Quantum Mechanics I: The Fundamentals 1482263378, 9781482263374
Quantum Mechanics I: The Fundamentals provides a graduate-level account of the behavior of matter and energy at the mole
382 58 4MB
English Pages 613 [611] Year 2014
Front Cover
Dedication
Contents
Preface
About the Authors
CHAPTER 1 - Why Was Quantum Mechanics Developed?
CHAPTER 2 - Schrödinger Equation and Wave Function
CHAPTER 3 - Operators, Eigenvalues and Eigenfunctions
CHAPTER 4 - Exactly Solvable Systems I: Bound States
CHAPTER 5 - Exactly Solvable Systems II: Scattering States
CHAPTER 6 - Matrix Mechanics
CHAPTER 7 - Various Pictures and Density Matrix
CHAPTER 8 - Heisenberg Uncertainty Principle
CHAPTER 9 - Momentum Representation
CHAPTER 10 - Wave Packet
CHAPTER 11 - Theory of Angular Momentum
CHAPTER 12 - Hydrogen Atom
CHAPTER 13 - Approximation Methods I: Time-Independent Perturbation Theory
CHAPTER 14 - Approximation Methods II: Time-Dependent Perturbation Theory
CHAPTER 15 - Approximation Methods III: WKB and Asymptotic Methods
CHAPTER 16 - Approximation Methods IV: Variational Approach
CHAPTER 17 - Scattering Theory
CHAPTER 18 - Identical Particles
CHAPTER 19 - Relativistic Quantum Theory
CHAPTER 20 - Mysteries in Quantum Mechanics
CHAPTER 21 - Numerical Methods for Quantum Mechanics
APPENDIX A - Calculation of Numerical Values of h and kB
APPENDIX B - A Derivation of the Factor hν/(ehν/kBT − 1)
APPENDIX C - Bose’s Derivation of Planck’s Law
APPENDIX D - Distinction Between Self-Adjoint and Hermitian Operators
APPENDIX E - Proof of Schwarz’s Inequality
APPENDIX F - Eigenvalues of a Symmetric Tridiagonal Matrix—QL Method
APPENDIX G - Random Number Generators for Desired Distributions
Solutions to Selected Exercises
Back Cover
Recommend Papers
![Quantum Mechanics I. The Fundamentals [2 ed.]
2022021033, 9780367769987, 9780367776367, 9781003172178, 9780367770006, 9780367776428, 9781003172192](https://ebin.pub/img/200x200/quantum-mechanics-i-the-fundamentals-2nbsped-2022021033-9780367769987-9780367776367-9781003172178-9780367770006-9780367776428-9781003172192.jpg)
File loading please wait...
Citation preview
Physics
“… these two books by Rajasekar and Velusamy will definitely tell you how to do quantum mechanics.” —Dr. K.P.N. Murthy, Professor, School of Physics, and Director, Centre for Integrated Studies, University of Hyderabad Quantum Mechanics I: The Fundamentals provides a graduate-level account of the behavior of matter and energy at the molecular, atomic, nuclear, and sub-nuclear levels. It covers basic concepts, mathematical formalism, and applications to physically important systems. The text addresses many topics not typically found in books at this level, including:
A collection of problems at the end of each chapter develops readers’ understanding of both basic concepts and the application of theory to various physically important systems. This book, along with the authors’ follow-up Quantum Mechanics II: Advanced Topics, provides readers with a broad, up-to-date introduction to quantum mechanics.
• Access online or download to your smartphone, tablet or PC/Mac • Search the full text of this and other titles you own • Make and share notes and highlights • Copy and paste text and figures for use in your own documents • Customize your view by changing font size and layout
K24364
Quantum Mechanics I The Fundamentals
Quantum Mechanics I
Bound state solutions of quantum pendulum Pöschl–Teller potential Solutions of classical counterpart of quantum mechanical systems A criterion for bound state Scattering from a locally periodic potential and reflection-less potential Modified Heisenberg relation Wave packet revival and its dynamics Hydrogen atom in D-dimension Alternate perturbation theories An asymptotic method for slowly varying potentials Klein paradox, Einstein-Podolsky-Rosen (EPR) paradox, and Bell’s theorem Numerical methods for quantum systems
The Fundamentals
• • • • • • • • • • • •
WITH VITALSOURCE ® EBOOK
Rajasekar • Velusamy
“Excellent, up-to-date… Quantum Mechanics I: The Fundamentals covers the canonical basics and Quantum Mechanics II: Advanced Topics covers a range of modern developments…I recommend this set highly.” Dr. Jonathan P. Dowling, Hearne Professor of Theoretical Physics, Louisiana State University
S. Rajasekar R. Velusamy
w w w. c rc p r e s s . c o m
K24364_cover.indd 1
11/5/14 1:23 PM
Quantum Mechanics I
Quantum Mechanics I The Fundamentals
S. Rajasekar R. Velusamy
CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2015 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20141118 International Standard Book Number-13: 978-1-4822-6338-1 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
To our wives.
Contents Preface
xvii
About the Authors
xxi
Chapter
1 Why Was Quantum Mechanics Developed?
1
1.1
INTRODUCTION
1
1.2
BLACK BODY RADIATION
2
1.3
PHOTOELECTRIC EFFECT
6
1.4
HYDROGEN SPECTRUM
10
1.5
FRANCK–HERTZ EXPERIMENT
12
1.6
STERN–GERLACH EXPERIMENT
14
1.7
CORRESPONDENCE PRINCIPLE
15
1.8
COMPTON EFFECT
17
1.9
SPECIFIC HEAT CAPACITY
21
1.10
DE BROGLIE WAVES
22
1.11
PARTICLE DIFFRACTION
23
1.12
WAVE-PARTICLE DUALITY
24
1.13
CONCLUDING REMARKS
27
1.14
BIBLIOGRAPHY
28
1.15
EXERCISES
29
Chapter
2 Schrödinger Equation and Wave Function
31
2.1
INTRODUCTION
31
2.2
CONSTRUCTION OF SCHRÖDINGER EQUATION
32
2.3
SOLUTION OF TIME-DEPENDENT EQUATION
34
∗
2.4
PHYSICAL INTERPRETATION OF ψ ψ
35
2.5
CONDITIONS ON ALLOWED WAVE FUNCTIONS
36
2.6
BOX NORMALIZATION
38 vii
viii Contents
2.7
CONSERVATION OF PROBABILITY
39
2.8
EXPECTATION VALUE
41
2.9
EHRENFEST’S THEOREM
45
2.10
BASIC POSTULATES
50
2.11
TIME EVOLUTION OF STATIONARY STATES
51
2.12
CONDITIONS FOR ALLOWED TRANSITIONS
52
2.13
ORTHOGONALITY OF TWO STATES
53
2.14
PHASE OF THE WAVE FUNCTION
55
2.15
CLASSICAL LIMIT OF QUANTUM MECHANICS
56
2.16
CONCLUDING REMARKS
58
2.17
BIBLIOGRAPHY
59
2.18
EXERCISES
60
Chapter
3 Operators, Eigenvalues and Eigenfunctions
65
3.1
INTRODUCTION
65
3.2
LINEAR OPERATORS
66
3.3
COMMUTING AND NONCOMMUTING OPERATORS
69
3.4
SELF-ADJOINT AND HERMITIAN OPERATORS
73
3.5
DISCRETE AND CONTINUOUS EIGENVALUES
76
3.6
MEANING OF EIGENVALUES AND EIGENFUNCTIONS
78
3.7
PARITY OPERATOR
80
3.8
ALL HERMITIAN HAMILTONIANS HAVE PARITY
83
3.9
SOME OTHER USEFUL OPERATORS
84
3.10
CONCLUDING REMARKS
85
3.11
BIBLIOGRAPHY
85
3.12
EXERCISES
87
Chapter
4 Exactly Solvable Systems I: Bound States
91
4.1
INTRODUCTION
91
4.2
CLASSICAL PROBABILITY DISTRIBUTION
92
4.3
FREE PARTICLE
93
4.4
HARMONIC OSCILLATOR
4.5 4.6
94 2k
PARTICLE IN THE POTENTIAL V (x) = x , k = 1, 2, · · · 105 PARTICLE IN A BOX
107
Contents ix
4.7
PÖSCHL–TELLER POTENTIALS
114
4.8
QUANTUM PENDULUM
115
4.9
CRITERIA FOR THE EXISTENCE OF A BOUND STATE
117
4.10
TIME-DEPENDENT HARMONIC OSCILLATOR
120
4.11
RIGID ROTATOR
121
4.12
CONCLUDING REMARKS
126
4.13
BIBLIOGRAPHY
127
4.14
EXERCISES
129
Chapter
5 Exactly Solvable Systems II: Scattering States 135
5.1
INTRODUCTION
135
5.2
POTENTIAL BARRIER: TUNNEL EFFECT
136
5.3
FINITE SQUARE-WELL POTENTIAL
143
5.4
POTENTIAL STEP
149
5.5
LOCALLY PERIODIC POTENTIAL
153
5.6
REFLECTIONLESS POTENTIALS
159
5.7
DYNAMICAL TUNNELING
162
5.8
CONCLUDING REMARKS
162
5.9
BIBLIOGRAPHY
163
5.10
EXERCISES
164
Chapter
6 Matrix Mechanics
167
6.1
INTRODUCTION
167
6.2
LINEAR VECTOR SPACE
168
6.3
MATRIX REPRESENTATION OF OPERATORS AND WAVE FUNCTION 170
6.4
UNITARY TRANSFORMATION
172
6.5
TENSOR PRODUCTS
174
6.6
SCHRÖDINGER EQUATION AND OTHER QUANTITIES IN MATRIX FORM 174
6.7
APPLICATION TO CERTAIN SYSTEMS
176
6.8
DIRAC’S BRA AND KET NOTATIONS
179
6.9
EXAMPLES OF BASIS IN QUANTUM THEORY
180
6.10
PROPERTIES OF KET AND BRA VECTORS
181
6.11
HILBERT SPACE
183
x Contents
6.12
PROJECTION AND DISPLACEMENT OPERATORS
184
6.13
CONCLUDING REMARKS
187
6.14
BIBLIOGRAPHY
188
6.15
EXERCISES
189
Chapter
7 Various Pictures and Density Matrix
191
7.1
INTRODUCTION
191
7.2
SCHRÖDINGER PICTURE
191
7.3
HEISENBERG PICTURE
195
7.4
INTERACTION PICTURE
198
7.5
COMPARISON OF THREE REPRESENTATIONS
199
7.6
DENSITY MATRIX FOR A SINGLE SYSTEM
199
7.7
DENSITY MATRIX FOR AN ENSEMBLE
202
7.8
TIME EVOLUTION OF DENSITY OPERATOR
204
7.9
A SPIN-1/2 SYSTEM
205
7.10
CONCLUDING REMARKS
206
7.11
BIBLIOGRAPHY
206
7.12
EXERCISES
207
Chapter
8 Heisenberg Uncertainty Principle
209
8.1
INTRODUCTION
209
8.2
THE CLASSICAL UNCERTAINTY RELATION
210
8.3
HEISENBERG UNCERTAINTY RELATION
211
8.4
IMPLICATIONS OF UNCERTAINTY RELATION
215
8.5
ILLUSTRATION OF UNCERTAINTY RELATION
217
8.6
THE MODIFIED HEISENBERG RELATION
220
8.7
CONCLUDING REMARKS
223
8.8
BIBLIOGRAPHY
224
8.9
EXERCISES
224
Chapter
9 Momentum Representation
227
9.1
INTRODUCTION
227
9.2
MOMENTUM EIGENFUNCTIONS
228
9.3
SCHRÖDINGER EQUATION
230
9.4
EXPRESSIONS FOR hXi AND hpi
231
Contents xi
9.5
TRANSFORMATION BETWEEN MOMENTUM AND COORDINATE REPRESENTATIONS 233
9.6
OPERATORS IN MOMENTUM REPRESENTATION
234
9.7
MOMENTUM FUNCTION OF SOME SYSTEMS
235
9.8
CONCLUDING REMARKS
238
9.9
BIBLIOGRAPHY
238
9.10
EXERCISES
239
Chapter
10 Wave Packet
241
10.1
INTRODUCTION
241
10.2
PHASE AND GROUP VELOCITIES
241
10.3
WAVE PACKETS AND UNCERTAINTY PRINCIPLE
245
10.4
GAUSSIAN WAVE PACKET
247
10.5
WAVE PACKET REVIVAL
254
10.6
ALMOST PERIODIC WAVE PACKETS
255
10.7
CONCLUDING REMARKS
256
10.8
BIBLIOGRAPHY
257
10.9
EXERCISES
259
Chapter
11 Theory of Angular Momentum
261
11.1
INTRODUCTION
261
11.2
SCALAR WAVE FUNCTION UNDER ROTATIONS
262
11.3
ORBITAL ANGULAR MOMENTUM
266
11.4
2
EIGENPAIRS OF L AND Lz
267 2
11.5
PROPERTIES OF COMPONENTS OF L AND L
268
11.6
EIGENSPECTRA THROUGH COMMUTATION RELATIONS
272
11.7 11.8 11.9
2
MATRIX REPRESENTATION OF L , Lz AND L±
276
WHAT IS SPIN?
278
SPIN STATES OF AN ELECTRON
283
11.10 SPIN-ORBIT COUPLING
285
11.11 ROTATIONAL TRANSFORMATION
286
11.12 ADDITION OF ANGULAR MOMENTA
288
11.13 ROTATIONAL PROPERTIES OF OPERATORS
294
11.14 TENSOR OPERATORS
295
xii Contents
11.15 THE WIGNER–ECKART THEOREM
298
11.16 CONCLUDING REMARKS
299
11.17 BIBLIOGRAPHY
300
11.18 EXERCISES
301
Chapter
12 Hydrogen Atom
305
12.1
INTRODUCTION
305
12.2
HYDROGEN ATOM IN THREE-DIMENSION
305
12.3
HYDROGEN ATOM IN D -DIMENSION
320
12.4
FIELD PRODUCED BY A HYDROGEN ATOM
321
12.5
SYSTEM IN PARABOLIC COORDINATES
324
12.6
CONCLUDING REMARKS
327
12.7
BIBLIOGRAPHY
327
12.8
EXERCISES
328
Chapter
13 Approximation Methods I: Time-Independent Perturbation Theory
331
13.1
INTRODUCTION
331
13.2
THEORY FOR NONDEGENERATE CASE
332
13.3
APPLICATIONS TO NONDEGENERATE LEVELS
339
13.4
THEORY FOR DEGENERATE LEVELS
342
13.5
FIRST-ORDER STARK EFFECT IN HYDROGEN
345
13.6
ALTERNATE PERTURBATION THEORIES
350
13.7
CONCLUDING REMARKS
353
13.8
BIBLIOGRAPHY
353
13.9
EXERCISES
354
Chapter
14 Approximation Methods II: Time-Dependent Perturbation Theory
359
14.1
INTRODUCTION
359
14.2
TRANSITION PROBABILITY
359
14.3
CONSTANT PERTURBATION
361
14.4
HARMONIC PERTURBATION
365
14.5
ADIABATIC PERTURBATION
369
14.6
SUDDEN APPROXIMATION
371
Contents xiii
14.7
THE SEMICLASSICAL THEORY OF RADIATION
373
14.8
CALCULATION OF EINSTEIN COEFFICIENTS
375
14.9
CONCLUDING REMARKS
376
14.10 BIBLIOGRAPHY
377
14.11 EXERCISES
377
Chapter
15 Approximation Methods III: WKB and Asymptotic Methods
381
15.1
INTRODUCTION
381
15.2
PRINCIPLE OF WKB METHOD
381
15.3
APPLICATIONS OF WKB METHOD
385
15.4
WKB QUANTIZATION WITH PERTURBATION
392
15.5
AN ASYMPTOTIC METHOD
394
15.6
CONCLUDING REMARKS
396
15.7
BIBLIOGRAPHY
396
15.8
EXERCISES
397
Chapter
16 Approximation Methods IV: Variational Approach
399
16.1
INTRODUCTION
399
16.2
CALCULATION OF GROUND STATE ENERGY
399
16.3
TRIAL EIGENFUNCTIONS FOR EXCITED STATES
404
16.4
APPLICATION TO HYDROGEN MOLECULE
407
16.5
HYDROGEN MOLECULE ION
410
16.6
CONCLUDING REMARKS
413
16.7
EXERCISES
413
Chapter
17 Scattering Theory
415
17.1
INTRODUCTION
415
17.2
CLASSICAL SCATTERING CROSS-SECTION
417
17.3
CENTRE OF MASS AND LABORATORY COORDINATES SYSTEMS
418
17.4
SCATTERING AMPLITUDE
422
17.5
GREEN’S FUNCTION APPROACH
424
17.6
BORN APPROXIMATION
427
xiv Contents
17.7
PARTIAL WAVE ANALYSIS
431
17.8
SCATTERING FROM A SQUARE-WELL SYSTEM
437
17.9
PHASE-SHIFT OF ONE-DIMENSIONAL CASE
440
17.10 INELASTIC SCATTERING
441
17.11 CONCLUDING REMARKS
441
17.12 BIBLIOGRAPHY
442
17.13 EXERCISES
443
Chapter
18 Identical Particles
445
18.1
INTRODUCTION
445
18.2
PERMUTATION SYMMETRY
446
18.3
SYMMETRIC AND ANTISYMMETRIC WAVE FUNCTIONS
449
18.4
THE EXCLUSION PRINCIPLE
451
18.5
SPIN EIGENFUNCTIONS OF TWO ELECTRONS
452
18.6
EXCHANGE INTERACTION
454
18.7
EXCITED STATES OF THE HELIUM ATOM
456
18.8
COLLISIONS BETWEEN IDENTICAL PARTICLES
458
18.9
CONCLUDING REMARKS
461
18.10 BIBLIOGRAPHY
461
18.11 EXERCISES
461
Chapter
19 Relativistic Quantum Theory
463
19.1
INTRODUCTION
463
19.2
KLEIN–GORDON EQUATION
464
19.3
DIRAC EQUATION FOR A FREE PARTICLE
468
19.4
NEGATIVE ENERGY STATES
476
19.5
JITTERY MOTION OF A FREE PARTICLE
477
19.6
SPIN OF A DIRAC PARTICLE
480
19.7
PARTICLE IN A POTENTIAL
482
19.8
KLEIN PARADOX
483
19.9
RELATIVISTIC PARTICLE IN A BOX
487
19.10 RELATIVISTIC HYDROGEN ATOM
490
19.11 THE ELECTRON IN A FIELD
492
19.12 SPIN-ORBIT ENERGY
493
Contents xv
19.13 CONCLUDING REMARKS
495
19.14 BIBLIOGRAPHY
495
19.15 EXERCISES
497
Chapter
20 Mysteries in Quantum Mechanics
499
20.1
INTRODUCTION
499
20.2
THE COLLAPSE OF THE WAVE FUNCTION
500
20.3
EINSTEIN–PODOLSKY–ROSEN (EPR) PARADOX
501
20.4
HIDDEN VARIABLES
506
20.5
THE PARADOX OF SCHRÖDINGER’S CAT
507
20.6
BELL’S THEOREM
508
20.7
VIOLATION OF BELL’S THEOREM
510
20.8
RESOLVING EPR PARADOX
515
20.9
CONCLUDING REMARKS
516
20.10 BIBLIOGRAPHY
516
20.11 EXERCISES
518
Chapter
21 Numerical Methods for Quantum Mechanics
519
21.1
INTRODUCTION
21.2
MATRIX METHOD FOR COMPUTING STATIONARY STATE SOLUTIONS 521
21.3
FINITE-DIFFERENCE TIME-DOMAIN METHOD
529
21.4
TIME-DEPENDENT SCHRÖDINGER EQUATION
535
21.5
QUANTUM SCATTERING
544
21.6
ELECTRONIC DISTRIBUTION OF HYDROGEN ATOM
548
21.7
SCHRÖDINGER EQUATION WITH AN EXTERNAL FIELD 551
21.8
CONCLUDING REMARKS
555
21.9
BIBLIOGRAPHY
557
21.10 EXERCISES
519
558
Appendix
A Calculation of Numerical Values of h and kB
561
Appendix
B A Derivation of the Factor hν/(ehν/kB T − 1)
563
xvi Contents
Appendix
C Bose’s Derivation of Planck’s Law
Appendix
D Distinction Between Self-Adjoint and Hermitian Operators 567
Appendix
E Proof of Schwarz’s Inequality
569
Appendix
F Eigenvalues of a Symmetric Tridiagonal Matrix—QL Method
571
G Random Number Generators for Desired Distributions
575
Appendix
565
Solutions to Selected Exercises
579
Index
587
Preface Quantum mechanics is the study of the behaviour of matter and energy at the molecular, atomic, nuclear levels and even at sub-nuclear level. This book is intended to provide a broad introduction to fundamental and advanced topics of quantum mechanics. Volume I is devoted to basic concepts, mathematical formalism and application to physically important systems. Volume II covers most of the advanced topics of current research interest in quantum mechanics. Both the volumes are primarily developed as texts at the graduate level and also as reference books. In addition to worked-out examples, numerous collection of problems are included at the end of each chapter. Solutions are available to confirmed instructors upon request to the publisher. Some of the problems serve as a mode of understanding and highlighting the significances of basic concepts while others form application of theory to various physically important systems/problems. Developments made in recent years on various mathematical treatments, theoretical methods, their applications and experimental observations are pointed out wherever necessary and possible and moreover they are quoted with references so that readers can refer them for more details. Volume I consists of 21 chapters and 7 appendices. Chapter 1 summarizes the needs for the quantum theory and its early development (old quantum theory). Chapters 2 and 3 provide the basic mathematical framework of quantum mechanics. Schrödinger wave mechanics and operator formalism are introduced in these chapters. Chapters 4 and 5 are concerned with the analytical solutions of bound states and scattering states respectively of certain physically important microscopic systems. The basics of matrix mechanics, Dirac’s notation of state vectors and Hilbert space are elucidated in chapter 6. The next chapter gives the Schrödinger, Heisenberg and interaction pictures of time evolution of quantum mechanical systems. Description of time evolution of ensembles by means of density matrix is also described. Chapter 8 is concerned with Heisenberg’s uncertainty principle. A brief account of wave function in momentum space and wave packet dynamics are presented in chapters 9 and 10, respectively. Theory of angular momentum is covered in chapter 11. Chapter 12 is devoted exclusively to the theory of hydrogen atom. Chapters 13 through 16 are mainly concerned with approximation methods such as time-independent and time-dependent perturbation theories, WKB method and variational method. The elementary theory of elastic scattering is presented in chapter 17. Identical particles are treated in chapter 18. The xvii
xviii Preface next chapter presents quantum theory of relativistic particles with specific emphasize on Klein–Gordon equation, Dirac equation and its solution for a free particle, particle in a box (Klein paradox) and hydrogen atom. Chapter 20 examines the strange consequences of role of measurement through the paradoxes of EPR and a thought experiment of Schrödinger. A brief sketch of Bell’s inequality and the quantum mechanical examples violating it are given. Considering the rapid growth of numerical techniques in solving physical problems and significances of simulation studies in describing complex phenomena, the final chapter is devoted for a detailed description of numerical computation of bound state eigenvalues and eigenfunctions, transmission and reflection probabilities of scattering potentials, transition probabilities of quantum systems in the presence of external fields and electronic distribution of atoms. Some supplementary and background materials are presented in the appendices. The pedagogic features volume I of the book, which are not usually found in textbooks at this level, are the presentation of bound state solutions of quantum pendulum, Pöschl–Teller potential, solutions of classical counter part of quantum mechanical systems considered, criterion for bound state, scattering from a locally periodic potential and reflectionless potential, modified Heisenberg relation, wave packet revival and and its dynamics, hydrogen atom in Ddimension, alternate perturbation theories, an asymptotic method for slowly varying potentials, Klein paradox, EPR paradox, Bell’s theorem and numerical methods for quantum systems. The volume II consists of 10 chapters. Chapter 1 describes the basic ideas of both classical and quantum field theories. Quantization of Klein–Gordon equation and Dirac field are given. The formulation of quantum mechanics in terms of path integrals is presented in chapter 2. Application of it to free particle and linear harmonic oscillator are considered. In chapter 3 some illustrations and interpretation of supersymmetric potentials and partners are presented. A simple general procedure to construct all the supersymmetric partners of a given quantum mechanical systems with bound states is described. The method is then applied to a few interesting systems. The next chapter is concerned with coherent and squeezed states. Construction of these state and their characteristic properties are enumerated. Chapter 5 is devoted to Berry’s phase, Aharonov–Bohm and Sagnac effects. Their origin, properties, effects and experimental demonstration are presented. The features of Wigner distribution function are elucidated in chapter 6. In a few decades time, it is possible to realize a computer built in terms of real quantum systems that operate in quantum mechanical regime. There is a growing interest on quantum computing. Basic aspects of quantum computing is presented in chapter 7. Deutsch–Jozsa algorithm of finding whether a function is constant or not, Grover’s search algorithm and Shor’s efficient quantum algorithm for integer factorization and evaluation of discrete logarithms are described. Chapter 8 deals with quantum cryptography. Basic principles of classical cryptography and quantum cryptography and features of a few quantum cryptographic
Preface xix systems are discussed. A brief introduction to other advanced topics such as quantum gravity, quantum Zeno effect, quantum teleportation, quantum games, quantum cloning, quantum diffusion and quantum chaos is presented in chapter 9. The last chapter gives features of some of the recent technological applications of quantum mechanics. Particularly, promising applications of quantum mechanics in ghost imaging, detection of weak amplitude objects, entangled two-photon microscopy, detection of small displacements, lithography, metrology and teleportation of optical images are briefly discussed. During the preparation of this book we have received great supports from many colleagues, students and friends. In particular, we are grateful to Prof. N. Arunachalam, Prof. K.P.N. Murthy, Prof. M. Daniel, Dr. S. Sivakumar, Mr. S. Kanmani, Dr. V. Chinnathambi, Dr. P. Philominathan, Dr. K. Murali, Dr. S.V.M. Sathyanarayana, Dr. K. Thamilmaran, Dr.T. Arivudainambi and Dr.V.S. Nagarathinam for their suggestions and encouragement. It is a great pleasure to thank Dr. V.M. Gandhimathi, Dr. V. Ravichandran, Dr. S. Jeyakumari, Dr. G. Sakthivel, Dr. M. Santhiah, Mr.R. Arun, Mr. C. Jeevarathinam, Mr. R. Jothimurugan, Ms. K. Abirami and Ms. S. Rajamani for typesetting some of the chapters. We thank the senior publishing editor Luna Han and Michael Davidson, production editor at Taylor & Francis for various suggestions and careful editing of the manuscript, and their team members for smooth handling of the publication process. Finally, we thank our family members for their unflinching support, cooperation and encouragement during the course of preparation of this work. Tiruchirapalli May, 2014
S. Rajasekar R. Velusamy
About the Authors Shanmuganathan Rajasekar was born in Thoothukudi, Tamilnadu, India in 1962. He received his B.Sc. and M.Sc. in Physics both from the St. Joseph’s College, Tiruchirapalli. In 1987, he received his M.Phil. in Physics from Bharathidasan University, Tiruchirapalli. He was awarded Ph.D. degree in Physics (Nonlinear Dynamics) from Bharathidasan University in 1992 under the supervision of Prof.M. Lakshmanan. He was a visiting scientist during 1992-93 at the Materials Science Division, Indira Gandhi Centre for Atomic Research, Kalpakkam and worked on multifractals and diffusion under Prof.K.P.N. Murthy. In 1993, he joined as a Lecturer at the Department of Physics, Manonmaniam Sundaranar University, Tirunelveli. In 2003, the book on Nonlinear Dynamics: Integrability, Chaos and Patterns written by Prof.M. Lakshmanan and the author was published by Springer. In 2005, he joined as a Professor at the School of Physics, Bharathidasan University. With Prof.M. Daniel he edited a book on Nonlinear Dynamics published by Narosa Publishing House in 2009. His recent research focuses on nonlinear dynamics with a special emphasize on nonlinear resonances. He has authored or coauthored more than 80 research papers in nonlinear dynamics. Ramiah Velusamy was born in Srivilliputhur, Tamilnadu, India in the year 1952. He received his B.Sc. degree in Physics from the Ayya Nadar Janaki Ammal College, Sivakasi in 1972 and M.Sc. in Physics from the P.S.G. Arts and Science College, Coimbatore in 1974. He worked as a demonstrater in the Department of Physics in P.S.G. Arts and Science College during 1974-77. He received M.S. Degree in Electrical Engineering at Indian Institute of Technology, Chennai in the year 1981. In the same year, he joined in the Ayya Nadar Janaki Ammal College as an Assistant Professor in Physics. He was awarded M.Phil. degree in Physics in the year 1988. He retired in the year 2010. His research topics are quantum confined systems and wave packet dynamics.
xxi
CHAPTER
1
Why Was Quantum Mechanics Developed?
1.1 INTRODUCTION Quantum mechanics is the study of matter and radiation at an atomic level. It is a mathematical tool for predicting the behaviors of microscopic particles1 . Classical physics produced satisfactory results for systems that are much larger than atoms and moving with much slower than the speed of light. But it failed for the systems with small size (atomic size) or with speed near that of light. In the early 20th century scientists observed that results of certain experiments were not satisfactorily explained by classical physics. Particularly, it failed to account the observations in the following phenomena. 1. Black body radiation: In the typical experiments with a black body the energy emitted per unit time and per unit wavelength at a temperature increases with increase in wavelength (λ), reaches a maximum and then decreases. Classical theory gave incorrect variation of emitted energy. 2. Photoelectric effect: Electrons emission from a light illuminated metal surface happens only above a threshold frequency of the incident light. This and other associated experimental results were not accounted by the classical theory. 3. Spectrum of hydrogen: Hydrogen atoms are found to emit radiations of certain wavelengths alone. But classical theory predicted emission of radiations of all wavelengths. 1 The microscopic scale is one where the lengths are of the order of several angstroms at the most. Macroscopic scale is the one where the length is not less than, say, 10−4 cm.
1
2 Quantum Mechanics I: The Fundamentals 4. X-ray scattering: The experimentally observed results with X-ray scattering have not been satisfactorily interpreted by the electromagnetic wave theory. 5. Variation of heat capacity at low temperatures: Classically, the atomic heat of solids is 3R, a universal constant. In experiments it decreases and approaches to zero as temperature T approaches 0K. In classical physics energy is a continuous variable. To obtain a radiation formula that fit the experimental data, Max Karl Ludwig Planck, in 1901, proposed a revolutionary idea of discrete energy levels of harmonic oscillator. He assumed that the absorption and emission of electromagnetic radiation happens in discrete quanta. His formula agreed remarkably with the experimental result. The importance of Planck’s idea was first appreciated by Albert Einstein. He applied it to the photoelectric effect and specific heat. The results of Einstein were subjected to experimental test by Robert Andrews Millikan, Maurice de Broglie and others. Their observations established the validity of predictions of Einstein. Next, Niels Henrik David Bohr applied the idea of Planck to the atomic structure and spectral lines. He proposed that the electron in hydrogen atom moves in a circular orbit with angular momentum assuming integral multiples of h/2π where h = 6.626 × 10−34 Js is Planck constant. In this way, Bohr connected h to the mechanics of the atom. The experiment of James Franck and Gustav Hertz gave the striking evidence for Bohr’s postulate, namely, atomic energy states are quantized and the amount of energy transfer to atomic electrons is indeed discrete. Then Arthur Compton applied it to the scattering of X-rays. His results were verified experimentally. In 1924 Louis de Broglie proposed that all matters exhibit wave-like properties. The wave-particle duality is supported by electron diffraction with crystals and double-slit experiment. In this chapter, we briefly discuss the above historical background of quantum theory.
1.2 BLACK BODY RADIATION James Clerk Maxwell established that radiation is a wave made of electric and magnetic fields and is characterized by its frequency ν and wavelength λ. At a given temperature, different materials emit radiation of different wavelengths. A solid need not be so hot to emit electromagnetic energy. In fact, all objects radiate energy continuously irrespective of their temperatures. However, the predominate frequency depends on the temperature. Most of the radiation at room temperature is in the invisible infrared part of the spectrum.
1.2.1
Black Body Spectrum
In 1859-60 Gustav Robert Kirchhoff (grandfather of quantum theory) introduced the concept of black body radiation. A black body is an object which
Why Was Quantum Mechanics Developed? 3
T = 1400K
E
T = 1200K T = 1000K
ν FIGURE 1.1
The spectral distribution of energy density in a black body.
absorbs and emits all the radiation incident on it. It is difficult to realize a perfect black body. However, a closed oven with a small hole is a good approximation to black body. Here, the radiation is admitted through the hole. The entered radiation is totally absorbed due to repeated reflections inside it. The radiation emitted can be studied using a spectrometer and the spectral distribution of the radiation can be obtained. Suppose the walls of the black body are at a constant temperature T . Then a thermodynamic equilibrium, a balance between emission and absorption of radiation, is reached. The spectrum obtained can be characterized by E(ν, T ) where E(ν, T )dν is the energy contribution by radiation per unit volume of the object in the interval ν to ν + dν. From the thermodynamic arguments E(ν, T ) is shown to depend only on T of the cavity. Since a black body absorbs all frequencies, we expect that when heated it should radiate all frequencies. But, experimentally, E(ν, T ) was found to be as shown in Fig. 1.1. Why does the black body spectrum have the shape shown in Fig. 1.1? Attempts were made to describe this by combining the concepts of thermodynamics and electromagnetic theory.
1.2.2
The Stefan, Wien and Rayleigh–Jeans’ Laws
Kirchhoff proved that E must be E(ν, T ) and does not depend on the particular object. In 1879 Josef Stefan calculated the total energy density u(T ) of thermal radiation by adding all frequencies and showed that it was ∝ T 4 and is known as Stefan’s fourth power-law of black body: Z ∞ u(T ) = E(ν, T )dν = σT 4 . (1.1) 0
In 1884, Ludwig Boltzmann presented a proof of this result with the idea of pressure exerted by the radiation. This law did not account correctly the energy distribution in the spectrum. In 1893, Wilheim Wien, through ther-
4 Quantum Mechanics I: The Fundamentals modynamic arguments, obtained the empirical formula E(ν, T ) = a ν 3 e−bν/T ,
(1.2)
where a and b are constants. It explained the spectrum for high value of ν (visible and ultraviolet regions) but not for low values of ν (infrared region). In 1900, Lord Rayleigh argued that radiation consisted of modes of vibration. Applying the equipartition energy theorem, he calculated the energy of each mode of vibration and the energy distribution. His result after an amendment by James Jeans in the year 1905 read as E(ν, T ) =
8πν 2 kB T , c3
(1.3)
where kB is called Boltzmann constant (= RN0−1 = 1.38 × 10−23 JK−1 where R = 8.3 × 103 J · kilomole−1 K−1 is the universal gas constant, and N0 = 6 × 1026 atoms · kilomole−1 ) and c is the velocity of light in vacuum. For the calculation of the value of kB see appendix A. Equation (1.3) is the Rayleigh– Jeans’ law . It did a good job only for low frequencies and failed at high frequencies. Further, it led to the total energy in the spectrum as Z ∞ Z 8πkB T ∞ 2 Edν = ν dν = ∞ . (1.4) c3 0 0 The spectrum diverges at high frequencies. Paul Ehrenfest called this phenomenon ultraviolet catastrophe. This divergence was not observed experimentally. The disagreement of theoretical predictions with the experimental observation hinted that certain basic assumptions of classical physics were somehow wrong.
1.2.3
Planck’s Radiation Formula
Max Planck [1] thought that the assumptions on which Wien’s and Rayleigh– Jeans’ laws were derived might be incorrect. He assumed that the assumption of continuous emission of radiation in classical mechanics led to incorrect predictions. Therefore, he made the following assumptions: 1. Changes of energy in radiators happen discontinuously and discretely. 2. The walls of the black body are made of harmonic oscillators. They have discrete energies En = nhν, where h is the Planck constant2 , n = 0, 1, · · · . (This is the origin of concept of quantization). That is, the 2 This is the first occurrence of the Planck constant in quantum mechanics. It appears in all quantum phenomena. It occurs in angular momentum, spin, momentum, commutator relations, Heisenberg uncertainty relation, etc. We cannot speak of quantum mechanics without h. h is the ratio of the frequency of radiation and the size of the quanta (h = 4.2 × 10−15 eV or h = 6.626 × 10−34 Js). For a determination of h from black body radiation refer refs.[2–5]. Also see appendix A.
Why Was Quantum Mechanics Developed? 5 oscillators have only discrete values of energy and are integral multiples of hν. (In the words of Born in this postulate there is an absurdity. Here the energy is that of a localized particle while frequency ν is usually associated with a wave). 3. The distribution of oscillators in discrete energy states is governed by the Boltzmann distribution law3 . If hE(ν, T )i denotes the average energy of an oscillator then E(ν, T ) =
8πν 2 hE(ν, T )i . c3
(1.5)
Planck started with dS 1 = = function of hE(ν, T )i , dhEi T
(1.6)
where S is the entropy4 of the oscillator. Now, for high ν (or small hEi) from Eqs. (1.2) and (1.5) we obtain 1 aνc3 1 1 = ln − lnhEi . (1.7) T bν 8π bν Then
dS 1 = ln dhEi bν
aνc3 8π
−
1 lnhEi , bν
d2 S 1 . =− dhEi2 bνhEi
(1.8)
From the Rayleigh–Jeans’ law, for small ν (or large hEi) we get dS kB = , dhEi hEi
d2 S kB =− . dhEi2 hEi2
(1.9)
Planck combined the second equations in Eqs. (1.8) and (1.9) as 1 kB d2 S =− =− . dhEi2 bνhEi + (hEi2 /kB ) hEi (hEi + kB bν)
(1.10)
Integration of Eq. (1.10) gives dS 1 kB bν = ln 1 + + C, . dhEi bν hEi
(1.11)
3 If there are N molecules in the system that can be in different energy states then the fraction of the molecules in the jth state (Nj /N ) is given by the Boltzmann distribution P law Nj /N = e−Ej /kB T / i e−Ei /kB T . 4 Quantitatively, entropy, S, is defined through the differential quantity dS = δQ/T , where δQ is the heat absorbed in an isothermal reversible process when the system undergoes transition from one state to another and T is the absolute temperature.
6 Quantum Mechanics I: The Fundamentals where C is a constant. Setting dS/dhEi = kB T for large hEi makes C = 0. Comparison of Eqs. (1.6) and (1.11) gives 1 kB hν = ln 1 + , (1.12) T hν hEi
where kB b = h, Planck constant. From the above equation we have
hν . (1.13) ehν/kB T − 1 (For an alternative simple derivation of the above equation see appendix B). Then Eq. (1.5) becomes hEi =
8πν 2 hν . (1.14) c3 ehν/kB T − 1 This is Planck’s radiation formula. The Planck’s formula agreed remarkably with the experiments. At low frequencies, hν ≪ kB T , (1.14) becomes the Rayleigh–Jeans’ law (1.3). Planck’s discovery led to the formulation of quantum mechanics. Planck received Nobel Prize for physics for the year 1918 for his work. Satyendra Nath Bose derived Planck’s law by considering photons as indistinguishable particles obeying Bose–Einstein statistics. His derivation is presented in appendix C. We note that the size of the energy quanta is proportional to the frequency. Radiation of low frequencies is easy to generate because it needs only small quanta of energy. But for high frequencies radiation requires large amounts of energy. This accounts the explanation for the spectrum of black body radiation. It is easy to verify that the ultraviolet catastrophe not occurs in the Planck’s formula. In the limit ν → ∞ the exponential term in the denominator of Eq. (1.14) diverges much faster than ν 3 in the numerator and E → 0. Planck’s idea of discrete energy of a harmonic oscillator implies that the frequencies for the absorbed and emitted radiation should be discrete. According to classical physics, a system can absorb or emit any amount of energy. In contrast, in the work of Planck the energy emitted by an oscillator is discrete or quantized. Planck coined the term quanta for the packets of energy. Bohr in 1913 proposed that like Planck’s oscillator, atoms can exist only in certain energy states. Essentially, in classical physics energy is a continuous variable while in the new theory it is quantized. Quantization of energy and certain other quantities such as angular momentum and spin in the new theory led to the name quantum physics or quantum mechanics. Max Born coined the term quantum mechanics. E(ν, T ) =
1.3 PHOTOELECTRIC EFFECT The electrons emitted in the photoelectric effect are called photoelectrons and the current formed by them is called photoelectric current. The substance exhibiting photoelectric effect is said to be photoemissive. The effect was discovered first by Willoughby Smith in 1873. In 1839, Alexandre Edmond Becquerel
Why Was Quantum Mechanics Developed? 7 observed the photoelectric effect with an electrode in a conductive solution exposed to light. In 1887 Heinrich Rudolf Hertz noticed the photoelectric effect in his experiments with a spark-gap generator. Hertz observed an increase in the sensitivity of the device when it was illuminated with ultra-violet or visible light. In 1897 Joseph John Thomson (Nobel Prize in 1906) showed that the observed increased sensitivity was because of the emission of electrons from the cathode by the incident light. Nikola Tesla described the photoelectric effect in 1901. He regarded such radiations as vibrations of aether of short wavelengths which ionized the atmosphere. In 1902 Phillip Edward Anton Von Lenard (Nobel Prize in 1905), an assistant of Hertz, studied photoelectric effect. His setup consisted a photocell connected to a circuit with a power supply, voltmeter and micro-ammeter. He illuminated the photoemissive surface with light of different frequencies and intensities. He found that the kinetic energy of the emitted electron was independent of the intensity. Classically, if a plate was illuminated with a highintensity light then the emitted photoelectron should have higher energy.
1.3.1
A Typical Experimental Study
We give briefly a typical experimental study and the common outcomes of the photoelectric effect. Consider a vacuum tube with two electrodes connected to a variable voltage source. Suppose a light incident on the surface of a cathode in the vacuum tube. The electrons released from the metal surface arrive at the other electrode and give rise the flow of current I in the circuit. There would be no flow of current until the potential difference V between the anode and the cathode is raised to a critical value, say, V0 . When V is increased beyond V0 the current I would increase and reach a limiting value Is called the saturation photoelectric current. At I = Is all the electrons that escaped from the cathode would reach the anode. The following are some of the important striking observations with the photoelectric effect. 1. Photoelectric effect occurs when the frequency of the incident light is greater than a certain frequency ν0 called threshold frequency. ν0 differs from metal to metal. 2. The kinetic energy of the photoelectrons emitted increases with increase in frequency, however, remains constant on further increasing the intensity of light. 3. Flow of photoelectric current takes place only above a certain critical applied voltage V0 . 4. The intensity of the incident light has no influence on the maximum kinetic energy of the photoelectrons. But the number of particles emitted decreases with decrease in the intensity of incident light. The velocity of the emitted particles remain the same.
8 Quantum Mechanics I: The Fundamentals 5. There is no delay between absorption of the incident light energy and the emission of photoelectrons. Classically, the electrons would require a certain time to absorb sufficient energy to escape. The above results were completely unexpected. These observations were not explained by classical physics. Einstein solved these mysteries in 1905 [6].
1.3.2
Einstein’s Photoelectric Equation
In 1905, Albert Einstein wrote a paper on photoelectric effect. (In the same year he wrote his paper on the special theory of relativity). He proposed that light is absorbed by matter in discrete quanta of electromagnetic radiation that were later called photons 5 . The electromagnetic radiation itself is quantized. All photons of monochromatic light have the same energy Eph = hν = hc/λ where λ is the wavelength of the light. Here Eph is the energy of the photon (a particle property) and ν is its frequency (a wave property). A conduction electron gets energy hν by absorbing a photon. The electron uses this energy to overcome the binding energy in the metal and to raise its kinetic energy. Let us denote φ0 as the minimum energy required to impart to an electron to escape it from a metal. φ0 is called work function because it is the work done by a photoelectron to escape from the metal. It is h times the critical frequency. The conservation of energy here is 1 mv 2 + φ0 , (1.15) 2 max where m and vmax are the mass and maximum velocity of a photoelectron respectively. When a photon of energy hν strikes an atom, hν − φ0 amount of energy is available for imparting kinetic energy (K) to an electron. Then hν =
1 mv 2 = |e|V0 = hν − φ0 . 2 max The above equation is called Einstein’s photoelectric equation. The notable features of Einstein’s equation are: Kmax =
(1.16)
1. Kmax = 0 when ν = ν0 = φ0 /h. If ν < ν0 (or hν < φ0 ) the incident photon will not have sufficient energy to remove the electron from the metal surface. Hence, there exists a threshold frequency (corresponding to Kmax = 0) of the incident photon above which only the photoelectric effect occurs. 2. Kmax , vmax and V0 depend only on the frequency of the incident light and φ0 . They are independent of the intensity of the incident light. By considering the instantaneous absorption of light we account for observed instantaneous emission of photoelectrons. 5 The
term photon was coined by the chemist Gilbert Lewis in 1926 [7].
Why Was Quantum Mechanics Developed? 9 The above features explained the experimental facts not explained by classical theory. For his work on photoelectric effect Einstein won the Nobel Prize in 1922. Experiments have been conducted by Robert Andrews Millikan (Nobel Prize in 1923), Arthur Llewelyn Hughes, Owen Willans Richardson (Nobel Prize in 1928), Arthur Holly Compton and de Broglie to verify Einstein’s photoelectric equation. Their experiments at both low and high frequencies ranges confirmed the validity of the equation. Though the predictions of the model were verified by Millikan and others it did not attract interest immediately. This is because the particle model of photon does not explain wave effects. Only when Compton in 1923 verified that X-rays were scattered by electrons as if they were particles the Einstein’s model got great interest. The photoelectric effect led to the invention of photoelectric cells, photomultipliers, photovoltaic cells and the image sensors in digital cameras. In the photoelectric effect, energy is transferred to electrons from photons. Is the inverse process, that is, conversion of a part of kinetic energy of a electron into a photon possible? An energetic electron passing through matters will loss a part of its kinetic energy in the form of photons. The process by which an electron emits photons is called bremsstrahlung 6 . The inverse photoelectric effect was observed by Wilhelm Konrad Roentgen in 1895. He found that a highly penetrating radiation (later called X-rays) is produced when fast electrons impinge on matter. He received Nobel Prize in 1902 for the discovery of X-rays.
Solved Problem 1: When a light of wavelength 210 nm falls on a metal surface the measured kinetic energy of ejected electrons is 1.914 eV. When a light of 250 nm falls on the same metal surface the maximum kinetic energy is 0.968 eV. Find the value of hc and the work function of the metal. The photoelectric equation is K.E. = hν − φ0 = (hc/λ) − φ0 . For λ = 210 nm and K.E. = 1.914 eV the above equation is 1.914 eV =
hc − φ0 . 210 nm
(1.17)
For λ = 250 nm and K.E. = 0.968 eV the photoelectric equation is 0.968 eV =
hc − φ0 . 250 nm
From the Eqs. (1.17) and (1.18) we get 1 1 0.946 eV = hc − . 210 nm 250 nm 6 Bremsstrahlung
(1.18)
(1.19)
is a German word meaning breaking radiation or deceleration radiation.
10 Quantum Mechanics I: The Fundamentals Then hc = 1242 eVnm. Substituting the value of hc in Eq. (1.18) we get 1242 − 0.968 eV = 4 eV . (1.20) φ0 = 250
1.4 HYDROGEN SPECTRUM Classical mechanics failed to explain the stability of a nuclear atom and the existence of spectral lines. Consider a hydrogen-like atom with electron of mass me and charge −e at a distance r from the nucleus of mass M and the charge Ze. The force between proton and electron is electrostatic and is (Coulomb interaction) −(1/4πǫ0 )Ze2 /r2 where ǫ0 = 8.85 × 10−12 c2 /Nm2 is the permittivity of free space. Due to the Coulomb force the electron will move towards the nucleus and collide with it in a fraction of second. We know that this is not happening. To stabilize the atom assume that the electron is orbiting the nucleus like the planets orbiting the sun. In a planetary system the motion of the planets and the centrifugal force prevents them colliding with the sun under the influence of gravity. With this analogy, Niels Bohr assumed that the electrons move along circular orbits around the nucleus. The Coulomb force gives a centrifugal acceleration of the electron given by v 2 /r where v is the velocity of the electron. The centrifugal force is F = mv 2 /r where m = me M/(me + M ) is the reduced mass and the Coulomb force is (1/4πǫ0 )Ze2 /r2 . The reduced mass is used because the motion of the electron is considered as relative to the centre-of-mass of the atom. In the orbit these forces are balanced so that 1 Ze2 mv 2 F = = (1.21) 2 4πǫ0 r r or 1 Ze2 mv 2 = . (1.22) 4πǫ0 r Now, the energy of the atom is E = K.E. + P.E. =
1 1 Ze2 Ze2 mv 2 − =− . 2 4πǫ0 r 8πǫ0 r
(1.23)
Classically, r can take any positive value and therefore E can take any value in the interval [−∞, 0]. An accelerated charge will emit electromagnetic radiation. Since the electron is revolving around the nucleus it will emit radiation. The emission will reduce the energy of the electron. Hence, the radius of the electron orbit will decrease. The frequencies of the emitted radiation will be continuous. Therefore, we must get a continuous spectrum. Further, the electron will spiral inside the nucleus and become unstable. But the experimentally observed energy spectrum is discrete and an atom is stable.
Why Was Quantum Mechanics Developed? 11 To explain the discrete spectrum of hydrogen, in 1914 Bohr [8] extended the discreteness of energy to atom and postulated that emission of radiation occurs when the atom jumps from a higher energy state Ei to a lower energy state Ef and the frequency of the emitted radiation is (1.24)
ν = (Ei − Ef ) /h .
Further, he assumed that an electron moves only in certain allowed orbits called stationary states with the orbital angular momentum7 mvr equal to an integer times ~ = h/2π: mvr = nh/(2π) = n~ ,
n = 1, 2, · · · .
(1.25)
Notice the presence of Planck constant in the above equation. (Here onwards we use ~ in place of h/2π). When the condition (1.25) holds the electron does not radiate energy and the orbit is stable. That is, according to Bohr, atoms in the above allowed orbits remain in them and could not radiate. There is a minimum angular momentum determining the closest orbit to the nucleus. Consequently, the electron cannot get into the nucleus. From Eq. (1.25) we have mv 2 = n2 ~2 /(mr2 ). Comparison of this with the Eq. (1.22) gives rn =
n2 r0 , Z
r0 =
4πǫ0 ~2 . me2
(1.26)
r0 (or a0 ) is called Bohr radius (r0 = 5.29167 × 10−11 m). The total energy is En =
1 1 Ze2 Z 2 e4 m mv 2 − =− . 2 4πǫ0 rn 32π 2 ǫ20 n2 ~2
(1.27)
Since n is discrete, the energy levels of the hydrogen atom are thus discrete. In view of Eq. (1.27) we rewrite Eq. (1.24) as ! Z 2 e4 m 1 1 ν = (Ei − Ef ) /h = − 2 . (1.28) 8ǫ20 h3 n2f ni Or ν = RH
1 1 − 2 n2f ni
!
,
RH =
Z 2 e4 m . 8ǫ20 h3
This expression agreed with the Rydberg equation 1 1 ν = RH − 2 , n2 m
(1.29)
(1.30)
where n and m are integers such that n < m and RH is the Rydberg constant (= 1.097373 × 107 m−1 ) deduced from experimental spectrum. This formula 7 In
three-dimension the orbital angular momentum L is r × p where p = mv.
12 Quantum Mechanics I: The Fundamentals was introduced by Johannes Rydberg in 1890. Equation (1.29) gives the frequencies of all transitions in the hydrogen atom. In this way Bohr accounted for the spectrum of hydrogen. Bohr shared the 1922 Nobel Prize (with Einstein) for his work on the investigation of the structure of atoms and of the radiation coming from them. The formula (1.30) with n = 2 was proposed few years earlier by Johann Jakob Balmer. In 1925, de Broglie gave an explanation for non-decay of an electron from a Bohr orbit. He hypothesized that electrons have wavelengths. If the wavelength of an electron orbiting a nucleus is an integral multiple of ~ (length of the orbit), a standing electron wave results. de Broglie’s equation is nλ = 2πr, n = 1, 2, · · · . Then the quantization of angular momentum means nλ = 2πr =
2πn~ nh = . mv mv
(1.31)
Thus, λ = h/(mv) = h/p. Although the theory of Bohr was developed for hydrogen atom it was successful for hydrogen-like ions such as He+ , Li++ which consist of a charged nucleus and a single orbiting electron. However, there were several unanswered questions. For example, some of the spectral lines have fine structures. Difficulties arose when the theory was extended to complex atoms. The model failed to account the effect of a magnetic field on the radiation emitted by certain atoms. It couldn’t account for the intensities of the spectral lines.
1.5 FRANCK–HERTZ EXPERIMENT In 1914 James Franck and Gustav Hertz [9] studied experimentally the impact of electrons on atoms. They observed that electrons moving through mercury vapour with an energy equal to or greater than a critical value near 4.9eV ◦ excited the 2536A line of the mercury spectrum. Their experiment supported the discrete energy states of atoms proposed by Bohr, and for this the 1925 Nobel Prize was awarded to Franck and Hertz. The experimental apparatus [9–11] consisted of a glass tube with mercury vapour and an electrode. A grid was placed between the anode and cathode. Electrons emitted from cathode were accelerated toward a positively charged grid by a voltage V . Acceleration gave kinetic energy to the electrons. This energy can be stated in electron volts (eV). For example, an electron accelerated by 5V will have 5eV of kinetic energy. A small potential difference V0 between the grid and the anode disallowed electrons with energies lower than a certain minimum amount reaching the anode. An ammeter connected to the anode measured the current’s flow. Franck and Hertz measured the current by changing the accelerating voltage. In the experiment as the voltage increased from zero the current increased. At about 4.9V the current decreased sharply indicating the occurrence of a new phenomenon wherein much energy of electrons were taken away. Consequently, the electrons were unable to reach the anode. Essentially, inelastic
Why Was Quantum Mechanics Developed? 13 collisions took place between the atomic electrons of the mercury atoms and the accelerated electrons. The increase of current up to 4.9V implied that energy less than 4.9eV was not absorbed by the mercury. Electrons with energy less than 4.9eV bounced elastically when they collided with mercury atoms and unable to excite electromagnetic radiation. The result of Franck–Hertz experiment is explained based on Bohr’s idea of quantized energy levels. The energy of first excited state of mercury is 4.9eV. If the kinetic energy of accelerated electron is lower than 4.9eV then the available energy becomes insufficient to excite an electron of mercury to next higher energy and thus no absorption energy. When the kinetic energy of an electron is, say, at least 4.9eV then it would able to transfer 4.9eV to an electron in mercury. The mercury can absorb this specific quanta of energy. Now, an electron of mercury jumps to the first excited state and at the same time the electron that has lost 4.9eV is unable to reach the collector. Consequently, current is reduced. In the experiment, when the accelerating voltage was raised just above 4.9V the mercury atoms absorbed only 4.9eV. Hence, the electrons had much energy to reach the anode. The current started increasing steadily. At 9.8V a second decrease in the current was noticed. At this voltage an electron with energy 9.8eV lost 4.9eV because of collision with one mercury atom. This electron moved with the energy of 4.9eV. When this electron collided with another mercury atom it lost the remaining 4.9eV. The electrons lost all their 4.9eV by two inelastic collisions and excited two mercury atoms. Then the electrons were with zero kinetic energy. As a result the current dropped. The above process repeated at intervals of 4.9V and each time the electrons underwent one more inelastic collision. Drops in the current occurred at multiples of 4.9V. This is an indication that the atoms absorbed energy only in specific amounts–quanta! In this process the atomic electrons made an abrupt transition to a discrete higher energy state. If the hitting electrons had energy less than the energy 4.9eV then no energy was absorbed by the atomic electrons and hence no transition. When they had this specific quanta of energy they lost it to the electrons of mercury. The absorbed energy got stored in the electron excited to higher states. This energy was released as a photon when the electrons came back to the lowest energy state (ground state). As described above the Franck–Hertz experiment confirmed quantized model of Bohr by demonstrating that atoms could only absorb specific amount of energy, namely those quantities which transfer the atom from one state to another stationary state.
Solved Problem 2: Atomic hydrogen is bombarded with electrons in a Franck–Hertz type experiment. Excitation potentials of 10.21 V and 12.10 V are measured and three spectral lines are found in the emission spectrum. Compute the wavelengths of these spectral lines.
14 Quantum Mechanics I: The Fundamentals Using the relation E = hν = hc/λ we write hc λ1 hc λ3
hc = 12.10 eV , λ2
(1.32a)
= (12.1 − 10.21) eV = 1.89 eV .
(1.32b)
= 10.21 eV ,
Since 1eV = 1.6 × 10−19 J we get hc = Then
6.626 × 10−34 eV × 3 × 108 m = 1242.375 eVnm . 1.6 × 10−19 λ1
=
λ2
=
◦ hc = 1216.8217A , 10.21 eV ◦ ◦ 1026.7562A , λ3 = 6573.4127A .
(1.33)
(1.34a) (1.34b)
1.6 STERN–GERLACH EXPERIMENT Equation (1.25) was realized as a special case of the Bohr–Sommerfeld quantum rule [12] I p dq = n~ , (1.35)
where q is a generalized coordinate and p is the corresponding canonically conjugate momentum. This condition is satisfied for periodic motion and the integral is taken over one period. Orbits with a specific direction for the angular momentum vector were alone considered. There is a further degeneracy with the various orientations for an angular momentum vector L with respect to any fixed axis. The angular momentum component parallel to the axis must be m~ with the quantum number m taking integer values only. This is called space quantization. When l is the quantum number characterizing the magnitude of the angular momentum then m = −l, −l + 1, · · · , l − 1, l. The first evidence of space quantization was shown experimentally by Otto Stern and Walther Gerlach in 1922 [13]. They demonstrated the magnetic nature of the electron. Suppose a beam of electrons is allowed to pass through a region where the magnetic field points in the z-direction. The strength of field increases with z. Such a field would make a pull upward on the south pole but downward on the north pole of the equivalent magnet of the electron. The total force on the electron will depend on its orientation. If the south pole of the electron is much higher in the field then the force will be upward. When its north pole is higher then the total force will be downward. Hence, an electron moving through this region would experience an downward or upward deflecting force and is directly proportional to Sz (the projection of its spin angular momentum8 on the z-axis). 8 The spin is usually said to be non-orbital, intrinsic or inherent angular momentum. For more details see chapter 11.
