.edu/~maloney/CygnusLoop.gif.) Right veil of matter between two of the Pleiade stars. The parallel wisps extending from lower left to upper right, discovered by G. Herbig and T. Simon in 1999, are due to radiation pressure of the strong starlight shining from Merope—outside frame—on the dust particles (Photo credit: NASA and Hubble Heritage Team http://heritage.stsci.edu/2000/36/big.html.)

13.4 Radioastronomy

333

One can understand from these examples that this type of observation is quite limited. In order to see such objects or mass distributions, they must either emit light or be illuminated by light sources, and furthermore they must not be hidden by other objects (after all, we see a horsehead, but there may also be woodlice or lobsters). The interstellar medium, which is cold, does not emit in the visible part of the spectrum, but it abundantly emits radiofrequency waves. The radiowaves emitted by the Milky Way, as well as by other galaxies, penetrate matter much more deeply than light waves, and they bring new information compared to what we can see directly. However, this radiation comes from sources that are of extremely low densities and the interstellar medium has a small brilliance compared to stars. In order to observe such emissions, one needs high-performance selective noiseless amplifiers. This is where a major technical contribution of quantum mechanics to astrophysics was made. Masers, which we described in Chap. 7, have precisely these features. The introduction of masers in astrophysics is due to Townes himself, who turned to astrophysics in the mid 1960 s and caused a true revolution. In fact, with masers one can observe the radiation coming from weak sources, that would be drowned in the background noise if one were to use traditional amplifiers. The masers used in astrophysics are solid-state masers, in particular, ruby masers that can deliver important amplification factors. The observation of the interstellar medium requires powerful radiotelescopes, and interferometric setups in order to reach an acceptable resolution (radio wavelengths are 100,000 times larger than optical ones). At present, a maximum resolution is obtained with the technique of very-large-baseline interferometry (VLBI). The data recorded with radiotelescopes located on different continents can be put together and synchronized with atomic clocks. The baseline is of the order of 10,000 km. Astrophysicists are contemplating putting space-radiotelescopes on large orbits in order to improve the resolution. Radioastronomical observations represent a formidable task. The total energy received up to now in radioastronomy is of the order of the kinetic energy of the ashes of a cigar falling from 1 m high. One aims at various directions in the sky. By a sophisticated analysis of the observed signal, one can reconstruct the position, the density, the velocity (by Doppler effect), and the composition of interstellar clouds. The pioneers of centimetric radioastronomy, which is fundamental in order to observe hydrogen, were Purcell and Van de Hulst in 1947.

13.5 The 21-cm Line of Hydrogen Coming “back home”, that is, to our galaxy and its present neighborhood, One frequency dominates the radiowave emission and plays a key role: 1420 MHz. It is on that frequency and only on that one, that hydrogen, the most abundant element in the Universe, emits radiation.

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This effect is one of the best-known phenomena of atomic physics. As we have seen, when the spins of the proton and the electron flip from one relative configuration to the other, there is emission or absorption on 1420 MHz, or at a wavelength of 21cm. (This is in a domain close to television frequencies, and it is protected by an international agreement.) Importance of the 21-cm Line The importance of the hydrogen 21-cm line in astrophysics comes from the following facts. • Hydrogen is very abundant. • It absorbs any emission in the visible part of the spectrum (the sky is opaque for hydrogen lines in the visible and ultraviolet regions). • On the other hand, cold interstellar atomic hydrogen emits this radiation abundantly. Furthermore, the interstellar medium and the entire galaxy are transparent for this radiowave. The temperature of interstellar clouds is typically 50–100 K. The corresponding thermal energy kT ∼ 10−2 eV is much smaller than E I , and the atoms cannot be excited appreciably from the 1s state to other levels of the Lyman series. But thermal transitions between the two hyperfine levels S = 1 and S = 0 are quite easy. The 21-cm emission corresponds to the spontaneous transition from the S = 1 state to the S = 0 state. This emission is very weak because the lifetime of the triplet state S = 1 is very long: τ ∼ 3, 5 × 1014 s ∼ 107 years!10 Nevertheless, there are enormous amounts of hydrogen in the interstellar medium and the signals received on earth are appreciable. • It is the only emission of hydrogen that can be collected, except for a few nearby stars. Its detection was predicted by Van de Hulst in 1947; the first observation is due to Purcell in 1951. Its observation gives information on galactic and extragalactic structure and dynamics, and on the formation of galaxies. The intensity distribution of the emission of the 21-cm line is shown in Fig. 13.6. We notice immediately that the emission extends in directions far beyond those of visible light (Fig. 13.3). The hydrogen clouds extend far beyond stars (i.e., condensed matter) in the sky. Their study is bound to bring new and abundant information. The observation of this atomic hydrogen line (and that due to carbon monoxide, less abundant but very luminous) profoundly changed our understanding of the interstellar medium.

10 This long lifetime is due to two effects: the energy difference between the two levels is small; and

it is a magnetic dipole transition much weaker than the electric dipole transitions. The advantage is that, symmetrically, these radiowaves are absorbed weakly.

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Fig. 13.6 Emission density of the 21-cm line in the sky. The emission is naturally brighter in the plane of the Milky Way where matter is closer and more dense on the average (Photo credit: J. Dickey (UMn), F. Lockman (NRAO), SkyView, http://antwrp.gsfc.nasa.gov/apod/ap010113. html.)

The Milky Way There are countless results. By probing our own galaxy, the Milky Way, (which is difficult because the sun lies in its plane), one has been able to reconstruct it structure and to show that it has a spiral structure, as does its sister galaxy, Andromeda, M31. It has been possible to show that it has a radius of 50,000 light-years, and that the sun lies at some 35,000 light-years from the center (see Fig. 13.4). Today, astronomers tend to think it has the structure of a barred spiral galaxy (see the right-hand side of Fig. 13.4). The measured average density of the interstellar medium is 0.3 atoms per cm3 , the temperature varies from 20 to 100 K, and there is on average one interstellar cloud every 1000 light-years along a line of sight. The thickness perpendicular to the galactic plane is roughly 1000 light-years. A strange discovery occurred in 2003. The Milky Way might be performing an act of cannibalism on its satellite galaxy, Canis Major. The detailed analysis of external hydrogen clouds, and infrared matter, reveals a closed winding filamentshaped structure that passes through a dwarf satellite galaxy of the Milky Way, Canis Major, as one can see in Fig. 13.7. Canis Major is at a distance of 25,000 light-years from the sun, it is the satellite galaxy closest to the center of the Milky Way. Therefore the Milky Way would be surrounded by a tidal stream of stars and matter. Such a feature is observed in other galaxies.

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Fig. 13.7 Left Global structure of atomic hydrogen and stars in and around the Milky Way. Right Residual filament going through Canis Major, the brighter region on the left, once the spiral arms nearest to the center of the Milky Way have been removed (Photo credit: R. Ibata, Strasbourg Observatory, Two-Micron All Sky Survey or “2MASS” experiment, http://astro.u-strasbg.fr/images_ri/ canm-e.html.)

13.6 The Intergalactic Medium; Star Wars A number of results have been obtained by observing outside our galaxy. First, we can observe our sister galaxy, Andromeda, at two million light-years from us, twice as massive as the Milky Way. The pictures in the visible part of the spectrum are shown in Fig. 13.8. One can see the spiral structure. In radiowave

Fig. 13.8 Andromeda nebula, M31. Left In visible light (Photo credit: NASA and Robert Gendler, http://antwrp.gsfc.nasa.gov/apod/ap021021.html.) Right In 21 cm radioastronomy (Photo credit: NRAO, Group of Astronomers at the Robert C. Byrd Green Bank Telescope (GBT) of the National Science Foundation, http://www.universetoday.com/am/publish/clouds_hydrogen_ swarm_andromeda. html 422004.)

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emission, one can see extended and powerful sources that do not correspond to any visible matter. Astrophysicists tend to think that these clouds are materials which have not been used up to now in the construction of galaxies, but that will play (or are playing) a major role in galactic dynamics. This reveals an interesting feature. In fact, atomic hydrogen extends much beyond stars in a galaxy. And that feature allows us to learn about the relationship of galaxies with respect to each other. This could not be guessed with usual telescopes on human time scales. In the same way as stars live in an interstellar medium, there exists an intergalactic medium. Galaxies possess links. Contrary to the interstellar medium, the intergalactic medium contains practically only primordial hydrogen and helium from the big bang (there is no intergalactic nucleosynthesis). The presence of intergalactic hydrogen clouds is obvious in the Cartwheel galaxy, which is 200 Mpc from us (1 pc = 3.26 light-years), Fig. 13.9, where the contour lines of hydrogen clouds are superimposed on the optical photograph. This galaxy survived a head-on collision with a smaller galaxy 300 million years ago. This provoked the annular structure of the Cartwheel, with a rim of the size of the Milky Way and a nucleus. Numerical simulations confirm this idea. The contour lines of the important hydrogen cloud surrounding all these galaxies are marked. It leaves some doubt about which galaxy is responsible for this hit-and-run offense.

Fig. 13.9 The Cartwheel galaxy, A 0035. On the left, the ring shape, with a wheel of the size of the Milky Way and a nucleus. A hydrogen cloud joins this structure to the right hand side galaxy, and also to the middle galaxy, which is not a spiral galaxy. It is thought that this latter object, which is now in the axis of the Cartwheel at 100 kpc, had a head-on collision with it 300 million years ago, which gave rise to the structure. The extension of the hydrogen cloud leaves some doubt on the object responsible for the collision which could be the upper-right galaxy (Photo credit: J. Higdon (NRAO), C. Struck, P. Appleton (ISU), K. Borne (Hughes STX), and R. Lucas (Stsci), NASA; http://antwrp.gsfc.nasa.gov/apod/ap970224.html.)

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Fig. 13.10 Group of galaxies around M81. In the center, one can see M81, which is a large spiral galaxy, on the right, M82 which was thought to be very irregular, and on the left NGC 3077 (Photo credit Robert Gendler, http://antwrp.gsfc.nasa.gov/apod/ap000209.html.)

A famous example, which is very rich and was one of the first to be analyzed is the group of three galaxies M81, M82, NGC 3077 shown in Fig. 13.10. This group is located in the constellation Ursa Major at 2.5 Mpc. These three galaxies (and other smaller ones) seem to orbit around each other quietly. M82 has an anomalous shape with a central prominence, perpendicular to its plane. The 21-cm radioastronomical observation transforms these impressions, as one can see in Fig. 13.11. One sees that the group bathes in hydrogen clouds. We notice that the spiral structure of M81 appears clearly, and that the emission at 21-cm, which is abundant in cold regions, is weak in the hot regions of galactic nuclei. These galaxies appear as floating islands in a common ocean of intergalactic hydrogen gas of a considerable extension. One observes the umbilical cords of hydrogen, in particular between M81 and NGC 3077, which show that these galaxies have a common history. One can see that the large galaxy M81 is in the process of performing cannibalism on a smaller galaxy, on its left, and of absorbing it. This phenomenon is very hard to see in visible light. Such observations are confirmed by numerical simulations. (We have mentioned interstellar organic molecules on this example in 10.) Finally, the Doppler picture of the 21-cm emission of M81 (Fig. 13.11, Left.) exhibits an unexpected feature. The plane of this galaxy rotates globally around an axis at roughly—30 ◦ . The upper part of the galaxy is redshifted and the lower part is blueshifted. At this level of accuracy, no individual motion appears in the spiral arms themselves, whereas the overall rotation of the plane of the galaxy is manifest.

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Fig. 13.11 Left The group of galaxies M81, M82, NGC3077 as seen on the 21-cm wavelength. The extension of the hydrogen cloud inside which the group is evolving is very large. One can see that M81 is eating up, on the left, a smaller galaxy, which is more difficult to see in visible light. We notice the absence of radio emissions in the vicinity of the hotter cores of galaxies. The orientation of this figure is not the same as the optical picture of Fig. 13.10; it is rotated by roughly 45◦ . Right Doppler shift of the top and bottom of M81. One can see that it corresponds to a global motion of the entire system around an axis and shows no motion of spiral arms (Photo credit: Greydon Moore http://www.cosmicastronomy.com/bodes4.htm.)

Spiral Arms, Birthplaces of Stars In other words, stars and gas clouds turn around the galactic nucleus, but the spiral arms themselves do not move! The spiral structure is a stationary situation of the gravitational field (the calculation of the field is a complicated nonlinear problem). Stars and interstellar matter turn around the center, and they slow down when they cross the spiral arms. This is why the spiral arms can be seen both in the visible part of the spectrum and in radiowaves. The spiral arms are brighter precisely because of the jamming phenomenon that happens in them, and matter is more abundant there. Spiral galaxies, an example of which is shown in Fig. 13.12, resemble huge rotating fireworks. But, as we have just seen, the spiral arms do not move. They do not turn around! Although it is not known at present how spiral arms are formed, one can show that if they exist, they are stable. It is a very difficult mathematical problem to explain how they arise. Nevertheless, this observation shows a fundamental type of mechanism of star formation. During their slowing down, stars and the interstellar medium are compressed

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Fig. 13.12 Left Spiral galaxy NGC 1232 (Photo credit: FORS1, 8.2 m VLT Antu, ESO, http:// www.star.ucl.ac.uk/~apod/apod/ap040125.html.) Right Schematic star formation mechanism by compression of molecular clouds in spiral arms

Fig. 13.13 The “Pillars of Creation” in the Eagle nebula, in the constellation Serpens. The two pictures of this star formation region are taken with different filters (and the prints do not have the same orientation). On the left, one can see the general shape of the molecular cloud in the nebula. The region of interest is a bit below the center (Photo credit: ESA, ISO, ISOGAL Team, http:// antwrp.gsfc.nasa.gov/apod/ap010914.html.) On the right, one can see a detailed picture of that region, where young bright stars are formed and ejected in the medium (Photo credit: Jean-Charles Cuillandre (CFHT), Hawaiian Starlight, CFHT http://antwrp.gsfc.nasa.gov/apod/ap030213.html.)

as sketched in Fig. 13.12. The compression results in an increase in temperature and in density. Locally, this leads to a gravitational collapse followed by an ignition of thermonuclear reactions and the birth of stars in the spiral arms. This is how many stars form, in particular, massive stars. Regions of star formation are among the most spectacular objects that one can see in the sky. In Fig. 13.13 one can see one of these “star cradles,” called the “Pillars of Creation” in the Eagle nebula, in the constellation Serpens. The compressed molecular cloud literally spits out bursts of young stars in the interstellar medium.

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Fig. 13.14 Zeeman splitting of the ground state of deuterium.

13.7 Exercises 1. Permutation operator s = Show that the permutation operator defined in (13.14) can be written as Pˆ12 (1 + σˆ 1 · σˆ 2 )/2.

2. The singlet state Consider two spins 1/2, and the eigenbasis {|±u ⊗ |±u } of the two operators S1 · u and S2 · u, u being any unit vector of R 3 . Show that the singlet state is written in that basis as: 1 √ (|+u ⊗ |−u − |−u ⊗ |+u ) . 2 3. Spin and magnetic moment of the deuteron We note Jˆ the total angular momentum observable of the electron cloud of an atom, and Iˆ the angular momentum of the nucleus. The respective magnetic moment ˆ and μ ˆ where g J and g I are dimenˆ I = g I μ N I/ observables are μ ˆ J = g J μ B J/ sionless factors. The magnetic interaction Hamiltonian of the electron cloud with the ˆ I where a is a constant which depends on the nucleus is of the form Wˆ = a μ ˆJ ·μ electron distribution around the nucleus. a. Suppose that the state of the nucleus (energy E I , square of the angular momentum I (I + 1)2 ) and the state of the electronic cloud (energy E J , square of the angular momentum J (J + 1)2 ) are both fixed. What are the possible values K (K + 1)2 of the total angular momentum Kˆ of the atom? 2 2 2 b. Express Wˆ in terms of Iˆ , Jˆ and Kˆ . Express the hyperfine energy levels of the atom in terms of I, J and K . c. Calculate the splitting between two consecutive hyperfine levels.

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d. When one applies a uniform weak magnetic field B on a deuterium atom, one observes that the two hyperfine levels (E K and E K  ) of the ground state are split as a function of B as shown on the Fig. 13.14. Knowing that the single electron of the atom is in its orbital ground state  = 0, what is the value of the deuteron spin? e. Assuming that the proton and neutron inside the deuteron have a zero orbital angular momentum, what is their spin state? f. We have a = −8μ0 /12πa13 where a1 is the Bohr radius and ε0 μ0 c2 = 1. Given that ge = 2 and g I = 0.86, to what frequency must a radiotelescope be tuned in order to detect deuterium in the interstellar medium?

Chapter 14

Identical Particles, the Pauli Principle

The origin of geometry, from Pythagoras to Euclid, lies in our environment, and in the observation that one can model the world in which we live by a space where each object is described by a point or a set of points. The concept of space itself came after the simpler concept of “place” of an object. The idea of space arose with the question as to whether the place exists independently of the fact that some object occupies it. In this context, by definition, two objects cannot have the same position at the same time. In this chapter, we address the quantum transposition of this problem. In the probabilistic quantum description, there is no reason a priori why the density probability for two particles to be at the same point in space should vanish, contrary to the classical observation. It is therefore legitimate to elevate the above question to state vectors (or wave functions) rather than positions. Can two particles be in the same state at the same time? Naturally, two particles of different kinds, such as an electron or a proton, will never be in the same state: even if their wave functions coincide, their mass difference implies differences in the values of various physical quantities and one can always tell them from each other. However, there exist in Nature identical particles: all electrons in the universe have the same mass, the same charge, and so on. Can such particles, whose intrinsic properties are all the same, be in the same state? The answer lies in one of the simplest, but most profound, principles of physics, whose consequences on the structure of matter are numerous and fundamental: the Pauli principle. The depth, the intellectual upheaval, and the philosophical implications of the Pauli principle are considerable. If this principle did not provoke the same interest as relativity among philosophers, and even among physicists, it is probably because it explained so many experimental facts (many more than relativity) that Fermi and Dirac had incorporated it immediately in the general theory of quantum mechanics. The result is the following. The state of a state of N identical particles is either totally symmetric or totally antisymmetric if one exchanges any two of these particles. It is symmetric if the spin of these particles is integer; it is antisymmetric if the spin is © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_14

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half-integer. This relation between the symmetry of states and the spin of particles is an experimental fact. This property can be proven theoretically but the proof cannot be explained in a simple way. We show in Sect. 14.1.1 that there is a genuine physical problem with the principles of Chap. 6. Some predictions are ambiguous, so that these principles do not suffice when one deals with systems containing identical particles. A new fundamental principle must be added in order to get rid of this ambiguity. The essential point is that, by definition, two identical particles can be interchanged in a physical system without modifying any property of this system. The mathematical tool that corresponds to the interchange of two particles is the exchange operator which we introduce in Sect. 14.2. In Sect. 14.3 we express the Pauli principle as an additional axiom. Finally, in Sect. 14.4, we discuss some consequences of the Pauli principle.

14.1 Indistinguishability of Two Identical Particles 14.1.1 Identical Particles in Classical Physics By definition, two particles are identical if all their intrinsic properties are the same. In classical mechanics, for a two-particle system, it is always possible to measure at a given time the position of each particle. At that instant, we can define which particle we call 1 and which one we call 2. It is also possible to follow the trajectory of particle 1 and that of particle 2. We can keep on distinguishing unambiguously each particle at any later time. For instance, in the collision of two billiard balls of the same color, we can unambiguously tell the difference between the two processes of Fig. 14.1. Therefore, for any system that is described by classical physics, two particles are always distinguishable, whether or not they are identical (the notion of identity of classical macroscopic objects is anyway an idealization).

Fig. 14.1 Collision between two identical particles

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14.1.2 The Quantum Problem The situation is different in quantum mechanics. At a given time we can still measure the positions of the particles and label them with the indices 1 and 2. However, because the notion of a trajectory is not defined, it may be impossible to follow the two particles individually as time goes on. For instance, one cannot tell the difference between the two processes sketched in Fig. 14.1 if the two wave functions of particles 1 and 2 overlap. It is impossible to know whether particle 1 has become particle 1 or particle 2 . In quantum mechanics two identical particles are indistinguishable. Here, physics falsifies the famous “principle of the identity of indistinguishables,” which is a basic principle in Leibniz’s philosophy, where two real objects are never exactly similar. We show that there exist cases where N identical particles can be in the same state (Bose– Einstein condensate) although they are not a single entity. The number N of these particles is a measurable quantity, although they are indistinguishable from each other.

14.1.3 Example of Ambiguities In the framework of the principles of Chap. 6, this indistinguishability leads to ambiguities in the predictions of physical measurements. Consider, for instance, two identical particles moving in a one-dimensional harmonic potential. We label the particles 1 and 2 and we assume that the Hamiltonian is: pˆ 2 1 1 pˆ 2 Hˆ = 1 + mω 2 xˆ12 + 2 + mω 2 xˆ22 = hˆ (1) + hˆ (2) . 2m 2 2m 2 For simplicity we suppose that the particles have no mutual interaction. Let (n + 1/2)ω and φn (x) (n = 0, 1, . . .) be the eigenvalues and eigenfunctions of the one-particle Hamiltonian hˆ = pˆ 2 /2m + mω 2 xˆ 2 /2. There is no problem in describing the physical situation where both particles are ˆ The corresponding state is: in the ground state of h. Φ0 (x1 , x2 ) = φ0 (x1 ) φ0 (x2 ), and its energy is E 0 = ω. On the contrary, the description of the first excited state of the system is ambiguous. This corresponds to one of the particles being in the first excited state of hˆ and the other in the ground state. The total energy is 2ω. One possible state is φ1 (x1 ) φ0 (x2 ); another possible state is φ0 (x1 ) φ1 (x2 ). Because these two states are possible candidates, then, according to the superposition principle, any linear combination: Φ(x1 , x2 ) = λ φ1 (x1 ) φ0 (x2 ) + μ φ0 (x1 ) φ1 (x2 ) also corresponds to an energy 2ω.

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Therefore there are several different states that appear to describe the same physical situation. This might not be a problem, provided no measurement could make the difference. Alas this is not true! These various states lead to different predictions concerning physically measurable quantities. Consider, for instance the product of the two positions, the observable xˆ1 xˆ2 , for which the labeling of the two particles is irrelevant. Its expectation value is: x1 x2  =

 Re(λ∗ μ). mω

This prediction depends on λ and μ. However, nothing in the theory that we have presented tells us the values of these parameters. Therefore there is a basic ambiguity in the predictions of our principles, and we need a prescription in order to fix the values of λ and μ. It is a remarkable fact of Nature that the only allowed values are λ = ±μ, and that the sign only depends on the nature of the particles under consideration. The allowed states for a system of identical particles are therefore restrictions of the most general states that one could imagine if the particles were distinguishable.

14.2 Two-Particle System; The Exchange Operator 14.2.1 The Hilbert Space for the Two-Particle System Within the framework that we have used up to now, we describe a two-particle system (distinguishable or not) by labeling these particles. A state of the system is therefore |ψ =



Ck,n |1 : k ; 2 : n.

(14.1)

k,n

Implicitly the vectors {|1 : k} form a basis of one-particle states. The states {|1 : k ; 2 : n} are factorized states where the first particle has the quantum numbers k and the second n. They form a basis of the two-particle Hilbert space.

14.2.2 The Exchange Operator Between Two Identical Particles The labeling of the particles that we used above has no absolute meaning if they are identical. Consequently, the predictions of experimental results must be independent of this labeling. In order to describe this property due to the exchange symmetry, we introduce the exchange operator Pˆ12 such that for any couple (k, n):

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Pˆ12 |1 : k ; 2 : n = |1 : n ; 2 : k.

(14.2)

One can verify that this operator is Hermitian and that it satisfies: 2 Pˆ12 = Iˆ.

(14.3)

Example a. For two spinless particles, we have (external) Pˆ12 ≡ Pˆ12 , that is, Pˆ12 Ψ (r 1 , r 2 ) = Ψ (r 2 , r 1 ).

b. For two spin 1/2 particles, then Pˆ12 exchanges both the orbital and the spin variables of the two particles: (external) ˆ (spin) P12 . Pˆ12 = Pˆ12

c. Permutation of two spin 1/2 particles. In this case, one can write down explicitly the action of Pˆ12 by using the representation (13.11), Pˆ12



Ψσ1 ,σ2 (r 1 , r 2 ) |1 : σ1 ; 2 : σ2  =

σ1 ,σ2



Ψσ1 ,σ2 (r 2 , r 1 ) |1 : σ2 ; 2 : σ1 

σ1 ,σ2

where σi = ± with i = 1, 2. In order to discuss the properties of this permutation, it is convenient to work with the eigenbasis of the square of the total spin Sˆ = Sˆ 1 + Sˆ 2 and Sˆ z |S = 1, m = 1 = |1 : + ; 2 : +, 1 |S = 1, m = 0 = √ (|1 : + ; 2 : − + |1 : − ; 2 : +) , 2 |S = 1, m = −1 = |1 : − ; 2 : −, 1 |S = 0, m = 0 = √ (|1 : + ; 2 : − − |1 : − ; 2 : +) . 2 We already noticed that: • The triplet states (S = 1) are symmetric under the interchange of σ1 and σ2 : (spin) Pˆ12 |S = 1, m = |S = 1, m,

• The singlet state (S = 0) is antisymmetric in this interchange: (spin) Pˆ12 |S = 0, m = 0 = − |S = 0, m = 0.

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14.2.3 Symmetry of the States How can one fulfill the requirement that the experimental results must be unchanged as one goes from |Ψ  to Pˆ12 |Ψ ? These two vectors must represent the same physical state, therefore they can only differ by a phase factor, that is, Pˆ12 |Ψ  = eiδ |Ψ . 2 = Iˆ, we have e2iδ = 1 and eiδ = ±1. Therefore: Because Pˆ12 Pˆ12 |Ψ  = ±|Ψ .

(14.4)

We reach the following conclusion. The only physically acceptable state vectors for a system of two identical particles are either symmetric or antisymmetric under the permutation of the two particles. Referring to (14.1), this implies Ck,n = ±Cn,k . The only allowed states are either symmetric under the exchange of 1 and 2: |Ψ S  ∝



Ck,n (|1 : k ; 2 : n + |1 : n ; 2 : k) ;

Pˆ12 |Ψ S  = |Ψ S 

(14.5)

Pˆ12 |Ψ A  = −|Ψ A ,

(14.6)

k,n

or antisymmetric: |Ψ A  ∝



Ck,n (|1 : k ; 2 : n − |1 : n ; 2 : k) ;

k,n

where the coefficients Ck,n are arbitrary. This restriction to symmetric or antisymmetric state vectors is a considerable step forward in order to solve the ambiguity pointed out in the previous section. For instance, the expectation value x1 x2  considered in Sect. 14.1.2 can now take √ only two values ±/(2mω), corresponding to the two choices λ = ±μ = 1/ 2. However, it is not yet sufficient and some essential questions are still open. a. Can a given species, such as electrons, behave in some experimental situations with the plus sign in (14.4) and in other situations with the minus sign? b. Assuming that the answer to the first question is negative, what decides which sign should be attributed to a given species? The answers to these two questions lead us to one of the simplest and most fundamental laws of physics: the Pauli principle. (The general formulation was derived from Pauli’s ideas by Fermi and Dirac.)

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14.3 The Pauli Principle 14.3.1 The Case of Two Particles All particles in Nature belong to one of the two following categories: • Bosons for which the state vector of two identical particles is always symmetric under the operation Pˆ12 . • Fermions for which the state vector of two identical particles is always antisymmetric under the operation Pˆ12 . All particles of integer spin (including 0) are bosons (photon, π meson, α particle, gluons, etc.). All particles of half-integer spin are fermions (electron, proton, neutron, neutrino, quarks, He3 nucleus, etc.). The state vectors of two bosons are of the form |Ψ S  (Eq. (14.5)), those of fermions are of the form |Ψ A  (Eq. (14.6)). The Pauli principle therefore consists of restricting the set of accessible states for systems of two identical particles. The space of physical states is no longer the (tensor) product of the basis states, but only the subspace formed by the symmetric or antisymmetric combinations. The Pauli principle also applies to composite particles such as nuclei or atoms, provided experimental conditions are such that they are in the same internal state (be it the ground state or a metastable excited state). For instance, hydrogen atoms in their electronic ground state have a total spin S = 0 or S = 1 and they behave as bosons. Deuterium, the isotope of hydrogen is a fermion (nuclear spin 1, electron spin 1/2, total spin half-integer). As we announced, this connection between the symmetry of states and the spin of the particles is an experimental fact. However, it is both a triumph and a mystery of contemporary physics that this property, called the “spin–statistics connection,” can be proven starting from general axioms of relativistic quantum field theory.1 It is a mystery because it is probably the only example of a very simple physical law for which the proof exists but cannot be explained in an elementary way. Example a. The wave function of two identical spin zero particles must be symmetric: Ψ (r 1 , r 2 ) = Ψ (r 2 , r 1 ). b. The state of two spin 1/2 particles must be of the form: 1 Markus Fierz (1939). “Über die relativistische Theorie Kräftefreier Teilchen mit beliebigem Spin”. Helvetica Physica Acta 12 (1): 337. doi:10.5169/seals-110930. Wolfgang Pauli (15 October 1940). “The Connection Between Spin and Statistics”. Physical Review 58 (8): 716722. doi:10.1103/ PhysRev.58.716. See also: I. Duck and E.C.G. Sudarshan, Pauli and the Spin-Statistics Theorem, World Scientific, Singapore, (1997).

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|Ψ  = Ψ0,0 (r 1 , r 2 ) |S = 0, m = 0 +



Ψ1,m (r 1 , r 2 ) |S = 1, m,

m

where Ψ0,0 and Ψ1,m are, respectively, symmetric and antisymmetric: Ψ0,0 (r 1 , r 2 ) = Ψ0,0 (r 2 , r 1 ) , Ψ1,m (r 1 , r 2 ) = −Ψ1,m (r 2 , r 1 ). Therefore the orbital state and the spin state of two identical fermions are correlated.

14.3.2 Independent Fermions and Exclusion Principle Consider a situation where two fermions, for instance, two electrons, are independent, that is, they do not interact with each other. The (effective) total Hamiltonian then ˆ ˆ reads Hˆ = h(1) + h(2). In such conditions, the eigenstates of Hˆ are products of ˆ |1 : n ; 2 : n  . We remark that, if n = n  , that is, if the eigenstates |n of h: two particles are in the same quantum state, the state |1 : n ; 2 : n is necessarily symmetric. This is forbidden by the Pauli principle, which results in the (weaker) formulation: Two independent fermions in the same system cannot be in the same state. If |n and |n   are orthogonal, the only acceptable state is the antisymmetric combination:  1  |Ψ A  = √ |1 : n ; 2 : n   − |1 : n  ; 2 : n . 2 In this simplified form, the Pauli principle appears as an exclusion principle.

14.3.3 The Case of N Identical Particles For a system of N identical particles, we proceed in a similar manner. We introduce the exchange operator Pˆi j of the two particles i and j. The indistinguishability imposes that Pˆi j |Ψ  leads to the same physical results as |Ψ . The general form of the Pauli principle is as follows. The state vector of a system of N identical bosons is completely symmetric under the interchange of any two of these particles. The state vector of a system of N identical fermions is completely antisymmetric under the interchange of any two of these particles.

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351

For instance, for N = 3, one has: Ψ± (u 1 , u 2 , u 3 ) ∝ ( f (u 1 , u 2 , u 3 ) + f (u 2 , u 3 , u 1 ) + f (u 3 , u 1 , u 2 )) ± ( f (u 1 , u 3 , u 2 ) + f (u 2 , u 1 , u 3 ) + f (u 3 , u 2 , u 1 )) , where f is any function of the three sets of variables u 1 , u 2 , u 3 . The plus sign (resp., minus sign) corresponds to a function Ψ that is completely symmetric (resp., completely antisymmetric). More generally, let us consider an orthonormal basis {|n} of the one-particle states, and the N ! permutations P of a set of N elements. We want to describe the following physical situation, “One particle in the state |n 1 , one particle in the state |n 2 , . . ., one particle in the state |n N .” In order to do this, we number in an arbitrary way the N particles from 1 to N . The Case of Bosons The state is: C  |Ψ  = √ |1 : n P(1) ; 2 : n P(2) ; . . . ; N : n P(N ) , N! P

(14.7)

 where P denotes the sum over all permutations. Notice that two (or more) indices n i , n j , . . . labeling the occupied states may coincide. The normalization factor C is expressed in terms of the occupation numbers Ni of the states |n i : C = (N1 ! N2 ! . . .)−1/2 . The Case of Fermions In the case of fermions, the result is physically acceptable if and only if the N states |n i  are pairwise orthogonal. The state |Ψ  is then: 1  ε P |1 : n P(1) ; 2 : n P(2) ; . . . ; N : n P(N ) , |Ψ  = √ N! P

(14.8)

where ε P is the signature of the permutation P: ε P = 1 if P is an even permutation and ε P = −1 if P is odd. This state vector is often written in the form of a determinant, called the Slater determinant:    |1 : n 1  |1 : n 2  . . . |1 : n N     1  |2 : n 1  |2 : n 2  . . . |2 : n N   |Ψ  = √  (14.9) . .. .. ..  N !  . . .   |N : n 1  |N : n 2  . . . |N : n N  

352

14 Identical Particles, the Pauli Principle

If two particles are in the same state, two columns are identical and this determinant vanishes. The set of states that can be constructed using (14.7) (resp., (14.8)) forms a basis of the Hilbert space of an N boson (resp., N fermion) system.

14.4 Physical Consequences of the Pauli Principle We give below a few of the many physical consequences of the Pauli principle that concern both few-body systems and the macroscopic properties of a large number of bosons or fermions.

14.4.1 Exchange Force Between Two Fermions Consider the helium atom and neglect magnetic effects, as we did for hydrogen in Chap. 11. We label the electrons 1 and 2, and the Hamiltonian is: pˆ 2 2e2 2e2 e2 pˆ 2 − + , Hˆ = 1 + 2 − 2m e 2m e rˆ1 rˆ2 rˆ12

with

rˆ 12 = rˆ 1 − rˆ 2 .

The eigenvalue problem is technically complicated and can only be solved numerically, but the results of interest here are simple (Fig. 14.2). The ground state (E 0 = −78.9 eV) corresponds to a symmetric spatial wave function, whereas the first two excited states E 1A = −58.6 eV and E 1S = −57.8 eV have, respectively, antisymmetric and symmetric spatial wave functions. The symmetry of the wave function implies a specific symmetry of the spin state: E 0 and E 1S are singlet spin states, and E 1A is a triplet spin state. In the ground state, the two spins are antiparallel. In order to flip one of them to make them parallel, a considerable amount of energy (∼20 eV) is necessary. This corresponds to a “force” that maintains the spins in the antiparallel state. It is not a magnetic coupling between the spins: this magnetic interaction can be calculated, and it corresponds to an energy of the order of 10−2 eV. The “force” that we are facing here has an electrostatic origin, the Coulomb interaction, and it

Fig. 14.2 The first three levels of the helium atom

14.4 Physical Consequences of the Pauli Principle

353

is transferred into a constraint on the spins via the Pauli principle. Such an effect is called an exchange interaction. The same effect is the basic cause of ferromagnetism (unpaired valence electrons in, e.g., iron, cobalt, nickel).

14.4.2 The Ground State of N Identical Independent Particles Consider N identical independent particles. The Hamiltonian is therefore the sum of N one-particle Hamiltonians: N  (14.10) hˆ (i) . Hˆ = i=1

ˆ hφ ˆ n = εn φn , Let {φn , εn } be the eigenfunctions and corresponding eigenvalues of h: where we assume that the εn are ordered: ε1 ≤ ε2 · · · ≤ εn · · · . From the previous considerations, we see that the ground state energy of a system of N bosons is E 0 = N ε1 , whereas for a system of fermions, we have E0 =

N 

εi .

i=1

In this latter case, the highest occupied energy level is called the Fermi energy of the system and it is denoted  F . The occupation of the states φn is represented on Fig. 14.3 both for a bosonic and a fermionic assembly. Consider, for instance, N independent fermions of spin s confined in a cubic box of size L. We choose here a basis of states corresponding to periodic boundary√conditions. Each eigenstate of the Hamiltonian hˆ is a plane wave φ p (r) = ei p·r/ / L 3 , associated with one of the 2s + 1 spin states corresponding to a well-defined component m s  of the spin on a given axis (m s = −s, −s + 1, . . . , s). The momentum p can be written p = (2π/L)n, where the vector n = (n 1 , n 2 , n 3 ) stands for a triplet of positive or negative integers. The ground state of the N fermion system is obtained by placing 2s + 1 fermions in each plane wave φ p , as long as | p| is lower than the Fermi momentum p F . This Fermi momentum p F is determined using: N=

 p ( p< p F )

(2s + 1).

354

14 Identical Particles, the Pauli Principle

Fig. 14.3 Ground state of a system of N independent identical particles. Left bosonic case, with all the particles in the ground state of the one-body Hamiltonian. Right fermionic case, where the first N /(2s + 1) states of the one-body Hamiltonian are occupied (here s = 1/2)

For a large number of particles, we can replace this discrete sum by an integral that yields  2s + 1 L 3 p 3F L3 3 d p = . (14.11) N  (2s + 1) (2π)3 p< p F 6π 2 3 This equation relates the density ρ = N /L 3 of the gas and the Fermi momentum, independently of the size of the box: ρ=

2s + 1 ( p F /)3 . 6π 2

(14.12)

The average kinetic energy per particle can also be easily calculated:  p2  1 = 2m N which leads to

 p ( p< p F )

(2s + 1)

p2 2s + 1 L 3  2m N (2π)3

 p< p F

p2 3 d p, 2m

 p2  3 p 2F = . 2m 5 2m

(14.13)

14.4.3 Behavior of Fermion and Boson Systems at Low Temperatures The difference between the ground states of N -fermion or N -boson systems induces radically different behaviors of such systems at low temperature.

14.4 Physical Consequences of the Pauli Principle

355

Fermions In a system of fermions at zero temperature and in the absence of interactions, we have just seen that all the energy levels of the one-body Hamiltonian are filled up to the Fermi energy  F . This simple model describes the conduction electrons in a metal remarkably well, and it accounts for many macroscopic properties of a solid, such as its thermal conductibility. Using the result (14.12), the Fermi energy  F = p 2F /2m e can be written in terms of the number density ρe of conduction electrons,  2/3 2 3ρe π 2 , F = 2m e where we used 2s + 1 = 2 for electrons. This energy can reach large values ( F = 3 eV for sodium). This is much larger than the thermal energy at room temperature (k B T  0.025 eV). This explains the success of the zero-temperature fermion gas model for conduction electrons. At room temperature, very few electrons participate in thermal exchanges. The application of the Pauli principle to fermionic systems has many consequences, ranging from solid-state physics to the stability of stars such as white dwarfs or neutron stars. In nuclear physics, the Pauli principle explains why neutrons are stable under β decay inside nuclei. An isolated neutron is unstable and decays through the process n → p + e + ν¯ with a lifetime of the order of 15 min. Inside a nucleus, a neutron can be stabilized if all the final states allowed by energy conservation for the final proton are already occupied. Therefore, the decay cannot occur because of energy conservation. Bosons Concerning bosonic systems, a spectacular consequence of quantum statistics is the Bose–Einstein condensation. In the absence of interactions between the particles, if the number density ρ = N /V is such that: ρΛ3T > 2.612

with

ΛT = √

h , 2πmk B T

(14.14)

a macroscopic accumulation of particles occurs in a single quantum state, i.e. the ground state of the confining potential of the particles. Such congregations of bosons contrast with the “individualistic” character of fermions. Until June 1995, the usual example of a Bose–Einstein condensation that was given in textbooks was the transition normal liquid → superfluid liquid of helium which happens at a temperature of T = 2.17 K. However the complicated interactions inside the liquid make the quantitative treatment of the superfluid transition quite involved, and different from the simple theory of the Bose–Einstein condensation of an ideal gas.