Why Was Quantum Mechanics Developed? 15
Electromagnet
Electron beam
Z - axis FIGURE 1.2
Splitting of electron beam in Stern–Gerlach experiment.
If we send an electron beam through a region with a magnetic field of the above system then the upward or downward forces on the electrons will make the beam to fan-out vertically. The vertical position of an electron, on a display screen, can be used as a direct measurement of Sz of that electron. Because of the random orientation of the incoming electrons we expect the electrons be at the screen to be with a continuous range of vertical positions. But in the experiment electrons were found to reach at only two positions as shown in Fig. 1.2. This implied that Sz could have one of the two possible values. The spin is thus quantized. The point is that the beam is bent by the field into two paths: one above and one below the z-axis. The force in the z-direction on a particle is Fz = µz ∂Bz /∂z where µz is the z-component of the magnetic moment. The separation of beams means that µz takes on two equal but opposite values. The atoms from the source are with their magnetic moments oriented at random, but when entered the field they take one of the two values in the z-direction. Because of the relationship between µz and Lz (z-component of the orbital angular momentum), Lz can take only one value. For a given value of the orbital quantum number l, there are 2l + 1 values of the z-component. Here the two values of the z-component means l = 1/2, which is in disagreement with l an integer. George Eugene Uhlenbeck and Samuel Abraham Goudsmit suggested that the electron has an intrinsic angular momentum (spin) with two values of the z-component ±1/2. If the electron is like a spinning ball then its collisions with other electrons would give spin angular momentum with different magnitudes. Hence, more than two parts could occur. But this is not observed. Thus, the spin of an electron appears as one of its intrinsic characteristics like mass or charge.
1.7 CORRESPONDENCE PRINCIPLE Bohr investigated the passage of α-particles in matter and studied the processes of ionization of the atoms. The classical ideas failed. Bohr thought that
16 Quantum Mechanics I: The Fundamentals the new theory (quantum theory) should match with the classical theory for low frequencies or large quantum numbers. This led to the famous correspondence principle. This principle is of great interest in statistical mechanics and electrodynamics. It played a significant role in the early formulation of quantum theory [14,15]. Planck stated that lim [quantum physics] = [classical physics] .
h→0
(1.36)
He illustrated this principle through the observation that (1.14) reduced to the classical Rayleigh–Jeans’ law (1.3) in the limit h → 0. He stated: The classical theory can simply be characterized by the fact that the quantum of action becomes infinitesimally small . Another form of this principle was given by Paul Ehrenfest and is known as Ehrenfest’s theorem which will be discussed in the next chapter. In the year 1913 Bohr observed that the frequency spectrum emitted by a hydrogen atom approaches the classical spectrum in the limit of large principal quantum number n. The emission frequency in this limit is the orbiting of the electron classical frequency which approach zero as n → ∞. This is called Bohr frequency correspondence principle. This is obtained when the quantum spectrum coalesces with that of the classical. Let us illustrate the correspondence principle [14,15] for linear harmonic oscillator and hydrogen atom. In the Planck’s idea energy of harmonic oscillator is discrete while in classical mechanics it is continuous. To understand this difference we calculate ∆E/E for any two adjacent energy levels of linear harmonic oscillator. Since En = (n + 1/2) ~ω, n = 0, 1, 2, · · · (see sec. 4.4) the quantity ∆E/E corresponding to n and n + 1 levels is 3 1 n+ ~ω − n + ~ω ∆E 1 2 2 . = (1.37) = 3 3 E ~ω n+ n+ 2 2 In the limit of n → ∞ we get limn→∞ (∆E)/E → 0. This is in agreement with the correspondence principle. In hydrogen atom the force on the electron due to the nucleus is F = −(1/4πǫ0 )e2 rˆ/r2 . This gives the centripetal force to the electron to follow a circular orbit. The radial acceleration is ar = v 2 /r. Applying Newton’s second-law we get 1/2 mv 2 e2 e2 = or v = . (1.38) r 4πǫ0 r2 4πǫ0 mr Then the classical frequency νc is ω v νc = = = 2π 2πr
e2 16π 3 ǫ0 mr3
1/2
.
(1.39)
Why Was Quantum Mechanics Developed? 17 We substitute rn from Eq. (1.26) in (1.39) and obtain νc =
2|E1 | , hn3
E1 = −
me4 , 32π 2 ǫ20 ~2
(1.40)
where E1 is the lowest energy of hydrogen atom (refer Eq. (12.36)). For the transition from (n + 1) to nth orbit we get from Eqs. (1.24) and (1.27) |E1 | 1 1 |E1 | (2n + 1) νBohr = − = . (1.41) 2 2 h n (n + 1) h n2 (n + 1)2 For large n 2|E1 | = νc . (1.42) hn3 Bohr predicted νBohr coincides with νc for large n. This led to the correspondence principle: For large quantum numbers the state of a system is governed by the laws of classical physics and we ignore the quantum effects. The results of quantum mechanics in the limit of large quantum numbers and h → 0 must be same as that of classical mechanics. The limit n → ∞ and h → 0 are not the same. Let us point out an example [15]. The energies of a particle of mass m bounded to a cubical box with impenetrable walls of length a are given by En = E1 n2x + n2y + n2z , E1 = h2 /(8ma2 ) (1.43) νBohr =
and nx,y,z = 1, 2, 3, · · · . The frequencies of emission νn are obtained from ′2 ′2 2 2 2 hνn = E1 n′2 (1.44) x + ny + nz − E1 nx + ny + nz . Since hνn /E1 is the difference of integers, it is an integer. Then E1 q = ν1 q , q − an integer . νn = h
(1.45)
In any limiting process with n these frequencies do not include the classical continuum. For high quantum number domain the adjacent frequencies are given by νn+1 = νn +
h , I
I − constant .
(1.46)
In the limit h → 0, νn+1 → νn . h → 0 leads to frequency correspondence but the limit n → ∞ does not.
1.8 COMPTON EFFECT The Franck–Hertz experiment gave an evidence of discrete energy levels of atoms. The particle nature of radiation is established by the Compton effect.
18 Quantum Mechanics I: The Fundamentals
Scattered photon Incident photon Electron
FIGURE 1.3
φ θ
Scattered electron
Scattering of a photon by an electron.
In 1923, both theoretically and experimentally, Arthur Holly Compton (Nobel Prize in 1927) showed that during propagation electromagnetic radiation manifests as a stream of particles (photons) [16]. Let us consider a collision between an electron and a photon. Assume that an X-ray photon of frequency ν and energy E(= hν) is made to strike a stationary electron. The photon is scattered, and the electron gets an impulse and moves as shown in Fig. 1.3. Classical theory of X-ray scattering predicts the following: 1. The wavelength of a scattered photon should have a continuous values. 2. The wavelength of incident and scattered photons are the same. 3. Scattering occurs in all directions. But experimental observations have shown discrepancy from the above results. Compton noticed that when X-rays were scattered by a graphite block the scattered radiation had a longer wavelength in addition to the original wavelength. Further, the longer wavelength component was found to increase with increase in the scattering angle. Considering the scattering as a result of an elastic collision of two particles (photon and electron) Compton accounted the experimental results. In the scattering process shown in Fig. 1.3 the kinetic energy of the scattered electron has to be from the incident photon. Therefore, the energy Esp of the scattered photon is lower than Eip , the incident photon energy. Thus, the frequency ν ′ of the scattered photon is lower than the frequency ν of the incident photon. The kinetic energy gained by the scattered electron is thus hν − hν ′ . In the collision process momentum must be conserved in the direction of the incident photon as well as in the direction perpendicular to it. The momenta of the photon and the electron before collision are p = hν/c and 0,
Scattered
P cos θ θ P sin θ
hν c Incident
φ hν cos φ c
hν sin φ c
h cν
Why Was Quantum Mechanics Developed? 19
Scattered
Vector components of momenta of the incident and scattered photons. FIGURE 1.4
respectively. After collision they become p′ = hν ′ /c and P respectively. From Fig. 1.4 we write the equations of conservation of momenta. Parallel to the direction of incident photon: hν hν ′ +0= cos φ + P cos θ . c c
(1.47)
Perpendicular to the direction of incident photon: 0=
hν ′ sin φ − P sin θ . c
(1.48)
We multiply Eqs. (1.47) and (1.48) by c and obtain P c cos θ = hν − hν ′ cos φ ,
P c sin θ = hν ′ sin φ .
(1.49)
θ can be eliminated by squaring the above two subequations and adding them. We obtain P 2 c2 = (hν)2 − 2hνhν ′ cos φ + (hν ′ )2 . (1.50)
The energy conservation equation is
Eip + Ere = Esp + Ese ,
(1.51)
where Ere = me c2 is the energy of the electron at rest and Ese = p 2 2 P c + m2e c4 is the energy of the scattered electron. Equation (1.51) takes the form 1/2 hν + me c2 = hν ′ + P 2 c2 + m2e c4 . (1.52) Eliminating square-root in Eq. (1.52) and then substituting Eq. (1.50) for P 2 c2 results in 2 h(ν − ν ′ ) + me c2 = m2e c4 + (hν)2 − 2hνhν ′ cos φ + (hν ′ )2 . (1.53)
20 Quantum Mechanics I: The Fundamentals Simplifying the above equation we obtain me c ν ν′ ν ν′ − = (1 − cos φ) . h c c c c In terms of the λ = c/ν, λ′ = c/ν ′ the above relation becomes me c 1 1 (1 − cos φ) − ′ = . h λ λ λλ′
(1.54)
(1.55)
Or
h (1 − cos φ) . (1.56) me c The shift in the wavelength △λ of the scattered photon is from its collision with the electron at rest and is called Compton effect . This shift in the wavelength is not detected by the classical physics. △λ depends on h, c, me and φ. The shift in the wavelength when the scattering is at an angle π/2 is called the Compton wavelength and is given by △λ = λ′ − λ =
λc =
h = 2.426 × 10−12 m . me c
(1.57)
From Eq. (1.56) we obtain λ′ = λ +
h (1 − cos φ) . me c
(1.58)
Substituting λ = c/ν, λ′ = c/ν ′ and simplifying the above equation we get ν′ =
ν , 1 + 2α sin2 (φ/2)
α=
hν . me c2
(1.59)
Then from Eqs. (1.47) and (1.48) we obtain tan θ =
1 ν ′ sin φ = ν − ν ′ cos φ (1 + α) tan(φ/2)
(1.60)
which gives the possible directions of scattering. From the above analysis the following conclusions were drawn: 1. Equation (1.56) shows that λ′ of the scattered radiation (photon) is greater than that of the incident photon. 2. △λ is independent of the wavelength of the incident photon and the nature of the target but depends on the angle of scattering. 3. The relation between θ and φ, Eq. (1.60), shows that as φ increases from 0 to π, θ decreases from π/2 to 0. That is, the photon is scattered in all directions while the target electron is scattered only in the forward direction.
Why Was Quantum Mechanics Developed? 21 4. From Eq. (1.57) we write me c2 = hc/λc = hνc . Therefore, λc is the wavelength of the light produced if the energy of the electron was converted into a light quanta. 5. Here p (or p′ ) is a particle property while λ is a wave property. There is a paradox in Compton’s experiment: Light manifests both particle-like and wave-like properties. Soon after Compton’s theoretical work, several scientists including Compton himself carried out experiments with X-rays. Their results confirmed the Compton’s predictions.
Solved Problem 3: Express the change in wavelength in terms of Compton wavelength of electron when light is scattered through an angle π/4 by an electron The change in wavelength of a scattered light due to a particle of mass m is h me (1 − cos φ) = λc (1 − cos φ) , mc m where λc = h/(me c) is the Compton wavelength. For an electron m = me and hence 1 ∆λ = λc (1 − cos(π/4)) = λc 1 − √ = 0.29289λc . 2 ∆λ
=
λ′ − λ =
1.9 SPECIFIC HEAT CAPACITY Due to thermal agitations, atoms in solids oscillate about their mean positions. When each atom is considered as a three-dimensional harmonic oscillator then its average thermal energy should be 3kB T where kB T is that of onedimensional harmonic oscillator. Then the thermal energy of a solid becomes 3kB T per atom or 3RT = 3N kB T per gram-atom where N is the Avogadro number. The rate of increase of thermal energy with T , per gram-atom (called atomic heat or specific heat) becomes 3R (= 3 × 8.3 × 103 J · kilomole−1 K−1 ), a universal constant. This was confirmed experimentally in solids at normal temperatures. But for sufficiently lower temperature, the specific heat decreases and approach to zero as T → 0K. In order to explain this Einstein [17] assumed that the energy of an atom in a solid takes only a discrete set of values: the energy of each component of oscillation is nhν, n = 0, 1, · · · . The mean energy per atom is obtained as hE(ν)i =
hν ehν/kB T
−1
.
Then the atomic heat is given by 2 d 3N hν hν ehν/kB T C= = 3R 2 . dT ehν/kB T − 1 kB T ehν/kB T − 1
(1.61)
(1.62)
22 Quantum Mechanics I: The Fundamentals C given by the above equation decreases with decrease in T . The variation of specific heat of solids is approximately accounted by Eq. (1.62).
1.10 DE BROGLIE WAVES According to special theory of relativity mass and energy are equivalent. Therefore, electromagnetic waves and electrons are considered forms of energy. Electromagnetic waves possess particle properties. When the postulates of Planck and Einstein shown that electromagnetic wave posses particle properties, at that time, wave properties of particles were not observed. To have a symmetrical picture Louis de Broglie, a young French prince, in his doctorate thesis (1924) attached waves to all moving particles [18]. For his idea of matter waves he was awarded the Nobel Prize in the year 1929. The matter waves proposed by de Broglie are called de Broglie waves. To know the wavelength of a de Broglie wave, let us attach a periodic travelling wave with a moving particle. Denote the wave solution as φ and let it be a0 sin(2πν0 t0 ) where a0 is the amplitude and ν0 is the frequency observed by an observer at rest with respect to the particle. Suppose the particle’s speed is v along the x-axis. Then taking according to the time-dilation
we have
t − (βx/c) t0 = p , 1 − β2
(1.63)
β = v/c
"
2πν0 (t − (βx/c)) p φ = a0 sin 1 − β2
#
.
Comparison of this with the standard solution h x i φ = A sin 2πν t − , u
where u is the speed of the wave along the x-direction, gives p u = c/β = c2 /v , ν = ν0 / 1 − β 2 .
(1.64)
(1.65)
(1.66)
From E = hν0 = mc2 we have ν0 = mc2 /h. Substituting this expression in Eq. (1.66) we obtain mc2 ν= p . (1.67) h 1 − β2
Now, the (de Broglie)wavelength λ is u/ν = h/p. λ depends on m and speed of the particle. It is shorter if the particle moves faster. The angular frequency ω and the wave number k of the de Broglie wave are 2πmc2 ω = 2πν = p , h 1 − β2
k=
2π 2πmv = p . λ h 1 − β2
(1.68)
Why Was Quantum Mechanics Developed? 23 Evidently, the de Broglie wavelength is extremely small for ordinary objects like bullets and balls because of the extreme smallness of h (= 6.626×10−34Js). For example, a cricket ball of 100g travelling with a speed 120km/h have a de Broglie wavelength of 1.9878 × 10−34 m which is too small to be detectable. But p for an electron accelerated through 150V the de Broglie wavelength is ◦ λ = 150/V × 10−8 cm = 1A which is significant in atomic scale. We note that matter waves are quite different from electromagnetic waves. The latter in vacuum always travels with velocity 3 × 108m/s while the former do not. de Broglie formula has found several applications. The wave length of electrons at high speed or high energy are much smaller than those of visible light. Because of this electron microscope have relatively much higher resolutions than those of optical microscopes.
Solved Problem 4: Find the de Broglie wavelength (λd ) associated with an electron of energy 20 MeV. We find λd = h/(mv), that is, λd
= = =
√
h 2mE 6.626 × 10−34
1/2
(2 × 9.11 × 10−31 × 20 × 106 × 1.6 × 10−19 )
m
◦
0.002744A .
1.11 PARTICLE DIFFRACTION Compton effect shows that light behaves like particles when it collides with electrons. Can a particle exhibit a wave phenomenon, supporting matter waves concept? In 1927 Clinton Joseph Davisson, Lester Halbert Germer [19] and George Paget Thomson9 [20] independently observed the wave nature of particles in their experiments with diffraction of a particle with a single crystal. Then Arthur Jeffrey Dempster in 1927 and Immanuel Estermann and Otto Stern in 1930 obtained diffraction effects with hydrogen and helium by reflecting them from crystal gratings. The results of their experiments could be explained only by attributing wave-like character to microscopic particles. According to classical physics the scattered electrons move in all directions with a moderate dependence of their intensity on angle of scattering and less on the energy of the incident electrons. These results were verified by Davisson and Germer using the target as a polycrystalline nickel solid. They heated the nickel to a high temperature and then cooled it. They got a single crystal 9 Joseph John Thomson got Nobel Prize for his discovery of electron and showing that it has particle character. His son George Paget Thomson received Nobel Prize for showing electron behaves as a wave.
24 Quantum Mechanics I: The Fundamentals with the atoms arranged in a regular lattice. They measured the intensity of scattering in various directions. The intensity of scattering increased from zero ◦ in the plane of the target to a highest value in co-latitude 20 . When the colatitude and the bombarding potential were fixed and measurement was done in azimuth angle, they observed a variation in the intensity of scattering. At ◦ 40V a slight hump occurred near 60 in the co-latitude curve for azimuth 111. This hump then developed rapidly with increasing voltage into a strong spur. ◦ The spurs were due to scattered electrons. At 54V and co-latitude 50 the graph of intensity of scattered electron with azimuth angle was shown a series of maxima and minima. When the incident electron beam was replaced by a monochromatic X-rays, phenomena noticed with the electron were observed. If the electrons were particles then the above result cannot be explained. Thus, the experiment of Davisson and Germer shown that a beam of electrons possesses wave-like characteristics. According to de Broglie hypothesis the electron waves were diffracted by the single crystal . The de Broglie wavelength λ of the diffracted electrons obtained from nλ = 2d sin θ agreed with the electron wavelength λ = h/(mv) thereby verifying the wave nature of moving bodies. Beams of neutrons, hydrogen atom and helium atom have produced diffraction patterns. The wave properties of electrons are utilized in many practical ways. For example, in an electron microscope, electrons replace waves due to their shorter wavelengths compared to the visible light and thus able to resolve much finer detail. Further, electron and neutron wave effects are useful to investigate defects in metals. A beam of neutron waves directed towards a target nuclei can be tuned in frequency to resonate with the natural interval frequencies of the nuclei.
1.12 WAVE-PARTICLE DUALITY In classical mechanics we have two types of phenomena: waves and particles. As we know waves spread in space and carry energy with them but do not transport mass. In contrast, particles are localized and have both energy and mass. In quantum mechanics these differences are unclear. In certain cases objects thought of as particles behave like waves. In some cases waves behave like particles. For example, the photoelectric effect can only be explained if the light has a particle nature. On the other hand, electrons can give rise to wave-like diffraction patterns. As suggested by de Broglie, all objects have both particle and wave aspects. The different aspects manifest according to the type of process it undergoes. This is known as the principle of wave-particle duality. In this section, we demonstrate it by considering the famous Young’s double-slit experiment. In 1803, Thomas Young (1773-1829) demonstrated that when a light beam passes through two narrow slits, the waves coming out from these slits inter-
Why Was Quantum Mechanics Developed? 25
y d
θ L
Slits FIGURE 1.5
Screen
Schematic arrangement of a double-slit experiment.
fered with one another and produce interference patterns on a photographic plate10 . This experiment exemplifies the wave and particle nature of light. Here, we takeup three experiments similar in concept with a double-slit [22, 23]. Suppose a light is sent to a screen through a double-slit (Fig. 1.5). A photographic film on the screen is used to extract information created by the incident light. We consider the following three cases: 1. Slit-1 is opened while the slit-2 is closed. 2. Slit-2 is opened and the slit-1 is closed. 3. Both the slits are opened. Image patterns are recorded on photographic film for each experiment. If we measure the intensity of the light on the film we get the results as indicated in Fig. 1.6. When the slit-1 or slit-2 alone is opened there is no interference. When both the slits are opened we observe interference pattern with maxima and minima but we expect only two bright lines aligned with the slits. The condition for the constructive interference patterns (bright bands) at the photographic film in the case of lights were in phase when they were passing through the slits is d sin θ = mλ ,
m = ±1, ±2, · · · .
The condition for destructive interference patterns (dark bands) is 1 λ , m = ±1, ±2 · · · . d sin θ = m + 2
(1.69a)
(1.69b)
10 Young’s double-slit experiment has been rated as the most beautiful experiment in physics by the readers of Physics World in 2002 [21].
26 Quantum Mechanics I: The Fundamentals
(a) (b) (c) (d) Slit-1 Slit-2
I1
I2
I3
P12
Intensity patterns in the double-slit experiment. (a) Photon intensity I1 on the screen with slit-1 only opened. (b) Photon intensity I2 with slit-2 only opened. (c) Interference pattern when both the slits were opened. (d) Probability distribution P12 of photons when both the slits were opened and when the film was replaced by an array of photon detectors. FIGURE 1.6
These dark and bright regions are called interference fringes. The spacing between successive dark and bright fringes is ∆ = λL/d. In optics, the wave amplitude is generally a complex number, φ. The quantity, φ∗ φ = |φ|2 is the intensity I of the light. Now, if φ1 and φ2 are the amplitudes of the waves through the slits 1 and 2 respectively then I1 = |φ1 |2 ,
I2 = |φ2 |2 ,
I3 = |φ1 + φ2 |2 = I1 + I2 + 2Re (φ∗1 φ2 ) .
(1.70)
I3 6= I1 + I2 is the characteristic of interference pattern. 2Re(φ∗1 φ2 ) describes the interference of the waves passing through the two slits. In the second experiment assume that the source releases the photons slowly so that photons hit the screen without interaction. We do not expect interference pattern. Surprisingly, we get the interference pattern as in Fig. 1.6(c). This means that a single particle is also interfering with itself. What will happen if we replace the photographic film by an array of photon detectors? In the experiment, we close the slit-2 first and let the photons pass through slit-1. We get the probability distribution P1 of photons similar to the pattern shown in Fig. 1.6a. We repeat the experiment by opening slit-2 and closing the other. We record the result. The probability distribution P2 becomes as in Fig. 1.6b. Finally, we open both the slits. Figure 1.6d shows a probability distribution. We observe that P12 = P1 + P2 . If we treat photon as a particle then Fig. 1.6d is not surprising. But in the case of wave nature I3 6= I1 + I2 .
Why Was Quantum Mechanics Developed? 27 In the first experiment, interference occurred because the same light wave went through the two slits. Interference took place in the second experiment also where the photons (particles) were sent one by one so that they did not interact. Here the photons behaved as a wave. The same individual photons behaved as particles when a detector was used. The above experiments clearly demonstrate the dual nature of photons. An interesting observation is that we have interference in the absence of observers and no interference pattern when observers are present. In other words, when we carry out a measurement, it disturbs the trajectories of photons and the results of the experiment satisfy with the classical physics – two bright lines on the film. But when we cease the measurement, the pattern become multiple lines with brightness and darkness. More precisely, particle cannot behave as a particle and a wave simultaneously. If an experiment is performed for particle character then particle property is observed. Similarly, if an experiment looks to observe wave nature then wave property is realized. To explain the wave-particle duality Bohr assumed that we are able to measure a wave property of light or particle property of light but not both simultaneously. This is called complementarity principle of light. In 1992 Indian physicists Partha Ghose, Dipankar Home and Girish S. Agarwal [24] proposed an experimental setup to observe both wave and particle properties of photon simultaneously. The experiment was then performed by the Japanese scientists Y. Mizobuchi and Y. Ohtake [25] using a pulsed laser. The experiment caught the wave and particle behaviors of photons simultaneously.
1.13 CONCLUDING REMARKS In this chapter, we considered some of the experimental observations which were not adequately explained by classical theory and outlined how they have been properly interpreted by the quantum theory. In the quantum theory, certain dynamical variables take on only discrete or quantized values whereas in classical theory they take on a continuum of values. The discreteness property gave rise to the name quantum mechanics or quantum theory. After the Planck’s idea of discrete energy values, initially, in the development of a new theory (quantum theory), it was thought that classical mechanics was correct but with the assumption that certain observables like energy and angular momentum take only discret values. This is often referred to as old quantum theory. For example, in 1905 in his work on the photoelectric effect Einstein postulated that the energy of a beam of light is quantized. In 1913 Bohr applied the idea of energy quantization to atoms and explained the various colours emitted by hydrogen glowing in a discharge tube. In 1911 Arnold Sommerfeld analysed the influence of energy quantization on photon and momentum. We wish to stress that in all the above work the concept of wave equation and wave function did not appear! Very soon after the de Broglie’s proposal of matter waves, Erwin Schrödinger, Werner Karl Heisenberg and Paul Adrian Maurice Dirac de-
28 Quantum Mechanics I: The Fundamentals veloped quantum theory with three different approaches. The approach of Heisenberg involved matrix calculus. Dirac introduced operator theory. In 1926 Schrödinger starting with the idea of matter waves, setup a wave equation known as Schrödinger wave equation and then the theory has seen tremendous fundamental developments. In this book we present various developments and essential features of the quantum theory. Quantum theory is an amazingly successful theory giving predictions in very good agreement with experimental observations in minute detail. The range of phenomena to which the quantum theory has been applied is enormous. It covers phenomena from elementary particle level to the early universe. Many modern technologies would be impossible without quantum physics. Lasers, semiconductors and superconductors are just few among the various applications of the most radical predictions of quantum mechanics. Many ordinary scale physical phenomena are also explained by quantum theory. These include the existence of solid bodies, physical properties of materials, phenomena of freezing and boiling, etc.
1.14 BIBLIOGRAPHY [1] M. Planck, Annalen der Physik 4:553, 1901. [2] R. L. Bobst and E. A. Karlow, Am. J. Phys. 53:911, 1985. [3] J. D. Barnett and H. T. Stokes, Am. J. Phys. 56:86, 1988. [4] R. E. Crandall and J. F. Delord, Am. J. Phys. 51:90, 1983. [5] J. Dryzek and K. Ruebenbauer, Am. J. Phys. 60:251, 1992. [6] A. Einstein, Annalen der Physik 17:132, 1905. [7] G.N. Lewis, Nature 118:874, 1926. [8] N. Bohr, Phil. Mag. 27:506, 1914. [9] J. Franck and G. Hertz, Verhandl. deut. Physik. Ges. 16:457, 512, 1914. [10] D.W. Preston and E.R. Dietz, The Art of Experimental Physics. John Wiley, New York, 1991. [11] G.F. Hanne, Am. J. Phys. 56:696, 1988. [12] A. Sommerfeld, Annalen der Physik 51:1, 1916. [13] W. Gerlach and O. Stern, Z. Phys. 8:110, 1922. [14] B.L. Van der Waerden, Sources of Quantum Mechanics. Dover, New York, 1967. [15] R.L. Liboff, Physics Today February pp. 50, 1984.
Why Was Quantum Mechanics Developed? 29 [16] A.H. Compton, Phys. Rev. 21:483, 1923; 22:409, (1923). [17] A. Einstein, Annalen der Physik 22:180, 1907. [18] L. de Broglie, Phil. Mag. 47:446, 1924; Annals d. Physique 3:22, 1925. [19] C.J. Davisson and L.H. Germer, Nature 119:558, 1927; Phys. Rev. 30:705, 1927. [20] G.P. Thomson, Nature 120:802, 1927; Proc. Roy. Soc. London A 117:600, 1928. [21] P. Rogers, Physics World September pp.19, 2002. [22] N.D. Mermin, Physics Today 46:9, 1993. [23] G. Venkataraman, The Quantum Revolution-I: The Breakthrough. University Press, Hyderabad, 1994. [24] P. Ghose, D. Home and G.S. Agarwal, Phys. Lett. A 168:95, 1992. [25] Y. Mizobuchi and Y. Ohtake, Phys. Lett. A 168:1, 1992.
1.15 EXERCISES 1.1 Calculate the energy of a photon of 253.6 nm light. State the energy in eV (1eV = 1.6 × 10−19 J). 1.2 Calculate the wavelength of a quantum of electromagnetic radiation with energy 4000 eV. 1.3 A light of wavelength 1000 nm produces photoelectrons from a metal. The threshold potential V0 is 0.5V. Calculate the work function φ0 . 1.4 Consider the previous problem. Calculate the threshold wavelength λc . 1.5 Calculate the frequency of light required to produce electrons of kinetic energy 5 eV from illumination of a material whose work function is 0.5 eV. 1.6 Light of wavelength 500 nm is incident on a metal with work function (φ0 ) of 0.75 eV. What is the maximum kinetic energy (Kmax ) of the ejected photoelectrons? 1.7 Suppose you are standing in front of a 40 W incandescent light bulb 5 m away. If the diameter of your pupils is about 2 mm, about how many photons, n, enter your eye every second? 1.8 Calculate the minimum voltage required for an electron to give all of its energy in a collision with a target and produce an electromagnetic radiation of wavelength 0.05 nm.
30 Quantum Mechanics I: The Fundamentals 1.9 Assume that the resolving power of a microscope is equal to the wavelength of the light used. If electrons are used as the light source calculate the kinetic energy of electrons needed to have the resolving power as 10−11 m. 1.10 Suppose a laser provides 100 TW of power in 1ns pulses at a wavelength of 0.26 µm. How many photons are contained in a single pulse? 1.11 Express the change in wavelength in terms of Compton wavelength of electron when light is scattered through an angle π/4 by a proton. ◦
1.12 A beam of X-rays of wavelength λ = 0.738A is Compton scattered by electrons through an angle π/2. Calculate the wavelength and the energy of the scattered radiation. 1.13 Show that the de Broglie wavelength λd associated with photons is equal to the electromagnetic wavelength λ. 1.14 A 5.78 MeV α-particle (assume that its velocity is ≪ c) is emitted in the decay of radium. If the diameter of the radium nucleus is 2 × 10−14 m, how many α-particle de Broglie wavelengths fit inside the nucleus? 1.15 A diatomic gas molecule consisting of two atoms of mass m separated by a fixed distance d is rotating about a central axis perpendicular to the line joining the two atoms. Assuming that its angular momentum is quantized as in the Bohr atom, determine the possible angular velocities (ω) and the possible quantized rotational energies (E).
CHAPTER
2
Schrödinger Equation and Wave Function
2.1 INTRODUCTION Till 1925 the idea of what the quantum theory would be was unknown. During 1925 three different but equivalent versions of quantum theory were proposed – Schrödinger-proposed wave mechanics; Heisenberg-developed matrix mechanics and Dirac-introduced operator theory. Considering the de Broglie’s matter waves Erwin Schrödinger, an Austrian physicist, argued that if a particle like an electron behaves as a wave then the equation of wave motion could be successfully applied to it. He postulated a function varying in both space and time in a wave-like manner (hence called wave function and denoted it as ψ). This function is generally complex and assumed to contain information about a system. Schrödinger set up a linear and time-dependent wave-like equation, called Schrödinger wave equation, to describe the wave aspect of a particle taking account of de Broglie’s relation for wavelength. Physically, |ψ(X, t)|2 , where ψ(X, t) is the solution of the Schrödinger equation, is interpreted as position probability density. That is, |ψ(X, t)|2 is the probability density of observing a particle at position X at time t. ψ does not give exact outcomes of observations but helps us to know all possible events and their probabilities. Further, the probability interpretation allows us to find the average or expected result of a set of measurements on a quantum system. There are several interesting features of the wave function ψ. The probability interpretation of |ψ|2 imposes certain conditions on meaningful ψ. Further, the |ψ|2 satisfies a conservation law, an equation analogous to the continuity equation of flow in hydrodynamics. To set R ∞the total probability unity the ψ must satisfy the normalization condition −∞ |ψ|2 dτ = 1. Knowing ψ we can compute expectation values of variables such as position, momentum, etc. In quantum mechanics the experimentally measurable variables are no longer dynamical variables but they become operators. The outcomes of experiments are the eigenvalues of the operators of the observables. An interesting result is 31
32 Quantum Mechanics I: The Fundamentals that the equations of the time evolution of expectation values of position and momentum operators (called Ehrenfest’s theorem) obey the same equation of motion of these variables in classical mechanics. On the other hand, the equation for the phase of the ψ function takes the form of Hamilton–Jacobi equation of classical mechanics. In this chapter, we set-up the Schrödinger equation and discuss the above-mentioned general properties of the solution of the Schrödinger equation.
2.2 CONSTRUCTION OF SCHRÖDINGER EQUATION In this section we construct the wave equation for a particle moving in a potential V (X, t). We start with a free particle then generalize the result for a particle in a potential V (X, t).
2.2.1
Schrödinger Equation for a Free Particle
Suppose a particle is free and moving along a one-dimensional line. Assume that it behaves like a wave and its wave function is of the form ψ(x, t) = aei(kx−ωt) .
(2.1)
It is the simplest plane monochromatic wave. ψ refers to the case of wave propagating along x-direction whereas the disturbance is in a plane perpendicular to the x-direction. ψ given by Eq. (2.1) is called a plane wave since its √ √ amplitude |ψ| = ψ ∗ ψ = a∗ a = |a| is a constant. Schrödinger combined the particle and wave properties in constructing the wave equation. The angular frequency ω of the above wave should correspond to the energy E of the particle while the wave number k should correspond to the momentum px = mv. According to Planck and Einstein the relation between E of a particle and ω = 2πν of the wave is E = ~ω . (2.2) According to de Broglie h h 2π = = ~k . (2.3) λ 2π λ The wave equation satisfying the solution Eq. (2.1) and the relations (2.2)(2.3) is constructed as follows. Differentiating Eq. (2.1) with respect to t we obtain px =
−i
∂ψ = −ωψ . ∂t
(2.4)
Differentiation of Eq. (2.1) twice with respect to x gives ∂ 2ψ = −k 2 ψ . ∂x2
(2.5)
Schrödinger Equation and Wave Function 33 Replacing k by px /~ (Eq. (2.3)) and px in terms of E using E = p2x /2 (free particle energy) and then E by ~ω (Eq. (2.2)) we get ~ ∂2ψ = −ωψ . 2m ∂x2
(2.6)
Comparison of Eqs. (2.4) and (2.6) leads to the equation i~
~2 ∂ 2 ψ ∂ψ =− . ∂t 2m ∂x2
(2.7)
Equation (2.7) is called a Schrödinger wave equation or ψ-equation for a free particle in one-dimension. For a free particle in three-dimensions the wave equation is 2 ∂ψ ~2 ∂ ∂2 ∂2 i~ =− + + ψ. (2.8) ∂t 2m ∂x2 ∂y 2 ∂z 2 Multiplying the equation E = p2 /(2m) by ψ on both sides we get Eψ =
1 2 p ψ. 2m
(2.9)
Comparison of this equation with Eq. (2.8) gives E = i~
∂ , ∂t
p = −i~∇ ,
px = −i~
∂ , ∂x
(2.10)
where ∇ = i∂/∂x + j∂/∂y + k∂/∂z.
2.2.2
Schrödinger Equation for a Particle in a Potential
Since E − V = p2x /(2m), the wave equation for V (x) 6= 0 can be obtained from Eq. (2.7) by simply replacing i~∂/∂t by i~∂/∂t − V (x): i~
∂ψ ~2 ∂ 2 ψ =− + V (x)ψ . ∂t 2m ∂x2
(2.11)
For a three-dimensional system the Schrödinger equation is1 ∂ψ ~2 2 =− ∇ ψ + V (X)ψ , i~ ∂t 2m
2
∇ =
∂2 ∂2 ∂2 + + ∂x2 ∂y 2 ∂z 2
.
(2.12)
Equation (2.12) describes the evolution of ψ(X, t) in the presence of a potential function V (X). Whenever we make a measurement on a quantum system, the results are dictated by ψ. The Schrödinger equation is used and accepted because it leads to results in agreement with observation wherever it is applied properly. 1 In 1987 the Schrödinger equation appeared on the first-day postmark of the Austrian stamp commemorating Schrödinger’s 100th anniversary [1].
34 Quantum Mechanics I: The Fundamentals
2.2.3
Schrödinger Equation in Operator Form
Equation (2.12) shows that −
∂ ~2 2 ∇ + V (X) = i~ . 2m ∂t
(2.13)
This operator is called the Hamiltonian operator . In classical mechanics, Hamiltonian is a basic dynamical variable from which we obtain Hamilton’s equation of motion describing the evolution of state of a system. In quantum mechanics the Hamiltonian operator of the system essentially determines the time variation of ψ. Note that to set-up the wave equation for a quantum mechanical system we require classical Hamiltonian. Then we obtain the wave equation by replacing the classical dynamical variables X, p, V (X) and E by X, −i~∇, V (X) and i~∂/∂t respectively and operating both sides of the equation E = H(X, p, t) on ψ.
2.3 SOLUTION OF TIME-DEPENDENT EQUATION Let us assume that the potential V is independent of time and search the solution of Eq. (2.12) of the form ψ(X, t) = φ(X)f (t) .
(2.14)
Substitution of the above solution in Eq. (2.12) gives i~
f˙ ~2 1 2 =− ∇ φ+V . f 2m φ
(2.15)
The left-side of Eq. (2.15) is function of t only while the right-side is function of space variables only. Therefore, both sides of Eq. (2.15) must be equal to a same constant. Comparing the right-side of Eq. (2.15) with Hφ = Eφ we write ~2 2 − ∇ φ + V φ = Eφ . (2.16) 2m Rewrite the above equation as ∇2 φ(X) +
2m [E − V (X)] φ(X) = 0 . ~2
(2.17)
Equation (2.17) is called time-independent Schrödinger equation or stationary state Schrödinger equation. Further, we have i~f˙/f = E whose solution is f (t) = a e−iEt/~ ,
(2.18a)
where a is arbitrary. Different values of E give different solutions. So, we write fn (t) = an e−iEn t/~ .
(2.18b)
Schrödinger Equation and Wave Function 35 Then ψ is given by ψ(X, t) =
∞ X
an φn (X)e−iEn t/~ +
n=0
Z
aE e−iEt/~ φ(X) dX ,
(2.19)
where the first term in the right-side corresponds to discrete values of E while the second term is for the continuous range of values of E. In Eq. (2.19) φn and φ are solutions of Eq. (2.17).
2.4 PHYSICAL INTERPRETATION OF ψ ∗ ψ In connection with the wave function ψ we ask: 1. Is ψ an experimentally measurable quantity? 2. How is ψ related to the physical properties of the quantum mechanical system? 3. How do we determine the properties of the system from the given ψ? Schrödinger thought that ψ described a particle; that is, ψ was in some sense regarded as matter waves like sound waves and water waves. But this interpretation had several difficulties. First, when we try to locate the particles using, say, detectors we find them concentrated in a small region – it is localized . But the wave function ψ(X, t) at a given time spreads out like a wave – it is nonlocalized . Second, suppose we interpret ψ as a wave in physical space and time of one particle. In the case of n particles the wave has to be represented in a 3n-dimensional space. The third point is that ψ is generally complex. We can give interpretation to real waves only. How do we account for complex values of ψ? Therefore, a new interpretation is essential. ψ is in some sense related to the presence of a particle. Therefore, we do not expect a particle in the regions of space where ψ = 0. On the other hand, ψ must be nonzero in the region where there is no particle. In the year 1926 Max Born (Noble Prize in 1954) considered the collision of two particles. Far before and far after the collision the two particles behave as free particles. Using a perturbation theory he obtained Z X ψscat (r → ∞) ≈ cf i φf , cf i = φ∗f V φi dτ . (2.20) f
Here φf is the solution of the stationary state Schrödinger equation and represents the state of the particle far after scattering (called final state) with momentum Kf . φi is the initial state solution (state far before collision) with the momentum Ki and V is the potential. In Eq. (2.20) suppose |cf i |2 is related to the number of particles scattered with final state φf and initial state φi . That is, |c1i |2 , |c2i |2 , · · · represent, far after scattering, the number of particles (or percentage of incident particles) in the states φ1 , φ2 , · · · respectively
36 Quantum Mechanics I: The Fundamentals with all their state far before interaction is φi . In this case ψ would be describing the particles themselves. Assume a barrier potential having the property that it would reflect 60% of the incident particles and scatter 40% of them. There is no difficulty with the above meaning of cf i and ψ. Suppose only one particle incident on the barrier. Then according to the above 60% of it is to be scattered and 40% is to be reflected. Born realized that since a single particle cannot divide itself, ψ could not describe the particle itself. Therefore, only one interpretation is possible for the coefficient cf i – it is the probability amplitude that the incident particle with momentum Ki to be scattered with momentum Kf . We have no answer to the question What is the state after the collision? but we have the answer to the question What is the probability of a given effect of the collision? Considering the above Max Born and Ernst Pascual Jordan interpreted ψ ∗ ψ as position probability density. They interpreted |ψ|2 as P dxdydz = ψ ∗ ψdxdydz = |ψ|2 dxdydz ,
(2.21)
where P dxdydz is the probability that the system will be in the volume element dxdydz at the position X. The basis for Eq. (2.21) is that in the classical theory of light, the intensity of light at each point is given by the square of the amplitude of a wave. This indicates that, in quantum mechanics a wave function can be introduced playing the role of a probability. Note that, ψ itself cannot be a probability. This is because ψ may be complex and negative but probability is always real and positive. ψ has no direct physical significance and is not experimentally measurable. It is not a physical entity existing in space such as a cricket ball or electric field. Its purpose is to calculate |ψ|2 . That is, it is a mathematical tool for computing the outcomes of observations. According to probability interpretation of ψ ∗ ψ, we can never say that quantum particle exists or not at a certain place2 . Born, Bohr and his groups in Copenhagen analysed the probability interpretation of ψ ∗ ψ in great depth. For this reason probabilistic interpretation of ψ ∗ ψ is also referred as the Copenhagen interpretation. Other interpretations have also been proposed. For example, sum-over histories [3,4] by Richard Feynman and many worlds interpretation [5,6] by Huge Everett and John Cramer’s transactional interpretation [7,8] have also received considerable interest. The point is that a given initial conditions can give a range of final outcomes, so quantum theory makes the future open and hinder our ability to know the past.
2.5 CONDITIONS ON ALLOWED WAVE FUNCTIONS The probability interpretation of ψ ∗ ψ leads to following conditions on ψ. 2 In connection with the above Robert Oppenheimer made a remark that [2] we must say “no” for whether the position of the electron remains the same or changes with time or the electron is at rest or it is in motion.
Schrödinger Equation and Wave Function 37 1. Integration of ψ ∗ ψ with respect to space variables over all space should be unity: Z ∞ ψ ∗ ψ dτ = 1 . (2.22) −∞
2. |ψ| → 0 as |X| → ∞. 3. ψ must be single-valued at each and every point. 4. ψ must not have the value of infinite at any point. 5. ψ, ψx , ψy and ψz must be continuous functions of X. The integral in Eq. (2.22) represents total probability. Since total probability must be one for any event ψ must satisfy the condition Eq. (2.22). Such a wave function is said to be normalized . The condition Eq. (2.22) is called normalization condition. If Z ∞ ψ ∗ ψ dτ = N 2 , N 2 = constant 6= 1, 6= ∞ (2.23) −∞
then ψ is said to be square-integrable and the constant N 2 is called norm of ψ. Such a wave function can always be normalized by multiplying it by the constant 1/N . This is possible because the Schrödinger equation is linear and therefore if ψ is a solution of it then αψ, where α is a constant, is also a solution. The finiteness requirement of the norm imposes the condition (2). This means that ψ ∗ ψ → 0 as |X| → ∞ and thus the particle is confined to a finite region of space. Such a state is said to be a bound state. It is called a bound state because the corresponding classical particle motion is bounded. If ψ(X, t) does not tend to 0 as |X| → ∞ then it is nonnormalizable. Such a state is called a scattering state. The motion of the corresponding classical particle is unbounded. Note that an overall phase eiα in a wave function is unobservable. ψ and iα e ψ both describe the same physical state. But if a constant phase appears as a relative phase between two ψ’s in a linear superposition as in ψ = ψ1 + eiα ψ2 then it is observable; it gives interference effects and a shift in α introduces a shift in interference fringes. Is ψ(X, t) a dimensionless quantity? If not, what is the dimension of it? The probability of finding a particle in a given volume dτ = dxdydz at a point X at a time t is |ψ(X, t)|2 dxdydz. Probability is a dimensionless quantity. Therefore, |ψ|2 dxdydz should also be dimensionless. dx, dy and dz have the −3/2 dimension of length. Therefore, the dimension of ψ(X, t) is (length) . This is for a single particle in three-dimensional space. For N -particles in a d−N d/2 dimensional space the dimension of the wave function is (length) .
Solved Problem 1: If ψ = sechax, calculate the probability of finding the particle right of the point x = 0.
38 Quantum Mechanics I: The Fundamentals First, the wave function must be normalized. For this purpose we write ψ = N sechax. The normalization condition becomes Z ∞ N2 2N 2 2 sech2 (ax) dx = 1=N tanh(ax)|∞ . (2.24) −∞ = a a −∞ p p Thus, N = a/2 and ψ = a/2 sechax. Now, the probability of finding the particle right of the point x = 0 is Z Z ∞ 1 a ∞ sech2 (ax) dx = . (2.25) P = ψ ∗ ψ dx = 2 0 2 0
Solved Problem 2: 2 2 1 If the normalized wave function of a particle is ψ = √ 1/4 e−x /(2a ) find the aπ region of space where the particle is most likely to be found.
The probability density is given by 2 2 1 |ψ|2 = √ e−x /a . a π
(2.26)
This is a simple function. We use the graph of |ψ|2 versus x to find the maximum value of |ψ|2 . It is maximum at x = 0 and decreases exponentially for |x| > 0. Thus, the particle is most likely to be found in a region of width a on either side of x = 0. For the complicated form of |ψ|2 = f (x) the conditions df /dx = 0 and d2 f /dx2 < 0 are to be used to determine a maximum of f (x).
2.6 BOX NORMALIZATION Normalization of a given wave function is possible only if the norm, the value R∞ of −∞ ψ ∗ ψ dτ , is finite. But for many physically interesting wave functions the value of this integral becomes infinite. For example, for the wave function given by Eq. (2.1) the value of the integral is infinite. Hence, it is nonnormalizable. However, such wave functions are widespread in quantum mechanics, specifically in scattering problems, and they have considerable mathematical advantages. An approach of handling a nonnormalizable ψ is the following. Assume that the particle is confined within a large box and that the ψ ∗ ψ is taken over the interior of the box only. Discard the infinite region of space outside the box. Now, the norm is finite; it can be made unity by multiplying a suitable constant. For any calculation use this normalized wave function. After completion of calculation, impose the limit of infinite volume for the box. This procedure is called box normalization. In this way the nonnormalizable ψ is visualized as a limit of realizable and normalizable wave function.
Schrödinger Equation and Wave Function 39 Let us perform the above scheme to the wave function, Eq. (2.1). Suppose the box is a cube with edges of length L parallel to the x, y, z axes and its centre is the origin. Then the normalization gives Z Z Z L/2 Z Z Z L/2 dxdydz = L3 |a|2 . (2.27) 1= ψ ∗ ψ dxdydz = |a|2 −L/2
−L/2
Hence, a = L−3/2 and the normalized wave function is ψ = L−3/2 e−i(k·X−ωt) .
(2.28)
Note that the condition ψ(X, t) → 0 as |X| → ∞ is not applicable to the above ψ. Impose that ψ is periodic with respect to L of the box. Then eikx L = eiky L = eikz L = 1
(2.29)
and hence kx , ky and kz must be integer multiples of 2π/L: k=
2π n, L
nx , ny , nz = 0, ±1, · · · .
(2.30)
Thus, the momentum vectors are discrete or quantized for the particle in a box. The box normalization discretizes k. When the values of k are continuous the so-called Dirac-delta normalization is useful. Consider the integral Z ∞ ZZZ ∞ ψk∗ (X)ψl (X) dτ = c∗ (k)c(l) ei(k−l)·X dxdydz . (2.31) −∞
We have Z ∞
−∞
e
i(kx −lx )x
dx = lim
−∞
Therefore,
Z
L→∞
Z
L
−L
ei(kx −lx )x dx = 2πδ (kx − lx ) .
ψk∗ ψl dτ = c∗ (k)c(l)(2π)3 δ(k − l) .
(2.32)
(2.33)
When l = k we have c(k) = 1/(2π)3/2 .
2.7 CONSERVATION OF PROBABILITY The probability interpretation of ψ ∗ ψ can be made consistently, provided the conservation of probability is guaranteed. If the probability of finding a particle in a region of space decreases as time increases then the probability of finding it outside this region must increase by the same amount so that total probability is always unity. Therefore, we expect ρ = ψ ∗ ψ to satisfy the conservation equation (also called continuity equation) ∂ρ +∇·J= 0 . ∂t
(2.34)
40 Quantum Mechanics I: The Fundamentals The continuity equation is necessary to prove that particles do not disappear locally. In hydrodynamics ρ in Eq. (2.34) is the fluid density and J represents the fluid current density. In electrodynamics ρ and J are the charge density and current density respectively. Let us derive a probability conservation law for ψ from the Schrödinger equation. Define ρ = ψ ∗ ψ as the position probability density. The probability P of R the particle in a volume V is V ρ dτ . The rate of change of P is Z Z ∂ ∂ψ ∂ψ ∗ ∂P = ρ dτ = ψ∗ +ψ dτ . (2.35) ∂t ∂t ∂t ∂t Replace the time derivative terms in Eq. (2.35) by equivalent space derivatives using Schrödinger equation. The Schrödinger equation for ψ ∗ is obtained from Eq. (2.12) as ∂ψ ∗ ~2 2 ∗ −i~ =− ∇ ψ + V ψ∗ . (2.36) ∂t 2m Multiplications of the Schrödinger Eq. (2.12) by ψ ∗ and Eq. (2.36) by ψ and subtraction of one from another gives ψ∗
∂ψ ∂ψ ∗ +ψ ∂t ∂t
= = =
where J=
~ ∗ 2 ψ ∇ ψ + ψ∇2 ψ ∗ 2mi ~ − ∇ · [ψ ∗ ∇ψ − ψ∇ψ ∗ ] 2mi −∇ · J , −
~ [ψ ∗ ∇ψ − ψ∇ψ ∗ ] . 2mi
(2.37)
(2.38)
J can also be expressed as J=
~ Re (−iψ ∗ ∇ψ) m
or J =
~ Im (ψ ∗ ∇ψ) . m
Using Eq. (2.37) in Eq. (2.35) we obtain Z ∂P =− ∇ · J dτ . ∂t V R Substituting P = V ρ dτ in Eq. (2.40) we have Z Z ∂ ρ dτ + ∇ · J dτ = 0 . ∂t V V Therefore,
(2.39)
(2.40)
(2.41)
∂ρ +∇·J =0 . (2.42) ∂t The interpretation of ψ ∗ ψ as a probability density leads to the satisfaction of continuity equation.
Schrödinger Equation and Wave Function 41 J is defined as a probability current density or probability flux vector . J at X is the amount of probability crossing at that point in unit time. It corresponds to the Poynting vector in classical electromagnetic theory. The increase (decrease) of the probability inside a finite volume V is due to the influx (outward flux) of the probability current through the surface A bounding this volume. What is the significance of ∇ · J = 0?
Solved Problem 3: Find the current density carried by the wave function Aeikx of a free particle and verify that it satisfies the equation of continuity. In one-dimension the probability current density J is given by ~ ∂ψ J = Im ψ ∗ . m ∂x
(2.43)
For Aeikx we get J=
~k 2 ~ Im A∗ e−ikx Aikeikx = |A| = v|A|2 . m m
(2.44)
We obtain ∂ρ/∂t = 0 and ∂J/∂x = 0 and hence ∂ρ/∂t + ∂J/∂x = 0.
2.8 EXPECTATION VALUE Suppose we measure the position of a quantum mechanical particle a large number of times. Even if the wave function of the particle is same before every observation we do not get same result. In this case we are interested to know the average or expected value of the set of measurements of position denoted as hxi. Knowing ψ, how do we compute hxi? To obtain the expression for hxi, instead of a quantum system, consider a classical system, for example a die with six sides numbered as 1, 2, 3, 4, 5 and 6 or say a1 , a2 , · · · , a6 respectively. Throw the die N (very large) number of times and record the outcome in each throw. Let, Ni denote the number of times the event ai occurred. Then the average or expectation value of throw of the die is X 1 houtcomei = (N1 a1 + N2 a2 + · · · + N6 a6 ) = ai P (ai ) N i or simply houtcomei =
X
aP (a) ,
(2.45)
a
where summation is over all possible values of a. Here the possible values of a are assumed as discrete. When the outcome of an experiment takes any value in an interval, say, [−∞, ∞] then Z ∞ houtcomei = aP (a) da . (2.46) −∞
42 Quantum Mechanics I: The Fundamentals In quantum theory P (X, t)dτ = ψ ∗ ψdτ , where dτ = dxdydz, gives the probability of finding a particle in a volume dτ centred at X and hence for position measurements the Eq. (2.46) becomes Z ∞ Z ∞ hXi = XP (X, t) dτ = Xψ ∗ ψ dτ −∞ −∞ Z ∞ = ψ ∗ (X, t)Xop ψ(X, t) dτ , (2.47) −∞
where ψ(X, t) is assumed to be normalized. For simplicity drop the subscript ‘op’ in the operator. Equation (2.47) gives the expectation value of position of a quantum mechanical particle. The expectation value of Xn is given by in terms of ψ or φ Z hXn i =
∞
ψ ∗ (X, t)Xn ψ(X, t) dτ .
(2.48)
−∞
In a similar way, the expectation value of any quantity which is a function of position (and time) only can be defined. For a function f (X, t) Z ∞ hf (X, t)i = ψ ∗ f ψ dτ , (2.49) −∞
where f in the integral is the operator form of the function f . How do we define expectation values of momentum and functions of momentum? First, consider hpi. The first-order time derivative of hXi is shown to be hpi/m (see the next section for its proof). That is, 1 d hXi = hpi . dt m
(2.50)
We start with the above equation and obtain Z ∞ d hpi = m ψ ∗ Xψ dτ dt −∞ Z ∗ ∂ψ ∂ψ ∗ = m Xψ + ψ X dτ ∂t ∂t Z i~ 2 ∗ = − ∇ ψ Xψ − ψ ∗ X∇2 ψ dτ 2 Z ∞ i~ = − (∇ψ ∗ Xψ − ψ ∗ ψ − ψ ∗ X∇ψ) + 2 ψ ∗ ∇ψ dτ . 2 −∞
Since ψ → 0 as |X| → ∞, we get Z hpi =
∞
ψ ∗ p ψ dτ ,
−∞
where p in the integral is operator form of momentum.
(2.51)
Schrödinger Equation and Wave Function 43 Next, find the expression for hp2 i. Since E = p2 /(2m) + V (X) we write Z ∞ Z ∞ 2 ∗ ∗ hp i = 2m ψ Eψ dτ − ψ V ψ dτ −∞ −∞ Z Z ∗ ∗ ∂ψ dτ − ψ V ψ dτ = 2m −i~ ψ ∂t Z = −~2 ψ ∗ ∇2 ψ dτ Z ∞ = ψ ∗ p2 ψ dτ . (2.52) −∞
Equations (2.51) and (2.52) suggest that Z ∞ hpn i = ψ ∗ pn ψ dτ .
(2.53)
−∞
Then hf (p)i =
Z
∞
ψ ∗ f (p)ψ dτ . (Why?).
(2.54)
−∞
In view of Eqs. (2.49) and (2.54) the expectation value of an arbitrary dynamical variable A(X, p, t) is Z ∞ hAi = ψ ∗ Aψ dτ . (2.55) −∞
The expectation values are functions of time only, since the space coordinates have to be integrated out. The expectation value may be real or complex depending on the nature of A. However, in an experiment we can measure only real value. Thus, only the dynamical variables whose expectation values are real to be considered directly measurable or observable. Therefore, if A is an observable then hAi = hAi∗ . That is, Z ∞ ∗ Z ∞ Z ∞ ψ ∗ Aψ dτ = ψ ∗ Aψ dτ (Aψ)∗ ψ dτ . (2.56) = −∞
−∞
−∞
To ensure the above requirement the operator A must have a certain property called self-adjointness which will be discussed in the next chapter. Equation (2.55) is defined for a noninteracting particle. For the more general case, such as a particle interacting with a heat bath or the environment, the expected value is given by [9] hA(t)i = Tr[ρ(t)A] ,
(2.57)
where ρ is the density matrix and is the solution of the von Neumann equation i~
∂ρ = [H, ρ] ∂t
(2.58)
44 Quantum Mechanics I: The Fundamentals with H now the Hamiltonian for this entire system and [H, ρ] = Hρ − ρH is called commutator of H and ρ (which will be discussed in the next chapter).