356

14 Identical Particles, the Pauli Principle

Fig. 14.4 Bose–Einstein condensation of a gas of 87 Rb atoms observed via the evolution of the momentum distribution of the particles in the x y plane. The atoms are confined in an anisotropic harmonic trap (ωx < ω y ) and cooled by evaporation. The trap is then turned down suddenly and the momentum distribution is measured using a time-of-flight technique. Left the temperature T is noticeably larger than the condensation temperature Tc . The momentum distribution is isotropic and close to a Maxwell–Boltzmann distribution (mvi2  = k B T for i = x, y). Middle images (T ≤ Tc ): a noticeable fraction of the atoms accumulate in the ground state of the magnetic trap. Right image (T Tc ): a very large fraction of the atoms is in the ground state of the trap. The momentum distribution of the trap reflects the trap anisotropy. (mvi2  = ωi /2). Strictly speaking one should take into account the interactions between atoms to explain quantitatively the details of this distribution (photographs by F. Chevy et K. Madison, ENS Paris)

We now have at our disposal experiments2 performed on gases of alkali atoms (lithium, sodium, rubidium) which are initially cooled by laser inside a vacuum chamber with a very low residual pressure (below 10−11 mBar). The atoms are then confined by an inhomogeneous magnetic field and they are cooled further down by evaporation until they reach Bose–Einstein condensation, at a temperature below 1 µK. The evaporative cooling technique consists in eliminating the more energetic atoms, in order to keep only the slower ones. The collisions between the trapped atoms maintain permanently the thermal equilibrium. Starting from 109 atoms, one can obtain, after evaporation, a situation where the 106 remaining atoms are practically all in the ground state of the system. Figure 14.4 shows the time evolution of the momentum distribution of a gas of bosons (rubidium 87 atoms) confined in a magnetic trap and cooled down to the condensation point. These Bose–Einstein condensates possess remarkable coherence and superfluid properties, and this has been a very active field of research in recent years.3 The 2001 Nobel prize in Physics was awarded jointly to Eric A. Cornell, Wolfgang Ketterle and Carl E. Wieman “For the achievement of Bose–Einstein condensation

2 See M.H. Anderson et al., Science 269, 198 (1995), where the first Bose–Einstein condensate is shown in an experiment carried out with rubidium atoms. 3 E. Cornell and C. Wieman, Bose-Einstein condensation, Scientific American, March 1998, p. 26; W. Ketterle, Experimental studies of Bose-Einstein condensation, Physics Today, December 1999, p. 30; K. Helmerson et al., Atom Lasers, Physics World, August 1999, p. 31; Y. Castin et al., BoseEinstein condensates make quantum leaps and bounds, Physics World, August 1999, p. 37; M. W. Zwierlein, C. A. Stan, C. H. Schunck, S. M. F. Raupach, S. Gupta, Z. Hadzibabic, and W. Ketterle (2003). Observation of BoseEinstein Condensation of Molecules, Physical Review Letters 91 (25): 250401.

14.4 Physical Consequences of the Pauli Principle

357

in dilute gases of alkali atoms, and for early fundamental studies of the properties of the condensates”.

14.4.4 Stimulated Emission and the Laser Effect Consider a system of independent bosons (cf. Sect. 14.4.2) on which we apply for a finite length of time the one-body potential Vˆ = i vˆ (i) . Each potential vˆ (i) acts only on the particle i and it can induce transitions between the various eigenstates of hˆ (i) . We want to show that the probability for a particle to reach a given final state |φl  is increased if this state is already occupied. Consider first the case where we are dealing with a single particle initially in the state |φk . If we assume that the effect of vˆ is weak, the probability that the particle ˆ l |2 . reaches the state |φl  under the action of vˆ is proportional to |vkl |2 = |φk |v|φ This result of time dependent perturbation theory will be proven in Chap. 16 (see e.g. Eq. (16.15)). Suppose now that the state |φl  is already occupied by N particles, but there is only one particle in the state |φk . The properly symmetrized initial state of the N + 1 boson system is: |Ψi  = √



1 N +1

|1 : φk ; 2 : φl ; . . . ; N : φl ; N + 1 : φl 

+ |1 : φl ; 2 : φk ; . . . ; N : φl ; N + 1 : φl  + · · · + |1 : φl ; 2 : φl ; . . . ; N : φl ; N + 1 : φk  and we are interested in the probability to reach the final state: |Ψ f  = |1 : φl ; 2 : φl ; . . . ; N : φl ; N + 1 : φl . The transition probability is now proportional to: |Vi f |2 = |Ψi |Vˆ |Ψ f |2 = (N + 1) |vkl |2 . The presence of N particles in the state |φl  increases by a factor N +1 the probability that the particle initially in |φk  reaches this state. The transition probability is the sum of the rate for a spontaneous transition, proportional to |vkl |2 and independent of N , and of the rate stimulated by the presence of the N bosons in the state |φl  and proportional to N |vkl |2 . This gregarious behavior also manifests itself for photons, which are massless bosons. This explains the phenomenon of stimulated emission of light which is at the basis of the principle of lasers. An excited atom decays preferentially by emitting a photon in the quantum state occupied by the photons already present in the laser

358

14 Identical Particles, the Pauli Principle

cavity. This leads to a chain reaction in the production of photons which is the key point in the mechanism of lasers.4

14.4.5 Uncertainty Relations for a System of N Fermions Consider N independent fermions of spin s, each placed in a potential V (r ) centered at the origin. The Hamiltonian is therefore: Hˆ =

N 

hˆ (i)

with

i=1

pˆ 2 hˆ = + V (ˆr ). 2m

We note εn the energy levels of hˆ and gn their degeneracies. The ground state E 0 of Hˆ is obtained by filling the lowest lying levels εn up to the Fermi energy  F , each with (2s + 1) particles per state: E 0 = (2s + 1)

k 

gn εn ,

n=0

where the number k is determined by the relation N = (2s + 1) harmonic potential V (r ) = mω 2 r 2 /2, we have:

k

n=0 gn .

For a

εn = (n + 3/2)ω, gn = (n + 2)(n + 1)/2. Therefore we find for N  1:

k

6N 2s + 1

1/3 ,

E 0  ξ N 4/3 ω,

with ξ = (3/4)61/3 (2s + 1)−1/3 . Consider now an arbitrary state |Ψ  of these N fermions. We define r 2  = ri2  and  p 2  =  pi2 , with i = 1, . . . , N . In this state, H  ≥ E 0 and consequently: H  = N

 p2  N + mω 2 r 2  ≥ ξ N 4/3 ω, 2m 2

4 C.H. Townes, How the Laser Happened: Adventures of a Scientist, Oxford University Press (1999); A. Siegman, Lasers (University Science Books, Mill Valley, 1986); O. Svelto, Principles of Lasers Plenum Press, New-York (1998).

14.4 Physical Consequences of the Pauli Principle

359

which gives:  p 2  + m 2 ω 2 r 2  − 2ξ N 1/3 mω ≥ 0

for N  1.

This second degree trinomial in mω is positive for all values of mω. Therefore we obtain that in any state of these N fermions: r 2   p 2  ≥ ξ 2 N 2/3 2

for N  1.

(14.15)

This relation is valid in particular in the center-of-mass frame, where  p = 0. If we choose the origin of space at the center of the cloud, we obtain: Δx Δpx ≥

ξ 1/3 N  3

for N  1.

(14.16)

For spin 1/2 particles, ξ/3 ∼ 0.36. A similar calculation for fermions placed in a 1/r potential leads to: 2 1 N 2/3 p  ≥ γ  r 2

2

for

N 1

(14.17)

with γ = 3−1/3 (2s + 1)−2/3 . The relation (14.17) plays an important role in the stability of self-gravitating systems, in particular neutron stars. The Pauli principle therefore modifies the uncertainty relations. If we place N identical fermions in a volume V ∼ (Δx)3 , each of these fermions must occupy a different quantum state. We can thus consider that the space accessible to each particle is a region of linear extension ∼(V /N )1/3 , so that the de Broglie wavelength of each particle is reduced by a factor N 1/3 . This brings a different, but equivalent, point of view on the physics of an N fermion system that we investigated in Sect. 14.4.3. In particular, consider again an ideal Fermi gas confined in a cubic box of size L at zero temperature. The position distribution in the box is uniform and the average momentum per particle can be deduced from (14.13) so that: Δx 2 = Δpx2

1 L



L/2 −L/2

x2 dx =

L2 , 12

Δp p2 2 N 2/3 = = F = 3 5 5L 2 2



6π 2 2s + 1

2/3 ,

from which one can check that (14.16) is well satisfied. Actually the product Δx Δpx calculated for the ideal Fermi gas confined in a square box exceeds by ∼10% the rigorous lower bound (14.16).

360

14 Identical Particles, the Pauli Principle

Fig. 14.5 An incident wave packet φ1 (r) or φ2 (r) crosses a 50–50 % beam splitter. After the crossing, this provides a coherent superposition of two outgoing wave packets φ3 (r) and φ4 (r)

Exercises 1. Identical Particles on a Beam Splitter Consider a particle prepared at initial time ti in a wave packet ψ(r, ti ) = φ1 (r) incoming on a 50–50 % beam splitter (Fig. 14.5). At a later time t f , the wave packet has crossed the beam√splitter and the state of the particle can be written ψ(r, t f ) = (φ3 (r) + φ4 (r)) / 2, where φ3 and φ4 correspond to normalized wave packets propagating in each of the output ports. We assume φ3 |φ4   0. a. We prepare the particle in the state ψ(r, ti ) = φ2 (r), which is deduced from φ1 (r) by symmetry with respect to the beam splitter plane. The state of the particle at time t f can then be written: ψ(r, t f ) = αφ3 (r) + βφ4 (r). Determine (within a global phase factor) the coefficients α and β. Make use of the fact that the interaction of the particle with the beam splitter is described by a Hamiltonian (which needs not to be explicitly written). We take φ2 |φ1  = 0. b. We prepare at time ti two fermions in the same spin state, one in the external state φ1 (r), the other one in the state φ2 (r). What is the final state of the system? Can one detect both fermions in the same output port? c. Consider the same problem with two bosons, also prepared in the same spin state, one in the external state φ1 (r), the other one in the state φ2 (r). Show that the two bosons always exit in the same port. This experiment has been performed with photons by C.K. Hong et al., Phys. Rev. Lett. 59, 2044 (1987). 2. Fermions in a Square Well Consider two spin 1/2 particles of mass m confined in a one dimension infinite square well of size L (Chap. 5, Sect. 5.3). a. We neglect the interactions between the particles. Determine the four lowest energy levels.

14.4 Physical Consequences of the Pauli Principle

361

b. We suppose that the particles interact together via a contact potential V (x1 −x2 ) = g δ(x1 − x2 ), where the coupling constant g does not depend on spin. Determine to first order in g the four lowest energy levels of this two particle system.

14.5 Problem: Discovery of the Pauli Principle Pauli understood what is known as his Principle in 1925. He proceeded by first gathering a whole set of independent and sometimes paradoxical pieces of experimental spectroscopic data which had accumulated in the beginning of the 1920s. Putting that together with his personal intuition the symmetry properties of the wave functions, he discovered a common feature in all those experimental results and he understood what was general in that discovery. As Pauli did, we shall focus mainly on the case of the helium atom. However, we shall assume that we know what has been developed in Chap. 12 about the existence of the electron spin 1/2 and its properties (which Pauli did not know completely...) Experimental Data In the beginning of the 1920s the following aspects of helium spectroscopy were known. a. The lines observed in the optical spectroscopy of this atom were measured. In the presence of a magnetic field B, some of them were split by the Zeeman effect with a given multiplicity. A non-split was calle a singlet, split levels were called triplets if they were split into three sub-levels. Now, spectroscopy had clearly observed a very intriguing fact,—early in the 20th century—. There were actually two sorts of helium, called orthohelium and parahelium which coexisted in all experiments and were characterized by different energy levels. All helium spectra exhibited both categories of lines, in variable proportions according to what kind of helium was studied. For instance, the Solar helium contained abundant parahelium. Figure 14.6 represents schematically the arrangement of the lowest lying levels in the two cases together with their multiplicities. All the parahelium states are singlets, all the orthohelium ones are triplets. b. One notices that the ortho and parahelium are stable with respect to one another: they do not transform into each other in experimentally accessible time scales (say a few months). However, in its ground state, the orthohelium has a much larger energy than that of the parahelium (about 20 eV: see Fig. 14.6). Why can’t it transform into parahelium? The reaction energy, 20 eV, is quite considerable. Pauli is deeply intrigued by that observation. c. In its ground state the parahelium has a practically vanishing magnetic susceptibility. It is not paramagnetic, it does not posses any internal magnetic moment, as opposed to orthohelium.

362

14 Identical Particles, the Pauli Principle

Fig. 14.6 Lowest lying energy levels of helium as seen experimentally. Only the three first ones (two singlets of parahelium and the triplet ground state of orthohelium) are represented. The transitions actually observed are represented by vertical arrows

Theory of the Helium Atom. Non-perturbed Description. Let r 1 and r 2 be the positions (x1 , y1 , z 1 ) and (x2 , y2 , z 2 ) of the two electrons with respect to the nucleus, which we assume fixed. In first approximation, we neglect the electrostatic repulsion e2 /r12 (r12 = |r 1 − r 2 |) between the two electrons, and the electron spins. The Hamiltonian of the two electron system of in the field of the nucleus of charge 2 is p2 2e2 (i = 1, or 2). Hˆ 0 = Hˆ 1 + Hˆ 2 , with Hˆ i = i − 2m ri If ψa and ψb are two normalized eigenfunctions of the hydrogen-like hamiltonian Hˆ i (with eigenenergies E a and E b ) the function ψ(r 1 , r 2 ) = ψa (r 1 )ψb (r 2 ) is an eigenfunction of Hˆ 0 . Since electrons (1) and (2) are not distinguishable, we introduce the wave functions 1 ψ ± = √ [ψa (r 1 )ψb (r 2 ) ± ψa (r 2 )ψb (r 1 )] 2

(14.18)

14.5 Problem: Discovery of the Pauli Principle

363

which are respectively symmetric and antisymmetric in the permutation of the two electrons. 0 corresponding to ψ + and ψ − ? What are the eigenenergies E a,b Introduction of the Perturbation e2 /r12 . We are now interested in the two hydrogen-like states |1s (n = 1,  = 0, m = 0) and |2s (n = 2,  = 0, m = 0) and therefore in the two following cases: ψa = ψb = ψ1s ψa = ψ1s ψb = ψ2s

on one side and, on the other side.

(14.19) (14.20)

(a) How can one evaluate, in first order perturbation theory, the influence of the 0 ? Coulomb repulsion e2 /r12 on the energy E a,b Note One will not take into account the states ψ2 p (n = 2,  = 1) in questions (a), (b), (c) and (d). (b) Show that once this perturbation is taken into account, the energies of the states ψ + and ψ − can be written ± 0 E a,b = E a,b + Ca,b ± Aa,b

where Ca,b and Aa,b are integrals that one will write but not calculate explicitly. (c) Show the the integral Ca,b is positive and give its simple physical meaning. (d) Show qualitatively the the integral Aa,b is most likely positive. Note One can remark that the term e2 /r12 takes large values if the electrons are close to one another. (e) The values of the following integrals are  I1 =

|ψ1s (r1 )|2 |ψ1s (r2 )|2

e2 3 d r1 d 3r2 = 34 eV r12

|ψ1s (r1 )|2 |ψ2s (r2 )|2

e2 3 d r1 d 3r2 = 15 eV r12

 I2 =  J=

∗ ∗ (r1 )ψ2s (r2 )ψ1s (r2 )ψ2s (r1 ) ψ1s

e2 3 d r1 d 3 r2 = 0.4 eV r12

0 Represent on the same diagram, starting from the levels E a,b which will be ± + − calculated, the levels E a,b of the two states ψ and ψ in first order perturbation theory in the cases (14.19) and (14.20) above, i.e., a = 1s, b = 1s and a = 1s, b = 2s. Note that in the case (14.19) only the state ψ + remains.

364

14 Identical Particles, the Pauli Principle

Introduction of the Electron Spins For each of the electrons, (1) and (2), we must now introduce two possible spin states. We use the notation of Chap. 13, Sect. 13.1.1 |σ1 , σ2  with σ1,2 = ±1. We consider the states of total spin S defined by |Σ +  = | + + √ |Σ 0  = (| + − + | − +)/ 2 |Σ −  = | − − √ | σ  = (| + − − | − +)/ 2

(14.21) (14.22)

a. To what values of the total spin S and of its projection Sz along Oz (in  units) do the states |Σ + , |Σ 0 , |Σ −  et |σ correspond to? b. Show that the states |Σ + , |Σ 0 , |Σ −  and |σ are eigenstates of the spin exchange sp sp operators Pˆ12 defined by Pˆ12 |σ1 , σ2  = |σ2 , σ1 , and give the corresponding eigenvalues. Zeeman Effect. Singlet and Triplet States When the two-electron system interacts with a magnetic field B (along Oz) its energy levels are modified proportionally to the projection on Oz of the magnetic moment, therefore of Sz . A system of total spin S = 1 with components Sz = +, 0, − has, in the presence of the field B, the three Zeeman interaction energies: +2μe B, 0, −2μe B

μe : Bohr magneton.

Draw the variation of the energy levels of a system of two electrons in a magnetic field B in the spin states |Σ + , |Σ 0 , |Σ −  and |σ. Justify the names triplet and singlet. Total Wave Function of the Helium Atom We now incorporate both the space variables and the spin variables. For instance, we note |ψ + , Σ 0  ≡ ψ + (r 1 , r 2 )|Σ 0  the state of the atom where the electrons are in the orbital state |ψ +  and in the spin state Σ 0 . Make a list of the a priori possible states of the atom in the two orbital states (14.19) and (14.20), without imposing any exclusion criterion (the problem is considered in 1924–1925). Note, for each of these states whether it is globally symmetric or antisymmetric under a permutation of the two electrons. Transition Probabilities We now consider the possible transitions between the states of this list of a priori possible states of the helium atom.

14.5 Problem: Discovery of the Pauli Principle

365

It can be shown that the transition probability amplitude (emission or absorption) of an atom between a state |i and a state | f  is proportional to the matrix element of a transition hamiltonian Hˆ int between these two states:  f | Hˆ int |i. In an electromagnetic transition of an atom, the transition hamiltonian is proportional to the electric dipole moment D = er. In our two-electron problem (1) and (2), the electric dipole moment is D = e(r1 + r2 ). Note that this hamiltonian is invariant under permutations of the two electrons. (a) Consider the matrix elements ψ  , Σ| Hˆ int |ψ, σ or ψ  , σ| Hˆ int |ψ, Σ where ψ and ψ  represent any orbital state and Σ one of the triplet states. Show that these matrix elements vanish. What conclusion can one draw on the triplet-singlet or singlet-triplet transitions? (b) Consider the non-zero matrix elements found in the previous question. Let ψ(r 1 , r 2 ) and ψ  (r 1 , r 2 ) be the orbital wave function of electrons in states ψ and ψ  . Express the matrix elements as integrals over r 1 and r 2 . Since the orbital wave functions are either symmetric or antisymmetric under the exchange of r 1 and r 2 , don’t some matrix elements also vanish? What can we deduce about the possible transitions in the helium atom? Confrontation Between Theory and Experiment We now go back to the set of experimental observations of Sect. 1, that we note respectively a, b and c. Show that a certain elimination among the states which are a priori possible in Sects. 2 and 5 allows to account for these experimental results in a consistent way. Give an explanation to the experimental result of figure (14.6) and indicate to which quantum states these levels correspond. Generalize that elimination in the form of a Principle.

14.5.1 Solution Theory of the Helium atom. 0 2.1 The eigenenergy of ψ + and ψ − is E a,b = Ea + Eb 2.2

(a) and (b) We have a perturbation of degenerate states. However in the basis {ψ + , ψ − } the perturbation is diagonal. In first order, the perturbation ΔE is the solution of C − A − ΔE 0 =0 0 C + A − ΔE with

 C=

ψa2 (r 1 )ψb2 (r 2 )

e2 3 3 d r1 d r2 r12

366

14 Identical Particles, the Pauli Principle

Fig. 14.7 Theoretical lowest energy levels of helium: left, case (14.19), right, case (14.20)

 A=

ψa∗ (r 1 )ψb∗ (r 2 )ψa (r 2 )ψb (r 1 )

e2 3 3 d r1 d r2 r12

Therefore ΔE ± = C ± A, i.e. E ± = E 0 + C ± A. (c) The integral C is positive since −eψa2 (r 1 )d 3r1 and −eψb2 (r 2 )d 3 r2 represent respectively the charge densities of electron (1) and of electron (2). The integral C is the electrostatic repulsion of the two electrons. (d) If r12 is small, ψa∗ (r 1 )ψa (r 2 ) is positive, since r1 is close to r2 . Similarly for ψb∗ (r 2 )ψb (r 1 ). Now, small values of r12 correspond to large values of the function to integrate. The integral A (called exchange integral) is most probably positive. (e) The energy diagram therefore has the following structure shown on Fig. (14.7), with: • Case (14.19), 1s 2 . E a + E a = 2 × 4 × (−13.5) eV  −108 eV, Caa = I1 = 34 eV + =⇒ E aa = −74 eV

• Case (14.20), 1s 2s. E a + E b = 5 × (−13.5) eV  −67.5 eV, Cab = I2 = 15 eV A = J = 0.4 eV + − + − =⇒ E ab − E ab = 0.8 eV , (E ab + E ab )/2 = 52.5 eV

2.3 a. The total spin S is equal to 1 (states |Σ) with projections (+, 0, −), or zero (state |σ). b. One has obviously sp sp Pˆ12 |Σ = |Σ , Pˆ12 |σ = −|σ .

14.5 Problem: Discovery of the Pauli Principle

367

The eigenvalues are ±1, the states |Σ are symmetric and the state |σ is antisymmetric. 2.4 From the previous results, the states |Σ and |σ give rise to the following Zeeman levels

The levels |Σ are associated in three levels and give a triplet, the state |σ is a singlet, insensitive to the magnetic field. 2.5 There are eight possible states in case (14.20), ψa  = ψb : |ψ + Σ +  , |ψ + Σ 0  , |ψ + Σ −  |ψ + σ

is antisymmetric (singlet),

|ψ − Σ +  , |ψ − Σ 0  , |ψ − Σ −  |ψ − σ

are symmetric (triplet),

are antisymmetric (triplet),

is symmetric (singlet).

In case (14.19) the function ψ − vanishes identically. Only four states are left; |ψ + Σ +  , |ψ + Σ 0  , |ψ + Σ −  |ψ + σ

symmetric (triplet),

antisymmetric (singlet).

2.6 (a) The operator Hˆ int does not involve spin variables. The orthogonality of |Σ and |σ : Σ|σ = 0, is such that the transitions singlet↔triplet are forbidden. (b) ψ  Σ| Hˆ int |ψΣ = ψ  σ| Hˆ int |ψσ  =

ψ ∗ (r 1 , r 2 )Hint (r 1 , r 2 )ψ(r 1 , r 2 )d 3r1 d 3r2 .

The integral of an antisymmetric function of r 1 and r 2 vanishes, we therefore have ψ + | Hˆ int |ψ −  = ψ − | Hˆ int |ψ +  = 0. The only allowed transitions take place between orbital states of same symmetry.

368

14 Identical Particles, the Pauli Principle

3. Confrontation Between Theory and Experiment If we look back at the states (2,5) and if we note their global symmetry as follows:

STATE |1s2s + , Σ |1s2s + , σ |1s2s − , Σ |1s2s − , σ |1s 2 , Σ |1s 2 , σ

Space sym sym antisym antisym sym sym

Spin sym antisym sym antisym sym antisym

GLOBAL S A A S S S

• the result 2.6 (a) says the transitions between singlet and triplet states are forbidden. If helium is a singlet, it stays so, it is parahelium. If it is a triplet it stays so, it is orthohelium. We recover the experimental result [1-2]. • The only allowed transitions between the above states are |1s2s + , Σ

↔

|1s 2 , Σ

and |1s2s + , σ

↔

|1s 2 , σ.

• The ground state of helium is a singlet (result [1-1]). We know (result [1-3] redundant with [1-1]) that this ground state does not have a magnetic moment. It is therefore the state |1s 2 , σ of parahelium. • The first para excited state must be |1s2s + , σ since one observes the transition from this state to the ground state. • The levels ortho |1s 2 , Σ and |1s2s + , Σ do not exist! Otherwise, one would see the transition |1s2s + , Σ → |1s 2 , Σ. The absence of that line in the spectrum of helium is what gave Pauli the idea of his exclusion principle! • The interpretation of Fig. 14.6 is therefore that the only states that appear are (increasing values of energy) |1s 2 , σ |1s2s − , Σ |1s2s + , σ

(ground state—para) (lowest energy ortho) (first excited state para)

which are all totally antisymmetric. The states that account for these experimental results are all totally antisymmetric. One generalizes that, after further investigations, by postulating that the only possible states of an electron system are totally antisymmetric.

14.6 Problem. Heisenberg Relations …

369

14.6 Problem. Heisenberg Relations for Fermions. The Way to Macroscopic Systems 14.6.1 Uncertainty Relations for N Fermions Consider N independent fermions of spin s, (N  1), each placed in a potential V (r ) centered at the origin. The Hamiltonian is therefore: Hˆ =

N 

hˆ (i)

pˆ 2 hˆ = + V (ˆr ). 2m

with

i=1

We note εn the energy levels of hˆ and gn their degeneracies. The ground state E 0 of Hˆ is obtained by filling the lowest lying levels εn up to the Fermi energy  F , each with (2s + 1) particles per state: E 0 = (2s + 1)

k 

gn εn ,

n=0

where the number k is determined by N = (2s + 1)

k 

gn .

n=0

In all this problem, we are interested in systems of a very large number of particles: N  1. Harmonic Interactions We first consider N independent fermions of spin s interacting with a harmonic potential centered at the origin, so that the total hamiltonian is H˜ N =

N N  pi2 mω 2  2 r . + 2m 2 i=1 i i=1

(14.23)

We have: εn = (n + 3/2)ω, gn = (n + 2)(n + 1)/2. a. Calculate k and the ground state energy E 0N of this system (for N  1). b. Consider an arbitrary state |Ψ  of these N fermions. In this state, we define the expectation values r 2  = ri2  and  p 2  =  pi2 , with i = 1, . . . , N . Using the

370

14 Identical Particles, the Pauli Principle

variational theorem, Ψ |H |Ψ  ≥ E 0 , ∀ω, show that r 2   p 2  ≥ ξ 2 N 2/3 2

for N  1,

(14.24)

where ξ is a parameter. Since the potential is centered at the origin, show that this leads to the uncertainty relation Δx Δpx ≥

ξ 1/3 N  3

for N  1.

(14.25)

Calculate ξ/3 for spin 1/2 particles. c. We now turn to pairwise interactions. Justify briefly the Lagrange identity, valid for N euclidean vectors { pi }, i = 1, N N

N  i=1

pi2

=

N 

( pi − p j ) + 2

i≤ j=1

 N 

2 pi

.

i=1

d. Using this identity, show that the previous hamiltonian of N independent spin 1/2 fermions can be rewritten as the sum N + Hˆ rNel , H˜ N = Hˆ cm

where

(14.26)

N N  P2 mω 2  2 N Hˆ cm = r i ) with P = pi + ( 2m N 2N i=1 i=1

is the center of mass hamiltonian, and Hˆ rNel =

N N 1  mω 2  2 ( p − pj) + (r i − r j )2 2m N i≤ j=1 i 2N i≤ j=1

is the translation invariant hamiltonian of the N pairwise interacting fermions. N and Hˆ rNel commute. – Justify that Hˆ cm – Calculate the ground state energy ErNel of Hˆ rNel in terms of E 0N . e. Setting the relative momentum of particle i and j as q i j = ( pi − p j )/2, q 2  = q i2j , and r 2  = (ri − r j )2 , and proceeding as before in Eqs. (14.24) and (14.25) prove the uncertainty relation Δx Δqx ≥

ξ  1/3 N  3

for N  1.

Calculate ξ  /3 for spin 1/2 particles. – Compare this inequality with (14.25) (for all values of N ).

(14.27)

14.6 Problem. Heisenberg Relations …

371

Gravitational Interactions We now turn to a gravitating system of N fermions of mass m and spin s with interactions centered at the origin: V (ri ) = −Gm 2 /(ri ) where G is Newton’s constant. The hamiltonian is Hˆ (N ) =

N 

pˆi 2 /2m +

i=1

N 

V (ri ).

i=1

The energy levels εn of the one-particle hamiltonian and their degeneracies gn are εn = −

m(Gm 2 )2 , gn = n 2 . 22 n 2

a. Calculate the ground state energy E 0 of Hˆ N , (N  1). b. Setting, as previously,  p2  =  pi2 , and (1/r ) = (1/ri ), use the above result to obtain a Heisenberg inequality of the form 2 1 N 2/3 p  ≥ γ  r 2

2

for

N  1.

(14.28)

and calculate the value of γ.

14.6.2 White Dwarfs and the Chandrasekhar Mass The sun burns quietly its hydrogen and transforms it into helium. In 5 billion years, the fuel will be exhausted and there will no longer exist a thermal pressure to balance the gravitational pressure. The system will implode until it reaches temperature and density conditions such that the fusion reactions of helium into carbon and oxygen occur: 3 4 He→ 12 C and 4 4 He→ 16 O. After this second phase, which is much shorter (∼108 years) than the previous one, a new effect appears which prevents further thermonuclear reactions to start, such as burning carbon and oxygen into 23 Na, 28 Si and 31 P, and these latter nuclei into 56 Fe, the most tightly bound nucleus. Actually, the density is so high that the quantum pressure of the degenerate electron gas can stop the gravitational collapse. The system then becomes a white dwarf star, whose only fate is to lose its heat by radiating. All stars do not end up as white dwarfs. For initial masses noticeably larger than M , the star can reach the nuclear stages 28 Si and 56 Fe where fusion can no longer take place. Beyond some critical value of the final mass of the star, called the Chandrasekhar mass, the pressure of the degenerate electron gas cannot compete any longer with gravitation, and this leads to a gravitational catastrophe, the explosion of a supernova and the formation of a neutron star in its center.

372

14 Identical Particles, the Pauli Principle

White dwarfs have masses of the order of the solar mass M , sizes comparable to the Earth radius (i.e. 0.01 R ∼ 104 km), and densities 106 g cm−3 (i.e. ∼ 106 larger than usual densities). Neutron stars are much more compact objects, with masses ∼ M and radii ∼ 10 km. The densities then reach 1015 g cm−3 . We wish to understand these orders of magnitude. White Dwarfs Consider N nuclei of mass Am p and charge Z , surrounded by N Z electrons. The complete calculation is somewhat involved5 but a simple and quite accurate estimate can be obtained because, to good approximation, the mass in the star has a constant density distribution inside a sphere of radius R. This holds for the N nuclei of mass A m p and charge Z (Z ∼ A/2 in our case), as well as for the distribution of the Z N electrons. The dominant terms in the total energy of the system are the following: • The potential energy E p is dominated by the gravitational attraction between these nuclei. For a spherical object of radius R and with a uniform spatial density, one has 3 G M2 , Ep = − 5 R where M = N A m p is the mass of the star. Note that the electrostatic interaction of the electrons is negligible since the local charge density inside the star is zero. Since the electron density is also uniform, one has, for the electrons, 2 3 1 1 so that E p = − G M 2 . (14.29)  = r 2R 5 r • The kinetic energy E k is that of the electrons, E k = p 2 /2m e . However, nuclear energies are far above the electron mass m e c2 ∼ 0.5 MeV. Therefore, the proper approach is to treat E k in its relativistic form6 : Ek =

NZ  

pi2 c2 + m 2e c4

⇒

 E k  ∼ N Z c  p 2  + m 2e c2 .

(14.30)

i=1

• Putting together (14.29) and (14.30) we obtain  2 2 1 + N Z c  p 2  + m 2e c2 , E = E p + Ek ∼ − G M 5 r

5 S.

(14.31)

Chandrasekhar, An introduction to the study of Stellar Structure, Dover, New York, 1967.  p 2 c2 + m 2 c4 in the Schrödinger equation, see J.L. Basdevant and S. Boukraa, Z. Phys. C28, 413 (1985); C30, 103 (1986), and further references therein.

6 Concerning the use and properties of the operator

14.6 Problem. Heisenberg Relations …

373

where we further make use of the Heisenberg inequality (14.28) in the form  p 2  ∼ 2

2 1 (N Z )2/3 r

for

(N Z )  1.

(14.32)

(We set the constant γ of (14.28) equal to 1. This simplifies calculations, and also has the advantage of slightly moving apart from the strict minimum of the inequality.) Of course, both terms are approximate. Inequality (14.28) has been   obtained in a non-relativistic approach, and  p 2 + m 2e c2   =  p 2  + m 2e c2 . However, it can be shown that for an ideal Fermi gas, the total energy derived with this approximation coincides to within 10 % with the exact result,7 in the two limits of non-relativistic and ultra-relativistic electron systems. As such this yields acceptable orders of magnitude (more complete procedures are available of course). a. Express the energy (14.31) as a function of the parameter  p 2 . b. Express the condition for the position of the minimum of the energy in the form  p2  = 2  p  + m 2e c2



M MCh

4/3

where M is the mass of the star, and MCh , called the Chandrasekhar mass, is expressed in terms of the dimensionless gravitation “structure” constant αG = Gm 2p /( c)  5.9 × 10−39 (where G = 6.7 × 10−11 m3 /kg/s2 is Newton’s constant), the proton mass m p = 1.67 × 10−27 kg, and Z and A. c. Calculate the numerical value of MCh . Compare with the solar mass m  = 2.0 × 1030 kg. d. Deduce that the Chandrasekhar mass MCh is an upper bound for the mass M of the star, beyond which the star collapses gravitationally. e. Give the expression of the equilibrium radius R. For M = M , compare with the solar radius R = 7 × 105 km and the earth’s radius R⊕ = 6.4 × 104 km.

14.6.3 Neutron stars At higher densities, it become energetically favorable for protons to capture electrons according to an inverse β process : p + e− → n + ν. The neutrinos escape from the star which forms a neutron star. Neutron stars were discovered in the middle 1960s as pulsars. They are gigantic nuclei, in the sense of nuclear physics. They are made of neutrons (electrically neutral) bound by the gravitational force, and packed up on one another not far from nuclear distances (∼10−15 m). The size of such objects is of the order of 10 km, 7 See

e.g. K. Huang, Statistical Mechanics, Chap. 11 (John Wiley, New-York, 1963).

374

14 Identical Particles, the Pauli Principle

their mass is of the order of a few solar masses M , and their density is of the order of 1014 –1015 g cm−3 . We only consider their main structure, and do not take into account the external proton and electron shells that stabilize them. Following a procedure very similar to the previous one, we shall assume that the dominant terms in the energy of a neutron star containing N neutrons are: • The potential energy E p dominated by the gravitational attraction (nuclear reactions are absent). Assuming for simplicity a uniform distribution of the N neutrons, we take as before: 3 G N 2 m 2n 2 1 ∼ − G N 2 m 2n , Ep = − 5 R 5 r • and the relativistic kinetic energy of the neutrons: Ek =

N  

pi2 c2 + m 2n c4

⇒

E k  ∼ N c

  p 2  + m 2n c2 .

i=1

Since neutrons are fermions, Eq. (14.32) which relates  p 2  and (1/r ) still holds. a. Express the total energy E p + E k in terms of  p 2 . b. Express the condition for the position of the minimum of the energy in the form  p2  = 2  p  + m 2n c2



N N1

4/3

and give the expression and the numerical value of N1 . Comment on the role of N1 . c. If the number of neutrons is below the critical number N1 , show that the radius R = (1/r ) of the star is given by:

R = R1

N1 N

1/3 

1−

N N1

4/3 1/2

d. Show that because of relativistic effects, the mass Ms of the star differs from the mass N m n of its constituents. Calculate the mass Ms using the total energy (Ms = (E p + E k )/c2 ). e. Calculate the maximum value of the mass Ms , the corresponding value of N and of the radius R.

14.6 Problem. Heisenberg Relations …

375

14.6.4 Mini-boson Stars To appreciate the power of the Pauli principle in generating “macroscopic” physical objects, it is interesting to consider the following (purely academic) problem. Consider a set of N bosons of equal mass m with pairwise gravitational interactions V (r i − r j ) and a hamiltonian Hˆ (N ) =

N  i=1

pˆi 2 /2m +

N 

V (r i − r j ).

i≤ j=1

a. Using the Lagrange identity, show that the N-body hamiltonian Hˆ (N ) can be written as ) Hˆ (N ) = Pˆ 2 /2N m + Hˆ r(N el , N where Pˆ = i=1 pˆ i is the total N-body momentum, and where the relative hamil(N ) ˆ tonian Hr el is a sum of two-body hamiltonians of the type Hˆ r2el (i, j) =

  4 (pi − p j /2)2 + V (r i − r j ) 2N m

where we write the kinetic energy in a form such that (pi − p j )/2 and (r i − r j ) have a canonical commutation relation. b. We place ourselves in the total c.m. system so that Pˆ 2 /2N m = 0. ) N Denoting by |Ω the normalized ground state of Hˆ r(N el , and E 0 the corresponding ground state energy. Show that

E 0N

≥

  N (N − 1) N (N − 1) 2 ˆ Ω| Hr el (i, j)|Ω ≥ E (2) (μ ), 2 2

(14.33)

where (i, j) are any two different indices, and E (2) (μ ) is the two-body groundstate energy corresponding to a mass mu  = m N /4. c. A boson star is, to first approximation, a system of N bosons of mass m which have pairwise gravitational interactions, i.e. V (r ) = −Gm 2 /r where G is Newton’s constant. Write an expression for the lower bound of E 0N = Ω| Hˆ (N ) |Ω in terms of the two-body energy E (2) (μ ) in the c.m.s. d. Deduce from that an inequality relating E 0N = Ω| Hˆ (N ) |Ω to the two-body energy E (2) (μ ) in the c.m.s (i.e. Pˆ 2 /2N m = 0).

376

14 Identical Particles, the Pauli Principle

e. Show that in the non-relativistic approximation this leads the Heisenberg inequality 2 1  p 2  ≥ 2 2 , (14.34) r where  p2  =  pi2  and 1/r  = 1/(|(r i − r j )|. f. We assume that in first approximation, the total energy of the mini-boson star is the sum of a potential term Ep = 

N 

V (r i − r j ) = −

i≤ j=1

N (N − 1) 1 Gm 2   2 r

and a relativistic kinetic term  E k = N c  p 2  + m 2 c2 . Using the inequality (14.34) express the total energy of the star in terms of  p 2 . g. Express the condition for the position of the minimum of the energy in the form  p2  = 2  p  + m 2 c2



N N1

2

and give the expression of N1 . h. What is the limiting upper value of N for which the star becomes gravitationally unstable and collapses in a mini-boson black hole? Calculate its value for m = m n . i. If the number of bosons is below the critical number N1 , show that the radius R = (1/r ) of the star is given by:

R = R1

N1 N

 

1−

N N1

2 1/2

j. Show that because of relativistic effects, the mass Ms of the star differs from the mass N m n of its constituents. Calculate the mass Ms using the total energy (Ms = (E p + E k )/c2 ). k. Calculate the maximum value of the mass Ms , the corresponding value of N and of the radius R. l. Calculate Ms and N1 in the academic case m = m n , and in the hypothetical case of a sneutrino of larger mass.