Solved Problem 4: Calculate hxi, hx2 i, hpx i, hp2x i if ψ = e−kx for x > 0 and 0 for x < 0. To normalize the wave function ψ assume that ψ = N e−kx for x > 0 and 0 for x < 0. The normalization condition gives Z ∞ N 2 −2kx ∞ N2 2 e−2kx dx = − 1=N e . (2.59) = 2k 2k 0 0 √ Therefore, N = 2k . Now Z ∞ 1 hxi = 2k x e−2kx dx = . (2.60) 2k 0 Z ∞ 1 hx2 i = 2k x2 e−2kx dx = 2 . (2.61) 2k Z0 ∞ d e−kx dx = i~k . (2.62) hpx i = 2k e−kx −i~ dx 0 Z ∞ 2 2 d hp2x i = 2k e−kx (−i~) e−kx dx = ~2 k 2 . (2.63) dx2 0
Solved Problem 5: The ground state (lowest energy state) wave function (in spherical polar coordinate) of a hydrogen atom is given by ψ = e−r/a0 where a0 is a constant. Calculate the average value of r. Writing ψ = N e−r/a0 the normalization condition gives Z Z ∞ Z π Z 2π 1 = ψ ∗ ψ dτ = N 2 e−2r/a0 r2 sin θ dr dθ dφ , 0
0
(2.64)
0
where r2 sin θdrdθdφ is the volume element dτ in the spherical polar coordinate system. Integrating with respect to φ and θ we get Z ∞ 1 = 4πN 2 r2 e−2r/a0 dr . (2.65) 0
Integrating by parts we obtain N = 1/(πa30 )1/2 . Then hri is Z ∞ Z π Z 2π hri = ψ ∗ rψr2 sin θ dr dθ dφ 0 0 0 Z ∞ 4 3 = r3 e−2r/a0 dr = a0 . 3 a0 0 2
(2.66)
Schrödinger Equation and Wave Function 45
2.9 EHRENFEST’S THEOREM In classical mechanics X and p satisfy the equations dX 1 = p, dt m
dp = F = −∇V . dt
(2.67)
What are the forms of these equations in quantum mechanics? In quantum mechanics rate of change of hX(t)i and hp(t)i are calculated using the Schrödinger equation as shown below.
2.9.1
Equations of Motion of hXi and hpi
We write
d d hXi = dt dt
Z
∞
−∞
ψ ∗ Xψ dτ =
Z
∞
−∞
ψ∗ X
∂ψ dτ + ∂t
Z
∞
−∞
∂ψ ∗ Xψ dτ . (2.68) ∂t
∗
Using the Schrödinger equations for ψ and ψ replace the time derivative terms ∂ψ/∂t and ∂ψ ∗ /∂t by space derivative terms. We obtain Z ∞ ∗ d i~ hXi = ψ X∇2 ψ − ∇2 ψ ∗ Xψ dτ . (2.69) dt 2m −∞
Integration of second integral in Eq. (2.69) by parts gives Z ∞ Z ∞ ∞ ∇2 ψ ∗ Xψ dτ = Xψ∇ψ ∗ − (∇ψ ∗ ) ∇ · (Xψ) dτ −∞ −∞ −∞ Z ∞ ∗ = − (∇ψ ) ∇ · (Xψ) dτ −∞ Z ∞ ∞ = −ψ ∗ ∇ · (Xψ) + ψ ∗ ∇2 (Xψ) dτ −∞ −∞ Z ∞ ∗ 2 = ψ 2∇ψ + X∇ ψ dτ . (2.70) −∞
Now,
d hXi = dt = =
Z ∞ ∗ i~ ψ X∇2 ψ − 2ψ ∗ ∇ψ − ψ ∗ X∇2 ψ dτ 2m −∞ Z i~ ∞ ∗ − ψ ∇ψ dτ m −∞ 1 hpi . m
(2.71a)
46 Quantum Mechanics I: The Fundamentals In a similar manner, we calculate the time rate of change of hpi as Z ∞ d d hpi = −i~ ψ ∗ ∇ψ dτ dt dt −∞ Z ∞ Z ∞ ∂ψ ∗ ∂ψ ∗ = −i~ ψ ∇ dτ + ∇ψ dτ ∂t −∞ −∞ ∂t Z ∞ = − ψ ∗ (∇(V ψ) − V ∇ψ) dτ −∞ Z ∞ = − ψ ∗ (∇V )ψ dτ =
−∞
h−∇V i .
(2.71b)
Equations (2.71) are analogous to the classical equations of motion, Eq. (2.67). Thus, the quantum mechanical hXi and hpi obey the classical equation of motion. This is known as Ehrenfest’s theorem. The theorem states that the Schrödinger equation leads to the satisfaction of Newton’s second-law of motion on the average. The Eqs. (2.71) are crucial for understanding the emergence of classical mechanics from quantum mechanics.
2.9.2
Verification of Ehrenfest’s Theorem
In this section we verify the Ehrenfest’s theorem in certain physically interesting systems. First, we solve the Eqs. (2.71) for linear harmonic oscillator and show that their solutions are identical to the solutions of classical equation of motion. Next, we consider a particle in an infinite square-well potential. We give the solutions of classical and quantum cases. We discuss the classical limit of quantum solutions. Then for a particle in a magnetic field we show that the expressions for hpi in classical and quantum cases are identical.
2.9.2.1 Linear Harmonic Oscillator For the linear harmonic oscillator with potential V (x) = ω 2 x2 /(2m) the classical equation of motion is dx 1 = px , dt m Its solution is
dpx = −mω 2 x . dt
(2.72)
px0 sin ωt , (2.73) mω are integration constants. The quantum equations for hxi x = x0 cos ωt +
where x0 and px0 and hpx i are or
d 1 hxi = hpx i , dt m d2 hxi dt2
d hpx i = −mω 2 hxi dt + ω 2 hxi = 0 .
(2.74) (2.75)
Schrödinger Equation and Wave Function 47 The solution of it is hxi = hxi0 cos ωt +
hpx i0 sin ωt mω
(2.76)
which is same as the Eq. (2.73).
2.9.2.2 Particle in an Infinite Square-Well Potential Let us consider a particle of mass m confined to one-dimensional infinite height square-well potential. The potential is V (x) = 0 for 0 < x < a and ∞ otherwise.
Classical Case The particle moves between the walls x = 0 and a with a uniform velocity v. Its velocity gets reversed when it hits the walls. This to-and-fro motion is given by [10] ( 2at/T, 0 < t < T /2 1 (2.77) x = px t = m 2a − 2at/T, T /2 < t < T, where px = mv = ma/(T /2) and T /2 is the time taken by the particle to go from one side of the wall to the other. A Fourier series of this solution is given by ∞ a 4a X cos(2r + 1)ωt xc = − 2 , ω = 2π/T . (2.78) 2 π r=0 (2r + 1)2
Quantum Solution The normalized eigenfunctions and energies are given by (refer to sec. 4.6) p ψn (x, t) = φn (x)e−iEn t/~ = 2/a sin (kn x) e−iEn t/~ , (2.79)
where
nπ n2 ~2 π 2 , En = . a 2ma2 P We write the wave function as ξ(x, t) = n cn ψn . Then Z a 2a hxi = ξ ∗ xξ dx = − 2 [I(t) + I ∗ (t)] , 2 π
(2.80)
kn =
(2.81a)
where I(t) =
∞ X ∞ X
c∗s cs+2r+1 e−i(2r+1)G
r=0 s=1
and G = (2s + 2r + 1)π 2 ~ t/(2ma2 ).
1 1 − 2 (2r + 1) (2s + 2r + 1)2
(2.81b)
48 Quantum Mechanics I: The Fundamentals
Classical Limit The de Broglie wavelength of a particle of momentum px is λ = h/px . In the potential well we approximately fit N such half-wavelengths where N=
a 2px a ma2 ω = = 2 ≫1. λ/2 h π ~
(2.82)
The energy En for this large quantum number N will correspond to the classical kinetic energy Ec = p2x /(2m). cn ’s are significant only over a narrow range N − δ ′ ≤ n ≤ N + δ ′′ of energy states so that NX +δ ′′
n=N −δ ′
|cn |2 = 1 with δ ′ , δ ′′ ≪ N .
(2.83)
In Eq. (2.81b) terms for which s ≈ N and r ≪ N alone contribute. G goes over to N π 2 ~ t/(ma2 ), that is to ωt. Neglect 1/(2s + 2r + 1)2 comparing to 1/(2r + 1)2 because s ≈ N is very large. Thus I(t) ≈
∞ X r=0
NX +δ ′′
s=N −δ ′
c∗s e−i(2r+1)ωt . (2r + 1)2
(2.84)
The sum over s goes to unity. Now, (2.81a) becomes hxi ≈
∞
a 4a X cos(2r + 1)ωt − 2 π 2 r=0 (2r + 1)2
(2.85)
which agrees with the classical result Eq. (2.78). Thus, we conclude that as N → ∞, the quantum hxi goes over to the corresponding classical trajectory.
2.9.2.3 Particle in a Magnetic Field Consider a charged particle moving in a uniform field B [11]. B is along the z-axis. The motion is then bi-dimensional in the (x, y) plane. Suppose Ax = −By, Ay = 0. The momentum p is mv + eA where e and m are, respectively, the charge and mass of the particle. The Hamiltonian of the problem H=
1 1 1 (p − eA)2 + p2x + p2y + mω 2 y 2 + ωypx . 2m 2m 2
Then the Schrödinger equation is ∂ψ ~2 ∂ 2 ψ ∂ 2 ψ 1 ∂ψ i~ + + mω 2 y 2 ψ − i~ωy =− . 2 2 ∂t 2m ∂x ∂y 2 ∂x
(2.86)
(2.87)
Schrödinger Equation and Wave Function 49
Classical Case In terms of X, v, f (X, v, t) and B the Liouville equation is ∂f ∂f e ∂f +v· + (v × B) · =0. ∂t ∂X m ∂v
(2.88)
We multiply Eq. (2.88) by X and integrate with respect to X then multiply by v and integrate again. After some integrations by parts we obtain [11] Z d hXi = hvi = vf dX dv , dt Z d e e hvi = hv × Bi = v × Bf dX dv . dt m m R Defining J(X, t) = e f v dv, Eq. (2.89) is rewritten as Z d d hXi = hvi , m hvi = J × B dX . dt dt
(2.89a) (2.89b)
(2.90)
When the field is uniform d hXi = hvi , dt
d e hvi = hvi × B . dt m
(2.91)
Quantum Mechanical Case For a particle in a magnetic field the equation for hXi and hvi are d hXi = dt
hvi ,
d e hvi = hv × B − B × vi , dt 2m
(2.92a)
where v
=
1 1 (p − eA) = (−i~∇ − eA) . m m
(2.92b)
From Eq. (2.92) we write d hXi = dt d m hvi = dt = = = =
Z 1 hvi = − ( i~ψ ∗ ∇ψ + eAψ ∗ ψ) dX, (2.93a) m e hv × B − B × vi 2m Z e − ψ ∗ [( i~∇ + eA) × B − B × ( i~∇ + eA)] ψ dX 2m Z Z e2 i~e − |ψ|2 A × B dX − ψ ∗ (∇ × (Bψ) + ∇ψ × B) dX m 2m Z i~ ∗ e2 − ψ ∇ψ − ψ∇ψ ∗ − |ψ|2 A × B dX 2m m Z J × B dX , (2.93b)
50 Quantum Mechanics I: The Fundamentals where the quantity inside the square-bracket is the current density J. Equations (2.93) are identical to the classical formulas, Eqs. (2.90). For a uniform field, the Eqs. (2.93) for the mean values become identical to Eqs. (2.91). Some other examples on the significance of Ehrenfest’s theorem are given in refs.[12,13]. Ehrenfest’s theorem is neither a necessary nor a sufficient condition for a quantum system to display classical behaviour. The harmonic oscillator satisfies Eqs. (2.71). But a quantum oscillator has discrete energy eigenvalues contradictory to the classical system. The system obeys Ehrenfest’s theorem but does not behave classically. This system is an example of lack of sufficiency. For an example of lack of necessity—that a system may behave classically even when Ehrenfest’s theorem does not apply—see ref.[12].
2.10 BASIC POSTULATES In classical mechanics we work with the position X, the momentum p, the total energy E, etc. These quantities are called dynamical variables. In quantum mechanics the dynamical variables are no longer variables but operators. An operator is operated on the wave function ψ to get various outcomes of the corresponding physical variable. From our discussion on Schrödinger equation and its wave function so far, we now list important basic postulates in quantum mechanics.
Postulate 1: For every dynamical system there exists a wave function (and its spatial derivatives). It is continuous, finite, single-valued function of the parameters of the system, space coordinate and time. From the wave function, predictions of physical properties of the system to be made. This postulate contains requirements that all physical waves such as sound waves, electromagnetic waves and water waves meet.
Postulate 2: The Hamiltonian operator of a quantum mechanical system determines the evolutions of the system. The classical Hamiltonian E = H = p2 /(2m)+V (X) is converted into the wave Eq. (2.12) by the following substitution of operators ˆ f (x) → fˆ(x), px → −i~∂/∂x, for dynamical variables: x → x ˆ, X → X, p → −i~∇ and E → i~∂/∂t.
Postulate 3: ψ ∗ ψ is interpreted as position probability density and an admissible wave function must satisfy the normalization condition (2.22) to make total probability unity.
Schrödinger Equation and Wave Function 51
Postulate 4: For every observable physical quantity, there exists a corresponding operator with real eigenvalues. The only measurable values of an observable are the possible eigenvalues of the corresponding operator.
Postulate 5: Quantum mechanics does not say with certainty the outcome an experiment. But it allows computing of probabilities for the various possible outcomes. Consequently, we are interested in the average value of a dynamical variable A and this is given by Eq. (2.55).
2.11 TIME EVOLUTION OF STATIONARY STATES Consider the time-independent Schrödinger equation. As stated earlier the state of a quantum mechanical system is represented by a wave function ψ. |ψ|2 is assumed to give the position probability density. From the solution φ(X) of time-independent Schrödinger Eq. (2.17), the solution of the timedependent Schrödinger equation is written as ψn (X, t) = φn (X)e−iEn t/~ .
(2.94)
We note that |ψn (X, t)|2 is independent of time. Further, the time dependence of ψn (X, t) is periodic with angular frequency ω = En /~. Such states are called stationary states. In quantum mechanics a stationary state does not mean that the system is strictly in a rest state or in a classical equilibrium point. In this state, expectation values of all dynamical variables Z ∞ Z ∞ ∗ hAi = ψn (X, t)Aψn (X, t)dτ = φ∗n (X)Aφn (X)dτ (2.95) −∞
−∞
are independent of time. It means that though ψ may vary with time, ψ ∗ ψ at any point in the configuration space is independent of time. The solution φn (X) of the time-independent Schrödinger Eq. (2.17) for a fixed value of En is obviously time-independent and thus φ∗n φn is independent of time. Therefore, we call Eq. (2.17) the stationary state Schrödinger equation and its solutions as stationary state solutions or eigenfunctions. This is an energy eigenvalue equation, Hφ = Eφ. Consequently, when the state of the particle is the solution of Eq. (2.17) then the energy of it has a definite value given by the eigenvalue E. Suppose a particle is initially in a stationary state represented by a ψn (X, 0) = φn (X). Then from Eq.(2.94) we write ψn (X, t) = ψn (X, 0)e−iEn t/~
(2.96)
and |ψn (X, t)|2 = |ψn (X, 0)|2 is independent of time. That is, If a system is initially in a stationary state then it remains in that state forever . Though
52 Quantum Mechanics I: The Fundamentals ψn (X, t) changes with time the energy and expectation values of observables remain the same. We note that the wave function given by Eq. (2.96) is a particular solution of the time-dependent Schrödinger equation. A combination of such particular solutions is given by the Eq. (2.19). Now we ask: Can the wave function given by Eq. (2.19) represent a stationary state? The answer is no, as shown below. Let us calculate ψ ∗ ψ of ψ given by Eq. (2.19). For simplicity assume that the energy values are discrete. We write ψ ∗ as X ψ∗ = a∗m φ∗m eiEm t/~ . (2.97) m
Then ψ∗ ψ
=
XX m
=
X n
a∗m an φ∗m φn e−i(En −Em )t/~
n
|an |2 |φn |2 +
XX
a∗m an φ∗m φn e−i(En −Em )t/~ . (2.98)
m6=n n
Since the second term in the above equation contains t, ψ ∗ ψ is dependent on time. Therefore, solution ψ given by the Eq. (2.19) does not represent a stationary state solution. From our above discussion we say that a stationary state solution of a quantum mechanical system is obtained by simply finding the solution of the time-independent Schrödinger Eq. (2.17) and then multiply it by e−iEt/~ , provided the potential is time-independent. Now an interesting question is: What are the allowed values of E? Can E be arbitrary? In other words, does there exist a (bounded)solution of Eq. (2.17) for all real values of E? The answer is no. Solutions exist only for certain specific values of E. This is because the solution of Eq. (2.17) satisfies the admissible conditions only for certain values of E. The allowed energy values are called the energy spectrum or eigenvalue spectrum or energy levels of the system. Can you recall some of the equations that are in the form of Aφ = λφ where A is a linear differential operator for which bounded solution exist only for certain discrete real values of λ? What are their eigenvalues and eigenfunctions?
2.12 CONDITIONS FOR ALLOWED TRANSITIONS Suppose we wish to know whether a particle is able to make a sudden transition from a given state to another state. Also we ask: Can a particle jump to an arbitrary state? If not, to which states can it jump? What does happen when the particle goes from one energy state to another? We will address these questions in this section. Bohr postulated that when a particle makes transition from a higher energy level Em to a lower energy level En then the frequency of the radiation emitted by the particle is νmn = (Em − En ) /~. Can a particle emit radiation when
Schrödinger Equation and Wave Function 53 it is in stationary energy state? The answer is no. To prove this consider a quantum system of a particle moving only in the x-direction and is in a state φn (x) with energy En . The time-dependent eigenfunction associated with this state is written as ψn (x, t) = φn (x)e−iEn t/~ . (2.99) Now, the expectation value of the position of the particle is Z ∞ Z ∞ hxi = xψn∗ (x, t)ψn (x, t)dx = xφ∗n (x)φn (x)dx −∞
(2.100)
−∞
which is independent of time. Thus, the hxi of the particle does not oscillate and no radiation occurs. Next, we calculate hxi for the transition from Em to En . The ψ of the particle is aφm + bφn where a∗ a and b∗ b are the probabilities for the particle to be in φm and φn respectively. We obtain Z ∞ hxi = cos (νmn t) x [ab∗ φm φ∗n + a∗ bφ∗m φn ] dx −∞ Z ∞ +i sin (νmn t) x [ab∗ φm φ∗n − a∗ bφ∗m φn ] dx . (2.101) −∞
If the particle is in a specific state Em or En then ν = 0 and hxi = constant, independent of time. When the particle makes a transition from the state Em to the state En then hxi oscillates with time and the frequency of radiation emitted is given by νmn = (Em − En ) /~. From Eq. (2.101) we obtain the condition for hxi 6= 0 as Z ∞ φ∗m x φn dx 6= 0 . (2.102) −∞
If the value of the above integral is nonzero then the transition from the level Em to En is allowed otherwise the transition is forbidden. The conditions on m and n for which the integral in Eq. (2.102) is nonzero are called selection rules. Since the electric dipole operator is given as ex, the transition based on the above selection rule is called electric dipole transition.
2.13 ORTHOGONALITY OF TWO STATES A wave function belonging to a specific energy (eigen)value En is commonly called an energy eigenfunction φn . An important property of eigenfunctions φn of stationary state Schrödinger equation is that two eigenfunctions belonging to two different energy values Em and En satisfy the integral equation Z ∞ φ∗m φn dτ = 0. (2.103) −∞
54 Quantum Mechanics I: The Fundamentals P In vector analysis two vectors u and v for which u · v = i ui vi = 0 are called orthogonal . (If the number of the components of u and v are very large and the increment is infinitesimally small then we replace the summation in the above by integral). In analogy with this φm and φn satisfying the condition (2.103) are called orthogonal functions. Orthogonal functions refer to states that are independent of one another. The different possible energy states of a particle are all orthogonal to one another. Let φm and φn are two eigenfunctions with energies Em and En respectively. These functions satisfy the equations (refer to Eq. (2.17)) 2m (Em − V ) φm ~2 2m ∇2 φn + 2 (En − V ) φn ~
∇2 φm +
=
0,
(2.104a)
=
0.
(2.104b)
The equation for φ∗m is ∇2 φ∗m +
2m (Em − V ) φ∗m = 0 . ~2
(2.104c)
Multiplying Eq. (2.104b) by φ∗m and (2.104c) by φn and subtracting one from another we obtain φ∗m ∇2 φn − φn ∇2 φ∗m +
2m (En − Em ) φ∗m φn = 0 . ~2
(2.105)
Integrating Eq. (2.105) over the space variables we get Z ∞ Z ∞ 2m ∗ 2 2 ∗ φ∗m φn dτ = 0 . (2.106) φm ∇ φn − φn ∇ φm dτ + 2 (En − Em ) ~ −∞ −∞
Using Green’s second identity Z Z f ∇2 g − g∇2 f dV = (f ∇g − g∇f ) dS V
(2.107)
S
Eq. (2.106) becomes Z ∞ Z 2m φ∗m φn dτ = 0 . (φ∗m ∇φn − φn ∇φ∗m ) dS + 2 (En − Em ) ~ −∞ S
(2.108)
Since φm , φn → 0 as |X| → ∞ the surface integral vanishes and we have Z ∞ (En − Em ) φ∗m φn dτ = 0 . (2.109) −∞
Because Em 6= En we have condition Eq. (2.103). When Em = En and φn is linearly independent of φm then the two states are said to be degenerate states R ∞ ∗having the same eigenvalue but different eigenfunctions. In such cases −∞ φm φn dτ need not be zero. But it is possible to choose a set of orthogonal
Schrödinger Equation and Wave Function 55 functions Rby using Schmidt’s orthogonalization procedure. The integral for ∞ m = n is −∞ φ∗n φn dτ = finite and it can be normalized to one by normalization condition. Hence, if each φ is normalized then Z ∞ φ∗m φn dτ = δmn . (2.110) −∞
Functions satisfying the Eq. (2.110) are called orthonormal since they are orthogonal to each other and also normalized.
Solved Problem 6: The state of a system is represented ’s are R ∞by ψ = c1 φ1 + c2 φ2 + c3 φ3 where ci √ constants and ψ is normalized. If −∞ φ∗i φj dτ = δij , c1 = 2/3 and c2 = 2/3 find the value of c3 . Using the normalization condition we obtain Z ∞ ψ ∗ ψ dτ 1 = −∞ Z ∞ = (c∗1 φ∗1 + c∗2 φ∗2 + c∗3 φ∗3 ) (c1 φ1 + c2 φ2 + c3 φ3 ) dτ −∞ |c1 |2 +
= |c2 |2 + |c3 |2 = 2/3 + |c3 |2 . √ Thus, c3 = 1/ 3 .
(2.111)
2.14 PHASE OF THE WAVE FUNCTION In 1926 Gregor Wentzel, Hendrik Anthony Kramers and Leon Nicolas Brillouin showed that the possibility of getting the classical Hamilton–Jacobi equation from the Schrödinger equation in the Planck’s limit h → 0. This is an another example for Planck’s correspondence principle. In sec. 2.9 we have seen that in quantum mechanics the expectation values of position and momentum operators obey the same equation of motion of the position and momentum variables in classical mechanics. In this section we show that in quantum mechanics the equation for phase of the complex wave function takes the form of the Hamilton–Jacobi equation of classical physics [14]. Assume the solution of a time-dependent Schrödinger equation in the form ψ(X, t) = A(X, t)eiS(X,t)/~ .
(2.112)
Substituting Eq. (2.112) in the Schrödinger Eq. (2.12) and separating real and imaginary parts we get A
∂S ~2 2 1 − ∇ A+ A(∇S)2 + V A = ∂t 2m 2m ∂A 1 + A∇2 S + 2∇A · ∇S = ∂t 2m
0,
(2.113a)
0.
(2.113b)
56 Quantum Mechanics I: The Fundamentals Defining ρ = |ψ|2 = A2 Eq. (2.113b) takes the form, with the substitution ∂A √ √ = ∂ρ/(2 ρ) and ∇A = ∇ρ/(2 ρ), ∂ρ 1 + ∇ · (ρ∇S) = 0 . ∂t m
(2.114)
This is the continuity equation for conservation of probability with the probability flux J = ρ∇S/m. Equation (2.113a) is rewritten as ∂S 1 + (∇S)2 + V + Vq = 0 , ∂t 2m
Vq = −
~2 ∇2 A . 2m A
(2.115)
Since V and Vq appears in the equation in the same way, Vq is called quantum potential . Vq is state dependent. Equation (2.115) is of the form of classical Hamilton–Jacobi equation. It is called quantum Hamilton–Jacobi equation. The observation is that the phase of ψ obey the same equation as the Hamilton principle function in classical mechanics in the limit Vq → 0. Introducing v(X, t) as J/ρ = ∇S/m and taking the gradient of Eq. (2.115) gives ∂v m + m (v · ∇) v + ∇ (V + Vq ) = 0 . (2.116) ∂t The first two terms in the above equation is mdv/dt. So, the system moves with the acceleration dv m = −∇ (V + Vq ) . (2.117) dt The force is given not only by the gradient of V but also by ∇Vq . If Vq → 0 as ~ → 0 then the particle trajectories will obey Newton’s equation of motion.
2.15 CLASSICAL LIMIT OF QUANTUM MECHANICS Classical mechanics has been verified in a wide variety of phenomena. It is thus natural to know the limit with which quantum mechanics agree with classical mechanics. We cannot simply define ~ → 0 as the classical limit. We know that special relativity reduces to classical Galilean relativity in the limit c (speed of light)→ ∞ or v/c ≪ 1 where v is the speed of a particle. In this limit we say that relativistic mechanics includes Newtonian mechanics as a special case. But quantum mechanics is formulated in terms of probabilities. It does not predict the individual observed possibility but predicts only the probabilities of the possibilities. If quantum mechanics has to give an individual trajectory in the classical limit, it is necessary for the probability distribution to become arbitrarily narrow in the limit ~ → 0. Certain special case of states behave like this. But many physically realistic states do not. Therefore, we cannot expect to get an individual classical trajectory when we let the classical limit of quantum mechanics. Instead we may expect the probability distribution of quantum mechanics to become equivalent to the probability distribution of an ensemble of classical trajectories.
Schrödinger Equation and Wave Function 57 Consider an ensemble of identical particles with an initial position x0 as F (x0 ) and with the identical initial momenta px0 [15]. In classical mechanics, the position probability density function P (x, t) of an ensemble is given by [16] 2 ∂ 2 S ∂S P (x, t) = F , (2.118) ∂px ∂x ∂px where S is the Hamilton principle function. An approximate solution of the time-dependent Schrödinger equation, up to terms linear in ~ obtained using WKB approximation method is [16] 2 1/2 ∂ S ∂S F . (2.119) ψ(x, t) = eiS/~ ∂px ∂x ∂px The |ψ(x, t)|2 obtained from Eq. (2.119) agrees with P (x, t) given by Eq. (2.118). Thus, if one restricts to the effects of the first-order in ~ then in the classical limit the time evolution of the quantum mechanical ensemble is similar to that of the classical ensemble. To get deeper insight of the classical limit of quantum mechanics let us introduce the following statement and call it as the classical limit postulate. In the classical limit the time development of a quantum mechanical ensemble is identical with that of an ensemble of a classical system whose initial distribution of position and momentum are same as that of the quantum mechanical ensemble at the initial instant. To test the above postulate, we take a classical limit in the construction of ψ(x, t) by suitably combining the eigenfunctions belonging to the continuum limit so that the time-dependent wave equation is exactly solvable. The evolution of the classical ensemble can be computed using the equations of motion applied to the phase space density function. According to the above postulate the evolution of two such ensembles should be the same if the initial position and momentum probability distribution functions are the same. In the following we verify this connection for a free particle. For the case of harmonic oscillator and a particle in a homogeneous potential field the reader may refer to ref.[15]. Let us choose ψ(x, 0) of an ensemble of free particles as the Gaussian function centred at the origin 1/2 1 ik0 x − x2 ψ(x, 0) = √ exp , (2.120) πσx 2σx2 where px0 = ~k0 . The position distribution function is 2 2 1 P (x, 0) = |ψ(x, 0)|2 = √ e−x /σx . π σx
Then ψ(x, t) =
ZZ
∗ ψE (ξ)ψE (x)ψ(ξ, 0)e−iEt/~ dEdξ,
(2.121)
(2.122)
58 Quantum Mechanics I: The Fundamentals where ψE (x) is the box normalized eigenfunction of a free particle. Evaluating the integrals we obtain P (x, t) = =
2
|ψ(x, t)| −1/2 ~ 2 t2 2 πσx + 2 2 m σx " # 2 (x − (~k0 t/m)) ~ 2 t2 × exp − + 2 2 . σx2 m σx
(2.123)
Next, consider the time evolution of a classical ensemble of a free particle with P given by Eq. (2.121). The classical phase space density function is D(x, k, 0) = |ψ(x, 0)|2 |φ(k, 0)|2 where k = px /~ and φ(k, 0) is the Fourier transform of ψ(x, 0) given by Eq. (2.120). Evaluating the Fourier transform we get 2 1 (k − k0 )2 x D(x, k, 0) = exp − 2 − , (2.124) π~ σx σk2 where σx σk = 1. Applying classical Liouville’s theorem we find (k − k0 )2 1 (x − (~kt/m))2 − D(x, k, t) = exp − . (2.125) π~ σx2 σk2 R Then P (x, t) = ~ D(x, k, t)dk. It is easy to verify that the resulting expression for P (x, t) is same as that given by Eq. (2.123). When the classical probability density for a particle of particular energy coincides with PQM = ψ ∗ ψ it is called configuration correspondence.
2.16 CONCLUDING REMARKS Note that the time evolution of ψ is described by the Schrödinger equation which is deterministic but the observable quantities associated with this ψ are stated in terms of probability only. The probability concept is also used in classical mechanics in some occasions. For most of the traditional experiments of quantum physics, it is not possible to set-up an individual quantum object, for example, an atom and probe it respectively. Instead, we can probe an ensemble of identically prepared objects then the ensemble average is treated as the quantum average. In an electron-atom collision experiment, thousands of different electrons knock electrons out of different atoms and for each individual case, the momenta of target electrons before the collision are determined by momentum conservation. The result is interpreted as the probability distribution of finding a given momentum in a single atom. Thus, even if the fundamental mechanics of atoms and molecules were classical, we still need to describe most experiments with atoms using probabilities. The spectacular advances in science in 20th century could not have occurred without the use of quantum theory and the Schrödinger equation.
Schrödinger Equation and Wave Function 59 Though the quantum theory gives the probability interpretation of the wave function, there is no other theory which gives a comparable quantitative description of the atom, molecule, nucleus, fields and even macroscopic matters like solids, liquids, conductors and semiconductors. Without quantum theory, phenomena like superconductivity, superfluidity, Bose–Einstein condensation cannot be explained. Every advanced technology makes use of the tools provided by quantum mechanics.
2.17 BIBLIOGRAPHY [1] J.P. McEvoy and O. Zarate, Introducing Quantum Theory. Icon Books, Cambridge, 1996. [2] F. Capra, The Tao of Physics. Bantam Books, New York, 1984 pp.139. [3] R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals. McGraw–Hill, New York, 1965. [4] J.Z. Justin, Path Integrals in Quantum Mechanics. Oxford University Press, Oxford, 2005. [5] H. Everett, Rev. Mod. Phys. 29:454, 1957. [6] M.A. Rubin, Found. Phys. Lett. 14:301, 2001. [7] J.G. Cramer, Rev. Mod. Phys. 58:647, 1986. [8] J.G. Cramer, Inter. J. Theor. Phys. 27:227, 1988. [9] V. Fano, Rev. Mod. Phys. 29:74, 1957. [10] J. Nag, V.J. Menon and S.N. Mukherjee, Am. J. Phys. 55:802, 1987. [11] G. Manfredi, S. Mola and M.R. Feix, Eur. J. Phys. 14:101, 1993. [12] L.E. Ballentine, Quantum Mechanics: A Modern Development. World Scientific, Singapore, 1998. [13] D. Sen, S.K. Das, A.N. Basu and S. Sengupta, Current Science 80:536, 2001. [14] R.A. Leacock and M.J. Pedgett, Phys. Rev. D 28:2491, 1983. [15] D. Home and S. Sengupta, Am. J. Phys. 51:265, 1983. [16] L.S. Brown, Am. J. Phys. 40:371, 1972.
60 Quantum Mechanics I: The Fundamentals
2.18 EXERCISES 2.1 Write the Schrödinger equation for a particle in a potential sin x. 2.2 For a particle carrying a charge q in a uniform time-dependent electric field E(t) the force experienced by it is F = −∇(−qE(t) · X). Write the Schrödinger equation of the system. 2.3 Write the Schrödinger equation for a charged particle in an electromag2 1 p − qc A(X, t) + qφ(X, t). netic field with the Hamiltonian H = 2m Rewrite the Schrödinger equation for A = (Bz, 0, 0) and φ = −Ez. 2.4 What is the Schrödinger equation of a system of two particles of masses m1 and m2 carrying charges q1 and q2 respectively with H = p21 /(2m1 )+ p22 /(2m2 ) + q1 q2 /r12 and r12 = |r1 − r2 |?
2.5 Starting from the equation for ψ, obtain the equation for ψ ∗ .
2.6 If E1 and E2 are the eigenvalues and φ1 and φ2 are the eigenfunctions of a Hamiltonian operator then find whether the energy corresponding to the superposition state φ1 + φ2 is equal to E1 + E2 or not. 2.7 Express the Schrödinger equation −(~2 /2m)∇2 + V (x, y, z) ψ(x, y, z) = Eψ(x, y, z) in the spherical polar coordinates defined by x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ and in the parabolic coordinates defined by ξ = r(1 − cos θ), η = r(1 + cos θ), φ = φ cos θ.
2.8 Find the condition under which both ψ(X, t) and ψ ∗ (X, −t) will be the solutions of the same time-dependent Schrödinger equation.
2.9 What is the major difference between real and complex wave functions? 2.10 What is the difference between the wave function ψ1 = ei(kx−ωt) and ψ2 = ei(k·X−ωt) ? 2.11 Write the Hamiltonian of a photon. 2.12 What is the physical meaning of hxi = 0? 2.13 Write the operators of kinetic energy and angular momentum L = r× p. 2.14 Consider the one-dimensional Schrödinger equation ψxx + (λ − g(x))ψ = 0, λ = 2mE/~2 and g = 2mV /~2 . Obtain the Schrödinger equation d under the change of variable u(x) = − dx ln ψ(x) [S.B. Haley, Am. J. Phys. 65:237, 1997]. 2.15 Find the conditions to be satisfied by the functions f and g such that under the transformation ψ = f (x)F (g(x)) the Schrödinger equation ψxx + (E − V )ψ = 0 can be written as Fgg + Q(g)Fg + R(g)F (g) = 0. Then show that ′′ ′ ′′ 2 1 Q2 g g ′ 2 E − V = (g ) R − Qg − + − . 2 4 2g ′ 2g ′
Schrödinger Equation and Wave Function 61 2.16 What are the effects of addition of a constant to a potential on the time-independent Schrödinger equation and the energy levels? 2.17 Which of the following wave functions are admissible in quantum mechanics? State the reasons. 2
−x (a) e−x . (b) sechx. (c) √ e 2 . (d) tanhx. (e) sin x, 0 < x a is normalized. 2.20 RNormalize the wave function ψ = e−|x| sin αx. It is given that ∞ −x e sin αx dx = α2 /(1 + α2 ). 0
2.21 A particle of mass m moves in a one-dimensional box of length L with origin as the centre. If the wave function of this particle is ψ(x) = N x(1 − x2 ) for |x| < L/2 and 0 otherwise find the factor N . 2.22 A quantum mechanical particle moving in one-dimension has the wave function ψ(x) = cxe−|x|/b , −∞ < x < ∞ where c and b are constants (b > 0). Find the probability that the position of the particle lies in the region −∞ < x ≤ b. 2.23 What is the probability current density corresponding to ψ(x) = Ae−αx where A and α are constants? 2.24 A free particle in one-dimension is in a state described by ψ = A ei(px x−Et)/~ + B e−i(px x+Et)/~ where px and E are constants. Find the probability current density. 2
2
2.25 If the wave function of a particle at t = 0 is ψ(x, 0) = N e−ikx−(x /2a ) calculate the probability density and current density. √ 1/2 2 2 2.26 For the Gaussian wave function ψ(x) = [1/(σ π)] e−x /2σ calculate the probability current density. 2.27 Verify whether the wave function ψ = N eikx−x nuity equation or not.
2
/(2a2 )
satisfies the conti-
2.28 The wave function of a linear harmonic oscillator with potential V = 2 mω 2 x2 /2 is ψ = xe−mωx /(2~) . Find its energy. p 2.29 If ψ = 1/L cos(πx/(2L)) for the system confined to the potential V (x) = 0 for |x| < L and ∞ for |x| > L then calculate E.
62 Quantum Mechanics I: The Fundamentals 2.30 p The wave function of a particle confined to a box of length L is 2/L sin(πx/L) in the region 0 < x < L and zero everywhere else. Calculate the probability of finding the particle in the region 0 < x ≤ L/2. 2.31 Write the law of conservation of energy H = T + V in terms of expectation values. 2.32 Are the wave functions 1/2 1/2 1 r 1 −r/a0 ψ1 = e and ψ = 2 − e−r/(2a0 ) 2 πa30 32πa30 a0 of the electron in hydrogen atom orthogonal? 2.33 Find the potential corresponding to the following wave functions: 2 1 (a) ψ = 1/4 e−x /2 , E = ~ω/2. π (b) ψ = A sin(πx/(2L)) for |x| < L and 0 for |x| > L, E = ~2 π 2 /(8mL2 ). (c) ψ = αxe−βx eiE/~ .
(d) ψ = (mV0 /~)e−kx for x > 0, (mV0 /~)ekx for x < 0 and E = −mV0 /(2~2 ), k 2 = mV0 /~2 . 2.34 A particle of mass m is confined in the infinite square-well potential V = 0 for 0 < x ≤ L and V = ∞ otherwise. It has the normalized stationary state eigenfunctions φn (x) and eigenvalues En = n2 ~2 π 2 /(2mL2 ).√Its wave function at time t = 0 is given by ψ(x, 0) = (φ1 (x) + φ2 (x))/ 2 . What is the smallest positive time τ for which ψ(x, t) will be orthogonal to ψ(x, 0)? 2.35 Show that hxpx + px xi is real. 2.36 Show that hAn i = hAin in its eigenstates. √ 2.37 Calculate hp2x i for ψ = ke−k|x| . 2.38 A particle of mass m in the one-dimensional energy well V (x) = 0 for 0 ≤ x ≤ and ∞ otherwise is in a state whose wave function is given by ψ(x) = N x(L − x) where N is the normalization constant. Determine hEi in this state. 2.39 Show that
d 2 2 i~ hx i = hxpx i − . dt m m
2.40 Show that
d2 2 2 2 hx i = 2 hp2x i + hxF i. dt2 m m
2.41 Show that
d 2 hp i = h2F px + px F i. dt x
2.42 Show that
d 1 hpx xi = hp2x i + hxF i. dt m
Schrödinger Equation and Wave Function 63 2.43 For a free particle show that
d 2 d 2 hpx i = 0 and hxpx + px xi = hp2x i. dt dt m
2.44 φ1 and φ2 are the only eigenfunctions of a system belonging to the energy eigenvalues E0 and −E0 respectively. In a measurement of the energy of the system, hEi is found to be E0 /2. Find the wave function of the system. 2.45 The H of a charged particle in uniform electric and magnetic fields is given by H = (p−eA)2 /(2m)−eE0 k.k where A = B0 (−y, x, 0)/2. Both electric field E and magnetic field B are applied along the z-direction. Applying Ehrenfest’s theorem, show that dhri/dt = hp − eAi/m and dhp − eAi/dt = eE0 k + (e/m)hp − eAi × ∇ × A. 2.46 A particle of mass m enclosed in a one-dimensional box of length L such that 0 < x < L, has energy eigenfunctions φn (x) = A sin(nπx/L), n = 1, 2, · · · and En = n2 π 2 ~2 /(2mL2 ). At t = 0, the particle has the wave function ψ(x) = B sin(2πx/L) cos(πx/L) where B is a constant. (i) If the energy of the particle is measured at t = 0, what are the possible results of the measurement? (ii) What is the expectation value of energy? 2.47 Given the normalized ground state wave function of hydrogen atom ψ100 = 1/(πa30 )1/2 e−r/a0 find the expectation value of its z-coordinate. 2.48 If Hφn (x) = En φn (x) find the expectation value of the Hamiltonian operator H in the normalized superposition state ψ(x, t) = P ∞ −iEn t/~ . n=0 Cn φn (x) e √ 2.49 Consider a system in a state ψ = (φ1 + φ2 )/ 2 where φ1 and φ2 are orthonormal eigenfunctions with the eigenvalues E1 and E2 respectively. What is the probability of finding the system in the energy E1 ? What is hEi? 2.50 Consider a spherically symmetric potential energy function given by V (r) = 0, for 0 < r < a and ∞ for r > a. Given the solution ψ(r) = A(sin kr)/r + B(cos kr)/r where k = (2mE/~2 )1/2 satisfying the Schrödinger equation, obtain the corresponding eigenvalues by applying proper boundary conditions, seen from the frame (x, t).
3
CHAPTER
Operators, Eigenvalues and Eigenfunctions
3.1 INTRODUCTION A basic quantity in quantum mechanics is the wave function ψ. Given a quantum mechanical system we write the Schrödinger equation for it. Its solution is the wave function. When introducing the wave function we stated that the maximum information about the state of a system can be obtained only from ψ. How do we extract observable properties of the system from the wave function? In quantum mechanics for every classical dynamical variable there exists an operator. For example, the operator of momentum p is −i~∇ while the operator of energy variable H = p2 /(2m) + V (X) is −(~2 /2m)∇2 + V (X). Operation of an operator on ψ gives another function. If the new function is λψ, then λ is the eigenvalue of the operator. For example, the energy of a particle in a particular state represented by ψ can be obtained by operating H on ψ. That is, we have Hψ = Eψ where E is the energy eigenvalue. The expectation value of A(X, p) can be obtained from the formula Z ∞ hAi = ψ ∗ Aψ dτ , (3.1) −∞
where ψ is the solution of the equation Hψ = Eψ. Operators in quantum mechanics which represent observables are of a special type: They are linear and Hermitian. Such operators not only form computation tools, but also form the most effective language in terms of which the theory can be formulated. The fascinating point is that using linear operators we can understand the quantum mechanics. Therefore, it is important to study the basic theory of linear operators. In this chapter, we present important properties of linear operators and their eigenfunctions. We discuss the features of various operators which are necessary for studying quantum
65
66 Quantum Mechanics I: The Fundamentals mechanical systems. This chapter will show how rich and beautiful the mathematical structure of quantum mechanics is.
3.2 LINEAR OPERATORS An operator Aop is a mathematical operation which, when it acts on a function f (x) gives another function g(x): Aop f (x) = g(x) .
(3.2)
For simplicity we drop the subscript ‘op’ in Aop . Mathematical operations such as addition, subtraction, differentiation, integration, etc. can be represented by appropriate operators. An example of operator is d/dx. Usually an operator is specified by its action. However, an operator essentially consists of its action and its domain. By action we mean what the operator does to the functions. On the other hand, the specified set of functions on which the operator acts are referred as the domain of the operator. Why should one define the domain of an operator? We will come to this question when discussing about self-adjoint and Hermitian operators in sec. 3.4.
3.2.1
Definition
An operator A is said to be a linear operator if it satisfies the rules A (φ1 + φ2 ) = Aφ1 + Aφ2 , A (cφ1 ) = cAφ1 ,
A (cφ2 ) = cAφ2 ,
(3.3a) (3.3b)
where φ1 and φ2 are given functions and c is any arbitrary complex constant. The operator d/dx is linear because d (φ1 + φ2 ) = dx d c1 φ1 = dx
d d φ1 + φ2 , dx dx d d d c1 φ1 , c2 φ2 = c2 φ2 . dx dx dx
(3.4a) (3.4b)
Operators do not generally have the property given by Eq. (3.3a). Suppose an operator A is defined by Aφ = sin φ. Then left-side of Eq. (3.3a) is A (φ1 + φ2 )
= sin (φ1 + φ2 ) .
(3.5a)
The right-side of Eq. (3.3a) is Aφ1 + Aφ2
=
sin φ1 + sin φ2 .
(3.5b)
Right-sides of Eqs. (3.5) are not same. That is, the operator A defined by Aφ = sin φ is not a linear operator.
Operators, Eigenvalues and Eigenfunctions 67
3.2.2
Eigenvalues and Eigenfunctions of Linear Operators
Suppose we assume that Aφ = λφ ,
(3.6)
where λ is a number. Equation (3.6) is called the eigenvalue equation of A. φ is called an eigenfunction of A. λ is called the eigenvalue of the operator A corresponding to the eigenfunction φ. Given an eigenvalue equation we ask: What are the values of λ for which nontrivial solutions which are finite everywhere even at x = ±∞ occur? In general, nontrivial solutions exist for certain values of λ only. The certain values are called eigenvalues 1 . The eigenvalues are the important values of λ because only for these specific set of values Eq. (3.6) has nontrivial solutions. These solutions, called eigenfunctions, are the important solutions because they are the only possible meaningful solutions. The set of eigenvalues is called a spectrum. Suppose A and B are two linear operators. The linear combination of them is C = αA + βB. Is C a linear operator? The answer is yes (see the problem 5 at the end of this chapter). That is, a linear combination of the two or more linear operators is also a linear operator. It is easy to show that product of two or more linear operators is also a linear operator. If Aφ = 0 then A is called a singular operator . When AB = BA = I then A−1 = B and B −1 = A. Now, the operators A and B are called nonsingular operators. Some of the properties of linear operators are summarized below: = B+A = (A + B) + C = A + (B + C)
(3.7a) (3.7b)
A(BC) (A + B)C
= (AB)C + ABC = AC + BC
(3.7c) (3.7d)
A(B + C) A2
= AB + AC = AA .
(3.7e) (3.7f)
A+B A+B+C
In quantum mechanics, the operators associated with the observables such as position, linear momentum, angular momentum and energy all satisfy the properties of linear operators.
Example 1: Let us see how eigenfunctions and eigenvalues of an operator can be calculated. For illustrative purpose we consider the operator x + d/dx. The eigenvalue equation for the given operator is d x+ φ = λφ . (3.8) dx 1 Eigen
is a German word which means certain.
68 Quantum Mechanics I: The Fundamentals Rewriting it as
dφ = (λ − x) dx φ
(3.9)
and integrating on both sides we obtain φ(x) = c eλx−x
2
/2
,
(3.10)
where c is a constant. The eigenfunction φ is continuous and well defined for any value of λ. Thus, the eigenvalues of the given operator form a continuous spectrum. However, in quantum mechanics a physically acceptable wave function should satisfy certain boundary conditions. When these boundary conditions are applied, then the eigenvalues may not take continuous values. Only for certain values of λ, the wave function may be physically acceptable. Under such situations, λ values will take discrete values. In the case of the Schrödinger equation Hψ = Eψ the boundary conditions on ψ lead to discrete energy eigenvalues and quantum numbers.
Example 2: There are a number of operators for which the eigenvalues are discrete. For example, consider the operator L = 1 − x2
d2 d − 2x . 2 dx dx
(3.11)
It is a linear operator. Its eigenvalue equation is Ly = λy. With −λ = l(l + 1) we can identify the equation Ly = λy as the Legendre differential equation. It has bounded solutions only for integer values of l and the associated solutions (called Legendre polynomials) are defined for x ∈ [−1, 1] [1,2]. The eigenvalues of L are certain integers and the corresponding eigenfunctions are Legendre polynomials. If λ is not chosen as −l(l+1) with l taking integer values, then we will get a diverging infinite series solution which is not an acceptable physical wave function. So, the condition λ = −l(l + 1) with l taking 0, 1, 2, · · · gives a finite polynomial solutions which are physically acceptable. An important property of eigenfunctions of linear operators is that a linear combination of two eigenfunctions belonging to two distinct eigenvalues of a linear operator A is not an eigenfunction of it. To prove this assume that φa and φb are two eigenfunctions of an operator A with distinct eigenvalues a and b respectively. The eigenvalue equations for φa and φb are Aφa = aφa ,
Aφb = bφb .
(3.12)
Suppose the linear combination ψ = αφa + βφb is an eigenfunction of A. Then we must have Aψ = λψ, that is A (αφa + βφb ) = λ (αφa + βφb ) ,
(3.13)
Operators, Eigenvalues and Eigenfunctions 69 where λ is the eigenvalue of A corresponding to ψ. Using Eq. (3.12) in Eq. (3.13) we obtain α(a − λ)φa + β(b − λ)φb = 0 .
(3.14)
This equation is true only if λ = a and λ = b. Since a and b are distinct eigenvalues the Eq. (3.14) is not true. Therefore, ψ = αφa + βφb cannot be an eigenfunction of A. Equation (3.14) implies that if φ1 and φ2 are two linearly independent eigenfunctions of a linear operator belonging to a same eigenvalue λ then a linear combination of φ1 and φ2 will also be an eigenfunction of A with same eigenvalue λ. One can easily verify that if φ is an eigenfunction of an operator A with the eigenvalue a then φ is also an eigenfunction of the operators A2 , An and eA . What are the corresponding eigenvalues? (See the problem 11 at the end of this chapter).
Solved Problem 1: Obtain the eigenvalues and eigenfunctions of the operator 1 d 2 d A= 2 r . r dr dr The eigenvalue equation of A is Aψ = λψ which becomes d2 ψ 2 dψ + = λψ . dr2 r dr
(3.15)
Introducing the change of variable ψ = y/r we can eliminate the second term in the above equation and obtain y ′′ + α2 y = 0,
λ = −α2 .
(3.16)
Its particular solutions are sin αr and cos αr. Then ψ=N
sin αr , r
N
cos αr . r
(3.17)
In the limit r → 0, (sin αr)/r = α while (cos αr)/r → ∞. Therefore, we discard the solution N cos αr/r and choose the eigenfunctions as N sin αr/r where N is the normalization constant. The eigenvalues are −α2 and form a continuous spectrum.
3.3 COMMUTING AND NONCOMMUTING OPERATORS In classical mechanics if we know some set of physical quantities such as (r, p) then certain other quantities of interest can be determined. For example, given r and p one can determine orbital angular momentum L = r × p. From the set of all dynamical variables many such compatible subsets can be chosen.
70 Quantum Mechanics I: The Fundamentals Their member is restricted by the Poisson brackets. If the coordinates of a particle are denoted by x1 , x2 , x3 and momenta by p1 , p2 , p3 then the Poisson bracket [3-5] of two functions f and g (which are functions of xi ’s and pi ’s) is defined as 3 X ∂f ∂g ∂f ∂g . (3.18) {f, g} = − ∂xi ∂pi ∂pi ∂xi i=1 In quantum mechanics the Poisson bracket conditions are replaced by commutator brackets.
3.3.1
Definition, Identities and Examples
The operator AB − BA is called the commutator of A and B and is denoted [A, B]. The properties of commutator brackets are similar to those of Poisson brackets. If [A, B] = 0, that is, AB = BA, then the two operators A and B are said to be commuting operators; otherwise noncommuting operators. For commuting operators their order of occurrence can be freely changed. When AB = −BA the operators A and B are said to be anticommuting operators. The following identities can be easily verified: = − [B, A] , = [A, C] B + A [B, C] ,
(3.19a) (3.19b)
= [A, B] C + B [A, C] , = α1 [A, C] + α2 [B, C] ,
(3.19c) (3.19d)
= α1 [A, B] + α2 [A, C] , ∞ X 1 n n A [A, A ] = 0 , e = A . n! n=0
(3.19e)
[A, B] [AB, C] [A, BC] [α1 A + α2 B, C] [A, α1 B + α2 C]
(3.19f)
The rule for the calculation of area of a rectangle of length l and breadth b is commutative; l×b = b×l. However, there are certain mathematical operations which are noncommutative. Let us consider the following two operations: 1. Move two steps in the forward direction. Call this operation as S. 2. Rotate by 90o to right-side. Call this operation as R. Suppose we stand in a room and perform the operations 1 and 2. Note down the final position and it be P12 . Next, we go to the starting position and perform the operations 2 and 1. The position of us is say P21 . Obviously, P21 is different from P12 . That is, symbolically SR 6= RS or SR − RS 6= 0. The operators S and R are noncommutative. For the coordinate and momentum operators we have [x, px ] φ = = =
(xpx ) φ − (px x) φ ∂φ ∂ −i ~ x − (xφ) ∂x ∂x i ~φ .
(3.20)
Operators, Eigenvalues and Eigenfunctions 71 Thus, [x, px ] = i~ which implies that the order of these operators cannot be changed. What is the consequence of this relation? A consequence of it is that x(px φ) 6= px (xφ) where φ is a state of the system. That is, a position measurement followed by a momentum measurement might give different result from a momentum measurement followed by a position measurement. Similarly, we obtain [y, py ] = [z, pz ] = i ~ , [px , x] = [py , y] = [pz , z] = −i ~ .
(3.21)
Note that the conjugate pairs (x, px ), (y, py ) and (z, pz ) do not commute. But for x, py we have [x, py ] = 0 so that the order of these operators can be changed without affecting the result. Further, [x, pz ] = 0 ,
3.3.2
[y, px ] = 0 ,
[y, pz ] = 0 ,
[z, px] = 0 ,
[z, py ] = 0 . (3.22)
Features of Commuting and Noncommuting Operators
There are three important features associated with commuting and noncommuting quantum mechanical operators. They are given below: 1. If the operators A and B of two observables do not commute then they cannot be measured simultaneously with arbitrary accuracy (for details see sec. 8.3). 2. If A and B are two commuting operators then the eigenfunctions of A (or B) are also the eigenfunctions of B (A). (For a proof see sec. 3.3.3). 3. The expectation value of an operator commuting with the Hamiltonian H is a constant of motion or time-independent. Later in sec. 7.2 we show that the time-independency of expectation value of an operator A is governed by the equation d ∂A i hAi = + h[H, A]i . (3.23) dt ∂t ~ If A has no explicit time-dependence then i d hAi = h[H, A]i . dt ~
(3.24)
When [H, A] = 0 we have dhAi/dt = 0. Hence, hAi of an operator commuting with the Hamiltonian is a constant of motion. Further, for any time-independent Hamiltonian operator dhHi/dt = 0 because of [H, H] = 0. So, for any system with time-independent Hamiltonian operator, energy is conserved. In classical mechanics with respect to generalized position and momentum coordinates q and p, the equation of motion is [3–5] d ∂f f (q, p, t) = + {H, f } . dt ∂t
(3.25)
72 Quantum Mechanics I: The Fundamentals Comparison of Eqs. (3.23) and (3.25) shows that the Poisson bracket of classical mechanics corresponds to the commutator multiplied by i/~.
Solved Problem 2: Compute [H, px ] for the linear harmonic oscillator with H =−
~2 d2 1 + kx2 , 2 2m dx 2
k = mω 2 .
(3.26)
We obtain [H, px ]φ =
=
H(px φ) − px (Hφ) ~2 d2 1 2 d − + kx −i~ φ 2m dx2 2 dx d ~2 d2 1 2 − −i~ − + kx φ dx 2m dx2 2 i~kxφ .
=
i~mω 2 xφ .
=
(3.27)
Thus, [H, px ] = i~mω 2 x for the linear harmonic oscillator. Since [H, px ] 6= 0, hpx i is not a constant of motion. For a free particle k = 0 and hence [H, px ] = 0, that is, hpx i is a constant of motion.
3.3.3
Simultaneous Eigenfunctions
Suppose A and B are two operators and φ is a function such that Aφ = aφ ,
Bφ = bφ
(3.28)
then φ is called a simultaneous eigenfunction of A and B belonging to the eigenvalues a and b respectively. The equation Aφ = aφ implies BAφ = aBφ = abφ .
(3.29)
Similarly, from Bφ = bφ we get ABφ = bAφ = baφ .
(3.30)
From Eqs. (3.29) and (3.30) we obtain (AB − BA)φ = [A, B]φ = 0 .
(3.31)
Therefore, if φ is a simultaneous eigenfunction of the operators A and B then [A, B] = 0. Thus, [A, B] = 0 is the necessary condition for φ to be a simultaneous eigenfunction of A and B. In fact, it is also to be the sufficient condition. For a free particle [H, px ] = 0. Hence, the operators H = −(~2 /2m)d2 /dx2 and px = −i~d/dx have same set of eigenfunctions. (Verify this with the solutions of the time-independent Schrödinger equation of the free particle).
Operators, Eigenvalues and Eigenfunctions 73
3.4 SELF-ADJOINT AND HERMITIAN OPERATORS In this section we introduce the adjoint, self-adjoint and Hermitian operators.
3.4.1
Definition
Consider the integral (φ1 , Aφ2 ) =
Z
∞
−∞
φ∗1 Aφ2 dτ
(3.32)
which involves two different functions, φ1 and φ2 , and A is an operator. For an operator A one can always find another operator A† (A-dagger) called the adjoint of A such that (φ1 , Aφ2 ) = A† φ1 , φ2 , (3.33)
R∞ ∗ R∞ that is, −∞ φ∗1 Aφ2 dτ = −∞ A† φ1 φ2 dτ . As far as the value of the integral is concerned it does not matter whether A acts on φ2 or its adjoint A† acts on other function φ1 . When A† = A the operator A is said to be self-adjoint. A self-adjoint operator with certain boundary conditions on the eigenfunctions is called Hermitian 2 . When an operator A is self-adjoint then
(3.34)
(φ1 , Aφ2 ) = (Aφ1 , φ2 ) .
An interesting feature of a self-adjoint (Hermitian) operator is that the expectation value of the corresponding dynamical variable is real as shown below. When A† = A and φ1 = φ2 = φ then using Eq. (3.33) and Eq. (3.34) we get Z ∞ ∗ Z ∞ Z ∞ ∗ ∗ † ∗ hAi = φ Aφ dτ = A φ φ dτ = φ (Aφ) dτ =
−∞
hAi∗ .
−∞
−∞
(3.35)
hAi is thus real. If the operator of a dynamical variable is self-adjoint then the possible values of the variable are only real. In an experiment, the result of any single measurement of a variable is necessarily a real number. Therefore, selfadjoint operators are suitable for representing observable dynamical variables. If (φ1 , Aφ2 ) = −(Aφ1 , φ2 ) then A is said to be anti-self-adjoint (or antiHermitian). In this case hAi = −hAi∗ − the expectation value of an anti-selfadjoint operator is always a pure imaginary. 2 Usually, self-adjoint operators are also called Hermitian. However, there is a distinction between these two operators. The difference lies in the domain of the operators. For details see appendix D.
74 Quantum Mechanics I: The Fundamentals
3.4.2
Examples
In an experiment position and momentum of a particle can be measured and hence their operators must be self-adjoint. The position operator is obviously self-adjoint. For the momentum operator we have Z ∞ (φ1 , px φ2 ) = φ∗1 (px φ2 ) dx −∞ Z ∞ dφ2 dx = −i~ φ∗1 dx −∞ Z ∞ dφ∗1 = −i~φ∗1 φ2 |∞ + i~ φ2 dx . (3.36) −∞ −∞ dx Since φ1 , φ2 → 0 as |x| → ∞ the above equation is rewritten as ∗ Z ∞ Z ∞ dφ1 ∗ (φ1 , px φ2 ) = −i~ φ2 dx = (px φ1 ) φ2 dx dx −∞ −∞ = (px φ1 , φ2 ) . (3.37) The adjoint of px is defined through the equation (φ1 , px φ2 ) = p†x φ1 , φ2 .
(3.38)
Comparison of Eqs. (3.37) and (3.38) gives p†x = px . Thus, px = −i~d/dx is a self-adjoint operator. But the operator A = d/dx is not a self-adjoint. For this operator we have ∗ Z ∞ Z ∞ Z ∞ dφ2 dφ1 dφ∗1 φ∗1 dx = φ∗1 φ2 |∞ φ dx = − φ2 dx. (3.39) − 2 −∞ dx dx −∞ −∞ −∞ dx Thus, the adjoint of d/dx is −d/dx and so it is not a self-adjoint. An eigenfunction of the form eikx corresponds to a momentum ~k along the It is not normalizable if x extends from −∞ to +∞ as R ∞ x-direction. ikx 2 |e | dx = ∞. Under this situation, p = −i~d/dx is not Hermitian −∞ as i~φ∗1 φ2 |∞ in Eq. (3.36) is not zero. In order to make p Hermitian, we −∞ use a box normalization by confining the system in an arbitrarily large box of length L and imposing the periodic boundary condition φ(x) = φ(x + L). Under this boundary condition φk (x) = eikx = eik(x+L) . Then k will take discrete ∞ L values 2πn/L, n = 0, ±1, · · · . Then i~φ∗1 φ2 |−∞ (in Eq. (3.36)) → i~φ∗1 φ2 |0 = 0. Hence, p is a Hermitian.
3.4.3
Properties of Self-Adjoint Operators
Some of the properties of self-adjoint operators are given below with necessary proofs.
Operators, Eigenvalues and Eigenfunctions 75 1. The eigenvalues of a self-adjoint operator are real. Proof: Let A be a self-adjoint operator and φa and φa′ be two eigenfunctions of it so that Aφa = aφa , Aφa′ = a′ φa′ . (3.40) The self-adjoint condition with φa and φa′ is (φa , Aφa′ ) = (Aφa , φa′ ), that is, Z ∞ Z ∞ φ∗a Aφa′ dx = (Aφa )∗ φa′ dx . (3.41) −∞
−∞
Using Eq. (3.40) in Eq. (3.41) we obtain Z ∞ (a′ − a∗ ) φ∗a φa′ dx = 0 .
(3.42)
−∞
When φa′ = φa Eq. (3.42) becomes Z ∞ ∗ (a − a ) φ∗a φa dx = 0 .
(3.43)
−∞
The integral in the above equation is the norm of φa and it should be nonzero and finite. Therefore, we have a∗ = a, a is real. 2. Any two eigenfunctions belonging to distinct eigenvalues of a self-adjoint operator are mutually orthogonal . Proof: Because of a∗ = a, Eq. (3.42) can be rewritten as Z ∞ ′ (a − a) φ∗a φa′ dx = 0 . If a′ 6= a then
R∞
−∞
(3.44)
−∞
φ∗a φa′ dx = 0 which completes the proof.
3. The product of the self-adjoint operators is self-adjoint if and only if they commute. Proof: Assume that A and B are two self-adjoint operators. If the product operator AB is a self-adjoint then (φ, ABφ2 ) = (ABφ1 , φ2 ): Z ∞ Z ∞ ∗ ∗ φ1 ABφ2 dx = (ABφ1 ) φ2 dx . (3.45) −∞
−∞
R∞
Now, consider −∞ φ∗1 ABφ2 dx. As A and B are self-adjoint we write Z ∞ Z ∞ Z ∞ ∗ ∗ φ∗1 ABφ2 dx = (Aφ1 ) Bφ2 dx = (BAφ1 ) φ2 dx . −∞
−∞
−∞
(3.46) From Eqs. (3.45) and (3.46) we find AB = BA. That is, [A, B] = 0.
76 Quantum Mechanics I: The Fundamentals 4. Because of [A, An ] = 0 and the property (3) any power of a self-adjoint operator is also a self-adjoint operator . 5. A linear combination of two or more self-adjoint operators is also a selfadjoint operator . The proof of it is left as an exercise to the reader. 6. Self-adjoint operators can be used to generate a unitary transformation, eiAθ . This will be discussed in secs.6.4 and 11.2. Operators of observables such as position, linear momentum, angular momentum and energy−all are self-adjoint and they satisfy the above properties. Real eigenvalues are not restricted to self-adjoint operators alone. There are nonself-adjoint operators with all eigenvalues being real. For example, it has been shown that the eigenvalues of the nonself-adjoint or non-Hermitian operator d2 1 A = − 2 − (ix)2M − α(ix)M−1 + l(l + 1) 2 , (3.47) dx x where M > 1, α and l are real, all are real [6-11]. For discussion on nonHermitian operators one may refer to refs.[6-11] and also in sec. 3.9 for some details.
Solved Problem 3: For a Hermitian operator A prove that hAA† i ≥ 0. We obtain hAA† i = = = = =
Z
∞
φ∗ AA† φ dτ
−∞
Z
Z
Z
∗ A† φ A† φ dτ
(Aφ)∗ A† φ dτ ∞
−∞ Z ∞
∗
(Aφ) Aφ dτ ψ ∗ ψ dτ ,
(3.48)
−∞
where ψ = Aφ. The above integral is norm of ψ which is positive. Therefore, hAA† i ≥ 0.
3.5 DISCRETE AND CONTINUOUS EIGENVALUES In order to treat a dynamical variable A as an observable the operator representing it must be Hermitian so that hAi =real. Another requirement is that the eigenfunction of the operator should form a complete set. The eigenfunctions {φa } of the self-adjoint operator A are said to form a complete set if
Operators, Eigenvalues and Eigenfunctions 77 any arbitrary wave function ψ of the system can be expressed into a linear combination Z X ψ= ca φa + ca φa da . (3.49) a∈C
a∈D
of the members of the set3 . In Eq. (3.49) the summation is over the discrete part of the eigenvalue spectrum and integration is over the continuous part. Any dynamical variable represented by a Hermitian operator having a complete set of eigenfunctions is called an observable. Let us normalize the wave function ψ assuming that all the eigenvalues of the operator A are discrete. We have Z ∞ Z ∞ XX X 2 ∗ ∗ 1= ψ ψ dτ = c a c a′ φ∗a φa′ dτ = |ca | , (3.50) −∞
a
−∞
a′
a
where δaa′ is known as Kronecker-delta function which is defined as δaa′ P = 1 for a = a′ otherwise 0. By suitably choosing the values of caRwe can make |ca |2 ∞ ∗ equal to 1. Thus, ψ is normalizable. Further, we have −∞ φa φa′ dτ = δaa′ . When a′ = a Z ∞ φ∗a φa dτ = 1 . (3.51) −∞
That is, φa ’s are normalized. Thus, the eigenfunctions belonging to discrete eigenvalues are normalizable. When all the eigenvalues form a continuous spectrum then the normalization condition becomes Z ∞ Z ∞ Z ∞ 1= c∗a da ca′ da′ φ∗a φa′ dτ . (3.52) −∞
−∞
−∞
Since a is continuous we write Z ∞ φ∗a φa′ dτ = δ (a, a′ ) = δ (a − a′ ) ,
(3.53)
−∞
where δ(a, a′ ) is Dirac-delta function given by Z
δ (a, a′ ) = ∞
δ(a, a′ ) da′
=
∞ for a′ = a otherwise 0,
(3.54a)
1.
(3.54b)
−∞
3 In the language of linear vector space, this complete set of vectors φ span the infinitea dimensional vector space known as Hilbert space. In this space any arbitrary vector ψ can be represented as the linear combination of these basis vector with the expansion coefficients being the coordinates of the vectors. These coordinates form column matrix which represents the vector ψ. Then any operation on ψ is represented as an infinite-dimensional matrix. Diagonalization of this matrix gives the eigenvalues and eigenvectors which are the basis of Heisenberg’s matrix mechanics.
78 Quantum Mechanics I: The Fundamentals Using the result
Z
∞
F (x′ ) δ (x, x′ ) dx′ = F (x)
(3.55)
−∞
Eq. (3.52) becomes 1=
Z
∞
−∞
2
|ca | da .
(3.56)
Therefore, ψ is normalizable. However, using Eq. (3.54) in Eq. (3.53) we get Z ∞ φ∗a φa dτ = ∞ . (3.57) −∞
The eigenfunctions are nonnormalizable. Thus, the eigenfunctions belonging to continuous eigenvalues are nonnormalizable. When we look for normalizable eigenfunctions we would end up with discrete eigenvalues. If the eigenvalues are continuous then the eigenfunctions are nonnormalizable. If the eigenvalue spectrum has both discrete and continuous parts then the normalization condition gives Z X |ca |2 + |ca |2 da = 1 . (3.58) a∈D
a∈C
In connection with the eigenvalue problems, the following questions may be raised. If we assign an arbitrary real value to a in the eigenvalue equation Aφa = aφa does there exist a corresponding eigenfunction φa ? The answer is negative. This is because the solution of the eigenvalue equation has to satisfy the physical admissibility conditions. So, only for specific values of a eigenfunctions exist. The next question to be discussed is whether there can be more than one eigenfunction for a particular eigenvalue. It is quite possible for a set of eigenfunctions to have same eigenvalue. If there exists only one eigenfunction belonging to a given eigenvalue, the eigenvalue is said to be nondegenerate; otherwise, it is degenerate. If φa and χa belong to the same eigenvalue a, for the operator A, then the linear combination ψ = c1 φa + c2 χa is also an eigenfunction of the operator A with the same eigenvalue a. We can construct an infinite number of eigenfunctions corresponding to any degenerate eigenvalue. If there are rlinearly independent eigenfunctions corresponding to a particular degenerate eigenvalue, we say that the eigenvalue is r-fold degenerate.
3.6 MEANING OF EIGENVALUES AND EIGENFUNCTIONS In connection with the eigenvalues (a), eigenfunctions (φa ) and the normalized wave function (ψ) we ask the following questions: What are actually a, φa , ca and ψ? What do they represent in a quantum mechanical system? Can we measure them in an experiment? The aim of this section is to give precise answers to these questions.
Operators, Eigenvalues and Eigenfunctions 79 Recall the experiment of throwing a die of six sides considered in sec. 2.8. Let it be thrown a very large number of times, say N times. The average or mean or expectation value of an outcome of a throw of the die is given by X houtcomei = P (a)a , a = 1, 2, · · · , 6, (3.59) where P (a) is the probability of getting a as an outcome. The formula, Eq. (3.59), is not only for the throw of a die but also for outcome of any experiment. In quantum mechanics for the position measurement we write Z ∞ X hxi = P (x)x = ψ ∗ xψ dx . (3.60) −∞
For an observable A we can write
Z
hAi = Substituting ψ =
∞
ψ ∗ Aψ dx .
(3.61)
−∞
P
ca φa in the above equation we get Z ∞X X c∗a φ∗a A hAi = ca′ φa′ dx −∞
=
Z
∞
−∞
=
X a
a′
c∗a φ∗a
a
XX a
=
a
X
X
a′ ca′ φa′ dx
a′
c∗a ca′ a′ δaa′
a′
2
|ca | a .
(3.62)
Comparing the Eqs. (3.59) and (3.62) we infer that a in the eigenvalue equation Aφa = aφa is a possible outcome of an experiment. Suppose we measure the energy of a system. We get a number. This number is one of the possible energy eigenvalues. In the experiment we say that the outcome of energy measurement is the energy value of the system. In the theory we call a possible energy value as an energy eigenvalue. The physical significance of the eigenvalues of an operator of an observable is that they are the possible results of measurement of an observable. Eigenvalues can be measured in an experiment. Comparing Eqs. (3.59) and (3.62) we find that |ca |2 = P (a). The absolute square of the coefficient ca in the expansion of ψ in terms of eigenfunctions of the operator A is the probability that a measurement of A will yield the eigenvalue a. ca is the probability amplitude. We note that quantum theory does not give exact value of an observable of a particle at a time. Rather, it gives various possible values of the observable and their corresponding probabilities. Each eigenfunction is associated with each possible value of the observable. That is, quantum states are described by the mathematical functions called eigenfunctions. Operating the
80 Quantum Mechanics I: The Fundamentals corresponding operator A of the observable on an eigenfunction φa we get the corresponding eigenvalue a. If the value of the observable is a′ then we say that the system is in an eigenstate φa′ . In quantum theory all possible (eigen)states are combined together with complex number ca weighings as X ψ= ca φa , (3.63) a
where the eigenvalues are assumed to be discrete. This collection of all possible states with complex number weighings is called the wave function and it describes arbitrary quantum state of the particle. Generally, φa is a particular state of the system and ψ is the arbitrary state of the system. Further, φa is a physically realizable state. In an experiment a system can be found in the stationary eigenstate φa . However, ψ is not a physically realizable stationary state. In fact, ψ given by the Eq. (3.49) or (3.63) is not a solution of the eigenvalue equation Aψ = λψ (why?). Then, what does ψ represent? ψ states the various possible eigenstates and their corresponding probabilities. What do you infer if ψ of a system is c1 φ1 + c2 φ2 while ψ of another system is c1 φ1 + c2 φ2 + c3 φ3 where φi ’s are stationary state eigenfunctions? ca can be called as the A-space wave function just ψ(x) is called the coordinate space or configuration space wave function. ca may be used instead of ψ for describing the state since ψ can be obtained from ca through the Eq. (3.63). That is, ca and ψ are different representations of the same state.
3.7 PARITY OPERATOR Suppose P is an operator such that for any function φ(x), P φ(x) = φ(−x). Then P 2 φ = P P φ(x) = P φ(−x) = φ(x) . (3.64) The operator P is called a parity operator . We find that P 2 = 1. The operator P reverses x to −x and p into −p etc. When P φ(X) = φ(X) then P reverses X to −X and is equivalent to a reflection in x − y plane followed by a rotation of π about the z-axis.
3.7.1
Properties of the Parity Operator
The operator P has the following properties: 1. Eigenvalues of P are ±1. 2. For a symmetric potential [P, H] = 0. 3. Expectation value of P is a constant of motion. 4. P is Hermitian and linear. 5. P xop P = −xop , P pop P = −pop .
Operators, Eigenvalues and Eigenfunctions 81 We now prove the above properties. To find the eigenvalues of P assume that φλ is an eigenfunction of P corresponding to the eigenvalue λ. Then P φλ = λφλ ,
P 2 φλ = P λφλ = λ2 φλ .
(3.65)
But according to Eq. (3.64) P 2 φλ must be φλ . Therefore, λ2 = 1 or λ = ±1. When λ = 1 we have P φλ = φλ and the operator P is called an even parity operator . When λ = −1 we have P φλ = −φλ and the operator P is called an odd parity operator . Any function which is even, that is φ(x) = φ(−x), is an eigenfunction of parity operator with eigenvalue +1. Similarly, any odd function of x, φ(x) = −φ(−x), is an eigenfunction of parity operator with eigenvalue −1. The spherical harmonics Ylm (θ, φ) are examples of eigenfunctions of a parity operator. For any symmetric potential V (x, y, z) = V (−x, −y, −z) the parity operator commutes with H. So, it is a constant of motion. Also P has even or odd parity. Therefore, a system with symmetric potential will be described by eigenfunctions with even or odd parity. In central force field problems p V (r) = V ( x2 + y 2 + z 2 ). The potential is symmetric with respect to parity operator. An eigenfunction for a central force problem in spherical polar coordinates will contain the radial part Rnl (r) and the spherical part Ylm (θ, φ) (see sec. 12.2). Since Rnl has even parity, the total parity of the state is given by the spherical harmonics Ylm (θ, φ). In spherical polar coordinates, the parity operator causes r → r, θ → π −θ, φ → π + φ. We can easily show that P Ylm = (−1)l Ylm . Eigenfunctions with φ(r) = φ(−r) and φ(r) = −φ(−r) are said to have even parity and odd parity respectively. The significance of even parity is that the waves traveling to the left and right described by ψ(r) have the same energy. One can easily verify that P commutes with Lz and L2 given by Eqs. (11.27c) and (11.28) respectively. Since Ylm (θ, φ) are the eigenfunctions of L2 and Lz (see sec. 11.4), they are also the eigenfunctions of P . We notice that parity is quantum mechanical and has no classical analogue as there is no classical dynamical variable from which it can be obtained by the operator correspondence r → rop and p → −i~∇.
3.7.2
Commutator of P and H
For a reflection symmetric potential V (x), P V (x) = V (x). Further, P Hφ(x) = Hφ(−x) = HP φ(x) which implies [P, H] = 0. Thus, for symmetric potentials P and H are commuting operators and they have simultaneous eigenfunctions. Further, dhP i/dt = 0 because [P, H] = 0 (see, Eq. (3.24)). Let us verify the relation [P, H] = 0 for the linear harmonic oscillator potential V (x) = kx2 /2.
82 Quantum Mechanics I: The Fundamentals We obtain P Hφ(x)
1 2 ~2 d2 φ(x) + kx = P − 2m dx2 2 ~2 d2 1 2 φ(−x) = − + k(−x) 2m d(−x)2 2 = Hφ(−x) = HP φ(x)
(3.66)
which implies [P, H] = 0. There is an another parity operator called time reversal parity which reverses time coordinate and is denoted as T . It does not satisfy the eigenvalue equation T φ(t) = αφ(t) = φ(−t) but T φ(t) = φ∗ (t). T transforms px into −px and x into x and i → −i. It preserves the relation [x, px ] = i~.
Solved Problem 4: Show that P is Hermitian. R∞ Consider the integral −∞ φ∗1 (x)P † φ2 (x)dx. We obtain Z
∞
−∞
φ∗1 (x)P † φ2 (x)dx
=
= = = =
Z
∞
h
P†
†
i∗ φ1 (x) φ2 (x)dx
Z−∞ ∗ [P φ1 (x)] φ2 (x) dx Z φ∗1 (−x)φ2 (x) dx Z φ∗1 (x)φ2 (−x) dx Z ∞ φ∗1 (x)P φ2 (x) dx .
(3.67)
−∞
Therefore, P † = P and P is a Hermitian.
Solved Problem 5: How do the parity operators act on the position and momentum operators? Suppose we operate P xop P on φ(x). We get (P xop P ) φ(x)
= (P xop ) φ(−x) = P (xφ(−x)) = −xφ(x) = −xop φ(x) .
(3.68)
Therefore, P xop P = −xop . Similarly, we can show that P pxop P = −pxop .
Operators, Eigenvalues and Eigenfunctions 83
3.8 ALL HERMITIAN HAMILTONIANS HAVE PARITY Let us consider a Hermitian Hamiltonian, for example, H=
1 2 1 3 1 4 p + x + x . 2 x 3 4
(3.69)
This Hamiltonian is symmetric under the operation of time reversal operator T . A parity operator for a Hermitian Hamiltonian can be constructed as follows [12]. Given a Hamiltonian like that in Eq. (3.69) we can in principle solve the time-independent Schrödinger equation Hφn (x) = En φn (x). Now, we consider a new operator P(x, y) given by P(x, y) =
∞ X
(−1)n φn (x)φ∗n (y) .
(3.70)
n=0
The operator P has the following four properties: 1. P is linear and Hermitian. 2. P commutes with H and hence it is conserved. 3. P 2 = 1, that is, in coordinate space Z ∞ P(x, z)P(z, y) dz = δ(x − y) .
(3.71)
−∞
4. φn is an eigenfunction of P with the eigenvalue (−1)n , that is, Z ∞ P(x, z)φn (z) dz = (−1)n φn (x)
(3.72)
−∞
by virtue of orthonormality. Because of the properties (3) and (4) the operator P exhibits the characteristics of the parity operator even though the Hamiltonian may not be symmetric under space reflection. If the potential V (x) of the Hamiltonian H = p2x /(2m) + V (x) is invariant under the transformation x → −x, then P(x, y) is the usual parity operator. We can follow the above procedure for constructing many differential operators that commute with H. For example, we can construct a tri-parity operator Q(x, y) ∞ X Q(x, y) = ω n φn (x)φ∗n (y) , (3.73) n=0
where ω = e±i2π/3 and ω 3 = 1. However, Q(x, y) is not an observable. Why?
84 Quantum Mechanics I: The Fundamentals
3.9 SOME OTHER USEFUL OPERATORS In addition to Hermitian and parity operators there are other operators which play important role in quantum theory. We define some of these operators in this section. An operator I is said to be identity if Iφ = φ. The effect of I on φ is to leave it as such. That is, the operator I does not change anything. The square of the parity operator, P 2 , is an identity operator. Suppose A and B are two operators such that Aφ1 = φ2 and Bφ2 = φ1 where φ1 and φ2 are two functions of x. We obtain BAφ1 = Bφ2 = φ1 ,
ABφ2 = Aφ1 = φ2 .
(3.74)
The effect of AB on φ2 is to leave it as such and so AB = I. Since we can write AA−1 = A−1 A = I and hence B = A−1 : the operator B is the inverse of the operator A and vice-versa. An operator U with the property U † = U −1 is called a unitary operator . Since U U −1 = 1 we have U U † = 1 or U † U = 1. In this case the operator U behaves like a transformation. For any two eigenfunctions we get (φ1 , φ2 ) = (φ1 , U † U φ2 ) = (U φ1 , U φ2 ) which shows that the action of U on eigenfunctions preserve their norm. That is, U φ and φ both have the same norm which means the net probability is unchanged by the unitary transformation U . Since P 2 φ = φ we write P 2 = 1 or P = P −1 . Because of U U † = 1 where U is a unitary operator we note that P † = P , hence P −1 = P † and P is a unitary. There exists a class of eigenvalue problems, the so-called parity-time reversal (PT )-symmetric non-Hermitian Hamiltonian problems. They have arisen in recent years in a number of physical contexts such as localization transitions, vortex pinning and population biology [13-15]. Such non-Hermitian Hamiltonians can be used to describe certain experimental phenomena. They have been used for Bose system of hard sphere [16], complex Toda lattice [17] and field theory models [18-21]. PT -symmetry has been noticed in certain optical systems, such as optical couplers, microwave billiard and large-scale temporal lattices [22]. An optical system with periodic complex refractive index is essentially equivalent to a time-periodic non-Hermitian quantum system. A necessary condition for an optical structure to be PT -symmetric is that its complex refractive index distribution becomes n(x) = n∗ (−x). In this case the real part of the index profile is symmetric in space while the imaginary component (gain-loss) is antisymmetric [23]. It has been shown that inductively coupled two LRC circuits exhibit all the salient features of non-Hermitian systems which commute with PT operator [24,25]. A PT -symmetric Hamiltonian is a Hamiltonian which is invariant under the product of the parity operator, that is, P(: x → −x), and the time reversal operation, that is, T (: i → −i). For example, the operator H=−
d2 + α x2 − (ix)2n+1 , 2 dx
(3.75)
Operators, Eigenvalues and Eigenfunctions 85 where α(x2 ) is a polynomial of degree n ≥ 1 with positive real coefficients, is PT -symmetric. A simple example is H = p2x /(2m) + ix3 . The parity operator P is linear and has the effect of px → −px , x → −x. The time-reversal operator T is anti-linear and has the effect px → −px , x → x, i → −i. An example of an operator which is not PT -symmetric is H=−
d2 + x − (ix)3 . dx2
(3.76)
If H = −d2 /dx2 +V (x) is PT -symmetric then V ∗ (−x) = V (x) and so ReV (x) is an even function and ImV (x) is an odd function. Hence, if V (x) is a polynomial then V (x) = P (x2 ) + ixQ(x2 ) for some real polynomials P and Q. It has been shown that the PT -symmetric Hamiltonian H = −d2 /dx2 + x2 − g(ix)3 has a discrete spectrum for g real and the eigenvalues are real if g is small enough [26-27]. There are some PT -symmetric Hamiltonians that have no eigenvalues [28] or no real eigenvalues [29]. For more details about PT symmetric Hamiltonians one may read refs.[13-32]. Some of the applications of non-Hermitian PT -symmetric Hamiltonian have been reported in refs.[16, 17, 33]. It is noteworthy to mention that PT -symmetric materials can exhibit power oscillations and nonreciprocity of light propagation [34], absorption enhanced transmission and unidirectional invisibility.
3.10 CONCLUDING REMARKS In this chapter we have seen that the time-independent Schrödinger equation is an eigenvalue equation. We have seen that the operators representing an observable must be a Hermitian operator and discussed few important properties of Hermitian operators. How do you obtain an anti-Hermitian operator from a Hermitian operator? Multiplying it by i. We note that the operator representing an observable should be (i) linear, (ii) Hermitian and (iii) its eigenfunctions must form a complete set. It is easy to verify that the operators L and H are Hermitian if the potential V (x, y, z) is real. Further, the parity operator P commutes with Lz and L2 . In chapters 4 and 5 we will consider some simple and physically interesting quantum mechanical systems and solve the eigenvalue equations associated with them. We will find that in some systems normalizable solutions occur with discrete eigenvalues while in certain systems the solutions are nonnormalizable with continuous eigenvalues. Classical mechanics describes the dynamical variables as functions in phase space. We have seen that in quantum mechanics, they are Hermitian operations. As all the operators in quantum mechanics are linear operators, the state of a system is represented by a superposition of eigenfunctions.
3.11 BIBLIOGRAPHY [1] A. Jeffrey, Advanced Engineering Mathematics. Academic Press, San Diego, 2003.
86 Quantum Mechanics I: The Fundamentals [2] E. Kreyszig, Advanced Engineering Mathematics. John Wiley, New York, 2001. [3] H. Goldstein, Classical Mechanics. Narosa, New Delhi, 1985. [4] V.B. Bhatia, Classical Mechanics. Narosa, New Delhi, 1997. [5] E. Corinaldesi, Classical Mechanics for Physics Graduate Students. World Scientific, Singapore, 1998. [6] P. Dorey, C. Dunning and R. Tateo, J. Phys. A 34:L391, 2001. [7] K.C. Shin, J. Math. Phys. 42:2513, 2001. [8] C.M. Bender, S. Boettcher and P.N. Meisinger, J. Math. Phys. 40:2201, 1999. [9] C.M. Bender, F. Cooper, P.N. Meisinger and V.M. Savage, Phys. Lett. A 259:224, 1999. [10] C.M. Bender and G.V. Dunne, J. Math. Phys. 40:4616, 1999. [11] C.M. Bender, D.C. Brody and H.F. Jones, Am. J. Phys. 71:1095, 2003. [12] C.M. Bender, P.N. Meisinger and Q. Wang, J. Phys. A 36:1029, 2003. [13] N. Hatano and D.R. Nelson, Phys. Rev. Lett. 77:570, 1996. [14] N. Hatano and D.R. Nelson, Phys. Rev. B 56:8651, 1997. [15] D.R. Nelson and N.M. Shnerb, Phys. Rev. E 58:1383, 1998. [16] T.T. Wu, Phys. Rev. 115:1390, 1959. [17] T. Hollowood, Nucl. Phys. B 384:523, 1992. [18] R.C. Brower, M.A. Furman and M. Moshe, Phys. Lett. B 76:213, 1978. [19] C.M. Bender and K.A. Milton, Phys. Rev. D 57:3595, 1998; J. Phys. A 32:L87, 1999. [20] C.M. Bender, K.A. Milton and V.M. Savage, Phys. Rev. D 62:85001, 2000. [21] C.M. Bender, P.N. Meisinger and H. Yang, Phys. Rev. D 63:045001, 2001. [22] J. Luo, J. Huang, H. Zhong, X. Qin, Q. Xie, Y.S. Kivshar and C. Lee, Phys. Rev. Lett. 110:243902, 2013. [23] A. Regensburger, M. Ali Miri, C. Bersch, J. Nager, G. Onishchukov, D.N. Christodoulides and U. Peschel, Phys. Rev. Lett. 110:223902, 2013.
Operators, Eigenvalues and Eigenfunctions 87 [24] J. Schindler, Z. Liu, J.M. Lee, H. Ramezani, F.M. Ellis and T. Kottos, J. Phys. A: Math. Theor. 45:444029, 2012. [25] J. Schindler, A. Li, M.C. Zheng, F.M. Ellis and T. Kottos, Phys. Rev. A 84:040101, 2011. [26] E. Caliceti, J. Phys. A 33:3753, 2000. [27] E. Caliceti, S. Graffi and M. Maioli, Commun. Math. Phys. 75:51, 1980. [28] G.A. Mezincescu, J. Phys. A 33:4911, 2000. [29] E. Delabaere and F. Pham, Phys. Lett. A 250:25, 1998. [30] P. Dorey, C. Dunning and R. Tateo, J. Phys. A 34:5679, 2001. [31] C.M. Bender, S. Boettcher and P.N. Meisinger, J. Math. Phys. 40:2201, 1999. [32] C.M. Bender, J. Brod, A. Refig and M.E. Reuter, J. Phys. A 37:10139, 2004. [33] C.M. Bender, G.V. Dunne and P.N. Meisinger, Phys. Lett. A 252:272, 1999. [34] MC. Zheng, D.N. Christodoulides, R. Fleischmann and T. Kottos, Phys. Rev. A 82:010103, 2010.
3.12 EXERCISES 3.1 Show that the following operators are linear. 2 d d d −x2 d 2 (a) x , (b) x , (c) x + and (d) e . dx dx dx dx 3.2 If Af = f 2 evaluate A(x3 + 5x). 3.3 Distinguish between Aφ = 5φ and Aφ not directly proportional to φ. 3.4 If A and B are two linear operators prove that AB and BA are also linear operators. 3.5 Show that linear combination of two linear operators A and B is also a linear operator. Also prove that A+B = B+A, A+B+C = (A+B)+C = A + (B + C), A(BC) = (AB)C = ABC, (A + B)C = AC + BC, A(B + C) = AB + AC and A2 = AA. 3.6 Find whether the operators defined through the equations Aφ = φ2 and Bφ = eφ are linear or nonlinear.
88 Quantum Mechanics I: The Fundamentals R∞
U (x−x′ )ψ(x′ ) dx′ where 0, x < 0 the function U (x) is the unit step-function U (x) = If the 1, x > 0. class of functions ψ is the class for which ψ and dψ/dx are integrable −1 for |x| → ∞, show that Ω = (d/dx) .
3.7 Let the operator Ω be defined by Ωψ(x) =
−∞
3.8 What is the effect of e−iHτ /~ on ψ(t)? 3.9 Show that eiapx /~ f (x) = f (x + a). 3.10 If φ is an eigenfunction of a linear operator A with an eigenvalue λ find the eigenfunctions and the eigenvalues of A2 and A3 . What do you infer from your result? 3.11 Verify that if φ is an eigenfunction of A then it is also an eigenfunction of An , eA and A−n . 3.12 Find the eigenvalues (λ) and the eigenfunctions (y) of the operator d2 /dx2 . Assume that the eigenfunctions are defined in the interval 0 ≤ x ≤ π and they vanish at x = 0, π, · · · . 3.13 For the class of functions satisfying proper boundary conditions show that the Hermitian condition (ψ, Aψ) = (Aψ, ψ) implies (φ1 , Aφ2 ) = (Aφ1 , φ2 ). 3.14 Write down the eigenvalue equations for position and momentum operators and solve them. Express the normalization conditions for the eigenfunctions of position and momentum operators. 3.15 Show that the standard deviation of the measurement of a dynamical variable in its eigenstate is zero. 3.16 Find the eigenfunctions and eigenvalues of a particle of mass m in the potential V (x) = V0 , (a constant), V0 > 0. 3.17 Show that product of two unitary operators is also a unitary operator. R ˆ (x) = ∂f /∂x and If ˆ I]f ˆ (x) = x f (x)dx find (i) [d, ˆ (x) and (ii) 3.18 Given df 0 ˆ [x, I]. 3.19 If X and Y are two operators and Z = eαX Y e−αX show that dZ/dα = [X, Z]. Hence, show that the Taylor expansion of Z (in powers of α) is eαX Y e−αX = Y + α[X, Y ] +
α2 α3 [X, [X, Y ]] + [X, [X, [X, Y ]]] + · · ·. 2! 3!
3.20 Show that if A and B are two linear operators then [A, B] = −[B, A] and [A, B −1 ] = −B −1 [A, B]B −1 . 3.21 Is sin x a Hermitian operator?
Operators, Eigenvalues and Eigenfunctions 89 3.22 If A and B are two linear operators with [A, B] = a where a is a constant, by induction, show that [An , B] = naAn−1 and [A, B n ] = naB n−1 . Then deduce the value of [xn , px ] and [x, pnx ]. 3.23 Show that [H, t] = i~. 3.24 Suppose the operators A and B commute with their commutator, show that (i) [A, B n ] = nB n−1 [A, B] and (ii) [An , B] = nAn−1 [A, B]. 3.25 If [A, B] = 0 and [B, C] = 0, are A and C commuting operators? Give examples. 3.26 What are the eigenvalues of the operators (i) [x, px ] and (ii) [x, py ]? 3.27 Suppose the operators A and B commute with their commutator, that is [B, [A, B]] = 0. Prove that (i) for any analytic function F (x), [A, F (B)] = [A, B]F ′ (B). (ii) eA eB = eA+B e[A,B]/2 . 3.28 Prove that [A, e−αB ] = −αe−αB [A, B], if A and B commutes with their commutation. 1 d d 3.29 Find [X, P ] for X = sinh z, P = − i~ cosh z + cosh z . 2 dz dz 3.30 Consider a particle in a box potential V (x) = 0 for |x| < a and ∞ elsewhere. Write the Hamiltonian of the system for |x| < a. Compute [H, x] and [H, p]. 3.31 Prove the Jacobi identity [[A, B], C] + [[B, C], A] + [[C, A], B] = 0. 3.32 If A and B are operators such that [A, B] = 1 and if Z = αA + βB, where α and β are numbers, verify that (i) [A, Z n ] = nBZ n−1 . Hence, show that (ii) (∂/∂α)Z n = nAZ n−1 − n(n − 1)βZ n−2 /2 and
(iii) (∂/∂α)eZ = (A − β/2)eZ .
(iv) By integrating the differential equation for eZ , with due attention to the order of operator factors, show that eαA+βB = eαA eβB e−αβ/2 .
3.33 Prove the following adjoint operators relations. (i) adj d2 /dx2 = d2 /dx2 , (ii) adj d3 /dx3 = −d3 /dx3 and
(iii) adj dn /dxn = (−1)n dn /dxn .
90 Quantum Mechanics I: The Fundamentals 3.34 What is the value of h[A, B]i, where A and B are linear operators, associated with any eigenstate of A or B? √ 3.35 Find the adjoint a† of the operator a = (x + d/dx) / 2. Is a a selfadjoint operator? Verify that a† a is self-adjoint. 3.36 Suppose A, B, C are three linear operators such that C = A + B. If A and C are Hermitian, is B necessarily Hermitian? 3.37 Is the operator A = [x, px ] Hermitian? Why? 3.38 A and B are two linear operators. For the product AB to become a Hermitian, are A and B necessarily Hermitian? 3.39 Is H = −(~2 /2m)d2 /dx2 + V (x), where V (x) is a polynomial in x, a Hermitian operator? Why? 3.40 Are the operators xd/dx and x2 d2 /dx2 Hermitian? Why? 3.41 Show that the operators L and H are Hermitian if the potential V (x, y, z) is real. 3.42 Show that if H1 and H2 are Hermitian then i[H1 , H2 ] is also a Hermitian. 3.43 Show that the anticommutator of two Hermitian operators is also a Hermitian. 3.44 Prove whether or not the following are eigenfunctions of the parity operator P . What are their corresponding eigenvalues? (i) x4 + 6x2 . (ii) x5 . (iii) sin x. (iv) cos x. (v) sin2 x. 2
(vi) x3 + 5x. (vii) x3 + x2 + x. (viii) e−x . (ix) e−x . 3.45 Show that (i) eigenfunctions of the parity operator belonging to different eigenvalues are orthogonal and (ii) every function can be expressed as a linear combination of the eigenfunctions of the parity operator. 3.46 What is the effect of time reversal on any time-independent operator A representing a dynamical variable? 3.47 Show that the Hamiltonian H = −d2 /dx2 − (ix)3 is not a Hermitian. 3.48 Show that H = p2x + i sin x is not a Hermitian but PT -symmetric. 3.49 Show that H = p2x + x2 + 2x is not PT -symmetric. 3.50 Show that the time reversal operator changes any number into its complex conjugate and hence it is not a linear operator.
CHAPTER
4
Exactly Solvable Systems I: Bound States
4.1 INTRODUCTION Analytical solutions to the Schrödinger equation serve a variety of purposes. They help us 1. to identify the features of potentials and state functions, 2. to construct appropriate potentials from the experimental data and 3. to get analytical insight of the system considered. The problem of solving a time-dependent Schrödinger equation for a given time-dependent potential is difficult. However, for some simple and physically interesting potentials the time-independent Schrödinger equation is solvable exactly and the solutions are obtainable in terms of well-known functions. Useful informations about a given system are extracted from the solutions of timeindependent Schrödinger equation. The time-independent Schrödinger equation gives stationary state solutions which are independent of time. For a particle in a one-dimensional potential V (x), the time-independent Schrödinger equation with H = p2x /(2m) + V (x) is given by ψxx +
2m (E − V (x))ψ = 0 , ~2
(4.1)
where the subscript x in ψ denotes differentiation with respect to x. The eigenvalues of energy eigenstates can be discrete or continuous. The energy eigenstates with discrete eigenvalues are called bound states because they correspond to the classical bound orbits confined to a finite region in space. The bound state eigenfunctions are normalizable. On the other hand, the energy eigenstates with continuous eigenvalues are called unbound states or scattering states and they correspond to the classical unbound orbits in which 91
92 Quantum Mechanics I: The Fundamentals the particle approaches from and recedes to infinite distance. The eigenfunctions of scattering states are nonnormalizable. The present chapter is primarily concerned with the study of bound states of some simple one-dimensional systems. Study of bound systems and their bound states is very important because atoms, molecules and other fundamental matters are bound systems and their characteristics are understood by quantum mechanics. We consider a one-dimensional linear harmonic oscillator, an anharmonic oscillator, a particle in a box and Pöschl–Teller potentials. Since free particle is the simplest quantum mechanical system and form the starting point of the study of quantum systems we include the study of it in this chapter even though its solutions are unbound. The quantum mechanical treatment of three-dimensional version of the above mentioned systems are easy to analyse and are given as problems at the end of this chapter. However, we will consider the important systems like rigid rotator and hydrogen atom. We solve the rigid rotator in this chapter while the hydrogen atom will be explored in chapter 12. Potentials giving scattering states will be analysed in the next chapter. The time evolution of stationary state eigenfunctions is easy to obtain. We illustrate this with reference to linear harmonic oscillator. When the Schrödinger equation of a potential is not exactly solvable we can study the solution by means of numerical techniques. In chapter 21 we solve the time-independent and time-dependent Schrödinger equations of certain potentials numerically.
4.2 CLASSICAL PROBABILITY DISTRIBUTION How do we establish the connection between the Schrödinger wave function and the classical motion of a particle for stationary state solutions? An approach is to compare the probability density PQM = |ψn (x)|2 with a classical probability distribution PCM (x) [1]. Then show that these probabilities approach each other in a locally averaged manner, in the limit of large quantum number n. Let us describe the calculation of PCM (x) and PCM (p) for a classical particle exhibiting periodic motion. Consider a particle exhibiting a bounded periodic motion (with period τ ) with E = H = p2x /(2m) + V (x). We have p2x /(2m) = E −V (x). Since p2x /(2m) must be greater than or equal to zero, the classical motion is confined to the regions where E−V (x) ≥ 0. The points at which E−V (x) = 0, (or E = V (x)), are called classical turning points. At these points the velocity of the particle vanishes and it turns back. Let us denote these points as a and b. Then the particle will move back and forth between a and b. The time taken by the particle for one traversal from a to b is half the period, τ /2. The time dt spend by the the particle in the small region, dx, near the point x is dt = dx/v(x) where v(x) is the speed. The probability of finding the particle in dx region is ratio of dt and the time taken for one traversal: PCM (x) dx =
dt 2 dx 2 = or PCM (x) = . τ /2 τ v τv
Exactly Solvable Systems I: Bound States 93 The particle will spend more time in regions where its speed is low. Further, Rb PCM given by Eq. (4.2) is normalized ( a PCM (x)dx = 1) . Using E = mv 2 /2+ V (x) for v in Eq. (4.2) we obtain [1] r m 2 . (4.2) PCM (x) = τ 2(E − V (x))
Solved Problem 1: Find an expression for PCM (px ). The probability of measuring the classical particle with momentum px is PCM (px ) dpx
dt τ /2 2 dpx . τ |dpx /dt|
= =
(4.3)
Since dpx /dt = F (x) = −dV /dx we have PCM (px ) =
2 . τ |F |
(4.4)
4.3 FREE PARTICLE For a free particle the force is zero and hence the potential is a constant: With v = 0. Constant potential does not influence the dynamics of the system. A free particle has significance in problems related to collision and metallic state. The Schrödinger equation for the free particle is ψxx + k 2 ψ = 0,
k 2 = 2mE/~2 .
(4.5)
Its solution is ψ = Ae±ikx ,
(4.6)
where A is a constant. This ψ(x) is always single-valued, finite and continuous for all values of k. Hence, there is no restriction on the values of k and E. They can take continuous values. But the ψ given by Eq. (4.6) is nonnormalizable! Why is this? What does it imply? For the free particle the energy is always positive and the expectation value of kinetic energy is E. What is the expectation value of momentum? Let us calculate the flux with ψ = Aeikx . We have i~ ∂ψ ∗ ~k|A|2 |A|2 ∗ ∂ψ J(x) = − ψ −ψ = = px . (4.7) 2m ∂x ∂x m m Classically, if ρ(x) is the spatial charge density of particles and v is the speed
94 Quantum Mechanics I: The Fundamentals of the particles then J(x) = vρ(x) = ρpx /m. Thus, ψ = Aeikx physically represents flow of particles. In Eq. (4.6) for a given k or E there are two independent eigenfunctions: eikx and e−ikx . The energy eigenvalue corresponding to these eigenfunctions is thus degenerate. However, for k 6= 0: Z ∞ Z ∞ −ikx ∗ ikx e dx = e2ikx dx ∝ δ(k) = 0 . (4.8) e −∞
−∞
The eigenfunctions eikx and e−ikx are orthogonal. How do we distinguish these degenerate eigenfunctions? Operating pˆx on them we have pˆx e±ikx = pˆx eipx x/~ = ±px e±ipx x/~ .
(4.9)
The two eigenfunctions have the same energy but different momenta. Instead of e±ikx , suppose consider the eigenfunctions sin kx and cos kx. Does pˆx distinguish them? The answer is no. Because pˆx sin kx = −i~k cos kx, sin kx is not a momentum eigenfunction. Therefore, pˆx is not useful to distinguish them. However, as sin kx and cos kx are odd and even functions of x respectively, we can distinguish them by the parity operator. Even though ψ = Ae±ikx is not normalizable we can find the constant A. The orthogonality condition for eigenfunctions with continuous eigenvalues is Z ∞ ψλ∗ ψλ′ dx = δ(λ′ − λ) , (4.10) −∞
where δ is Dirac-delta function. Hence, for the free particle eigenfunctions the above condition is Z ∞ ψk∗ (x)ψk′ (x)dx = δ(k ′ − k) . (4.11) −∞
Substituting for ψk and ψk′ we obtain Z ∞ ′ |A|2 ei(k −k)x dx = δ(k ′ − k) .
(4.12)
−∞
That is, 2π|A|2 δ(k ′ − k) = δ(k ′ − k) (see problem 3 at the end of √ this chapter). Integrating this with respect to k ′ we get 2π|A|2 = 1 or A = 1/ 2π. Hence, 1 ψk (x) = √ e±ikx . 2π
(4.13)
4.4 HARMONIC OSCILLATOR Consider the linear harmonic oscillator, a particle of mass m bounded to the origin by a restoring force F = −kx. Here k is the restoring force constant
Exactly Solvable Systems I: Bound States 95 and x is the displacement from equilibrium. The sign of the force is opposite to the sign of x. The force is always directed towards the equilibrium point. The potential of the system is V = kx2 /2. The Hamiltonian is p 1 2 1 2 1 2 1 H= px + kx or H = px + mω 2 x2 , ω = k/m . (4.14) 2m 2 2m 2 The classical equation of motion is 1 px , p˙x = −mω 2 x . (4.15) m The Schrödinger equation of the quantum mechanical linear harmonic oscillator is 1 2 2m (4.16) ψxx + 2 E − kx ψ = 0 , ~ 2 where E is the energy of the particle with respect to the bottom of the potential well. x ¨ + ω 2 x = 0 or x˙ =
4.4.1
Physical Systems Modelled by the Linear Harmonic Oscillator
The linear harmonic oscillator (LHO) is a fundamental system solved exactly in both classical and quantum mechanics. It is used in the modelling of subnuclear quarks to cosmic systems. Any system with small amplitude vibration about its equilibrium point is approximately described by LHO. Vibration of diatomic molecules is approximately a LHO. Electrons restricted in threedimensional electromagnetic traps are found to behave as quantized harmonic oscillators. The LHO description is used in the study of probability distribution of hadronic matter. The model is found to give a simple explanation for a wide range of hadronic phenomena noticed in high energy physics experiments. For a very general potential Vc (x) and for small oscillations of a system about a stable state x∗ = a we consider Taylor series expansion of Vc (x) as 1 Vc (x) = V (a) + V ′ (a)(x − a) + V ′′ (a)(x − a)2 + · · · . 2
(4.17)
Because V ′ (a) = 0 and V ′′ (a) > 0 the system behaves approximately as a harmonic oscillator with k/2 = V ′ (a). An oscillating electromagnetic field is similar to a LHO [2]. An electromagnetic wave at a point in space can be considered as oscillating electric and magnetic fields or a collection of photons. Consider a standing electromagnetic wave in a one-dimensional enclosure without a source. An example is a resonant cavity of a laser. Suppose the enclosure is kept along the x-axis. Its volume is, say V . The electric field is plane polarized parallel to the y-axis and its angular frequency is ω. The strength of electric and magnetic fields are given by s r ωx ωx 2N ω 2 2µ0 Ey (x, t) = E(t) sin , Bz (x, t) = B(t) cos , (4.18) ǫ0 V c NV c
96 Quantum Mechanics I: The Fundamentals where N is the normalization factor. E and B are given by the Maxwell’s equations 1 dB dE = B, = −N ω 2 E. (4.19) dt N dt The total energy obtained integrating the energy density over the enclosure is 1 2 1 B + N ω2E 2 . (4.20) 2N 2 Equations (4.19) and (4.20) are similar to the Eqs. (4.15) and (4.14) respectively. There are two standard approaches to determine the eigenvalues and eigenfunctions of quantum LHO – one is differential and another is algebraic. First we take up the former. U=
4.4.2
Transformation of Schrödinger Equation to Hermite Equation
Equation (4.16) is a variable coefficient and linear ordinary differential equation. We do not have a systematic method to construct exact analytical solutions of such equations. However, the Schrödinger Eq. (4.16) can be transformed into the Hermite differential equation by introducing a suitable change of variable and parameters. Writing ξ = αx, we have ψx = ψξ dξ/dx = αψξ and ψxx = α2 ψξξ where α = dξ/dx. Then Eq. (4.16) becomes 2mE mk 2 ψξξ + − 2 4ξ ψ = 0 . (4.21) ~2 α2 ~ α Defining m2 ω 2 mk , α = 2 = ~ ~2 we rewrite the Eq. (4.21) as 4
2mE 2E λ= 2 2 = ~ α ~
r
ψξξ + λ − ξ 2 ψ = 0 .
m 2E = k ~ω
(4.22)
(4.23)
With the change of variable
ψ = H(ξ) e−ξ
2
/2
(4.24)
Eq. (4.23) takes the form Hξξ − 2ξHξ + (λ − 1)H = 0 .
(4.25)
Redefining λ = 2n + 1 the above equation becomes the well-known Hermite differential equation Hξξ − 2ξHξ + 2nH = 0 . (4.26)
Equation (4.26) is not an exactly solvable equation. However, it admits a power series solution. We are seeking bound state solutions of (4.16): ψ(x) → 0 as x → ±∞. The particular solutions which are finite in the interval −∞ < ξ < ∞ and decay to zero as ξ → ±∞ exist for integer values of n and are called Hermite polynomials [3-5].
Exactly Solvable Systems I: Bound States 97
4.4.3
Eigenfunctions and Eigenvalues
The Hermite polynomial solutions of Eq. (4.26) are given by [3–5] Hn (ξ) = (−1)n eξ
2
∂ n −ξ2 e . ∂ξ n
(4.27)
The first few Hermite polynomials are H0 = 1,
H1 = 2ξ,
H2 = 4ξ 2 − 2,
H3 = 8ξ 3 − 12ξ .
(4.28)
From Eq. (4.24) the eigenfunctions are obtained as ψn (ξ) = Nn Hn (ξ) e−ξ
2
/2
,
(4.29)
where Nn ’s are normalization constants. The substitution ξ = αx in Eq. (4.29) gives 2 2 ψn (αx) = ψn (x) = Nn Hn (αx) e−α x /2 . (4.30) Thus, the eigenfunctions of LHO are Hermite polynomials times a Gaussian function and are called Weber functions. The normalization condition gives Z ∞ Z ∞ 2 2 2 1 = ψn∗ (x)ψn (x) dx = Nn2 Hn (αx) e−α x dx =
−∞ 2 |Nn |
−∞
√ n π 2 n!/α .
Thus, Nn =
(4.31)
√
α π 2n n!
1/2
.
(4.32)
The eigenfunction ψn with a particular n is called a quantum state n. The number n is called quantum number . Some of the features of a LHO are summarized below. 1. Because of λ = 2n + 1, Eq. (4.22) gives 1 ~ω , n = 0, 1, 2, · · · (4.33) En = n + 2 p where ω = k/m is the classical angular frequency of the oscillator. En ’s are the only stationary state energy levels of the oscillator. En ’s are discrete because n is an integer and En depend on n. Hence, the quantum LHO has a discrete energy spectrum. This is not true for the classical LHO since an arbitrary amount of energy can be imparted to the classical system. We note that the energy spectrum of a free particle is continuous. 2. The states represented by the eigenfunctions ψn are normalized and hence they are bound states.
98 Quantum Mechanics I: The Fundamentals
-6
0
x
1
-1
ψ -6
FIGURE 4.1
0
x
6
ψ -6
0
x
1
n=3
0
0
-1
6
0
x
6
-6
0
x
1
n=4
-6
n=2
0
-1
6
0
-1
1
n=1
ψ
0
-1
ψ
1
n=0 ψ
ψ
1
6
n=5
0
-1
-6
0
x
6
First few eigenfunctions of LHO where k = 1.
3. The energy difference between two successive energy levels is ~ω, a fixed value, and thus the energy levels are equally spaced. This is due to the potential energy function V (x) = kx2 /2. This potential expands in such a manner that, as the amplitude increases with energy the levels remain equally spaced. 4. The ground state energy obtained by substituting n = 0 is ~ω/2 and is nonzero. This is known as zero-point energy. The classical minimum energy is zero. In contrast, quantum mechanically it is ~ω/2. This is because ψ0 (x) does not minimize V (x) alone but rather the sum of kinetic and potential energies. 5. What would happen if En = 0? If En = 0 then Z ∞ 1 2 k En = 0 = ψn∗ Hψn dx = hpx i + hx2 i 2m 2 −∞
(4.34)
would imply hx2 i = hp2x i = 0. For a statistical variable x, which is not always zero, hx2 i should be greater than zero. 6. Figure 4.1 shows first few eigenfunctions of LHO. ψn (−x) = ψn (x) for even values of n while ψn (−x) = −ψn (x) for odd integer of n. Thus, the parity associated with ψn is P = (−1)n . 7. ψn given by Eq. (4.30) are the eigenfunctions of the Hamiltonian operator H given by Eq. (4.14) with the eigenvalues En . H is a Hermitian operator (verify). Earlier, in sec. 3.4 we have shown that for a Hermitian operator eigenfunctions belonging to distinct eigenvalues are mutually orthogonal. Therefore, the eigenfunctions of LHO are orthogonal. We can verify the above using the ψn ’s given by Eq. (4.30) also.
Exactly Solvable Systems I: Bound States 99
4.4.4
Classical and Quantum Mechanical Probabilities
The solution of the classical equation of motion (4.15) is x(t) = A sin(ωt + φ) where A and φ are integration constants. The solution is periodic with period τ = 2π/ω. The maximum amplitude of the solution is A and the turning points are +A and −A. The classical motion is bounded between x = −A and +A, that is, −A ≤ x(t) ≤ A. Now, for the LHO Eq. (4.2) becomes v 1 2ω u m u = r PCM (x) = . (4.35) 1 2 2π t 2E 2 2 E− kx π −x 2 k
We express E in terms of the turning point A. At x(t) = ±A the velocity dx/dt is zero and E = V (x = ±A) = kA2 /2 or 2E/k = A2 . Then PCM (x) =
1 √ , π A2 − x2
(4.36a)
where A2 =
2En 2 = k k
1 2 1 n+ ~ω = 2 n + . 2 α 2
(4.36b)
Figure 4.2 depicts PQM = |ψn (x)|2 for n = 0, 1, 2, 3 and 4. In this plot |ψ| is scaled into |ψ|2 + E and k = 1. Figure 4.3 shows PCM (x) and PQM (x) for n = 10 with k = 1. The vertical dashed lines indicate classical turning points. Region within the vertical dashed lines is classically allowed region and outside region is classically forbidden. From the Figs. 4.2 and 4.3 we note that quantum mechanically there is a small probability in the classical forbidden region. It is clear from these figures that the position probability densities |ψn |2 associated with the stationary state eigenfunctions have little resemblance to the corresponding densities of the classical oscillation. The agreement between classical and quantum probability densities improves with increase in n. The agreement is fairly good, on the average for large n, the principal discrepancy being the rapid oscillation of |ψn |2 . 2
Solved Problem 2: What are the probabilities of the LHO system in the ground state to be inside and outside the classical regions? For simplicity, consider the ground state. The classical allowed region is obtained from Eq. (4.36b) with n = 0. It is (−A, A) → (−1/α, 1/α). The PQM
100 Quantum Mechanics I: The Fundamentals
5
E, |ψ|2
4
3
2
1
0
-5
0
5
x
|ψ|2 versus n for LHO. From the bottom curve to top the values of n are 0, 1, 2, 3 and 4. k is chosen as 1. E is in units of ~ω. The potential is also shown. FIGURE 4.2
in this region is PQM
=
Z
1/α
2
−1/α
= = =
2α √ π
Z
|ψ0 | dx 1/α
2
e−α
x2
dx
0
Z 1 2 2 √ e−y dy π 0 0.83 .
(4.37)
The probability of finding the particle in the classically allowed region is 0.83 (obtained numerically computing the value of the integral in Eq. (4.37)). The probability in the outside region is 0.17.
4.4.5
Selection Rules
It is possible to find the selection rules for the allowed transitions for LHO. We know from classical electrodynamics, any oscillatory electric or magnetic multipole emits or absorbs radiation. We can show that the presence of any electric
Exactly Solvable Systems I: Bound States 101
0 .3
n = 10
|ψ|2
0 .2
0 .1
0
-6
-3
0
x
3
6
FIGURE 4.3 PCM (dashed curve) and PQM (x) (continuous curve) versus x for n = 10, k = 1 of LHO. The two vertical dashed lines represent classical turning points.
or magnetic multipole matrix elements between any two states results in induced absorption or emission of the electromagnetic wave. The electric dipole transition is the predominant one. So, normally selection rules are worked-out by considering the matrix element of electric dipole operation er between the states involved in the transition. As shown in sec. 2.12 a transition between two states characterized by the quantum numbers m and n is possible only if the condition (2.102) is satisfied. Selection rules for allowed transitions are obtained starting from the recurrence relation of the Hermite polynomials (4.38)
Hn+1 − 2ξHn + 2nHn−1 = 0 . Using Eq. (4.29) the above equation is expressed in terms of ψn as √ √ √ n + 1 ψn+1 − 2 ξψn + n ψn−1 = 0 .
(4.39)
Multiplication of Eq. (4.39) by ψn∗ and integrating with respect to ξ we get Z ∞ Z ∞ √ Z ∞ ∗ √ √ n+1 ψn∗ ψn+1 dξ − 2 ψn ξ ψn dξ + n ψn∗ ψn−1 dξ = 0 . −∞
−∞
−∞
(4.40) The first and last integrals in Eq. (4.40) become zero because of the orthogonality of the eigenfunctions. Then Z ∞ ψn∗ ξ ψn dξ = 0 . (4.41) −∞
102 Quantum Mechanics I: The Fundamentals ∗ Similarly, multiplication of Eq. (4.39) with ψn+1 and integrating with respect to ξ give Z ∞ p ∗ ψn+1 ξ ψn dξ = (n + 1)/2 . (4.42) −∞
Replacing n by n − 1 in Eq. (4.42) we have Z ∞ p ψn∗ ξ ψn−1 dξ = n/2 .