14.6 Problem. Heisenberg Relations …

377

14.6.5 Solution Uncertainty Relations for N Fermions Harmonic Interactions a. We find for N  1:

k

6N 2s + 1

1/3 ,

E 0  ξ N 4/3 ω,

with ξ = (3/4)61/3 (2s + 1)−1/3 . b. In the state |Ψ , H  ≥ E 0 , therefore: H  = N

 p2  N + mω 2 r 2  ≥ ξ N 4/3 ω, 2m 2

which gives:  p 2  + m 2 ω 2 r 2  − 2ξ N 1/3 mω ≥ 0

for N  1.

This second degree trinomial in mω is positive for all values of mω. Therefore we obtain that in any state |Ψ  of these N fermions: r 2   p 2  ≥ ξ 2 N 2/3 2

for E 0N

(14.35)

This relation is valid in particular in the center-of-mass frame, where  p = 0. If we choose the origin of space at the center of the cloud, we obtain: Δx Δpx ≥

ξ 1/3 N  3

for N  1.

(14.36)

For spin 1/2 particles, ξ/3 ∼ 0.36. c. The identity is simple: the diagonal terms pi2 have the same coefficient N , and the crossed terms pi · p j cancel. d. The decomposition follows from the Lagrange identity. N ˆN and H since [ P, (r i − r j )] = 0 and vice versa. – Hˆ cm √r el commute  √ – Since P/ N and r i / N have canonical commutation relations, the grounds N N is E cm = (3/2)ω. state energy of Hˆ cm N we obtain Therefore, since the ground state energy of H˜ N is E 0N = ErNel + E cm N ErNel = E 0N − E cm  ξ N 4/3 ω − (3/2)ω ∼ E 0N for N  1.

378

14 Identical Particles, the Pauli Principle

e. The definitions are such that q i j = ( pi − p j )/2 and (ri − r j ) have canonical commutation relations. Therefore, we can proceed as previously and obtain: Δx Δqx ≥

ξ  1/3 N  3

for N  1.

(14.37)

Where ξ and ξ  coincide for large N , since the term (3/2)ω is negligible. The two inequalities are equivalent for large N . The Pauli principle modifies profoundly the uncertainty relations. Gravitational Interactions The energy levels εn of the one-particle hamiltonian and their degeneracies gn are εn = −

m(Gm 2 )2 , gn = n 2 . 22 n 2

a. We therefore obtain that there exists an integer k such that: N  (2s + 1)

k3 3

E 0  −(2s + 1)

m(Gm 2 )2 k. 22 n 2

Therefore E 0 and N are related by: E0  −

m(Gm 2 )2 (2s + 1)2/3 (3N )1/3 22

b. Consider an arbitrary state |Ψ  of this N fermion system, we therefore obtain 2 1 p  ≥ γ  N 2/3 r 2

2

for

N 1

(14.38)

with γ = 3−1/3 (2s + 1)−2/3 . Again, the Pauli principle modifies profoundly the uncertainty relations when fermions are involved. White Dwarfs and the Chandrasekhar Mass The potential energy E p is dominated by the gravitational attraction between the nuclei, we assume a spherical object of radius R with a uniform density, and Ep = −

3 G M2 , 5 R

where M = N A m p is the mass of the star.

14.6 Problem. Heisenberg Relations …

We have

1 3  = r 2R

379

so that E p = −

2 1 G M2 . 5 r

(14.39)

We consider the kinetic energy E k in its relativistic form Ek =

NZ  

pi2 c2 + m 2e c4

⇒

 E k  ∼ N Z c  p 2  + m 2e c2 .

(14.40)

i=1

Putting together (14.29) and (14.40) we obtain E = E p + Ek ∼ −

 2 1 G M2 + N Z c  p 2  + m 2e c2 , 5 r

(14.41)

We use a simplified Heisenberg–Pauli inequality  p 2  ∼ 2

2 1 (N Z )2/3 r

for

(N Z )  1.

(14.42)

a. The energy of the star, as a function of  p2  is therefore E =−

  2 G(N A)2 m 2p c(N Z )−1/3  p 2  + N Z c  p 2  + m 2e c2 5 c

b. The radius R of the star and the average squared momentum  p 2  are now obtained by minimizing the total energy E p + E k with respect to  p 2 . This leads to:  p2  =  p 2  + m 2e c2



M MCh

4/3 ,

(14.43)

where we have introduced the Chandrasekhar mass MCh : MCh ∼

4 3/2

αG

Z2 m p. A2

(14.44)

Where αG = (Gm 2p )/(c) = 5.9 × 10−39 is dimensionless. c. The numerical value of the Chandrasekhar mass is, in these approximations, MCh ∼ 3.7 × 1030 kg ∼ 1.8 M . The exact value, which requires solving a non-linear differential equation, is ∼1.4 M . d. We notice that the Eq. (14.43) has a solution only if the mass M is smaller than the Chandrasekhar mass. For masses larger than the Chandrasekhar mass, the Fermi

380

14 Identical Particles, the Pauli Principle

pressure of the electron gas cannot compete with the gravitational pressure: the system is unstable and it undergoes a gravitational collapse. e. For a mass lower than the Chandrasekhar mass, the equilibrium radius is given in our model by:

R = RCh

MCh M

1/3 

1−

M MCh

4/3 1/2 ,

(14.45)

where we have put: RCh ∼ 2.5

 Z 1 √ A αG m e c

(∼6300 km for A = 2Z ).

This result is in good agreement with predictions obtained from more elaborate treatments. The predicted value for MCh is then 1.4 M . For a relatively low mass (M MCh ), we obtain the scaling law R ∝ N −1/3 of the non relativistic treatment:

 M −1/3 R ∼ 7800 km × for A = 2Z . (14.46) M The star of van Maanen, was one of the first white dwarfs to be discovered. Its radius is ∼8900 km (78 times smaller than the sun radius) and its mass is 0.68 M , in good agreement with (14.46). The equilibrium radius (14.45) is a decreasing function of the mass of the white dwarf, and, owing to relativistic effects, it shrinks to zero when the mass approaches the Chandrasekhar mass. Therefore, the most massive white dwarfs correspond to ultra relativistic electrons and they all have the same mass MCh . For M = M , compare with the solar radius R = 7 × 105 km and the earth’s radius R⊕ = 6.4 × 104 km. Neutron Stars At higher densities, it become energetically favorable for protons to capture electrons according to an inverse β process : p + e− → n + ν. The neutrinos escape from the star which forms a neutron star. The neutron stars were discovered in the middle 1960s as pulsars. They are gigantic nuclei, in the sense of nuclear physics. They are made of neutrons (electrically neutral) bound by the gravitational force, and packed up on one another at nuclear distances ∼10−15 m. The size of such objects is of the order of 10 km, their mass is of the order of a few solar masses M , and their density is of the order of 1014 –1015 g cm−3 . The dominant terms in the energy of a neutron star containing N neutrons are: • The potential energy E p dominated by the gravitational attraction. Assuming for simplicity a uniform distribution of the N neutrons, we take as before:

14.6 Problem. Heisenberg Relations …

381

Ep =

3 G N 2 m 2n . 5 R

• The relativistic kinetic energy of the neutrons: Ek =

N  

pi2 c2 + m 2n c4

⇒

E k  ∼ N c

  p 2  + m 2n c2 .

i=1

a. Express the total energy E p + E k in terms of  p 2 . The total energy is E p + Ek = −

  2 G N 2 m 2n c(N )−1/3  p 2  + N c  p 2  + m 2n c2 5 c

b. Taking  p 2  as the variable, the minimum energy is obtained for:  p2  =  p 2  + m 2n c2



N N1

4/3 with N1 ∼

4 3/2

αG

∼ 9 × 1057 .

As for the case of a white dwarf star, this equation has a solution only if the number of neutron is below a critical number N1 . c. In this case, the radius of the star is given by:

R = R1

N1 N

with: R1 ∼ 2.5

1/3 

1−

N N1

4/3 1/2

 1 ∼ 7 km . √ m n c αG

d. Because the neutrons have an average velocity close to c, the mass Ms of the star differs from the mass N m n of its constituents. The mass Ms is obtained using the total energy (E p + E k = Ms c2 ), which yields:  Ms = N m n 1 −



N N1

4/3 1/2 .

e. The mass is maximal when N ∼ 0.7 N1 and it takes the value: Msmax ∼ 6.5 × 1030 kg ∼ 3 M ,

(14.47)

corresponding to a radius ∼5 km. Beyond the critical number N1 , a gravitational catastrophe happens: the system can lose energy by radiating neutrinos, it can fall into this “catastrophic” ground

382

14 Identical Particles, the Pauli Principle

state where a new kind of physics applies, and it becomes a black hole. One can refine this extremely simple model, in particular by taking into account the inhomogeneous density profile in the neutron star. The theory, which is due to Landau, Oppenheimer and Volkov, leads to a critical mass very similar to (14.47). Again, the collapse comes from a simple property of degenerate Fermi gases. In the non-relativistic regime, the pressure P of a Fermi gas is related to the density ρ by P ∝ ρ5/3 . This can always balance the gravitational inward pressure which is Pgrav ∝ ρ4/3 , provided the density is large enough. If the conditions are such that the Fermi gas is relativistic, then its pressure is P ∝ ρ4/3 , and there is a value of the mass such that the gravitational pressure prevails and the system collapses. Mini-boson Stars To appreciate the power of the Pauli principle in generating “macroscopic” physical objects, it is interesting to consider the following problem which is yet purely academic. Consider a set of N bosons of equal mass m with pairwise gravitational interactions V (r i − r j ) and a hamiltonian Hˆ (N ) =

N 

pˆi 2 /2m +

i=1

N 

V (r i − r j ).

i≤ j=1

a. The expression follows immediately from the Lagrange identity: ) Hˆ (N ) = Pˆ 2 /2N m + Hˆ r(N el ,

N ˆ i is the total N-body momentum, and where the relative where Pˆ = i=1 p ) is a sum of two-body hamiltonians of the type hamiltonian Hˆ r(N el ) Hˆ r(N el =

N  i≤ j=1

Hˆ r2el (i, j) =

  N  4 (pi − p j /2)2 + V (r i − r j ) 2N m i≤ j=1

where we write the kinetic energy in a form such that (pi − p j )/2 and (r i − r j ) have a canonical commutation relation. b. We place ourselves in the total c.m. system so that Pˆ 2 /2N m = 0. By the variational principle, we know that if E (2) (μ = n M/4) is the two-body ground state energy, then the expectation value   4 (pi − p j /2)2  + V (r i − r j ) ≥ E (2) (μ = N m/4) 2N m holds for any state. Therefore, since all two-body sub-systems are the same, it follows immediately that

14.6 Problem. Heisenberg Relations …

383



  N (N − 1) N (N − 1) Ω| Hˆ r2el (i, j)|Ω ≥ E (2) (μ = n M/4), 2 2 (14.48) c. From the above expression, we have E 0N =

E (2) (μ ) = − and thus for N  1, E 0N ≥ −

1 N mc (Gm 2 )2 2 4 2 c 2

N 3 2 (Gm 2 )2 . mc 16 (c)2

d. Therefore, since  N 

2

pˆi /2m +

i=1

N 

 V (r i − r j ) ≥ E 0N

i≤ j=1

which amounts to N2 N 3 mc2 (Gm 2 )2 1 N  p2  . − Gm 2 ≥− 2m 2 r 16 (c)2 e. Since this holds whatever G, we obtain the inequality:  p 2  ≥ 22

2 1 r

(14.49)

where  p 2  =  pi2  and 1/r  = 1/(|(r i − r j )| . f. Assuming that in first approximation, the total energy of the mini-boson star is the sum of a potential term Ep = 

N 

V (r i − r j ) = −

i≤ j=1

N (N − 1) 1 Gm 2   2 r

and a relativistic kinetic term  E k = N c  p 2  + m 2 c2 . Using the inequality (14.49) the total energy of the star can be expressed as a function of  p 2  as 

 N (Gm 2 )  2 2 2 2  p  + m c − 3/2  p  /c2 E = Nc 2 (c)

384

14 Identical Particles, the Pauli Principle

g. The minimum of the energy occurs for  p2  =  p 2  + m 2 c2



N N1

2 with N1 ∼

c 23/2 Gm 2

.

h. N1 is the limiting upper value of N for which the star becomes gravitationally unstable and collapses in a mini-boson black hole? For m = m n , we obtain N1 = 6 × 1037 which is very small compared to the fermionic neutron case (for 3/2 instance we have N1 m n ∼ 1011 kg). This is due to the effect of the term αG instead of αG . The power 3/2 is characteristic of fermion systems. It enhances enormously the number of constituents, the sizes and the masses of fermion systems as compared to what they would be in “similar” boson systems. i. If the number of bosons is below N1 , the radius R = of the star is given by:

R = R1

N1 N

 

1−

N N1

2 1/2

 j. The mass Ms = (E p + E k )/c2 is Ms = N m 1 − (N /N1 )2 and the radius √  2 N1  R= 1 − (N /N1 )2 . mc N √ k. The maximum value of Ms occurs for (N /N1 )2 = 1/2 : M max = N m/ 2 and R = 2/(mc). l. In the academic case m = m n , N1 = 6 × 1037 , Ms ∼ 7 × 1010 kg, and R ∼ 0.6 fm (a nuclear size!). The hypothetical case of a sneutrino of larger mass can be calculated easily. Such a mini boson-star or black hole (for N ≥ N1 ) would indeed be a terribly small (and comparatively massive) stellar object. Notice that the formation of such an object would imply some mechanism of energy loss by radiation (of photons, neutrinos, gravitons etc.).

Chapter 15

Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

The Lorentz force q v × B acting on a particle of charge q moving with a velocity v in a magnetic field B, does not derive from a potential. Therefore, the formulation of quantum mechanics we have used up to now does not apply. The aim of this Chapter is to generalize the correspondence principle, in order to obtain the form of the Hamiltonian in such a problem. Naturally, it would suffice to simply introduce this Hamiltonian with no other justification than the fact that it accounts for observed phenomena. However, it is instructive to first develop some considerations on classical mechanics. In fact, the development of analytic mechanics in the 18th and 19th centuries, which is due to the works of d’Alembert, Bernoulli, Euler, Lagrange, Hamilton, etc., had brought out an amazing geometric structure of the theory, based on a variational principle, the principle of least action. It was one of the first remarkable discoveries of Dirac in 1925 and 1926, to understand that the same basic structure underlies quantum mechanics. Starting from this observation, the correspondence principle can be stated in a much more profound manner, which enables one to treat complex problems that would be difficult to attack without this analysis. In Sect. 15.1 we shall recall the basic elements of the Lagrangian formulation of mechanics, based on the principle of least action. In Sect. 15.2, we shall present the “canonical” formulation of Hamilton and Jacobi, which will allow us to exhibit, in Sect. 15.3, the parallelism between classical and quantum mechanics. The word Hamiltonian which we have used so often will then take its full significance. In Sect. 15.4, we shall give the Lagrangian and Hamiltonian formulations of the problem of interest, i.e. the motion of a charged particle in a magnetic field. Finally, in Sect. 15.5, we shall transpose the result to quantum mechanics and we will also take into account the possibility that the particle has spin.

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_15

385

386

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Fig. 15.1 Examples of trajectories starting from x1 at time t1 and arriving at x2 at time t2 . Among all these possible trajectories, the trajectory actually followed by the particle is the one for which the action S is extremal

15.1 Lagrangian Formalism and the Least Action Principle In his Méchanique analytique published in 1787, one century after Newton’s Principia, Lagrange proposed a new way to consider mechanical problems. Instead of determining the position r(t) and the velocity v(t) of a particle at time t knowing its initial state {r(0), v(0)}, Lagrange asks the following equivalent but different question. What is the trajectory actually followed by the particle if, leaving r1 at time t1 , it reaches r2 at t2 ? Least Action Principle In order to simplify the discussion, consider first a one-dimensional problem. Among the infinite number of possible trajectories (see Fig. 15.1) such that: x(t1 ) = x1

x(t2 ) = x2 ,

(15.1)

what is the law that determines the good one? Lagrange made use of the “principle of natural economy”,1 which is an expression due to Fermat, that was adopted by Maupertuis and Leibniz (who called it the principle of “the best”). Lagrange’s prescription is the following:

a. Any mechanical system is characterized by a Lagrange function, or Lagrangian L(x, x˙ , t),which depends on the coordinate x,on its derivative with respect to time x˙ = dx/dt,and possibly on time. The quantities x and x˙ are called state variables. For instance, considering a particle in a one-dimensional potential, one has: L=

1 2 m˙x − V (x, t). 2

(15.2)

b. For any trajectory x(t) satisfying (15.1), one defines the action S :

1 In

full rigor, the variational principle as we present it here was formulated by Hamilton in 1828. In order to simplify the presentation, we have shrunk parts of History.

15.1 Lagrangian Formalism and the Least Action Principle



t2

S=

387

L(x, x˙ , t) dt.

(15.3)

t1

The principle of least action states that the physical trajectory X(t) is such that S is minimal, or, more generally, extremal. Lagrange Equations Let X(t) be the physical trajectory. Consider a trajectory x(t) infinitely close to X(t), and also starting from x1 at time t1 , and arriving at x2 at time t2 : ˙ + δ x˙ (t), δ x˙ (t) = d δx(t), x(t) = X(t) + δx(t), x˙ (t) = X(t) dt

(15.4)

with, by assumption: δx(t1 ) = δx(t2 ) = 0.

(15.5)

To first order in δx, the variation of S is: 

t2

δS =



t1



∂L ∂L δx(t) + δ x˙ (t) ∂x ∂ x˙

dt.

Integrating the second term by parts and taking into account (15.5), we obtain:  δS = t1

t2



d ∂L − ∂x dt



∂L ∂ x˙

 δx(t) dt.

(15.6)

It follows from the principle of least action that δS must vanish whatever the infinitesimal function δx(t). Therefore, the equation which determines the physical trajectory is the Lagrange equation:   d ∂L ∂L = . (15.7) ∂x dt ∂ x˙ We can check readily that we recover the usual equation of motion m¨x = −dV / dx for a point particle placed in the potential V (x, t) if we consider the Lagrangian (15.2). The generalization to s degrees of freedom: {xi , x˙ i }, i = 1, . . . , s (with, for instance, s = 3N for N particles in a three-dimensional space) is straightforward. One uses a Lagrangian L({xi }, {˙xi }, t) and obtains the set of Lagrange equations: d ∂L = ∂xi dt



∂L ∂ x˙ i

 i = 1, . . . , s.

(15.8)

388

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Remarks (1) The Lagrange equations keep the same form in any coordinate system. This is particularly useful in order to make changes of variables, e.g. to get from cartesian coordinates (x, y, z) to spherical coordinates (r, θ, ϕ). The xi are called generalized coordinates. (2) It is remarkable that the laws of mechanics can be derived from a variational principle, which states that the physical trajectory minimizes a certain quantity, here the action. Almost all physical laws can be formulated in terms of variational principles like the Fermat principle in geometrical optics. The situations which are physically encountered appear to result as “optimizing” the effects of various “conflicting” contributions. (3) In the absence of forces (i.e. a uniform potential in (15.2)), S is minimum for x˙ = constant, corresponding to a linear uniform motion. The presence of the potential can be visualized as a property of space which bends the trajectories. Forces and inertia appear to be in conflict. The particle follows a trajectory of minimal “length”, this length being measured by the action S. Therefore one can understand that a mechanical problem is reduced to a geometric problem. The motion of a particle in a field of force deriving from a potential in a flat Euclidean space, can be transformed into the free motion of a particle in a curved space, where it follows geodesics. Einstein had this idea in mind in 1908 when he started to construct General Relativity. It took him seven years to elaborate the mathematical structure of the theory. The elaboration of the fundamental concepts and principles of mechanics was performed during the 17th century. Copernicus had given the notion of reference frames. Galileo had understood the principle of inertia: uniform linear motion is a state relative to the observer, and not a process. It is the modification of the velocity which constitutes a process. The final lines were written by Newton. After the Newtonian synthesis and the publication in 1687 of the Philosophiae Naturalis Principia Matematica, the 18th and the 19th centuries were marked by a fascinating endeavor. Through the impetus of d’Alembert, Maupertuis, the Bernoulli brothers (in particular Daniel), of Euler and of Lagrange, the basic structure of mechanics, i.e. a geometric structure, was discovered. A large class of problems could be reduced to problems of pure geometry. D’Alembert, who was the first to understand the importance of the abstract concepts of mass and of momentum, attacked the concept of force introduced by Newton. For d’Alembert, motion is the only observable phenomenon, whereas the “causality of motion” remains an abstraction. Hence the idea to study not a particular trajectory of the theory, but the set of all motions that it predicts (to characterize a force by the set of all its effects is actually a very modern point of view.) In 1787, one century after the Principia, Lagrange published his Méchanique analytique, and gave a new formulation of mechanics where the geometric and global structure of the theory is emphasized. The first formulation of a physical law in terms of the least action principle originated from a dispute between Fermat and Descartes, around 1640, about the notion of proof (Descartes’s “proof” of Snell’s laws for refraction was actually wrong). Fermat, who was a mathematician, and who knew little, if any, physics, got interested in the laws of geometrical optics, in particular the equality between the angles of incidence and of reflection. He proved the results as being a geometrical property of the optical length of the light rays (in the case of reflection, this had been understood by Heron of Alexandria in 100 a.d.). The Snell–Descartes laws predicted which path would be followed by a light ray with given initial properties. In the more general point of view of Fermat, one determines the path effectively followed by a light ray which goes from A to B.

15.1 Lagrangian Formalism and the Least Action Principle

389

At the end of his life, in 1661, Fermat stated his “principle of natural economy” which led to numerous fields of research, still very active nowadays, on variational principles.

Energy We assume the system is isolated, i.e. ∂L/∂t = 0, and we calculate the time evolution of the quantity L(x, x˙ ) along the physical trajectory x(t): ∂L ∂L d dL (x, x˙ ) = x˙ (t) + x¨ (t) = dt ∂x ∂ x˙ dt



 ∂L x˙ (t) , ∂ x˙

where we have transformed the first term using the Lagrange equation (15.7). Therefore:   d ∂L x˙ (t) − L = 0. dt ∂ x˙ For an isolated system, the quantity: ∂L −L, E = x˙ (t) ∂ x˙

 resp. E =

s  i=1

 ∂L x˙ i (t) −L ∂ x˙ i

(15.9)

is conserved. It is a constant of the motion, called the energy of the system. In the simple case (15.2) we recover E = m˙x 2 /2 + V (x).

15.2 Canonical Formalism of Hamilton and Jacobi

Conjugate Momenta The quantity: p=

∂L ∂ x˙

 resp. pi =

 ∂L , ∂ x˙ i

(15.10)

which appears in the definition of the energy (15.9), is called the conjugate momentum, or generalized momentum, of the variable x. In the simple case (15.2), it reduces to the linear momentum p = m˙x , but this is no longer true in non-Cartesian coordinates or, as we shall see, when the forces are velocity dependent. We notice that Eq. (15.7) implies:   ∂L ∂L , resp. p˙ i = . (15.11) p˙ = ∂x ∂xi

390

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Canonical Equations The description of the state of the particle (or the system) by x and the conjugate momentum p, instead of x and the velocity x˙ , has some advantages. We assume that we can invert Eq. (15.10) and calculate x˙ in terms of the new state variables x and p. The equations of motion are obtained by performing what is called a Legendre transformation. Let us introduce the Hamilton function, or Hamiltonian:    pi x˙ i − L . (15.12) H(x, p, t) = p˙x − L resp. H(xi , pi , t) = i

We write the total differential of H: dH = p d x˙ + x˙ dp −

∂L ∂L ∂L dx − d x˙ − dt. ∂x ∂ x˙ ∂t

If we take into account (15.10) and (15.11), the first and fourth term cancel, and the third term is nothing but −˙p dx, therefore: dH = x˙ dp − p˙ dx −

∂L dt. ∂t

(15.13)

This gives the equations of motion: ∂H ∂H , p˙ = − x˙ = ∂p ∂x



∂H ∂H , p˙ i = − resp. x˙ i = ∂pi ∂xi

 (15.14)

which are called the canonical equations of Hamilton and Jacobi. They are first order differential equations in time, and they are symmetric in x and p (up to a minus sign). They have the big technical advantage to present the time evolution of the state variables directly in terms of theses state variables. More generally, if we note X = (r, p) the coordinates of the system in phase space, these equations have the form X˙ = F(X). Such a problem, called a dynamical system, is of considerable interest in many fields, including mathematics. Notice that in quantum mechanics, the Ehrenfest theorem derived in Chap. 7 gives the time evolution of expectation values as:  ∂ Hˆ d xi  = dt ∂ pˆ i

 ∂ Hˆ d pi  = − , dt ∂ xˆ i

where we can see some similarity with the canonical Hamilton–Jacobi equations.

15.2 Canonical Formalism of Hamilton and Jacobi

391

Poisson Brackets Consider two functions f and g of the state variables x, p and possibly of time, for instance two physical quantities. One defines the Poisson bracket of f and g as the quantity: 

∂f ∂g ∂f ∂g − {f , g} = ∂x ∂p ∂p ∂x

s  ∂f ∂g ∂f ∂g − {f , g} = ∂xi ∂pi ∂pi ∂xi i=1

 .

(15.15)

We find immediately: {f , g} = −{g, f }

{x, p} = 1,

(15.16)

and more generally: {xi , xj } = 0

{pi , pj } = 0

and: {x, f } =

∂f ∂p

{xi , pj } = δij ,

{p, f } = −

∂f . ∂x

(15.17)

(15.18)

Let us now calculate the time evolution of a quantity f (x, x˙ , t): df ∂f ∂f ∂f f˙ = = x˙ + p˙ + . dt ∂x ∂p ∂t

(15.19)

Using Hamilton’s equations (15.14), we obtain: ∂f f˙ = {f , H} + . ∂t

(15.20)

In particular, the canonical equations (15.14) are written in the completely symmetric way: x˙ = {x, H} p˙ = {p, H}. (15.21) In the canonical formalism, the Hamiltonian governs the time evolution of the system. If a physical quantity f does not depend explicitly on time, i.e. ∂f /∂t = 0, then its time evolution is obtained through the Poisson bracket of f and the Hamiltonian: f˙ = {f , H}. If this Poisson bracket vanishes, f is a constant of the motion.

15.3 Analytical Mechanics and Quantum Mechanics The results derived in the previous section reveal an amazing property. There is a strong analogy in the structures of analytical mechanics and quantum mechanics. Let ˆ the observable A˙ˆ such that by definition: us associate to any quantum observable A,

392

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

˙ˆ ≡ d a  A dt for any state of the system. The Ehrenfest theorem then implies: ∂ Aˆ 1 ˆ ˆ H] + A˙ˆ = [A, i ∂t

(15.22)

to be compared with (15.20). Similarly, the canonical commutation relations: [ˆxj , pˆ k ] = iδjk

(15.23)

are highly reminiscent of Eq. (15.17). This identity in the structures of the two kinds of mechanics was one of the first great discoveries of Dirac. Of course, the mathematical nature and the physical interpretation of the objects under consideration are different. But the equations which relate them are the same provided we make the following correspondence, which was understood by Dirac during the summer of 1925: Quantization rule. One replaces the Poisson brackets of analytical mechanics by the commutators of the corresponding observables, divided by i: Analytical mechanics {f , g} −→

1 ˆ [f , g] ˆ Quantum mechanics i

(15.24)

This is the genuine form of the correspondence principle. In general, for complex systems (large number of degrees of freedom, constraints, etc.), the systematic method to obtain the form and the commutation relations of observables consists in referring to the Poisson brackets of the corresponding classical systems. We will see an example below when we treat the Lorentz force in quantum mechanics. One can now understand why the name of Hamilton (1805–1865) appears so often in quantum mechanics, although Hamilton lived one century before its invention. Hamilton is one of the great geniuses of science. He made decisive contributions to analytical mechanics, and he invented vector analysis; he also invented the same year as Cayley and Grassmann (1843), noncommutative algebras and matrix calculus (the elements of Hamilton’s quaternions are called ... Pauli matrices in quantum mechanics). He is the author of the synthesis of the geometrical and wave theories of light. He proved in what limit the first is an approximation of the second. Hamilton was fascinated by variational principles, in particular by the similarity between Maupertuis’s principle in mechanics and Fermat’s principle in optics. In 1830 he made the remarkable statement that the two formalisms of optics and of mechanics were basically the same and that Newtonian mechanics corresponds to the same limit as geometrical optics, which is only an approximation. This remark was ignored by his contemporaries, and the mathematician Felix Klein said in 1891 that it was a pity. It is true that, in 1830, no experimental fact could reveal the existence of Planck’s constant. However, in many ways, Hamilton can be considered as a precursor of quantum mechanics. Louis de Broglie refers to Hamilton’s work in his thesis.

15.4 Classical Charged Particle in an Electromagnetic Field

393

15.4 Classical Charged Particle in an Electromagnetic Field Classically a particle of charge q, placed in an electromagnetic field, is subject to the Lorentz force: f = q (E + v × B) . This force is velocity dependent and it does not derive from a potential. Furthermore, the magnetic force qv × B does no produce any work and the energy of the particle is E = mv 2 /2 + qΦ, where Φ is the scalar potential associated with the electric field E. The Hamiltonian is certainly different from p2 /2m + qΦ. Otherwise the equations of motion would be strictly the same as in the absence of a magnetic field. We can use the previous developments in order to determine the correct form of the Hamiltonian H. Maxwell’s equations, specifically the couple of equations: ∇·B=0,

∇×E =−

∂B , ∂t

(15.25)

allow to express the fields E and B in terms of the scalar and vector potentials Φ and A: ∂A B=∇×A, E = −∇Φ − . (15.26) ∂t Consider a particle of mass m and charge q placed in this electromagnetic field. We note r and r˙ = v the position and the velocity of this particle. A possible Lagrangian for this particle can be written in terms of the potentials A and Φ: L=

1 2 m˙r + q r˙ · A(r, t) − q Φ(r, t). 2

(15.27)

Indeed, starting from the Lagrange equations and using: ∂A ∂A ∂A ∂A d A(r, t) = + x˙ + y˙ + z˙ , dt ∂t ∂x ∂y ∂z we can check that we obtain the desired equation of motion: m

dv = q(E + v × B). dt

Consider now the conjugate momentum p. From the definition (15.10) we have: p = m˙r + qA(r, t).

(15.28)

394

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

In other words the conjugate momentum p no longer coincides with the linear momentum, i.e. the product of the mass times the velocity m˙r! Equation (15.28) is easily inverted: r˙ = (p − qA(r, t))/m, which gives the Hamiltonian we are looking for: H=

1 (p − qA(r, t))2 + qΦ(r, t). 2m

(15.29)

As the Lagrangian, it is expressed in terms of the potentials A and Φ, and not the fields E and B.

15.5 Lorentz Force in Quantum Mechanics 15.5.1 Hamiltonian We now follow Dirac’s quantization rules of Sect. 15.3. The Hamiltonian of a charged particle in an electromagnetic field is: 1 (ˆp − qA(ˆr, t))2 + qΦ(ˆr, t) Hˆ = 2m

(15.30)

where the position and the conjugate momentum operators rˆ and pˆ satisfy the canonical commutation relations: [ˆxj , xˆ k ] = 0

[ˆpj , pˆ k ] = 0

[ˆxj , pˆ k ] = i δjk .

In the wave function formalism, we can still choose pˆ = −i∇. The velocity observable is no longer pˆ /m, but: 1 (15.31) vˆ = (ˆp − qA(ˆr, t)). m Note that two components of the velocity (e.g. vˆx and vˆy ) do not, in general, commute in presence of a magnetic field. Using the Ehrenfest theorem, one can verify that (15.30) provides the appropriate structure of the equations of motion for the expectation values.

15.5.2 Gauge Invariance One thing, though, seems surprising. The potentials Φ and A are not unique. Two sets (Φ, A) and (Φ  , A ) related to each other by a gauge transformation:

15.5 Lorentz Force in Quantum Mechanics

A = A + ∇χ(r, t)

395

Φ = Φ −

∂χ , ∂t

(15.32)

where χ(r, t) is an arbitrary function, correspond to the same electric and magnetic fields E and B. Since the energy observable Hˆ is expressed in terms of A and Φ, the energy seems to depend on the gauge. However, we know that physical results should not depend on the gauge! The answer to this problem is simple and remarkable. In a gauge transformation, the wave function also changes: ψ(r, t) → ψ  (r, t) = eiqχ(r,t)/ ψ(r, t).

(15.33)

One can check that if ψ is a solution of the Schrödinger equation for the choice of potentials (A, Φ), then ψ  is a solution for the choice (A , Φ  ). For time independent problems, this guarantees that the energy spectrum of the Hamiltonian for the choice (A, Φ) coincides with the one obtained with (A , Φ  ). The transformation (15.33) does not modify the probability density: |ψ(r, t)|2 = |ψ  (r, t)|2 , which is of course crucial. It simply affects the phase of the wave function by an amount which depends on the point in space. One can verify more generally that the expectation values of all measurable quantities

 are gauge invariant. Consider, for instance, the velocity operator vˆ = pˆ − qAˆ /m. We find:

   pˆ − qAˆ ψ  = −i∇ − qAˆ − q∇ χˆ eiqχ/ ψ



 = eiqχ/ −i∇ − qAˆ ψ = eiqχ/ pˆ − qAˆ ψ,



from which we deduce:



  ψ ∗ pˆ − qAˆ ψ  = ψ ∗ pˆ − qAˆ ψ. This proves that the probability current is the same in both gauges. If we integrate this relation over space, we find that the expectation value of the velocity is also gauge independent. On the contrary, the momentum pˆ is not a gauge invariant physical quantity. If one postulates that the laws of physics are invariant under all gauge transformations (15.33) where χ(r, t) is arbitrary, one can derive that the Hamiltonian has the structure (15.30). In quantum field theory, gauge invariance plays a crucial role in the physics of fundamental interactions and elementary constituents of matter.

The fact that the Hamiltonian (15.30) depends on the potentials and not on the fields can be verified experimentally following a suggestion of Aharonov and Bohm

396

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Fig. 15.2 The Bohm–Aharonov effect: in a Young slit experiment performed with charged particles, the interference pattern is shifted when a current runs in the solenoid. However, the magnetic field created by this solenoid is zero everywhere except inside the solenoid itself. The corresponding phase shift has no classical counterpart (in terms of a force acting on the particles for instance) and one refers to it as a topological phase shift

in 1956. In a Young slit interference device, one places between the two slits a solenoid of small diameter, parallel to the slits (Fig. 15.2). When a current flows in the solenoid, one can observe a modification of the system of fringes. However, the magnetic field is zero everywhere outside the solenoid, in particular near the slits. Conversely, the vector potential is non zero outside the solenoid. This experiment has been performed and it has confirmed the quantum mechanical predictions.2

15.5.3 The Hydrogen Atom Without Spin in a Uniform Magnetic Field We place a hydrogen atom in a constant uniform field B, which derives from the vector potential A = B × r/2, and we neglect spin effects for the moment. The Hamiltonian 1 ˆ 2 + V (ˆr ), (ˆp + qA) Hˆ = 2me where V (r) = −q2 /4πε0 r and −q is the electron charge, can be expanded as:  2 q ˆ ˆ2 ˆ · pˆ + q A pˆ · A + A Hˆ = Hˆ 0 + 2me 2me

2 A.

Tonomura et al., Phys. Rev. Lett 56, 792 (1986).

with

H0 =

pˆ 2 + Vˆ (r). 2me

15.5 Lorentz Force in Quantum Mechanics

397

The first term Hˆ 0 is simply the Hamiltonian studied in Chap. 11. The second term is ˆ =A ˆ · pˆ in this gauge, we called the paramagnetic term. We remark that, since pˆ · A can rewrite this term as:

q q B × rˆ · pˆ = rˆ × pˆ · B = −γ0 Lˆ · B = −µ ˆL ·B 2me 2me

(15.34)

with γ0 = −q/2me . We recover the magnetic dipole interaction term introduced in Chap. 10 (Eq. 10.41). ˆ 2 /2me is called the diamagnetic term. One can check that The third term q2 A for the lowest lying levels En of the hydrogen atom, and for magnetic fields below 1 Tesla, the diamagnetic term is negligible: it is much smaller (by a factor of ∼10−4 ) than the paramagnetic term, which is itself small (by a factor of ∼10−4 ) compared to |En |.

15.5.4 Spin 1/2 Particle in an Electromagnetic Field Consider a spin 1/2 particle, charged or not, with an intrinsic magnetic moment µ ˆ S = γS Sˆ where Sˆ is the spin observable. If we place this particle in an electromagnetic field, and possibly in another potential V (r), its Hamiltonian is:

2 1 ˆ S · B(ˆr, t) pˆ − qA(ˆr, t) + qΦ(ˆr, t) + V (ˆr) − µ Hˆ = 2m

(15.35)

where q is the charge of the particle, and A and Φ are the electromagnetic potentials. This Hamiltonian is called the Pauli Hamiltonian. It acts on the Hilbert space Eexternal ⊗ Espin described in Chap. 12. For an electron, the form (15.35) is directly obtained as the nonrelativistic limit of the Dirac equation, which predicts γS = 2γ0 = −q/me .

15.6 Exercises 1. The Lorentz force in quantum mechanics In this exercise, we want to check that with the prescription: 1 (ˆp − qA(ˆr))2 Hˆ = 2m for the Hamiltonian of a charged particle in a magnetic field, one recovers, using the Ehrenfest theorem, the classical equations of motion.

398

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

Fig. 15.3 Study of the Bohm–Aharonov effect: classical paths in a two-hole Young interference experiment

a. We assume in all what follows that the field B is constant, uniform and directed along the z axis. We set B = |B|. We introduce the vector potential A = B × r/2. Check that this choice gives the appropriate value of B. b. Write the classical equation of motion of a particle of charge q and mass m in this field. Give the expression of the energy E of the particle. Describe the characteristics of the motion of the particle. ˆ = A(ˆr). Here pˆ is the usual momenc. Consider the observable uˆ = pˆ − qAˆ with A ˆ p. tum operator, i.e. pˆ = −i∇ (which yields [ˆx , pˆx ] = i). Show that pˆ .Aˆ = A.ˆ Write the commutation relations [ˆux , uˆ y ], [ˆuy , uˆ z ] and [ˆuz , uˆ x ] for the three comˆ ponents of the observable u. uˆ 2 . Calculate d. We assume that the quantum Hamiltonian has the form Hˆ = 2m dr / dt and du / dt. Compare the result with the classical equations of motion. e. Deduce from this result the form of the velocity observable vˆ . f. Can the three components of the velocity be defined simultaneously in a magnetic field? Write the corresponding uncertainty relations. 2. The Aharonov–Bohm effect Consider the two-hole Young interference experiment shown in Fig. 15.3. A solenoid whose axis is perpendicular to the plane of the figure is placed between the two holes B and B . One sends a particle from the source point O at time t1 and one detects the impact of this particle on the detection screen at a later time t2 . We shall assume that the probability amplitude A(C) to detect the particle in C is approximately given by3 : 

A(C) = AOBC + AOB C ∝ eiS/ + eiS  , where S et S  are the classical actions calculated along the paths OBC et OB C respectively. a. In absence of current in the solenoid, recover the fringe spacing xs found in Chap. 1 (Sect. 2.2). 3 This prescription can be deduced from the formulation of quantum mechanics based on path integrals; see R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals (McGrawHill, New-York, 1965).

15.6 Exercises

399

b. Determine the change in the interference signal when a current flows in the solenoid. Express the result in terms of the total magnetic flux πr 2 B (where r is the radius of the solenoid and B the magnetic field inside the solenoid) and the magnetic flux quantum h/q.