(4.43)
−∞
From Eqs. (4.41) and (4.43) we write Z ∞ ψn∗ ξ ψm dξ 6= 0 for n = m + 1 .
(4.44a)
−∞
Interchanging n and m we get Z ∞ ∗ ψm ξ ψn dξ 6= 0 for m = n + 1 .
(4.44b)
−∞
Therefore, R ∞m −∗ n = ±1 or ∆mn = ±1. Hence, the electric dipole matrix exψn dx is not equal to zero if m − n = ±1. Thus, transition element −∞ ψm can occur between adjacent energy levels. Transitions between energy levels involve emission or absorption of quanta of energy by the oscillator. When transition from level n to level n−1 occurs then ∆E = En−1 −En = −~ω which is negative and hence energy is emitted by the oscillator. For the transition from level n to level n+1 we have ∆E = ~ω and therefore absorption of energy by the oscillator takes place. A quantum mechanical system in its ground state cannot radiate energy because there is no lower energy state for it to go.
4.4.6
Abstract Operator Method to Linear Harmonic Oscillator
Let us solve the LHO problem by the operator method. To determine the eigenspectra of H we introduce two operators a and a† as 1 a= √ (mωx + ipx ), 2mω~
a† = √
1 (mωx − ipx ) . 2mω~
(4.45)
These we rewrite as 1 i d 1 i d a= √ αx + , a† = √ αx − , (4.46) α dx α dx 2 2 p where α = mω/~ . From Eqs. (4.46) x and px are expressed in terms of a and a† as i~α 1 x= √ a + a† , p x = − √ a − a† . (4.47) 2α 2
Exactly Solvable Systems I: Bound States 103 † † What is the commutation relation of† a, a ? To find [a, a ] consider [x, px ]. We obtain [x, px ] = xpx − px x = i~ a, a . Since [x, px ] = i~ we get (4.48) a, a† = aa† − a† a = 1 .
We express the H of LHO in terms of a and a† as
~2 α2 mω 2 1 † † 2 † 2 H=− + = ~ω a a + a−a a+a . 2m 4α2 2
(4.49)
Thus, the problem of finding the eigenvalues of H is reduced to the problem of finding the eigenvalues of a† a.
4.4.6.1 Eigenvalues We define an operator n ˆ = a† a and denote ψn as the eigenfunction of n ˆ with the eigenvalue n: n ˆ ψn = nψn . (4.50) We find n ˆ , a† = a† a, a† = a† a, a† + a† , a† a = a† . From this we have n ˆ a† − a† n ˆ = a† .
(4.51)
From Eq. (4.51) we obtain n ˆ a† ψn = a† + a† n ˆ ψn = (n + 1)a† ψn .
Next, n ˆ a† In general
2
2 ψn = a† + a† n ˆ a† ψn = (n + 2) a† ψn . n ˆ a†
Similarly,
m
ψn = (n + m) a†
m
ψn .
n ˆ am ψn = (n − m)am ψn .
(4.52)
(4.53) (4.54) (4.55)
The news from (4.54) is that if ψn is the eigenfunction of n ˆ with the eigenvalue n then a† ψn , (a† )2 ψn , · · · are the eigenfunctions of n ˆ belonging to the eigenvalues (n + 1), (n + 2), · · · respectively. On the other hand, Eq. (4.55) implies that, aψn , a2 ψn , · · · are the eigenfunctions of n ˆ corresponding to the eigenvalues (n − 1), (n − 2), · · · respectively. The effects of a† and a are the raising and lowering, respectively, of the eigenvalue by one. a† essentially adds one quantum of energy while a removes one quantum of energy. For this reason a† and a are called raising and lowering operators or simply ladder operators. They are also called creation and annihilation operators respectively. These operators appear particularly in fields of applied physics where quantum mechanics is of primary importance. The areas include lasers, nanometer scales devices and studies of electrical conduction in a variety of devices.
104 Quantum Mechanics I: The Fundamentals Now, we find the range of eigenvalues of n ˆ . Note that a† a is Hermitian irrespective of whether a is Hermitian or not. Further, Z ∞ Z ∞ hˆ ni = ha† ai = ψ ∗ a† aψ dx = (aψ)∗ aψ dx ≥ 0. (4.56a) −∞
−∞
because the value of the last integral is the norm of aψ which should be > 0. If the system is in the state ψλ then the eigenvalues of n ˆ = a† a is λ. This gives Z ∞ Z ∞ ∗ hˆ ni = ψλ n ˆ ψλ dx = λ ψλ∗ ψλ dx = λ . (4.56b) −∞
−∞
Comparison of (4.56a) and (4.56b) gives λ > 0. Thus, the eigenvalues of n ˆ must be nonnegative. Therefore, (n − 1), (n − 2), · · · must be nonnegative and aψn , a2 ψn , · · · must terminate. Suppose ψ0 is the last eigenfunction in the above set. Since the effect of a is to decrease an eigenvalue by one, we have aψ0 = 0 and a† aψ0 = 0 or n ˆ ψ0 = 0 . (4.57) ψ0 is the eigenstate of a† a or n ˆ with the eigenvalue 0. The other eigenfunctions are constructed by the repeated application of a† on ψ0 . Thus, ψ0 , a† ψ0 , (a† )2 ψ0 , · · · are the eigenfunctions belonging to the eigenvalues 0, 1, 2, · · · respectively. The values of n are 0, 1, 2, · · · . Then the eigenvalues of H are En = n + 21 ~ω.
4.4.6.2 Eigenfunctions
The norm (a† ψn , a† ψn ) is written as a† ψn , a† ψn = ψn , aa† ψn = (n + 1) (ψn , ψn ) .
(4.58)
If ψn and ψn+1 are normalized to unity then (ψn , ψn ) = (ψn+1 , ψn+1 ) and a† ψn , a† ψn = (n + 1) (ψn+1 , ψn+1 ) √ √ n + 1 ψn+1 , n + 1 ψn+1 (4.59) = which implies
a† ψn =
√
n + 1 ψn+1 .
(4.60)
Denoting the nth eigenfunction as |ni (and √ its complex conjugate as hn|) the above√equation is expressed as a† |ni = n + 1 |n + 1i. a† on the nth state gives n + 1√times the next higher state. Similarly, it is easy to show that a on |ni gives n times the next lower state, except when n = 0. Replacement of n by n − 1 in Eq. (4.60) gives a† ψn = √ ψn−1 . n
(4.61)
Exactly Solvable Systems I: Bound States 105 Repeated use of Eq. (4.60) in Eq. (4.61) leads to a† (a† )n a† ψn = √ √ ψn−2 = · · · = √ ψ0 , n n−1 n!
(4.62)
where ψ0 is the solution of aψ0 = 0. Integrating the equation 1 d 1 αx + ψ0 = 0 aψ0 = √ α dx 2 2
we get ψ0 (x) = Ae−α
x2 /2
ψn (x)
(4.63)
√ . Normalization gives A = [α/(2n n! π)]1/2 . Then
= =
α √ 2n n! π α √ n 2 n! π
1/2 1/2
a†
n 2
e−α
2
e−α
x2 /2
x2 /2
(4.64)
Hn (αx) .
From Eq. (4.49) we write ~ωa† a = H + E0 , where E0 = ~ω/2 is the ground state energy. a† a measures the number of energy quanta (photons) present above the ground state and hence is called number operator . The energy eigenstates are also eigenstates of this operator and are thus often called number states. There is a straightforward interpretation of the above formalism for the radiation field. Each quantum of energy, ~ω, represents a photon. The eigenstates ψn represents a system with n photons. The ground state is the vacuum state (no photons) with nonzero energy ~ω/2. The operators a† and a represent the physical processes by which photons are absorbed or emitted. Therefore, a† and a are called creation and annihilation operators respectively. Does an operator of an observable of a conservative system contain a or a† alone? Why?
4.5 PARTICLE IN THE POTENTIAL V (x) = x2k , k = 1, 2, · · ·
We consider a particle of mass m, moving along the x-axis with the potential energy function V (x) = Axn , where A is a constant [6]. As n increases V (x) becomes more and more flat-bottomed and the sides become steeper and steeper near the turning points. For very large n the V (x) approximates a square-well or box potential. It is of interest to know the change in the energy levels are due to deformation of V (x). For convenience, we rewrite the potential V (x) as x 2k V (x) = α , k = 1, 2, · · · (4.65) a where α, a and p are parameters. Defining θ2 =
2ma2 α, ~2
λ = θ2k/(k+1)
E , α
x′ = θ1/(k+1)
x a
(4.66)
106 Quantum Mechanics I: The Fundamentals and dropping x′ for convenience. The Schrödinger equation for the problem takes the form (4.67) ψxx + λ − x2k ψ = 0 .
The boundary conditions are ψ(0) =finite and ψ → 0 as |u| → ∞. When k = 1, Eq. (4.67) becomes the LHO Eq. (4.23) For k → ∞, the boundary conditions become ψ(u) = 0 for |u| ≥ 1 and the Eq. (4.67) reduces to ψuu + λψ = 0, |u| ≤ 1 (4.68)
Its solution subjected to the boundary conditions ψ(±1) = 0, ψ(0) = 1 is ( √ cos λ0 u, if |u| < 1 ψ0 (u) = (4.69) 0, if u ≥ 1 , √ where λ0 = π/2. Though the solution of Eq. (4.67) for arbitrary p is not known, we get approximate solution for large k. For this purpose w = 1/(k+1) uk+1 /(k + 1) or u = [(k + 1)w] . We assume solution of the form ψ(u) = e−w f (u) .
(4.70)
Substitution of Eq. (4.70) in Eq. (4.67) gives fuu − 2uk fu + λ − kuk−1 f = 0 .
(4.71)
For large p and |u| < 1, the terms containing uk and uk−1 are small. In this case fuu + λf ≈ 0 whose solution is given by Eq. (4.69). Thus, √ ψ(u) = e−w cos λ u (4.72) is an approximate solution of Eq. (4.67). For large k, e−w is almost constant at 1 until u is close to 1, so the variation of ψ is described by the cosine function. For u ≫ 1, e−w in Eq. (4.72) is dominant and the eigenfunction very rapidly decays to zero. Near u ≈ 1 we obtain h√ i ψ(w) = e−w cos λ k(k + 1)1/(k+1) w1/(k+1) . (4.73)
For k ≫ 1, w1/(k+1) is ≈ 1. Therefore, ψ(ω) will vanish if √ λ (k + 1)1/(k+1) = π/2 .
(4.74)
Therefore, the eigenvalue condition is λ1 =
π 2 2
1 , (k + 1)2/(k+1)
where λ1 is the ground state eigenvalue of Eq. (4.67).
(4.75)
Exactly Solvable Systems I: Bound States 107
V (x) ∞
−a FIGURE 4.4
∞
0
a
x
Sketch of an infinite height potential, Eq. (4.76).
4.6 PARTICLE IN A BOX Another interesting exactly solvable quantum mechanical system is a single particle of mass m confined within a potential well V given by ( 0, |x| < a (4.76) V (x) = ∞, |x| > a and is depicted in Fig. 4.4. Realization of this kind of potentials arises for molecules of a dilute gas in a long thin tube with a rigid stoppers at the ends. The vibration of a string clamped at two points is modelled by the above potential. The thermodynamic equilibrium of a cold white dwarf star is modelled by balancing the dynamics of electrons, represented as particles in a box, with the gravitational forces. This model has predicted the stellar mass threshold for final collapse of a white dwarf to a black hole [7]. The electrons in metals are also modelled by particles in a three-dimensional box potential. This model explains the motion of conduction electrons and their contribution to magnetism and specific heat. It is now possible to manufacture a box trapping electrons in a few nanometers-wide region. The box potential is a good description for one-dimensional bound-state problems, when the confining potential is much steeper compared to the parabolically curved LHO potential.
4.6.1
Classical Result
The particle is confined between the two walls of the potential and bounces elastically when it incidents on the walls. The problem is a bound state problem. The total energy is E = mv 2 /2. The classical equation of motion is x ¨ = 0, |x| < a. Its solution is x(t) = At + B. The choice x(0) = 0 gives x(t) = At and A is the velocity v of the particle. The particle starting from x = 0 moves
108 Quantum Mechanics I: The Fundamentals
a x(t) 0 -a A
t v(t)
0 -A
t
Classical trajectory and velocity of a particle in the infinite height potential, Eq. (4.76). FIGURE 4.5
towards x = a with velocity v. Its trajectory is a straight-line as shown in the subplot of Fig.4.5. When the particle reaches x = a, it turns back with the velocity −v and proceeds towards −a. At x = −a, it turns back towards +a with +v and so on. The turning points are −a and a. The distance the particle traverses to come back to the origin with velocity v is 4a. Therefore, the period of motion is τ = 4a/|v|. The fundamental frequency f (in Hertz) is 1/τ (and ω = 2πf , where ω is in rad/sec). The solution is written as [1] x(t) = −a +
4a |t|, τ
or 4a x(t) = − 2 π
∞ X
for − τ /2 < t < τ /2
(4.77)
1 cos(2πnt/τ ) . n2
(4.78)
n=1 odd
Figure 4.5 shows classical trajectory and velocity. Obtain PCM (x) from Eq. (4.2). Substitution of V (x) = 0, E = mv 2 /2, τ = 4a/|v| in Eq. (4.2) gives PCM (x) = 1/(2a).
4.6.2
Quantum Mechanical Result
The Schrödinger equation for |x| ≤ a is given by ψxx + k 2 ψ = 0,
k 2 = 2mE/~2 .
(4.79)
For |x| > a, ψ(x) = 0. The solution of Eq. (4.79) is ψ(x) = A sin kx + B cos kx .
(4.80)
Exactly Solvable Systems I: Bound States 109 This solution must satisfy the boundary conditions ψ(a) = 0, ψ(−a) = 0. Applying the boundary conditions the following equations are obtained: A sin ka + B cos ka = 0 ,
−A sin ka + B cos ka = 0 .
(4.81)
For nontrivial solutions (A 6= 0 or B = 6 0) the condition is the determinant of the coefficient matrix must be zero: sin ka cos ka =0. (4.82) − sin ka cos ka This gives sin 2ka = 0. In other words k=
nπ , 2a
n = 1, 2, · · · .
(4.83)
Thus, the allowed values of ka are discrete. Since k is related to E in Eq. (4.79) E becomes discrete. Since ~k is the momentum we write px n = ~kn = nπ~/(2a). The news is that px is discrete. For a free particle px is continuous. From the Eqs. (4.81) with kn = nπ/(2a) we obtain 2A sin(nπ/2) = 0,
2B cos(nπ/2) = 0 .
(4.84)
When n is an odd integer (n = 1, 3, · · · ) A = 0 while B is arbitrary. For even integer values of n, n = 2, 4, · · · , A is arbitrary and B = 0. Thus, Eqs. (4.81) lead to the following two cases. Case 1: A = 0, B 6= 0, ka =
nπ , n = 1, 3, · · · (odd n) . 2
(4.85)
A 6= 0, B = 0, ka =
nπ , n = 2, 4, · · · (even n) . 2
(4.86)
Case 2:
(n = 0 is not acceptable. Why? ). The discreteness of n has an immediate significance. For example, for a string fixed at the ends x = ±a we have 2π/λ = nπ/(2a), giving λ = 4a/n, n = 1, 2, · · · . The result is certain wavelengths alone are allowed. In other words, only certain frequencies of vibrations are allowed. This is the quantization of frequency. The waves oscillate more rapidly as n increases. For a particle in a box, the de Broglie wavelengths are λn = 4a/n, n = 1, 2, · · · . The lowest energy of the particle in the box potential is not zero. If it is zero v = 0 and hence λ = h/mv = ∞. But there is no way to reconcile an infinite wavelength to the particle in a box. The features of a particle in a box are listed in the following: 1. From Eqs. (4.79) and (4.83) the energy levels are obtained as En =
n2 ~2 π 2 , 8ma2
n = 1, 2, · · · .
(4.87)
110 Quantum Mechanics I: The Fundamentals There are infinite number of energy levels. However, the energy levels are discrete or quantized. The lowest energy is E1 = ~2 π 2 /(8ma2 ) which is nonzero. The particle will make to and fro motion in the box in the region −a to a even at absolute zero temperature. E1 is the zero-point energy. 2. The energy difference between the successive levels is △En = En+1 − En =
(2n + 1)~2 π 2 . 8ma2
(4.88)
The energy levels are not equally spaced. En are proportional to n2 so they are close at lower energy levels than at higher levels. 3. The momentum of the particle is px = ~πn/(2a) and En ∝ 1/a2 . Therefore, squeezing a particle increases its momentum and the total energy. 4. Because of discreteness of energy spectrum we discretize ψ, k, A and B and rewrite the Eq. (4.80) as ( An sin kn x, kn = nπ/(2a), n = 2, 4, · · · (4.89) ψn (x) = Bn cos kn x, kn = nπ/(2a), n = 1, 3, · · · . These eigenfunctions are normalizable.RApplying the normalization cona dition for ψn with even n we get 1 = −a A2n sin2 kn x dx = A2n a. Thus, √ √ An = 1/ a . Similarly, for odd n Bn = 1/ a . 5. For |x| < a the values of x at which ψ vanishes is called a node. For odd n the nodes are x = ±a/m, m = 1, 3, · · · , n. For even n the nodes are x = 0, ±2a/m, m = 2, 4, · · · , n. For n = 1 the nodes are x = ±a, ±a/3 and for n = 2 they are 0, ±a, ±a/2. For a given state the number of nodes is n + 1. 6. Because cos(−x) = cos x, the eigenfunction for an odd n is symmetric with respect to x = 0. Since sin(−x) = − sin x, ψ is anti-symmetric about x = 0 for even n. Adjacent states have opposite symmetry. For even n, P ψn (x) = ψn (x). Therefore, sin(nπx/2a) and cos(nπx/2a) are also the eigenfunctions of the parity operator. 7. Figure 4.6 shows first few energy levels and eigenfunctions. The corresponding probability distribution 1 1 πx 2 2 PQM (x) = |ψn (x)| = cos n− (4.90) a 2 a is depicted in Fig. 4.6. For large value of n, the cos2 term in Eq. (4.90) averages to 1/2 showing that PQM approaches the classical limit 1/2a.
20 15 10
E,|ψ|2
E,ψ
Exactly Solvable Systems I: Bound States 111
5 0 -2
-1
0
x
1
2
20 15 10 5 0 -2
-1
0
x
1
2
First few energy levels and the eigenfunctions of infinite height potential (left subplot). Here a = 1 and ~2 and m are set to unity. ψn (x) is scaled into ψn (x)+En . From bottom to top the values of n are 1, 2, 3 and 4. |ψn |2 versus x (right subplot) for the eigenfunctions shown in the left subplot. FIGURE 4.6
8. A transition between a pair of states with the quantum numbers n and n′ is allowed if one of the numbers is even and the other is odd, that is △n = ±1. Transitions between two odd or even quantum numbers states is forbidden. 9. Since V (x) = V (−x) we have [P, H] = 0 where P is the parity operator. The parity of the eigenfunctions of particle in a box will not change in time. 10. Let us describe the time dynamics in the infinite height potential with an analogy to the classical wave P propagation on a stretched string of length L. Assume that ψ(x) = cn ψn (x) be the initial profile of a p stretched ¯n /m string which then propagate without dispersion at speed v = 2E ¯ where En is the mean energy level. The elastic collisions at the infinitely high potential barriers are analogous to perfect reflections from the fixed ends of the string. When a wave form propagating on a string is reflected at the string ends, it travels in the opposite direction with a change of phase π in its amplitude. Its analogous behaviour is found in the box potential [8]. The eigenstates given by the Eqs. (4.87) and (4.89) are the stationary states of the system with energy eigenvalue being constant, independent of the space variable x. A nonstationary but normalizable eigenfunction of the system is ( N x2 − a2 , |x| < a ψ(x) = (4.91) 0, |x| > a ,
112 Quantum Mechanics I: The Fundamentals where N is the normalization constant. But the energy of this eigenfunction is space dependent. In the domain of our everyday experience we do not feel the quantum effects, since on macroscopic scale they are imperceptible. Suppose a ball of mass 10 g is rolling on a smooth floor of 10cm wide box with a speed of 3.3 cm/sec. The permitted energies for the ball according to Eq. (4.87) are En = 5.49 × 10−57 n2 erg. The kinetic energy of the ball is 54.4 ergs. This corresponds to the energy level of quantum number n = 1029 . This is very large. Therefore, it is practically impossible to find whether the ball is in the energy states n = 1029 or 1029 + 1, 1029 + 2, · · · , etc. (discrete). Therefore, we say that continuous range of energy is possible.
Solved Problem 3: The size of a nucleus is say, 10−14 m. Assuming it as a one-dimensional box calculate the ground state energy of an electron in the nucleus. Show why the electron does not exist in the nucleus. The lowest energy is E1
= = =
π 2 ~2 2ma2 2 π 2 × 1.0544 × 10−34
2
2 × 9.11 × 10−34 × (10−14 )
Js
6.023 × 10−10 Js .
(4.92)
The Coulombic attraction energy is calculated as Ec = −
e2 = −2.3 × 10−14 Js . r
(4.93)
Since Ec is very weak compared to E1 the nucleus cannot hold an electron.
4.6.3
Abstract Operator Method to a Particle in a Box
In sec. 4.4 we obtained eigenspectra for LHO by operator method. The problem of a particle in a box is also solved in this approach [9]. We define the number operator n ˆ and its inverse n ˆ −1 as n ˆ ψn = nψn ,
n ˆ −1 ψn = n−1 ψn .
(4.94)
Suppose a† and a are the creation and annihilation operators such that a† ψn = nψn+1 ,
aψn = (n − 1)ψn−1 .
(4.95)
For convenience we write the width of the box as 2L and z = πx/(2L) so that the Hamiltonian is 2 2 1 ~π d Hz = − . (4.96) 2m 2L dz 2
Exactly Solvable Systems I: Bound States 113 The choice [9] a† = (cos z)ˆ n + sin z with
d a = (cos z)ˆ n − sin z n ˆ −1 (ˆ n − 1) dz
d , dz
1 ψn = √ sin nz, L
z=
πx 2L
(4.97)
(4.98)
gives a† ψn = nψn+1 ,
aψn = (n − 1)ψn−1 .
(4.99)
†
Equations (4.99) are simply the Eqs. (4.95). That is, a and a given by the Eqs. (4.97) are the creation and annihilation operators of the eigenstates of the Hamiltonian (4.96). The commutator [a, a† ] is calculated as aa† − a† a = (2n − 1). We write [a, a† ] = 2ˆ a0 , a ˆ0 = n ˆ − 1/2. Also aa† ψn = n2 ψn and 2 ~π 1 Hz ψn = aa† ψn . (4.100) 2m 2L
ˆ = aa† . We have [N ˆ , a† ] = 2ˆ ˆ , a† ] = Now, define the operator N a0 a† and [N † ˆ (2n + 1)a . Similarly, we obtain [N , a ˆ] = −2 (n − 1/2) a. ˆ (a† ψn ) = (n + 1)2 (a† ψn ). Further, N ˆ (a† )2 ψn = (n + Next, we find N 2 † 2 2) (a ) ψn . In general, ˆ (a† )m ψn N ˆ (a)m ψn N
= =
(n + m)2 (a† )m ψn , (n − m)2 (a)m ψn .
(4.101) (4.102)
ˆ with Equations (4.101) and (4.102) imply that if ψn is an eigenfunction of N eigenvalue n2 then a† ψn , (a† )2 ψn , · · · are the eigenfunctions with the eigenvalues (n + 1)2 , (n + 2)2 , · · · respectively. Also aψn , (a)2 ψn , · · · are the eigenfunctions belonging to the eigenvalues (n − 1)2 , (n − 2)2 , · · · respectively. From Eq. (4.94) we note that n cannot be zero. Suppose ψ1 is the eigenˆ ψ1 = function for the lowest state. Then Eq. (4.94) is satisfied. Further, N ˆ with the eigenvalue aa† ψ1 = ψ1 . Therefore, ψ1 is the eigenstate of aa† or N 1. The other eigenfunctions are constructed by the repeated application of a† on ψ1 . Thus, ψ1 , a† ψ1 , (a† )2 ψ1 , · · · , (a† )n ψ1 are the eigenfunctions belonging to the eigenvalues 1, 2, 4, · · · , n2 . The eigenvalues of Hz are 2 1 ~π Ez = n2 . (4.103) 2m 2L From Eq. (4.95) we write ψn+1 = (a† /n)ψn or ψn
=
1 (a† )n−1 ψ1 , (n − 1)!
n 6= 0 .
(4.104)
From Eqs. (4.100) and (4.103) the equation for ψ1 is obtained as ψ1xx + α2 ψ1 = 0, α = π/(2L) . √ Its solution is ψ1 (x) = (1/ L) sin(πx/2L).
(4.105)
114 Quantum Mechanics I: The Fundamentals 5000
4000
c
V (x)
3000
2000
b 1000
0
a 0
0 .2
0 .4
x
0 .6
0 .8
1
Pöschl–Teller potentials, Eq. (4.106), for (λ, κ) = (2, 2)(a), (2, 32)(b) and (2, 64)(c). FIGURE 4.7
4.7 PÖSCHL–TELLER POTENTIALS A class of potentials which are exactly solvable and possessing bound states are the Pöschl–Teller potentials given by [10-12] 1 λ(λ − 1) κ(κ − 1) V (x) = V0 + , 0 ≤ x ≤ πa (4.106) 2 cos2 (x/2a) sin2 (x/2a) where λ, κ > 1 and V0 is a constant. These potentials were first introduced in a molecular physics context. Figure 4.7 shows the potentials V (x) for a = 1/π, V0 = 1 and (λ, κ) = (2, 2), (2, 32), (2, 64). When λ = κ = 1 the potential V (x) becomes the box potential. At the turning points of the classical motion E = V (x). Solving this equation for the turning points we obtain α−β p 1 ± △ , △ = 1 − (α + β) + (α − β)2 , (4.107a) x± = a cos−1 2 4
where
α = λ(λ − 1)V0 /E ,
β = κ(κ − 1)V0 /E .
(4.107b)
Therefore, the classical motion is possible only if E>
2 p 1 p V0 λ(λ − 1) + κ(κ − 1) . 2
(4.108)
Exactly Solvable Systems I: Bound States 115 The solution x(t) is given by [12] (α − β) p + △ cos ωt , x(t) = a cos−1 (4.109) 2 p x(0) = x− and x(0) ˙ = 0. The period of the solution is where ω = 2E/m/a, p τ = 2π/ω = 2πa m/(2E) . Note that τ is independent of V0 , λ and κ. For the quantum mechanical problem, we impose ψ(x) = 0 for x ≥ πa and x ≤ 0, ψ(0) = ψ(πa) = 0. The stationary state Schrödinger equation is 2 2m ~2 (λ + κ)2 d + 2 E − V (x) + ψ(x) = 0 , 0 ≤ x ≤ πa , (4.110) dx2 ~ 8ma2 where for convenience the Hamiltonian of the system is shifted by the amount −~2 (λ + κ)2 /(8ma2 ). Without loss of generality choose V0 = ~2 /(4ma2 ) in V (x). The normalized eigenfunctions and eigenvalues are worked out as [12] x λ x κ 1 2 x ψn (x) = N cos sin , 2 F1 −n, n + λ + κ; κ + ; sin 2a 2a 2 2a (4.111) where N is a normalization factor, 2 F1 is a hypergeometric function and En =
~2 n(n + λ + κ) , 2ma2
λ, κ > 1, n = 0, 1, · · · .
(4.112)
4.8 QUANTUM PENDULUM A rigid pendulum is a point particle restricted to a circular radius and subjected to a uniform gravitational or electric field. Classical and quantum versions of this system are interesting. At low energies the potential energy of a quantum pendulum is approximated by that of a harmonic oscillator and for high energies it is thought of as the free rotor [13]. The dynamics of the quantum pendulum are closely related to molecular motions known as hindered rotations. The system is used to describe almost free rotations in ethene (C2 H2 ) and low energy torsional vibrations in K2 PtCl6 (potassium hexachloroplatinate) [14].
4.8.1
Classical Result
Consider a point particle of mass m restricted to a circle of radius l. It has a rotational inertia I = ml2 . The system is subjected to a uniform gravitational force (mg) (or electric force QE) in the vertical direction. The corresponding potential energy function is V (θ) = −V0 cos θ where V0 = mgl (or QEl). The classical Hamiltonian is H =E=
I ˙2 θ − V0 cos θ . 2
(4.113)
116 Quantum Mechanics I: The Fundamentals The classical equation of motion is g θ¨ + sin θ = 0 . l
(4.114)
˙ Assuming θ(0) = 0 when θ(0) = θmax where θmax = cos−1 (−E/V0 ) the solution of Eq. (4.114) is given by [15] h p i , K = sin (θmax /2) , (4.115) θ(t) = 2 sin−1 Ksn t g/l , K where sn is the Jacobian elliptic function. From Eq. (4.113) we write r ml2 dθ √ dt = . (4.116) 2 E + V0 cos θ
For −V0 < E < V0 the motion will be periodic with θ˜ = ± cos−1 (−E/V0 ) being the turning points. Then the classical period is
τ (E) = 2
r
ml2 2
Zθ˜
−θ˜
dθ √ . E + V0 cos θ
(4.117)
For low energies (small oscillations), with −V0 ≤ E ≪ V0 or θ˜ p ≪ 1, the motion is oscillator-like. In this case the time period is τ = 2π/ V0 /ml2 which is independent of energy. For the case of unbounded motion for which E > V0 one period corresponds to one revolution and is given by τ (E) =
r
ml2 2
Zπ
−π
dθ √ . E + V0 cos θ
(4.118)
In the higher-energy (rotor-like) limit, where E = Iω 2 /2 ≫ V0 we have r ml2 2π 2π √ = τ→ . (4.119) 2 ω E
4.8.2
Quantum Result
The Schrödinger equation for the quantum pendulum is given by −
~ ψθθ − V0 cos θ ψ = Eψ . 2I
(4.120)
Introducing the change of variable θ = 2x and defining a = 8IE/~2 and q = −4IV0 /~2 Eq. (4.120) becomes ψxx + (a − 2q cos x)ψ = 0 .
(4.121)
Exactly Solvable Systems I: Bound States 117 Equation (4.121) is identified as angular Mathieu equation. Since θ = 2x the boundary condition is ψ(2x) = ψ(2(x + π)). The eigenfunction has to be periodic in x with period π. Because of the intrinsic parity of the potential, there are two types of solutions. Even solutions cer (x, q) (cosine-elliptic) for r = 0, 2, 4, · · · and odd solutions, ser (x, q) (sine-elliptic) for r =1, 3, · · · [4]. cer and scr are expanded in terms of Fourier series. There are given below [16]: ce2m (x, q) ce2m+1 (x, q) se2m+2 (x, q) se2m+1
= = = =
∞ X
k=0 ∞ X
k=0 ∞ X
k=0 ∞ X
A2k (q) cos 2kx ,
(4.122a)
A2k+1 (q) cos(2k + 1)x ,
(4.122b)
B2k+2 (q) sin(2k + 2)x ,
(4.122c)
B2k+1 (q) sin(2k + 1)x ,
(4.122d)
k=0
where m = 0, 1, 2, · · · . The recurrence relations between the coefficients are found by substituting the above series in Eq. (4.121). For example, for ce2m the recurrence relations are [16] aA0 h i a − (2k)2 A2k
= qA2 , (a − 4)A2 = q (2A0 + A4 ) ,
(4.123a)
= q (A2k−2 + A2k+2 ) ,
(4.123b)
where k ≥ 2. When q = 0 the solutions (after normalization) are √ ce0 = 1/ 2, cer = cos rx, r ≥ 1, ser = sin rx, r ≥ 1.
(4.124)
For fixed values of q, periodic solutions with period π or 2π exist only for specific values (characteristic values) of a and are denoted by ar for the even solutions. For odd solutions the characteristic values are, say, br . Mathematical packages, such as Mathematica, are useful to evaluate the values of a and b. It is quite difficult to compute Mathieu functions. For discussion on computation of Mathieu functions one may refer to refs.[17-19]. Figure 4.8 shows the plot of first few eigenvalues (a) as a function of q [14]. The eigenvalues are doubly-degenerate when q = 0 and partially small values of q. Observe the similarity of the probability density functions of the first three energy levels [14] shown in Fig. 4.8 with the harmonic oscillator.
4.9 CRITERIA FOR THE EXISTENCE OF A BOUND STATE Is it possible to know whether a given potential admits at least one bound state? The answer is not positive for arbitrary potential. However, for certain
30 a4 20
a
a2
0
-20 0
(a)
5
10
15
q
|ψ3|2
b2
a0
|ψ2|2
b4
10
-10
|ψ1|2
118 Quantum Mechanics I: The Fundamentals
20
-3
(b)
0
x
3
Eigenvalues a as a function of q and probability density function for the first three energy levels of the Mathieu’s equation. FIGURE 4.8
potentials the conditions for existence of a bound state have been obtained [20]. In this section, we consider the one-dimensional potential not vanishing identically but vanishing everywhere outside some interval: V (x) 6= 0 and V (x) = 0 for |x| > a .
(4.125)
For this type of potential the conditions for the existence of a bound state are obtained [20]. For a bound state Z ∞ .Z ∞ hHi = ψ ∗ Hψ dx ψ ∗ ψ dx < 0 , (4.126) −∞
−∞
where ψ is unnormalized and integrations are performed over the interval −∞ < x < ∞. Denoting the above quotient by N/D and since D is positive, the sufficient condition (4.126) becomes Z ∞ DhHi = N = −ψψxx + V ψ 2 dx < 0 , (4.127) −∞
where we have set ~ = m = 1 for simplicity. Integration of the first term in the integral by parts gives Z ∞ DhHi = N = ψx2 + V ψ 2 dx < 0 . (4.128) −∞
Exactly Solvable Systems I: Bound States 119 Let us choose the normalizable trial eigenfunction as (4.129)
ψ(x) = P (x) + λV (x) ,
where λ is a real parameter, −∞ < λ < ∞, and P (x) is the Pyramid function if 0 ≤ |x| ≤ a 1, P (x) = 1 − (|x| − a)/L, if a < |x| ≤ L + a (4.130) 0, if L + a < |x| < ∞ .
ψ given by Eq. (4.129) is normalizable for L > 0. Substituting Eq. (4.129) in Eq. (4.128) we obtain DhHi = Aλ2 + Bλ + C + where A = C
2 Z
=
Z
∞
V
′2
+V
3
−∞ ∞
V dx
dx ,
2 , L
B=2
(4.131a)
Z
∞
V 2 dx > 0 ,
−∞
(4.131b)
−∞
and V ′ = dV /dx. Since A, B and C are independent of L, for large L, the condition (4.131a) is written as DhHi = Aλ2 + Bλ + C < 0 .
(4.132)
Therefore, a bound state must exist if the above condition is satisfied for some values of λ. From Eqs. (4.131) and (4.132) we infer the following: 1. When λ = 0, ψ becomes P (x) and the sufficiency condition is [21,22] Z ∞ V (x) dx < 0 . (4.133) −∞
2. If A ≤ 0, then Eq. (4.132) is satisfied for some λ of sufficiently large magnitude. 3. If A > 0 then Eq. (4.132) is satisfied if C < B 2 /(4A). Combining the above results, the Hamiltonian H will have a bound state if one of the following conditions is realized: Z ∞ (1) V ′2 + V 3 dx ≤ 0 , A < 0 . (4.134a) −∞
(2)
Z
∞
−∞
(3)
Z
∞
−∞
V dx ≤ Z
Z ∞
∞
−∞
2 V dx 2
(V ′2 + V 3 ) dx
, C
0 . (4.134b) 4A
−∞
V dx ≤ 0 .
(4.134c)
120 Quantum Mechanics I: The Fundamentals The third condition is due to the fact that either (4.134a) is true or the rightside of (4.134b) is positive. Further, it includes the case of equality. Thus, for any one-dimensional potential, V (x) 6= 0, which vanishes R ∞ outside some interval, the system must have at least one bound state if −∞ V (x) dx = 0, (the potential is zero on the average). The condition (4.134c) is weaker. It has been pointed out that a weak attractive potential in two-dimensions can generate a bound state [23]. The problem of the existence of bound states in one-dimension and two-dimensions for N -particle systems has also been examined [24]. The mathematical and physical properties for the occurrence and nonoccurrence of bound states in a square-well potential have been discussed by Sahu et al. [25].
4.10 TIME-DEPENDENT HARMONIC OSCILLATOR In sec. 4.4 we studied the time-independent Schrödinger equation of LHO. Now, we solve the time-dependent Schrödinger equation of it given by i~
∂ψ ~2 ∂ 2 ψ 1 2 =− + kx ψ . ∂t 2m ∂x2 2
(4.135)
Let us write the solution as ψ(x, t) =
∞ X
n=0
an φn (x) e−iEn t/~ = e−iωt/2
X
an φn (x) e−inωt ,
(4.136)
where φn (x) is the solution of the time-independent Schrödinger equation and an are the constants to be determined. Assume that at t = 0, the solution is 1/2 X 2 2 α e−α (x−b) /2 , (4.137) ψ(x, 0) = an φn (x) = √ π
where the centre of the wave packet is shifted by an amount b, b > 0. To calculate am , multiply Eq. (4.137) by φ∗m and integrate over x. The result is 1/2 Z ∞ 2 2 2 α ∗ am = √ Nm e−(αx−αb) /2 Hm (αx) e−α x /2 dx , (4.138) π −∞ where we substituted the expression for φm (x). Introducing the change of variable αx = ξ and αb = ξ0 Eq. (4.138) becomes 1/2 Z ∞ 2 2 α ∗ √ am = N m Hm (ξ) e−(ξ +(ξ−ξ0 ) )/2 dξ . (4.139) π −∞ To evaluate the integral in the above equation consider the generating function Z ∞ Z ∞ X 2 1 2 sn ∞ −s2 +2sξ −(ξ 2 +ξξ0 + 21 ξ02 ) e e dξ = Hn (ξ) e−(ξ −ξ0 ξ+ 2 ξ0 ) dξ . n! −∞ −∞ n=0 (4.140)
Exactly Solvable Systems I: Bound States 121 Using the relation √
πe
− 14 ξ02 +sξ0
∞ √ − 1 ξ2 X (sξ0 )n 0 4 = πe n! n=0
(4.141)
and equating the coefficient of sn on both sides we obtain 2
ξ n e−ξ0 /4 an = 0n 1/2 . (2 n!)
(4.142)
Now, Eq. (4.136) becomes ψ(x, t) =
α √ π
1/2
Since
Now,
2
− 14 ξ02 − 12 iωt
n ∞ X Hn (ξ) 1 ξ0 e−iωt . n! 2 n=0
∞ X 2 Hn (ξ)sn = e−s +2sξ n! n=0
we obtain ψ(x, t)
1
e− 2 ξ
(4.143)
(4.144)
1 2 −2iωt 1 2 1 2 1 −iωt = exp − ξ − ξ0 − iωt − ξ0 e + ξ0 ξ e 2 2 2 4 1/2 α 1 2 √ = exp − (ξ − ξ0 cos ωt) π 2 ωt 1 + ξ0 ξ sin ωt − ξ02 sin 2ωt −i . (4.145) 2 4 α √ π
1/2
2 2 α P (x, t) = ψ ∗ ψ = √ e−α (x−b cos ωt) . π
(4.146)
We find that ψ(x, t) is an oscillating wave packet. It oscillates about x = 0 without change of shape.
4.11 RIGID ROTATOR Having studied the bound states of certain one-dimensional systems we now examine the three-dimensional rigid rotator. Consider two particles of masses m1 and m2 joined about an axis passing through a point O. The particle of mass m1 is at a distance r1 from O and the other particle is at a distance r2 from O and r1 + r2 = R (Fig. 4.9). Such a system is called a rigid rotator because the distances r1 and r2 are fixed. The magnitude of the vibration of a diatomic molecule is generally small compared with the equilibrium bond length so the rigid rotator is a good first approximation.
122 Quantum Mechanics I: The Fundamentals
z
m2
r2
B
θ r1
O m1
y
φ
A
x FIGURE 4.9
4.11.1
Parameters of the rigid rotator AB.
Hamiltonian and Schrödinger Equation
Let us denote the coordinates of the two particles with respect to the origin O as (x1 , y1 , z1 ) and (x2 , y2 , z2 ) respectively. The classical equation of motion for the rigid rotator is Kinetic Energy (K.E) = =
1 1 m1 v12 + m2 v22 2 2 1 m1 r12 + m2 r22 ω 2 , 2
(4.147)
where v1 = r1 ω, v2 = r2 ω and ω is the angular frequency of rotation. Thus, K.E. = Iω 2 /2 where I = m1 r2 + m2 r22 is the moment of inertia. The potential of the system is zero because there is no external force on the system and hence H = Iω 2 /2. The system is spherically symmetric; therefore, it is convenient to deal it in spherical polar coordinates. In this coordinates system the coordinates of the particles are (a, θ, φ) and (b, π − θ, π + φ) respectively. We have x1
= a sin θ cos φ,
x2 = b sin θ cos φ,
(4.148a)
y1
= a sin θ sin φ,
y2 = b sin θ sin φ,
(4.148b)
z1
= a cos θ,
z2 = −b cos θ.
The component of velocity of the particles are dθ dφ dx1 = a cos θ cos φ − sin θ sin φ , v1x = dt dt dt dy1 dθ dφ v1y = = a cos θ sin φ + sin θ cos φ , dt dt dt dz1 dθ v1z = = −a sin θ . dt dt
(4.148c)
(4.149a) (4.149b) (4.149c)
Exactly Solvable Systems I: Bound States 123 Then
2 2 2 (4.150) v12 = v1x + v1y + v1z = r12 θ˙2 + sin2 θ φ˙ 2 . Similarly v22 = r22 θ˙2 + sin2 θ φ˙ 2 . Then the total kinetic energy of the two particles is written as 1 K.E. = T1 + T2 = m1 r12 + m2 r22 θ˙2 + sin2 θφ˙ 2 2 1 ˙2 = (4.151) I θ + sin2 θφ˙ 2 . 2 For a particle of mass I moving on the surface of a sphere of radius r the kinetic energy is given by 1 1 K.E. = I x˙ 2 + y˙ 2 + z˙ 2 = Ir2 θ˙2 + sin2 θ φ˙ 2 . (4.152) 2 2 Equations (4.151) and (4.152) are identical if r = 1. Therefore, the kinetic energy given by Eq. (4.151) is identical to the kinetic energy of a particle of mass I moving on the surface of a sphere of radius unity. For a particle of mass I moving with a velocity v 1 2 ~2 1 2 Iv = p = − ∇2 . 2 2I 2I 2 The Laplacian ∇ in spherical polar coordinates is 1 ∂ 1 ∂ ∂ 1 ∂2 2 2 ∂ ∇ = 2 r + 2 sin θ + 2 2 . r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 K.E. =
For the rigid rotator r is constant so 1 ∂ ∂ 1 ∂2 ∇2 = sin θ + . sin θ ∂θ ∂θ sin2 θ ∂φ2
(4.153)
(4.154)
(4.155)
The Hamiltonian is
1 ∂ ∂ 1 ∂2 ~2 sin θ + . H =− 2I sin θ ∂θ ∂θ sin2 θ ∂φ2
(4.156)
The orientation of the rigid rotator depends only on θ and φ; therefore, its ψ depends only on θ and φ and is independent of r. The Schrödinger equation Hψ = Eψ takes the form ~2 1 ∂ ∂ 1 ∂2 − sin θ + ψ(θ, φ) = Eψ(θ, φ) (4.157) 2I sin θ ∂θ ∂θ sin2 θ ∂φ2 or ∂ψ 1 ∂2ψ 2I 1 ∂ sin θ + + 2 Eψ = 0 . (4.158) sin θ ∂θ ∂θ ~ sin2 θ ∂φ2 The operator in the left-side of Eq. (4.157) is identified as L2 /2I where L2 is the square of the orbital angular momentum (see chapter 11). Equation (4.157) implies that the eigenvalues E of the rigid rotator are (1/2I) times the eigenvalues of the operator L2 .
124 Quantum Mechanics I: The Fundamentals
4.11.2
Eigenfunctions and Eigenvalues
We show that the solution of Eq. (4.158) is simply the spherical harmonics. We obtain the solution by variable separable method. Assume the solution of the form ψ(θ, φ) = Θ(θ)Φ(φ) . (4.159) Substituting Eq. (4.159) in Eq. (4.158) and multiplying by sin2 θ/(ΘΦ) we get dΘ 2IE 1 d2 Φ sin θ d sin θ + 2 sin2 θ = − . (4.160) Θ dθ dθ ~ Φ dφ2 Since θ and φ are two independent variables the above equation is true only if both sides are equal to the same constant. We choose the constant as m2 . Then 1 d2 Φ2 − = m2 . (4.161) Φ dφ2 The solution of it is Φm = Ae±imφ .
(4.162) R 2π
A is determined from the normalization condition 0 Φ∗ Φ dφ = 1. A is ob√ tained as 1/ 2π . Equation (4.160) is now written as 1 d sin2 Θ dΘ 2IE m2 − − − Θ=0. (4.163) sin θ dθ sin θ dθ ~2 sin2 θ Substituting cos θ = x,
2IE/~2 = l(l + 1)
(4.164)
in Eq. (4.163) we have 1 − x2 Θ′′ − 2xΘ′ + l(l + 1) −
m2 Θ=0 1 − x2
(4.165)
which is the associated Legendre differential equation. Let us find a solution of Eq. (4.165) of the form Θ(x) = 1 − x2 Now Eq. (4.165) becomes 1 − x2
d2 X dx2
− 2(m + 1)x
m/2
X(x) .
(4.166)
dX + [l(l + 1) − m(m + 1)] X = 0 . dx
(4.167)
Differentiating the Legendre differential equation 1 − x2
d2 Pl dPl − 2x + l(l + 1)Pl = 0 2 dx dx
(4.168)
Exactly Solvable Systems I: Bound States 125 m times we get 1 − x2
d2 dx2
dm Pl dxm
− 2(m + 1)x
+ [l(l + 1) − m(m + 1)]
d dx
dm Pl dxm
dm Pl =0. dxm
(4.169)
Comparing the Eq. (4.169) with Eq. (4.167) we write X(x) =
dm Pl (x) dxm
(4.170) (m)
provided m assumes only positive integers. Then the solution Θl (m)
Θl
(x) = 1 − x2
|m|
|m|/2 d|m| |m| Pl (x) = Pl . dx|m|
is (4.171)
Pl are called associated Legendre polynomials. We note that since Eq. (4.165) does not depend on sign of m, because Pl (x) are polynomials of degree l, we have d|m| Pl /dx|m| = 0 if |m| > l. Therefore, |m| ≤ l. The |m| orthogonality relation for the associated Legendre polynomials Θl is Z
1
|m|
−1
|m|
Θl Θl′ dx =
(l − |m|)! 2 δll′ (l + |m|)! 2l + 1
(4.172)
|m|
which gives the normalization factor for Θl . Now |m|
Θ l Φm =
(2l + 1)(l − |m|)! 4π(l + |m|)!
1/2
|m|
Pl
(cos θ) eimφ .
(4.173)
The right-side of Eq. (4.173) are the so-called spherical harmonics and are the eigenfunctions of the rigid rotator. What are the allowed energy eigenvalues of the system? From Eq. (4.164) ~2 l(l + 1) , l = 0, 1, · · · . (4.174) 2I The energy levels are thus discrete. The separation between two successive energy levels are given by El =
△El
= El+1 − El =
~2 (l + 1) . I
(4.175)
The distance between successive energy levels increases with an increase in l. For l = 0 only one value of m is possible: m = 0. For other values of l there are (2l + 1) allowed values of m: −l, −l + 1, · · · , 0, · · · , l − 1, l. Therefore, each energy level is (2l + 1)-fold degenerate. It can be shown that the allowed transitions between rotational states obey △l = ±1 and △m = 0, ±1.
126 Quantum Mechanics I: The Fundamentals
Solved Problem 4: A rigid body freely rotates in the x − y plane. At t = 0 the wave function is assumed as ψ(0) = C sin2 φ where φ is the angle between the x-axis and 1 the rotator axis. Express ψ(0) in terms of the eigenfunctions Φm = √ eimφ , 2π m = 0, ±1, ±2, · · · . In terms of Φm we can write an arbitrary wave function as X ψ(0) = Cm Φm m
= C0 Φ0 + C1 Φ1 + C−1 Φ−1 + C2 Φ2 + C−2 Φ−2 + · · · . (4.176)
We can rewrite the given ψ(0) as ψ(0)
= C sin2 φ C = (1 − cos 2φ) 2 C i2φ C − e + e−i2φ = 2 4 √ C C C Φ0 − Φ2 − Φ−2 . 2π = 2 4 4
(4.177)
Comparison of the above two equations for ψ(0) gives √ √ p C π C π C0 = C π/2, C1 = 0, C−1 = 0, C2 = − √ , C−2 = − √ (4.178) 2 2 2 2 and other Ci ’s are zero.
4.12 CONCLUDING REMARKS Apart from the potentials considered in this chapter many other physically interesting potential problems have also been solved exactly. Solving some of them are given as exercises in the problem section. For example, a particle in a symmetric linear potential and the case of a constant acceleration in onedimension with V (x) = F x, F > 0 are solvable exactly and the solutions of the corresponding Schrödinger equation are expressed in terms of Airy function [1]. Another interesting potential is given by V (x) = ∞ for x < 0 and mgx for x > 0 and is equivalent to a bouncing ball. The solutions to this problem are the odd states of the linear potential [1,26]. Particle in a box with a deltafunction potential [27] and in the presence of a static electric field [28] are solved exactly. Ladder operators for Pöschl–Teller potentials [29] have been constructed. The stationary one-dimensional Schrödinger equation with a real polynomial
Exactly Solvable Systems I: Bound States 127 potential V (q) of degree > 2 on the real axis is studied [30,31]. There exists a class of potentials called quasi-exactly solvable [32-34] which depends on a parameter J. For these systems it is possible to find J number of eigenstates. A method P of constructing analytical solutions of the Schrödinger equation for V (x) = m Vm f m , where Vm are constants and f (x) is a function satisfying certain conditions are proposed [35]. The dynamical group structure of the one-dimensional quantum system with the infinitely deep square-well potential and its creation and annihilation operators are studied [36]. The influence of distant boundaries on quantum mechanical energy levels is also investigated. For a simple and explicit formula for the ground state energy change with application to LHO and hydrogen atom the readers may refer the ref.[37]. There exists some potentials tending to zero at infinity and yet possessing eigenfunctions with positive energy. For details about such potentials and a method to construct them see for example ref.[1]. There has been a considerable interest in recent years in obtaining exact and quasi-exact solutions of the position-dependent mass Schrödinger equation for various potentials and mass functions by using various methods [38]. A Green’s function procedure to construct ground state and excited state solutions has been illustrated in refs.[39,40]. A variational method has been developed for obtaining energy eigenvalues of N -dimensional nonpolynomial potential systems [41]. The connection between the quantized energy eigenvalue spectrum of a bound state and the classical motions of the corresponding classical system have become increasingly interesting as the ability to experimentally probe the quantum-classical interface has dramatically improved. Methods such as periodic orbit theory [42,43] are very useful to explore this. Can degenerate bound states occur in one-dimensional quantum systems? It has been pointed out that bound states with degeneracy in energy but differing in parity can occur even if the potential is nonsingular in an finite domain [44].
4.13 BIBLIOGRAPHY [1] R.W. Robinett, Am. J. Phys. 63:823, 1995. [2] I.D. Johnston, Am. J. Phys. 64:245, 1996. [3] I.N. Sneddon, Special Functions of Mathematical Physics and Chemistry. Longman, London, 1980. [4] M. Abramowitz and I.A. Stegun, Hand Book of Mathematical Functions with Formulas, Graphs and Mathematical Tables. Dover, New York, 1966. [5] G.B. Arfken and H.J. Weber, Mathematical Methods for Physicists. Academic Press, San Diego, 2005. [6] L.S. Salter, Am. J. Phys. 58:961, 1990.
128 Quantum Mechanics I: The Fundamentals [7] S. Connell, www.src.wits.ac.za/pages/teaching/Connell/phys284/2005. [8] D.L. Aronstein and C.R. Stroud Jr., Phys. Rev. A 55:4526, 1997. [9] S.H. Dong and Z.Q. Ma, Am. J. Phys. 70:520, 2002. [10] S. Flügge, Practical Quantum Mechanics-I., Springer, Berlin, 1971. [11] G. Pöschl and E. Teller, Z. Phys. 83:143, 1933. [12] J.P. Antoine, J.P. Gazeau, P. Monceau, J.R. Klauder and K.A. Penson, J. Math. Phys. 42:2349, 2001. [13] M.A. Doncheski and R.W. Robinett, Ann. Phys. 308:578, 2003. [14] G.L. Baker, J.A. Blackburn and H.J.T. Smith, Am. J. Phys. 70:525 , 2002. [15] T. Davis, Introduction to Nonlinear Differential and Integral Equations. Dover, New York, 1962. [16] J.C.G. Vega, R.M.R. Dagnino, M.A.M. Nava and S.C. Cerda, Am. J. Phys. 71:233, 2003. [17] A. Lindner and H. Freese, J. Phys. A 27:5565, 1994. [18] D. Frenkel and R. Portugal, J. Phys. A: Math. Gen. 34:3541, 2001. [19] F.A. Alhargan, ACM Trans. Math. Software 26:390, 2001. [20] K.R. Brownstein, Am. J. Phys. 68:160, 2000. [21] W.F. Buell and B.A. Shadwick, Am. J. Phys. 63:256, 1995. [22] C.A. Kocher, Am. J. Phys. 45:71, 1977. [23] M.L. Goldberger and B. Simon, Ann. Phys. 108:69, 1977. [24] F.A.B. Coutinho, C.P. Malta and J.F. Perez, Phys. Lett. A 97:242, 1983. [25] B. Sahu, M.Z. Rahmankhan, C.S. Shastry, B. Dey and S.C. Phatak, Am. J. Phys. 57:886, 1989. [26] I. Gea-Banaclocin, Am. J. Phys. 67:776, 1999. [27] I.R. Lapidus, Am. J. Phys. 55:172, 1987. [28] D. Nguyen and T. Odagaki, Am. J. Phys. 55:466, 1987. [29] S.H. Dong and R. Lemus, Int. J. Quan. Chem. 86:265, 2002. [30] A. Voros, J. Phys. A 32:5993, 1999.
Exactly Solvable Systems I: Bound States 129 [31] P. Dorey and R. Tateo, J. Phys. A 32:L419, 1999 and references therein. [32] C.M. Bender, G.V. Dunne, J. Math. Phys. 37:6, 1996. [33] A. Turbiner, Sov. Phys. JETP 67:230, 1988. [34] A.G. Ushveridze, Quasi-exactly Solvable Models in Quantum Mechanics. Institute of Physics, Briston, 1993. [35] L. Skala, J. Cizek, J. Dvorak and V. Spirko, Phys. Rev. A 53:2009 , 1996. [36] S.H. Dong and Z.Q. Ma, Am. J. Phys. 70:520, 2002. [37] G. Barton, A.J. Bray and A.J. Mckane, Am. J. Phys. 58:751, 1990. [38] B. Midya and B. Roy, J. Phys. A: Math. Gen. 42:285301, 2009 and references therein. [39] R.G. Winter, Am. J. Phys. 45:569, 1977. [40] H.T. Williams, Am. J. Phys. 70:532, 2002. [41] N. Saad, R.L. Hall and H. Ciftci, J. Phys. A: Math. Gen. 39:7745, 2006. [42] M. Gutzwiller, The Interplay Between Classical and Quantum Mechanics., American Assoc. Phys. Teachers College Park, Maryland, 2001. [43] M. Brack and R.K. Bhaduri, Semiclassical Physics. Addison–Wesley, Reading, Massachusetts, 1997. [44] S. Kar and R.R. Parwani, Europhys. Lett. 80:30004, 2007.
4.14 EXERCISES 4.1 Why does not a quantum system in its stationary energy states radiate energy? 4.2 Find the wave function and the energy eigenvalue for a free particle with potential V = a. 4.3 Using the representation of the δ-function δ(x) = lima→∞ (sin ax)/(πx) R∞ ′ show that −∞ ei(k −k)x dx = 2πδ(k ′ − k).
4.4 Write the time-independent Schrödinger equation for a free particle in three-dimensions. Solve it by the variable separable method. 1/4
2
4.5 If ψ = (1/~π) e−x /2~ is an eigenfunction of the one-dimensional linear harmonic oscillator with potential V (x) = x2 /2 then obtain the expression for corresponding energy. Assume the mass of the particle as unity.
130 Quantum Mechanics I: The Fundamentals 4.6 For a linear harmonic oscillator ground state eigenfunction compute hxi and hpi. 4.7 For the linear harmonic oscillator with Hamiltonian H = p2x /(2m) + mω 2 x2 /2 calculate hxi, hx2 i, hpx i, hp2x i and (△x)(△px ) in its nth state. 4.8 Find the expectation value of energy E when the state of the linear harmonic√oscillator is described by the wave function ψ = [φ0 (x, t) + φ1 (x, t)]/ 2 where φ0 and φ1 are the eigenfunctions of the ground state and first excited state respectively. 4.9 For a linear harmonic oscillator using the expectation values of x2 and p2 show that E ≥ ~ω/2. 4.10 Calculate the probability of finding a linear harmonic oscillator with first excited state eigenfunction in a classically allowed region. Assume R √3 2 −y2 that 0 y e dy ≈ 0.39366.
2 2 4.11 Show that the Hamiltonian H = mq˙2 /2 + mω p q /2, by† means of the −iωt † iωt + ae and [a, a ] = 1, can be transformation q = ~/2mω ae brought into the form H = ~ω aa† + 1/2 .