15.7 Problem. Landau Levels Energy Levels in a Magnetic Field In this problem we want to determine the energy levels of a spinless particle of charge q and mass m, which is placed in a constant and uniform magnetic field B = B uz . We use here the Landau gauge A(r) = Bx uy . ˆ The eigenfunction is a. Write the eigenvalue equation for the Hamiltonian H. denoted Ψ (r) and the corresponding eigenvalue Etot . b. We look for particular solutions which are factorized: Ψ (x, y, z) = eikz z ψ(x, y). Show that ψ(x, y) is solution of the eigenvalue equation: −2 2m



∂2 + ∂x 2



∂ qB −i x ∂y 

2  ψ(x, y) = E ψ(x, y)

(15.36)

where we set E = Etot − 2 kz2 /2m. c. Equation (15.36) describes the motion of the charge in the xy plane. We look for particular solutions of this equation which are also factorized with respect to x and y: ψ(x, y) = eiky y χ(x) (i) Write the equation which determines χ(x). To which physical problem does it correspond? One will introduce the cyclotron angular frequency ωc = qB/m. (ii) Show that the possible eigenvalues for the energy E are: 

1 E = n+ 2

 ωc .

(15.37)

Do the eigenvalues depend on the wavevector ky ? The corresponding energy levels are called the Landau levels. d. We now determine the degeneracy of a given Landau level, assuming that the xy motion of the particle is confined in a rectangle [0, X] × [0, Y ]. We shall neglect any edge effect, assuming that a0 = (2/qB)1/2 X, Y and we restrict to relatively low values of the quantum number n.

400

15 Lagrangian and Hamiltonian, Lorentz Force in Quantum Mechanics

(i) We choose periodic boundary conditions for the motion along the y axis. Show that the wave vector ky is quantized: ky = 2πj/Y , where j is an integer. (ii) What are the relevant values of j, such that the wave function ψ(x, y) is localized in the rectangle X × Y (and is thus physically acceptable)? (iii) Express the degeneracy of a Landau level as a function of the flux BXY and of the magnetic flux quantum h/q. The Lowest Landau Level (LLL) We now consider the quantum motion of a particle of charge q and mass m in a uniform magnetic field B = B uz . Here we choose the symmetric gauge A = B × r/2. We restrict ourselves to the motion of the particle in the xy plane (kz = 0) and we set as above ωc = qB/m. a. Write the eigenvalue equation for the energy. One will introduce Lˆ z = xˆ pˆ y − yˆ pˆ x . b. Consider the Landau level with the lowest energy ELLL = ωc /2 (see Eq. (15.37)). Show that the functions: ψ (x, y) = (x + iy) e−(x

2

+y2 )/(2 a02 )

,

where  is an arbitrary integer and a0 = (2/qB)1/2 are all energy eigenstates for the eigenvalue ELLL . c. Recover for the LLL the degeneracy calculated in the previous exercise, assuming that the particle is confined in a disk centered in x = y = 0, with a radius R a0 . This eigenstate basis plays an important role for the study of the fractional quantum Hall effect, which was discovered for a two-dimensional electron gas placed in a magnetic field.

15.7.1 Solution a. The eigenvalue equation for the energy reads: −2 2m



∂2 + ∂x 2



∂ qB −i x ∂y 

2

∂2 + 2 ∂z

 Ψ (x, y, z) = Etot Ψ (x, y, z)

b. This eigenvalue equation is separable and one can check immediately that the functions Ψ (x, y, z) = eikz z ψ(x, y) satisfying (15.36) with E = Etot − 2 kz2 /2m are eigenfunctions of the energy. As for the classical case, the motion along z is linear and uniform. c. (i) The substitution ψ(x, y) = eiky y χ(x) leads to the following equation for χ(x): −

2 d 2 χ 1 + mωc2 (x − xc )2 χ = Eχ 2m dx 2 2

15.7 Problem. Landau Levels

401

where we set xc = ky /(qB). This is the Schrödinger equation for a one dimension harmonic oscillator of frequency ωc /2π, centered in xc . (ii) The energy eigenvalues are E = (n + 1/2) ωc , where n is a non negative integer. These eigenvalues do not depend on ky . d. (i) The periodic boundary conditions for the y axis entail eiky Y = 1, i.e. ky = 2πj/Y , where j is an integer which can be a priori positive or negative. (ii) In order to have a wave function localized in the desired rectangle, the center of the oscillator corresponding to the motion along the x axis has to be between 0 and X: qBXY 0 < xc < X ⇒ 0 < j < jmax = 2π Since the extension of the wave function along x is of the order of a few a0 for the first Landau levels, the hypothesis a0 X implies that the particle is indeed localized with a probability close to 1 in the rectangle X × Y . (iii) The number of independent states corresponding to a given Landau level is jmax = Φ/Φ0 , where we set Φ = BXY and Φ0 = 2π/q. This is the degeneracy of the level. The functions ψn (x, y) = eiky y χn (x) constitute a basis for this level. Another possible basis is obtained by exchanging the roles of x and y, using the gauge A(r) = −By ux . A third possible gauge choice is the symmetric gauge, studied in the next exercise, and it leads to a third possible basis for each of the Landau level (for simplicity, the next exercise is actually restricted to the lowest Landau level). The Lowest Landau Level (LLL) a. For the gauge chosen in the text, the eigenvalue equation for the motion in the xy plane is: 

−2 2m



∂2 ∂2 + 2 2 ∂x ∂y



 −



1 ωc ˆ Lz + mωc2 x 2 + y2 2 8

Ψ (x, y) = E Ψ (x, y)

b. We introduce the polar coordinates ρ, θ in the xy plane. The functions ψ (x, y) = 2 2 2 2 2 (x + iy) e−(x +y )/(2 a0 ) = ρ eiθ e−ρ /(2a0 ) are eigenstates of Lˆ z = −i ∂ /∂θ with the eigenvalue . Inserting this result into the above eigenvalue equation, one reaches the desired result, after a relative long, but basic, calculation. c. The state ψ (x, y) is relevant if this wave function is essentially localized inside the 2 2 disk of radius R. The probability density |ψ (x, y)|2 ∝ ρ2 e−ρ /a0 has a maximum located a distance 1/2 a0 from the origin, with a width 1/4 a0 (for  1). Therefore the quantum numbers  must be located between 0 and max = R2 /a02 , which can also be written max = Φ/Φ0 , where Φ0 = 2π/q and Φ = πR2 B represents the field flux across the accessible surface. We recover indeed the degeneracy found in the previous exercise.

Chapter 16

The Evolution of Systems

Any experimental observation or any practical use of quantum phenomena relies on processes where a system evolves from a know initial state, and one performs measurements on it at some later time. It is therefore important to understand the various types of evolution a system can have. Up to now, we have been mainly interested in isolated systems whose nature did not change, i.e. there was no creation or annihilation of particles. In this Chapter, we present two characteristic processes: the oscillatory behavior of a two-state system under the influence of an external field, constant or oscillating, and the irreversible decay process of a system coupled to a continuum. In Sect. 16.1, we introduce the notion of a transition probability and a basic tool: time-dependent perturbation theory. In Sect. 16.2, we consider the atomic transitions induced by an external electromagnetic field, i.e. absorption and induced emission. In Sect. 16.3, we consider the problem of the decay of a system, such as an excited atom or an excited nucleus. We show how the exponential decay law emerges, and how one can calculate the lifetime of a system. We also introduce the notion of width of an unstable system. Finally, in Sect. 16.4, we discuss a few aspects of the time-energy uncertainty relation, ΔE Δt ≥ /2, which differs quite radically from the uncertainty relations we established in Chap. 8, and which illustrates the special role played by time in non-relativistic quantum theory.

16.1 Time-Dependent Perturbation Theory Transition Probabilities Consider a system whose evolution can be derived from the Hamiltonian: Hˆ = Hˆ 0 + Hˆ 1 (t),

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_16

(16.1)

403

404

16 The Evolution of Systems

where Hˆ 0 is time-independent. The eigenvectors |n and eigenvalues E n of Hˆ 0 are assumed to be known: Hˆ 0 |n = E n |n. (16.2) The operator Hˆ 1 (t) is an interaction term which may depend explicitly on time, and which a priori does not commute with Hˆ 0 . This term can induce transitions between two eigenstates |n and |m of Hˆ 0 . Our aim is the following. Assuming the system is prepared at time t0 in a given state |ψ(t0 ) = |n, we want to calculate the probability P n→m (t) = |m|ψ(t)|2

(16.3)

to find the system in the eigenstate |m of Hˆ 0 at a later time t. Example: A Collision Process In addition to the examples that we shall meet in this Chapter, a collision is a typical situation where this problem appears. Consider two particles a and b, each of which is prepared in a wave packet sharply defined in momentum p. At initial time the centers of these two wave packets propagate towards each other. In the absence of interaction the particles propagate freely:  pa  and  pb  are constants of the motion. If we take into account the interaction potential Hˆ 1 between the particles, a scattering process takes place. A measurement of the distribution of final momenta of the particles will give useful information on the forces themselves. The problem is to calculate the probability distribution of final momenta, knowing the interaction potential. Evolution Equations At any time, the state of the system |ψ(t) can be expanded in the basis {|n} of eigenstates of Hˆ 0 :  (16.4) |ψ(t) = γn (t)e−i En t/ |n. In this expression, we explicitly write the time evolution factors e−i En t/ that would be present for Hˆ 1 = 0. This simplifies the evolution equations of the coefficients γn (t). Using the Schrödinger equation we obtain: i



(γ˙ n (t) −

n

 i γn (t)e−i En t/ ( Hˆ 0 + Hˆ 1 )|n, E n γn (t))e−i En t/ |n =  n

and therefore: i



γ˙ n (t)e−i En t/ |n =

n



γn (t)e−i En t/ Hˆ 1 |n.

(16.5)

n

Multiplying by k|, we get: iγ˙ k (t) =

 n

γn (t) e−i(En −Ek )t/ k| Hˆ 1 |n.

(16.6)

16.1 Time-Dependent Perturbation Theory

405

The problem is completely determined by this set of coupled differential equations and by the initial condition which specifies |ψ(t0 ). Perturbative Solution In general, this set of equations does not have an analytic solution. In order to make progress, we assume as in Chap. 9 that Hˆ 1 is “small” compared to Hˆ 0 . In other words, we consider the Hamiltonian Hˆ λ = Hˆ 0 + λ Hˆ 1 and we assume that the corresponding coefficients γk (t) are analytic functions of λ around the origin including at λ = 1: ( p)

γk (t) = γk(0) (t) + λ γk(1) (t) + · · · + λ p γk (t) + · · ·

(16.7)

Inserting this expansion in (16.6), we identify the coefficients of each power of λ and we obtain: to order 0 : iγ˙ k(0) (t) = 0,  to order 1 : iγ˙ k(1) (t) = γn(0) (t) e−i(En −Ek )t/ k| Hˆ 1 |n ,

(16.8) (16.9)

n



to order r : iγ˙ k(r) (t) =

γn(r−1) (t) e−i(En −Ek )t/ k| Hˆ 1 |n.

(16.10)

n

This system can be solved by iterations. The terms γk(0) (t) are determined by the knowledge of the initial state of the system. Inserting these into (16.9) we can calculate the terms of order 1, γk(1) (t), which in turn give the terms of order 2 through Eq. (16.10), and so on. One therefore determines successively all the terms in the expansion. First Order Solution: The Born Approximation The zeroth order equation is solved immediately. We find that γk(0) (t) is a constant. If we choose the initial condition |ψ(t0 ) = |i, we obtain: γk(0) (t) = δk,i .

(16.11)

Inserting this in (16.9), we have for f  = i: i(E f −E i )t/ iγ˙ (1)  f | Hˆ 1 |i. f (t) = e

(16.12)

Taking into account the assumption γ (1) f (t0 ) = 0, this gives: γ (1) f (t)

1 = i



t

ei(E f −Ei )t/  f | Hˆ 1 |i dt.

(16.13)

t0

In this approximation, called the Born approximation, the transition probability from an initial state |i to a final state | f  (with f  = i) is given by:

406

16 The Evolution of Systems 2 Pi→ f (t) = |γ (1) f (t)| .

This approximation is acceptable if Pi→ f  1 (necessary condition). Particular Cases Constant Perturbation We suppose the perturbation Hˆ 1 is “switched on” at t0 = 0 and “switched off” at a later time T . We suppose also that it is constant between 0 and T . Setting ω0 = E f − E i , we obtain: γ (1) f (t ≥ T ) =

1 eiω0 T − 1 ,  f | Hˆ 1 |i i iω0

(16.14)

1 | f | Hˆ 1 |i|2 y(ω0 , T ). 2

(16.15)

and consequently: Pi→ f (t ≥ T ) =

We shall often make use of the above function y(ω, T ), defined as: y(ω, T ) =

sin2 (ωT /2) (ω/2)2



+∞

with −∞

y(ω, T ) dω = 2πT.

(16.16)

Its graph is given on Fig. 16.1. Sinusoidal Perturbation Consider a coupling Hˆ 1 (t) such that Hˆ 1 (t) = H˜ 1 e−iωt for 0 < t < T and Hˆ 1 (t) = 0 otherwise. A simple calculation gives: Pi→ f (t ≥ T ) =

Fig. 16.1 Graph of the function y(ω, T )

1 | f | H˜ 1 |i|2 y(ω − ω0 , T ). 2

(16.17)

16.1 Time-Dependent Perturbation Theory

407

There is a resonance if the angular frequency ω of the perturbation is equal to the Bohr frequency ω0 = (E f − E i )/ of the system. The resonance curve giving the variation of Pi→ f as a function of ω has a full width at half maximum Δω ∼ 2π/T . It gets sharper as the interaction time T increases. Perturbative and Exact Solutions We already met this in the case of a two-state system when we studied in Chap. 12 the magnetic resonance of a spin 1/2 placed in a rotating magnetic field. For this particular problem we know the exact solution of the evolution equations and it is instructive to compare it with the approximate result we have derived above. Consider the specific example of the Rabi experiment. We note T the time spent by the molecular beam inside the cavity where the magnetic field rotating at frequency ω/2π is applied. The time-dependent coupling Hˆ 1 is (cf. Eq. (12.53)): +| Hˆ 1 |− =

ω1 −iωt e 2

(ω1 = −γ B1 ).

The exact formula obtained by Rabi is (cf. (12.62)): P+→− (T ) =

ω12 sin2 (Ω T /2) Ω2

with

1/2  Ω = (ω − ω0 )2 + ω12

and the approximation (16.17) gives: P+→− (T ) =

ω12 sin2 ((ω − ω0 )T /2). (ω − ω0 )2

We notice that both formulas nearly coincide in two cases. • If the excitation frequency is sufficiently far from resonance: |ω − ω0 | ω1 . In this case Ω  |ω − ω0 | and the two results coincide for all times. • If the excitation is close to resonance (|ω − ω0 |  ω1 ) and if the interaction time is short enough: ω1 T /2  1.

16.2 Interaction of an Atom with an Electromagnetic Wave Electromagnetic transitions play a central role in atomic and molecular physics. Three basic processes are involved. Under the influence of an electromagnetic wave, an atom or a molecule can absorb energy. If it is in an excited state, it can also decay into a lower energy state either by spontaneous emission of radiation, or through the emission of radiation induced by an external electromagnetic wave. These three processes were introduced in 1917 by Einstein who understood, in a remarkable intuition, how a collection of atoms and photons can reach thermal equilibrium.

408

16 The Evolution of Systems

Here we want to study the behavior of an atom in a monochromatic wave whose electric field is: (16.18) E(r, t) = E0  cos(ωt − k · r). This plane travelling wave has an amplitude E0 , a wave vector k, and a polarization  orthogonal to k. We want to calculate the probabilities for the two processes of absorption and induced emission of light by the atom. Spontaneous emission cannot be treated quantitatively in this book, because the proper approach requires the quantization of the electromagnetic field. We will simply make a few qualitative remarks concerning this latter process (see also Sect. 16.3 and Exercise 2 at the end of this chapter).

16.2.1 The Electric Dipole Approximation We assume that we know the energy levels of the atomic system. We note respectively |1 and |2 the ground state of energy E 1 and an excited state of energy E 2 . We want to study here the absorption of light which results in the transition of the atom from the initial state |1 to the final state |2. The induced emission of light can be calculated in the same way, assuming the initial state is |2 and the final state is |1. In order to describe this phenomenon, we consider the simple case of a oneˆ = q rˆ the electric dipole moment operator, which is proelectron atom. We note D portional to the position of the external electron with respect to the core of the atom. We treat the atom as infinitely heavy, and we note R0 the position of the core. The coupling between the atom and the electric field (16.18) is given by: ˆ Hˆ 1 (t) = − D.E(R 0 , t).

(16.19)

This coupling is called the electric dipole interaction Hamiltonian.

16.2.2 Justification of the Electric Dipole Interaction The complete interaction of an atom with an external electromagnetic field (E, B) deriving from the potentials ( A, Φ) is obtained using the considerations developed in Chap. 15. Let rˆ i and pˆ i be the position and momentum operators of the electrons (i = 1, . . . , Z ). Assuming the nucleus of charge Z is fixed, and omitting the spin magnetic interactions, the Hamiltonian of the system in presence of external fields is:

16.2 Interaction of an Atom with an Electromagnetic Wave

Hˆ = +

409

N  2 1  Zqe2 pˆ i − qe A( rˆ i , t) + qe Φ( rˆ i , t) − 2m 4πε0 rˆi i=1

qe2 1  . 2 i j=i 4πε0 | rˆ i − rˆ j |

(16.20)

As such, t his expression is much too complicated. In practice, one must expand (16.20) and make approximations. In a systematic expansion of the Hamiltonian (16.20), there exist terms due to the electric field of the incident wave, and others due to the magnetic field. We neglect this second type of interactions. In fact, one has |B| = |E|/c for a plane wave in vacuum. Since the typical velocity of an external electron in an atom is of the order of αc ∼ c/137, i.e. much smaller than the velocity of light, the Lorentz force, and the ensuing magnetic effects, are very small compared with the electric part. If we were considering X rays and internal electrons, these magnetic effects would be comparable to the electric ones. Even if we limit ourselves to the electric dipole interaction of a one-electron atom, we should keep in full rigor the dependence of the incident field on r. However, the typical extension of the electron orbit is the atomic scale (r  ∼ 1 Å). This is much smaller than the wavelengths of a radiation corresponding to the infrared, the visible or the ultraviolet part of the spectrum, (λ = 2π/k ≥ 103 Å). Consequently the variation of E with r is negligible and it is legitimate to replace E(R0 + r, t) by E(R0 , t). To summarize, the simple expression we choose for Hˆ 1 in the case of a oneelectron atom, is the dominant term of the interaction between the electromagnetic field (E, B), and the charge and current density inside the atom. It is the first term of a multipole expansion which contains smaller effects, of magnetic and/or relativistic origin.

16.2.3 Absorption of Energy by an Atom In order to simplify the notations, we assume that the center of mass of the atom is at R0 = 0. At time t, the atomic state is:  |ψ(t) = γ1 (t) e−i E1 t/ |1 + γ2 (t) e−i E2 t/ |2 + γn (t) e−i En t/ |n n=1,2

with the initial conditions: γ1 (0) = 1 and γ2 (0) = · · · = γn (0) = 0. Injecting the expression (16.19) into the general result (16.13), we find: qe E0 γ2 (t) = 2| rˆ · |1 2



ei(ω0 +ω)t − 1 ei(ω0 −ω)t − 1 + ω0 + ω ω0 − ω

 (16.21)

410

16 The Evolution of Systems

with ω0 = E 2 − E 1 . A resonance phenomenon appears for ω ∼ ω0 . In the above expression, the first term is of the order of 1/ω = T0 /2π, where T0 is the period of the exciting field (T0 ∼ 10−15 s in the optical domain). For ω = ω0 the second term increases linearly with the interaction time t. If t T0 , we can neglect the first term compared to the second and we obtain: P1→2 (t) =

qe2 E02 |2| rˆ · |1|2 y(ω − ω0 , t). 42

(16.22)

In this expression, the presence of the square of the matrix element |2|ˆr |1|2 is of great importance in order to determine which transitions are allowed, as we shall see. We also remark the presence of the function y(ω − ω0 , t) (16.16). This transition probability has a resonant behavior in the vicinity of ω = ω0 , and the width of the resonance is of the order of 1/t. Contribution of Spontaneous Emission At resonance, the time t must be sufficiently small so that |γ2 (t)|  1, which is a necessary condition for the perturbative approach to be valid. Also the time t has to be much smaller than the lifetime τ of the level |2 due to spontaneous emission. Otherwise this process has to be taken into account in the above calculation and it gives a finite width to the resonance line (see Sect. 16.3 and Exercise 2 at the end of this chapter). We shall see in Sect. 16.2.5 that τ T0 , so that it is possible to fulfill simultaneously the conditions t T0 , so that (16.22) holds, and t  τ so that spontaneous emission can be neglected. The Concept of Photon We remark on the result (16.22) that the transitions are important only when the frequency of the light wave is close to a Bohr frequency of the atom: ω = E 2 − E 1 . This phenomenon is analogous to the photoelectric effect: an electron jumps from a state to another provided the incoming frequency is tuned to a Bohr frequency. In the case of the photoelectric effect, an electron is emitted and the final state belongs to the continuum of {ionized atom + electron} states. Contrary to a common prejudice due to the chronology of the discoveries, this provides an explanation of the photoelectric effect although we have not quantized the electromagnetic field and we have not introduced the concept of a photon. This concept becomes necessary when one studies the properties of radiation itself and the spontaneous emission of radiation. Validity of the Perturbative Treatment The electrostatic Coulomb field seen by an electron in an atom is of the order of ∼1011 V/m, which is enormous compared to the electric field of a “standard” light wave. In order to compete with the Coulomb field, one must use laser beams with an intensity of ∼1015 W/cm2 , which is considerable. In most usual situations, the use of perturbation theory is justified, i.e. the external field appears as a very small fluctuation compared with the Coulomb field.

16.2 Interaction of an Atom with an Electromagnetic Wave

411

16.2.4 Selection Rules We now derive from (16.22) the selection rules for electric dipole absorption and induced emission. Consider the matrix element: 2|ˆr |1 ≡ n 2 , 2 , m 2 |ˆr |n 1 , 1 , m 1 . In spherical coordinates, we have z = r cos θ, x ± i y = r sin θe±iϕ , i.e. the coordinates of r are expressed linearly in terms of r Y1,m (θ, ϕ). In the above matrix element, the contribution of interest is the angular integral: 



∗ Y2 ,m 2 (Ω) Y1,m (Ω) Y1 ,m 1 (Ω) d 2 Ω.

Owing to the properties of spherical harmonics seen in Chap. 10, this integral is non-zero if and only if: 2 = 1 ± 1

and

m 2 − m 1 = 1, 0, −1.

(16.23)

This is the case for instance for the Lyman α line of hydrogen: 2 p → 1s, or the resonance line of sodium: 3 p → 3s; in both cases, 1 = 1 and 2 = 0. For a pair of levels which does not fulfill (16.23), the transition is forbidden. An example is the transition corresponding to the 21 cm line of hydrogen, for which both levels have zero orbital angular momentum (1 = 2 = 0). The dominant coupling between these two levels is a magnetic dipole interaction, whose matrix element is much smaller than for an electric dipole coupling.

16.2.5 Spontaneous Emission The complete calculation of spontaneous emission requires the quantization of the electromagnetic field and we shall not treat it here. However it is interesting to give the main results and to discuss them. Consider an excited atomic state |i which is coupled by an electric dipole transition to a state | f  with a lower energy. The atom prepared in the state |i may decay to the state | f  by emitting spontaneously a photon with an energy ωi f = E i − E f . One can show that the probability d Pi→ f that the decay takes place during an arbitrarily short time interval dt is proportional to dt. Therefore one defines a probability per unit time d Pi→ f /dt, which is independent of dt and which is given by the formula: ωi3f dPi→ f ˆ f |2 , |i| D| = dt 3πε0 c3

(16.24)

412

16 The Evolution of Systems

ˆ is the electric dipole moment introduced above. Since each photon carries where D an energy ωi f , the energy radiated per unit time d I /dt is therefore: ωi4f dI ˆ f |2 . |i| D| = dt 3πε0 c3 We notice that these transitions follow the same selection rules as found in Sect. 16.2.4, since the same matrix element is concerned. We can compare this result with the classical formula giving the total intensity radiated per unit time by an electric dipole of moment p(t) = P cos ωt: 1 dI , | p¨ (t)|2 . = dt 6πε0 c3 After a time average over a period 2π/ω, we obtain: dI ω4 P 2. = dt 12πε0 c3 We notice the analogy between the classical and quantum formulas, with the correspondence: Classical Quantum Frequency ω → ωi f = (E i − E f )/ ˆ f | Amplitude P → 2 |i| D| This substitution was made by Heisenberg in 1925. It was a basic ingredient of his matrix mechanics. Lifetime of an Atomic Level; Orders of Magnitude Consider an assembly of N0 atoms all in state |i at time 0. Since the probability that a given atom decays in a time step dt is proportional to dt, the number of atoms N (t) still in the state |i at time t follows an exponential decay law: N (t) = N0 e−t/τ

with

dPi→ f 1 = . τ dt

The quantity τ is called the lifetime of the level |i. For a monovalent atom, we know that the size a of an outer electron orbit is of the order of 2 /me2  e2 /ω, which gives: ωi3f 1 q 2 a 2 ∼ ωi f α3 ∼ τ 3πε0 c3 e

with

α=

e2 1  . c 137

(16.25)

16.2 Interaction of an Atom with an Electromagnetic Wave

413

For an optical radiation, the order of magnitude of the lifetime τ of atomic levels is 10−7 to 10−9 s. This is much longer than a typical Bohr period 2π/ωi f , owing to the smallness of the coefficient α3 entering into (16.25).

16.3 Decay of a System After studying the resonant or quasi-resonant coupling between two levels, we turn to another class of problems where an initial state is coupled to a continuum of final states, i.e. a collection of states whose energies are very close and can be considered as a continuum. This is a central problem in collision physics and in the description of the decay of a system.

16.3.1 The Radioactivity of 57 Fe In order to present in a concrete way the issues addressed in this section, we consider the specific case of a radioactive nucleus. We start with cobalt 57. This has the peculiarity that the isolated nucleus 57 Co is stable, but the atom is not. A proton of the nucleus can absorb an electron of the internal K shell. This gives rise to what is called an electronic capture, or, equivalently, an inverse β reaction: 57

Co + e− →

57

Fe∗∗ + ν .

The atom 57 Co has a lifetime of 270 days. The excited Fe nucleus 57 Fe∗∗ produced in this reaction emits a first photon γ1 , of energy 123 keV, with a very short lifetime τ1  10−10 s. This leaves the nucleus in another excited state 57 Fe∗ . Then the 57 Fe∗ emits a second photon γ2 , of energy 14 keV, with a lifetime τ2  1.4 10−7 s, leaving the 57 Fe nucleus in its ground state: 57 57

Fe∗∗ → 57 Fe∗ + γ1 Fe∗ → 57 Fe + γ2

ω1 = 123 keV ω2 = 14 keV

It is possible to measure for each decay the time interval between the emissions of the two photons γ1 and γ2 . By convention, we denote t0 = 0 the time when the photon γ1 is detected. We want to calculate the probability P(t) that the nucleus decays and falls back in its ground state between the time 0 and t. Experimentally the answer to this question is well known; the decay 57 Fe∗ → 57 Fe + γ obeys the exponential law: P(t) = 1 − e−t/τ ,

414

16 The Evolution of Systems

with in the present case τ  1.4 10−7 s. For t  τ , the probability that the system decays between 0 and t is proportional to t: t τ

→

P(t) 

t . τ

(16.26)

The Hilbert Space of the Problem For this example, we must consider the Hilbert space which describes the state of an iron nucleus, accompanied by a certain number of photons. This is a situation different from those we have met up to now. Strictly speaking it requires the formalism of quantum field theory. Here we simply assume that some matrix elements between the relevant states exist, but we will not attempt to calculate them explicitly. There are two types of states to be considered: • The initial state |i ≡ |Fe∗  prepared at time t = 0 in the absence of photons (the photon γ1 , which is the signature that the Fe∗ is prepared, is absorbed by the detector at t = 0). • The possible final states | f  ≡ |Fe + γ, E f , representing the Fe nucleus in its ground state accompanied by one photon. E f represents the sum of the energies of the γ photon and of the nucleus in its ground state. In full rigor, we must specify also the direction of propagation of the outgoing photon and its polarization, in order to define | f  completely. The states |i and | f  are eigenstates of the Hamiltonian Hˆ 0 which describes nuclear forces on one hand, and freely propagating photons on the other hand. These states are not eigenstates of the coupling Hˆ 1 between the nucleus and the quantized electromagnetic field. In particular, a nucleus prepared in the state |i will not remain in this state indefinitely. We want to calculate the evolution of the system assuming we know the matrix elements  f | Hˆ 1 |i. Density of Final States For simplicity, we neglect the recoil of the nucleus. The energy of the 57 Fe is therefore fixed. The emitted photon can be in a whole series of energy states, which form a discrete set if we assume that the system is contained in a finite volume. Consider some energy band d E. Inside this band, there is a number d N of photon states. We define the density of states ρ(E): ρ(E) =

dN . dE

(16.27)

This allows us to replace a discrete sum over the possible final states, by an integral over the final state energy E f , which is much easier to handle:  f

 −→

ρ(E f ) d E f .

16.3 Decay of a System

415

Although we have not defined precisely the relation between photons and the electromagnetic field, all we need to know here is that photons are massless particles. The corresponding density of states can then be easily calculated. It is sufficient to make use of the fact that photons of momentum p have an energy E = c | p|. As for the non relativistic particles that we studied in Chap. 4, the momenta of the photon are quantized if we suppose that the experiment takes place in a (arbitrarily large) cubic box of size L, with periodic boundary conditions: p=

2π n, L

n = (n 1 , n 2 , n 3 )

n 1 , n 2 , n 3 integers.

(16.28)

Combining E = cp and (16.28) we obtain: ρ(E) =

L3 E2 , 2π 2 3 c3

to be compared with (16.27), for non relativistic massive particles. It remains to be checked in the end of the calculation that the predictions for any measurable quantity does not depend on the volume L 3 of the fictitious box that we have introduced. In the present case this is ensured by the expression for the matrix element  f | Hˆ 1 |i, which scales as L −3/2 .

16.3.2 The Fermi Golden Rule We now come back to our decay problem. The nucleus 57 Fe∗ can decay in a continuous set of Fe + γ states. We are not interested in the probability that it decays in a specific state, but in the probability that it decays to some domain D f of final states, characterized by their direction Ω (within a small solid angle d 2 Ω). We must therefore sum the formula which gives Pi→ f (16.15) on all possible final states of the domain D f : 1  | f | Hˆ 1 |i|2 y(ωi f , t) 2 f ∈D f  d 2Ω 1 | f | Hˆ 1 |i|2 y(ω f i , t) ρ(E f ) d E f = 2 ,  Df 4π

d 2 Pi→D f (t) =

(16.29)

with ω f i = (E f − E i )/. We now use the fact that, as the time t increases, the quantity y considered as a function of E f becomes more and more peaked in the vicinity of E f = E i . Using (16.16) we obtain: 1 y(ω f i , t) ≈ δ(ω f i ) =  δ(E f − E i ). 2πt

(16.30)

416

16 The Evolution of Systems

We can therefore neglect the variations of ρ(E) and of the matrix element  f | Hˆ 1 |i in the integral over E f . In other words, we extract the matrix element  f | Hˆ 1 |i and the density of states ρ from the integral, and evaluate them at the central point E f = E i . Using (16.16), this leads to: d 2 Pi→D f (t) =

2π d 2Ω | f, E f = E i | Hˆ 1 |i|2 ρ(E i ) t.  4π

(16.31)

We recover the linear dependence in time observed experimentally for short times (cf. (16.26)). Let us assume for simplicity that the matrix element  f, E f = E i | Hˆ 1 |i does not depend on the direction Ω of the emitted photon. Summing over all solid angles, we deduce the lifetime of state |i: 2π 1 = | f, E f = E i | Hˆ 1 |i|2 ρ(E i ). τ 

(16.32)

The fundamental relation (16.31) is called the Fermi golden rule. The range of times t for which it can be applied is limited by two constraints: • The time t should be short enough so that Pi→all f (t)  1: t  τ.

(16.33)

This is a necessary condition for the validity of first order perturbation theory. • The time t should be long enough so that the frequency width ∼1/t of the function y(ωi f , t) in (16.29) is much smaller than the typical scale of variation of the two other terms,  f | Hˆ 1 |i and ρ. Denoting κ this scale of variation in the frequency domain, the second constraint reads: t −1  κ.

(16.34)

In any problem where the Fermi golden rule is used, one must check that there exists a time interval during which these two constraints are simultaneously satisfied.

16.3.3 Orders of Magnitude We have already given in (16.25) the scaling laws for the lifetime of an atomic excited level, which can decay by spontaneous emission with an electric dipole transition. Aside from geometric factors, this decay rate reads: 3 ωi3f a12 ωi3f 1 = , ∼α τ c2 m 2e e2 c3

(16.35)

16.3 Decay of a System

417

where m e is the electron mass, a1 = 2 /m e e2 is the Bohr radius and ωi f is the energy of the emitted photon. We can discuss the consistency of the Fermi Golden Rule on this example. The frequency scale κ for the variations of  f | Hˆ 1 |i and ρ is typically κ ∼ ωi f . Therefore (16.33) and (16.34) can be simultaneously verified if: ωi−1 f τ

⇒

ωi2f m 2e c3

 1.

A typical Bohr frequency is E I /, where E I = m e e4 /22 is the ionization energy of the hydrogen atom. The consistency of our approach then reads: α3  1. Since α  1/137  1, this inequality is well satisfied. The smallness of the fine structure constant guarantees that the perturbative treatment of the effect of electromagnetic interactions onto the atomic levels is a good approximation. In going from atomic systems to nuclear systems, considering the expression 1/τ ∼ α a12 ωi3f /c2 , we expect that the electric dipole decay rates should be (i) reduced by a factor of order 10−10 owing to the change of size (10−15 m instead of 10−10 m), (ii) enhanced by a factor of order 1018 owing to the change of energy scale (1 MeV instead of 1 eV). We can therefore transpose (16.35) to the nuclear scale using R ∼ r0 A1/3 for the radius of a nucleus, where r0 ∼ 1.2 fm and where A is the number of nucleons. We obtain: 1 r 2 A2/3 ω 3 . (16.36) ∼α 0 2 τ c One can check that the energies and lifetimes of the 57 Fe excited states agree acceptably with this estimate. In particular we verify immediately that τ2 /τ1 ∼ (ω1 /ω2 )3 ∼ 103 . In the case of nitrogen 13, there exists an excited state with ω = 2.38 MeV and a lifetime τ ∼ 10−15 s. Using these parameters and (16.36), we obtain τ ∼ 2 10−15 s, a good order of magnitude.

16.3.4 Behavior for Long Times We have just found how the notion of lifetime for an excited atomic or nuclear level emerges using the short-time approximation of the decay law. For longer times, first order perturbation theory no longer applies since we no longer have Pi→all f  1. In this case, one can recover the measured exponential decay law using another approximation due to Wigner and Weisskopf.

418

16 The Evolution of Systems

In order to illustrate this, we consider the following simple model. We assume that the only non-vanishing matrix elements are i| Hˆ 1 | f  and  f | Hˆ 1 |i: i| Hˆ 1 |i =  f | Hˆ 1 | f  = 0. The initial state is |ψ(0) = |i. Using the above form of the coupling, we can write the state of the system at a later time t as: |ψ(t) = γi (t) e−i Ei t/ |i +



γ( f, t) e−i E f t/ | f  ρ(E f ) d E f .

(16.37)

Here we assume for simplicity that the energy is the only quantum number which characterizes the final states. We set H ( f ) ≡ i| Hˆ 1 | f  (H ( f ) is simply a function of E f ), and the Schrödinger equation gives:  iγ˙ i (t) =

ei(Ei −E f )t/ H ( f ) γ( f, t) ρ(E f ) d E f ,

iγ( ˙ f, t) = ei(E f −Ei )t/ H ∗ ( f ) γi (t),

(16.38) (16.39)

with the initial conditions γi (0) = 1 , γ( f, 0) = 0. We integrate formally (16.39): γ( f, t) =

H ∗( f ) i



t



ei(E f −Ei )t / γi (t  ) dt  ,

(16.40)

0

and we insert this result into (16.38). We then obtain the integro-differential equation: γ˙ i (t) = −

1 2





t

d E f ρ(E f )



ei(Ei −E f )(t−t )/ |H ( f )|2 γi (t  ) dt  ,

(16.41)

0

which can be rewritten as: 

t

γ˙ i (t) = −

N (t  ) γi (t − t  ) dt  ,

0

with

1 N (t ) = 2  



ei(Ei −E f )t



/

|H ( f )|2 ρ(E f ) d E f .

The function N (t  ) is proportional to the Fourier transform of the function of the final energy G(E f ) = |H ( f )|2 ρ(E f ). By definition of a continuum, the width of the function G(E f ) is large. Therefore N (t  ) has a narrow width and it is non-vanishing only if t  is close enough to 0. We note t  = τc the characteristic time above which the integrand oscillates so rapidly that N (t  ) is negligible. We make the approximation (to be justified a posteriori in each case) that γi (t − t  ) varies slowly in the time

16.3 Decay of a System

419

interval 0 < t  < τc . We can then replace γi (t − t  ) by γi (t) in the right hand side of the integro-differential equation and we obtain: 

t

γ˙ i (t) = −γi (t)

N (t  ) dt  .

0

For times t large compared to τc , the upper bound of the integral can be extended to infinity. Finally we use the relation: 

+∞



dt  ei(ω−ω0 )t = πδ(ω − ω0 ) + iPP



0

1 ω − ω0

 ,

where PP is the principal value integral, and we obtain: 

1 γ˙ i (t) = − + iδωi 2τ

 γi (t)

with 2π 1 = |H ( f )|2 ρ(E i ) and δωi = PP τ 



|H ( f )|2 ρ(E f ) d E f Ei − E f

 . (16.42)

The differential equation which gives the evolution of γi (t) is integrated immediately. This gives the probability that the system has decayed at time t: P(t) = 1 − |γi (t)|2 = 1 − e−t/τ ,

(16.43)

i.e. the exponential law. One can check that the value for τ derived in (16.42) coincides with the value (16.32) calculated previously using perturbation theory. The quantity δωi corresponds to an energy shift of the excited state due to the coupling of the nucleus and the electromagnetic field. This shift is exactly the same as what one obtains in second order time-independent perturbation theory (cf. (9.21)). Note that to first order the energy-shift vanishes because of our assumptions concerning the diagonal elements of Hˆ 1 . In the case of atomic levels, this second order shift is called the Lamb shift (see Chap. 13, Sect. 13.2.1). We now insert the result for γi (t) in the Eq. (16.40) giving γ( f, t). We obtain the energy distribution of the final states: p(E f ) = |γ( f, t = ∞)|2 = |H ( f )|2

1 , ¯ (E f − E i )2 + Γ 2 /4

(16.44)

where we have set E¯ i = E i + δωi and Γ = /τ . If we assume that |H ( f )|2 varies slowly, this probability law is a Lorentz function, centered at E¯ i , with a full width at half maximum Γ = /τ (see Fig. 16.2). In other words the energy of final states is on the average E¯ i with a dispersion ΔE:

420

16 The Evolution of Systems

Fig. 16.2 Energy distribution of the photon γ2 emitted in the decay of 57 Fe∗

ΔE = Γ /2 = /2τ .

(16.45)

This dispersion in energy of the final state is characteristic of any unstable system: beta decay of nuclei, radiative decay of atomic states, etc. It originates from the fact that the initial state |i is an eigenstate of the Hamiltonian in the absence of interaction, but it is not an eigenstate of the full Hamiltonian. Therefore, this initial state does not have a well-defined energy.