4.12 Show that the eigenfunctions φ1 and φ2 of linear harmonic oscillator are orthogonal. 4.13 If u(x) is such that ψ ′ + uψ = 0 then using the Schrödinger equation of linear harmonic oscillator show that u is the solution of the Riccati equation u′ − u2 − λ + g(x) = 0, λ = 2mE/~2 and g = m2 ω 2 x2 /~2 . Express ψ in terms of u. If u = mωx/~ find E and ψ of the harmonic oscillator. 4.14 Consider the one-dimensional linear harmonic oscillator in an applied electric field. The Hamiltonian of the system is He = p2x /(2m) + mω 2 x2 /2 − eEx. Obtain the eigenfunction and energy eigenvalues of the system. 4.15 Show that a linear harmonic oscillator with the same energy and same amplitude as a particle in a box has frequency that is 2/π times the frequency of the particle in the box. 4.16 Show that the eigenvalues of n ˆ = a† a of the linear harmonic oscillator are 0 and 1 if {a, a} = 0 where { } denotes anticommutator bracket. 4.17 The ground state energy and eigenfunction of a one-dimensional particle 2 of unit mass in a potential V (x) are ~/2 and φ0 = (1/~π)1/4 e−x /2~ respectively. Calculate the potential. 4.18 The eigenfunction of a linear harmonic oscillator is of the form φ = 2 N (a + x2 )e−bx /2 . Determine a, b, N and the energy of the system.
Exactly Solvable Systems I: Bound States 131 4.19 Show by induction (a† )n g(x), where a† is the raising operator of the linear harmonic oscillator, is proportional to dn 2 2 2 2 eα x /2 n e−α x /2 g(x) dx for any arbitrary function g(x). Then establish the relation 1/2 α 2 2 √ e−α x /2 Hn (αx). ψn (x) = n!2n π 4.20 The time-dependent wave √ function PN +sof a linear harmonic oscillator at t = 0 is given by ψ(0) = 1/ 2s n=N −s φn , N ≫ s ≫ 1. 1/2
(a) Show that for large n hx(t)i = [2~N/(mω)]
cos ωt.
(b) Compare the classical amplitude of the oscillation with the quantum mechanical amplitude. 4.21 Obtain the energy levels of three-dimensional linear harmonic oscillator described by the Hamiltonian H = p2 /(2m) + m(ω12 x2 + ω22 y 2 + ω32 z 2 )/2. Show that the energy levels are degenerate when ω1 = ω2 = ω3 and the degree of degeneracy is (n + 1)(n + 2)/2 where n = n1 + n2 + n3 . 4.22 An electron of mass m = 9.109 × 10−31 kg is confined within a box with perfectly rigid walls. The width of the box is 2 nm(= 2 × 10−9 m). If the electron is in the ground state, compute the energy. 4.23 Calculate the lowest energy gap for an electron of mass m = 9.109 × 10−31 kg moving in a box of length 2a. What would be the total energy of the system for the energy spacing between two adjacent levels to be 1J? 4.24 Obtain the electric dipole selection rule for allowed transitions for a particle in a box with the potential V (x) = 0 for |x| < a and ∞ otherwise. 4.25 For a particle in the box potential V (x) = 0 for |x| < a and ∞ otherwise calculate hHi. 4.26 For a particle in a box potential V (x) = 0 for |x| < a and ∞ otherwise calculate hxi, hx2 i, hpx i and hp2x i. 4.27 The normalized wave function of a particle in a box potential V (x) = 0 for |x| < a and ∞ otherwise at time t = 0 is πx h πx i 1 ψ(x, 0) = √ cos 1 + 6 sin , |x| < a 2a 2a 10a and zero otherwise. Find the probability of finding the particle in the region 0 ≤ x ≤ a. 4.28 Consider the wave function given in the previous problem.
132 Quantum Mechanics I: The Fundamentals (a) Obtain the wave function at t = t0 6= 0.
(b) Calculate the average energy of the particle at t = 0. 4.29 The wave function of a particle in an infinite square-well potential of width 2a in the interval |x| < a is given by ψ(x) = C [cos(πx/2a) + sin(πx/a)/4]. (a) Determine the coefficient C. (b) What are the possible energy eigenvalues obtained when energy is measured? 4.30 The even eigenfunctions of a particle in a box with sides at x = −a h πx i 1 , n = 1, 2, · · · . An even norand x = a are φn (x) = √ cos (2n − 1) a 2a 1/2 15 malized wave function is given by ψ(x) = a2 − x2 , |x| ≤ a 5 16a and ψ(x) = 0 for |x| > a. Expand ψ(x) on the complete basis of eigenfunction. Also compute hEi. πx 1 3πx 1 and φ3 = √ cos of 4.31 Show that the eigenfunctions φ1 = √ cos a 2a a 2a a particle in the box potential V (x) = 0 for |x| < a and ∞ otherwise, are orthogonal. 4.32 For the infinite height potential V (x) = 0, |x| < a and ∞ otherwise obtain the probability PQM (b < x < b + c) in the limit of the quantum number n → ∞. 4.33 The wave function of a particle in a box is ψ(x) = N (a2 − x2 ) for |x| < a and 0 for |x| > a. Normalize the wave function. Also calculate the probability that a measurement of energy yields the eigenvalue En . 4.34 An electron moves in a one-dimensional potential V (x) = 0 for |x| < a and ∞ otherwise. At t = 0 the wave function is assumed to be ψ(x, 0) = N (2ax − x2 ) for |x| √ < a. Find N . Express ψ(x, t) in terms of the basis eigenfunctions (1/ a) sin(nπx/2a), n = 2, 4, · · · . 4.35 An electron of mass m = 9.109×10−31kg is moving in a one-dimensional box of size 1A◦ . Determine the probability of finding it in the interval 0 to 0.5A◦ for the states with the quantum numbers n = 1 and n = 2. 4.36 An electron in a box potential V (x) = 0 for |x| < a = 0.5 nm and ∞ otherwise absorbs the incident radiation and excited to the energy level n = 2 from n = 1. Calculate the wave length of the incident radiation. 4.37 Consider a gas molecule of molecular weight 50 in a box of size 1 cm. Show that there is no quantization of energy.
Exactly Solvable Systems I: Bound States 133 4.38 The wave function of a quantum mechanical particle of mass m in a square-well potential V (x) = −V0 for 0 < x < a and otherwise 0 is given by ψ = sin(πx/a). Calculate hHi at t = 0. 4.39 A particle in a box is found to be in the ground state. Suddenly the walls of the box at x = −a and a are shifted to x = −2a and x = 2a respectively. (a) What is the probability that the particle to be found in the ground state of the expanded box? (b) What would happen if the shifting of the walls is adiabatic? 4.40 Consider the system in the previous problem. (a) Show that the probability for finding the particle in the first excited state (n = 2) of the expanded box is zero. (b) Show that the probability for the transition to the second excited state of the expanded box is 64/(25π 2). 4.41 Find the energy eigenvalues and eigenfunctions of a particle in a threedimensional box-potential. 2
4.42 Assuming the solution of the form ψ(x, t) = e−α(t)x −β(t) for the timedependent linear harmonic oscillator equation determine α(t) and β(t). 4.43 Consider the solved problem 4. Find ψ(t). 4.44 The Schrödinger equation for a rigid body that is constrained to rotate about a fixed axis with the moment of inertia I is i~∂ψ/∂t = −(~2 /2I)∂ 2 ψ/∂φ2 where ψ(φ, t) is a function of time and the angle of rotation about the axis. What boundary condition must be applied to the solution of this equation? Find the normalized energy eigenfunctions and eigenvalues. Is there any degeneracy? 4.45 Find the selection rules for electric dipole transition between the rotational levels of a rigid rotator. 4.46 An electron of mass m moves in a potential V (x) = −a/x, x > 0, a =constant, V (x) = ∞, x ≤ 0. Construct the ground state eigenfunction and calculate ground state energy. 4.47 Consider the harmonic oscillator with the potential V− (x) = −kx2 /2, k > 0. The difference between the linear harmonic oscillator potential V (x) = kx2 /2, k > 0 and the potential V− (x) is that the former has a stable equilibrium point while the latter has an unstable one. In other words, the motion of the particle is stable for V (x) while it is unstable for V− (x). Solve the classical equation of motion and the Schrödinger equation. Also obtain PCM and PQM .
134 Quantum Mechanics I: The Fundamentals 4.48 The one-dimensional linear harmonic oscillator potential in the presence of a dipole-like interaction is written as [G. Palma and V. Raff, Am. J. Phys. 71:247, 2003] V (x) = (mω 2 /2)x2 + (2mα/~2 )(1/x2 ) with α > 0. Determine the energy eigenfunctions and eigenvalues. 4.49 Solve the Schrödinger equation with √ the potential V (x) √ = F x, F > 0 and then show that PCM (x) ∝ 1/ x and PQM (x) ∝ 1/ x. 4.50 Find the unnormalized wave function of the Schrödinger equation with the potential V (x) = F |x|. Also obtain the energy eigenvalues.
CHAPTER
5
Exactly Solvable Systems II: Scattering States
5.1 INTRODUCTION In the previous chapter we focused our discussion on certain exactly solvable quantum mechanical systems possessing bound states. In the present chapter we wish to concentrate on some fascinating and interesting one-dimensional systems where the quantum states are essentially scattering states. For scattering states energy eigenvalues are continuous and energy eigenfunctions are nonnormalizable. An interesting feature of many scattering potentials is the occurrence of tunneling. It is the most striking qualitative difference between classical mechanics and quantum mechanics. In tunneling a quantum mechanical particle passes through a region where the potential energy exceeds the total energy of the particle. This phenomenon is impossible in classical mechanics. Classically, a particle is completely reflected from a barrier. Quantum tunneling has many technological applications. In the present chapter we consider the following potentials and discuss the features of scattering states of them: 1. Barrier potential 2. Square-well potential 3. Step potential 4. Locally periodic potentials 5. Reflectionless potentials – sech2 potentials
135
136 Quantum Mechanics I: The Fundamentals
V V0
R1
R2 -a
FIGURE 5.1
R3 a
x
One-dimensional barrier potential.
We will construct the reflected and transmitted waves for the above potentials subjected to appropriate boundary conditions. We obtain the expressions for reflection and transmission amplitudes and analyse their dependence on the energy of the incident particle and the parameters of the potentials. The square-well and sech2 potentials admit bound states also. We construct the bound state eigenfunctions and the energy eigenvalues for these potentials. Finally, we point out a new kind of tunneling called dynamical tunneling which refers to a motion allowed by quantum mechanics but forbidden by classical mechanics.
5.2 POTENTIAL BARRIER: TUNNEL EFFECT As a first intriguing example of a scattering system, we consider a particle in the presence of a square-barrier potential (Fig. 5.1) ( V0 , for |x| < a V (x) = (5.1) 0, for |x| > a .
5.2.1
Classical Solution
A classical particle moving towards the barrier will be completely reflected from it if E < V0 . For the particle in region R1 moving towards the barrier from a position x0 with velocity v, the position x(t) is given by ( x0 + vt, 0 < t < t1 = (|x0 | − a)/v x(t) = (5.2) −a − |v|(t − t1 ), t > t1 . Similar solution can also be written for the region R3 . The motion of the particle in the regions R1 and R3 is not confined to a finite region of space.
Exactly Solvable Systems II: Scattering States 137 It does not exhibit periodic oscillation. The motion is classically forbidden in the barrier region |x| < a. The problem thus turns out to be scattering of the particle by the potential barrier.
5.2.2
Solution of the Wave Equation for E < V0
The Schrödinger equations in the regions R1 , R2 and R3 are given by ψ1xx + k02 ψ1 2
ψ2xx − k ψ2
ψ3xx +
k02 ψ3
=
0,
=
0,
=
0,
x < −a |x| < a
x>a
(5.3a) (5.3b) (5.3c)
where k02 = 2mE/~2 , k 2 = 2m(V0 − E)/~2 . We treat the quantum cases with E < V0 and E > V0 separately. The solutions of Eqs. (5.3) are ik0 x + Be−ik0 x ψ1 = Ae ψ(x) = ψ2 = F e−kx + Gekx (5.4) ik0 x −ik0 x ψ3 = Ce + De .
The functions eik0 x and e−ik0 x in ψ1 represent waves moving towards the barrier (from x < −a) and moving away from the barrier (towards x = −∞) respectively. They are the incident and reflected waves. In the region R3 the functions eik0 x and e−ik0 x are the waves travelling away from the barrier towards x = ∞ and moving towards the barrier. The unknowns in Eqs. (5.4) are determined by applying the appropriate boundary conditions. Note that the wave functions given by Eqs. (5.4) are nonnormalizable. The boundary conditions are set by requiring that ψ’s and their first derivative are continuous at x = a and −a. The boundary conditions are at x = −a : at x = a :
ψ1 = ψ2 and ψ1x = ψ2x , ψ2 = ψ3 and ψ2x = ψ3x .
(5.5a) (5.5b)
At x = −a the boundary conditions lead to a set of equations which are given in matrix form as ka −ka −ik a ik a e e 0 0 e e A F ik ik = . (5.6) B G e−ik0 a eik0 a eka − e−ka k0 k0 F From the boundary conditions at x = a we obtain an expression for . G Substituting this in Eq. (5.6) we get C A α11 α12 , (5.7a) = B α21 α22 D
138 Quantum Mechanics I: The Fundamentals where α11
=
α21
=
ǫ =
iǫ iη e2ik0 a cosh 2ka + sinh 2ka , α12 = sinh 2ka , (5.7b) 2 2 iǫ iη − sinh 2ka , α22 = e−2ik0 a cosh 2ka − sinh 2ka , (5.7c) 2 2 2 2 2 2 k − k0 k + k0 , η= . (5.7d) kk0 kk0
For a particle incident from left D = 0 and hence A = Cα11 and B = −Cα21 . Setting the amplitude A of the wave function of the incident particle as 1 gives C = 1/α11 . Since Ceik0 x represents transmitted wave function, the probability for the particle to be transmitted, called transmission amplitude, is given by T = CC ∗ . It is worked out as −1 ǫ2 sinh2 2ka . (5.8a) T = 1+ 1+ 4 The reflection amplitude R is R = BB ∗ =
1 2 η sinh2 2ka T . 4
(5.8b)
Since the total probability must be 1 we require that T + R = 1. This can be easily verified from Eqs. (5.8). |T |2 is called tunneling probability.
5.2.3
Solution of the Wave Equation for E > V0
For E > V0 we obtain ǫ η α11 = cos 2ka − sin 2ka e2ik0 a , α12 = − sin 2ka , (5.9a) 2 2 −k 2 + k02 −k 2 − k02 1 ǫ = , η= , C= , B = −Cα12 , (5.9b) ikk0 ikk0 α11 −1 ǫ2 T = 1− 1+ sin2 2ka , R=1−T . (5.9c) 4 For a particle with energy E > V0 , Eq. (5.9c) predicts complete transmission when sin2 2ka = 0. In this case, Tmax = 1 and 2ka = nπ, n = 1, 2, · · · . Such resonances in transmission correspond to the situation in which the scattered waves originating from x = −a and x = a interfere and exactly cancel any reflection from the potential barrier.
5.2.4
Quantum Tunneling
A classical particle would be reflected from the barrier for E < V0 . In contrast, in quantum theory there is a finite transmission probability. That is, if a
Exactly Solvable Systems II: Scattering States 139 microscopic particle is incident on the barrier with energy relatively lower than the height of the potential barrier then it will not necessarily be reflected by the barrier. But there is always a finite nonzero probability for the particle to cross the barrier and continue its motion in the classically forbidden region R3 . This phenomenon of crossing the barrier is called tunnel effect . (When a metal surface is subjected to a large electric field the potential at the surface changes and some of the free electrons in the metal tunnel through the resulting barrier and are emitted. This is known as field emission). The tunneling effect is one of the phenomena for which the difference between classical and quantum physics is most striking. It describes the propagation of a particle in a classically forbidden domain. (What are the kinetic energy and the velocity of the particle in this zone? ) Without tunneling, tunnel diode would not exist. The probability of tunneling depends in general on (i) the energy E of the particle, (ii) the height V0 of the barrier and (iii) the width 2a of the barrier. Figure 5.2 illustrates the Re ψ for a few values of E where V0 = 1 and a = 6. The behaviour of Re ψ inside the barrier is not always an exponential decay. For E = 0.6, Re ψ exhibits exponential decay with x; for E = 0.9 it displays a power-law type variation; for E = 1.0001 (that is ≈ V0 ) it varies linearly; for E = 1.1 and 2 it shows oscillation. Figure 5.3 shows T versus E/V0 for two fixed values of V0 a2 . For E < V0 (E/V0 < 1), T increases with increase in E. For E > V0 , an interesting feature is that T (as well as R) varies in an oscillatory fashion. It shows a series of resonances peaked at specific E/V0 values where T = 1, the maximum allowed value. The resonances come from the multiple scattering effects within the barrier. (What will happen if we consider a double-barrier or a finite number of square-barrier? For some details see sec. 5.5). For 2ka ≫ 1 Eq. (5.8a) becomes 16 −4ka T ≈ e . (5.10) 4 + ǫ2 The exponential factor comes from the exponential decay of the wave function from the incident side to the transmitted side a distance 2a away. The factors in front of the exponential term come from the matching conditions. T decreases exponentially with a and quantum tunneling is extremely sensitive to small changes in a. This dependence of T on a has been experimentally confirmed in the phenomenon of vacuum tunneling [1]. We wish to know whether the oscillatory behaviour of T is peculiar to the rectangular shape or characteristic of other potential barriers also. Oscillatory behaviour of T and R has been observed in the truncated Gaussian potential [2] ( 2 2 V0 e−β x , |x| < a V (x) = (5.11) 0, |x| > a
140 Quantum Mechanics I: The Fundamentals
Re ψ
3
E = 0.6
2 1 0 -1 -20
-10
x
0
Re ψ
3
10
E = 0.9
2 1 0 -1 -20
-10
x
0
Re ψ
3
10
E = 1.0001
2 1 0 -1 -20
-10
x
0
Re ψ
3
10
E = 1.1
2 1 0 -1 -20
-10
x
0
Re ψ
3
10
E=2
2 1 0 -1 -20
-10
x
0
10
Re ψ is plotted for 5 values of E for the barrier potential (5.1) with V0 = 1 and a = 6. For illustration Re ψ is shifted as Re ψ+E. FIGURE 5.2
Exactly Solvable Systems II: Scattering States 141
1 T
0.5 0
0
1
2
3 E/V0
4
5
Variation of transmission coefficient T as a function of E/V0 for V0 a = 1 (dashed curve) and V0 a2 = 8 (continuous curve) for the barrier potential (5.1). For E/V0 < 1, Eq. (5.8a) is used to compute T . For E/V0 > 1, T is calculated using Eq. (5.9c). FIGURE 5.3 2
and truncated parabolic potential 2 V 1 − x , 0 b2 V (x) = 0,
|x| < a
(5.12)
|x| > a,
where V0 , β and a are constant parameters. When the width of the potential barrier is increased the magnitude of oscillations in T for E < V0 is found to decrease. The results of the potentials given by Eqs. (5.1), (5.11) and (5.12) suggest that the discontinuous change in the potential and its slope essentially determine the relatively large variations in T that occur for E > V0 . We expect from these results that the smaller the abrupt change in a potential barrier, the smaller will be the variations in T for E > V0 .
5.2.5
Physical Systems Modelled by the Square-Barrier Potential
In 1838 Heinrich Friedrich Emil Lenz observed the freezing of water into ice on a bismuth-antimony junction when an electric current passed through the junction in one direction. When the direction of the current was reversed the melting of the ice into water was found. Consequently, a thermo-electric power generator works as a refrigerator. The fundamental quantum process of an electron moving across a potential barrier plays the vital role in thermionic power generation and refrigeration. Consider two parallel metal electrodes with a vacuum gap between them acting as a barrier. The charge accumulation only affects the magnitude of the thermionic current. Because of the nature of the quantum transmission probability and the temperature behaviour of the Fermi distribution under
142 Quantum Mechanics I: The Fundamentals certain conditions the thermionic current reverses its direction. This effect occurs in a system of two identical metallic conducting wires separated by a layer of insulating material [3]. There is no bias applied across the wires. The wires are connected to two thermal reservoirs kept at a higher temperature Th and a lower temperature Tc . The difference in chemical potentials µh and µc is due to the different temperatures. Such a system can be realized with the semiconductor hetero-structures fabrication technology. The barrier potential is also useful to model metal-oxide-semiconductor (MOS) and field-effect-transistors (FETs). A barrier can arise if a metal is heated to give rise to thermionic emission. Lothar Wolfgang Nordheim and Ralph Howard Fowler (1920s) realized that the barrier penetration could be used to understand the emission of electrons from a metal in a strong electric field.
5.2.6
Importance of Quantum Tunneling
There are many phenomena in physics where tunneling plays a central role. Five Nobel Prizes in physics were awarded for research concerned with tunneling in semiconductor and for the invention of scanning tunneling microscopy. Tunneling is crucial for nucleo-synthesis in stars and in the evolution of early universe. Molecular spectra, electron emission from metals, cold emission of electrons from metals, current-voltage characteristics of transistors, magnetization in certain metals, microwave absorption of ammonia and alpha decay are explained through the tunneling phenomenon. Some of the examples and applications of tunneling are given below: 1. In tunnel diodes, the tunneling effect is a source for a region in the current-voltage characteristic where the differential resistance is negative. α-decay occurs due to the tunnel effect. 2. In 1973, Brain David Josephson was awarded the Nobel Prize together with Leo Esaki and Ivar Giaever for an invention of a junction called Josephson junction. In a Josephson junction transition regions between two superconducting materials are separated by a thin insulating layer. In these devices pairs of electrons known as cooper pairs tunnel from one superconducting region to another through thin insulating layer. This effect gives rise to currents flowing without an applied voltage. 3. A scanning tunneling microscope (STM) working by quantum tunneling between the surface of an electrical conductor and a fine conducting probe was invented by Gerd Binning and Heinrich Rohrer. They received the Nobel Prize in 1986 for this invention. In a STM a thin needle moves across the surface of a sample. If the needle and the sample are close to each other then electrons can tunnel from the needle to the sample. In this way the picture of the surface of the sample is obtained. 4. Friedrich Hund (1927) was the first to use the quantum mechanical bar-
Exactly Solvable Systems II: Scattering States 143 rier penetration in the theory of molecular spectra. Internal structure of complex molecules and the rotationally induced dissociation of a diatomic molecule from an excited state can be studied by the tunneling phenomenon. 5. A series of nuclear reactions in the sun produces light energy. In this scenario, two protons come very close and form a bound consisting of a proton, a neutron, a positron and a neutrino. First two protons come closer. Since they have same electric charge they repel each other. According to classical mechanics this reaction would take place very slowly and no light would come out from the sun. But quantum mechanics predicts that one proton tunnels through the barrier of repulsion separating the two protons and allows the reaction to continue [4].
Solved Problem 1: A beam of electrons incident on the square-barrier potential, Eq. (5.1), from left. Find the fraction of electrons reflected and transmitted if the energy of √ the incident electrons is V0 /2 and 2a mV0 /~ = 3/2. From Eq. (5.8a) the transmission amplitude T is given by 1 ǫ2 =1+ 1+ sinh2 2ka , T 4
(5.13a)
where ǫ=
k 2 − k02 V0 − 2E . = p kk0 E(V0 − E)
Substituting E = V0 /2 we get ǫ = 0. Further s 2a V0 3 2ka = 2m V0 − = . ~ 2 2 Writing sinhx = (ex − e−x ) /2 we get 3/2 2 3 e − e−3/2 sinh2 2ka = sinh2 = = 4.534 . 2 2
(5.13b)
(5.14)
(5.15)
Then 1/T = 1 + 4.534 = 5.534. Therefore, T = 0.181 and R = 0.819.
5.3 FINITE SQUARE-WELL POTENTIAL Consider a potential which is zero within a certain interval and V0 elsewhere. Such a potential is called a square-well potential and is represented as ( 0, for |x| < a V (x) = (5.16) V0 , for |x| > a.
144 Quantum Mechanics I: The Fundamentals
V
R2
R3
V0
R1 -a FIGURE 5.4
0
x a
Finite square-well potential.
The potential is schematically represented in Fig. 5.4. An interesting remarkable feature of this potential is that it admits both bound and scattering states. Further, the bound states of square-well potential are finite in number .
5.3.1
Physical Systems
Certain materials crystallize in an almost similar manner and possess exactly the same lattice constants. Therefore, they may be grown together as a single crystal. A junction in a single crystal between two materials whose lattices match but are otherwise different is known as heterojunction. In a double heterojunctions, two junctions are used in order to sandwich a layer of a narrow band gap material between wide band gap materials. In this way, two heterostructure interfaces form a quantum well. Such double-heterostructures are very common for light-emitting diodes and lasers used for optical communication. This potential is used to model emission of electrons in metals and for short-range forces.
5.3.2
Solution of the Wave Equation for E < V0
For a classical particle with kinetic energy less than the potential energy, its motion is bounded within the region R1 . Its dynamics are similar to that of infinite height potential discussed in the sec. 4.6. The particle in R1 cannot enter into the region R2 or R3 . When the kinetic energy is greater than the potential energy the particle is able to overcome the potential barrier and enter into the region R2 or R3 . Then the motion is not bounded. Now we focus on quantum dynamics of the particle. The Schrödinger equation for |x| < a and |x| > a are given by ψxx + k 2 ψ ψxx − α2 ψ
= 0, = 0,
where k 2 = 2mE/~2 , α2 = 2m(V0 − E)/~2 .
|x| < a |x| > a
(5.17a) (5.17b)
Exactly Solvable Systems II: Scattering States 145 Evidently, Eq. (5.17a) is a free particle equation. α2 is positive for 0 < E < V0 and negative for E > V0 . The nature of the solution of Eq. (5.17b) is different for α2 < 0 and α2 > 0. Therefore, these two cases must be studied separately.
5.3.2.1 Bound States: 0 < E < V0 , α2 > 0 The solution of Eq. (5.17a) (for the region R1 ) is ψ1 = A sin kx + B cos kx ,
|x| < a .
(5.18a)
The solution of Eq. (5.17b) for the regions R2 and R3 is ψ = Ce−αx +Deαx . In the region R2 when x < −a, as x → −∞ the first term in the above solution tends to ∞. To avoid this, set C = 0 so that ψ2 = Deαx ,
(5.18b)
x < −a .
In the region R3 as x → ∞ the second term in ψ tends to ∞. Therefore, we choose D = 0. Now, the solution is ψ3 = Ce−αx ,
(5.18c)
x>a.
The solutions ψ1 , ψ2 and ψ3 must satisfy certain appropriate boundary conditions.
5.3.2.2 Boundary Conditions The boundary conditions arise due to the continuity of the solutions and their derivatives at x = ±a. Matching the solutions (5.18) and their derivatives at x = ±a give a system of four equations for the four unknowns A, B, C and D. The compatibility condition for these equations leads to an equation for the energy En . The boundary conditions are ψ1 = ψ3 , ψ1x = ψ3x ,
(5.19a)
at x = −a, ψ1 = ψ2 , ψ1x = ψ2x .
(5.19b)
at x = a,
These conditions lead to the following set of equations: A sin ka + B cos ka = Ak cos ka − Bk sin ka =
−A sin ka + B cos ka = −Ak cos ka − Bk sin ka =
Ce−αa , −αCe−αa ,
De−αa , −αDe−αa .
(5.20a) (5.20b) (5.20c) (5.20d)
Adding Eq. (5.20a) with (5.20c) and (5.20b) with (5.20d) we get 2B cos ka =
(C + D)e−αa ,
2Bk sin ka =
−αa
α(C + D)e
(5.21a) .
(5.21b)
146 Quantum Mechanics I: The Fundamentals Eliminating (C + D)e−αa in Eq. (5.21a) using Eq. (5.21b) we have k 2B 1 − tan ka = 0 . α
(5.22)
Equation (5.22) is satisfied if the following two conditions are satisfied: (i) B = 0, k is arbitrary. (ii) B 6= 0, tan ka = α/k.
Next, subtracting Eq. (5.20c) from Eq. (5.20a) and Eq. (5.20d) from Eq. (5.20b) we obtain 2A sin ka = 2Ak cos ka =
(C − D)e−αa ,
−α(C − D)e
−αa
(5.23a) .
Elimination of (C − D)e−αa in Eq. (5.23a) gives k 2A + tan ka = 0 α
(5.23b)
(5.24)
which implies either (iii) A = 0, k is arbitrary or (iv) A 6= 0, tan ka = −k/α.
Because k 2 = 2mE/~2 , arbitrariness of k means E is arbitrary. From the results (i)-(iv) we note that E is arbitrary only if both A and B are zero. In this case the wave function ψ1 is zero which we do not want. For both A 6= 0, B 6= 0 the condition is k k 2 tan ka = − = −1 . (5.25) α α For real ka this is not possible and hence the conditions (iii) and (iv) are not satisfied simultaneously. The news is, A 6= 0, B 6= 0 is not possible. Therefore, either A = 0 or B = 0. Now, we have the following two cases: Case 1: A = 0, B 6= 0, tan ka = k/α.
Case 2: A 6= 0, B = 0, tan ka = −k/α.
Since k is not arbitrary, E is not arbitrary and it can take only specific sets of values. This leads to discreteness of energy levels. For the cases 1 and 2 we C = D = Beαa cos ka ,
C = −D = Aeαa sin ka .
(5.26)
Further, from Eq. (5.17c) we obtain k 2 + α2 =
2mV0 = constant . ~2
(5.27)
Exactly Solvable Systems II: Scattering States 147
5.3.2.3 Calculation of En and ψn Case 1: For the case 1, since tan(k/a) > 0, we should consider only the values of ka in the interval 2r
π π ≤ ka ≤ (2r + 1) , 2 2
r = 0, 1, · · · .
(5.28a)
Then Eq. (5.27) gives 2mV0 a2 = k 2 a2 sec2 ka . ~2
(5.28b)
Similarly, for the case 2 we need to consider (2r − 1)
π π ≤ ka ≤ 2r , 2 2
r = 1, 2, · · ·
(5.29a)
and consequently 2mV0 a2 = k 2 a2 cosec2 ka . ~2
(5.29b)
Equations (5.28) and (5.29) are satisfied only for certain discrete values of k. Denote these values as kn . Then the energy values from k 2 = 2mE/~2 are En =
~2 2 k , 2m n
n = 0, 1, 2, · · · .
(5.30)
The energy levels are discrete. The energy levels with n = 0, 2, 4, · · · correspond to the case 1. The eigenfunctions are αn a cos kn a eαn x , x < −a Be ψ = B cos kn x , (5.31) −a < x < a αn a Be cos kn a e−αn x , x>a. Case 2:
The energy levels with n = 1, 3, · · · are for the case 2. In this case αn a sin kn a eαn x , x < −a −Ae ψ = A sin kn x , (5.32) −a < x < a αn a Ae sin kn a e−αn x , x>a.
The constants A and B are to be determined by the normalization condition. Next, we determine the eigenfunctions for specific values of V0 and a. Let us fix V0 = 35, a = 1 and ~2 = m = 1. Using the Newton–Raphson method the roots of Eq. (5.28b) are found to be 1.4024, 1.7859, 4.1881, 5.4166 and
148 Quantum Mechanics I: The Fundamentals
n=0
-1
0 -1
-2
-1
0
x
1
2
n=4
1
ψ
0
n=2
1
ψ
ψ
1
0 -1
-2
-1
0
x
1
2
-2
-1
0
x
1
2
Three eigenfunctions for the case 1 of the finite square-well potential with V0 = 35, a = 1 and ~2 = m = 1. The eigenfunctions are unnormalized (B = 1 in Eq. (5.31)). FIGURE 5.5
n=1
-1
0 -1
-4
-2
0
x
2
4
n=5
1
ψ
0
n=3
1
ψ
ψ
1
0 -1
-4
-2
0
x
2
4
-4
-2
0
x
2
4
Three eigenfunctions for the case 2 of the finite square-well potential with V0 = 35, a = 1 and ~2 = m = 1. The eigenfunctions are unnormalized (A = 1 in Eq. (5.32)). FIGURE 5.6
6.8870. Among these 5 values, the values of k satisfying the condition (5.28a) are k0 = 1.4024, k2 = 4.1881 and k4 = 6.8870. The notable result is that there are only three discrete energy levels for the case 1. The corresponding unnormalized eigenfunctions are depicted in Fig. 5.5. For the case 2 we obtain k1 = 2.8003, k3 = 5.5568 and k5 = 8.1048. In this case also there are only three discrete energy levels. The unnormalized eigenfunctions of the discrete energy levels are shown in Fig. 5.6. The energy difference between V0 and the top-most bound state energy is called the work function φ0 . An electron bound in a metal would be dislodged only if the incident photon energy exceeds φ0 . The φ0 of the metals Ni, Ta, Cd and Ag are 4.2, 4.5, 4.6 and 4.8eV respectively. In the regions R2 and R3 the total energy E is less than V0 . Classically, a particle in R1 cannot enter the regions R2 and R3 . When it reaches x = ±a, it turns back and hence the motion is confined within |x| < a. In contrast to this, in quantum mechanics |ψ2 |2 and |ψ3 |2 are nonzero in R2 and R3 and hence the probability for the presence of particle in the classically forbidden regions is finite and nonzero. Let us summarize the significant features of the square-well potential. 1. The potential admits only a few discrete energy eigenvalues and nor-
Exactly Solvable Systems II: Scattering States 149 malizable eigenfunctions. All the allowed discrete energy values are less than the depth of the well. 2. The number of discrete states increases with increase in the depth of the well. However, since α increases with increase in V0 , the extent of the tail of the eigenfunctions in the classically forbidden region decreases (refer to Eq. (5.32)). 3. From Eq. (5.32) it is clear that the eigenfunctions are oscillatory in the classically allowed region (−a < x < a) and exponentially decaying in the classically forbidden regions (|x| > a). 4. From Figs. 5.5 and 5.6 we note that ψn has n nodes. 5. As a consequence of the symmetry in V , V (x) = V (−x), the eigenfunctions have alternating even (Eq. (5.31)) and odd (Eq. (5.32)) parity.
Solved Problem 2: Construct the solution of the wave equation of the square-well potential system for E > V0 . When E > V0 , writing α = i|α| Eq. (5.17b) becomes ψxx + |α|2 ψ = 0. Its solution is ψ2,3 = C2,3 sin |α|x + D2,3 cos |α|x . (5.33) Solutions in the regions R1 , R2 and R3 are now in sinusoidal form. Applying the boundary conditions we get a set of four equations. But the number of unknowns are six. We have fewer numbers of equations than the unknowns. However, we can fix the values of two unknowns and then use the four equations to solve for the other four unknowns. In this way an infinite number of solutions can be generated. For the case E < V0 , the boundary conditions provided four equations for four unknowns. The conditions for nontrivial solutions lead to the discrete values of E. In contrast to this, for the case E > V0 since we have less numbers of equations than the unknowns, conditions like Eq. (5.22) or (5.24) do not arise. Thus, for E > V0 , E assumes continuous distribution of energy.
5.4 POTENTIAL STEP As an example of another interesting scattering state system we consider the problem of a homogeneous beam of free particles moving along a straight-line crossing one region to another of a different potential. The two regions are separated by a plane surface. Suppose the potential of the above system is of the form ( 0, x 0
150 Quantum Mechanics I: The Fundamentals
V V0
0 FIGURE 5.7
x
Step potential.
which is depicted in Fig. 5.7. The potential is zero for x < 0 and V0 for x > 0. The potential given by Eq. (5.34) is called a single-step potential barrier.
5.4.1
Physical Systems with Step Potential
Heterojunctions are commonly grown with semiconductors, which may be undoped or doped n or p type. Consider two conductors A and B with conduction and valence band edges Ec and Ev respectively and with different magnitudes of forbidden gap [5]. An electron at conduction band edge in A will have lower energy than in material B. When junction is made between A and B, a potential barrier is developed at the interface with region x < 0 and x > 0 representing the materials A and B respectively. This restricts the motion of electrons incident from the A (left) side. This type of situation is realized in GaAs and Alx Ga1−x As junctions. The electron in the GaAs encounters a barrier to enter in AlGaAs part of the structure. It is reflected back by the potential.
5.4.2
Particle Energy Above the Potential Step: E > V0
Let us first mention the dynamics of the classical particle. The total energy is greater than the potential energy. Further, the kinetic energy is positive. Therefore, a classical particle moving towards x = 0 from x < 0 will not be reflected at x = 0. In the quantum case, the Schrödinger equation corresponding to the potential, Eq. (5.34), is ψ1xx + k02 ψ1
=
0,
ψ2xx + k 2 ψ2
=
0,
2mE , x0. ~2 k02 =
(5.35a) (5.35b)
When E > V0 , k02 and k 2 are positive. k 2 is negative for E < V0 . We analyse these two cases separately.
Exactly Solvable Systems II: Scattering States 151 The solutions of the Eqs. (5.35) are ψ1
=
Aeik0 x + Be−ik0 x ,
x0
(5.36b)
where A, B, C and D are constants to be determined. In Eq. (5.36a) eik0 x represents the incident wave moving towards positive x-direction and e−ik0 x represents the reflected wave moving towards negative x-direction. In Eq. (5.36b) eikx represents the transmitted wave moving in positive x-direction and e−ikx represents the reflected wave moving towards negative x-direction. Since discontinuity occurs only at x = 0 there is no reflected wave for x > 0 and so D becomes zero. Further, we set the amplitude of the incident wave as unity. We write the Eqs. (5.36) as x < 0 : ψ1 = ψinc + ψref , x > 0 : ψ2 = ψtra ,
(5.37a) (5.37b)
where ψinc = eik0 x , ψref = Be−ik0 x and ψtra = Ceikx . Here ∗ ψinc ψinc = 1 ,
∗ ψref ψref = |B|2 ,
∗ ψtra ψtra = |C|2 .
(5.37c)
The boundary conditions ψ1 (0) = ψ2 (0) and ψ1x (0) = ψ2x (0) lead to 1 + B = C and k0 (1 − B) = kC. Solving the above set of equations for B and C gives B=
1 − (k/k0 ) , 1 + (k/k0 )
C=
2 . 1 + (k/k0 )
(5.38)
Then
2 1 − (k/k0 ) k 4k/k0 R = |B| = , T = |C|2 = . (5.39) 1 + (k/k0 ) k0 (1 + k/k0 )2 p Here T + R = 1 and k/k0 = 1 − (V0 /E) . Since V0 > 0 we have k < k0 , or k/k0 < 1. As a result the probability of finding the particle anywhere in the region x > 0 is greater than the probability of finding it anywhere in the region x < 0. We account for this from the classical argument that the probability of finding a particle is more in the region where its velocity is less. Since the velocity of the particle in the region x > 0 is less than its velocity in x < 0, the probability is more in the region x > 0. Figure 5.8 shows the reflection (R) and transmission (T ) probabilities as a function of k/k0 for the step potential. Continuous and dashed curves represent R and T respectively. 2
5.4.3
Particle Energy Below the Potential Step: 0 < E < V0
Next, consider the case of the energy E of the particle is smaller than the height V0 of the potential step. The classical particle in the left region will
152 Quantum Mechanics I: The Fundamentals
R,T
1 0.5 0 0
0.5
1
k/k0 Variation of R (continuous curve) and T (dashed curve) of the step potential. FIGURE 5.8
remain in the left region. The particle moving towards the right will be completely reflected at x = 0. In classical mechanics finding the particle in x > 0 is impossible because in this region the total energy is less than the potential energy, so that the kinetic energy p2x /(2m) < 0 and the momentum px is imaginary. Defining k 2 = 2m(V − E)/~2 , the Schrödinger equation takes the form ψ1xx + k02 ψ1 ψ2xx − k 2 ψ2
= 0, = 0,
x0.
(5.40a) (5.40b)
The solutions of Eqs. (5.40) are ψ1 ψ2
= =
eik0 x + Be−ik0 x , Cekx + De−kx ,
x0
(5.41a) (5.41b)
In the limit x → ∞, we have ekx → ∞, therefore, choose C = 0. The probability of finding the particle in the region x > 0 decreases rapidly with x as e−2kx . But there is a finite nonzero probability for finding the particle in the region x > 0. The penetration into the classically forbidden region is a striking p prediction of quantum mechanics. The penetrating distance is ∆x = ~/(2 2m(V0 − E)). As done earlier for E > V0 , we determine B and D using the boundary conditions at x = 0. We obtain B=
k0 − ik , k0 + ik
D=
2k0 . k0 + ik
(5.42)
Further, R = |B|2 = 1 while T = 0. Even though T is zero in the region x > 0, the wave function is not zero! For the linear harmonic oscillator and the particle in a box, the stationary state eigenfunctions are normalizable and the energy eigenvalues are discrete.
Exactly Solvable Systems II: Scattering States 153 The eigenfunctions for the cases of the free particles, square-barrier and step potentials are nonnormalizable and their energy eigenvalues are continuous. Figure 5.9 shows Reψ (actually Reψ + E) for a few chosen values of E for the step potential with V0 = 1. For E < V0 , the eigenfunction is decaying for x > 0. For E > V0 , it is oscillatory for x > 0. The eigenfunction does not fall abruptly to zero at x = 0 but diminishes exponentially with x. The probability of finding the particle in x > 0 decreases exponentially with increasing x.
Solved Problem 3: A beam of electrons incident on an energy barrier of height V0 eV of the step potential. Find the fraction of electrons reflected and transmitted at the barrier if the energy of the incident electrons is (a) 3V0 /2eV, (b) V0 and (c) V0 /4eV. Case (a): E = 3V0 /2 eV T and R can be calculated from Eq. (5.39). In this case E=
3 V0 ev , 2
k02 =
2m(E − V0 ) mV0 3mV0 , k2 = = 2 . 2 2 ~ ~ ~
(5.43)
Using these in Eq. (5.39) we obtain R
√ 2 √ 3mV0 − mV0 √ √ = = 3mV0 + mV0
!2 √ 3−1 √ ≈ 0.0718 . 3+1
(5.44)
Then T = 1 − R ≈ 0.9282.
Case (b): E = V0 eV
When E = V0 , k is 0 and hence R = 1 and T = 0. That is, all the incident electrons are reflected back. Case (c): E = V0 /4 eV From Eq. (5.42) we have T = 0 and R = 1.
5.5 LOCALLY PERIODIC POTENTIAL In this section we consider scattering by a potential consisting of N -identical cells. The potential is locally periodic, that is, it is periodic on a finite interval and zero elsewhere. It has been shown that the N -cell transmission and reflection probabilities can be expressed in terms of single cell probabilities [6]. The N -delta function and square-barrier potentials exhibit band structure, a striking special character of transmission probability [6-8]. This feature is observed even for quite small number of unit cell. First, we define the transfer matrix M [6] associated with the scattering by a single cell. Next, we show the determination of the transmission probability for multiple scattering using M . Then we obtain a simple closed form
154 Quantum Mechanics I: The Fundamentals
Re ψ
4 2 0 -2 -20
E = 0.3 -10
0
x
10
-10
0
x
10
-10
0
x
10
-10
0
x
10
-10
0
10
Re ψ
4 2 0 -2 -20
E = 0.8
Re ψ
4
0
E = 1.0
4
Re ψ
20
2
-2 -20
20
2 0 -2 -20
E = 1.1
4
Re ψ
20
20
2 0 -2 -20
E = 1.5
x
20
Re ψ is plotted for a few values of E for the step potential with V0 = 1, ~2 = 1 and m = 1. For illustration Re ψ is shifted as Re ψ + E.
FIGURE 5.9
Exactly Solvable Systems II: Scattering States 155 expression for transmission probability for arbitrary Dirac-delta and rectangular barrier cells.
5.5.1
M Matrix
Let us consider a locally periodic potential consisting of three regions: Region−I Region−II Region−III
−∞ < x < a a≤x≤b b (N − 1)s + a. (5.54b)
For the nth cell, using Eqs. (5.47), (5.51) and (5.53) we obtain −iks An An+1 e 0 =P , P =M . Bn Bn+1 0 eiks From Eq. (5.55) we have A0 AN N =P , B0 BN
detP = 1.
(5.55)
(5.56)
We need to find P N . Making use of Cayley–Hamilton theorem [9,10] we find a simple formula for P N . The characteristic equation for P is det(P − IP ) = 0, that is, P 2 − P tr(P ) + detP = P 2 − 2ξP + I = 0 , (5.57) where ξ = tr(P )/2 . According to the Cayley–Hamilton theorem any matrix satisfies its own characteristic equation P 2 − 2P ξ + I = 0 .
(5.58)
The point is that any higher power of P can be reduced to a linear combination of P and I: P N = P UN −1 (ξ) − IUN −2 (ξ) , (5.59)
Exactly Solvable Systems II: Scattering States 157 where UN (ξ) is a polynomial of degree N in ξ. We can easily find U ’s. Multiplying Eq. (5.59) by P and using Eq. (5.58) for P 2 we obtain P N +1 = (2P ξ − I) UN −1 (ξ) − P UN −2 (ξ) .
(5.60)
On the other hand, replacing N by N + 1 in Eq. (5.59) we get P N +1 = P UN (ξ) − IUN −1 (ξ) .
(5.61)
Comparison of Eqs. (5.60) and (5.61) gives UN +2 (ξ) − 2ξUN +1 (ξ) + UN (ξ) = 0 .
(5.62)
Equation (5.62) is the recursion relation for Chebychev polynomials. For N = 2, Eq. (5.59) becomes P 2 = P U1 (ξ) − IU0 (ξ) . (5.63) Equating this expression with Eq. (5.58) we have U0 = 1 and U1 = 2ξ. Using these U2 can be computed from Eq. (5.62). Similarly, other U ’s can be computed. Now, ikN s e 0 N MN = P 0 e−ikN s WN ZN = (5.64) ∗ ZN WN∗ where WN Zn
=
W e−iks UN −1 − UN −2 eikN s ,
= ZUN −1 e
−ik(N −1)s
(5.65) (5.66)
with |WN |2 − |ZN |2 = 1. For a single cell (with D = 0), from Eq. (5.47) C/A is found to be 1/α11 . Therefore, 2 C 1 1 = . (5.67) T = = A |α11 |2 |W1 |2 Hence,
TN = where ξ=
1 1 1 = = , |WN |2 1 + |ZN |2 1 + |Z|2 (UN −1 (ξ))2
(5.68)
1 1 trP = W e−iks + W ∗ eiks = Re(W ) cos ks + Im(W ) sin ks . (5.69) 2 2
From (5.52) we find W = 1 + iβ, Z = iβ and thus ξ = cos ks + β sin ks. Then TN =
1 1 + β 2 (UN −1 (ξ))
2
.
(5.70)
1
1
0 .5
0 .5
0
T
T
158 Quantum Mechanics I: The Fundamentals
N =0 0
200
N =2 0
200
400
600
T N =8 0
200
400
600
800
E
400
600
800
E
N =4 0
200
400
600
800
E
1
0 .5
0
200
0 .5
0
800
E
1
N =1 0
1
0 .5
0
0
800
T
T
600
E
1
T
400
0 .5
0
N = 16 0
200
400
600
800
E
FIGURE 5.10 Transmission probability as a function of energy [7,8]. Here s = 0.5, V0 = 4 and m = ~2 = 1. N = 0 corresponds to one barrier.
In the limit E → ∞, k is also → ∞, β → 0 and hence T → 1. Figure 5.10 shows the plots of T against E for N = 0 (one barrier), N = 1, 2, 4, 8, 16 with V0 = 4 and s = 0.5. Some of the special features of T versus E are summarized below: 1. For more than one barrier the graph of T has number of energy bands in which T (N ) ≈ 1. 2. The bands are separated by valleys where T (N ) approaches zero for large N . They can be regarded as the precursors of the band structure of periodic potentials. The bands occur even for small numbers of barriers. 3. Each band contains N ripples. 4. As N increases the valleys become more and more narrow. The number of valleys that are flatter and deeper increases with an increase in N . 5. For a fixed energy value, T changes significantly as the number of barriers is increased. This is shown in Fig. 5.11 where T is plotted as a function of N for E = 500 and 725.
Exactly Solvable Systems II: Scattering States 159
T
1 0.5 E=500 00
FIGURE 5.11
E=725 20
N
40
Transmission probability T versus N for two values of E.
For a detailed discussion on multiple rectangular barrier one may refer to ref.[8]. For discussion on quantum mechanical tunneling in neural networks refer to refs.[11,12].
5.6 REFLECTIONLESS POTENTIALS In this present chapter so far we studied scattering of particles by some simple potentials. In the case of square-barrier oscillating T is observed. For very large E greater than V0 we have T ≈ 1. For the Dirac-delta function potential T is found to increase with E and approach to the limiting value 1. Now, we ask an interesting question: Are there potentials for which T = 1 for arbitrary energy? In other words, do we have a potential which perfectly transmits incoming waves with any energy? The answer is yes. There are infinite number of potentials with the above property. Such potentials are called reflectionless potentials [13-16] and they can be constructed using inverse scattering transform [13]. A class of one-dimensional reflectionless potentials with their bound and scattering states expressed in terms of elementary functions are the sech2 potentials, called modified Pöschl–Teller potentials, given by [14-16] Vl (x) = −
~2 2 V l(l + 1)sech2 βx , 2m 0
l = 0, 1, · · · .
(5.71)
These potentials have two remarkable properties: (i) Bound states at zero energy and (ii) reflectionless scattering. Introducing the change of variables x′ = βx, k 2 = 2mE/(~2 V02 ) and dropping the prime the Schrödinger equation of the given problem takes the form d2 ψ 2 + k + l(l + 1)sech2 x ψ = 0 . 2 dx
(5.72)
Making a new change of variable y = cosh2 x and writing ψ = y (l+1)/2 v(y) the
160 Quantum Mechanics I: The Fundamentals Eq. (5.72) becomes 1 3 − (l + 2)y v ′ − (l + 1)2 + k 2 v = 0 . y(1 − y)v ′′ + l + 2 4
(5.73)
Equation (5.73) is the hypergeometric differential equation and its solution is given by [17,18] p 1 1 1 3 v(y) = A 2 F1 a, b, ; 1 − y + B 1 − y 2 F1 a + , b + , ; 1 − y (5.74) 2 2 2 2
where a = (l + 1 + ik)/2, b = (λ + 1 − ik)/2 and 2 F1 (a, b, c; x) is the hypergeometric function with the usual notation. We look for a fundamental system of two real solutions with one is odd and the other is even in x. The even solution is obtained with A = 1, B = 0: 1 (l+1) 2 ψe = cosh x 2 F1 a, b, ; −sinh x . (5.75) 2 The odd solution is obtained with A = 0, B = −i: 1 3 1 2 (l+1) (l+1) ψo = cosh x sinh x 2 F1 a + , b + , ; −sinh x . 2 2 2
(5.76)
When l = 0, the Hamiltonian is that of the free particle. For l > 0, px = 0 at the turning points of classically bounded solutions. When px = 0 the Hamiltonian becomes Hl = −l(l + 1)sech2 x∗ where x∗ is a turning point. Since sech2 x∗ is an even function of x∗ we have E < 0. Therefore, E < 0 corresponds to bound states and E > 0 corresponds to scattering states. We now consider the cases E < 0 and E > 0 separately.
Case 1: Bound States, E < 0 We write k = iǫ, ǫ > 0 so that El = −ǫ2 . By performing an asymptotic analysis of (5.75), we obtain the eigenvalues for even eigenstates from the relation ǫ = l − 2n, n = 0, 1, · · · . For the odd states ǫ = l − 1 − 2n, n = 0, 1, · · · . Then the energy eigenvalues are given by (n)
El
= (l − n)2 ,
n≤l
(5.77)
where n = 0, 1, · · · . Since n ≤ l for a given l, there are only a finite number of bound states. From Eq. (5.77) we note that El = 0 is an energy eigenvalue for any integer value of l.
Case 2: Scattering States From the asymptotic form of ψo and ψe we infer the following. When E > 0, a and b are complex conjugates and the general solution for |x| → ∞
Exactly Solvable Systems II: Scattering States 161 is written as ψ
=
and ψ
=
˜ e AC eiφe e−ikx + e−iφe eikx 2 ˜ o BC − eiφo e−ikx + e−iφo eikx , 2 ˜ e AC eiφe eikx + e−iφe e−ikx 2 ˜ o BC + eiφo eikx + e−iφo e−ikx , 2
x → −∞
(5.78a)
x→∞
(5.78b)
where the constants Ce , Co , φe and φo are to be determined by making an ˜ are asymptotic analysis of the solutions at x → ±∞. The constants A˜ and B found to be arbitrary while φo and φe are determined as [15] φe
=
φo
=
Γ(ik)e−ik log 2 , Γ[(l + 1 + ik)/2]Γ[(−l + ik)/2] Γ(ik)e−ik log 2 . arg Γ[(l + ik)/2]Γ[(−l + ik)/2] arg
(5.78c) (5.78d)
Now, we look for the specific asymptotic form (|x| → ∞) of the solution ( eikx + Re−ikx , x → −∞ (5.79) ψ= T eikx , x → ∞, where R and T are reflection and transmission coefficients. Comparison of Eqs. (5.79) and (5.78) results in R
=
T
=
1 [(cos 2φe + cos 2φo ) + i (sin 2φe + sin 2φo )] , 2 1 [(cos 2φe − cos 2φo ) + i (sin 2φe − sin 2φo )] . 2
Then |R|2 =
1 , 1 + s2
|T |2 =
s2 , 1 + s2
(5.80a) (5.80b)
(5.81)
where s = sinhπk/ sin πl. When l is an integer |s| = ∞ and hence |R|2 = 0 and |T |2 = 1. That is, a particle incident on the potential is transmitted with unit probability. As a result the potentials given by Eq. (5.71) are known as reflectionless potentials. The reflectionless potential systems can also be studied by means of creation and annihilation operators [19].
162 Quantum Mechanics I: The Fundamentals
5.7 DYNAMICAL TUNNELING The tunneling phenomenon encountered in the study of potential barrier is a process permitted by quantum theory but disallowed by classical mechanics on energy consideration. A new kind of tunneling called dynamical tunneling was found. It refers to a behavior allowed by quantum mechanics but forbidden by classical mechanics. It involves no potential barrier. However, a constant of motion (other than energy) forbids classically the quantum allowed phenomenon. Dynamical tunneling has been observed in 1. a periodically driven undamped pendulum, 2. ultracold atoms forming a Bose–Einstein condensate in an amplitudemodulated optical standing wave and 3. a ball bouncing between two semicircular mirrors. In the third example, it is easy to notice that classically the ball cannot escape from the region between the mirrors. Even though escape is possible energetically, and potential barrier would not prevent it, but still the classical motion of the ball disallows it to escape. Here escape is prevented not by energy considerations. Note that escape is not ruled out by the quantum formalism. The ball would be able to escape by dynamical tunneling slowly. Another example, is the formaldehyde molecule H2 CO. In classical picture, oxygen in H2 CO molecule spinning around the C-O axis always points out in one direction. In quantum mechanics it can oscillate between pointing up and down without violating the conservation of energy or angular momentum. For details on dynamical tunneling see refs.[20-26].
5.8 CONCLUDING REMARKS In the systems we studied in this chapter the bound states and continuum states, (scattering states) are well separated in energy. However, there are quantum systems with bound states embedded in continuum (BSECs) [2733]. Such states are regarded as exotic quantum states and they do not have classical analogues. BSECs have been observed experimentally [28]. Scattering by a double square-well potential [34], double delta-function potential [35, 36], a Coulomb field in two-dimensions [37], asymmetrical potential [38] and a finite periodic potential [39] have also been analysed. There are number of modern opto-electronic devices involving low-dimensional heterostructure. The unbound states of semiconductor quantum wells and quantum dots play significant role in their optical properties. Quantum mechanical transmission and reflection of electrons at potential energy steps can be applied to the design of nonequilibrium transistors.
Exactly Solvable Systems II: Scattering States 163
5.9 BIBLIOGRAPHY [1] G. Binnig, H. Rohrer, Ch. Gerber and E. Weibel, Physica B+C 109 & 110:2075, 1982. [2] J.D. Chalk, Am. J. Phys. 56:29, 1988. [3] M. Larsson, V.B. Antonyuk, A.G. Mal’shukov, Z. Ma and K.A. Chao, J. Phys. A 35:L531, 2002. [4] D.F. Styer, The Strange World of Quantum Mechanics. Cambridge University Press, Cambridge, 2000. [5] M. Ramrakhiani and S.K. Pandey, Physics News, 2000 pp.20. [6] D.W.L. Sprung, H. Wu and J. Martorell, Am. J. Phys. 61:1118, 1993. [7] D.J. Griffiths and N.F. Taussig, Am. J. Phys. 60:883, 1992. [8] D.J. Griffiths and C.A. Steinke, Am. J. Phys. 69:137, 2001. [9] L.A. Pipes and L.R. Harvill, Applied Mathematics for Physicists. McGraw–Hill, Singapore, 1984. [10] T.L. Chow, Mathematical Methods for Physicists. Cambridge University Press, Cambridge, 2000. [11] M. Dugic and D. Rakovic, Eur. J. Phys. B 13:781, 2000. [12] D. Rakovic, Informatica 21:507, 1997. [13] S. Novikov, S.V. Manakov, L.P. Pitaevskii and V.E. Zakharov, Theory of Solitons: The Inverse Scattering Method. Consultants Bureau, New York, 1984. [14] J. Lekner, Am. J. Phys. 75:1151, 2007. [15] S. Flügge, Practical Quantum Mechanics–I. Springer, Berlin, 1971. [16] N. Kiriushcheva and S. Kuzmin, Am. J. Phys. 66:867, 1998. [17] J. Mathews and R.L. Walker, Mathematical Methods of Physics. Pearson Education, New Delhi, 2004. [18] M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. Applied Mathematics Series No.55, 1964. [19] R.L. Jaffe, An algebraic approach to reflectionless potentials in onedimension. Unpublished, 2009. [20] M.J. Davis and E.J. Heller, J. Chem. Phys. 75:246, 1981. [21] E.J. Heller, Nature 412:33, 2001.
164 Quantum Mechanics I: The Fundamentals [22] D.A. Steck, W.H. Oskay and M.G. Raizen, Science 293:274, 2001. [23] W.K. Hensinger, H. Haffner, A. Browaeys, N.R. Heckenberg, K. Helmerson, C. McKenzie, G.J. Milburn, W.D. Phillips, S.L. Rolston, H. Rubinsztein-Dunlop and B. Upcroft, Nature 412:52, 2001. [24] A. Ramamoorthy, R. Akis, J.P. Bird, T. Maemoto, D.K. Ferry and M. Inoue, Phys. Rev. E 68:026221, 2003. [25] M. Henseler, T. Dittrich and K. Richter, Europhys. Lett. 49:289, 2000. [26] G. Hackenbroich and J.U. Nöckel, Europhys. Lett. 39:371, 1997. [27] J. von Neumann and E. Wigner, Phys. Z. 30:467, 1929. [28] F. Capasso, C. Sirtori, J. Faist, D.L. Sivco, S.N.G. Chu and A.Y. Cho, Nature 358:565, 1992. [29] B.A. Arbuzov, S.A. Shichanin and V.I. Savrin, Phys. Lett. B 275:144 , 1992. [30] J.R. Spence and J.P. Vary, Phys. Lett. B 271:27, 1991. [31] J. Pappademos, U. Sukhatme and A. Pagnamenta, Phys. Rev. A 48:3525, 1993. [32] T.A. Weber and D.L. Pursey, Phys. Rev. A 50:4478, 1994. [33] A.A. Stahlhofen, J. Phys. A 27:8279, 1994. [34] B. Stec and C. Jedrzejek, Eur. J. Phys. 11:75, 1990. [35] I.R. Lapidus, Am. J. Phys. 50:663, 1982. [36] P. Senn, Am. J. Phys. 56:916, 1988. [37] Q.G.Lin, Am. J. Phys. 65:1007, 1997. [38] Y. Nogami and C.K. Ross, Am. J. Phys. 64:923, 1996. [39] D.W.L. Sprung, H. Wu and J. Martorell, Am. J. Phys. 61:1118, 1993.
5.10 EXERCISES 5.1 Express the transmission probability for the potential barrier as a function of energy for E < V0 and E > V0 . 5.2 For a symmetrical barrier potential of energy V0 and thickness 2a find the minimum transition probability for E > V0 .
Exactly Solvable Systems II: Scattering States 165 5.3 For the potential barrier −1 ǫ2 T η2 T = 1+ 1+ sinh2 2ka , R= sinh2 2ka. 4 4 Using the above equations prove that T = 0 and R = 1 if V0 = ∞. 5.4 A beam of electrons is incident on the square-barrier potential V (x) = V0 for |x| < a and 0 otherwise from left. Find the fraction of electrons reflected √ and transmitted if the energy of the electrons is 3V0 /4eV. Assume that 2mV0 a/~ = 3/2. 5.5 For the square-barrier potential find the probability current, Jx , through the barrier for the incident particles. 5.6 An object of mass 10 kg is moving towards a 120 J potential barrier of 2m wide with a speed 2 m/s. Compute the probability for it to tunnel through the barrier. 5.7 For the barrier potential V (x) = V0 for 0 ≤ x ≤ a and 0 otherwise, write down the expression for transmission probability (T ) for E < V0 and E > V0 . 5.8 A beam of electrons is incident on an energy barrier of height V0 eV of the step potential. Compute the energy of the incident electrons in order to have T = R = 0.5. 5.9 Show that the energy eigenspectra of a particle in a box can be obtained from the expression for the finite square-well potential in the limit V0 → ∞. 5.10 Find the relation between the energy of the electrons incident on the step potential barrier and the strength of the potential V0 in order to have 75% of the incident particles to be transmitted. 5.11 In the treatment of step potential the mass is assumed as constant in both the regions x < 0 and x > 0. Assuming that the masses are different (which is the case in LEDs and laser diodes) in the two regions determine the reflection and transmission coefficients. Also find the condition for complete transmission. 5.12 Estimate the penetration distance ∆x for (i) a small dust particle of mass 4 × 10−14 kg moving at a very low velocity v = 10−2 m/sec and (ii) an electron moving at a velocity 107 m/sec if they are incident on a potential step of height equal to twice their kinetic energy (use nonrelativistic energy relation). 5.13 For the reflectionless potential system with the Hamiltonian Hl = (p2x /~2 ) − l(l + 1)sech2 x introducing the operators al = (px /~) − il tanh x and a†l = (px /~) + il tanh x determine the ground state eigenfunctions
166 Quantum Mechanics I: The Fundamentals and eigenvalues for l = 0 and l = 1. Also construct the continuum eigenstates for l = 1 and 2 and obtain the reflection and transmission amplitudes. 5.14 Consider the one-dimensional delta-function potential V (x) = V0 δ(x). Obtain the bound state eigenvalues and eigenfunctions for V0 < 0 and reflection and transmission amplitudes for scattering states for V0 > 0. 5.15 For the Dirac-delta potential of the previous problem compute the required energy of the incident particles to make T = R = 1/2.
CHAPTER
6
Matrix Mechanics
6.1 INTRODUCTION The quantum theoretical description of microscopic systems discussed so far is based on differential operators and the solution of the underlying eigenvalue equations. This approach is called Schrödinger approach or wave mechanics. Heisenberg developed another equivalent approach called matrix mechanics. Because the trajectory of an electron is unobservable and moreover line spectra implied discrete allowed states, Heisenberg discarded the idea of electron orbits and formulated a theory for line spectra of excited systems based on observables. When Heisenberg arranged the allowed set of transitions in a matrix form, to his surprise, each element of the matrix was found to be related to the probability of a particular transition. This gave the birth to matrix mechanics. Then Heisenberg, Born and Jordan worked out the rules of these matrices. In the Heisenberg formulation the wave function ψ becomes a vector while the operators are represented by matrices [1,2]. The transformations from one vector into another is described by matrices. It is thus possible to represent the operators as matrices. The differential equations of the Schrödinger’s wave mechanics became matrix equations. For example, consider the differential equation Aφ = φ′ , where φ and φ′ are two functions and A is a linear differential operator. In matrix method the functions φ and φ′ are column (or row) vectors and A is a square matrix. The wave mechanical approach and matrix approach, although mathematically different, they are found to give equivalent solutions to the same problems1 [3,4]. In matrix mechanics all observables are represented by matrices. All possible eigenvalues of a matrix representing an observable are the possible outcomes of experimental measurement on that observable. Since observables take only real values and because the eigenvalues of Hermitian matrices are real 1 D.F. Styer described the equivalence of the two theories as very much as the process of adding algebraic numerals is very different from the process of adding roman numerals but the two processes nevertheless always give the same result[5].
167
168 Quantum Mechanics I: The Fundamentals they are the most suitable matrices to represent observables. More precisely, in matrix mechanics the properties of the matrices are of interest and not the space on which these matrices operate. In contrast to this, in wave mechanics we study the properties of the elements of this space. In this chapter first we give the preliminary concepts relative to the use of the matrix method. Next, we present the various elements of matrix mechanics. Then we discuss the matrix representation of operators of certain systems. We introduce Dirac’s notation of state vectors, Hilbert space, projection operator, displacement operators and the generators of translations.
6.2 LINEAR VECTOR SPACE The quantum description of a state of a system is in terms of a vector. To understand the implications of this kind of representation of a system, in this section let us introduce the notion of linear vector space. A linear vector space is defined as set of vectors A, B, C, · · · satisfying the following: 1. For every pair of vectors A and B there exists a C: C = A + B. 2. Vector addition is commutative and associative: A+B = B+A ,
(A + B) + C = A + (B + C) = A + B + C . (6.1)
3. The origin of the vector space is the zero vector, 0. 4. Multiplication is distributive and associative: a(A+ B) = aA+ aB ,
(a+ b)A = aA+ bA ,
a(bA) = (ab)A , (6.2)
where a and b are complex numbers. 5. For every vector A there exists −A such that A + (−A) = 0. If a set of vectors A1 , A2 , · · · , AN obey the relation c1 A1 + c2 A2 + · · · + cN AN = 0 ,
(6.3)
where c1 , c2 , · · · , cN are not all zero then A’s are said to be linearly dependent . When no such c’s exist then the A’s are linearly independent. The maximum number of linearly independent vectors in a vector space is the dimension of the vector space. The dimension can be finite or infinite. A linearly independent set of vectors An satisfying ψ=
N X
cn An
(6.4)
n=1
for every vector ψ in the linear vector space of dimension N is called a coordinate system or bases vectors or simply bases for the vector space. Here any
Matrix Mechanics 169 arbitrary vector in the space may be written as a linear combination of the basis vectors in one and only one way. cn in Eq. (6.4) are the components of ψ. c1 , c2 , · · · , cN are called vectors and are represented in matrix form as c1 c2 (6.5) .. or [c1 c2 · · · cN ] . . cN
Thus, a vector can be expressed as either a row or a column matrix. A1 , A2 , · · · , AN are called orthonormal if (An , Am ) = An · A†m = δnm ,
n, m = 1, 2, · · · , N .
(6.6)
From a set of N linearly independent vectors we can construct an orthogonal set employing the Schmidt orthogonalization procedure [6,7]. If the basis set An has N vectors then the space is N -dimensional. We can choose any set of N linearly independent vectors which can span the same vector space. Any vector represented in two different bases will have different components and the column matrices will be different. We can write the basis vectors of one representation as a linear combination of the basis vectors of a second representation. Hence, any two representations of the same linear vector space will be related by a transformation and the transformation matrix will be unitary if the norm of the vector is same in any representation. Consider a vector OA shown in Fig. 6.1. Using the base vectors (A1 , A2 ) the vector OA is expressed as OA = aA1 + bA2 . On the other hand, in terms of the base vectors (A′1 , A′2 ) we write OA = a′ A′1 + b′ A′2 . The change from one basis set to another is called transformation of the base vectors. Suppose a problem is defined in (A1 , A2 ). Assume that solving it in terms of (A′1 , A′2 ) is much simpler than with (A1 , A2 ). Then we can go from the set (A1 , A2 ) to (A′1 , A′2 ). This is similar to (q, p) → (Q, P ) in classical mechanics. Suppose a set of N orthonormal vectors {φn } forms a basis for a N dimensional vector space. Then we express any vector in the space as N X
cn = (φn , ψ) , φ1 = [1 0 · · · ], φ2 = [0 1 · · · ]
(6.7)
and so on. The inner product of two vectors ψ1 and ψ2 is given by DX E X hψ1 , ψ2 i = cn φn , dm φm
(6.8)
ψ=
n=1
cn φn ,
n
m
which is a number. We have hψ1 , ψ2 i = hψ2 , ψ1 i∗ where asterisk denotes the complex conjugate. If hψ1 , ψ2 i = 0 then ψ1 and ψ2 are said to be orthogonal . If the basis vectors {φn } are orthonormal then hφn , φm i = δnm . Then in an orthonormal basis we have X X X hψ1 , ψ2 i = c∗n dm hφn , φm i = c∗n dm δnm = c∗n dn . (6.9) n,m
n,m
n
170 Quantum Mechanics I: The Fundamentals
j
j’
A
b b'
i’
a’ O(0,0)
i
a
Illustration of representation of a vector OA in two different base vectors (A1 , A2 ) and (A′1 , A′2 ). FIGURE 6.1
The square of norm of ψ1 is
Solved Problem 1:
P
n
|cn |2 . If ψ1 is normalised then hψ1 , ψ1 i = 1.
Determine whether the following set of vectors are linearly independent or dependent. ψ1 = 3i + 2j − k, ψ2 = 3i + 6j − 3k, ψ3 = i + 2j − k. Three vectors A1 , A2 and A3 are linearly dependent if at least one Ci , i = 1, 2, 3 is nonzero in C1 A1 + C2 A2 + C3 A3 = 0. For the given vectors we write C1 ψ1 + C2 ψ2 + C3 ψ3
=
(3C1 + 3C2 + C3 )i + (2C1 + 6C2 + 2C3 )j
=
−(C1 + 3C2 + C3 )k 0.
(6.10)
= 0 , 2C1 + 6C2 + 2C3 = 0 ,
(6.11)
= 0.
(6.12)
From the above we have 3C1 + 3C2 + C3 C1 + 3C2 + C3
Solving the above three equations we get C1 = 0, C2 = 1, C3 = −3. Since not all the Ci ’s are zero the three vectors are linearly dependent.
6.3 MATRIX REPRESENTATION OF OPERATORS AND WAVE FUNCTION An operator A operating on a vector ψ1 in a linear vector space changes it into another vector ψ2 in the same space. Rewrite ψ2 = Aψ1 as X X X dn φn = A cm φm = cm Aφm . (6.13) n
m
m
Matrix Mechanics 171 Then dn =
P
m hφn , Aφm icm .
Defining
we have dn =
X
Anm cm ,
m
Z
∞
φ∗n Aφm dτ
(6.14)
n = 1, 2, · · · , N .
(6.15)
Anm = hφn , Aφm i =
−∞
The numbers Anm form a square-array and set up a matrix. The equation ψ2 = Aψ1 in a matrix form is c1 d1 A11 · · · A1N c2 d2 A21 · · · A2N . (6.16) ψ1 = . , ψ2 = . , [Amn ] = ··· .. .. AN 1 · · · AN N cN dN
Note that any valid arithmetical relation between two or more operators is also applicable between the corresponding matrices. The eigenvalues of an operator A are also the eigenvalues of the matrix representation of the same operator. Similar result holds for eigenfunctions also. Hermitian operators are represented by Hermitian matrices. A matrix A is Hermitian if A = A† where (Amn )† = (Anm )∗ . When A = −A† it is antiHermitian. The following properties are easy to verify: 1. The eigenvalues of a Hermitian matrices are real. 2. The eigenvectors of a Hermitian matrix corresponding to different eigenvalues are orthogonal. 3. The product of Hermitian matrices is a Hermitian if and only if they commute. 4. Commuting Hermitian matrices have simultaneous eigenvectors. There exists an orthonormal basis φ1 , φ2 , · · · , φN for a Hermitian operator A in a linear vector space of N -dimensions. In relative to φ’s the operator A is represented by a diagonal matrix α1 0 ··· 0 0 α2 · · · 0 , A= (6.17) ··· 0 0 · · · αN
where α’s are real numbers. φ’s and α’s are solutions of Aφ = αφ. Then Z ∞ Anm = φ∗n αm φm dτ = αm δnm . (6.18) −∞
Notice that any operator in its own eigenvector basis is represented by a
172 Quantum Mechanics I: The Fundamentals diagonal matrix. The diagonal elements of A are the eigenvalues. Any vector ψ can be represented as a linear combination of the eigenvectors of A, as X ψ= cn φn , Aφn = αn φn . (6.19) n
6.4 UNITARY TRANSFORMATION The representations in different basis system are related by unitary transformation as shown below. Suppose the set of functions {φn (x)} are orthonormal and A is an operator. Then the matrix representation of A in the basis {φn (x)} is defined as (6.20)
Anm = (φn , Aφm ) . Any arbitrary state ψ(x) is represented as X ψ(x) = cn φn (x) , cn = (φn , ψ) .
(6.21)
n
Now, consider a second orthonormal basis {φ′n (x)}. The operator A and ψ(x) in this new basis system are X A′nm = (φ′n , Aφ′m ) , ψ(x) = c′n φ′n (x) , c′n = (φ′n , ψ) . (6.22) n
What is the relationship between the representations φn (x) and φ′n (x)? Let us expand φ′n (x) in terms of {φn (x)} as Z ∞ X ′ ′ φn (x) = Umn φm (x) , Umn = (φm , φn ) = φ∗m (x)φ′n (x) dx . (6.23) m
−∞
P
∗ Umn Um ′ n is calculated as Z ∞ X XZ ∞ ∗ Umn Um′ n = dx dyφ∗m (x)φ′n (x)φm′ (y)φ′∗ n (y)
The quantity
n
n
n
= = P
Z
∞
−∞
−∞
dx
δmm′ .
Z
∞
−∞
−∞
dyδ(x − y)φm′ (y)φ∗m (x)
∗ n Unm Unm′ †
(6.24) †
†
Further, = δmm′ . Thus, U U = U U = I where U is unitary. Hence det(U U ) = detU 2 = 1. Therefore, detU = ±1. Now, we write X X c′n = (U † )nm cm , cm = Umn c′n . (6.25) m
In matrix form Eqs. (6.25) take the form ′ c1 c1 c′2 = U † c2 , .. .. . .
n
c1 c2 =U .. .
c′1 c′2 . .. .
(6.26)
Matrix Mechanics 173 The proof of Eq. (6.26) is left as an exercise to the reader. For the matrix form of the operator we have X ∗ A′nm = Uln Alk Ukm or A′ = U † AU . (6.27) l,k
Thus, the representations in different basis systems are related by unitary transformations. We have TrAB = TrBA, detAB = (detA)(detB) and TrA′ = Tr(U † AU ) = Tr(U † U )A = TrA.
(6.28)
So, the trace and determinant of any operator are invariant under unitary transformations. Hence, no operator has unique matrix representation. Any operator will have different matrix representation in different basis. But all these matrices will have same eigenvalues. Any matrix representation of an operator on an arbitrary basis can be brought into a diagonal form by means of a unitary transformation of the arbitrary basis to the eigenvector basis. There are many ways of expressing a unitary operator U in terms of a Hermitian operator. The most common one is U = eiQ =
∞ X (iQ)n , n! n=0
(6.29)
†
where Q is Hermitian so that U † = e−iQ = e−iQ and U U † = U † U = I. Another useful form is U = (1 + ik)/(1 − ik) where k is Hermitian.
Solved Problem 2: What will happen to a unitary operator under a unitary transformation? Let A′ = U AU † . Consider A′† A′ . We get A′† A′
=
(U AU † )† (U AU † )
= =
(U A† U † )(U AU † ) U A† (U † U )AU †
= =
U (A† A)U † UU†
=
I.
(6.30)
Similarly, we can show that A′ A′† = I. Thus A′ A′† = A′† A′ = I. Thus, a unitary operator remains unitary under a unitary transformation.
174 Quantum Mechanics I: The Fundamentals
6.5 TENSOR PRODUCTS The tensor product is a way of making vector spaces together to form larger vector spaces. The construction is crucial for understanding the quantum mechanics of multiparticle system. Suppose V and W are vector spaces of dimensions m and n respectively. The V ⊗ W (read V tensor W ) is an mndimensional vector space. The elements of V ⊗ W are linear combination of tensor products |vi ⊗ |wi of elements of |vi of V and |wi of W . What sorts of linear operators act on the space V ⊗ W ? Suppose |vi and |wi are vectors in V and W and A and B are linear operators in V and W respectively. Then we define a linear operator A ⊗ B in V ⊗ W by the equation (6.31)
(A ⊗ B(|vi ⊗ |wi)) = A|vi ⊗ B|wi . Suppose A is an m × n matrix and B is a matrix representation A11 B A12 B A21 B A 22 B A⊗B = Am1 B Am2 B For example, the tensor product of σx =
σx ⊗ σy =
0 × σy 1 × σy
1 × σy 0 × σy
p × q matrix. Then we have the A1n B A2n B . Amn B
··· ··· .. . ··· 0 1
0 0 = 0 i
1 0
0 0 −i 0
(6.32)
0 and σy = i 0 −i i 0 . 0 0 0 0
−i 0
is
(6.33)
6.6 SCHRÖDINGER EQUATION AND OTHER QUANTITIES IN MATRIX FORM Suppose the energy eigenfunctions of a Hamiltonian operator are φ1 , φ2 , · · · , φn and the corresponding energy eigenvalues are E1 , E2 , · · · , En so that Hφn = En φn . Now, Hnm = hφn , Hφm i = Em hφn , φm i = Em δnm ,
(6.34)
where φn ’s are an orthonormal set of basis functions. H is a diagonal square matrix with the diagonal elements Em . Let us write the time-dependent Schrödinger equation in matrix form. We write X ψ(X, t) = ci (t)φi (X) . (6.35) i
Matrix Mechanics 175 Then Hψ = Eψ is given by X
Hci (t)φi (X) = i~
i
X d ci (t)φi (X) . dt i
(6.36)
Multiplying Eq. (6.36) by φ∗j (X) and integrating over all space we obtain Z ∞ Z ∞ X X d ci (t) φ∗j φi dτ (6.37) ci (t) φ∗j Hφi dτ = i~ dt −∞ −∞ i i
or
X
ci (t)Hji = i~
i
X d d ci (t)δji = i~ cj (t) . dt dt i
(6.38)
Interchanging i and j gives X
cj (t)Hij = i~
j
d ci (t) . dt
(6.39)
R∞ P In matrix form the normalization condition 1 = −∞ ψ ∗ ψ dτ = n |cn |2 is written as c1 Z ∞ c2 ψ ∗ ψ dτ = [c∗1 c∗2 · · · c∗N ] . = c† c = I . (6.40) .. −∞ cN
The orthogonality condition 0 =
R∞
Z
ψa∗ ψb dτ = 0 in matrix form is
−∞ ∞
XX
−∞ n
=
X
=
m
c∗an cbn
n
=
c∗an φ∗n cbm φm dτ
[c∗a1 c∗a2 · · · c∗aN ]
c†a cb
cb1 cb2 .. . cbN
.
In matrix form the expectation value of an observable A is Z ∞X X XX cm φm dτ = c∗n Anm cm . hAi = c∗n φ∗n A −∞ n
m
n
m
What is hHi if ψ is any one of the eigenfunctions, say, φk ?
(6.41)
(6.42)
176 Quantum Mechanics I: The Fundamentals
6.7 APPLICATION TO CERTAIN SYSTEMS In this section we work out the matrix forms of operators of certain quantum mechanical systems. We start with the linear harmonic oscillator.
6.7.1
Linear Harmonic Oscillator
R∞ The matrix elements of x are determined from xnm = −∞ φ∗n xφm dx. The eigenfunctions of linear harmonic oscillator are given by Eq. (4.30). Substituting the expressions for φ∗n and φm in the above integral and introducing the change of variable x = y/α we get Z 2 Nn Nm ∞ xnm = Hn (y)yHm (y)e−y dy . (6.43) 2 α −∞ The value of the integral in the above equation is given by Z ∞ √ n−1 2 Hn (y)yHm (y)e−y dy = π 2 n!δm,n−1 + 2n (n + 1)!δm,n+1 −∞ √ n−1 n!, for m = n − 1 √π 2 π 2n (n + 1)!, for m = n + 1 (6.44) = 0, for m 6= n ± 1 . p When m p = n + 1 we have xnm = (n + 1)/2/α. For m = n − 1 we get xnm = n/2/α. Then p for m = n − 1 pn/2/α , xnm = (6.45) (n + 1)/2/α , for m = n + 1 0, for m 6= n ± 1 , where n, m = 0, 1, · · · . The matrix form of the operator x is √ 1 √0 0 √0 r 1 x00 x01 · · · 0 2 0 √ √ ~ 0 x = x10 x11 · · · = 2 √0 3 2mω ··· ··· ··· 0 0 3 0 · · · ·
··· ··· ··· ··· ···
. (6.46)
The matrix elements of the momentum operator px = −i~d/dx are computed as follows. We obtain Z ∞ 2 2 2 2 d pxnm = −i~Nn Nm Hn (αx)e−α x /2 Hm (αx)e−α x /2 dx dx −∞ Z ∞ 2 = i~Nn Nm Hn Hm y e−y dy −∞ Z ∞ ′ −y 2 − Hn Hm y e dy . (6.47) −∞
Matrix Mechanics 177 The value of the first integral in Eq. (6.47) is given by Eq. (6.44) while that of the second integral is Z ∞ √ 2 ′ Hn (y)Hm (y)e−y dy = 2m π2n n!δn,m−1 , (6.48) −∞
′ where we have used the relation Hm (y) = 2mHm−1 (y). Now Eq. (6.47) becomes √ pxnm = i~Nn Nm π 2n−1 n!δm,n−1 + 2n (n + 1)!δm,n+1 −2m2n n!δn,m−1 ] . (6.49)
Or, pxnm
√ Nn Nn+1 √π [2n (n + 1)! − 2(n + 1)2n n!] , Nn Nn−1 π 2n−1 n!, = i~ 0,
Using the values of Nn and Nn+1 we get p n/2 , α p pxnm = i~ −α (n + 1)/2 , 0,
m=n+1 m = n − 1 (6.50) m 6= n ± 1 .
for m = n − 1 for m = n + 1 for m 6= n ± 1 .
(6.51)
Comparison of Eqs. (6.45) and (6.51) indicates that the matrix of px is analogous to the matrix of x with 1/α in Eq. (6.45) is replaced by −i~α. The matrix form px is √ 0 − 1 0 0 0 ··· √ √ r 1 − 2 0 ··· √0 √0 m~ω 0 . (6.52) px = i 2 0 − 3 0 · · · √ √ 2 0 0 3 0 − 4 ··· · · · · · ··· The operators a and a† given by Eq. (4.46) are obtained as √ 0 1 √0 0 0 ··· 0 1 0 2 √0 0 ··· (mωx + ipx ) = a= √ 0 0 0 3 0 ··· 2mω~ · · · · · ···
(6.53)
and
a =√ (mωx − ipx ) = 2mω~ †
1
√0 1 0 0 ·
0 0 √ 2 0 ·
0 0 √0 3 ·
0 0 0 0 ·
0 0 0 0 ·
··· ··· ··· ··· ·
. (6.54)
178 Quantum Mechanics I: The Fundamentals The Hamiltonian operator H = ~ω(a† a + aa† )/2 is 1 0 0 0 0 2 3 0 0 0 0 H = ~ω 2 5 0 0 0 0 2 · · · · ·
6.7.2
then worked out to be ··· ··· . (6.55) ··· ···
Particle in a Box
We consider the infinite height potential V (x) = 0 for p 0 < x < a and ∞ elsewhere. The normalized eigenfunctions are φn (x) = 2/a sin(nπx/a), n = 1, 2, · · · . The matrix elements of xnm are obtained as Z Z 2 a nπx mπx 2a π xnm = sin x sin dx = 2 y sin ny sin my dy . a 0 a a π 0
The value of the integral in the above equation is obtained as Z π Z π a xnm = y cos(n − m)y dy − y cos(n + m)y dy π2 0 0 a 1 1 n−m n+m = (−1) −1 − (−1) −1 π 2 (n − m)2 (n + m)2 4anm n−m = − 1 , n 6= m, n, m = 1, 2, · · · . (6.56) 2 (−1) 2 2 2 π (n − m )
When n = m we find xnm
a = 2 π
Z
0
π
y dy −
Z
π
y cos 2ny dy =
0
a . 2
(6.57)
Combining Eqs. (6.56) and (6.57) we write −8anm/(π 2 (n2 − m2 )2 ) , for m 6= n, n − m = odd a/2 , for m = n xnm = 0, for m 6= n, n − m = even .
The matrix form of x is π2 −32/9 a −32/9 π2 x= 0 −96/25 2π 2 · ·
0 −96/25 π2 ·
−64/225 0 −192/49 ·
··· ··· . ··· ···
(6.58)
(6.59)
Next, we evaluate the matrix elements of the momentum operator px . We have Z 2i~mπ a nπx mπx pxnm = − sin cos dx 2 a a a 0 2i~nm = (−1)n+m − 1 , n 6= m . (6.60) a(n2 − m2 )
Matrix Mechanics 179 When m = n pxnn = −
2i~n a
Z
π
(6.61)
sin ny cos ny dy = 0 .
0
Combining the Eqs. (6.60) and (6.61) we write −4inm~/(a(n2 − m2 )) , for n + m = odd pxnm = 0, otherwise . The matrix form of px is 0 i~ −8/3 px = 0 a ·
8/3 0 −24/5 ·
0 24/5 0 ·
Since the energy eigenvalues are given by En the matrix form of the Hamiltonian is 1 0 0 π 2 ~2 0 4 0 H= 0 9 2ma2 0 · · ·
16/15 0 48/7 ·
··· ··· . ··· ···
(6.62)
(6.63)
= π 2 ~2 n2 /(2ma2 ), n = 1, 2, · · · 0 0 0 ·
··· ··· . ··· ···
(6.64)
6.8 DIRAC’S BRA AND KET NOTATIONS As mentioned earlier the quantum state of a system can be represented in terms of a vector. The values of the wave functions ψ(X) at various points X are the components of a vector. The components depend on the choice of coordinate system or basis in the vector space. Instead of specifying one or another set of infinitely many components, Dirac introduced a vector notation for the state vector [8]. A state vector is represented by a symbol | i called a ket . For example, ψ(X) is written as |ψi. Corresponding to every vector |ψi there is a vector hψ| called bra vector, the complex conjugate of |ψi. Recall from our discussion in earlier sections of this chapter that any vector can be written in terms of a basis vector. What are the basis of a quantum system? A basis is essentially constituted by the set of vectors with each possible outcome of a measurement of a particular observable of the system. In other words, the eigenvectors of the operator of an observable can always be used as the basis for the vector space. In a three-dimensional vector (in physical space) v = ai + bj + ck where i, j and k are unit vectors along the x, y and z-axis respectively and a, b and c are the components of v along the x, y and z-axes respectively. In the same way, a quantum state |ψi is written as |ψi = c1 |φ1 i + c2 |φ2 i + · · · ,
(6.65)
180 Quantum Mechanics I: The Fundamentals where the unit vectors |φ1 i, |φ2 i, · · · , etc are like the unit vectors i, j, k · · · and the coefficients c1 , c2 · · · are like a, b, c · · · . However, ci ’s are generally complex. We write Eq. (6.65) as X X |ψi = ci |φi i or ci |ii , (6.66) i
i
where |ii denotes a unit vector or a base vector |φi i. Then the bra state is X hψ| = c∗i hi| . (6.67) i
The number of base vectors required in quantum mechanics depends on the situation. In matrix form c1 |ψi = c2 , hψ| = [c∗1 c∗2 · · · ] . (6.68) .. .
6.9 EXAMPLES OF BASIS IN QUANTUM THEORY In this section we give a few examples of basis in quantum mechanics [9].
1. Energy representation Let the discrete energy eigenvalues of a particle are E1 , E2 , · · · . A measurement of energy of the particle would yield one of Ei ’s. The set of basis vectors in this case is |E1 i, |E2 i, · · · . The number of basis vectors is infinite.
2. Position representation
Consider a one-dimensional particle. A measurement of its position would give a real number, say q. Denote |qi the state of the particle for which the outcome of position measurement is q. The basis states are taken as the set {|qi} where −∞ < q < ∞. As the possible values of q are infinite, the number of basis vectors is infinite.
3. Momentum representation Instead of position values we may measure the momentum p of the particle. In this case the set {|pi} where −∞ < p < ∞ forms the basis states. In this representation also the number of basis vectors is infinite.
4. Spin representation Consider a particle whose spin measurement in any direction gives ±~/2. If the spin measurement in a direction e yields ~/2 then the corresponding state is, for example, labeled as |1/2, ei. If the outcome is −~/2 then the state is | − 1/2, ei. The state of the spin component in any direction is then represented in terms of the states |1/2, ei and | − 1/2, ei. Consequently, the number of basis vectors is only two.
Matrix Mechanics 181 The basis vectors in energy, position and momentum representations are ψn (x) ψq (x)
→ |ψn i → |ni or |ii , ψE (x) → |Ei , → |qi , ψp (x) → |pi ,
(6.69a) (6.69b)
respectively. In terms of the basis vectors the state vector in these three representations in ket notation are X X ψ(x) → |ψi = cn |ni or ci |ii , (6.70a) Z ∞ ψ(x) → |ψi = cq |qi dq , (6.70b) −∞ Z ∞ ψ(x) → |ψi = cp |pi dp . (6.70c) −∞
To determine the complex coefficients cn in the linear vector space, define inner product between ψm and ψn as Z ∞ (ψn , ψm ) = ψn∗ ψm dτ = hψn |ψm i or hn|mi (6.71) −∞
and is a number δnm . Observe that when the bra is on the left of a ket, that is, (hψn |)(|ψm i), then they contract to give a number. hψn | and |ψm i are said to be orthogonal if hψn |ψm i = 0. The basis vectors are orthonormal if hφi |φj i = δij . On taking the inner product of Eq. (6.66) successively with |φi i and on using hφi |φj i = δij we obtain ci = hφi |ψi which gives the value of the component of |ψi along |φi i. The c’s in Eq. (6.70) are given by cn = hn|ψi ,
cq = hq|ψi ,
cp = hp|ψi .
(6.72)
If λi is the eigenvalue corresponding to the eigenvector |φi i of an operator Λ then the probability of getting λi as an eigenvalue in a measurement is P (λi ) = |ci |2 = |hφi |ψi|2 . (6.73) P 2 Since the total probability is 1 we have |ci | = 1. If a system is in the state |φi i then the state |ψi no longer exists. If |ψi is normalized to unity then X X X hψ|ψi = hψ|φi ihφi |ψi = |hφi |ψi|2 = |ci |2 = 1 . (6.74)
6.10 PROPERTIES OF KET AND BRA VECTORS
An operator by definition acts on one ket to produce another ket. In Dirac’s notation there is a useful way of representing linear operators, known as the outer product representation. For example, the inner product hv|wi is a complex number whereas |vihw| called the outer product is an operator. If |w′ i is some ket then the action of |vihw| on |w′ i is defined by |vihw|w′ i = hw|w′ i|vi. What is the rule to obtain Hermitian conjugate of an expression? The rule contains
182 Quantum Mechanics I: The Fundamentals 1. reversing the order of the terms in the expression and 2. changing (a) the numbers into their complex conjugates (b) the operators into their adjoints, (c) the kets into bras and bras into kets. The Hermitian conjugate of λ|ψ2 ihψ1 |AB † C is λ∗ C † BA† |ψ1 ihψ2 |. Some of the properties of the ket and bra vectors are summarized below where A and B are Hermitian operators: A|ψi = hAi =
hψm |ψn i = (A + B)|ψi =
hq|q ′ i =
|ψ ′ i , hψ|A = hψ ′ | , hψ|A|ψi , Anm = hψn |A|ψm i ,
(6.75a) (6.75b)
hψn |ψm i∗ , hψn |ψm i = δnm , A|ψi + B|ψi ,
(6.75c) (6.75d)
δ(q − q ′ ) ,
(6.75e)
hp|p′ i = δ(p − p′ ) .
The time-dependent Schrödinger equation is written as ∂ |ψi = H|ψi . (6.76) ∂t The time-independent Schrödinger equation for |ψn i is H|ψn i = En |ψn i. Then i~
|ψn (t)i = e−iEn t/~ |ψn (0)i .
(6.77)
The arbitrary state |ψi is expressed as X |ψi = cn e−iEn t/~ |ψn i .
(6.78)
n
Solved Problem 3: Compare two-dimensional ordinary vectors and quantum state vectors. The following table gives a comparison of ordinary and quantum state vectors. S.No.
Ordinary Vectors
1.
X = ai + bj
2.
i·j=0 i·i= j·j= 1
3. 4.
5.
A·B= B·A
A = a1 i + a2 j B = b1 i + b2 j A · B = a1 b 1 + a2 b 2
X · i = a, X · j = b, X = i(X · i) + j(X · j)
Quantum State Vectors P |ψi = cn |ni hn|mi = 0 hn|ni = hm|mi = 1
hx|ψi = hψ|xi∗ P |ψi = P cn |ni |χi = P dn |ni hχ|ψi = cn dn |ψi =
P
|nihn|ψi
Matrix Mechanics 183
6.11 HILBERT SPACE Suppose a physical system is described by a set of linearly independent finite norm states denoted by |φ1 i, |φ2 i, · · · . The set {|φn i} represents all the possible outcomes of measurements of an observable A. Assume that 1. there is no liner combination of these vectors vanishing except when all the coefficients are zero and P 2. the set is complete so that |ψi = cn |φn i.
Then the space of the ket vectors |ψi is called a Hilbert space [10]. The basis of the Hilbert space are the set |φn i. The number of cn ’s are the component of |ψi and gives a representation of |ψi. Thus, we write the cn ’s in a column (or row) vector such that a state function is represented by a column (or a row) vector in the Hilbert space. In this space we imagine an axis for each φn (x) and treat cn the component of the vector in the direction of that axis. The quantum states in Hilbert space are called state vectors. Suppose |φ1 i and |φ2 i are two states in Hilbert space. The linear superposition of themm is |ψi = |φ1 i + |φ2 i. We cannot form a new state by superposition of a state with itself. The state c1 |φ1 i + c2 |φ1 i is not a new state. This corresponds to the same state that |φ1 i does. That is, in the Hilbert space, we can multiply a state vector by a nonzero complex number, √ however, the physical state is unchanged. The vectors |ψi, 5|ψi, −|ψi, − 2|ψi, i|ψi and (2 + i3)|ψi all represent the same physical state. Their magnitudes or lengths are physically unimportant. In this sense state vectors differ from conventional vectors. A single point of Hilbert space represents the quantum state of the system. The origin does not represent a state. A state vector is represented by a line passing through the origin. Since |ψi and 5|ψi are equivalent, a line passing through the origin represents all possible multiples of a state vector. Each dimension of the Hilbert space corresponds to one of the rows of the one-column matrix describing the state. That is, each dimension corresponds to one of the various independent physical states of a quantum state. Essentially, the quantum description of a system with observable assuming continuous values requires a continuous states. In this case the dimension of the Hilbert space is infinite. Thus, the dimension of the Hilbert space for energy, position and momentum representations is infinite. However, there are many systems with discrete and finite sets of values of observables so that the dimension of the associate Hilbert space is finite. In spin representation, for s = 1/2 the dimension is two. Another p example of an observable with finite values is angular momentum J = j(j + 1)~. The corresponding dimension of the Hilbert space is 2j + 1. Some mathematical features of finite-dimensional Hilbert space for the position and momentum of a particle on a one-dimensional lattice is discussed in ref.[11]. The formulation of quantum mechanics in a finite-dimensional Hilbert space has found applications in quantum cryptography, quantum computers and quantum optics.
184 Quantum Mechanics I: The Fundamentals Let us see how position states and momentum states are represented in a same Hilbert space. Consider a Hilbert space with each coordinate axis corresponds to a possible position state of a single particle. The momentum state can also be represented in the same Hilbert space. Momentum states are expressed as combination of position states. Therefore, each momentum state corresponds to an axis going off-diagonally tilted with respect to the position-space axes. The set of all momentum states forms a new set of axes. Transformation from the position-space axes to the momentum-space axes involves a rotation in the Hilbert space.
Solved Problem 4: Compare the Hilbert space with the N -dimensional vector space. The following table gives a comparative study of N -dimensional vector space and Hilbert space. S.No. Vector Space
Hilbert Space
1.
The basis vectors of a vector space are unit vectors.
2.
The orthogonality condition for the basis vectors is vi · vj = δij , i, j = 1, 2, · · · , N where δ is the Kroneckerdelta function.
The basis vectors of the Hilbert space are the functions φn (x), n = 1, 2, · · · or φk (x), −∞ < k < ∞.
3.
The completeness property for aPvector F is given by N F = i=1 fi vi where fi ’s are the N components of F.
4.
The projection formula is fi = vi · F
The condition is R ∞ orthogonality ∗ φ (x)φ (x) dx = δ(m − n) n m −∞ where δ(m−n) is the Dirac-delta function. For a function F (x) the completeness property is F (x) = R∞ f (k)φ k (x) dk if the basis is −∞ continuous. P For discrete basis F (x) = n fn φn (x).
The R ∞ projection formula is f (k) = φ (x)F (x) dx. −∞ k
6.12 PROJECTION AND DISPLACEMENT OPERATORS Let the eigenkets |ii of an operator Ω have eigenvalues wi . Since the setPof kets |ii form a complete set, any arbitrary ket |αi is expanded as |αi = i ai |ii where ai = hi|αi is a number. We write X X |αi = |iihi|αi = Pi |αi , (6.79) i
i
Matrix Mechanics 185 where Pi = |iihi| is called the projection operator [12] for the ith state. In (6.80)
Pi |αi = |iihi|αi
the operator Pi projects only the particular eigenstate |ii of Ω among |αi. Hence it is called a projection operator. From (6.79) and (6.80) we find that X X Pi |αi = |iihi|αi = |αi . (6.81) i
The result is
P
i
Pi = I. What does it state? We have Pi2 |αi = |iihi|iihi|αi = |iihi|αi = Pi |αi .
(6.82)
This gives Pi2 = Pi . Further, Pj Pi |αi = |jihj|iihi|αi = 0. Therefore, Pj Pi = 0. Combining Pi2 = Pi and Pj Pi = 0 we get Pi Pj = Pj Pi = δij Pi . Any operator A can be written as A = IAI. Therefore, the completeness relation is X X XX XX A= |iihi|A| |jihj| = |iihi|A|jihj| = |iihj|Aij , (6.83) i
j
i
j
i
j
where Aij is the matrix element hi|A|ji. Thus, A is written as a linear combination (with coefficients Aij ) of the operator |iihj|. If A = Ω, then hi|Ω|ji = ωj hi|ji = ωj δij and XX X Ω= |iihj|ωj δij = Pi ωi . (6.84) i
j
i
So, any observable can be written as a weighted sum of projection operators to their own eigenstates. The weighting factors More P are the eigenvalues. P generally, for any function of Ω, we have f (Ω) = i f (ωi )|iihi| = f (ωi )Pi . When the eigenstates of an observable are used as bases for representations then the observable is defined in relation to a coordinate system, say, S : oxyz in ordinary space. A change in the observable arises if there is a change in the coordinate system. This results in a transformation of the basis in the Hilbert space. Suppose the basis transformation is made by a translation (shifting the origin without altering the orientation of the axes). Let us investigate the effect of translation on a wave function of a system. Let us consider that a one-dimensional system is displaced through a distance along the x-axis. The final wave function is unknown because it can be multiplied by a phase factor. Since superposition relations between eigenstates are unchanged under the displacements we write ψUD = φaUD + φbUD ,
(6.85)
where ψUD is the wave function of the undisplaced system. Here assume that the system has only two eigenstates φaUD and φbUD . Equation (6.85) for the
186 Quantum Mechanics I: The Fundamentals displaced system is ψD = φaD +φbD . Now, define an operator called displacement operator [13], the effect of it on ψUD is to give ψD : DψUD = ψD . An example is Dψ(x) = ψ(x + δ). The displacement operator is widely used in quantum computation. Let us calculate hD† Di. We obtain Z ∞ Z ∞ ∗ ∗ hD† Di = ψUD D† DψUD dx = ψD ψD dx = 1 . (6.86) −∞
−∞
Since this is valid for any eigenfunction we have DD† = 1, D is a unitary. Suppose γ is an operator such that γφaUD = φbUD and γd φaD = φbD where γd is the displaced operator [13]. We have γd φaD = DφbUD = DγφaUD = DγD† φaD .
(6.87)
Thus, γd = DγD† . For an infinitesimal displacement δ of the system we write lim
δ→0
DφaUD − φaUD (D − 1)φaUD (φaD − φaUD ) = lim = lim . δ→0 δ→0 δ δ δ
Defining
(6.88)
D−1 (6.89) δ where θ is a phase angle we write
d = lim
δ→0
and since D can be replaced by Deiθ
Deiθ − 1 D − 1 + iθ = lim = d + iγ , δ→0 δ→0 δ δ lim
(6.90)
where γ = limδ→0 (θ/δ). Is the operator d Hermitian? To answer this, for small δ we write D = 1 + δd. Then DD† = 1 becomes (1 + δd† )(1 + δd) = 1. Neglecting δ 2 we obtain d† + d = 0 or d† = −d. Thus the displacement operator d is anti-Hermitian.
Examples:
Assume that the system is displaced along x-direction by an amount, say, δx . The old and new wave functions are ψ(x) and ψ(x − δx ) respectively. Then xd = x − δx . Replacing x by γ we have γd = γ − δx or (γd − γ)/δx = −1. Now, the above equation becomes [d, γ] = [dx , x] = −1. That is, xdx − dx x = 1. Multiplication on both sides by i~ gives xi~dx − i~dx x = i~[x, dx ] = i~ .
(6.91)
Comparison of this equation with xpx − px x = i~ gives dx = −d/dx. Thus, the result is that the displacement operator in the x-direction is proportional to the momentum conjugate to x. Hence, px is the generator of translations along the x-axis. Since any finite translation along the x-axis can be constructed from a series of many infinitesimal translations we write N iδx px = e−ipx δx /~ . (6.92) D(δx ) = lim 1 − N →∞ N~
Matrix Mechanics 187 Alternatively, we write ψ(x + δx ) as a Taylor series 1 ψ(x) + δx ψ ′ (x) + δx2 ψ ′′ (x) + · · · 2! # "∞ X 1 ipx δx n ψ(x) n! ~ n=0
ψ(x + δx ) = = =
D(δx )ψ(x) .
(6.93)
In general, for a three-dimensional wave function we write D(δj ) = e−ipj δj /~ ,
D(δj )ψ(xj ) = ψ(xj + δj ) ,
(6.94)
where j = x, y or z. Next, consider the change of representation induced by a rotation of coordinate axes, say, from S : oxyz to S ′ : ox′ y ′ z ′ about oz through an angle θ. We have x′ = x cos θ + y sin θ,
y ′ = −x sin θ + y cos θ,
z′ = z .
(6.95)
Since the wave function is independent of the orientation of the coordinate axes we write ψ ′ (X′ ) = ψ(X). Substituting Eq. (6.95) and assuming θ to be infinitesimally small so that cos θ ≈ 1 and sin θ ≈ θ we obtain ψ ′ (x + yθ, y − xθ, z) = ψ(x, y, z) .
(6.96)
Replacing x by x − yθ and y by y + xθ in Eq. (6.96) we get ψ ′ (x, y, z) = =
ψ(x − yθ, y + xθ, z) ∂ ∂ ψ(x, y, z) + θ x −y ψ(x, y, z) , ∂y ∂x
(6.97)
where higher-order terms in θ are neglected. The terms in the square-bracket of Eq. (6.97) is the z-component of orbital angular momentum without −i~. Then iθ ′ (6.98) ψ (X) = 1 + Lz ψ(X) . ~ Thus, the operator Lz /~ plays the role of the generators of infinitesimal rotations. For a rotation about an arbitrary axis along the unit vector n we have iθ ψ ′ (X) = 1 + n · L ψ(X) . (6.99) ~
6.13 CONCLUDING REMARKS In this chapter we outlined the matrix mechanics for quantum mechanical systems. In this theory the behaviour of a quantum mechanical system is
188 Quantum Mechanics I: The Fundamentals determined by the matrices of p and X. The Hamiltonian is a diagonal matrix having diagonal elements En , the energy eigenvalues. To solve Hψ = Eψ as a matrix problem, we need to find an appropriate linear vector space. We note that instead of characterizing a vector ψ by its components with respect to a particular coordinate system, it is convenient to use the coordinate independent vector. We can characterize a vector on some basis {φn }. In this case the vector ψ itself does not change, only its components with respect to the coordinate axes change under the transformation from one basis to another basis, for example, say from {φn } to {φn′ }. We have shown that unitary transformation relates the representations in different basis systems. A unitary transformation from one representation of |ψi to another corresponds to a rotation of axis in the Hilbert space without change in the state vectors. For example, the energy representation corresponds to choosing axes in such a way that a state vector oriented along one of these axes is an eigenstate of the Hamiltonian. Though the wave mechanics and matrix mechanics are logically distinct, they are equivalent in all respects.
6.14 BIBLIOGRAPHY [1] W. Heisenberg, Z. Phys. 33:879, 1925. [2] M. Born, W. Heisenberg and P. Jordan, Z. Phys. 35:557, 1926. [3] E. Schrödinger, Annalen der Physik 79:734, 1926. [4] C. Eckart, Phys. Rev. 28:711, 1926. [5] D.F. Styer, The Strange World of Quantum Mechanics. Cambridge University Press, Cambridge, 2000 pp.127. [6] G.B. Arfken, Mathematical Methods for Physicists. Academic Press Florida, 1985. [7] G.H. Golub and C.F. Van Loan, Matrix Computations. Johns Hopkins, Baltimore, 1996. [8] P.A.M. Dirac, The Principles of Quantum Mechanics. Oxford, New York, 1958. [9] R.R. Puri, Physics News January−March 2002, pp.26-35. [10] R. Courant and D. Hilbert, Methods of Mathematical Physics. Volume V. Interscience, New York, 1953. [11] A.C. de la Torre and D. Goyeneche, Am. J. Phys. 71:49, 2003. [12] L.I. Schiff, Quantum Mechanics. McGraw-Hill, Singapore, 1968. [13] R. Fitzpatrick, Quantum Mechanics. Unpublished, 2010.
Matrix Mechanics 189
6.15 EXERCISES 6.1 Find out whether the set of vectors ψ1 = 3i − j + 2k, ψ2 = −2i + j, ψ3 = j + 2k are linearly independent. P P 6.2 If |φi = i Ci |ii, |χi = i Di |ii and A is an operator write down hχ|A|φi. 6.3 If A|φi i = λi |φi i, i = 1, 2, · · · , n write the expression for hψ|A|ψi where |ψi is a general state. 6.4 The wave function of a system is given by |ψ(t)i = a1 (t)|Ai + a2 (t)|Bi, t ≥ t1 where a1 (t) = e−iα(t−t1 ) cos[ω(t − t1 )/2], a2 (t) = −i e−iα(t−t1 ) sin[ω(t − t1 )/2] with α and ω being two constants. Check whether the wave function is normalised. 1 and |v2 i = 6.5 Suppose V is a vector space with basis vectors |v1 i = 0 0 . If A is a linear operator such that A|v1 i = |v2 i and A|v2 i = |v1 i 1 find the matrix representation of A. 6.6 (a) Show that a two-dimensional vector space can be spanned by 1 0 1 1 ′ the two set of basis |v1 i = , |v2 i = and |v1 i = √ , 2 1 0 1 1 1 ′ |v2 i = √ . 2 −1 (b) Find the transformation matrix between the two basis. 2 (c) Express the vector |vi = in both the basis. −3 6.7 Show that the basis (1, x, x2 , · · · ) in the domain −1 ≤ x ≤ 1 is not orthogonal. Find an orthogonal basis using Gram–Schmidt procedure. 6.8 Express the Schrödinger equation i~∂|ψi/∂t = H|ψi in terms of H and U. 6.9 If |ni and |mi are two arbitrary state vectors and |ii are the complete P set of basis vectors show that hn|mi = i hn|iihi|mi.
6.10 For a set of three operators A, B and C if AB = C then what is [A][B]? 6.11 Show that if A is an anti-Hermitian matrix then eA is unitary. 6.12 State the condition for a Hermitian matrix to become unitary.
190 Quantum Mechanics I: The Fundamentals 6.13 Express the definition of Hermitian operator in Dirac notation. 6.14 Write the Hermitian conjugate of λA|ψ1 ihψ2 |B † . P 2 6.15 Prove that hAm i = n am n |cn | where a’s are eigenvalues of A.
6.16 If λi is the eigenvalue P of an operator Λ corresponding to the eigenvector |φi i and if |ψi = i ci |φi i then find the probability of getting λi as an eigenvalue in a measurement. 6.17 Show that the eigenvalues of an operator represented in two different basis connected by a transformation matrix U are identical. P θσz where σz is the Pauli matrix 6.18 Using f (Ω) = i f (ωi )|iihi| find e 1 0 . 0 −1 6.19 Find the Pauli’s spin eigenkets of the matrices forspin−1/2 systems: 0 1 0 −i 1 0 , σy = and σz = . Find σx = 1 0 i 0 0 −1 the unitary transformation matrices from eigenket basis of (a) σz to σx , (b) σz to σy and (c) σx to σy .
1 1 (mωx + ipx ), a† = √ (mωx − ipx ) being the 6.20 With a = √ 2mω~ 2mω~ lowering and raising operators, respectively, of the linear harmonic oscillator, show that the matrix elements hm|A|ni vanish unless r−s = m−n if A is an operator which contains r factors of a† and s factors of a in any order. Also calculate hn|x4 |ni.
CHAPTER
7
Various Pictures and Density Matrix
7.1 INTRODUCTION In classical mechanics the state of a system is defined by the numerical values of a set of dynamical variables such as the position coordinates and the velocities. In quantum mechanics the state of a system at an instant is represented by a unit vector in a Hilbert space, where a set of axes are the eigenvectors of a complete set of observables. Here the observable are not the state vectors but the expectation values of a set of self-adjoint or Hermitian operators corresponding to the dynamical variables. The variation of hAi with time is seen as arising in any one of the following ways: 1. The state vector changes with time, but the operator A remains unchanged. 2. The operator changes with time, whereas the state vector remains unchanged. 3. Both the operator A and the state vector change with time. These three cases are called Schrödinger , Heisenberg and interaction pictures respectively. The time evolution of ensembles is described by means of a density matrix. We discuss the above mentioned pictures in this chapter.
7.2 SCHRÖDINGER PICTURE The description of quantum dynamics where the wave function is timedependent and the operators describing the observables are time-independent is known as the Schrödinger picture.
191
192 Quantum Mechanics I: The Fundamentals
7.2.1
Matrix Form of Schrödinger Equation
The time-dependent Schrödinger equation in Dirac’s notation is given by Eq. (6.76). Its Hermitian adjoint is −i~
∂ hψ| = hψ|H . ∂t
(7.1)
Since the dependence of |ψi on coordinate or other variables is not explicit we replace ∂/∂t by d/dt. If H is independent of time t then the solution of Eq. (7.1) is |ψ(t)i = e−iHt/~ |ψ(0)i and hψ(t)| = hψ(0)|eiHt/~ .
(7.2)
e−iHt/~ is also an operator (because H is an operator) and write it as U (t) = e−iHt/~ = 1 +
(−iHt) 1 (−iHt) (−iHt) + + ··· . ~ 2 ~ ~
(7.3)
It is possible to represent U (t) as a square matrix. What is the effect of U (t) = e−iHt/~ on |ψ(0)i? Its significant effect is to change |ψ(0)i into |ψ(t)i. Therefore, the operator U (t) is termed as the evolution operator . We note † that if H is Hermitian then U † U = ei(H −H)t/~ = e0 = 1 implying that U is unitary. Equation (7.2) shows that the time evolution is represented by a unitary transformation. Because the origin of T has no significance U depends on time differences only. From Eq. (7.2) for t = t1 we write |ψ(t1 )i = e−iHt1 /~ |ψ(0)i
(7.4)
|ψ(0)i = eiHt1 /~ |ψ(t1 )i .
(7.5)
|ψ(t)i = e−iH(t−t1 )/~ |ψ(t1 )i .
(7.6)
and Then In Eqs. (6.76) and (7.2) the operator is time-independent whereas the wave function is time-dependent. This representation of a quantum mechanical system is referred as a Schrödinger picture. In this picture, all operators are functions of (the time-independent operators) xS and pxS where the subscript S is used to denote the Schrödinger representation.
7.2.2
Equation of Motion in the Schrödinger Picture
We proceed to obtain the evolution equation of an operator. The expectation value of an observable AS is hAS i = hψS (t)|AS |ψS (t)i .
(7.7)
Various Pictures and Density Matrix 193 The time-dependent kets are given by Eq. (7.2). Differentiation of Eq. (7.7) with respect to time gives ∂AS dψS d d hAS i = hψS |AS |ψS i + ψS ψS + ψS AS . (7.8) dt dt ∂t dt
Using Eqs. (6.76) and (7.1) in the above equation we get d hAS i = dt = = =
1 [−hψS |HAS |ψS i + hψS |AS H|ψS i] + i~ 1 ∂AS hψS |AS H − HAS |ψS i + i~ ∂t 1 ∂AS hψS | [AS , H] |ψS i + i~ ∂t ∂AS 1 + h [AS , H] i , ∂t i~
∂AS ∂t
(7.9)
where [AS , H] is the commutator of AS and H. The first term in the rightside of this equation arises from the explicit time-dependence in AS and the second term is from the time development of the states. If AS has no explicit time-dependence then d 1 i hAS i = h [AS , H] i or h [H, AS ] i . dt i~ ~
(7.10)
This is the equation of motion of the expectation value of a dynamical variable AS in the Schrödinger representation and is called the Schrödinger equation of motion. What do we infer if [AS , H] = 0? In this case dhAS i/dt = 0 and the operator is said to be an invariant. What is the significance of an invariant property of an operator? We deduce the following two major results from Eq. (7.10). 1. The expectation value of an observable whose operator commutes with the Hamiltonian is time-independent. 2. If the system is initially in an energy eigenstate then hAi will become time-independent. The system will remain in that eigenstate forever. For convenience define the total time derivative of any operator as dAS ∂AS 1 = + [AS , H] . dt ∂t i~
(7.11)
It is easy prove that dA dB d (AB) = B+A , dt dt dt
d df dA (f (t)A) = A+f dt dt dt
(7.12)
194 Quantum Mechanics I: The Fundamentals These imply that the total derivative satisfies the same algebraic rules as ordinary derivatives. Because of this the total time derivative is not defined as a rate of change of anything (Does it matter? ). We illustrate below the use of the total time derivative for a free particle and linear harmonic oscillator. For a few other interesting systems the reader may refer to ref.[1].
Solved Problem 1: Obtain the evolution equations for hxS i and hpxS i and their solutions for a free particle with H = p2x /(2m). Making use of the relations [xS , P (pxS )] = i~
∂P , ∂pxS
[pxS , X(xS )] = −i~
∂X ∂xS
(7.13)
from Eq. (7.10) we write 1 1 ∂ p2xS 1 d hxS i = h[xS , H(pxS )]i = i~ = hpxS i . dt i~ i~ ∂pxS 2m m
(7.14a)
Further, d 1 1 hpxS i = h[pxS , H]i = dt i~ i~
∂H − i~ ∂xS
=0.
(7.14b)
Thus, hpxS i is independent of time. The solution of Eqs. (7.14) is hxS (t)i =
1 hpxS (0)it + hxS (0)i , m
hpxS (t)i = hpxS (0)i .
(7.15)
This solution is similar to the classical solution xS (t) = x0 + v0 t, v(t) = v(0).