16.4 The Time-Energy Uncertainty Relation One of the great controversial questions in the 1930’s concerned the time-energy uncertainty relation: ΔE Δt ≥ /2. (16.46) Although this relation is commonly accepted, its interpretation varies considerably from one author to the other. It is indeed quite different from the uncertainty relations that we derived in Chap. 8. The relation Δx Δp ≥ /2 for instance, follows directly from the principles and the commutation relation of the operators xˆ and p. ˆ It is therefore an intrinsic property of any system. On the opposite, in the Schrödinger equation, time is not an operator, but a parameter which has a well defined value in the equations. Although we can measure it physically, time is not an observable. We will not give an exhaustive review of all points of view, neither will we adopt one attitude rather than another.1 We simply wish to make a few observations which can be used as a starting point for further reflection.

1 See

for instance the article of Aharonov and Bohm, Physical Review, vol. 122, p. 1649, (1961).

16.4 The Time-Energy Uncertainty Relation

421

16.4.1 Isolated Systems and Intrinsic Interpretations We recall some results presented in the first Chapters for systems whose Hamiltonians do not depend on time. Stationary States These are eigenstates of the energy, whose time evolution reduces to a multiplicative global phase factor: |ψ(t) = e−i Et/ |ψ(0). If the system is prepared in such a state, the expectation value a of any observable Aˆ does not change with time. This agrees with the relation (16.46): An isolated system whose energy is well-defined (ΔE = 0) does not evolve from t = −∞ to t = +∞. Evolution of a System The state of a system |ψ(t) can be a superposition of two or more energy eigenstates. For instance in Chap. 2, we constructed a wave packet as an infinite sum of stationary states. Such a system does not have a well defined value of the energy, and the expectation values of observables evolve with time, except if they correspond to a conserved quantity. Consider a physical quantity A associated with the system, such as the position of the needle on a wrist watch. We denote at and Δa the mean position and the mean-square deviation of this quantity at time t. Let v = dat /dt be the velocity associated to at . The characteristic time τ it takes the “wave packet” to cross a certain point, for instance a = 0, is τ = Δa/|v|. In Chap. 8 we proved the following properties: • Δa is related to the energy dispersion ΔE by: Δa ΔE ≥

1 ˆ Hˆ ]|ψ|, |ψ|[ A, 2

where |ψ is the state of the system at time t. • v is given by: 1 dat ˆ Hˆ ]|ψ. = ψ|[ A, v= dt i Combining these two relations, we obtain: τ ΔE =

 Δa ΔE ≥ . |v| 2

We recover a relation similar to (16.46), and it appears here as an intrinsic property of the quantum system: the larger the dispersion ΔE on the energy, the shorter the characteristic time of evolution of any quantity. This formulation is due to Mandelstamm and Tamm.

422

16 The Evolution of Systems

Decay of an Unstable System We have seen in Sect. 16.3.4 above that when a system is unstable and decays, its energy is not well defined. The energy distribution of the final products is peaked around some value with a dispersion related to the lifetime τ by ΔE = /(2τ ). This is also of the form (16.46), and it is again an intrinsic property of the system.

16.4.2 Interpretation of Landau and Peierls This interpretation2 comes from the analysis of the measurement of the energy E of a system. In order to perform such a measurement, we must couple the system of Hamiltonian Hˆ s , to a detector of Hamiltonian Hˆ d . The detector is initially in a state of known energy d , and the coupling takes place for a lapse of time T . When the coupling is switched off, the state of the set system + detector is a superposition of eigenstates of Hˆ s + Hˆ d , with an average energy E  + d close to E + d , up to an uncertainty /T :  |E  + d − E − d | ∼ . T This results from the shape of the function y(ω = E/, t) introduced in Sect. 16.1. Suppose that we know precisely the initial and final energies of the detector d and d . We can therefore deduce the uncertainty on E  − E: Δ(E  − E)  /T . In other words, even if the system is in a well-defined energy state before the measurement, an observer has access to this value only up to an uncertainty /T .

16.4.3 The Einstein–Bohr Controversy In 1930, Einstein presented the following argument. A clock is placed in a box, hanging on a spring. It is set to open a shutter at time t1 and to close it at time t2 = t1 + T , the interval T being determined with great accuracy. A radiation escapes from the box when the shutter is open and we measure the corresponding energy E by weighing the box before and after the experiment (E = δm c2 ). Since we have all our time to do this weighing, it can be very precise. Therefore this procedure seems to be a counter-example to the relation ΔE T ≥ /2. Bohr disproved the argument in the following way: (1) The position of the box which contains the clock is defined up to some quantum uncertainty Δz. Since the clock is placed in a gravitational field, its rate depends on the gravitational potential and, owing to general relativity, there is an uncertainty:

2 See, for instance, L. Landau and E. Lifshitz, Quantum Mechanics (Pergamon Press, Oxford, 1965).

16.4 The Time-Energy Uncertainty Relation

423

ΔT g Δz = 2 T c

(16.47)

on how long the shutter stays open. (2) At time t2 , the determination of the decrease of the weight δm g = Eg/c2 of the box is made by measuring the momentum the box acquires during this time interval T: Eg pz = δm g T = 2 T. c Here we assume that T is shorter than the oscillation period of the spring; a similar argument could be elaborated in the reverse case. Owing to the quantum uncertainty Δpz on the initial momentum of the box, the accuracy on the measurement of the energy is: c2 (16.48) Δpz . ΔE = gT Combining the two Eqs. (16.47) and (16.48) and using Heisenberg’s inequality Δz Δpz ≥ /2 for the position and the initial momentum of the box, we recover the desired inequality. The story says that Bohr, who had spent an entire night to find this counter argument, was quite proud of using Einstein’s General Relativity to solve the problem.

16.5 Exercises 1. Excitation of an Atom with Broad Band Light Consider n two-level atoms driven by the electric field E(t) = E 0 ez f (t) cos ωt, where f (t) is a function which is zero outside the interval [−τ , τ ]. We consider an electric dipole coupling between the atoms and the field. The ground and excited atomic states are denoted a and b respectively, and we set by convention E a = 0 and E b = ω0 . We suppose that (i) the typical scale of variation of f (t) is very large compared with the period 2π/ω and (ii) the excitation frequency ω/2π is close to the Bohr frequency ω0 /2π. One will neglect the contribution of non resonant terms. a. We define Ω1 = −d E 0 with d = b| D · ez |a (Ω1 is supposed to be real) and we denote g(Ω) the Fourier transform of f (t). Using perturbation theory, calculate the average number of excited atoms at time τ . This number is denoted n b (τ ). b. The electric field now consists in a succession of wave packets: E(t) = E 0 ez

∞ 

f (t − t p ) cos(ω(t − t p ))

with t1 < t2 < · · ·

p=1

Consider T such that t + τ < T < t+1 − τ . Calculate n b (T ).

424

16 The Evolution of Systems

c. We suppose that the successive wave packets arrive in a random way, with γ wave packets per unit time in average. We note n¯ b (T ) the statistical mean of n b (T ). Calculate n¯ b (T ) for γT 1. Show that one can define a transition probability per unit time from level a to level b. This quantity will be denoted Γa→b . d. We put w(ω + Ω) = (0 c/2) E 02 γ |g(Ω)|2 and we denote Φ the incident flux of energy. Relate w and Φ. Express Γa→b in terms of w(ω0 ). e. We suppose now that all atoms are initially in the state b. How can on transpose the previous reasoning? f. Write the evolution equations for the mean populations n a (t) and n b (t). What is the steady state of the system? 2. Atoms in Equilibrium with Black-Body Radiation We consider again the model of the previous exercise and we suppose that the atomic assembly is irradiated by the radiation of a black body of temperature T . We recall that one has in this case: ω3 , w(ω) = μ ω/k T B e −1 where μ depends only on fundamental constants. What must one add to the previous model in order to ensure the consistency of Statistical Physics (Einstein, 1917)? 3. Ramsey Fringes A neutron (which is a spin 1/2 particle) propagates along the z axis. We denote |± the eigenstates of the operator Sˆ z , projection of the neutron spin on the z axis. The neutron is initially prepared in the state |+ and it crosses two radio-frequency cavities of length L, separated by a distance D L (Fig. 16.3). One applies in each cavity a rotating magnetic field: B 1 = B1 (cos ωt u x + sin ωt u y ). The whole experimental setup is placed in a constant and uniform magnetic field B 0 parallel with the z axis. The motion of the neutron is treated classically as a uniform linear motion with velocity v. We are interested only in the quantum evolution

Fig. 16.3 Experimental setup for the observation of Ramsey fringes

16.5 Exercises

425

of the spin state of the neutron. The magnetic moment operator of the neutron is denoted μ ˆ = γ Sˆ and we set ω0 = −γ B0 et ω1 = −γ B1 . Calculate at first order in B1 the probability amplitude to find the neutron in the spin state |− at the output of the device. Show that the spin flip probability varies rapidly with the detuning ω − ω0 , and that one can determine the resonance frequency ω0 with a much better precision than if only one of the two cavities is used.

16.6 Problem. Molecular Lasers Preliminaries

Laser operation is based on population inversion, i.e. a situation where an excited energy level e of a quantum system is more populated than a lower level g. This can indeed lead to the amplification of radiation resonant with the e − g transition. We study here a possible way to achieve population inversion in a molecule. The physical principle of the inversion is based on the large difference between the time scale for the relaxation of the vibrations of the molecule, and the lifetime of its electronically excited state. a. Consider a one-dimensional problem and a wave function ψ(x) which can be ˆ , where d is expanded in a Taylor series. Show that the operator Tˆ (d) = e−id p/ a length and pˆ is the momentum operator, is such that: Tˆ (d) ψ(x) = ψ(x − d)  ˆ n /n! is mathematically legitimate. Note: the expansion ei uˆ = ∞ n=0 (i u) b. Consider a one dimensional harmonic oscillator of mass M and frequency Ω/(2π) centered at x = 0. Its energy eigenstates are denoted |n. Show that Tˆ (d) = ˆ aˆ † ) where aˆ † and aˆ are the usual creation and annihilation operators. Calculate eα(a− the real constant α. c. Let |nd be the eigensates of the same oscillator, but centered at x = d. Show that the ground state |0d can be expanded on the states |n as |0d = e

−λ2 /2

∞  λn √ |n n! n=0

with λ = d

The following identity is useful: ˆ aˆ ) = e−α eα(a− †

2

/2

e−αaˆ eαaˆ . †

MΩ . 2

(16.49)

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16 The Evolution of Systems

Molecular Lasers We consider a molecule formed by a core of two nuclei, each accompanied by internal electrons, surrounded by a single external electron. The position r = {x, y, z} of the electron is taken with respect to the centre of gravity of the core. In good approximation, the external electron is placed in an effective potential V (r) and has a series of energy levels n and corresponding wave functions ψn (r). The core, consisting in the two nuclei and the internal electrons, has both vibrational and rotational motions. For simplicity we consider here only the vibrational dynamics, that can be approximated by a harmonic one-dimensional motion. We denote u the distance between the nuclei, M the reduced mass and Ω/(2π) the oscillation frequency. The key point in the following is that the mean distance of the two nuclei depends on the state of the outer electron. In the electronic ground state of energy 1 , the eigenfunctions of the molecular Hamiltonian are ψ1 (r) φn (u), where φn (u) is a solution of the Schrödinger equation −

2 d 2 1 φn (u) + MΩ 2 (u − b)2 φn (u) = E n φn (u). 2 2M du 2

Similarly, in the first excited state of the outer electron, of energy 2 , the eigenfunctions are ψ2 (r)χm (u) with −

1 2 d 2 χm (u) + MΩ 2 (u − c)2 χm (u) = E m χm (u). 2M du 2 2

As mentioned above, the constants b and c, which can be determined experimentally or in a more sophisticated theoretical approach, are different. a. What are the total energy levels (i.e. electronic plus vibrational energy) of the molecule, corresponding to the above states? What is the relation between the functions φn (u) and χn (u)? b. We now study the electromagnetic transitions between the above molecular states. The molecule is placed in an oscillating electric field F, polarized along the z axis, of angular frequency ω, and of amplitude F. The dominant part of the interaction Hamiltonian is Hˆ 1 = −q zˆ F cos ωt, i.e. the dipole electric interaction with the external electron. This interaction does not depend on the variable u. Assume the molecule is initially (t = 0) in its ground state (both electronic and vibrational). Using first order perturbation theory, write the transition probability to an arbitrary final state at time t. Show that a discrete number of transitions are allowed. Give the value ωn of the angular frequency of the external field corresponding to these allowed transitions, and give the wave functions of the corresponding excited states. c. Show that the probability P(ωn ) for a field of frequency ωn /(2π) to excite the molecule factorizes into the product of an electronic excitation probability, and a probability pn to excite the vibrational states of the nuclei. Show that

16.6 Problem. Molecular Lasers

427

 pn = | An,0 |2

where

An,0 =

χ∗n (u) φ0 (u) du.

Making use of the results obtained in the first section, calculate An,0 . What type of probability law is pn ? What is the value n 0 = n for which the excitation of the molecular vibration occurs on the average? What is the root mean square deviation Δn? d. Calculate n 0 and Δn for a molecule where M = 4 10−27 kg, Ω = 2 1014 s−1 , b = 2 Å, c = 3 Å. e. In molecular physics, one frequently uses a principle based on semi-classical arguments, called the Franck–Condon principle, which states that when a transition occurs between two different electronic levels, “the distance between the nuclei does not change”. To be more specific, the electromagnetic transition is sufficiently fast so that, if the nuclei were classically bound objects, neither their momenta nor their positions would change during the time interval of the transition. Consequently, only the potential energy of the nuclei changes. This can be taken into account in a quantum mechanical calculation by assuming that, when the transition occurs, the parameter b suddenly changes into c, the motion remaining harmonic. Within this model, calculate in terms of b and c the classical variation of the vibration energy of the nuclei, assuming they are at rest initially. Compare the result with the calculation of question 3, and show that this comparison provides a justification of the Franck–Condon principle. f. Suppose there exists a very rapid relaxation mechanism in the excited electronic level so that the electronically excited molecules fall in the state of smallest vibrational energy ψ2 (r)χ0 (u). Towards which sub-levels ψ1 (r)φn (u) will electric dipole transitions occur preferentially? What are the corresponding emission angular frequencies ωn ? Let n 0 = n  ; calculate with the data of question 4 the Boltzmann factor N (E n 0 )/N (E 0 ) = exp(−(E n 0 − E 0 )/kT ) at room temperature (kT ≈ 0.025 eV). Assuming that 2 > 1 + E n 0 , and that roughly one or two percent of the molecules are excited by the oscillating field from the ground state of 1 to the excited states of 2 , can you explain why such a system is well suited for achieving a laser source?

16.6.1 Solution Section 1 a. The proof is straightforward. We have: pˆ =

 ∂ i ∂x

⇒

∂ Tˆ (d) = e−d ∂x .

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16 The Evolution of Systems

Therefore Tˆ (d) ψ(x) =

 n ∞  (−d)n ∂ ψ(x), n! ∂x n=0

which is simply the Taylor expansion of ψ(x − d). b. The creation and annihilation operators are defined as 1 1 ˆ ˆ aˆ † = √ ( Xˆ − i P) aˆ = √ ( Xˆ + i P), 2 2 √ with Pˆ = p/ ˆ MΩ. Therefore: pˆ = −i



 MΩ  MΩ † † and Tˆ (d) = exp −d aˆ − aˆ (aˆ − aˆ ) 2 2

which is the desired result, with α = −d c. We have

MΩ . 2

2 † ˆ aˆ † ) |0 = e−α /2 e−αaˆ eαaˆ |0. |0d = Tˆ (d)|0 = eα(a−

with α = −λ. In addition: a|0 ˆ = 0 ⇒ eαaˆ |0 = |0 Therefore: |0d = e

−λ2 /2

(aˆ † )n |0 =

√ n! |n.

 † n ∞  aˆ λn 2 λ |0 = e−λ /2 √ |n n! n! n=0 n=0

∞ 

n

Section 2 a. The energy levels of the total system (core + external electron) are E n(1) = ε1 + (n + 1/2)Ω E n(2) = ε2 + (n + 1/2)Ω. Let { f n (u)} be the wave functions of a harmonic oscillator centered at the origin u = 0. We have φn (u) = f n (u − b) therefore

χn (u) = f n (u − c)

φn (u) = χn (u − (b − c))

16.6 Problem. Molecular Lasers

429

φn (u) = Tˆ (b − c) χn (u). b. The transition probability from the initial state i, of wave function ψ1 (r) φ0 (u), to a final state f is at lowest order in F:   0 sin2 ω−ω t q2 F2 2 2 | f |ˆz |i| Pi→ f (t) =  ω−ω0 2 42 2

where we have set ω0 = (E f − E i )/. Consider the matrix element  f |ˆz |i. It can be written as   f |ˆz |i =



ψ ∗f (r) z ψ1 (r) d 3r

ζ ∗f (u) φ0 (u) du

where ζ f is some vibrational state of the nuclei. Therefore, the transitions inside the same electronic level are doubly forbidden: • First, |ψ1 (r)|2 z d 3r = 0 since the electronic wave function has generally a well defined symmetry (odd or even) with respect to the z axis. • Secondly one has: φ∗n (u) φ0 (u) du = δn,0 . The only transitions allowed a priori are those between the electronic levels ε1 and ε2 . The corresponding values of ωn are given by ωn = ε2 − ε1 + nΩ

n = 0, 1, . . . .

The wave function correponding to the excited level ωn is ψ2 (r) χn (u). c. We have at resonance: Pi→ f (t) = Pe pn t 2 with

2



q 2 F 2

∗ 3 (r) z ψ (r) d r ψ 2 1

42  and pn = |An,0 |2 with An,0 = χ∗n (u) φ0 (u) du. Pe =

From the results of the first section, we obtain φ0 (u) = χ0 (u − d) = e−λ with d = b − c and:

2

/2

∞  λn √ χn (u) n! n=0

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16 The Evolution of Systems

λ = (b − c)

MΩ . 2

Since the χn are orthogonal, we therefore obtain An,0 = e−λ

2

/2

λn √ , n!

pn = | An,0 |2 = e−λ

2

λ2n n!

which is a Poisson law. The mean value and root mean square deviation are n 0 = n = λ2 = √ (b − c)2 MΩ/(2) and Δn = n 0 = |λ|. d. For c − b = 1 Å= 10−10 m, M = 4 10−27 kg, Ω = 2 1014 s−1 , we obtain n 0  38

Δn  6.

e. A classical oscillator centered on u 0 has an energy: E=

p2 1 + MΩ 2 (u − u 0 )2 2M 2

(u 0 = b or c).

The initial state is: p = 0, u = u 0 = b, E i = 1 , and the final state, if p and u have not changed, has an energy: E f = 2 +

1 MΩ 2 (b − c)2 . 2

From the Franck–Condon principle, the energy of the final state should be E f . In the quantum calculation, we have E f  = 2 + (n 0 + 1/2)Ω and therefore: 1 Ω E f  = 2 + (b − c)2 MΩ 2 + 2 2 which is precisely, up to the zero-point energy term 1/2, the classical result. Note that since n 0 and Δn are large, the 1/2 does not play any crucial role. f. The calculation of Sects. 16.2 and 16.3 can be transposed symmetrically for emission. The probability to emit on the transition ψ2 (r)χ0 (u) −→ ψ1 (r) φn (u) at the frequency ωn = ε2 − ε1 − nΩ, is proportional to | An,0 |2 and it is maximum for the sub-energy levels n 0 ∼ 38 of the electronic ground state. The Boltzmann factor N (E 38 )/N (E 0 ) ∼ 10−87 is extremely small at room temperature. The population of the sublevels n ∼ 38 of the electronic ground state at room temperature is negligible.

16.6 Problem. Molecular Lasers

431

Fig. 16.4 Principle of a molecular laser: the excitation by the oscillating field followed by a rapid relaxation mechanism to ψ2 (r) χ0 (u) generates the required population inversion

The process under consideration therefore allows to achieve a population inversion between the vibrational ground state of the electronically excited manifold E 0(2) and the excited states of the electronic ground state manifold E n(1) (with n ∼ 38) since an appreciable fraction of the molecules will have been excited to E 0(2) by the incident radiation (see Fig. 16.4). This population inversion can be used to generate stimulated emission, and, therefore, to create a laser oscillation.

Chapter 17

Entangled States. The Way of Paradoxes

The way of paradoxes is the way of truth. To test Reality we must see it on the tight-rope. When the Verities become acrobats, we can judge them. Oscar Wilde, The portrait of Dorian Gray

At the end of Chap. 3 we mentioned Einstein’s revolt against the probabilistic aspect of quantum mechanics and the uncertainty relations. As we said, Einstein was worried about two aspects. One is the notion of a complete description of reality. He thought that a complete description is possible in principle, and that the probabilistic description is simply easier to handle. The other aspect is the notion of determinism: same causes produce same effects. Einstein wrote the worldwide famous: The theory produces a good deal but hardly brings us closer to the secret of the Old One. I am at all events convinced that He does not play dice.1

which is usually contracted to “God does not play dice.” Because this theory works, it must be an intermediate step toward an underlying, more sophisticated theory (which could, for instance, involve “hidden variables”2 ) that is not available at present, and where we have to perform averages that lead to the present version of quantum mechanics.

1 “Die

Theorie liefert viel, aber dem Geheimnis des Alten bringt sie uns kaum näder. Jedenfalls bin ich überzeugt, dass Der nicht würfelt.” 2 Again, Einstein never used that word. © Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_17

433

434

17 Entangled States. The Way of Paradoxes

17.1 The EPR Paradox In 1935, Einstein, Podolsky, and Rosen pointed out, in a celebrated paper,3 a paradox that has stirred the world of physics since then. Its starting point is a “gedanken experiment.” The spirit of the original version can be presented in a relatively simple form. Consider a particle that decays into two particles. One of them comes in my direction, the other in yours. One can manage to make the momentum P = p1 + p2 of the initial particle, that is, the total momentum, as well defined and as small as one wishes. For simplicity, we assume it vanishes. We therefore do not know exactly where the particle is, but this is not a problem; we assume we use large enough detectors. Similarly, we can measure with all wanted accuracy the relative position of the final particles r = r 1 − r 2 , inasmuch as P and r commute. On the other hand, one can prove quite generally that in the decay, there is conservation of the total momentum. If P = 0, the two final particles have exactly opposite momenta. Let’s check this experimentally. On a large number N of decays (all of which are such that both counters are activated), we place detectors sufficiently far from each other so that no information can be transmitted between both detectors when they measure their respective signals. This amounts to saying that the relative position is large enough. We proceed and perform, say, 1000 measurements. We notice afterwards, by comparing the set of time-ordered data, that systematically if you find a value p of the momentum of your particle, I always find − p for my particle. We have checked momentum conservation. We become convinced that we can repeat the operation as many times as we wish; we will always come to the same conclusion. After a while, say event number 1001, I’m fed up and I let you measure the momentum of your particle while I measure the position of mine with as great an accuracy as I wish. After that measurement, I call you on the phone; you tell me the value p1 of the momentum that you have found. Therefore, I know exactly both the position r 1 , measured by me and the momentum − p1 , measured by you, of my particle with as great an accuracy as I wish, and this is in contradiction with Heisenberg’s inequalities. Notice, and it is important, that this scheme works because I have used a system such that I have information on the quantum state of the set of the two particles in order to obtain physical information from you on my particle. One says that the state of the particles is entangled or correlated. This means that information on one of the particles is directly connected to information on the other, wherever they are located in space. However, EPR did not coin the word entanglement, nor did they generalize the special properties of the state they considered. Following the EPR paper, Erwin Schrödinger wrote a letter (in German) to Einstein in which he used the word Verschränkung (translated by himself as entanglement) to describe the correlations between two particles that interact and eventually get separated in space as in the EPR experiment. 3 A. Einstein, B. Podolsky, and N. Rosen, “Can quantum-mechanical description of physical reality be considered complete?” Phys. Rev. 47, 777 (1935).

17.1 The EPR Paradox

435

The following assertion is a key statement of the EPR article: If, without in any way disturbing a system, we can predict with certainty (i.e. with a probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity. Notice the implicit and fundamental role of locality4 in the argument. Your measurement, which is done at a time and distance such that one cannot transmit any information to my particle, cannot by any means perturb the results of my measurements. Einstein, Podolsky, and Rosen claim that no reasonable definition of reality can allow the contrary. Their conclusion is that the description of reality by quantum mechanics is not complete.

17.2 The Version of David Bohm There are, however, some loopholes in the gedanken experiment of EPR, and the experimental verification seemed problematic. What leads to measurable criteria is the remarkable version of the EPR argument given by David Bohm5 in 1952. This presentation is more convenient to work with and to treat mathematically than the initial version, although it is basically equivalent from the conceptual point of view. Suppose we prepare two spin 1/2 particles a and b in the singlet spin state: 1 |Ψs  = √ (|a : +z ; b : −z − |a : −z ; b : +z) . 2

(17.1)

Particle a is detected by Alice, who measures the component of its spin along an axis of unit vector ua (Fig. 17.1); similarly, particle b is detected by Bob who measures its spin component along an axis of unit vector ub . Alice and Bob’s measurements are strongly correlated. If they both agree to make only one measurement of the projection of the spin of their particles along a common axis, each of them has a probability 1/2 to find +/2 or −/2. Here, the law of interest is angular momentum conservation. The total angular momentum is zero in the singlet state. All its components are zero. Let us assume first that Alice and Bob both choose the same axis z to do their measurements: ua = ub = ez . The argument presented in the introduction applies: 4 In

the present context, locality means that some action at a point in space can have a detectable effect only at some other point in space within the light cone of that action. At a distance r , one must wait for a time at least equal to r/c in order to observe such an effect, it cannot be immediate. 5 Bohm, D. (1951). Quantum Theory, Prentice-Hall, Englewood Cliffs, page 29, and Chap. 5 Sect. 3, and Chap. 22 Sect. 19.

436

17 Entangled States. The Way of Paradoxes

Fig. 17.1 Gedanken experiment corresponding to the EPR argument. Two spin 1/2 particles a and b are prepared in the singlet state. Alice measures the component of the spin of particle a along an axis ua . Bob measures the component of the spin of particle b along an axis ub

with a probability 1/2, Alice will find +/2 and Bob will find −/2, and with the same probability 1/2, Alice will find −/2 and Bob will find +/2. Alice and Bob can never obtain the same result. There is a perfect correlation, or rather anticorrelation, of the two results. This will occur if they decide to make their measurements along any axis (same for both) since |Ψs  is a state of total spin zero, invariant under rotations. For instance in the basis of spin eigenstates along the x-axis. Using 1 | ± z = √ (| + x ± | − x) , 2

(17.2)

the singlet state (17.1) is written as 1 |Ψs  = √ (|a : +x ; b : −x − |a : −x ; b : +x) . 2

(17.3)

Such correlations appear frequently in daily life. Suppose we have two cards, one is red and the other one is yellow. We place each of them in a sealed envelope, we mix the envelopes at random in a closed box, and we give one of them to Alice and the other one to Bob. When Alice opens her envelope, she sees the color of her card (red with a probability 1/2, yellow with a probability 1/2). There is obviously a perfect anticorrelation with Bob’s subsequent result. If Alice’s card is red, Bob’s card is yellow and vice versa. There is no paradox in these anticorrelations: the fact that Alice looks at the color of her card does not affect the color of Bob’s card. According to the EPR claim above, there is an element of physical reality associated with the color of Bob’s card, because, without perturbing it in any manner, one can determine the color of this card by simply asking Alice what her result is. Similarly, there is an element of physical reality associated with the component Sbz , because without perturbing in any manner particle b, one can determine the value of Sbz that one would measure in an experiment: it is sufficient to ask Alice to measure the component Saz and to tell Bob the result. If Alice finds +/2, Bob is sure to find −/2 by measuring Sbz , and vice versa. Actually, the EPR argument goes one step further. We can transpose the argument about the z-axis to the x-axis, therefore there must also exist an element

17.2 The Version of David Bohm

437

of physical reality associated with the component Sbx of particle b. If one prepares a two-particle system in the singlet state, Bob can determine the component Sbx without “touching” particle b. It is sufficient for him to ask Alice to measure the component Sax and to tell him her result. Although the term “element of physical reality” is somewhat vague up to this point, we feel that we are reaching interesting grounds. In fact, the observables Sˆbx and Sˆbz do not commute. How can they simultaneously possess this element of physical reality? (Notice that in this debate, no attention is paid to what one could do with that information experimentally in subsequent operations.) Obviously, the above argument is contrary to the basic principles of quantum mechanics. When particles a and b are in an entangled state, such as the singlet state, it is risky to claim that one doesn’t “act” on particle b when performing a measurement on a. Taken separately, particles a and b are not in well-defined states; only the global system a + b is in a well defined quantum mechanical state. It is only for factorized states, of the type |Φ = |a : +α ; b : −α ≡ |a : +α |b : −α

(17.4)

that the EPR argument can be applied safely. However, in that case there is no paradox: a measurement on a gives no information on a measurement that would be performed on b. At this stage, we can have either of the following attitudes. We can stick to the quantum description that bears this paradoxical nonlocal character: the two particles a and b, as far as they may be from each other (a on Earth, b on the moon), do not have individual realities when their spin state is an entangled state. It is only after Alice (on the Earth) has measured Saz that the quantity Sbz (for the particle on the moon) acquires a well-defined value.6 However, if a measurement on Earth affects instantaneously a measurement on the moon, there is something we do not really understand in the theory. On the contrary, we can adopt the point of view of Einstein, and hope that some day one will find a more “complete” theory than quantum mechanics. In that theory, the notion of locality will have the same meaning as it has in classical physics, and so will the notion of reality.

17.2.1 Bell’s Inequality In 1964, John Bell, an Irish physicist working at CERN, made a decisive theoretical breakthrough.7 This allowed to carry the debate between two radically antagonistic conceptions of the physical world onto experimental grounds. 6 One can check that this formulation does not allow the instantaneous transmission of information.

In order to see the correlations with Alice’s result, Bob must ask Alice what her result is, and the corresponding information travels (at most) at the velocity of light. 7 J.S. Bell, Physics 1, 195 (1964).

438

17 Entangled States. The Way of Paradoxes

Bell’s formulation is the following. Suppose the super-theory that Einstein was hoping for exists, and suppose it involves “hidden variables” in the following sense. For any pair (a, b) of the EPR problem described above, that theory involves a parameter λ that determines completely and ahead of time the results of the measurements of Alice and Bob. For the moment, we know nothing about the parameter λ, which is absent in an orthodox quantum description. We denote Λ the manifold in which the parameter λ evolves. In the super-theory framework, there must exist a function A(λ, ua ) = ±/2 for Alice and a function B(λ, ub ) = ±/2 for Bob, which give the results of their measurements. These results therefore depend on the value of λ: for instance, if λ pertains to some subset Λ+ (ua ), then A(λ, ua ) = /2; if λ is in the complementary subset Λ − Λ+ (ua ), then A(λ, ua ) = −/2. Locality plays a crucial role in the previous assumptions. In fact, we have assumed that the function A depends on the value of λ and on the direction of analysis ua chosen by Alice, but not on the direction of analysis ub chosen by Bob. The parameter λ of the super-theory varies from one pair (a, b) to another, whereas in quantum mechanics, all pairs are prepared in the same state |Ψs  and nothing can make any difference between them. This parameter is therefore not accessible to a physicist who uses quantum mechanics: it is a hidden variable. All the beauty of Bell’s argument is to prove that there exist strong constraints on the theories with local hidden variables, and that these constraints can be established without any further assumptions than the ones given above. Notice that all correlations encountered in daily life can be described in terms of hidden variable theories. On the previous example of cards with different colors, the hidden variable comes from the shuffling of the cards. If a careful observer memorizes the motion of the cards in this shuffling, they can predict with probability 1 the result of Alice (red or yellow) and that of Bob (yellow or red). In order to get to Bell’s result, we introduce the correlation function E(ua , ub ). Quite generally, considering two random variables x and y, corresponding to a probability density p(x, y), one defines the linear correlation coefficient r (x, y) by r=

x y − xy . Δx Δy

If the two variables are correlated, that is x = ay + b, then r = ±1 (according to the sign of a) and, if they are independent, that is p(x, y) = p1 (x) p2 (y), then r = 0. The function E(ua , ub ) is equal to the expectation value of the product of the results of Alice and Bob, for given directions of analysis ua and ub , divided by 2 /4 in order to have a dimensionless quantity. Whatever the underlying theory, one has the following property: (17.5) |E(ua , ub )| ≤ 1. Indeed, for each pair the product of Alice’s and Bob’s results is ±2 /4. For a hidden variable theory, the function E(ua , ub ) can be written as

17.2 The Version of David Bohm

E(ua , ub ) =

439

4 2

 P(λ) A(λ, ua ) B(λ, ub ) dλ,

(17.6)

where the function P(λ) describes the (unknown) distribution law of the variable λ. The only constraints on P are:  for any λ , P(λ) ≥ 0 , and

P(λ) dλ = 1.

(17.7)

Here we assume that the function P(λ) does not depend on the directions of analysis ua and ub . Indeed these directions can be chosen by Alice and by Bob after the pair with hidden parameter λ has been prepared. (The derivation given here is actually due to Clauser, Horne, Shimony, and Holt.8 ) In the framework of quantum mechanics, one can check that the value of the function E(ua , ub ) is: E(ua , ub ) =

4 Ψs |( Sˆ a .ua ) ( Sˆ b .ub )|Ψs  = −ua .ub . 2

(17.8)

Bell’s theorem can be stated in the following way. Theorem 8 1. For a local hidden variable theory, the quantity: S = E(ua , ub ) + E(ua , ub ) + E(ua , ub ) − E(ua , ub )

(17.9)

always satisfies the inequality: |S| ≤ 2.

(17.10)

2. This inequality can be violated by the predictions of quantum mechanics. We first prove the inequality satisfied by hidden variable theories. We introduce the quantity: S(λ) = A(λ, ua ) B(λ, ub ) + A(λ, ua ) B(λ, ub ) + A(λ, ua ) B(λ, ub ) − A(λ, ua )B(λ, ub ), which enters into the definition of S: S=

8 J.F.

4 2

 P(λ) S(λ) dλ.

Clauser, M.A. Horne, A. Shimony, and R.A. Holt, Phys. Rev. Lett. 23, 880 (1969).

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17 Entangled States. The Way of Paradoxes

This quantity S(λ) can be rewritten as   S(λ) = A(λ, ua ) B(λ, ub ) + B(λ, ub )   + A(λ, ua ) B(λ, ub ) − B(λ, ub ) ,

(17.11)

which is always equal to ±2 /2. Indeed the quantities B(λ, ub ) and B(λ, ub ) can only take the two values ±/2. Therefore they are either equal or opposite. In the first case, the second line of (17.11) vanishes, and the first one is equal to ±2 /2. In the second case, the first line of (17.11) vanishes, and the second term is ±2 /2. We then multiply S(λ) by P(λ) and we integrate over λ in order to obtain the inequality we were looking for. Concerning the second point of Bell’s theorem, it suffices to find an example for which the inequality (17.10) is explicitly violated. Consider the vectors ua , ua , ub , and ub represented on Fig. 17.2: 1 ub .ua = ua .ub = ub .ua = −ub .ua = √ . 2 Using (17.8) we find:

√ S = −2 2,

(17.12)

(17.13)

which obviously violates the inequality (17.10). After this remarkable step forward due to Bell, which transformed a philosophical discussion into an experimental problem, experimentalists had to find the answer. Is quantum mechanics always right, even for a choice of angles such as in Fig. 17.2, which would eliminate any realistic and local hidden variable super-theory, or, on the contrary, are there experimental situations where quantum mechanics can be falsified, which would allow for a more complete theory, as Einstein advocated?

Fig. 17.2 Choice of directions of measurements of Alice and Bob that leads to a violation of Bell’s inequality

17.2 The Version of David Bohm

441

17.2.2 Experimental Tests The first experimental attempts to find a violation of Bell’s inequality started at the beginning of the 1970s. These experiments were performed on photon pairs rather than on spin 1/2 particles, because it is experimentally simpler to produce a twophoton entangled state of the type (17.1). The previous argument can be transposed with no difficulty to photon pairs. The spin states | + z and | − z are replaced by the polarization states of the photon | ↑ and | →, corresponding to vertical and horizontal polarizations. The states | + x and | − x, which are symmetric and antisymmetric combinations of |± : z, are replaced by photon states linearly polarized at ±45 degrees from the vertical direction: 1 1 |  = √ (| ↑ + | →) , |  = √ (−| ↑ + | →) . 2 2

(17.14)

The first experimental tests, in the early 1970s in the United States and in Italy, led to contradictory results concerning the violation of Bell’s inequality. The experiments of Freedman and Clauser in Berkeley in 1972,9 of Fry and Thompson in Texas in 1976,10 and particularly the “two channel” experiments of Aspect and his group in Orsay between 1980 and 1982, led to the undeniable violation of Bell’s inequality in a situation close to the gedanken experiment presented above.11 The experiments of Aspect use pairs of photons emitted in an atomic cascade of calcium atoms (Fig. 17.3). These calcium atoms are prepared by lasers in an excited state e1 . This excited state has a lifetime of 15 ns, and decays toward an excited state e2 by emitting a photon a, of wavelength λa = 551 nm. This latter level e2 has a lifetime of 5 ns, and decays itself to the ground state f by emitting a second photon b, of wavelength λb = 422 nm. The initial level e1 and the final level f have zero angular momentum, whereas the intermediate level e2 has angular momentum 1. Under these conditions, one can show that the polarization state of the emitted photon pair is: 1 |Ψ p  = √ (|a :↑ ; b :↑ + |a :→ ; b :→) . 2

(17.15)

This entangled state leads to the same type of correlations as the singlet spin state considered above. The transposition of Bell’s argument shows that some quantity S  , involving correlation functions between the polarizations of detected photons, must verify |S  | ≤ 2 for any local hidden variable theory. The Orsay result, S  = 2.697 ± 0.015 violates this inequality, but it is in agreement with the quantum mechanical 9 S.J. Freedman, J.F. Clauser (1972), “Experimental test of local hidden-variable theories”, Phys. Rev. Lett. 28 (938): 938–941. 10 E.S. Fry and R.C. Thompson (1976) Phys. Rev. Lett. 37, 465. 11 A. Aspect, P. Grangier, and G. Roger, Phys. Rev. Lett. 49, 91 (1982); A. Aspect, J. Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804 (1982).

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17 Entangled States. The Way of Paradoxes

Fig. 17.3 Left levels of atomic calcium used in order to produce photon pairs with correlated polarizations. Right the photons a and b are first filtered in frequency: Fa transmits photons a and stops photons b, and vice versa for Fb . They are then detected on photomultipliers P Ma+ , P Ma− , P Mb+ , and P Mb− . The polarizing cubes are analogues of Stern–Gerlach devices. They transmit photons with the polarization | ↑ towards the detectors P Ma+ , P Mb+ , and they deflect photons with polarization | → toward the detectors P Ma− , P Mb− . Because the photons are emitted nearly isotropically, only a small fraction (∼10−5 ) of the emitted pairs is actually used

prediction S  = 2.70. Therefore the universal realistic local hidden variable supertheory which was supposed to replace quantum mechanics, as Einstein believed, cannot exist, at least for this system. Physicists must learn to live with the genuine indeterminism of quantum mechanics. The truth is that many physicists were disappointed with these results. They were of course hoping that the proof of the existence of a super-theory would give exciting research themes. But what is more disconcerting is that the result confirms the wave packet reduction. Something does happen instantaneously at a distance. We were discussing this with my great friend Albert Messiah on February 28, 2001 and he told me Entangled states and Bell’s inequality are what make quantum mechanics unbearable! When we observe long distance entanglement, we are not sure that we really understand what’s going on. When Newton faced the problem of instantaneous action at a distance, he called upon God’s finger. Quantum entanglement causes a similar reaction. Bell, with his inequalities, did hope to falsify quantum mechanics. The problem with that theory, is that it always manages to accommodate any situation, even with consequences that seem unbearable for us!12

Contrary to what many people say, Einstein was quite right when he thought and said that the interpretation of quantum mechanics causes problems. Here, the local hidden variables assumption is contradicted by experiment. This is a confirmation of Feynman’s premonitory claim of 1965: “I think I can safely say that nobody understands quantum mechanics”.

12 Albert

Messiah actually used the term “undrinkable”, which is much stronger than unbearable for a Frenchman.

17.3 The GHZ Experiment

443

17.3 The GHZ Experiment Needless to say that Bell’s initial discovery, which can be considered as one of the great intellectual achievements in the history of science, generated a number of subsequent works, some of which have already been mentioned. But in 1998, Greenberger, Horne and Zeilinger (GHZ), found an even more spectacular example of quantum entanglement which reveals unambiguously the non-locality of the quantum world. They first analysed systems of four particles but, together with Shimony, and upon a suggestion of David Mermin, they subsequently applied their arguments to measurements involving three observers.13 The key point which makes that approach starkling is that it does not involve the evaluation of any correlation, but, as we shall see, a simple inspection of the sets of individual results recorded by the observers, after the experiment is finished. The experimental technique was presented in 1998.14 The final conclusive experimental results were published in 2000 by Jian-Wei Pan, D. Bouwmeester, M. Daniell, H. Weinfurter and A. Zeilinger15 . Zeilinger and his collaborators worked with entangled photon states, measuring polarizations. The original GHZ three photon state is |ΨG H Z  =

| ↑, ↑, ↑ + | →, →, → √ 2

(17.16)

There are several excellent papers which explain in a simple way the issue of the GHZ experiment, among which that of Wilczek16 and that of Herbert Bernstein.17 Here, we choose, for simplicity, to discuss the case of three spin 1/2 particles18 which has been considered by Herbert Bernstein. We deal with three spin 1/2 particles

13 D.M. Greenberger, M. Horne, and A. Zeilinger, in Bells Theorem, Quantum Theory, and Concep-

tions of the Universe, edited by M. Kafatos, (Kluwer, Dordrecht, 1989); D.M. Greenberger, M.A. Horne, A. Shimony, and A. Zeilinger, Bell’s theorem without inequalities, Am. J. Phys. 58, 1131 (1990); N.D. Mermin, Phys. Today 43 (6),9 (1990). 14 Jian-wei Pan and Anton Zeilinger, Greenberger-Horne-Zeilinger-state analyzer, Phys. Rev. 57 (3), 1998. 15 Jian-Wei Pan, D. Bouwmeester, M. Daniell, H. Weinfurter and A. Zeilinger (2000). “Experimental test of quantum nonlocality in three photon GHZ entanglement”. Nature 403 (6769): 515–519; Jian-Wei Pan and Anton Zeilinger, (2002) Multi-Photon Entanglement and Quantum Non-Locality https://vcq.quantum.at/fileadmin/Publications/2002-12.pdf. 16 Frank Wilczek, Entanglement made simple, Quanta Magazine, April 28, 2016. 17 Herbert J. Bernstein,Simple Version of the Greenberger-Horne-Zeilinger (GHZ) Argument Against Local Realism, Foundations of Physics 29 (4):521–525 (1999). 18 I am deeply gratefull to James Rich for explaining to me the whole issue, and I reproduce his presentation.

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17 Entangled States. The Way of Paradoxes

named respectively a, b, c and we are interested in their spin components along an orthonormal reference frame (x, y, z). Quantum Situation Consider the GHZ state: 1 1 |G H Z  ≡ √ |sza = +, szb = +, szc = + − √ |sza = −, szb = −, szc = − 2 2 (17.17) where we omit the /2 in the values of these spin components. We know (see Chap. 12) that we can decompose the sz components into sx and s y components of a spin 1/2 in the following ways √ √ |sz = + = (|sx = + + |sx = −)/ 2 = (|s y = + + |s y = −)/ 2 and √ √ |sz = − = (|sx = + − |sx = −)/ 2 = (|s y = + − |s y = −)/(i 2) There are several ways to decompose the |G H Z  state. a. We can decompose the sz components into one sx and two s y components and obtain, after some simple arithmetics, the following decomposition of the |G H Z  state, Eq. (17.17), such as 1 a (|s = +, s yb = +, s yc = + + |sxa = +, s yb = −, s yc = − 2 x + |sxa = −, s yb = +, s yc = − + |sxa = −, s yb = −, s yc = + )

|G H Z  =

We remark that if one measures, on this state, two s y ’s and one sx , one always finds and odd number of +/2. (All four terms have that property.) b. Alternatively, we can also decompose the sz components into three sx components and obtain: 1 a (|s = +, sxb = +, sxc = − + |sxa = +, sxb = −, sxc = + 2 x + |sxa = −, sxb = +, sxc = + + |sxa = −, sxb = −, sxc = − )

|G H Z  =

We remark that if one measures 3 sx ’s, one always finds an even number of +/2. (All four terms have that property.)

17.3 The GHZ Experiment

445

Notice that in both cases, if b and c find the same result then sxa = −, and if they have opposite results then sxa = + which means that sxa has an EPR element of reality. A similar property holds for b and for c. Local Realistic Situations Suppose that pairs of particles (a, b) possess local predetermined values of measurements of (sx , s y ): ⎡ a a⎤ sx s y ⎦ ⎣ sxb s yb There are, a priori, 16 types of pairs:

++ ++ −+ ++





++ +− −+ +−





+− ++ −− ++





For example, for a pair of type

+− +− −− +−





++ −+ −+ −+

−− −+





++ −− −+ −−





+− −+ −− −+





+− −− −− −−





a measurement of (sxa , s yb ) yields (− +). Turning now to particle triplets, there are 64 kinds of them, represented below: ⎛

+ +

⎞ ⎡

+ +

⎤ ⎡

+ +

⎤ ⎡

+ −

⎤ ⎡

+ +

⎤ ⎡

+ −

⎤ ⎡

+ −

⎤ ⎛

+ −

⎞

⎝+ +⎠ ⎣+ +⎦ ⎣+ −⎦ ⎣+ +⎦ ⎣+ −⎦ ⎣+ +⎦ ⎣+ −⎦ ⎝+ −⎠ + +

⎡

+ +

+ −

⎤ ⎡

+ +

+ +

⎤ ⎡

+ +

+ +

⎤ ⎡

+ −

+ −

⎤ ⎡

+ +

+ −

⎤ ⎡

+ −

+ +

⎤ ⎡

+ −

+ −

⎤ ⎡

+ −

⎤

⎣+ +⎦ ⎣+ +⎦ ⎣+ −⎦ ⎣+ +⎦ ⎣+ −⎦ ⎣+ +⎦ ⎣+ −⎦ ⎣+ −⎦ − +

⎡

+ +

− −

⎤ ⎡

+ +

− +

⎤ ⎡

+ +

− +

⎤ ⎡

+ −

− −

⎤ ⎡

+ +

− −

⎤ ⎡

+ −

− +

⎤ ⎡

+ −

− −

⎤ ⎡

+ −

⎤

⎣− +⎦ ⎣− +⎦ ⎣− −⎦ ⎣− +⎦ ⎣− −⎦ ⎣− +⎦ ⎣− −⎦ ⎣− −⎦ + +

⎡

− +

+ −

⎤ ⎡

− +

+ +

⎤ ⎡

− +

+ +

⎤ ⎡

− −

+ −

⎤ ⎡

− +

+ −

⎤ ⎡

− −

+ +

⎤ ⎡

− −

+ −

⎤ ⎡

− −

⎤

⎣+ +⎦ ⎣+ +⎦ ⎣+ −⎦ ⎣+ +⎦ ⎣+ −⎦ ⎣+ +⎦ ⎣+ −⎦ ⎣+ −⎦ + +

+ −

+ +

+ +

+ −

+ −

+ +

+ −

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17 Entangled States. The Way of Paradoxes

⎡

+ +

⎤ ⎡

+ +

⎤ ⎡

+ +

⎤ ⎛

+ −

⎞ ⎛

+ +

⎞ ⎡

+ −

⎤ ⎡

+ −

⎤ ⎡

+ −

⎤

⎣− +⎦ ⎣− +⎦ ⎣− −⎦ ⎝− +⎠ ⎝− −⎠ ⎣− +⎦ ⎣− −⎦ ⎣− −⎦ − +

⎡

− +

− −

⎤ ⎡

− +

− +

⎤ ⎛

− +

− +

⎞ ⎡

− −

− −

⎤ ⎡

− +

− −

⎤ ⎛

− −

− +

⎞ ⎡

− −

− −

⎤ ⎡

− −

⎤

⎣+ +⎦ ⎣+ +⎦ ⎝+ −⎠ ⎣+ +⎦ ⎣+ −⎦ ⎝+ +⎠ ⎣+ −⎦ ⎣+ −⎦ − +

⎡

− +

− −

⎤ ⎛

− +

− +

⎞ ⎡

− +

− +

⎤ ⎡

− −

− −

⎤ ⎡

− +

− −

⎤ ⎡

− −

− +

⎤ ⎛

− −

− −

⎞ ⎡

− −

⎤

⎣− +⎦ ⎝− +⎠ ⎣− −⎦ ⎣− +⎦ ⎣− −⎦ ⎣− +⎦ ⎝− −⎠ ⎣− −⎦ + +

⎡

− +

+ −

⎤ ⎡

− +

+ +

⎤ ⎡

− +

+ +

⎤ ⎡

− −

+ −

⎤ ⎡

− +

+ −

⎤ ⎡

− −

+ +

⎤ ⎡

− −

+ −

⎤ ⎡

− −

⎤

⎣− +⎦ ⎣− +⎦ ⎣− −⎦ ⎣− +⎦ ⎣− −⎦ ⎣− +⎦ ⎣− −⎦ ⎣− −⎦ − +

− −

− +

− +

− −

− −

− +

− −

The types of triplets in curved brackets ( ) are the ones for which a measurement of 2 s y ’s and 1 sx always yields an odd number of +/2. But for these eight types, if one measures 3 sx ’s, one always finds an odd number of +/2. Therefore there is no particle triplet that imitates the GHZ state. In other words, there exists no predetermined values of the three spins that agree with quantum mechanical measurements. Notice that no reference to hidden variables is made here. The superposition of states is compulsory to explain the experimental results. In some sense, we are facing a situation which has similarities with Schrödinger’s cat as was pointed out by several authors.

17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment The ambition of cryptography is to transmit a message from an issuer (Alice) to a receiver (Bob) and to minimize the risk that a spy may intercept and decipher the message. In order to do that, classical cryptography uses sophisticated methods that cannot be “broken” in a reasonable amount of time, with the present capacities of computers. Quantum cryptography is based on a somewhat different principle. It allows Alice and Bob to make sure no spy has intercepted the message before actually sending it! The principle of this technique is to profit from the fact that in quantum mechanics a measurement perturbs the state of the system, in particular with entangled states. One therefore devises a procedure that will prove the existence of a spy before sending the actual message!

17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment

447

The Communication Between Alice and Bob A message can always be coded in binary language, that is, by a succession of 0 and 1. Each number, 0 or 1, represents a piece of elementary information, or bit. In order to transmit her message, we assume that Alice sends Bob a beam of spin 1/2 particles in a well-controlled sequence, and that Bob detects these particles one after the other in a Stern–Gerlach type of apparatus. Each particle carries a bit coded through its spin state. Suppose first that Alice sends each particle in the state | + z or | − z. By convention, | + z represents the value 1 and | − z the value 0. Bob orients his Stern– Gerlach apparatus also along the z-axis, he measures the spin states of the particles that arrive, and he reconstructs Alice’s message. Such a procedure has no quantum feature and it is simple to spy upon. The spy just sits between Alice and Bob and he places his own Stern–Gerlach apparatus in the z-direction. He measures the spin state of each particle, and re-emits it toward Bob in the same spin state. He therefore reads the message and neither Alice nor Bob can detect his presence. The situation changes radically if Alice chooses at random, for each particle she is sending, one of the four states | + z, | − z, | + x, or | − x, without telling anyone which axis she has chosen (x or z) for a given particle. Suppose Alice sends Bob a series of particles without trying, for the moment, to give it any intelligible form. There are 16 particles in the examples shown in Figs. 17.4 and 17.5, but in practice one works with much larger numbers. It is only at the end of the procedure, as we show, that Alice will decide which particles should be taken into account in order to construct the message she wants to transmit. What can Bob do in this situation? He can orient the axis of his Stern–Gerlach apparatus in an arbitrary way, x or z. On average, for half of the particles, his choice is the same as Alice’s, in which case the bit he detects is significant. Indeed, if Alice sends a particle in the state | + x and if Bob chooses the x-axis, he does measure + with probability 1. For the other half of the particles, Alice and Bob choose different axes and Bob’s results are useless: if Alice sends | + x and if Bob chooses the z-axis, he will detect + with probability 1/2 and − with probability 1/2. In order to make sure that no spy has intercepted the transmission, Bob announces openly the set of axes he has chosen, x or z, for each event. He also says what results he has obtained (i.e., + or −) for a fraction of the particles. For instance, in the case of 16 particles shown in Figs. 17.4 and 17.5, Bob announces publicly his 16 choices of axes, and his first 8 results. Alice examines the results, and she can detect whether a spy has operated. Her argument is the following. The spy does not know the directions x or z she has chosen for each particle. Suppose that the spy orients his Stern–Gerlach apparatus in a random way along x or z, and that he re-emits a particle whose spin state is the same as what he has measured. If he chooses the x-axis and he gets the result +, he sends Bob a particle in the | + x state. This operation is detectable, because it induces errors in Bob’s observations. Consider, for instance, the case where Alice has sent a particle in the state | + z, and Bob has also oriented his detector along the z-axis, but where the spy has oriented his own Stern–Gerlach apparatus along the x-axis. The spy can measure + with a probability 1/2 and − with a probability 1/2. According to his result, he re-emits to

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17 Entangled States. The Way of Paradoxes

Number of the particle Axis chosen by Alice (kept secret) State chosen by Alice (kept secret) Axis chosen by Bob (broadcast openly) State measured by Bob (broadcast openly) Useful measurement ?

1 z

2 z

3 x

4 z

5 6 7 z x x

8 z

+ − + − − − + − z

x x

z

x z

x x

+ − + − − + + + yes no yes yes no no yes no

Fig. 17.4 Detection of a possible spy among Bob’s results done along the same axis (particles 1, 3, 4, and 7), Alice looks for a possible difference that would mean a spy has operated. No anomaly appears here. In practice, in order to have a sufficient confidence level, one must use a number of events much larger than 8 Fig. 17.5 After making sure that there is no spy, Alice chooses among the useful measurements those that allows communication of the message. For instance, to communicate the message “1,1”, that is, “+, +”, she openly asks Bob to look at the results of his measurements 11 and 15

Number of the particle Axis chosen by Alice (kept secret) State chosen by Alice (kept secret) Axis chosen by Bob (broadcast openly) State measured by Bob (kept secret) Useful measurement ?

9 10 11 12 13 14 15 16 x z x z z x z z + − + + − − + − z

z

x x

z

z

z

x

− − + + − + + + no yes yes no yes no yes no

Bob a particle in the state | + x or | − x. In both cases, because Bob’s detector is oriented along z, Bob can measure + with a probability 1/2 and − with a probability 1/2. If the spy had not been present, Bob would have found + with probability 1. Therefore, among all the results announced by Bob, Alice looks at those where her own choice of axes is the same as Bob’s (Fig. 17.4). If no spy is acting, Bob’s results must be identical to hers. If a spy is present, there must be differences in 25 % of the cases. Therefore, if Bob announces publicly 1000 of his results, on average 500 will be useful for Alice (same axes), and the spy will have induced an error in 125 of them (on average). The probability that a spy is effectively present, but remains undetected by such a procedure is (3/4)500 ∼ 3 × 10−63 , which is negligible. Once Alice has made sure that no spy has intercepted the communication, she tells Bob openly which measurements he must read in order to reconstruct the message she wants to send him. She simply chooses them among the sequence of bits for which Bob and her have made the same choice of axes, and for which Bob did not announce his result openly (Fig. 17.5). The Quantum Non-cloning Theorem In the previous paragraph, we have assumed that the spy chooses at random the axis of his detector for each particle, and that he sends to Bob a particle in the state

17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment

449

corresponding to his measurement. One may wonder whether this is his best strategy in order to remain unseen. In particular if the spy could clone each incident particle sent by Alice into two particles in the same state, it would be possible to send one of them to Bob and to measure the other one. The spy would then become undetectable! Fortunately for Alice and Bob, this cloning of an unknown state is impossible in quantum mechanics.19 One cannot generate in a reliable way one or several copies of a quantum state unless some features of this state are known in advance. In order to prove this result, let us note |α1  an initial quantum state which we want to copy. The system on which the copy will be “printed” is initially in a known state, which we note |φ (the equivalent of a blank sheet of paper in a copying machine). The evolution of the total system original + copy during the cloning operation must therefore be: cloning :

|original : α1  ⊗ |copy : φ −→ |original : α1  ⊗ |copy : α1  . (17.18)

The evolution is governed by some Hamiltonian which we need not specify, but which cannot depend on |α1  since this state is unknown by assumption. For an other state |α2  of the original, orthogonal to |α1 , we must also have: cloning :

|original : α2  ⊗ |copy : φ −→ |original : α2  ⊗ |copy : α2  . (17.19)

The impossibility of cloning is then obvious if we consider the initial state: 1 |α3  = √ (|α1  + |α2 ) . 2

(17.20)

If the cloning were successful for this state, we would find: cloning : |original : α3  ⊗ |copy : φ −→ |original : α3  ⊗ |copy : α3  . (17.21) However, the linearity of the Schrödinger equation imposes, by linear superposition of (17.18) and (17.19): |original : α3  ⊗ |copy : φ −→ 1 √ (|original : α1  ⊗ |copy : α1  + |original : α2  ⊗ |copy : α2 ) . 2 This final state is an entangled state. It is therefore different from the desired state (17.21).

19 W.K.

Wooters and W.H. Zurek, Nature 299, 802 (1982).

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17 Entangled States. The Way of Paradoxes

The inspection of this proof allows to understand the contribution of quantum mechanics to cryptography. If we limit ourselves to a two-state transmission, |α1  = | + z and |α2  = | − z, then the spy can remain invisible as we have explained in the previous section. The two operations (17.18) and (17.19) are possible; we simply need to measure the spin state of the incident particle along the z axis, and to re-emit one or more particles in the same state. It is the fact that we can use simultaneously the states |α1 , |α2  and linear combinations of these states |α3,4  = |± : x which makes the originality of quantum cryptography, and forbids any reliable duplication of a message intercepted by a spy. Present Experimental Setups As for experimental tests of Bell’s inequality, current physical setups use photons rather than spin 1/2 particles. Various methods can be used to code information on photons. We only consider the coding of polarization, which is effectively used in practice. Alice uses four states that define two non-orthogonal bases, each of which can code the bits 0 and 1, for instance in the form: | ↑ : 1 ; | → : 0 ; |  : 1 ; |  : 0.

(17.22)

In present quantum cryptography devices, the challenge is to obtain sufficiently large distances of transmission. One currently reaches distances of the order of ten kilometers, by using optical telecommunication techniques, in particular photons in optical fibers. An important point is the light source. The non-cloning theorem, which is crucial in order to provide the safety of the procedure, only applies to individual photons. On the contrary, usual light pulses used in telecommunications contain very large numbers of photons, typically more than 106 . If one uses such pulses for coding the polarization, the noncloning theorem no longer applies. Indeed it is sufficient for the spy to remove a small part of the light in each pulse and to let the remaining part propagate to Bob. The spy can measure in this way the polarization of the photons of the pulse without modifying the signal noticeably. In order to guarantee the safety of the procedure, each pulse must contain a single photon. This is a difficult condition to satisfy in practice and one uses the following alternative as a compromise. Alice strongly attenuates the pulses so that the probability p of finding one photon in each pulse is much smaller than one. The probability of having two photons will be p 2  p, which means that there will be very few two (or more) photon pulses. Obviously, most of the pulses will contain no photons, which is a serious drawback of the method because Alice must code the information redundantly. In practice, a value of p between 0.01 and 0.1 is considered to be an acceptable compromise. Once this basic question is solved, the essential part of the system uses optical telecommunication technologies. The source is a strongly attenuated pulsed laser, and the coding polarization takes place directly in the optical fiber using integrated modulators. The attenuated pulses are detected with avalanche photodiodes, which transform a single photon into a macroscopic electrical signal by an electron multiplication process. In order to identify unambiguously the photons emitted by Alice and

17.4 Quantum Cryptography; How to Take Advantage of an Embarrassment

451

detected by Bob, electric pulses synchronized with the laser pulses are sent to Bob by conventional techniques, and they play the role of a clock. Finally, a computerized treatment of the data, involving a large number of pulses, fulfills the various stages of the procedure described above, in particular to test the absence of a possible spy on the line. At present the systems that have been built are more demonstration prototypes rather than operational systems. Several relevant parameters have been tested, such as the distance and the transmission rate, the error rate, and so on. Actually, developing these systems has for the moment a prospective character, because conventional (nonquantum) cryptographic systems are considered to be very reliable by civilian or military users. This confidence was a little bit shaken in 1994, as we shall see in the next section.

17.5 The Quantum Computer The Quantum Bits, or “q-bits” In the previous section, we have seen that one can code a bit of information (0 or 1) with two orthogonal states of a spin 1/2 particle or with a polarized photon. At this stage of a quantum mechanics course, a question rises naturally: in terms of information theory, what is the significance of a linear superposition of these two states? In order to account for this possibility, one introduces the notion of “q-bit” which, contrary to a classical bit, allows the existence of such intermediate states. The notion of a q-bit in itself is not very rich; however it has interesting implications if one considers a quantum computer, based on the manipulation of a large number of q-bits. We take the very simplified definition of a computer as a system which is capable of performing operations on sets of N bits called “registers”. The content of a register is a binary word, which represents a number memorized by the computer. For N = 3, we therefore have 8 possible words: (+, +, +) (+, +, −) (+, −, +) (+, −, −) (−, +, +) (−, +, −) (−, −, +) (−, −, −)

Consider now a q-register, made of a set of N q-bits. The 2 N possible states of the classical register will define a basis of the space of states of the q-register, which can itself be in a linear superposition of all the basis states: |Ψ  =

  

Cσ1 ,σ2 ,σ3 |σ1 , σ2 , σ3 

for N = 3.

σ1 =± σ2 =± σ3 =±

Suppose that the computer calculates, i.e. it performs an operation on the state of the q-register. Since this operation is performed on a linear superposition of states, we can consider that it is done “in parallel” on the 2 N classical numbers. This notion of

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17 Entangled States. The Way of Paradoxes

quantum parallelism is the basis of a gain in efficiency of the computer. The gain may be exponential if the 2 N calculations corresponding to N q-bits are indeed performed simultaneously. Naturally, many questions arise. From the fundamental point of view, what kind of calculations can one perform and what kind of algorithms can one use with such a device? In practice, how can one construct it? The Algorithm of Peter Shor In the previous section, we referred to non-quantum cryptography. The corresponding systems are often called algorithmic protocols. One of these protocols is based on the fact that some arithmetic operations are very easy to perform in one way, and very difficult in the reverse way. For instance it is simple to calculate the product of two numbers, but it takes much more time to factorize a number in its prime divisors. If one considers the product P of two large prime numbers, one must perform √ approximately P divisions in order to identify the factors. The computing time increases exponentially with the number of digits (or of bits) of P: the factorization operation becomes impossible in practice for numbers with more than 300 digits, while the product operation leading to P can still be performed easily with a small computer. This “non reversibility” is at the origin of a cryptographic method due to Rivest, Shamir and Adelman (RSA), which is commonly used at present (credit cards, electronic transactions, etc.), and which is considered to be extremely reliable. This is why the 1994 paper of Peter Shor20 created a shock in the community. Shor showed that a quantum computer could factorize the product of two prime numbers with a number of operations reduced exponentially as compared to known algorithms running on classical computers! The turmoil has now calmed down and the present situation is the following. The algorithm proposed by Shor is correct in its principle, and it does have the expected gain in efficiency. However the practical elaboration of a quantum computer seems outside the range of present technology, although no physical law forbids it. Principle of a Quantum Computer We will not attempt to explain Shor’s algorithm here, and we will only give some intuitive ideas on how a quantum computer can perform a calculation. The basic principle is that the calculation must reduce to the evolution of a system with a Hamiltonian. It starts from some initial state and it ends by a “measurement” which determines the state of the q-register, and which interrupts the evolution. According to the principles of quantum mechanics, the value found in the measurement is one of the eigenvalues corresponding to the eigenstates of the measured observable. In this context, it corresponds to the state of a classical register, i.e. a binary word. In order to perform the successive operations, the Hamiltonian of the system is timedependent and it evolves under the action of a clock which determines the rhythm of the calculation. At first sight, the determination of this Hamiltonian for a given 20 Peter

Shor Polynomial-time algorithms for prime factorization and discrete logarithms on a quantum computer, SIAM Journal on Computing 26 (1997), 1484–1509.

17.5 The Quantum Computer

453

practical calculation seems to be a formidable task. Actually, one can show that the construction of the Hamiltonian can be done relatively easily. An actual calculation can be decomposed into a succession of simple operations which affect only one or two bits. These simple operations are performed by logical gates, such as the well known classical NOT, AND, OR. The quantum gates required for the Shor algorithm must have some particular features: • They must be reversible, since they follow from a Hamiltonian evolution of the initial bits. • They must handle q-bits, on which one can perform certain logical operations which are classically inconceivable. √ A simple example of a quantum gate is the gate N OT . It transforms the q-bits 0 and 1 into the symmetric and antisymmetric linear superpositions of 0 and 1 (it is a rotation of π/2 for a spin 1/2). If one applies this gate twice, √ one inverts 0 and 1, which corresponds to the gate NOT, hence the name of gate N OT . One could think that it would be sufficient to let the computer evolve towards a onecomponent state, which would be the desired value. Actually, very few algorithms give rise to such a simple manipulation. In general, the final state of the computer is still a linear superposition, and the result of the calculation is therefore probabilistic. For instance in Shor’s algorithm, the result is rather an indication of a possible result. It is easy to check by a conventional method whether the answer is correct, and if not, to pursue the calculation. Peter Shor has proven that this trial and error procedure gives the correct answer with a probability arbitrarily close to 1, using a number of trials which increases linearly (and not exponentially) with the number of digits of the number we want to factorize. Decoherence The principle of a quantum computer is compatible with the laws of physics, and it seems possible to construct such a computer, at least if one only considers simple calculations with a small number of gates. If one wants to attain large computing sizes, the global state of the computer has to be a quantum superposition of a large number of states, whose evolution must be controlled in such a way that all the properties of a linear superposition are preserved. It is not clear at present whether such a system can be devised. Studies are underway in two main directions: • First the register which evolves must be extremely well protected from the outside environment. Indeed the coupling with this environment will induce a decoherence effect, which may destroy the interferences between the various terms of the linear superposition. • Second, in case perturbations occur, one must prepare error correction codes, in order to place the computer in the same state where it was before the external perturbation occurred. These two directions—the choice of the system and the correction codes—are under intensive investigation and the questions raised have been stimulating both for algorithmic and for experimental quantum mechanics. At present it is very difficult

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17 Entangled States. The Way of Paradoxes

to predict the outcome of theses investigations. However one spin-off is that it is quite possible that simple logical operations will, in the mean run, lead to applications in systems of quantum cryptography.

17.6 Quantum Teleportation Quantum entanglement allows an amusing operation, “quantum teleportation.” Suppose Alice has a spin 1/2 particle A in the spin state: α|+ + β|− , with |α|2 + |β|2 = 1, that she wants to teleport, or fax, to Bob in a simple manner without measuring α and β. Bell States Consider a system of two particles of spin 1/2, whose spin state is written: α|+; + + β|+; − + γ|−; + + δ|−; −

(17.23)

with |α|2 + |β|2 + |γ|2 + |δ|2 = 1. The probability of finding in a measurement (+/2, +/2) is |α|2 , the probability of finding (+/2, −/2) is |β|2 , and so on. The four states of the Bell basis are defined as follows, 1 |Ψ+  = √ (|+; + + |−; −) 2 1 |Ψ−  = √ (|+; + − |−; −) 2

1 |Φ+  = √ (|+; − + |−; +) 2 1 |Φ−  = √ (|+; − − |−; +) . 2

The projectors PΨ = |Ψ Ψ | on one of these states |Ψ  can be called occupation operators of one of these states. In fact, the eigenvalues of these operators are 1 (occupied state) and 0 (vacant state). A direct calculation gives the probabilities of finding (17.23) in each Bell state. State |Ψ+  occupied: probability |α + δ|2 /2 , State |Ψ−  occupied: probability |α − δ|2 /2 , State |Φ+  occupied: probability |β + γ|2 /2 , State |Φ−  occupied: probability |β − γ|2 /2 . Because the basis of Bell states is orthonormal, the sum of these four probabilities is indeed 1.

17.6 Quantum Teleportation

455

Fig. 17.6 Principle of the teleportation of the quantum state of a particle

Teleportation In order for Alice to “teleport” the state α|+ + β|− of particle A to Bob, the procedure is as follows. Alice and Bob also have in common a pair of spin 1/2 particles B and C, prepared in the singlet state (Fig. 17.6): 1 √ (|+; − − |−; +) . 2 a. The state of the system of three spins (A, B, C) is: β α β α |Ψ  = √ |+; +; −;  + √ |−; +; −;  − √ |+; −; +;  − √ |−; −; +; . 2 2 2 2 This state can be decomposed on the Bell basis of the spins A and B; the basis for spin C remains |±: |Ψ  =

1 1 |Ψ+  (α|− − β|+) + |Ψ−  (α|− + β|+) 2 2 1 1 − |Φ+  (α|+ − β|−) − |Φ−  (α|+ + β|−) . 2 2

If Alice measures the spin state of the couple AB that projects this state on one of the four vectors of the Bell basis of AB, the probability of finding the pair AB in each of the Bell states is the same (|α|2 + |β|2 )/4 = 1/4. b. If Alice finds the pair AB in the state |Φ− , the state of the spin C is the original state α|+ + β|−. c. In order to teleport the a priori unknown state α|+ + β|− of particle A to particle C, Alice must not try to measure this state. She must simply make a Bell measurement on the pair AB and tell the result to Bob. When she finds that the state |Φ−  is occupied (in 25 % of the cases), Bob need not do anything. The spin state of C after the measurement is the state of A before the measurement. In all other cases, Bob can reconstruct the initial state by a simple transformation. For instance, if Alice finds the pair AB in the Bell state |Φ+ , the spin state of C

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17 Entangled States. The Way of Paradoxes

is α|+ − β|−, which can be transformed into the initial state α|+ + β|− by performing a rotation of π around the z-axis. d. One cannot use this method to transmit information more rapidly than with usual means. As long as Alice doesn’t tell Bob the result of her Bell measurement, Bob has no available information. It is only after he knows Alice’s results, and he has rejected or reconstructed the fraction of experiments (75 %) which do not give |Φ−  that he can make profit of the “teleportation” of the spin state of particle A. Further Reading • On the EPR problem: Quantum theory and measurement, edited by J. A. Wheeler and W. H. Zurek (Princeton University Press, 1983). The conceptual implications of Quantum Mechanics, Workshop of the Hugot fundation of Collège de France, J. Physique 42, colloque C2 (1981). • J. S. Bell, Physics 1, 195 (1964); see also J. Bell, Speakable and unspeakable in quantum mechanics, Cambridge University Press (1993). Franck Laloë (2012), Do We Really Understand Quantum Mechanics, Cambridge University Press, ISBN 978-1-107-02501-1 • C. Bennett, G. Brassard et A. Ekert, Quantum Cryptography, Scientific American, October 1992; R. Hughes and J. Nordholt, Quantum cryptography takes the air, Physics World, May 1999, p. 31. • S. Lyod, Quantum mechanical computers, Scientific American, November 1995, p. 44; S. Haroche and J.-M. Raimond, Quantum computing: dream or nightmare, Physics Today, August 1996, p. 51; Special issue of Physics World on Quantum Information, March 1998; N. Gershenfeld and I.L. Chuang, Quantum computing with molecules, Scientific American, June 1998, p. 50; J. Preskill, Battling decoherence: the fault-tolerant quantum computer, Physics Today, June 1999, p. 24.

17.7 Concluding Remarks In writing the previous edition of this book, my last chapter was devoted to the developments of astrophysics and cosmology. Needless to say that these remain at the forefront of physics. The recent observation of gravitational waves, detected on September 14, 2015 by both of the twin Laser Interferometer Gravitationalwave Observatory (LIGO) detectors, located in Livingston, Louisiana, and Hanford, Washington, USA, and operated by Caltech and MIT is undoubtedly a major scientific event of our time. There is no doubt that enormous progress and impressive new results are to be expected in the field. I have decided to end this book with an introductory chapter of what is becoming an impressive byproduct of quantum mechanics: Quantum information. A simple inspection of the literature on the subject shows that it is becoming a major scientific and technological field of research. Quantum information science includes impressive theoretical as well as experimental topics in quantum physics. It is fascinating to

17.7 Concluding Remarks

457

see the developments of Quantum computing, of Quantum cryptography, Quantum communication Quantum teleportation etc. One of the very immediate proofs of that lies in the Springer collection “Quantum Information Processing”, whose Volume 1 appeared in 2002, and whose Volume 15, January 2016 - June 2016, was just issued (12 issues per year). The list of titles of the contributions is impressive, as is their number and diversity. I hope everyone will forgive me for ducking in front of an exhaustive review of that part of quantum physics. All those developments are impressive by themselves, their topics and their goals. But, what is perhaps more impressive is that their roots lie on purely intellectual questions raised by Einstein in 1935, and pursued by a few jewelers such as John Bell. Huge technological progress is made because of intellectual paradoxes! Things exploded after the 1970s with experiments which were devised to get an answer to the question: do we understand Quantum Mechanics? Answer which, in my opinion, is still lacking. Is that really crucial? Above all, everyone is convinced that quantum mechanics is a good theory, even a very impressive theory, given all the efforts and all the means that mankind  has made to find its limits (aside from saying that before the Planck time of, say, (G/c5 ) ∼ 5.4 10−44 s, “something else” replaced it). So one should learn and use quantum physics! But, as so many people say or have said: we are not at all sure that we understand it. In some sense, I have a tendency to feel that we should consider that quantum mechanics is a “complete” theory, with a beginning and an end, just as electromagnetism was after 1864, Maxwell’s equations and the discovery of electromagnetic radiation by Hertz. The invariance of the velocity of light under Galilean transformations was inscribed in Maxwell’s equation. Some nasty jerk could (and may) have claimed that all that was bullshit because of that stupid remark, which actually took so many years for Michelson to verify experimentally. Of course, nowadays, it seems quite natural for us to imagine that Einstein, Lorentz, Poincaré and others just thought that one should question the unshakeable notion of the nature of time, that mankind had always taken for granted. And then, was relativity the end of the story? Not at all, since the same Einstein imagined the photon etc. etc. and people went on with quantum mechanics and QED. This time, so much serious and hard work has been done by so many competent people on “thinkable” approaches or alternatives, that the answer, once it comes, will be unexpected, be it simple or terribly subtle. In any case fascinating.

Chapter 18

Solutions to the Exercises

Chapter 1 Probabilities 1. Distribution of impacts Going to polar coordinates gives: f (ρ) = One can check that

 +∞ 0

1 2 2 ρ e−ρ /2σ (ρ ≥ 0). 2 σ

f (ρ) dρ = 1.

2. Is this a fair game?  2  3 The probability that 6 does not appear is 56 ; that it appears once, 3 × 56 ×    2  3 1 ; twice, 3 × 56 × 16 ; thrice 16 . The respective gains are −1, 1, 2, 5, the 6 expectation value of the gain is therefore:  15 1  3 −5 + 3 × 52 + 2 × 3 × 5 + 5 = − . 3 6 216 Don’t play ! In order for the game to be fair, the third gain should be 20 euros! 3. Spatial distribution of the molecules in a gas We start with the simple alternative: either a molecule is inside v, (p = v/V ) or it is not (q = 1 − v/V ). The probability for finding k molecules in v is therefore:  PN (k) =

N k



v k  v N−k . 1− V V

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7_18

459

460

18 Solutions to the Exercises

√ The expectation is σ = Npq ∼ √ value is k = Np = Nv/V , and the dispersion √ Nv/V = k. In our numerical example k ∼ 3 × 1016 , therefore σ ∼ 1.7 × 108 . The relative dispersion σ/k is very small (∼10−8 ). The probability for finding a number of molecules outside the interval k ± 2σ is 5 % (Gaussian law). Chapter 2 1. Spreading of a wave packet of a free particle a. Using a procedure similar to the previous one, we obtain from the Schrödinger equation:   ∂ψ i ∂ψ ∗ dx 2 t = x ψ − ψ∗ dx. dt m ∂x ∂x b. The Schrödinger equation gives: 2 dA = dt 2m2



  ∂ 3 ψ∗ ∂ 2 ψ ∂ψ ∗ x ψ − dx + c.c. ∂x 3 ∂x 2 ∂x

Using  2 ∗  2 ∂ 2 ψ ∂ψ ∗ ∂ψ ∂ψ ∗ ∂ ∂ ψ ∂ 3 ψ∗ ∗∂ ψ − +ψ −2 + c.c. = ψ ψ ∂x 3 ∂x 2 ∂x ∂x ∂x 2 ∂x 2 ∂x ∂x and an integration by part, we obtain: dA 2 =− 2 dt 2m



2 ∂ 2 ψ∗ ∗ ∂ ψ + ψ ψ ∂x 2 ∂x 2



2 dx + 2 m



∂ψ ∂ψ ∗ dx, ∂x ∂x

and a second integration by parts of the first member of the right hand side yields the result for B(t). c. Using the Schrödinger equation again, one gets: i3 dB = 3 dt m



∂ψ ∂ 3 ψ ∗ ∂ 3 ψ ∂ψ ∗ − ∂x 3 ∂x ∂x ∂x 3

 dx,

and an integration by part shows that this is zero. The coefficient B is a constant and we shall put in the following B = 2v12 , where v1 has the dimension of a velocity. d. The integration of the equation of evolution for A(t) yields A(t) = 2v12 t + ξ0 and x 2 t = x 2 0 + ξ0 t + v12 t 2 . e. Using xt = x0 0 + v0 t, we obtain for Δxt2 = x 2 t − x2t the wanted result. Note that these results can be recovered in a simple way using Ehrenfest’s theorem (Chap. 7).

18 Solutions to the Exercises

461

2. The Gaussian wave packet a. For t = 0, we obtain: 1/4 ip0 x/ −x2 σ2 /2  e e ψ(x, 0) = σ 2 /π and: x0 = 0 p0 = p0 Δx0 =

1 σ √ Δp0 = √ . σ 2 2

Hence the result Δx0 Δp0 = /2. The Heisenberg inequality is saturated in the case of a Gaussian wave packet. b. At time t, the wave function ψ(x, t) is the Fourier transform of 2 e−ip t/(2m) ϕ(p), which is still the exponential function of a second order polynomial in the variable p (with complex coefficients). The general results concerning the Fourier transform of Gaussian functions apply and one obtains after a rather tedious calculation:

   1 1 p0 t 2 2 , |ψ(x, t)| = √ exp − x − m 2Δx 2 (t) Δx(t) 2π with Δx 2 (t) given by (3.24). One recovers on this particular case the general results of Chap. 2 and of the previous exercise: propagation of the center of the wave packet at the velocity p0 /m, quadratic variation of the variance of the wave packet. 3. Characteristic size and energy in a linear or quadratic potential For a quadratic potential, the relation (3.21) with γ = 1 yields as a typical energy E for the ground state: 1 2 + mω 2 Δx 2 E= 2m Δx 2 2 √ which is minimum for Δx = /mω, with E = ω in this case. This is indeed the correct order of magnitude for the extension and the energy of the ground state: the exact result is ω/2 (Chap. 4). For the linear potential α|x|, the order of magnitude of the energy of the ground state is obtained as the minimum of E=

2 + αΔx 2m Δx 2

 1/3 when the extension Δx varies. It corresponds to Δx = 2 /mα , and the corre 1/3 sponding energy is (3/2) 2 α2 /m .

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18 Solutions to the Exercises

4. Laplacian operator in three dimensions We notice that for r = 0, then Δ(1/r) = 0. Using an integration by parts, we find for any ϕ in S: Δ

    1 1 ϕ(r) d 3 r = − ∇ · ∇ϕ(r) r 2 dr d 2 Ω r r   ∞ 1 ∂ ϕ(r) r 2 dr = + d2Ω 2 r ∂r 0 2 = − d Ω ϕ(0) = −4π ϕ(0),

which demonstrates the very useful identity Δ(1/r) = −4πδ(r). 5. Fourier transform and complex conjugation The Fourier transform of f ∗ (k) is g ∗ (−x). If f (k) is real, then g ∗ (−x) = g(x). If f (k) is even, then g(x) is also even. We can then conclude: f (k) real and even ←→ g(x) real and even. Chapter 3 1. Expectation values and variances x = a/2, Δx = a 1/12 − 1/2π 2 , p = 0, Δp = π/a, therefore: Δx Δp = π 1/12 − 1/2π 2 ∼ 0.57. 2. The mean kinetic energy is positive Using an integration by parts, we find: p2x 

= −

2

∂2ψ ψ (x) dx = 2 ∂x 2 ∗



∂ψ 2

∂x dx ≥ 0.

3. Real wave functions If ψ is real, we have: p =

 i



+∞ −∞

ψ(x)

∂ψ   2 +∞ =0 dx = ψ −∞ ∂x 2i

since ψ 2 vanishes at infinity for a physical state.

18 Solutions to the Exercises

463

4. Translation in momentum space Using the general properties of the Fourier transform, we obtain p = p0 + q and Δp = σ. 5. The first Hermite function We use: d 2  −x2 /2  d  −x2 /2  2 2 e = −x e−x /2 and e = (x 2 − 1) e−x /2 , dx dx 2 hence the result. Chapter 4 1. Uncertainty relation for the harmonic oscillator The functions ψn (x) are either odd or even, so that the density probability |ψn (x)|2 is even, hence x = 0. From this well defined parity of the ψn ’s, we also deduce p = 0, since the operator ∂/∂x changes the parity of the function. We have x  = Δx = 2

2

+∞

−∞

ψn∗ (x) x 2 ψn (x) dx.

Therefore, by applying (5.16), x 2  = Δx 2 = (n + 1/2) /(mω). In order to calculate p2 , we can just refer to the initial eigenvalue equation (5.9) and notice that p2 /2m + m ω 2 x 2 /2 = En , therefore p2  = Δp2 = (n + 1/2) mω. Altogether, one finds that in the eigenstate n: Δx Δp = (n + 1/2). For n = 0, Δx Δp = /2: the eigenfunction is a Gaussian and the Heisenberg inequality is saturated. 2. Time evolution of a one-dimensional harmonic oscillator The initial wavefunction of the system is given as a linear combination of the eigenfunctions of the Hamiltonian, which makes the calculations quite simple. a. The wave function at time t is: ψ(x, t) = cos θ φ0 (x) e−iωt/2 + sin θ φ1 (x) e−3iωt/2 . b. One deduces the expectation values: E = (cos2 θ + 3 sin2 θ) ω/2

E 2  = (cos2 θ + 9 sin2 θ) 2 ω 2 /4,

and the variance ΔE 2 = sin2 (2θ) 2 ω 2 /4. The expectation values of functions of the energy are all time independent as a consequence of energy conservation (Ehrenfest’s theorem, see Chap. 7).

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18 Solutions to the Exercises

c. For the position distribution, we get:  x =

 cos ωt sin 2θ 2mω

x 2  =

 (1 + 2 sin2 θ). 2mω

3. Three-dimensional harmonic oscillator a. The reasoning is similar to the one for the three dimensional square well, and we obtain: En1 ,n2 ,n3 = (n1 + n2 + n3 + 3/2) ω, where the ni ’s are non negative integers. The eigen-energies can therefore be written En = (n + 3/2) ω, where n is a non negative integer. The degeneracy gn of the level En is the number of triplets (n1 , n2 , n3 ) whose sum n1 + n2 + n3 is equal to n; one finds that gn = (n + 1)(n + 2)/2. b. If the φn ’s are the eigenfunctions of the one-dimensional harmonic oscillator with angular frequency ω, we have: Φn1 ,n2 ,n3 (r) = φn1 (x) φn2 (y) φn3 (z). c. If the oscillator is not isotropic, the energy levels read: En1 ,n2 ,n3 = (n1 + 1/2) ω1 + (n2 + 1/2) ω2 + (n3 + 1/2)ω3 , and the corresponding wavefunctions are: (2) (3) Φn1 ,n2 ,n3 (r) = φ(1) n1 (x) φn2 (y) φn3 (z),

where φ(i) n represents the nth eigenstate in the potential of frequency ωi . These energy levels are generally not degenerate, except if the ratio of two frequencies ωi and ωj is a rational number. 4. One-dimensional infinite potential well a. For symmetry reasons, we get x = a/2. For x 2 , we find: x 2  =

2 a



a

x 2 sin 0

therefore: Δx 2 =

nπx dx = a2 a



1 1 − 2 2 3 2n π

  a2 6 1− 2 2 . 12 n π

 ,

18 Solutions to the Exercises

465

b. We first normalize the wavefunction ψ(x) = Ax(a − x):

a

|ψ| dx = |A| 2

a

2

0

x 2 (a − x)2 dx = |A|2

0

a5 . 30

hence |A|2 = 30/a5 . The probabilities are pn = |αn |2 with: 



2 αn = A a

a 0

√ nπx 4 15 x(a − x) sin dx = 3 3 (1 − (−1)n ) . a n π

For symmetry reasons, αn = 0 if n is even. We find: ∞ 

∞

1 960  =1 π6 (2k + 1)6

pn =

n=1

k=0

1 2 π 2 960  52 = 2ma2 π 6 (2k + 1)4 ma2 4 4   1  π 960 304 E 2  = En2 pn = = 4m2 a4 π 6 (2k + 1)2 m 2 a4 √ 2 5 ΔE = . ma2 

E =

En pn =

c. We can calculate directly the average value and the variance of the kinetic energy operator. For the average value we recover the result: E = −

2 2m



a

d2ψ 2 a 2 dx = |A| dx 2 m

ψ(x)

0



a

x(a − x) dx =

0

52 . ma2

In order to calculate E 2  one meets a difficulty since, in the sense of functions (but not of distributions), d 4 ψ/dx 4 = 0. This would lead to the absurd result E 2  = 0 and to a negative variance σ 2 = E 2  − E2 . In fact, in this problem, the Hilbert space is defined as the C∞ periodic functions, of period a, which vanish at x = 0 and x = a. For such functions, one has:

a

φ1 (x)

0

d 4 φ2 dx = dx 4



a 0

d 2 φ1 d 2 φ2 dx dx 2 dx 2

for all pairs of functions φ1 (x), φ2 (x). This relation can serve as a definition of the operator Hˆ 2 in this problem. With this prescription, one obtains: E 2  =

4 4m2



a 0



∂2ψ ∂x 2

2 dx =

A2 4 a 304 = 2 4. 2 m m a

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18 Solutions to the Exercises

Actually, the correct general definition of the operator Eˆ K2 consists in using  the spectral representation Eˆ K2 ψ(x) ≡ K(x, y) ψ(y) dy where the kernel K is defined as: ∞ 2 2 E sin(nπx/a) sin(nπy/a). K(x, y) = a n=1 n This amounts to using the probabilities pn as above in the calculation of the expectation values of powers of the energy E q . 5. Isotropic states of the hydrogen atom a. The Schrödinger equation is: − Therefore: −

2  A ψ − ψ = Eψ. 2m x

  2 x 2 − + 2 e−x/a − Ae−x/a = E x e−x/a . 2m a a

Equating the terms in e−x/a , we obtain: A = 2 /ma

i.e.

a = 2 /mA = /mcα.

The terms in x e−x/a then give: E = −2 /2ma2 , i.e. E = −mc2 α2 /2. b. E = −13.6 eV, a = 5.3 × 10−2 nm. ∞ c. The condition 0 ψ 2dx = 1 entails C 2 a3 /4 = 1, i.e. C = 2/a3/2 . ∞ d. We find 1/x = C 2 0 x e−2x/a dx = 1/a and therefore: 1 V  = −A  = −mc2 α2 = 2E. x The expectation value p2 /2m can be calculated directly by noticing that in the state under consideration: E=

A p2 1 p2  −   therefore   = mc2 α2 = −E. 2m x 2m 2

We obtain the relation 2p2 /2m = +A/x, which is also true in classical mechanics if the averaging is performed on a closed orbit (virial theorem). 6. δ-function potentials a. (i) One obtains lim (ψ  (+ε) − ψ  (−ε)) = (2mα/2 )ψ(0). ε→0

18 Solutions to the Exercises

467 −1/2 −|x|/λ0

(ii) There is only one bound state, with ψ(x) = λ0 and E = −2 /(2mλ20 ). b. The wave function can be written:

e

. One has Kλ0 = 1

for x < −d/2 ψ(x) = B eK(x+d/2) −K(x+d/2) + C  eK(x−d/2) for − d/2 ≤ x ≤ +d/2 ψ(x) = C e for x > d/2 ψ(x) = B e−K(x−d/2) The solutions can be classified in terms of their symmetry with respect to x = 0: • Symmetric solution ψS : B = B , C = C  , • Antisymmetric solution ψA : B = −B , C = −C  . The quantization condition is (Kλ0 − 1)2 = (e−Kd )2 . The energy levels E± are therefore obtained by solving the equation Kλ0 = 1 ± e−Kd . E− corresponds to ψS and E+ to ψA . If λ0 > d, there is only one bound state ψS ; if λ0 < d, there are two bound states ψS and ψA . One can compare this with the model of the ammonia molecule, in particular if Kd  1. Chapter 5 1. Translation and rotation operators a. The proof is straightforward. We have: pˆ =

 d i dx

Therefore: Tˆ (x0 ) ψ(x) =

⇒ Tˆ (x0 ) = e−x0 dx . d

  ∞  (−x0 )n d n ψ(x), n! dx n=0

which is simply the Taylor expansion of ψ(x − x0 ) around point x. b. Similarly, one finds ˆ R(ϕ) ψ(r, θ) =

  ∞  (−ϕ)n ∂ n ψ(r, θ), n! ∂θ n=0

where we recognize the Taylor expansion of ψ(r, θ − ϕ). 2. The evolution operator The formula giving the derivative of the exponential of a function remains valid ˆ ) and Hˆ for the operator Uˆ since the Hamiltonian is time independent. Hence U(τ commute. We have therefore:

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18 Solutions to the Exercises

d ˆ i ˆ − t0 ). U(t − t0 ) = − Hˆ U(t dt  ˆ − t0 ) |ψ(t0 ) is a solution of the Schrödinger with the This entails that |ψ(t) = U(t ˆ ˆ The unitarity of Uˆ is a direct consequence proper initial condition since U(0) = 1. ˆ of the fact that H is Hermitian: ˆ ˆ ) = Uˆ −1 (τ ). Uˆ † (τ ) = eiHτ / = U(−τ

3. Heisenberg representation We have by definition: ˆ ˆ a(t) = ψ(t)|A|ψ(t) = ψ(0)|Uˆ † (t)Aˆ U(t)|ψ(0). ˆ = Uˆ † (t)Aˆ U(t), ˆ We can define A(t) i.e. ˆ ˆ ˆ = eiHt/ Aˆ e−iHt/ , A(t)

which is indeed a solution of the differential equation given in the text of the exercise ˆ and which is such that a(t) = ψ(0)|A(t)|ψ(0). 4. Dirac formalism with a two-state problem ˆ 1  = E1 ψ2 |ψ1  = E2 ψ2 |ψ1 , therefore (E1 − E2 )ψ2 |ψ1  = 0 and a. ψ2 |H|ψ ψ2 |ψ1  = 0. 2 b. E = (E1 + E2 )/2, ΔE 2 = ((E1 − E√ 2 )/2) , ΔE = ω/2. −iE1 t/ −iE2 t/ |ψ1  − e |ψ2  / 2. c. |ψ(t) = e d. a = ±1. √ e. |ψ±  = (|ψ1  ± |ψ2 ) / 2. f. p = |ψ− |ψ(t)|2 = cos2 (ωt/2). Chapter 6 1. Linear three-atom molecule a. The three energy levels and the corresponding eigenstates are: ⎛ ⎞ 1 1 ⎝ ⎠ 0 E1 = E0 , |ψ0  = √ 2 −1 and

⎛ ⎞ 1 √ 1⎝ √ ⎠ E± = E0 ± a 2, |ψ± = ∓ 2 . 2 1

18 Solutions to the Exercises

469

b. The probabilities are PL = PR = 1/4 and PC = 1/2. c. One has: 1 1 1 |ψL  = √ |ψ0  + |ψ+  + |ψ−  2 2 2 hence E = E0 and ΔE = a. 2. Crystallized Violet and Malachite Green a. The Hamiltonian is:

b.

c.

d. e.

⎛

⎞ 0 −A −A Hˆ = ⎝ −A 0 −A ⎠ . −A −A 0

As in the case of NH3 , the off-diagonal elements of the matrix H written in the basis of classical configurations represent the quantum effects, i.e., the passage by tunnelling from one configuration to the another one. ˆ 1  = −2A, E 2 1 = 4A2 , therefore ΔE 2 = 0 in the state |φ1 . E1 = φ1 |H|φ ˆ 2  = +A, E 2 2 = A2 , therefore ΔE 2 = 0 in the state Similarly, E2 = φ2 |H|φ |φ2 . Consequently, these two states are energy eigenstates with respective eigenvalues −2A and +A. Naturally, this can be seen directly by letting Hˆ act on these two states. ˆ it suffices to look for a vector orthogonal to both Knowing two eigenvectors of H, √ of them. One finds |φ3  = (2|1 − |2 − |3)/ 6. Altogether, √ we obtain: eigenvalue λ = −2A eigenvector |φ1  = (|1 + |2 +√|3)/ 3; eigenvalue λ = +A, eigenvector |φ2  = (|2 − |3)/ 2; √ eigenvalue λ = +A, eigenvector |φ3  = (2|1 − |2 − |3)/ 6. The eigenvalue λ = A has a degeneracy equal to 2. This eigenbasis is not unique. Absorption of light occurs with ΔE = 3A = 2.25 eV, in the yellow part of the spectrum. The ion appears with the complementary color, i.e. violet. The Hamiltonian is now: ⎛ ⎞ Δ −A −A Hˆ = ⎝ −A 0 −A ⎠ −A −A 0 and the eigenvalue equation can be written:   ˆ = −(λ − A) λ2 + (A − Δ)λ − A(Δ + 2A) , det(Hˆ − λI) hence the eigenvalues:   E0 = A, E± = (Δ − A) ± (Δ + A)2 + 8A2 /2. • If Δ  A : • If Δ  A :

E0 = A , E+ ∼ A + 2Δ/3 , E− ∼ −2A + Δ/3. E0 = A , E+ ∼ Δ , E− ∼ −A.

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18 Solutions to the Exercises

f. There are two possible transitions from the ground state E− : hν1 = hc/λ1 = hc/(450 nm) ∼ 2.75 eV =

(Δ + A)2 + 8A2

which corresponds to a value Δ = 1 eV, and an absorption in the violet part. With this value of Δ one obtains hν2 = E0 − E− = (3A − Δ)/2 + 1/2 (Δ + A)2 + 8A2 = 2 eV. This corresponds to an absorption in the red-orange part of the spectrum and to a wavelength λ2 = hc/hν2 = 620 nm in good agreement with experimental observation. Chapter 7 1. Commutator algebra The first relation is immediate. To show the second relation, one can start with: Aˆ Bˆ n − Bˆ n Aˆ =

Aˆ Bˆ n − Bˆ Aˆ Bˆ n−1 + Bˆ Aˆ Bˆ n−1 − Bˆ 2 Aˆ Bˆ n−2 + Bˆ 2 Aˆ Bˆ n−2 − . . . ˆ + Bˆ n−1 Aˆ Bˆ − Bˆ n A,

ˆ B] ˆ Bˆ n−s−1 with s = and each line of the right hand side can be written as Bˆ s [A, 0, . . . , n − 1, hence the result. Finally the Jacobi identity is derived by expanding all commutators, and checking that the twelve terms which appear cancel each other. 2. Classical equations of motion for the harmonic oscillator For the harmonic oscillator, the Ehrenfest theorem gives: p dx = dt m

dp = −mω 2 x dt

hence the result: d 2 x / dt 2 + ω 2 x = 0. 3. Conservation law For a system of n interacting particles, the total Hamiltonian reads: Hˆ =

n  pˆ 2i 1  + V (ˆri − rˆj ). 2mi 2 i j =i i=1

18 Solutions to the Exercises

471

 The total momentum of the system is Pˆ = i pˆ i . This operator commutes both with the kinetic energy term of the Hamiltonian and with the interaction term: [Pˆ x , V (ˆri − rˆj )] = [px,i , V (ˆri − rˆj )] + [px,j , V (ˆri − rˆj )]  ∂V  ∂V = (ˆri − rˆj ) − (ˆri − rˆj ) = 0. i ∂x i ∂x Therefore P is a constant of motion. 4. Hermite functions The expression of the position and momentum operators in terms of creation and annihilation operators are:  xˆ =

   aˆ + aˆ † 2mω

  pˆ = i mω/2 aˆ † − aˆ

hence the result. 5. Generalized uncertainty relations a. Applying the commutator to any function Ψ (r), one has for all Ψ : [ˆpx , fˆ ] Ψ (r) =

 i



∂Ψ ∂ (f Ψ ) − f ∂x ∂x



x = −i f  (r) Ψ (r) r

hence the relation [ˆpx , fˆ ] = −i(ˆx /ˆr ) f  (ˆr ). b. • The square of the norm of Aˆ x |ψ is:  Aˆ x |ψ 2 = ψ|(ˆpx + iλˆx fˆ )(ˆpx − iλˆx fˆ |ψ = ψ|ˆp2x + λ2 xˆ 2 fˆ 2 − iλ[ˆpx , xˆ fˆ ]|ψ. By a direct calculation, one has: [ˆpx , xˆ fˆ ] = −i(f +

x2  f ), r

 Aˆ x |ψ 2 = p2x  + λ2 x 2 f 2  − λf +

therefore

x2  f . r

• Adding the analogous relations for Aˆ y and Aˆ z , we obtain for any state |ψ p2  + λ2 r 2 f 2  − λ3f + rf   ≥ 0. This second degree trinomial in λ must be non-negative for all λ. Therefore, the discriminant is negative or zero, i.e. 4p2 r 2 f 2  ≥ 2 3f + rf  2 .

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18 Solutions to the Exercises

c. For f = 1, we obtain p2  r 2  ≥ (9/4)2 ; for f = 1/r, p2  ≥ 2 r −1 2 ; for f = 1/r 2 , p2  ≥ (2 /4) r −2 . d. Harmonic oscillator For any state |ψ, one has E ≥ 2 /(2mr 2 ) + mω 2 r 2 /2. Minimizing with respect to r 2 , we find the lower bound E ≥ (3/2)ω for the energy of the oscillator. Since there is a value of λ and a corresponding value of r 2 , for which the trinomial has a double root and vanishes, this means there exists a state for which this lower bound is attained. It is therefore the ground state since no state can have a lower energy than the ground state. We have Aˆ x |φ = 0 → (ˆpx + iλˆx )|φ = 0. In terms of wave functions, this corresponds to:  

 ∂ + λx φ(r) = 0, ∂x

and a similar equation for y and z. The solution of this set of three equations is: φ(r) = N exp(−λr 2 ) = N exp(−3r 2 /4r02 ), with r 2  = r02 , N −1 = (2πr02 /3)3/4 and λ = 3/4r02 . e. Hydrogen atom • Similarly, the lower bound for the energy of the hydrogen atom is Emin = −me e4 /(22 ), and this bound is attained, therefore it is the ground state energy. • The equation Aˆ x |ψ = 0 leads to the differential equation: 

x ∂ψ + λ ψ(r) = 0 ∂x r

and a similar equation for yand z. The solution of this equation is ψ(r) = N exp (−r/r0 ) with N = 1/ πr03 , r −1  = 1/r0 , and λ = 1/r0 . This is indeed the ground state of the hydrogen atom. 7. Time-energy uncertainty relation ˆ H]|ψ|/2. ˆ Uncertainty relations give Δa ΔE ≥ |ψ|[A, The Ehrenfest theorem gives ˆ H]|ψ ˆ da/dt = (1/i)ψ|[A, hence the inequality and the result. 8. Virial Theorem a. We have:

 2  pˆ ∂V ˆ ˆ ˆ [H, xˆ pˆ ] = [H, xˆ ]ˆp + xˆ [H, pˆ ] = i − + xˆ m ∂x

and, for the potential under consideration:

18 Solutions to the Exercises

473

 2  pˆ ˆ [H, xˆ pˆ ] = i − + nV (ˆx ) . m ˆ xˆ pˆ ]|ψα  = 0. Hence: b. For an eigenstate |ψα  of Hˆ one has ψα |[H, ψα | −

pˆ 2 + nV (ˆx )|ψα  = 0 m

⇒

2T  = nV .

The harmonic oscillator corresponds to the case n = 2: the average kinetic and the potential energies are equal when the system is prepared in an eigenstate |n of the Hamiltonian. c. We obtain in three dimensions:   2 p ˆ [H, r · p] = i − + r · ∇V (r) m therefore 2T  = nV  as obtained previously. This applies to the Coulomb problem (V (r) = −e2 /r). In this case, n = −1 and we have 2T  = −V . For the harmonic oscillator, one has T  = V . d. From the equation obtained above, we have for a central potential V (r): rˆ · ∇V (r) = r

∂V ∂r

hence 2T  = r

∂V . ∂r

Chapter 8 1. Calculations of energy levels Consider a particle of mass m placed in the isotropic 3D potential V (r) ∝ r β . We choose the normalized Gaussian test function: ψa (r) = (a/π)3/4 exp(−ar 2 /2). We find in this state: p2  =

3 2 a 2

r β  = a−β/2

Γ (3/2 + β/2) Γ (3/2)

This gives an upper bound for the ground state of: • The harmonic potential (β = 2) for which we recover the exact result. • The Coulomb potential V (r) = −e2 /r; one finds: E0 = −

4 me4 1 me4 to be compared with the exact result: − . 3π 2 2 2

• The linear potential V (r) = gr; one finds:

(E.1)

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18 Solutions to the Exercises

 E0 =

81 2π

1/3 

2 g 2 2m



1/3  2.345

2 g 2 2m

1/3

to be compared with the coefficient 2.338 of the exact result. 2. Uncertainty relations Using the inequality (9.24) for systems whose ground state is known, we can derive uncertainty relations between p2  and r α , where α is a given exponent. a. The r 2  p2  uncertainty relation. Consider a one-dimensional harmonic oscillator, whose ground state is ω/2. Whatever the state |ψ we have: ω p2  1 + mω 2 x 2  ≥ 2m 2 2

⇒

p2  + m2 ω 2 x 2  − mω ≥ 0.

We recognize a second degree polynomial inequality in the variable mω. The necessary and sufficient condition for this to hold for all values of mω is: p2 x 2  ≥

2 . 4

(E.2)

In three dimensions, noting r 2 = x 2 + y2 + z2 , we obtain in the same manner: p2 r 2  ≥

92 . 4

(E.3)

b. The 1/r p2  uncertainty relation. The hydrogen atom Hamiltonian is H = pˆ 2 /2m − e2 /ˆr and its ground state energy is E0 = −me4 /(22 ) (see Chap. 11). Consequently we have for all |ψ: p2  1 me4 − e2   ≥ − 2 . 2m r 2 We have again a second degree polynomial in the variable me2 which is always positive, from which we deduce: 1 p2  ≥ 2  2 . r

(E.4)

3. Comparison of the ground states of two potentials We denote |ψi  (i = 1, 2) the ground state of the Hamiltonian Hˆ = pˆ 2 /2m + Vi (r) with energy Ei . Since V2 (r) > V1 (r) for any r we have:

18 Solutions to the Exercises

475

ψ2 |V2 (ˆr)|ψ2  > ψ2 |V1 (ˆr)|ψ2  pˆ 2 pˆ 2 ⇒ E2 = ψ2 | + V2 (ˆr)|ψ2  > ψ2 | + V1 (ˆr)|ψ2 . 2m 2m The second step of the reasoning consists in noticing that, because of the theorem at the basis of the variational method, one has: ψ2 |

pˆ 2 pˆ 2 + V1 (ˆr)|ψ2  ≥ E1 = ψ1 | + V1 (ˆr)|ψ1 , 2m 2m

hence the result. 4. Existence of a bound state in a potential well To show the existence of a bound state in a one-dimensional potential V (x) which is negative everywhere and tends to zero at ±∞, we use the variational method with Gaussian trial functions: ψσ (x) = (σ 2 /(2π))1/4 exp(−σ 2 x 2 /4). The mean kinetic energy is Tσ = 2 σ 2 /(8m). This positive quantity tends to zero quadratically when σ tends to zero. The mean potential energy is: σ V σ = √ 2π



V (x) e−σ

x /2

2 2

dx.

By assumption this quantity  ∞ is negative. When σ tends to zero, it tends to zero linearly with σ if the integral −∞ V (x) dx converges, or it may even diverge if the integral is itself divergent. In any case, there exist a range of value for the variable σ such that |V σ | > Tσ . For these values of σ, the total mean energy Eσ = Tσ + V σ is negative. The ground state energy is necessarily lower than Eσ and it is also negative: the corresponding state is a bound state. This result cannot be extended to three dimensions. If one still takes Gaussian 2 2 functions e−σ r /4 as a trial set, the kinetic energy varies as σ 2 as before, while the negative potential energy term now scales as σ 3 . When σ tends to zero, the potential energy tends to zero faster than the kinetic energy, and the total energy may be always positive so that one cannot infer the existence of a bound state. Actually one can find simple three-dimensional potentials negative or zero everywhere which have no bound state. Consider for instance an isotropic square potential well: V (r) = −V0 for r < r0 (with V0 > 0) and V (r) = 0 otherwise. One finds that there is no bound state if V0 < 2 π 2 /(8mr02 ). (To show this result, consider first states with zero angular momentum, and then generalize to arbitrary angular momenta using the previous exercise).

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18 Solutions to the Exercises

5. Generalized Heisenberg inequalities For any state |ψ, we have: 1 2 p  + gr α  − ε0 g 2/(α+2) 2m



2 2m

α/(α+2) ≥ 0.

Minimizing with respect to g, one obtains the result mentioned in the text (it is safe to treat separately the case α > 0, g > 0 in which case ε0 > 0, and the case α < 0, g < 0 in which case ε0 < 0). One recovers the usual results for α = 2 and α = −1. For the linear potential, α = 1, we have ε0 = 2.33811 hence the uncertainty relation: p2 r2 ≥

4 3 2 ε  ∼ 1.894 2 . 27 0

Chapter 9 1. Rotation invariant operator ˆ Lˆ x ] = 0 and [A, ˆ Lˆ y ] = 0 ⇒ [A, ˆ [Lˆ x , Lˆ y ]] = i[A, ˆ Lˆ z ] = 0. [A, 2. Commutation relations for rˆ and pˆ A straightforward calculation gives: [Lˆ z , xˆ ] = −ˆy[ˆpx , xˆ ] = iˆy [Lˆ z , yˆ ] = xˆ [ˆpy , yˆ ] = −iˆx [Lˆ z , zˆ ] = 0, [Lˆ z , pˆ x ] = pˆ y [ˆx , pˆ x ] = iˆpy [Lˆ z , pˆ y ] = −ˆpx [ˆy, pˆ y ] = −iˆpx [Lˆ z , pˆ z ] = 0. ˆ B] ˆ B], ˆ Bˆ 2 ] = [A, ˆ Bˆ + B[ ˆ A, ˆ we obtain [L, ˆ pˆ 2 ] = [L, ˆ rˆ 2 ] = 0. Therefore, since [A, 3. Rotation invariant potential If V (r) depends only on r = |r|, then Hˆ = pˆ 2 /2m + V (ˆr ) (which is rotation invariant) commutes with the angular momentum Lˆ which is a constant of the motion (Ehrenfest theorem). ˆ = 0 ⇒ dL = 0. ˆ L] [H, dt This is not true if V (r) depends not only on r, but also on θ and ϕ.

18 Solutions to the Exercises

477

4. Unit angular momentum a. The action of Lˆ z on the basis set | = 1, m is Lˆ z |1, m = m |1, m: ⎛

⎞ 10 0 Lˆ z =  ⎝0 0 0 ⎠ . 0 0 −1 The action of Lˆ x and Lˆ y is obtained using the operators Lˆ + and Lˆ − , whose matrix elements are deduced from the recursion relation (10.16): √ √ Lˆ + |1, 1 = 0 Lˆ + |1, 0 =  2 |1, 1 Lˆ + |1, −1 =  2 |1, 0 √ √ Lˆ − |1, 1 = 2 |1, 0 Lˆ − |1, 0 =  2 |1, −1 Lˆ − |1, −1 = 0 Hence the matrix of Lˆ x = (Lˆ + + Lˆ − )/2 and Lˆ y = i(Lˆ − − Lˆ + )/2: ⎛ ⎞ 010  Lˆ x = √ ⎝1 0 1⎠ 2 010

⎛ ⎞ 0 −i 0  Lˆ y = √ ⎝ i 0 −i⎠ . 2 0 i 0

b. The eigenvectors of Lˆ x are:  √ 1 eigenvalues ±  |1, 1 ± 2|1, 0 + |1, −1 2 1 |1, 0x = √ (|1, 1 − |1, −1) eigenvalue 0. 2

|1, ±1x =

The eigenfunction corresponding to mx = +1 is: 1 1 1 (Y1 (θ, ϕ) + Y1−1 (θ, ϕ)) + √ Y10 (θ, ϕ) 2 2  3 = (cos θ − i sin θ sin ϕ). 8π

ψ(θ, ϕ) =

Therefore: I(θ, ϕ) = |ψ(θ, ϕ)|2 =

3 (1 − sin2 θ cos2 ϕ). 8π

5. Commutation relations for Jˆx2 , Jˆy2 and Jˆz2 a. Since [Jˆ 2 , Jˆz ] = 0, we have also [Jˆ 2 , Jˆz2 ] = 0. Hence [Jˆx2 + Jˆy2 , Jˆz2 ] = 0, and therefore [Jˆz2 , Jˆx2 ] = [Jˆy2 , Jˆz2 ]. The third equality is obtained using a circular permutation of x, y and z.

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18 Solutions to the Exercises

One can also calculate explicitly these commutators: [Jˆx2 , Jˆy2 ] = i{Jˆx , {Jˆy , Jˆz }} =

i ˆ ˆ ˆ ({Jx , {Jy , Jz }} + {Jˆy , {Jˆz , Jˆx }} + {Jˆz , {Jˆx , Jˆy }}). 3

ˆ B} ˆ ˆ = Aˆ Bˆ + Bˆ A. where we set {A, b. For j = 0 the result is obvious since |0, 0 is an eigenstate of all components with eigenvalue zero. For j = 1/2, Jˆx2 , Jˆy2 and Jˆz2 are proportional to the unit 2 × 2 matrix, with eigenvalue +2 /4. They obviously commute. c. For j = 1, we consider the matrix elements: 1, m2 |[Jˆx2 , Jˆz2 ]|1, m1  = (m12 − m22 )1, m2 |Jˆx2 |1, m1 . For m12 = m22 , this is obviously zero. We only have to consider the cases m1 = 0, m2 = ±1. Since Jˆx |1, 0 ∝ (|1, 1 + |1, −1) and Jˆx |1, ±1 ∝ |1, 0 the corresponding scalar products, i.e. the matrix elements under consideration, vanish. Owing to the (x, y, z) symmetry, the common eigenbasis is {|j = 1, mx = 0, |j = 1, my = 0, |j = 1, mz = 0} where |j = 1, mi = 0 is the eigenvector of Jˆi (i = x, y, z) associated with the eigenvalue 0: 1 |j = 1, mx = 0 = √ (|1, 1 − |1, −1) , 2 1 |j = 1, my = 0 = √ (|1, 1 + |1, −1) , 2 |j = 1, mz = 0 = |1, 0. In spherical coordinates, the corresponding (angular) wave functions are x/r, y/r and z/r, with a normalization coefficient (4π/3)−1/2 . Chapter 10 1. Expectation value ofr for the Coulomb problem ∞ a. Using 0 |un, (ρ)|2 dρ = 1 and ε = 1/n2 , we obtain the first identity. b. To show the second identity we use:

   ρ2 un, un, dρ = ρ un, un, dρ  dρ = −1/2 ρ2 un, un, dρ = −ρ,

 un, dρ = 0 un,  ρ un, un,



which can be shown by integrating by parts.

18 Solutions to the Exercises

479

c. Summing the two equations derived before, we obtain the desired expression for ρ. 2. Three-dimensional harmonic oscillator in spherical coordinates a. The radial equation is 

1 ( + 1) 2 2 1 d 2 r + mω 2 r 2 +  − En, − 2m r dr 2 2 2mr 2

 Rn, (r) = 0.

√ We set ρ = r mω/ and  = E/ω and we obtain (11.30). b. The energy levels are indeed of the form En = (n + 3/2)ω with n = 2n + , and the corresponding eigenstates can be labelled as |n, , m. There exists a degeneracy in , but it is different from the case of the hydrogen atom. For a given value of the energy, i.e. of n,  has the same parity as n. Therefore the successive levels correspond alternatively to even and odd angular momenta: n=0

E = 3ω/2

=0

n=1 n=2 n=3

E = 5ω/2 E = 7ω/2 E = 9ω/2

=1  = 0, 2  = 1, 3

etc.

For a given value of n, the (n + 1)(n + 2)/2 states |n, , m with ( = 0, 2 . . . n), or ( = 1, 3 . . . n), are linear combinations of the (n + 1)(n + 2)/2 states |n1 ; n2 ; n3  with n1 + n2 + n3 = n. c. For n = 1, three orthogonal wave functions are ϕ100 (r) = Cxe−αr

2

/2

, ϕ010 (r) = Cye−αr

2

/2

, ϕ001 (r) = Cze−αr

2

/2

where C is a normalization constant. We therefore obtain, by expressing x, y, z in terms of r and Y1,m (θ, ϕ): |n = 1,  = 1, m = 0 = |n1 = 0; n2 = 0; n3 = 1, 1 |n = 1,  = 1, m = ±1 = ∓ √ |n1 = 1; n2 = 0; n3 = 0 2 i − √ |n1 = 0; n2 = 1; n3 = 0. 2

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18 Solutions to the Exercises

3. Relation between the Coulomb problem and the harmonic oscillator a. The change of variable ρ → x and of unknown function u(ρ) → f (x) leads to the following equation for the Coulomb problem: d2f 2α − 1 df + + 2 dx x dx



 4E 2 α(α − 2) − 4( + 1) + 8Z + x f (x) = 0. x2 EI

The choice α = 1/2 eliminates the term df /dx and leads to an equation with the same structure as the radial equation for the harmonic oscillator: 

 (2 + 1/2)(2 + 3/2) 4E 2 d2 − + 8Z + x f (x) = 0. dx 2 x2 EI

b. The correspondence between the parameters of, respectively, the harmonic oscillator and the Coulomb problem, is: harm. ↔ (2coul. + 1/2)

K 2 ↔ −4Ecoul. /EI

Eharm. /ω ↔ 4Z.

In other words, the roles of the coupling constant and of the energy eigenvalue are interchanged! The shift in  ensures the proper -degeneracy of the hydrogen levels. c. From the result of exercise 2, we know that the eigenvalues for the energy are:   Eharm = K 2n + harm + 3/2 ω. Using the correspondence that we just found, this yields:  4Z =

 −4Ecoul.   2n + 2coul. + 2 EI

which can also be written Ecoul. = Z 2 EI /(n + coul. + 1)2 . We recover indeed the energy levels of the Coulomb problem. Notice that this provides an expression of the Laguerre polynomials in terms of the Hermite polynomials. 4. Confirm or invalidate the following assertions ˆ = 0 one has [H, ˆ Lˆ 2 ] = 0. We consider an ˆ L] ˆ Lˆ z ] = 0 and [H, a. True. In fact, if [H, 2 ˆ = 0 implies that ˆ L] ˆ Lˆ and Lˆ z , |E,m , , m. Since [H, eigenbasis common to H, ˆ Lˆ ± ] = 0, we obtain: [H, ˆ ,m , , m = E,m Lˆ ± |E,m , , m = Hˆ Lˆ ± |E,m , , m. Lˆ ± H|E

18 Solutions to the Exercises

481

We know that Lˆ ± |E,m , , m is an eigenstate of Lˆ z with the eigenvalue (m + 1). Therefore the set of states {|E,m , , m, m = − . . . } are eigenstates of Hˆ with the same eigenvalue E,m ≡ E . b. Wrong. The square of the angular momentum commutes with the Hamiltonian and the energy levels can be labelled with . It is only the Coulomb and harmonic potentials which have special symmetry properties which produce degeneracies in . 5. Centrifugal barrier effects Consider the Hamiltonians: pˆ 2 ( + 1)2 + V (r) where pˆ 2r = −2 Hˆ  = r + 2m 2mr 2



1 ∂ r r ∂r

2

which act only on the variable r. The state |n = 0, , m ≡ |ψ , which is an eigenstate of: Lˆ 2 pˆ 2 + V (r), Hˆ = r + 2m 2mr 2 is an eigenstate of Hˆ  with the eigenvalue E , which is the smallest eigenvalue of Hˆ  . We obviously have: ( + 1)2 . Hˆ +1 = H + mr 2 Taking the expectation value of this expression in the state |ψ+1  = |n = 0,  + 1, m, we obtain: ( + 1)2 . E+1 = ψ+1 |Hˆ  |ψ+1  +  mr 2 We have ψ|Hˆ  |ψ ≥ E for all ψ, and ( + 1)2 /mr 2 is a positive operator: its expectation value in any state is positive. Therefore, E+1 ≥ E , and more quantitatively: E+1 − E ≥

( + 1)2 1 ψ+1 | 2 |ψ+1 . m r

6. Algebraic method for the hydrogen atom + a. The expression for A−  A is: + A−  A =

d2 ( + 1) 2 1 − + − . 2 2 dρ ρ ρ ( + 1)2

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18 Solutions to the Exercises

Therefore the radial equation can be written as 

+ A−  A





1 u =  − ( + 1)2

 u .

− b. One finds for A+  A : − A+  A =

d2 ( + 1)( + 2) 2 1 − + − . 2 2 dρ ρ ρ ( + 1)2

By multiplying (11.33) by A+  , we obtain   + − + A A A u =  −

1 ( + 1)2



A+  u ,

which can also be written:   2 ( + 1)( + 2) 2 d + − + A+  u (ρ) = ε A u (ρ). dρ2 ρ2 ρ Therefore, A+  u (ρ) satisfies the radial equation with the same eigenvalue ε but for an angular momentum  =  + 1. c. Similarly one can write the equation for the angular momentum  as: 

− A+ −1 A−1





1 u =  − 2 

 u .

One then obtains that A− −1 u (ρ) satisfies the radial equation with the same eigenvalue ε but for an angular momentum  =  − 1. d. By multiplying (11.33) by u∗ and integrating over ρ we find: 0

∞

u∗ (ρ)





+ A−  A u (ρ)



1 dρ = ε − ( + 1)2



∞

|u (ρ)|2 dρ.

0

We integrate by parts the left hand side of this equation:

∞ 0

  + u∗ (ρ) A−  A u (ρ) dρ = −



∞ 0

2 |A+  u (ρ)| dρ

We deduce that the quantity ε − 1/( + 1)2 is necessarily negative: ε≤

1 . ( + 1)2

18 Solutions to the Exercises

483

e. The argument is then analogous to the case of the harmonic oscillator or to the quantization of angular momentum. By repeatedly applying A+  one can increase the value of  by an integer. This is limited from above since ε ≤ 1/( + 1)2 and there is a maximum value max of  such that: ε=

1 1 ≡ 2. 2 (max + 1) n

The function A+ max umax (ρ) is identically zero. Therefore, umax satisfies: 

 n 1 d − + umax (ρ) = 0. dρ ρ n

f. The energy levels are En = −EI /n2 . The solution of the above equation is umax (ρ) ∝ ρn exp−ρ/n , up to a normalization factor. Coming back to the func√ tion R , we recover (11.23) for  = n − 1 (i.e. n = 0): Rmax (ρ) ∝ ρn−1 e−ρ ε . By repeatedly applying A− −1 one obtains the other solutions of the same energy ε = 1/n2 for  = n − 2, n − 3, . . . , 0. 7. Molecular potential a. The radial equation is: 

( + 1)2 + 2me A B 2 1 d 2 r + − − 2me r dr 2 2me r 2 r

 R(r) = E R(r).

We define S as the positive real number such that S(S + 1) = ( + 1) + 2me A/2 , i.e.: 1 1 S=− + (2 + 1)2 + 8me A/2 . 2 2 Note that S is generally not an integer. We set as usual a1 = 2 /Bme , E = −εmB2 /22 and r = ρa1 . The radial equation becomes: 



S(S + 1) 2 1 d2 ρ− + −ε ρ dρ2 ρ2 ρ

R(ρ) = 0.

As in the case of hydrogen, the normalisable solutions are labelled by an integer √ n ≥ 0 and they are of the form: R(ρ) = e−ρ ε ρS Pn , (ρ), where Pn , (ρ) is a polynomial of degree n . and one must have ε = (n + S + 1)−2 . The energies are then: 1 B2 me . En , = − 2 2 (n + S + 1)2

484

18 Solutions to the Exercises

Note that this potential is quantitatively very different from a molecular potential even though it has the same global features (attractive at long distances and repulsive at short distances). The long range attractive force in a molecule is not a Coulomb force and the repulsion at short distances is much stronger than a r −2 potential. Chapter 11 1. Determination of the magnetic state of a silver atom a. It is always possible to write the coefficients α et β as α = cos(θ/2), β = eiϕ sin(θ/2) (if necessary, one can multiply the state (12.65) by a global phase factor in order to have a real value for α). Consider now a Stern–Gerlach magnet, which is oriented along the unit vector u, defined by the polar angles θ, ϕ: u = sin θ cos ϕ ex + sin θ sin ϕ ey + cos θ ez . This gives:

 u·μ ˆ = μ0

cos θ e−iϕ sin θ iϕ cos θ e sin θ

 .

ˆ One can check easily that cos(θ/2)|+ + eiϕ sin(θ/2)|− is an eigenstate of u · μ with the eigenvalue +μ0 . b. Bob has to choose an axis u for his Stern–Gerlach measurement, but he does not know the value of u. His measurement yields a binary answer ±μ0 , and the state of the system is then |±u . It is different from the initial state and subsequent measurements will not bring any new information on the state sent by Alice. Consequently, Bob cannot determine the initial state from a measurement performed on a single magnetic moment. The only certainty that Bob can have when he obtains the result +μ0 (resp. −μ0 ) in his measurement with the u axis is that the initial state was not |−u (resp. |+u ), which is a very poor information. c. If Alice sends a large number N of magnetic moments, all prepared in the same unknown state (8.67), Bob can split this ensemble into three sets. For the first N/3 magnetic moments, he measures μz . From the relative intensity of the two spots μz = ±μ0 , he deduces |α|2 and |β|2 . For the second set of N/3 magnetic moments, he measures μx . The relative intensities of the two spots corresponding to μx = ±μ0 yield |α ± β|2 . Finally Bob measures μy for the last set of N/3 magnetic moments, which yields |α ± iβ|2 . From these three sets of results, Bob deduces α and β, within a global phase factor which is unimportant. Of course, this determination of α and β is only approximate and the relative statistical error is N −1/2 . 2. Results of repeated measurements; quantum Zeno paradox a. The energy levels of Hˆ are E± = ±ω/2. b. The state |ψ(0) is an eigenstate of μˆ x with eigenvalues +μ0 . Therefore, one obtains +μ0 with probability 1 in a measurement of μx .

18 Solutions to the Exercises

485

c. The evolution of the state under consideration is: √ |ψ(T ) = (|+e−iωT /2 + |−eiωT /2 )/ 2. d. The corresponding probability is: P(T ) = |x +|ψ(T )|2 = |ψ(0)|ψ(T )|2 = cos2

ωT . 2

e. After a measurement giving the result μx =√ +μ0 , the system is again in the same state as initially |ψ(0) = (|+ + |−)/ 2. The probability for all the N successive measurements to give the same result +μ0 is therefore PN (T ) = cos2N (ωT /2N). f. In the (mathematical) limit N → ∞, we obtain     ωT PN = exp N ln cos2 2N    2 2  ω T ω2 T 2 ∼ exp 2N ln(1 − ) ∼ exp − → 1. 8N 2 4N This result may seem paradoxical: observing the system prevents it from evolving! Some people claim that watching water prevents it from boiling. However, the solution of the “Quantum Zeno Paradox” lies in the fact that any measurement has a finite extension both in space and in time. In practice, one cannot divide T in infinitely small parts except by interacting permanently with the system, which is another problem. 3. Products of Pauli matrices One first checks that σˆ j2 = 1. For j = k, a direct calculation yields the result (12.66). 4. Algebra with Pauli matrices  We first develop the products: (σ · A)(σ · B) = jk σj σk Aj Bk . The result (12.66) of the previous exercise then yields the desired formula. 5. Spin and orbital angular momentum ∞ a. One must have 0 |R(r)|2 r 2 dr = 1/2 since the Ym ’s are orthonormal and (|ψ+ |2 + |ψ− |2 ) d 3 r = 1. b. We find for Sz : p(+/2) = 2/3, p(−/2) = 1/3, and for Sx : p(+/2) = 1/3, p(−/2) = 2/3. c. Lz =  and Lz = 0 with p(+) = 1/6, p(0) = 5/6.

486

18 Solutions to the Exercises

Chapter 12 1. Permutation operator We use the eigenbasis of Sˆ 1z and Sˆ 2z . In this basis we find: 

 σˆ 1x σˆ 2x + σˆ 1y σˆ 2y |σ; σ = 0   σˆ 1x σˆ 2x + σˆ 1y σˆ 2y |σ; −σ = 2| − σ; σ   1 + σˆ 1z σˆ 2z |σ; σ = 2|σ; σ   1 + σˆ 1z σˆ 2z |σ; −σ = 0, where σ = ±1. Hence the result:  1 1 + σˆ 1 σˆ 2 |σ; σ = |σ; σ, 2

 1 1 + σˆ 1 σˆ 2 |σ; −σ = | − σ; σ. 2

2. The singlet state This invariance of the decomposition of the singlet state when u varies comes from the fact that the state is of angular momentum zero, and therefore that it is rotation invariant. 3. Spin and magnetic moment of the deuteron a. The eigenvalues of Kˆ 2 are K(K + 1)2 with: K = J + I, J + I − 1, J + I − 2, . . . , |J − I|. 2

ˆ 2 where we set A = agI gJ μB μN . One has Jˆ · Iˆ = (Kˆ − b. We have Wˆ = AJˆ · I/ 2 2 Jˆ − Iˆ )/2, hence the expression of W . c. EI,J,K = EI + EJ + A (K(K + 1) − J(J + 1) − I(I + 1)) /2. d. EI,J,K − EI,J,(K−1) = AK. e. The electron has a total angular momentum 1/2, its spin. The possible values of K are therefore I ± 1/2. One of the levels is split into 4 sub-levels (angular momentum 3/2), while the other is split in two (angular momentum 1/2). Therefore I = 1. f. Triplet state (S = 1). g. One obtains for the hyperfine splitting in deuterium ΔE =

1.72 4 3A = α me c2 ∼ 1.36 × 10−6 eV. 2 1836

Hence a wavelength and a frequency of the emitted radiation λ ∼ 91 cm , ν ∼ 328 MHz.

18 Solutions to the Exercises

487

The experimental value is λ = 91.5720 cm. In a more accurate calculation, one must incorporate a more accurate value of gI (0.8574), and take into account reduced mass effects and relativistic effects. Chapter 13 1. Identical particles on a beam splitter 2 2 a. The final wave packet must be √ normalized: |α| + |β| = 1. In addition the two final states (φ3 (r) + φ4 (r)) / 2 and αφ3 (r) + βφ4 (r) must be orthogonal since the two initial states φ1 (r)√and φ2 (r). This entails α + β = 0. We shall take in the following α = −β = 1/ 2. b. The initial state for two fermions reads:

1 |Ψ (ti ) = √ (|1 : φ1 ; 2 : φ2  − |1 : φ2 ; 2 : φ1 ) . 2 We neglect the interaction between the fermions when they cross the beam splitter. The final state is then obtained by linearity: 1 |Ψ (tf ) = √ (|1 : φ3  + |1 : φ4 ) ⊗ (|2 : φ3  − |2 : φ4 ) 2 2 1 − √ (|1 : φ3  − |1 : φ4 ) ⊗ (|2 : φ3  + |2 : φ4 ) . 2 2 It can also be written: 1 |Ψ (tf ) = √ (|1 : φ4 ; 2 : φ3  − |1 : φ3 ; 2 : φ4 ) . 2 The two fermions never come out on the same side of the beam splitter, which is direct consequence of the exclusion principle. c. The initial state for two bosons is: 1 |Ψ (ti ) = √ (|1 : φ1 ; 2 : φ2  + |1 : φ2 ; 2 : φ1 ) , 2 hence the final state: 1 |Ψ (tf ) = √ (|1 : φ3  + |1 : φ4 ) ⊗ (|2 : φ3  − |2 : φ4 ) 2 2 1 + √ (|1 : φ3  − |1 : φ4 ) ⊗ (|2 : φ3  + |2 : φ4 ) 2 2 1 = √ (|1 : φ3 ; 2 : φ3  + |1 : φ4 ; 2 : φ4 ) . 2

488

18 Solutions to the Exercises

The two bosons are always detected on the same side of the beam splitter. This surprising conclusion results from the destructive interference between the two quantum paths:   φ1 → φ4 φ1 → φ3 and , φ2 → φ4 φ2 → φ3 √ which would both lead to the final state (|1 : φ3 ; 2 : φ4  + |1 : φ4 ; 2 : φ3 )/ 2, corresponding to one boson in each output port. 2. Fermions in a square well a. The energy levels for the one body Hamiltonian are En = n2 E1 with E1 = π 2 2 /(2mL 2 ). The four lowest levels correspond to: • • • •

the state |1+, 1−, of energy 2 E1 , the four states |1±, 2±, of energy 5 E1 , the state |2+, 2−, of energy 8 E1 , the four states |1±, 3±, of energy 10 E1 .

b. One must diagonalize the restriction of the potential to each of the eigensubspaces found above. For the non degenerate subspaces, one just needs to calculate the matrix element V1 = α+, α − |Vˆ |α+, α−,, which gives after a simple algebra V1 = 3g/(2L). Each level |α+, α− is displaced by this amount. The case of the eigensubspaces Eα,β , spanned by the four vectors |α±, β± with α = β, is more subtle. The problem can be simplified by noticing that Vˆ does not affect the spin variables. The diagonalization of Vˆ inside Eα,β then amounts to three distinct eigenvalue problems: • The dimension 1 subspace corresponding to |α+, β+ is not coupled to the other vectors of Eα,β . A simple calculation yields: α+, β + |Vˆ |α+, β+ = 0. This energy level is not displaced by Vˆ . • The same conclusion is reached for |α−, β−. • The restriction of Vˆ to the two dimensional subspace spanned by |α+, β− and |α−, β+ reads:   g 1 −1 , L −1 1 whose eigenvalues are 2g/L and 0. To summarize, an energy level corresponding to a 4 dimension eigensubspace is split into two sublevels: one sublevel is 3 fold degenerate and it is not shifted by Vˆ , the other level is not degenerate and its shift is 2g/L.

18 Solutions to the Exercises

489

Chapter 14 1. The Lorentz force in quantum mechanics a. One checks easily that ∇ × A = B. b. The classical equations of motion m¨r = f and r˙ = v give: r¨ = (q/m)˙r × B. The Lorentz force does not produce any work. The energy boils down to the kinetic energy E = mv 2 /2, which is a constant of the motion. The longitudinal velocity, parallel to B, is constant. The transverse velocity, perpendicular to B, rotates around B with the angular velocity ω = −qB/m. The trajectory is a helix of axis z: the motion along z is uniform and the motion in the xy plane is circular. c. We have Aˆ x = −Bˆy/2, Aˆ y = Bˆx /2, Aˆ z = 0. Therefore: [ˆpx , Aˆ x ] = [ˆpy , Aˆ y ] = [ˆpz , Aˆ z ] = 0

⇒

ˆ p. pˆ .Aˆ = A.ˆ

The commutation relations between the components of uˆ are: [ˆux , uˆ y ] = [ˆpx − qAˆ x , pˆ y − qAˆ y ] = −q[ˆpx , Aˆ y ] − q[Aˆ x , pˆ y ] = iqB and [ˆux , uˆ z ] = [ˆuy , uˆ z ] = 0. ˆ Cˆ 2 ] = [B, ˆ C] ˆ Cˆ + C[ ˆ B, ˆ C]. ˆ Therefore: d. One has [B, ˆ = [ˆr, H]

i ˆ = i u, ˆ (ˆp − qA) m m

and, by applying the Ehrenfest theorem, 1 u d ˆ r = ψ|[ˆr, H]|ψ = . dt i m On the other hand, one calculates the following commutators: ˆ = [ˆux , H]

1 qB [ˆux , uˆ y2 ] = i uˆ y 2m m

ˆ = [ˆuy , H]

1 qB [ˆuy , uˆ x2 ] = −i uˆ x , 2m m

ˆ = 0. In other words: and [ˆuz , H] ˆ = i q uˆ × B ˆ H] [u, m

⇒

d q u = u × B. dt m

Altogether, we obtain: q d d2 r = r × B, 2 dt m dt which is identical to the classical equations of motion. Note that one recovers the classical equations of motion identically, and not approximately, because the

490

18 Solutions to the Exercises

Hamiltonian is a second degree polynomial in the dynamical variables. ˆ Setting vˆ = u/m, the observable vˆ corresponds to the velocity operator. In terms of that observable, the Hamiltonian can be written as Hˆ = mvˆ 2 /2 which is the indeed the kinetic energy. Therefore, in a magnetic field, the linear momentum mvˆ does not coincide with the conjugate momentum pˆ . These are related by ˆ which corresponds to the classical relation (15.28). mvˆ = pˆ − qA, e. The commutation relations between the components of uˆ show that in the presence of a magnetic field, the various components of the velocity cannot be defined simultaneously. One can define simultaneously the longitudinal component and one of the transverse components. The two transverse components satisfy the uncertainty relation:  |qB| . Δvx Δvy ≥ 2 m2 2. The Aharonov–Bohm effect a. We denote respectively D and D the distances OBC and OB C. If the vector potential is zero, the classical action corresponds to a uniform motion at velocity D/Δt (or D /Δt) between O and B (or B ) and then between B (or B ) and C: S0 =

mD2 2Δt

S0 =

mD 2 , 2Δt

with Δt = t2 − t1 . For a point C located at a distance x from the center of the screen O , we find D2 − D 2  2D0 ax/L, where D0 represents the distance OBO = OB O , a is the distance between the holes, and L the distance between the plane pierced by the two holes and the detection screen (we assume x  L). The quantity D0 /Δt represents the average velocity v of the particles and we set λ = h/(mv). We find:

2

  

|A(C)|2 ∝ eiS0 / + eiS0 / ∝ 1 + cos (S0 − S0 )/ = 1 + cos(2πx/xs ), which corresponds to the usual signal found in a Young double slit experiment, with the fringe spacing xs = λL/a. b. When a current flows in the solenoid, the vector potential is not zero anymore, and the classical action is changed because of the term q˙r · A(r) in the Lagrangian (15.27). The classical trajectories are not modified since no force acts on the particle. We have thus: q r˙ · A(r) dt

S = S0 + OBC

S  = S0 +

OB C

and the intensity in C is:   |A(C)|2 ∝ 1 + cos (S0 − S0 )/ + Φ

q r˙ · A(r) dt

18 Solutions to the Exercises

491

where the phase Φ reads:   q r˙ · A(r) dt − r˙ · A(r) dt Φ=  OBC OB C    q q = A(r) · dr − A(r) · dr = A(r) · dr.  OBC  OB C The last integral is calculated along the closed contour OBCB O. Its value does not depend on the position of C. It is equal to the flux of the magnetic field inside this contour, i. e. πr 2 B0 , where r is the radius of the solenoid and B0 the magnetic field inside the solenoid. Therefore the current induces a global shift of the interference pattern, corresponding to the phase change Φ = πr 2 B0 q/. Chapter 15 1. Excitation of an atom with broad band light a. The state vector of an atom reads at time t: α(t) |a + β(t) e−iω0 t |b. The evolutions of α and β are given by: Ω1 f (t) eiδt β(t) 2 Ω1 f (t) e−iδt α(t), β˙ = −iΩ1 f (t) cos(ωt) eiω0 t α(t)  −i 2

α˙ = −iΩ1 f (t) cos(ωt) e−iω0 t β(t)  −i

where we neglected the non resonant terms and define δ = ω − ω0 . The initial conditions are α(0) = 1, β(0) = 0. The solution of the second equation is, at lowest order in Ω1 : Ω1 τ β(τ ) = −i f (t) e−iδt dt. 2 −τ The bounds of the integral can be extended to ±∞ since f (t) is zero outside the interval [−τ , τ ]. This gives: nb = n |β(τ )|2 = n

π 2 Ω |g(−δ)|2 . 2 1

b. We generalize the preceding calculation and we obtain:

2





π nb = n Ω12 |g(−δ)|2

eiω0 tp

. 2

p=1



2

c. We calculate the statistical average of p=1 eiω0 tp :

492

18 Solutions to the Exercises

2



   

 iω t  0 p

e = eiω0 (tp −tp ) .

p=1

p=1 p =1 

Since the various times tp are uncorrelated, the statistical average of eiω0 (tp −tp ) is zero unless p = p , in which case this term is equal to 1. Therefore there are  ∼ γT non zero terms in this sum and we get: n¯ b (T ) = n

π 2 Ω |g(−δ)|2 γT . 2 1

The average number of atoms in state b increases linearly with time and we can define a transition rate from a to b: Γa→b =

π 2 Ω |g(−δ)|2 γ. 2 1

d. The energy contained in a wave packet is (0 c/2) E02 energy flux reads: Φ=

0 c 2 E γ 2 0

f 2 (t) dt =

0 c 2 E γ 2 0



f 2 (t) dt, and the mean



|g(Ω)|2 dΩ =

w(ω + Ω) dΩ,

where we used the Parseval–Plancherel equality. The function w(ω) is the energy spectral density. It is related to Γa→b by: Γa→b =

πd 2 w(ω0 ). 2 0 c3

e. The preceding reasoning can be transposed to the case where the atoms are initially in the state b. One defines in this way a transition rate from b to a: Γb→a = Γa→b . f. For this incoherent excitation, the evolution of the atom numbers na and nb is obtained by adding simply the two transition rates that we just found (this can be proven rigorously using the density operator formalism. n˙ a = −Γa→b na + Γb→a nb

n˙ b = Γa→b na − Γb→a nb .

The steady state is simply na = nb = n/2 since Γb→a = Γa→b . 2. Atoms in equilibrium with black-body radiation Within the framework of the previous exercise, we predict that the atom numbers na and nb are equal. On the opposite, Statistical Physics imposes the result nb /na = exp(−ω0 /kB T ). Indeed one knows that a system (here the atomic assembly) in contact with a thermostat at temperature T (the black body radiation) must reach a thermodynamical equilibrium characterized by the same temperature T .

18 Solutions to the Exercises

493

Einstein’s hypothesis consists in adding a second decay process from level b to level a, which creates a asymmetry between the populations of these levels. Suppose  . The evolutions of the atom that this second process is characterized by the rate Γb→a numbers na and nb are now:  n˙ a = −Γa→b na + (Γb→a + Γb→a )nb

 n˙ b = Γa→b na − (Γb→a + Γb→a )nb ,

and the equilibrium state is: nb Γa→b =  na Γb→a + Γb→a

avec

Γa→b = Γb→a .

Let us impose the value exp(−ω0 /kB T ) to this ratio. This leads to the value of the  : rate Γb→a    Γb→a = Γb→a eω0 /kB T − 1 . We have obtained in the previous exercise the relation between Γb→a and the energy spectral density w(ω). For the case of black body radiation, this relation entails:  = Γb→a

 ω0 /kB T  πd 2 πd 2 w(ω ) e − 1 = μ . 0 2 0 c3 2 0 c3

 is independent of temperature, hence of the state of the We find that the rate Γb→a electromagnetic field. It corresponds to the spontaneous emission process described qualitatively in Chap. 16, Sect. 16.3. Einstein’s reasoning gives an account for two important characteristics of the spontaneous emission rate. It is proportional to the square of the average dipole d of the transition, and it varies as the cube of the Bohr frequency ω0 of this transition (cf. Eq. (16.24)).

3. Ramsey fringes The state vector of the neutron is γ+ (t) e−iω0 t/2 |+ + γ− (t) eiω0 t/2 |−. Inside the cavities, the evolutions of the coefficients γ± are given by: iγ˙ + =

ω1 i(ω0 −ω)t γ− (t) e 2

iγ˙ − =

ω1 i(ω−ω0 )t γ+ (t). e 2

At the entrance of the first cavity, one has γ+ (0) = 1 and γ− (0) = 0. At the exit of the first cavity (t1 = L/v), γ− (t1 ) is given by: γ− (t1 ) =

  ω1 1 − ei(ω−ω0 )t1 , 2(ω − ω0 )

where we restrict to first order term in B1 . The coefficient γ− does not evolve anymore until the neutron enters the second cavity at time T = D/v. The evolution equation

494

18 Solutions to the Exercises

of γ− during the crossing of the second cavity can be integrated similarly and we obtain finally: γ− (T + t1 ) =

    ω1 1 − ei(ω−ω0 )t1 1 + ei(ω−ω0 )T , 2(ω − ω0 )

hence the spin flip probability: P+→− =

ω12 sin2 ((ω − ω0 )t1 /2) cos2 ((ω − ω0 )T /2). (ω − ω0 )2

When one varies ω around the resonance frequency ω0 , one gets a resonance with a width ∼ π/T . This is much narrower than the width that one would obtain with a single cavity (∼ π/t1 ). This setup, which allows to accurately measure the resonance frequency ω0 , is currently used in metrology and high resolution spectroscopy.

Index

A Abasov, A.I. et al., 181 Abdurashitov, J.N. et al., 181 Absorption, 170 Action, 386, 398 Addition of angular momenta, 313 Adjoint operators, 126 Aharonov, Y., 395, 398, 490 Ahmad, Q.R. et al., 182 Alley, 167 Ammonia molecule, 92 Andromeda, 335 Ångström, A.J., 269 Angular distributions, 229 Angular momentum spin, 397 Anharmonic oscillator, 223, 481 Annihilation operator, 194, 273 Anomalous Zeeman effect, 250, 297 Anselmann, P. et al., 181 Antineutrino, 171 Antiparticle, 171 Antisymmetric state, 351 Approximation Born, 405 electric dipole, 408 Wigner–Weisskopf, 417 Aspect, A., 52, 441 Atmospheric neutrinos, 174 Atom cold, 356 helium, 254, 352 hydrogen, 202, 260 Atomic clock, 165 Avogadro’s number, 12

B Balmer, J., 265, 269 Balmer series, 265 Beam splitter, 360, 487 Bell basis, 454 Bell, John, 51 Bell state, 454 Benzene molecule, 208 Bernoulli, D., 385, 388 Bernoulli sequence, 31 β decay, 171 Bienaymé-Tchebycheff (inequality), 32 Binomial law (probability), 30 Black body, 424, 492 Black hole, 382 Bloch, F., 306 B mesons, 157 Bohm, David, 51, 398, 435, 490 Bohr frequency, 410 magneton, 12, 248, 289, 308 radius, 12, 262 Bohr–Einstein controversy, 422 Bohr, N., 17, 254, 269, 422 Bohr–Sommerfeld theory, 254 Boltzmann constant, 12 Bormann, E., 295 Born approximation, 405 Born, M., 44, 137, 201 Bose–Einstein condensate, 356 Bose–Einstein condensation, 355 Boson, 349 Bottom quark, 271 Bound state, 84, 228 Broad band field, 423, 491 Broglie (de), Louis, 19, 44, 392 Broglie (de), waves, 26

© Springer International Publishing Switzerland 2016 J.-L. Basdevant, Lectures on Quantum Mechanics, Graduate Texts in Physics, DOI 10.1007/978-3-319-43479-7

495

496 Bruno, Giordano, 282 C Cambridge, 185 Canonical commutation relation, 392 Canonical equation, 390 Cartan, Elie, 231, 309 Cartwheel galaxy, 337 Cascade (atomic), 441 Castin, Y., 356 Cauchy, A., 222 Cayley, 392 Central potential, 256 Centrifugal barrier, 257 Chandrasekhar (mass), 371 Classical configurations, 153 Clauser, J.F., 52, 441 Clebsch–Gordan coefficient, 315 Closure relation, 128 Coherent state, 204, 205 Cold atoms, 7, 356 Collision, 404 Commutation relation, 183, 392 CO molecule, 241 Complete set of commuting observables (CSCO), 197 Completely prepared quantum state, 198 Composite materials, 11 Compton wavelength, 12, 262 Conjugate momentum, 389, 394 Conservation of angular momentum, 256 of energy, 190, 422 of momentum, 255 of probability, 190 Conservation laws, 190 Constant of motion, 389 Constant perturbation, 406 Contact interaction, 326 Contact potential, 361 Continuous spectrum, 135 Continuum of final states, 413 Copernicus, 388 Cornell, E., 356 Correspondence principle, 68, 293, 392 Coulomb interaction, 260 Creation operator, 194, 273 Cryptography, 446 D d’Alembert, 385, 388 Dalibard, J., 52

Index Darwin term, 324 Davis, R., 182 Davisson, C., 20 Decay, 413 Decoherence, 81, 453 Degeneracy, 129 Density of states, 414 photon, 415 Descartes, R., 388 Destruction operator, 273 Deuterium, 341 Diamagnetic term, 397 Diatomic molecule, 239 Dirac equation, 324 function, 53 Dirac formalism, 124 Dirac, P.A.M., 119, 183, 348, 385, 392 Directional quantization, 295 Dispersion, 38 Distribution theory, 55 DONUT, 271 Double well, 92 Dye molecule, 169, 469

E Ehrenfest, P., 188, 189, 239, 296 Ehrenfest theorem, 190, 249, 290, 390, 392 Eigenfunction, 72 Eigenstate, 72 Eigenvalue, 72, 73, 129 Eigenvector, 129 Eigler, D., 102 Einstein, Albert, 6, 16, 296, 407, 422, 424 Einstein–Podolsky–Rosen, 51, 434 Electric dipole approximation, 408 Electric dipole transition, 259, 331 selection rule, 411 Electromagnetic transition, 407 Electron, 12 Electron gas, 355 Electron magnetic moment, 308 Elementary magneton, 300 Emission spontaneous, 407, 411 stimulated (or Induced), 357, 407 Energy, 74, 389 conservation, 422 Fermi, 353 width of a level, 419 Energy levels, 83 Energy quantization, 84

Index Entangled state, 434 Entanglement, 434 EPR paradox, 51 Equation canonical, 390 Lagrange, 387 Equilibrium thermodynamic, 424, 492 Error function, 32 Euler, 385, 388 Evolution operator, 144, 467 Exchange force, 352 Exchange operator, 346 Exclusion principle, 298, 350 Expectation value, 38

F Fermat, 386, 388 Fermi energy, 353 golden rule, 415 hyperfine structure, 328 Fermi, E., 348 Fermion, 349 Ferromagnetism, 353 Feynman path integrals, 398 Feynman, R.P., 16, 45, 52 Fine structure constant, 12 monovalent atom, 323 Fine structure constant, 261, 325, 417 First order perturbation theory, 221 Flavor, 174 Flux quantum, 399 Force exchange, 352 Lorentz, 393 Fourier transformation, 42, 58, 65 Fraunhofer, 268 Freedman, S.J., 52, 441 Fry, E.S., 441 Fukuda, Y. et al., 181 Fullerene, 242 Function Dirac, 53 Heaviside, 57

G GaAs, 88 Galileo, 282, 388 GALLEX Collaboration, 181

497 Garcia, C.P., 5 Gauge invariance, 394 Gaussian law (probability), 30 Gaussian wave packet, 53 Gedanken experiment, 70 Geometric law (probability), 32 Germer, L., 20 Glycine, 244 Göttingen, 137 GPS, 165 Grangier, Ph., 52 Grassmann, 392 Gravitational catastrophe, 371 Greeberger, D.M., 443 Ground state, 91, 202, 224 Gyromagnetic ratio, 246, 307

H Hale–Bopp comet, 244 Hall effect (quantum), 400 Hamilton, 385, 392 Hamiltonian, 75, 390 Hamilton–Jacobi, 390 Harmonic oscillator, 85 quasi-classical state, 204, 205 three dimensional, 105, 272, 464, 479 Haroche, S., 82 Heaviside function, 57 Heisenberg inequality, 227, 228, 474, 476 pressure, 50 representation, 144 uncertainty relations, 46 Heisenberg, W., 46, 136, 201, 358, 369 Helium atom, 254, 352, 361 Herbig, G., 333 Hermite functions, 122, 202 Hermitian operators, 127 Hidden variable, 438 Hilbert, D., 119 Hilbert space, 120 Horne, M., 443 Huggins, W., 269 Huygens probe, 166 Hydrogen 21-cm line, 333 Hydrogen atom, 105, 202, 260 spectrum, 264 Hydrogen maser, 328 Hyperfine structure, 325

498 I Identical particles, 343 Incoherent field, 423, 491 Indistinguishability, 344 Induced emission, 407 Inequality (Heisenberg), 227, 474 Interaction contact, 326 hyperfine, 325 spin-orbit, 323 Interstellar medium, 331 Interstellar molecules, 241 Invariance Gauge, 394 rotation, 250, 476 Inversion of the ammonia molecule, 98

J Jacobi, C.G.J., 200 Jacobi identity, 201 Jian-Wei Pan, 443 Jordan, P., 201

K K2K collaboration, 181 KamLAND collaboration, 173, 181 Ketterle, W., 356 Klein, Felix, 75, 392 Koshiba, Masatoshi, 172, 182

L Lagrange, 222, 385, 386, 388 equation, 387 Lagrangian, 386 Laguerre, 262 Lamb shift, 324, 419 Landau, L., 382, 422 Landau level, 399, 400 Langevin, P., 44 Laplace, P., 222 Laplacian, 57 Larmor frequency, 288 Larmor precession, 247, 249, 288, 301 Laser, 358 Least action principle, 386 Leibniz, 345, 386 LEP, 171, 271 Leptons, 8, 171 Lie algebra, 231, 309 group, 231, 309

Index Lie, Sophus, 231 Lifetime, 413 atomic level, 412 Line (21 cm), 325 Linear momentum, 394 Linear operators, 126 Logical gate (quantum), 453 Lorentz force, 393 Lorentzian line shape, 419 Lyman series, 265 M M81, 242, 338 M82, 242, 338 Magnetic dipole transition, 331 Magnetic moment, 245, 287 spin, 285 Magnetic resonance, 300, 302 nuclear, 306 Magnetism, 245, 300 Magneton Bohr, 248 Maser, 333 hydrogen, 328 Mass Boson star, 376, 384 Chandrasekhar, 371 neutron star, 374, 381 Matrices, 138 Maupertuis, 386, 388 McDonald, Arthur B., 172, 182 Mean square deviation, 32 Measurement, 291 repeated, 310, 484 Meson, 171 Messiah, Albert, 442 Metal, 355 Microlasers, 5 Milky Way, 331, 335 Miller, Stanley, 244 Moebius, 309 Molecular binding, 101 Molecule, 169, 468, 484 Benzene, 208 Momentum, 45 conjugate, 389, 394 linear, 394 Momentum conservation, 255 Momentum observable, 67 Muon, 171, 269 Muonic atom, 269 Muonium, 326 Murchison, 244

Index N Nanotechnologies, 6, 102 Neutrino, 170, 269, 381, 413 Neutrino masses, 172 Neutrino oscillations, 147, 170 Neutron, 12 Neutron magnetic moment, 308 Neutron star, 373, 380 Newton, I., 386, 388 NGC3077, 242, 338 Nuclear magnetic resonance, 306 magnetism, 300 magneton, 12, 308 Nucleus unstable, 413, 417

O Observable, 66, 151 Operator annihilation (or destruction), 273 creation, 273 Hermitian, 127 Laplacian, 57 rotation, 143, 208, 467 self-ajoint, 127 translation, 143, 467 Oppenheimer, 382 Optical lattice clocks, 329 Origin of life, 244 Orion nebula, 242, 243 Oscillation lengths, 175 Oscillator, 164 anharmonic, 223

P Paramagnetic term, 397 Parseval-Plancherel (theorem), 59 Particle in a box, 91 Pauli effect, 295 Hamiltonian, 397 matrices, 282 verbot, 298 Pauli principle, 298, 343, 361 Pauli, W., 139, 280, 348, 361 Peierls, R., 422 Pellegrini, V., 5 Penzias, A., 164 Perl, M., 271 Permutation, 351

499 Perturbation constant, 406 sinusoidal, 406 time dependent, 403 time independent, 219 Photoelectric effect, 410 Photon, 140, 357, 410 density of states, 415 Physical quantities, 35, 66 Physical system, 35 Pillars of Creation, 340 Planck constant, 12 Planck, M., 7, 16 Poisson, 222 Poisson bracket, 391 Poisson distribution, 205 Poisson law (probability), 32 Polarization of light, 140 Polarization of the photon, 157 Polaroid, 141 Population inversion, 160 Position observable, 67 Positronium, 326 Potential anharmonic, 223, 481 contact, 361 delta function, 106, 466 harmonic, 105 linear, 53, 461 scalar, 393, 408 vector, 393, 398, 408 Potential barriers, 101 Prebiotic chemistry, 244 Principal quantum number, 259 Principle correspondence, 392 of quantization, 134 of spectral decomposition, 134 of wave packet reduction, 134 symmetrization, 350 Probabilities (Notions on), 27 Probability, 130 density, 28 Probability amplitudes, 26, 64 Proton, 12 Proton magnetic moment, 308 Purcell, E., 306 Pythagorean theorem, 123

Q Quantization of energy, 83 Quantization rule, 392

500 Quantum bit (or q-bit), 446, 451 Quantum computer, 451 Quantum Hall effect, 400 Quantum logic, 142 Quantum superposition, 81 Quark, 8, 271 Quasi-classical state, 204, 205 Quaternions, 392

R Rabi experiment, 304 formula, 304 oscillations, 162, 304 Rabi, I., 271, 300 Radial quantum number, 258 Radial wave function, 257 Radiation, 407 Radioactivity, 413 Radius Bohr, 12 Ramsauer effect, 106, 107 Ramsey fringes, 424, 493 Random variable, 29 Reduced mass, 254 Reduced wave function, 257 Reduction of the wave packet, 73 Relative motion, 254 Relativistic effect, 324, 328 Restriction of an operator, 222 Result, 64 Riesz, F., 125, 131 Roger, G., 52 Rotating field, 302 Rotation invariance, 256 Rotation operator, 143, 208, 467 Rotation spectrum, 229 Rutherford, 269 Rydberg, 269 Rydberg constant, 12

S SAGE collaboration, 181 Saint Augustine, 76 Scalar potential, 393, 408 Scanning tunneling microscope, 102 Scattering, 84 Scattering state, 84 Schrödinger equation, 27, 38, 134 Schrödinger’s cat, 78 Schrödinger, E., 44, 119

Index Second order perturbation theory, 223 Secular equations, 222 Selection rule, 259, 411 Self-ajoint operators, 127 Shimony, A., 443 Shor, Peter, 452 Simon, T., 333 Singlet state, 318, 347, 435 Sinusoidal perturbation, 406 Slater determinant, 351 Smart materials, 11 Smy, M.B., 182 Snell’s laws, 388 SO(3) group, 309 Sodium yellow line, 323 Sodium spectrum, 259 Solvay congress, 50, 71 Sommerfeld, A., 254 Space curved, 388 Spectral decomposition, 132 theorem, 131 Spectroscopic notation, 259 Spherical coordinates, 256 Spherical harmonics, 237 Spin, 279 total, 316 Spin 1/2, 157, 229, 397 Spin 1/2 particle, 283 Spin magnetic moment, 285 Spin-orbit interaction, 323 Spin-statistics connection, 349 Spinor, 309 Spiral arms, 339 Spontaneous emission, 160, 407, 411 atom, 417 nucleus, 417 Square integrable functions, 121 Stability of matter, 48 Standard model, 8 Star neutron, 373, 380 white dwarf, 372, 378 State, 35 coherent (or quasi-classical), 204, 205 singlet, 318, 347, 435 stationary, 421 triplet, 318, 347 unstable, 413 State variable, 386 Stationary states, 77, 190, 421

Index Sterile neutrino, 180 Stern–Gerlach experiment, 287 Stern, O., 306 Stimulated emission, 161, 357, 407 Strange neutral mesons, 157 Strange quark, 271 Structure fine, 323 hyperfine, 325 SU(2) group, 309 Super-Kamiokande, 175 Supernovae, 172 Superposition principle, 40, 133 Suzuki, A., 174 Symmetric state, 351 Symmetrization principle, 350 Symmetry, 89 System, 35

T τ lepton, 171 Teleportation, 454 Tensor product, 196, 315 Theorem non cloning, 448 Parseval-Plancherel, 59 virial, 203 Thermodynamic equilibrium, 424, 492 Thomas precession, 299 Thompson, R.C., 441 Three-dimensional harmonic oscillator, 105, 464 Time dependent perturbation, 403 Time-energy uncertainty relation, 160, 203, 420 Top quark, 271 Total spin, 316 Transition electric dipole, 331 electromagnetic, 407 magnetic dipole, 331 Transition probability, 357, 403 Translation invariance, 255 Translation operator, 143, 467 Triplet state, 318, 347 Tunnel effect, 83, 94 Two-state system, 148

U Uncertainty relation, 152, 186, 227, 474 fermions, 358, 369

501 time-energy, 203, 420 Unitary operator, 144 Unstable state, 413

V Vacuum polarization, 328 Variable random, 29 Variance, 32 Variational method, 224 Vector potential, 393, 398, 408 Verifications of general relativity, 167 Vessot, 167 Virial theorem, 203, 267 VLBI, 333 Volkov, 382

W Wave function, 37 mechanics, 35 packet, 41 packet reduction, 73 Wavelength Compton, 12 Wave packet Gaussian, 53 spreading, 52 Waves de Broglie, 26 Weak interactions, 171 Weiss, P., 300 White dwarf, 372, 378 Wiemann, Carl E., 356 Wigner–Weisskopf approximation, 417 Wilson, K., 261 Wilson, R.W., 164 Wineland, D.J., 82

Y Young (Interferences), 396, 398

Z Zeeman effect (anomalous), 249, 297, 329, 342 Zeilinger, A., 20, 443 Zeno paradox, 310, 484 Zero point energy, 86, 104