Solved Problem 2: For the linear harmonic oscillator with H = p2xS /(2m) + kx2S /2 find the equations of motion for hxS i and hpxS i. Also, determine the energy eigenvalues. The equations of motion for hxS i and hpxS i are obtained as d 1 1 ∂H 1 hxS i = h[xS , H]i = i~ = hpxS i , (7.16a) dt i~ i~ ∂pxS m d 1 1 ∂H hpxS i = h[pxS , H]i = − i~ = −mω 2 hxS i , (7.16b) dt i~ i~ ∂xS p where ω = k/m . We rewrite the above equations in terms of hxS i alone as d2 hxS i + ω 2 hxS i = 0 . dt2
(7.17)
Various Pictures and Density Matrix 195 Its solution is hxS (t)i = hxS (0)i cos ωt +
1 hpxS (0)i sin ωt . mω
(7.18)
To obtain energy levels of the oscillator we calculate Hn′ n given by Hn′ n =
1 1 p2 (0) n′ n + mω 2 x2S (0) n′ n 2m xS 2
(7.19)
together with the eigenfunctions (4.30). We have Z ∞ 2 2 ψn∗ ′ ψn dx pxS (0) n′ n = −~ −∞ 1 1 2 2 = 2m n + ~ωδn′ n − mω xS (0) n′ n . (7.20) 2 2 Hence, Hn′ n = (n + 1/2) ~ωδn′ n and Hnn = En = (n + 1/2) ~ω, n = 0, 1, · · · .
7.3 HEISENBERG PICTURE Consider the circumstances where an operator is time-dependent and the wave function is time-independent.
7.3.1
Operators and Kets in a Heisenberg Picture
Consider the expectation value of the operator A given by (7.21)
hAS i = hψS (t)|AS |ψS (t)i . Using Eq. (7.2) we have hAS i
= ψS (0) eiHt/~ AS e−iHt/~ ψS (0) = hψS (0)|AH (t)|ψS (0)i ,
(7.22a)
where AH (t) = eiHt/~ AS e−iHt/~ .
(7.22b)
Suppose we define |ψH (t)i such that hAH (t)i = hψH (t)|AH |ψH (t)i .
(7.22c)
Comparison of Eqs. (7.22a) and (7.22c) gives hAH (t)i = hAS i provided |ψH (t)i = |ψH (0)i ,
AH (t) = eiHt/~ AS e−iHt/~ .
(7.23)
In this case the ket |ψH (t)i is time-independent while AH is time-dependent. This representation is called Heisenberg representation. Here hAS i = |AH (t)i = |ψS (0)|AH (t)|ψS (0)i .
(7.24)
196 Quantum Mechanics I: The Fundamentals The result is that the expectation value of a time-independent operator AS in a time-dependent state is same as the expectation value of a time-dependent operator AH (t) in the time-independent state |ψS (0)i. Though, the operator in Eq. (7.22b) is written as AH (t), it is actually function of x(0), p(0) and t. It should be noted from Eq. (7.22b) that H and any operator that commutes with H are same in both Schrödinger and Heisenberg pictures. This is because AH = eiHt/~ AS e−iHt/~ = eiHt/~ e−iHt/~ AS = AS . In the Heisenberg picture the kets and operators are labeled with the subscript H. Because the kets are time-independent we simply write |ψH i. The kets are given by |ψH (t)i = |ψH (0)i = |ψS (0)i. From this and from Eq. (7.2) it is clear that the kets in the Heisenberg picture are simply the kets in the Schrödinger picture at t = 0. What about the operators in these two representations? Do they coincide at all times t or only at t = 0?
7.3.2
Equations of Motion
Differentiation of AH = eiHt/~ AS e−iHt/~ with respect to t gives d AH dt
=
=
i ∂AS −iHt/~ HeiHt/~ AS e−iHt/~ + eiHt/~ e ~ ∂t i − eiHt/~ AS He−iHt/~ ~ i iHt/~ ∂AS −iHt/~ e e + eiHt/~ (HAS − AS H) e−iHt/~ . (7.25) ∂t ~
Frequently AS is function of x and px only. If it does not contain t explicitly then ∂AS /∂t = 0 so that d AH dt
= = = =
1 iHt/~ e (AS H − HAS ) e−iHt/~ i~ 1 iHt/~ e AS e−iHt/~ H − HeiHt/~ AS e−iHt/~ i~ 1 (AH H − HAH ) i~ 1 [AH , H] . i~
(7.26)
If AS has explicit time dependence then d AH dt
=
∂AH 1 + [AH , H] . ∂t i~
(7.27)
Equation (7.27) governs time evolution of an operator in Heisenberg representation and is called the Heisenberg equation of motion. What is the connection between the total time derivative of operators in Schrödinger and Heisenberg pictures? We obtain dAS dAH −iHt/~ = eiHt/~ e , dt dt
dAH dAS iHt/~ = e−iHt/~ e . dt dt
(7.28)
Various Pictures and Density Matrix 197 Thus, the total time derivative of an operator is the Schrödinger picture version of the time derivative of its Heisenberg picture version. Because the canonical operators qi and pi satisfy the rules [qi , pj ] = i~δij , we can prove that [q, G(p)] = i~
∂G , ∂p
[p, F (q)] = −i~
∂F . ∂q
(7.29)
Equation (7.26) gives dqi dt dpi dt
= =
1 ∂H [qi , H] = , i~ ∂pi 1 ∂H [pi , H] = − . i~ ∂qi
(7.30a) (7.30b)
Equations (7.30) have the same form of Hamilton’s classical canonical equation of motion. If the potential V has only a constant or linear or quadratic term corresponding to the cases of no force, constant force and harmonic motion, respectively, then it is easy to establish that hqi i and hpi i in any state evolve exactly like those of classical mechanics.
Solved Problem 3: For the linear harmonic oscillator obtain the evolutions equations of xH and pxH . The operators xH and pxH are given by xH (t) = eiHt/~ xS e−iHt/~ ,
pxH (t) = eiHt/~ pxS e−iHt/~ .
(7.31)
The p Hamiltonian in the Heisenberg picture is H = p2xH /(2m) + kx2H /2 where ω = k/m . The evolution equations of xH and pxH are obtained as follows: i~
dxH dt
=
[xH , H]
=
eiHt/~ xS e−iHt/~ H − HeiHt/~ xS e−iHt/~
=
eiHt/~ [xS , H] e−iHt/~ .
Then dxH dt
= = =
−
i iHt/~ 2 e pxS , xS e−iHt/~ 2m~
1 iHt/~ e pxS e−iHt/~ m 1 pxH (t) . m
(7.32a)
198 Quantum Mechanics I: The Fundamentals Next, dpxH dt
= = = = = =
1 [pxH , H] i~ 1 iHt/~ e pxS e−iHt/~ H − HeiHt/~ pxS e−iHt/~ i~ 1 iHt/~ pxS , p2xS /2m e−iHt/~ + eiHt/~ pxS , kx2S /2 e−iHt/~ e i~ 1 iHt/~ pS , kx2S /2 e−iHt/~ e i~ 1 ∂ 2 −iHt/~ − keiHt/~ x e 2 ∂xS S −k 2 xH . (7.32b)
Equations (7.32) are analogous in structure to the classical ones.
7.4 INTERACTION PICTURE Now, we take up the case of both the operators and the wave function varying with time.
7.4.1
Kets and Operators
Suppose the Hamiltonian of a system is split into two parts H0 and H ′ , where H0 is independent of time and H ′ is time-dependent. Such could be the case if a system is subjected to an external time-dependent field. In this case, the Schrödinger equation is i~
∂ |ψS i = (H0 + H ′ )|ψS i . ∂t
(7.33)
Let us introduce the interaction representation as |ψint i = eiH0 t/~ |ψS i .
(7.34)
Replacing |ψS i using Eq. (7.34) in Eq. (7.33) we get i~e−iH0 t/~
∂ |ψint i = H ′ e−iH0 t/~ |ψint i . ∂t
(7.35)
Premultiplying the above equation by eiH0 t/~ gives i~
∂ ′ |ψint i = eiH0 t/~ H ′ e−iH0 t/~ |ψint i = Hint |ψint i , ∂t
(7.36a)
where ′ Hint = eiH0 t/~ H ′ e−iH0 t/~ .
(7.36b)
Various Pictures and Density Matrix 199 For all operators we write Aint = eiH0 t/~ AS e−iH0 t/~ .
(7.37)
′ The time evolution of ket is governed by the part Hint while that of the operator is by the unperturbed part H0 . All operators are treated as functions of xS (0), pxS (0) and t. If an operator commute with H0 commute Aint = AS . That is, it is identical to the operator in the Schrödinger and Heisenberg pictures. From Eqs. (7.34) and (7.37) we see that the time dependence is associated with |ψint i and also with Aint . In this way interaction picture is a sort of intermediate between Schrödinger and Heisenberg pictures.
7.4.2
Equation of Motion
Differentiation of Eq. (7.37) with respect to t leads to ∂Aint 1 dAint = + [Aint , H0,int ] , dt ∂t i~
H0,int = eiH0 t/~ H0 e−iH0 t/~ .
(7.38)
The evolution equations of xint and pxint for a free particle and linear harmonic oscillator can be obtained easily (given as problems at the end of this chapter).
7.5 COMPARISON OF THREE REPRESENTATIONS The difference between the Schrödinger and Heisenberg pictures is best understood as the difference in the basic set of observables. The basic set of observables for Heisenberg picture is defined so that no explicit time dependence appears in the state vectors. In this picture all operators are written as functions of xS , pxS and t. In the Schrödinger picture, all operators are functions of xS and pxS . An operator A which has no explicit time dependence in the Schrödinger picture, appears with an explicit time dependence in the Heisenberg picture. Table 7.5 gives the comparison of the three pictures.
7.6 DENSITY MATRIX FOR A SINGLE SYSTEM Generally, the many-body wave function ψ(q1 , · · · , q3N , t) is too large to calculate for a macroscopic system. In this case we may leave the information about individual systems and wish to know the average properties of an ensemble of identical and noninteracting systems. That is, we aim to use the concept of ensembles to derive the expectation value of a dynamical variable A without directly finding the wave function. In this case we may speak of density operator and its matrix representation called density matrix . This formalism was introduced by John von Neumann and independently by Lev Landau and Felix Bloch. Consider a single member of an ensemble and let it be in the state |ψi. We
Comparison of Schrödinger, Heisenberg and interaction pictures.
No.
Quantity
Schrödinger picture
Heisenberg picture
Interaction picture
1.
Hamiltonian
H
H
Hint = H0 + H ′
2.
State
Time-dependent ∂ i~ |ψS (t)i = H|ψS (t)i ∂t
Time-independent ∂ |ψH (t)i = 0 ∂t
Time-dependent
|ψS (t)i = e−iHt/~ |ψS (0)i
|ψH (t)i = |ψS (0)i
i~
All operators are written as functions of xS and pxS .
All operators are written as functions of t and position and momentum operators at t = 0.
All operators are written as functions of t and position and momentum operators at t = 0.
AS = AS (xS , pxS )
AH (t) = eiHt/~ AS e−iHt/~
Aint (t) = eiH0 t/~ AS e−iH0 t/~
d 1 AS = [AS , H] dt i~
d 1 AH = [AH , H] dt i~
d 1 Aint = [Aint , H0,int ] dt i~
hAS i = hψS (t)|AS |ψS (t)i
hAH i = hψS (0)|AH (t)|ψS (0)i
hAint i = hψint (t)|Aint |ψint (t)i
3.
Operators
4.
Equation of motion
5.
hAS i
|ψint (t)i = eiH0 t/~ |ψS (0)i ∂ |ψint (t)i = Hint |ψint (t)i ∂t
200 Quantum Mechanics I: The Fundamentals
TABLE 7.1
Various Pictures and Density Matrix 201 expand |ψi as |ψi =
X n
cn |φn i =
X n
X
cn |ni,
|cn |2 = 1,
(7.39)
where |φn i are orthonormal set of vectors. We define the trace of an operator A in matrix representation with the basis set {|φn i} as X X trA = hφn |A|φn i = hn|A|ni . (7.40) n
n
Then hAi is given by hAi = hψ|A|ψi =
XX m
n
cm c∗n hφn |A|φm i .
(7.41)
We define a matrix ρ with elements ρmn given by ρmn = cm c∗n . Writing Anm = hφn |A|φm i we have XX X hAi = ρmn Anm = (ρA)nn = tr(ρA) . m
n
(7.42)
(7.43)
n
The matrix ρ with the elements given by Eq. (7.42) is called a density matrix of the single system. There exists an operator called a density operator corresponding to the density matrix; it is defined as ρ = |ψihψ| .
(7.44)
The density operator is the quantum analog of the classical density function. It is easy to verify that the elements of ρ defined by Eq. (7.44) are simply given by Eq. (7.42). Considering {|φn i}, ρmn are given by hφm |ρ|φn i. We have hφm |ρ|φn i = cm c∗n hφm |φm ihφn |φn i = cm c∗n = ρmn .
(7.45)
When m = n we obtain ρnn = |cn |2 which implies that the diagonal matrix elements of the density operator represent the probabilities of the system in the base states.
Solved Problem 4: For the density operator ρ determine (i) tr(ρA), (ii) Tr(ρ) = 1, (iii) Tr(ρ2 ) and (iv) whether ρ is Hermitian. (i) Starting from tr(ρA) we get X X tr(ρA) = hn|ρA|ni = hn|ψihψ|A|ni = hψ|A|ψi = hAi . n
n
(7.46)
202 Quantum Mechanics I: The Fundamentals (ii) We obtain tr(ρ) = tr(ρ1) = hψ|1|ψi = 1.
(7.47)
(iii) First, we find ρ2 . Using the definition of ρ we get ρ2 = |ψihψ|ψihψ| = |ψihψ| = ρ .
(7.48)
Since ρ2 = ρ we have tr(ρ2 ) = tr(ρ) = 1. (iv) From Eq. (7.42) we have ∗
∗
ρmn = cm c∗n = (c∗m cn ) = (cn c∗m ) = ρ∗nm .
(7.49)
Hence, ρ is a Hermitian.
7.7 DENSITY MATRIX FOR AN ENSEMBLE Suppose we consider an ensemble of systems with N members. If all the members are in the same state |ψi then the ensemble is said to be a pure ensemble and the state is a pure state. In a mixed ensemble at least one member must be in a state different from |ψi. Let Nn be the number of identical members out of N members. P If an is the eigenvalue of an operator of Nn identical members with |ψi = n cn |φn i then the probability |cn |2 is given by |cn |2 = lim
N →∞
Nn . N
(7.50)
The expectation value h A i of an observable taken over the ensemble is hAi =
X n
1 X N n an . N →∞ N
|cn |2 an = lim
(7.51)
Instead of identical members, an ensemble may have members with distinct states. When there are Ni members in the ith state |ψi i then the probability P of finding a member in the ith state |ψi i is given by pi = Ni /N and pi = 1. Then h A i is hAi =
X X 1 X N i ai = p i ai = pi hψi |A|ψi i . N i i i
(7.52)
P h A i can be expressed in terms ρ of the ensemble where ρ = i pi |ψi ihψi |. Let us enumerate the properties of the density operator ρ. 1. h A i = tr(ρA) P Tr(ρA) = n hn|ρA|ni becomes XX XX tr(ρA) = pi hn|ψi ihψi |A|ni = pi hn|ψi ihψi |A|ni . (7.53) n
i
n
i
Various Pictures and Density Matrix 203 Since hn|ψi i and hψi |A|ni are numbers we rewrite Eq. (7.53) as XX X tr(ρA) = pi hψi |A|nihn|ψi i = pi hψi |A|ψi i = hAi . (7.54) n
i
i
Thus h A i depends only on ρ. Once ρ is known we need not consider the individual systems to calculate ensemble average. 2. ρ is a Hermitian Defining ρ(i) is the density matrix of a member in the ith state we have ∗ X X X (i)∗ (i) ρmn = pi ρ(i) pi c(i) = pi c(i)∗ = ρnm . (7.55) mn = m cn m cn i
i
i
3. Tr(ρ) = 1 When A = 1 from Eq. (7.54) we obtain tr(ρ) = 1. 4. ρ2 6= ρ
Using the definition of ρ we obtain X ED X ED ρ2 = pi ψi ψi pj ψj ψj i
XX
=
i
X
=
i
j
j
pi pj |ψi iδij hψj |
p2i |ψi ihψi | .
(7.56)
If all pi ’s are zero except one, say pa , then pa = 1 and ρ2 = |ψi ihψi | = ρ. On the other hand, if more than one pi ’s are nonzero then ρ2 6= ρ. 5. Tr ρ2 < 1 We obtain
Tr ρ2
X
=
n
hn|ρ2 |ni
XXX
=
n
i
XX
=
i
X
=
j
pi pj hn|ψi ihψi |ψj ihψj |ni 2
pi pj (δij )
j
p2i
.
(7.57)
i
Since 0 < pi < 1 and
P
i
pi = 1 for a mixed state we have tr ρ2 < 1.
204 Quantum Mechanics I: The Fundamentals
Solved Problem 5: Show that the eigenvalues of h ρ i are real and nonnegative. For each |ψi i we have hρi = =
hψi | ρ |ψi i X pi hψi |ψi ihψi |ψi i i
=
X i
pi |hψi |ψi i|2 .
(7.58)
≥ 0. Defining ai as the Because of pi and |hψi |ψi i|2 are positive we have h ρ iP eigenvalue of ρ associated with |ψi i we write h ρ i = i pi ai . Comparison of this with the Eq. (7.58) indicates that ai = |hψi |ψi i|2 and hence the eigenvalues of hρi are nonnegative and the matrix ρ is positive.
7.8 TIME EVOLUTION OF DENSITY OPERATOR The time evolution of the density operator ρ can be predicted making use of the Schrödinger equations ∂ ∂ |ψi i = H|ψi i , −i~ hψi | = hψi |H . ∂t ∂t The time derivative of ρ is given by ∂ρ ∂ X 1 = pi |ψi ihψi | = [H, ρ] . ∂t ∂t i i~ i~
(7.59)
(7.60)
Equation (7.60) is called a von Neumann equation and is the quantum mechanical analog of the Liouville equation of classical statistical mechanics. Equation (7.60) is valid for time-dependent Hamiltonian also. Let us consider the Schrödinger equation with H(t). We write ∂ |ψ(t)i = H(t)|ψ(t)i . ∂t Its solution can be represented as i~
|ψ(t)i = U (t, 0)|ψ(0)i,
(7.61)
(7.62)
where U expresses the time development of |ψ(0)i. Substituting Eq. (7.62) in Eq. (7.61) we get i~∂U/∂t = HU . Assuming U (0, 0) = 1 the solution of the above equation is U (t, 0) = e−iHt/~ . In the Heisenberg picture ρH (t) = e−iHt/~ ρ(0) eiHt/~ = U (t, 0) ρ(0) U † . The equation of motion of h A i is obtained as d d ∂A . i~ h A i = i~ tr ρ A = tr ρ A, H + i~ dt dt ∂t
(7.63)
(7.64)
Various Pictures and Density Matrix 205
7.9 A SPIN-1/2 SYSTEM Let us consider a spin-1/2 system whose spin states are labeled as | ↑i and | ↓i. Suppose we are dealing with electron beams. The density matrix of an electron in the spin state | ↑i is ρ↑ = | ↑ih↑ | while the density matrix for an electron in the spin state | ↓i is ρ↓ = | ↓ih↓ |. Suppose the two beams are mixed. For a pure state |ψi = c1 | ↑i + c2 | ↓i .
(7.65)
The condition hψ|ψi = 1 gives |c1 |2 + |c2 |2 = 1. The density matrix is ρP
= |ψihψ| = (c1 | ↑i + c2 | ↓i) (c∗1 h↑ | + c∗2 h↓ |)
= |c1 |2 | ↑ih↑ | + c1 c∗2 | ↑ih↓ | + c2 c∗1 | ↓ih↑ | + |c2 |2 | ↓ih↓ | . (7.66) In matrix representation ρmn = hn|ρ|mi , and
ρP =
|c1 |2
c2 c∗1
n, m =↑, ↓ c1 c∗2 |c2 |2
(7.67)
.
(7.68)
p1 + p2 = 1 .
(7.69)
We can easily verify that trρP = 1, ρ2P = ρP and hψ| A|ψi = tr(ρP A ) = tr( A ρP ) for any operator. For a mixed state ρM = p1 | ↑ih↑ | + p2 | ↓ih↓ | , Then
ρM =
p1
0
0
p2
(7.70)
and ρ2M = p21 | ↑ih↑ | + p22 | ↓ih↓ | 6= ρM . Since any 2 × 2 matrix can be written as a linear combination of the unit matrix and the Pauli matrices σx , σy , σz and because of trI = 2 and trσx,y,z = 0 we can write the general spin density matrix as ρ = (P0 I +P·σ)/2. We have Px − iPy 1 1 P0 + Pz . (7.71) ρ = [P0 I + Px σx + Py σy + Pz σz ] = 2 2 Px + iPy P0 − Pz P0 is determined from the condition trρ = 1. We obtain P0 = 1. Then Px − iPy 1 1 1 + Pz . ρ = (I + P · σ) = (7.72) 2 2 Px + iPy 1 − Pz
206 Quantum Mechanics I: The Fundamentals The eigenvalues of ρ must be real and nonnegative. The eigenvalue equation for the matrix ρ is det(ρ − λI) = 0 which gives λ2 − λ +
1 1 − P2 = 0 . 4
(7.73)
Then λ1,2 = 1 ± P2 /2. Since λ1 , λ2 > 0, we have P2 ≤ 1. We can easily obtain the expectation value of a spin operator hSi where S = (~/2)σ. We write 1 hσx i = Tr(σx P ) = Tr (σx + Px + iσz Py − iσy Pz ) = Px . 2
(7.74)
Similarly, we obtain hσy i = Py and hσz i = Pz . Thus, we can write hσi = P and hSi = (~/2)hσi = (~/2)P. We note that P gives the average spin vector of the ensemble and is called the polarization vector . The spin is completely polarized (a pure state) if P2 = 1. The spin is said to be in a pure state corresponding to the eigenstate of P · S with the eigenvalue ~/2. P = 0 corresponds to an unpolarized spin. Further, we obtain Trρ2 = Tr
1 1 1 + P2 + 2σ · P = 1 + P2 . 4 2
(7.75)
Therefore, Trρ2 is maximum when P2 = 1. In other words, Trρ2 = 1 represents a full polarization and Trρ2 = 1/2 corresponds to null polarization.
7.10 CONCLUDING REMARKS In this chapter we presented the description of quantum dynamics in three different pictures. We have shown the connection between the total time derivative of operators in the various pictures. We have compared the three representations. The Heisenberg picture is better suited to bring out fundamental features, such as symmetries and conservation laws, and it is indispensable in systems with many degrees of freedom, like those dealt within quantum field theory and statistical physics. No matter which picture is applied to explain the dynamics of a quantum system, the laws of motion of the micro-system are all identical. As the three pictures are equivalent and connected by unitary transformations, we must choose the picture in which the physical properties of a system are more evident and the calculations are simple. We discussed the features of density matrix. For more details on density matrix formalism and its applications refer to refs.[2-4].
7.11 BIBLIOGRAPHY [1] M. Andrews, Am. J. Phys. 71:326, 2003. [2] K. Blum, Density Matrix Theory and Applications. Springer, Berlin, 1996.
Various Pictures and Density Matrix 207 [3] J. Vanier, Basic Theory of Lasers and Masers: A Density Matrix Approach. Taylor & Francis, New York, 1971. [4] D.A. Mazziotti, Reduced Density Matrix Mechanics with Applications to Many-Electron Atoms and Molecules. Wiley-Interscience, New Jersey, 2007.
7.12 EXERCISES 7.1 Show that if the Hamiltonian in a Schrödinger picture is independent of time then it is independent of time in a Heisenberg picture also. d ˙ + AB˙ where A and B are operators. Then 7.2 Show that (AB) = AB dt d find (f (t)A). dt 7.3 If T and V are kinetic energy and potential energy operators and the time average of the time derivative of the quantity r·p is 0, then establish the virial theorem 2hT i = hr · ∇V i. 7.4 Find (a)
d d 2 hx i and (b) hxpx i. dt dt
7.5 Show that for the Hamiltonian H = p2x /2 + (px x + xpx )/2 + x2 /2 the total time derivative of x and px are x˙ = px + x, p˙ x = −px − x. 7.6 In the previous problem suppose for any wave packet we define X = x − hxi and Px = px − hpx i. Then find (i) X˙ and (ii) P˙x . 7.7 Prove the following: (a) The expectation value of a dynamical variable whose operator commutes with the Hamiltonian is time-independent regardless of initial state. (b) If the system is initially in an energy eigenstate, the expectation value of any dynamical variable will be time-independent. (c) If the system is initially a superposition of energy eigenstates (nonstationary state) then the expectation value consists of oscillatory terms with Bohr’s frequencies. 7.8. For H = p2x /2 + f (t)px + g(t)x obtain x˙ and p˙ x . 7.9. A(x, px ) and B(x, px ) can be expressed as power series in x and px and [x, px ] = i~. Show by purely matrix method that lim~→0 (1/i~)[A, B] = {A, B} where the right-side is the Poisson bracket calculated as if x and px were classical variables. 7.10. Starting from the equation dAH /dt = (1/i~)[AH , H] + ∂AH /∂t find dAS /dt.
208 Quantum Mechanics I: The Fundamentals 7.11. For a free particle the Hamiltonian in a Schrödinger picture is H = p2x /(2m). Can you write i~x˙ H = [xH , HH ] where HH is the Hamiltonian in a Heisenberg picture? 7.12. Determine the evolution equation for xint and pxint in the interaction picture for a free particle. 7.13. Obtain the evolution equations for xint and px,int in the interaction picture for the harmonic oscillator Hamiltonian H = p2x /(2m) + kx2 /2. 7.14. In terms of the ladder operators a and a† the Hamiltonian of the harmonic oscillator can be written as H = ~ωaa† + ~ω/2. Obtain the equation of motions for a and a† and solutions of them. 7.15. A particle is in a one-dimensional harmonic oscillator potential. Determine the conditions for the expectation value of an operator to depend on time, if (a) the particle is initially in a momentum eigenstate and (b) the particle is initially in an energy eigenstate. 7.16. If the Hamiltonian of a particle of a charge q and mass m subjected to a uniform electrostatic field E is H = p2 /(2m) − qE · r then find (a) dx/dt, (b) dpx /dt and (c) md2 r/dt2 . 7.17. The Hamiltonian operator of a charged particle subjected to a magnetic 2 field is H = [p − (e/c)A(r)] /(2m). Show that in the Heisenberg picture ˙ m˙r = p − (e/c)A, m¨r = (e/2c)(˙r × B − B × r). 7.18. Prove the Baker–Hausdorff lemma (iλ)2 eiAλ B e−iAλ = B + iλ[A, B] + [A, [A, B] ] + · · · . 2! Then, for a linear harmonic oscillator find x(t). 7.19. Using the equation ρ(t) = U (t, 0)ρ(0)U † show that trρ2 (t) is independent of time. 7.20. Consider two quantum systems–the first one in the state |0i with probability 3/4 and in the state |1i with probability 1/4 and the second one must be prepared in the state |ai with probability 1/2 p p and in 3/4 |0i + 1/4 |1i, the state |bi with probability 1/2 where |ai = p p |bi = 3/4 |0i − 1/4 |1i. Determine the density matrices of the above two different ensembles of quantum states.
CHAPTER
8
Heisenberg Uncertainty Principle
8.1 INTRODUCTION One of the most striking deviations of quantum mechanics from classical mechanics is that any measurement of a microscopic system involves the interaction with the measuring instruments. The interaction cannot be neglected and in fact leads to uncontrollable disturbance on the system. Let a measurement of an observable A yields the value a and is the eigenvalue of the operator A belonging to the eigenfunction φa . The individual measurements will deviate from the average value (hai) computed over large number (N ) of measurements. Denote the deviation of a from hai as a ˜. What would be the value of average deviation? We obtain h˜ ai =
1 X (a − hai) = 0 . N
(8.1)
This is because some of the values of a are below hai while the others are above hai. Therefore, h˜ ai is not useful for describing how much an individual value of an observable deviates from the expected or mean value. The mean-square deviation or variance (σ 2 ), that is, the average of the squares of the deviations would be nonzero. The variance is given by σ 2 = (∆a)2 =
1 X (a − hai)2 = ha2 i − hai2 . N
(8.2)
The square-root of mean-square deviation is called standard deviation or dispersion. This quantity could be taken as a measure of the spread in the measured values. When most of the values a are near hai then the distribution of a peaks about hai and in this case σ is small. If most of the values of a are found over a wide range about hai then the distribution spreads over a wide range of a with a large value of σ. This spread, denoted as ∆A, in the measured values of A, is called the uncertainty in the measurement of A. 209
210 Quantum Mechanics I: The Fundamentals In 1927, Werner Karl Heisenberg in his celebrated paper [1] (Nobel Prize in 1932) established the quantitative expression for unavoidable momentum disturbance caused by any position measurement. Essentially, any measurement of the position x of a mass with ∆x would cause a disturbance ∆px on the momentum (px ) measurement and ∆x and ∆px should always obey the relation ∆x∆px ≥ ~/2. Heisenberg explained the physical intuition underlying the above expression through the gamma-ray microscope thought experiment. In this chapter, first we derive the classical uncertainty relation. Next, we give a proof of the Heisenberg uncertainty relation. Then we present a modified form of it.
8.2 THE CLASSICAL UNCERTAINTY RELATION Let A be a classical statistical variable with mean hAi. The uncertainty in A, ∆A, is defined by 2 (∆A)2 = σA =
The above can be rewritten as
1 X (A − hAi)2 = h(A − hAi)2 N
(∆A)2 = hA2 i − hAi2 .
(8.3)
(8.4)
This is the variance of A. The variance between two variables A and B is given by [2,3] σAB Let us define
=
h(A − hAi)(B − hBi)i = hABi − hAihBi .
(8.5)
A˜ = A − hAi ,
(8.6)
˜ = hBi ˜ = 0 and which implies hAi (∆A)2 = hA˜2 i ,
˜ = B − hBi B
˜2i , (∆B)2 = hB
˜ . σAB = hA˜Bi
(8.7)
For a statistical variable x that is not always 0 we have hx2 i ≥ 0. When ˜ we find x = A˜ + λB ˜ 2 i + 2λhA˜Bi ˜ ≥0. hx2 i = hA˜2 i + λ2 hB
(8.8)
˜ = 0. The nature of the An equality sign will be realized if and only if A˜ + λB roots of λ is determined from the quantity ˜ 2 − 4hA˜2 ihB ˜2i . α = 4hA˜Bi
(8.9)
When α is 0 the roots of λ are real and identical. That is, there is one real root of λ. Roots are complex for α < 0 while they are real and distinct if α > 0.
Heisenberg Uncertainty Principle 211 Thus, the condition for at most one real root of λ is (What would happen if there are two real roots of λ?) ˜ 2 − hA˜2 ihB ˜2i ≤ 0 . hA˜Bi
(8.10)
2 When the Eqs. (8.7) are used the above equation becomes σAB − 2 2 (∆A) (∆B) ≤ 0 or (∆A)(∆B) ≥ |σAB | . (8.11)
This is the uncertainty principle for classical statistics [2,3].
8.3 HEISENBERG UNCERTAINTY RELATION Let us consider two sets of complex numbers: A1 , A2 , · · · , An and B1 , B2 , · · · , Bn . We have XX |Ai Bj − Aj Bi |2 ≥ 0 . (8.12) i
j
An equality sign holds when these sets are identical. Expanding the left-side of Eq. (8.12) we get XX 0 ≤ (Ai Bj − Aj Bi )∗ (Ai Bj − Aj Bi ) i
≤ ≤ or
j
XX i
2
j
hX
|Ai |2 |Bj |2 − A∗i Bi Aj Bj∗ − A∗j Bi∗ Ai Bj + |Aj |2 |Bi |2
|Ai |2
X
X
|Ai |2
X X i |Bi |2 − A∗i Bi Ai Bi∗
X
2 X |Bi |2 ≥ A∗i Bi .
(8.13)
If A and B are continuous functions of x then the Eq. (8.13) takes the form Z ∞ Z ∞ Z ∞ ∗ ∗ A A dx B B dx ≥ A∗ B dx . (8.14) −∞
−∞
−∞
In a quantum state ψ, the uncertainty in an observable A is given by (∆A)2 = h(A − hAi)2 i = hA2 i − hAi2 .
(8.15)
Then i1/2 Z 2 ˜ ∆A = hA i =
1/2
=
Since A˜ is Hermitian we have Z ∞ 1/2 Z ∗ ˜ ˜ ∆A = (Aψ) (Aψ) dτ =
∞
h
∞
−∞
−∞
ψ A˜2 ψ dτ ∗
Z
∞
˜ Aψ) ˜ dτ ψ A( ∗
−∞
−∞
ψ1∗ ψ1
dτ
1/2
1/2
= ||ψ1 || ,
. (8.16)
(8.17)
212 Quantum Mechanics I: The Fundamentals where Similarly,
˜ = (A − hAi)ψ . ψ1 = Aψ
(8.18)
˜ = (B − hBi)ψ . ψ2 = Bψ
(8.19)
If ψ1 and ψ2 are two states then from Eq. (8.14) we write Z ∞ ||ψ1 || · ||ψ2 || ≥ ψ1∗ ψ2 dτ .
(8.20)
−∞
This is known as Schwarz’s inequality. The proof of this inequality is presented in appendix E. We rewrite the Eq. (8.20) as Z ∞ ||ψ1 || · ||ψ2 || ≥ Im ψ1∗ ψ2 dτ . (8.21) −∞
Suppose we write Z ∞ ψ1∗ ψ2 dτ = R + iI , −∞
Z
∞
−∞
ψ2∗ ψ1 dτ = R − iI ,
(8.22)
where R and I are real numbers. Then I is given by I = Im(R + iI) =
1 [(R + iI) − (R + iI)∗ ] . 2i
In terms of the integrals in Eq. (8.22) the Eq. (8.23) is written as Z ∞ Z ∞ Z ∞ 1 ∗ ∗ ∗ Im ψ1 ψ2 dτ = ψ ψ2 dτ − ψ2 ψ1 dτ . 2i −∞ 1 −∞ −∞ Then
That is,
Z Z ∞ 1 ∞ ψ ∗ ψ2 dτ − ψ2∗ ψ1 dτ . ||ψ1 || · ||ψ2 || ≥ 2i −∞ 1 −∞
∆A∆B
≥ ≥ ≥
(8.23)
(8.24)
(8.25)
Z Z ∞ ∞ 1 1 ∗ ˜ ∗ ˜ ˜ ˜ (Aψ) (Bψ) dτ − (Bψ) (Aψ) dτ 2 i −∞ −∞ Z Z ∞ ∞ 1 1 ˜ dτ − ˜ Aψ ˜ dτ ψ ∗ A˜Bψ ψ∗ B 2 i −∞ −∞ D h i E 1 1 ˜ B ˜ . A, (8.26) 2 i
˜ B] ˜ is shown to be The commutator [A, h i ˜ B ˜ ˜ −B ˜ A˜ A, = A˜B =
=
(A − hAi)(B − hBi) − (B − hBi)(A − hAi) [A, B] .
(8.27)
Heisenberg Uncertainty Principle 213 Thus, ∆A∆B ≥ or
1 1 h[A, B]i 2 i
(8.28)
1 (8.29) (∆A)2 (∆B)2 ≥ − h[A, B]i2 . 4 Equation (8.29) is known as Heisenberg uncertainty principle 1 . It is the uncertainty principle for any pair of observables A and B. This equation with equality sign is commonly referred as the minimum uncertainty product. In Eq. (8.29) (∆A)2 = hA2 i − hAi2 (8.30)
is variance in A. Similarly (∆B)2 is defined. For a canonically conjugate pair of operators A and B we have [A, B] = i~. Hence, (∆A)2 (∆B)2 ≥ ~2 /4 (8.31) or (∆A)(∆B) ≥ ~/2 .
(8.32)
Thus, the statement of the uncertainty principle is the following: The value of both the observables of a canonically conjugate pairs of a quantum mechanical system cannot be specified precisely and simultaneously. The crucial word here is simultaneously. The uncertainty relation gives the lower bound on the product of the uncertainties of two pairs of variables. But it says nothing about the accuracy with which individual observables can be measured. An important point is that the uncertainty relation has nothing to do with the precision of the measuring instrument or technique. Rather, it is an inherent characteristic of the quantum mechanical systems. The uncertainty relation is not obtained based on experimental data. Without the uncertainty principle, quantum mechanics becomes a deterministic theory. Because [x, px ] = i~, the position-momentum uncertainty principle is (∆x)(∆px ) ≥ ~/2 ,
(8.33)
where 2
(∆x)2 = hx2 i − hxi2 , 2
(∆px )2 = hp2x i − hpx i2 .
(8.34)
(∆x) and (∆px ) are the variance of position and momentum of a particle respectively. Note that only the mean values of x and x2 are needed to compute ∆x. The uncertainty principle always has members of a conjugate pair. In momentum space p = ~k and hence the uncertainty relation is (∆x)(∆k) ≥ 1/2 .
(8.35)
From the equality sign we have ∆x = 1/(2∆k). This implies that if ∆x is 1 This
principle was enunciated by Heisenberg when he was just 25 years old.
214 Quantum Mechanics I: The Fundamentals small then ∆k must be large and vice-versa. This accounts for the waveparticle paradox. We cannot make ∆x and ∆k small simultaneously (beyond the restriction ∆x∆k ≥ 1/2), that is, both a wave and a particle. A similar condition we can have for the energy and time measurements: (∆E)(∆t) ≥ ~/2 .
(8.36)
For example, suppose we wish to measure the time t at which a particle arrives a particular position with an error ∆t. As per the uncertainty principle, the uncertainty ∆E of the measured energy of the particle at that position cannot be smaller than ~/(2∆t). Another result of (8.36) is that a system which is not in a steady state but changing with time, cannot have a welldefined energy. The physical meaning of energy-time uncertainty is different from that of position-momentum uncertainty. Position and momentum are observables to which Hermitian operators are assigned and which are measured at a particular time. But time appears only as a parameter in quantum theory.
Solved Problem 1: With an example show that the uncertainty principle is insignificant to larger objects or particles and significant to microscopic particles. Suppose a cricket ball of mass 300gm is moving with a speed of 20m per second. The error in momentum measurement is, say, 0.1%. Then the uncertainty in the position measurement is ∆x ≥
~ = 2∆px
1.0544 × 10−34 = 8.78667 × 10−33 m . 0.1 2 × 0.3 × 20 × 100
(8.37)
∆x is of the order of 10−33 m. As this quantity is very small we cannot realize the uncertainty principle. This is true for heavier bodies. Next, consider an electron with velocity 2 × 107 m. Let the uncertainty in its velocity measurement be 0.1%. We calculate the uncertainty in position measurement. We obtain ∆x
= =
~ 2∆px ~ 2m∆v
1.0544 × 10−34 m 0.1 2 × 9.109 × 10−31 × × 2 × 107 100 = 2.89384 × 10−9 m .
=
(8.38)
The size of the nucleus in any atom is ≈ 10−14 m. The calculated ∆x is very much greater than the size of the nucleus and is thus matter in atomic scales.
Heisenberg Uncertainty Principle 215
Solved Problem 2: Calculate the ground state energy of a harmonic oscillator (H = p2x /(2m) + kx2 /2) making use of the Heisenberg’s uncertainty principle assuming that hxi = hpx i = 0. For the linear harmonic oscillator (∆x)2 2
(∆px )
= hx2 i − hxi2 = hx2 i , =
hp2x i
2
− hpx i =
hp2x i
(8.39) .
(8.40)
From the relation (∆x)2 (∆px )2 ≥ ~2 /4 we write (∆px )2 ≥
~2 ~2 = 2 4(∆x) 4hx2 i
(8.41)
For the linear harmonic oscillator hHi is hHi = E
= = ≥
1 2 1 hpx i + mω 2 hx2 i 2m 2 1 1 2 (∆px ) + mω 2 hx2 i 2m 2 ~2 1 + mω 2 hx2 i . 8mhx2 i 2
(8.42)
The ground state energy is the minimum of E. The conditions for minimization of E are dE/dhx2 i = 0 and d2 E/dhx2 i2 > 0. The first condition gives hx2 i = ~/(2mω). With this hx2 i the equality sign in the equation for hHi gives Emin = ~ω/2.
8.4 IMPLICATIONS OF UNCERTAINTY RELATION The uncertainty relation has a wide-ranging significance2 . In the following we point out some of them. 1. Suppose we wish to find the position and momentum of a moving particle. For this purpose we allow light to fall on the particle. The particle will scatter part of the incident light which will give details about the position of it. Note that the position cannot be found more accurately than the wavelength of the incident light. However, we can improve the accuracy using short wavelength wave. But according to Planck’s hypothesis we need to use at least one quantum. This will affect the particle and alter its momentum in a way that cannot be determined. Further, to measure the position more precisely we have to use shorter wavelength 2 Hans Bethe once said: Without the uncertainty principle there could not exist any atoms and there could not be any certainty in the behaviour of matter whatever. So, it is really the certainty principle [I. Goodwin, Physics Today, June 1995, pp.39].
216 Quantum Mechanics I: The Fundamentals light. In such a case, energy of a single quantum (hc/λ) is higher and hence the disturbance to the velocity of the particle will be large. Consequently, the more accurately we attempt to measure the position, the less accurately we can measure the speed and vice versa. 2. Consider the measurement of radius r0 of the electron orbit in a ground state of the hydrogen atom. Theoretically predicted value of r0 is 5.3 × 10−11 m. According to quantum mechanics this value is the most probable radius. This means in experiment, most trials will give a different value, larger or smaller than 5.3 × 10−11 m. 3. An important implication of the uncertainty principle is that quantum systems do not follow classical trajectories which conflict with our everyday experience. 4. Another consequence is that quantum mechanical systems in their lowest energy state (ground state) could not be at rest. For example, the electron of an atom in its ground state could not be at rest. What will happen if it is at rest? In this case its velocity is exactly known to be zero and its position would be known precisely. This would violate the Heisenberg uncertainty relation. 5. Suppose ψ(x) is a delta-function at x = x0 . The position of the particle is determined exactly. What about the corresponding momentum wave function? The momentum wave function C(p) is a sinusoidal function extending over all space. Thus, the momentum uncertainty is infinity. Similarly, if ψ(x) is sinusoidal then the position uncertainty is infinity whereas because the corresponding momentum wave function is deltafunction the uncertainty in momentum is zero. 6. It gives a resolution limit for the determination of the spectrum of a signal in terms of the measurement [4]. It also states resolution limits for imaging systems in terms of their aperture size. It serves as the basis of one of the standard measure of laser beam quality [5]. The uncertainty principle indicates the existence of quantum fluctuations in physical states that cannot be controlled completely. Particularly, the fluctuations of energy and momentum go arbitrarily large when the space-time interval becomes shorter and shorter. 7. The interaction between two particles occurs by means of exchange of a quantum of the interacting field. For example, Yukawa’s strong nuclear force between nucleons is due to the exchange of pions. The mass of the exchanged quantum is found to have an inverse relation with the range of the interaction. This can be accounted only by the uncertainty relation. If R is the range and m is the mass of the exchanged quantum then ∆t ≈ R/c and ∆E ≈ mc2 . The energy-time uncertainty relation gives m ≈ ~/(2Rc). If R ≈ 10−15 m then m ≈ 200me which is roughly the mass of pions.
Heisenberg Uncertainty Principle 217
A
B θθ gamma ray photon FIGURE 8.1
P
x-axis
Position measurement of an electron.
8. What prevents an electron from falling into nucleus? Suppose the electron was assumed to be in the nucleus. In this case we exactly know its position. Then according to uncertainty principle the momentum would be very large and hence its energy would be very large. This large energy would then able to break away the electron from the nucleus. 9. An important application of Heisenberg uncertainty pointed out by John Gribbin is that in nuclear fusion reactions the quantum uncertainty allows atomic nuclei that are not close enough to touch one another to overlap with one another to combine. Some of such nuclear reactions keep the stars hot. That is, without uncertainty principle the sun would not shine the way it does.
8.5 ILLUSTRATION OF UNCERTAINTY RELATION Let us explain the uncertainty relation with two examples.
8.5.1
Gamma-Ray Microscope
We can measure the x-coordinate of an electron using a gamma-ray microscope by scattering a photon off it. Here, to measure a position as accurate as possible, we need to use the light of short wavelength. The accuracy in the measurement is given by ∆x = (λ/2) sin θ , (8.43) where θ is the half-angle of the cone formed in Fig. 8.1 and λ is the wavelength of the gamma-ray used. In order to see the electron through the microscope the gamma-rays scattered by the electron has to enter the objective of the microscope. In this
218 Quantum Mechanics I: The Fundamentals
e
ν
Before interaction
v e
ν' FIGURE 8.2
v'
After interaction
Doppler effect.
process, the Compton effect occurs, namely, a decrease in energy or increase in the wavelength of a photon happens. Let λ represent the wavelength of the photon before interaction, and λ′ is the wavelength of it after interaction. In the process the electron undergoes a recoil and it can be set minimum by using minimum gamma-ray energy. However, at least one gamma-ray photon need to be used. In this case the scattered photon with momentum h/λ′ may enter the objective. If its path is along P A then its momentum component along the direction of the incident ray will be (h/λ′ ) sin θ. The momentum that is transferred to the electron is p1 =
h h − ′ sin θ . λ λ
(8.44)
In the case of the photon scattered along P B, the momentum given to the electron will be h h p2 = + ′ sin θ . (8.45) λ λ Thus, the value of the momentum transferred to the electron will be between p1 and p2 . Then 2h ∆px = ′ sin θ . (8.46) λ Hence, from Eqs. (8.43) and (8.46) ∆x∆px = h = 2π~ .
8.5.2
(8.47)
Doppler Effect
The Doppler effect refers to the change in the frequency of waves noticed by a detector or receiver due to the motion of the source of waves and the detector/observer. Let us consider the problem of finding momentum of an electron using Doppler effect (Fig. 8.2). We have 1 + v/c ′ ν =ν . (8.48) 1 − v/c
Heisenberg Uncertainty Principle 219 When v ≪ c,
2v ν ≈ν 1+ c ′
or
v ν′ − ν = . c 2ν
(8.49)
The electron momentum is p = mv = Then
mc(ν ′ − ν) . 2ν
∆p ∆(ν ′ − ν) = p ν′ − ν
(8.50)
(8.51)
gives the error in p. The accuracy of the measurement of ν ′ − ν given by ∆(ν ′ − ν) determines the limitation on p. The point is that the details of the particle’s position will be lost due to the time interval ∆t needed to measure ν ′ − ν. In the above considered process at least one photon must be scattered. After the scattering the velocity of the electron is changed. We do not know at what exact instant during ∆t, the scattering took place. This results in an uncertainty in the velocity of the electron and hence in its position.
Solved Problem 3: Determine the uncertainty product for an electron passing through a diffraction grating of infinite length. Diffraction grating is a device used to produce spectra by diffraction. The well known grating formula is d sin θ = nλ,
(8.52)
where d is the spacing between grating lines, θ is the diffraction angle of the diffracted maximum, n is the order of the spectral lines and λ is the wavelength of the light. The above equation is the condition for diffraction to take place. Let us consider the uncertainty in x-direction. After the diffraction we have d sin θ = 2nπ/kx or kx sin θ = n
2π . d
(8.53)
We do not know where the entrance of electron took place. Therefore, if the grating has m slits each of width d then the uncertainty in the position coordinate along x-direction is ∆x = md. Initially the electrons are moving along x-axis and hence no component of the momentum along kx axis. When they are diffracted they have momentum component along kx -direction. So the uncertainty in kx component of momentum is ∆kx =
2π ∆n . d
(8.54)
220 Quantum Mechanics I: The Fundamentals Taking ∆n = 1/m we get ∆kx = 2π/∆x. That is, ∆x∆kx = 2π. Multiplying both sides by ~ we obtain ∆x∆px = 2π~ > ~/2.
Solved Problem 4: An electron of energy 300 eV is allowed to move through a circular aperture of radius 10−4 cm. Calculate the uncertainty in the angle of divergence. The uncertainty in the angle of divergence is ∆θ ≈ ∆px /px . We calculate px and ∆px in eV as (1eV= 1.602 × 10−19 J) √ 2mE px = 1/2 = 2 × 9.109 × 10−31 × 300 × 1.602 × 10−19 kg/sec = 9.3571207 × 10−24 kg/sec
(8.55)
and ∆px
≈ = =
~/∆x 1.0544 × 10−34 kg/sec 2 × 10−6 5.272 × 10−29 kg/sec .
(8.56)
Then ∆θ ≈
5.272 × 10−29 = 5.6342 × 10−6 radians . 9.3571207 × 10−24
(8.57)
8.6 THE MODIFIED HEISENBERG RELATION First, we show the need to modify the standard uncertainty relation, Eq. (8.28). In order to hold the Heisenberg inequality, the state of a quantum system must possess certain criteria. Before pointing out these restrictions let us verify the Heisenberg uncertainty relation for the canonical pair Lz and φ.
8.6.1
What is (∆φ)(∆Lz )?
The components of orbital angular momentum L are (see sec. 11.3) ∂ ∂ Lx = ypz − zpy = −i~ y −z , (8.58a) ∂z ∂y ∂ ∂ Ly = zpx − xpz = −i~ z −x , (8.58b) ∂x ∂z ∂ ∂ Lz = xpy − ypx = −i~ x −y . (8.58c) ∂y ∂x In spherical polar coordinates we have (see sec. 11.3 for details) Lz = −i~∂/∂φ. Lz and the azimuthal angle φ form a canonical pair. It is easy to verify that
Heisenberg Uncertainty Principle 221 [φ, Lz ] = i~. From this relation we expect ∆φ∆Lz ≥ ~/2. Let us find ∆φ∆Lz for the eigenstate 1 (8.59) ψ(φ) = √ eimφ 2π of Lz . We obtain Z 2π Z 2π 4 hφi = ψ ∗ φψ dφ = π , hφ2 i = ψ ∗ φ2 ψ dφ = π 2 , (8.60a) 3 0 0 Z 2π Z 2π hLz i = ψ ∗ Lz ψdφ = m~, hL2z i = ψ ∗ L2z ψdφ = m2 ~2(8.60b) 0
0
and therefore (∆φ)2
=
(∆Lz )2
=
π2 , 3 hL2z i − hLz i2 = 0 .
hφ2 i − hφi2 =
(8.61a) (8.61b)
Thus, ~ . (8.62) 2 What went wrong? The reason is that eigenstates of Lz do not satisfy the criteria required for the Heisenberg principle. Specifically, φ is not a proper dynamical variable and hence does not represent a quantum mechanical operator. This is because φ converts an acceptable eigenfunction ψ to an inadmissible one when it operates on ψ. When φ is limited to the domain 0 to 2π then after it has increased to 2π, it jumps discontinuously down to zero. Hence, φψ will also have a discontinuous jump after φ taking 2π. As φψ is discontinuous, it is not physically acceptable. Therefore, φ is not a good operator and the uncertainty relation between φ and Lz does not exist. In the following we describe these criteria and show why ψ of Lz do not satisfy them. (∆φ)(∆Lz ) = 0
0 are the zeros of the Airy function. The zeros and the normalization factors are found to be 1/6 2/3 3π 1 2π 2 xn ≈ n− , Nn ≈ , n = 1, 2, · · · . 1 2 4 3 n− 4 (10.43) The solution of the time-dependent Schrödinger equation is then written as X ψ(x, t) = Cn e−iEn t/~ φn (x) . (10.44) n
The time evolution of initially Gaussian form wave packet is found to be nontrivial [19,21]. Assume the initial wave packet as ψ(x, 0) =
2 πσ 2
1/4
e−(x−x0 )
2
/σ2
.
(10.45)
The time evolution of this is given by Eq. (10.35) where 4(x0 − xn ) 8(x0 − xn )3 −(x0 −xn )2 /σ2 + e . σ4 3σ 6 (10.46) Figure 10.5 shows the plot of the wave packet at three values of time, t = 0, T /2 and T where x0 = 10, and σ = 2 [21]. In this figure a painted circle represents the position of the classical particle. This figure shows the first bounce: (a) is the initial state, (b) is the particle just reaching the floor and (c) is at t = T . The localized wave packet solution disperses or spreads in time, however, at a later time, it reforms and appears as its initial state. This is called wave packet revival . The behaviour of localized quantum wave packet has lately become an object of much interest to forge a link between classical and quantum mechanics. We can consider the wave packets synthesized by the superposition of a great Cn ≈ Nn
2 πσ 2
1/4
1−
254 Quantum Mechanics I: The Fundamentals
(a)
-15
-10
-5
0 -15
(c)
(b)
-10
-5
0 -15
-10
-5
0
Wave packet of a quantum bouncing ball at (a) t = 0, (b) t = T /2 and (c) t = T where x0 = 10, and σ = 2 [21]. FIGURE 10.5
many bound states centred around a rather large mean value n ¯ of the quantum number n. Such a wave packet may begin by mimicking classical dynamics, but after a short while, quantum effects set in, causing, at first a spreading and, at much later times, a partial or total restoration of the original form at a point which is not necessarily the same as that where the packet was initially localized.
10.5 WAVE PACKET REVIVAL If the probability density P (x, t) = |ψ(x, t)|2 of a nonstationary state satisfies the condition P (x, t0 ) = P (x, t0 + T ) for some finite T the wave packet is said to display exact revivals at intervals of T . Any wave packet ψ(x, t) can be expanded in terms of φn (x), the eigenfunctions of Hamiltonian H, as Z ∞ X ψ(x, t) = Cn φn (x) e−iEn t/~ , Cn = φ∗n (x)ψ(x, 0) dx . (10.47) −∞
n
For the linear harmonic oscillator, introducing the classical period T = 2π/ω we write the above equation in the form ψ(x, t) = e−iπt/T
∞ X
Cn φn (x) e−i2nπt/T
(10.48)
n=0
There are two interesting results from Eq. (10.48): (i) ψ(x, t + T ) = −ψ(x, t) ,
(ii) ψ(x, t = π/2) = −iψ(−x, 0) .
(10.49)
For the infinite square-well system, the general expansion (10.47) becomes ψ(x, t) =
∞ X
n=1
2
Cn φn e−i2n
πt/T
(10.50)
Wave Packet 255 with T = 2π~/E1 = 4mL2 /(π~) which leads to the conclusion ψ(x, t + T ) = ψ(x, t) ,
ψ(x, t = T /2) = −ψ(−x, 0) .
(10.51)
Then from (10.49) and (10.51) we have P (x, t + T ) = P (x, t) ,
P (x, t = T /2) = P (−x, 0) .
(10.52)
This means that any wave packet |ψ(x, t)| for the linear harmonic oscillator and the infinite square-well systems is periodic with period T and that the time T is the shortest interval after which a complete reconstruction of the initial probability density is guaranteed. Similarly, the time t = T /2 is the first instant at which the probability density coincides with the mirror image of the initial probability density. The wave packet is said to have exact revival at t = T and a mirror revival at t = T /2. Revivals in quantum systems were first studied in the Jaynes–Cumming model, describing a two-level atom interacting with a resonant monochromatic field [22,23]. Revivals have been observed experimentally in a micromaser cavity with rubidium atom [24], in Rydberg electron wave packet [25] and vibrational wave packets in Na2 [26]. The occurrence of wave packet revivals has been studied in systems such as rigid rotator, weakly anharmonic potentials, infinite square-well potential, circular billiards etc. [27-32]. An exact fractional wave function revival is said to occur when the wave function of a particle become a superposition of translated initial wave functions. Fractional revivals have been observed in many quantum systems such as Jaynes–Cummings model, Morse-like anharmonic potentials, Na2 etc. [33-40].
10.6 ALMOST PERIODIC WAVE PACKETS As seen earlier the Gaussian wave packet of a free particle, spreads in time while that of the linear harmonic oscillator oscillates with time. Wave packet revival also occurs in certain systems. There are a number of systems for which the time-dependent wave function does not spread in time. The occurrence of nonspreading wave function is found in certain one-dimensional system having discrete energy eigenvalues En [40]. The wave function of such systems exhibit almost periodic behaviour in time t. For a periodic function f (t) with period τ we write f (t + τ ) = f (t). By almost periodic function we mean that there is no values of τ strictly satisfying f (t + τ ) = f (t), however, there are infinite number of τn , n = 1, 2, · · · , so that f (t + τ ) ≈ f (t) is satisfied. Suppose for a particle in a one-dimensional confined potential the stationary state discrete nondegenerate eigenvalues and the eigenfunctions are En and φn respectively. Then the general time-dependent wave function is written as ψ(x, t) =
∞ X
n=0
Cn e−iEn t/~ φn (x) ,
(10.53a)
256 Quantum Mechanics I: The Fundamentals where X n
|Cn |2 = 1 ,
Cn =
Z
∞
−∞
φ∗n (x)ψ(x, 0) dx .
(10.53b)
It is easy to show that the wave function given by Eq. (10.53) is almost periodic in t [40]. Let the allowed tolerance in f is ǫ, that is |f (t + τn ) − f (t)| < ǫ. Almost periodic functions is viewed as a generalizationPof periodic functions since any almost periodic function is written as f (t) = n An eiωn t , where An are real or complex while ωn are real. For a strictly periodic function ωn = nω1 . According to the Riesz–Fischer theorem [41] a trigonometric series is the P Fourier series of an almost periodic function as long as n |An |2 converges. Applying this to Eq. (10.53) we say that ψ(x, t) is almost periodic in P theorem t provided |Cn |2 |φn |2 converges. We require X X |Cn |2 |φn (x)|2 ≤ |φm (x)|2 |Cn |2 = |φm (x)|2 , (10.54) n
where |φm (x)| is the largest of all of the φn (x) for the particular value of x. Hence, for a discrete energy eigenvalue spectrum ψ(x, t) is an almost periodic function of time t. For a linear harmonic oscillator En = (n + 1/2)~ω. Writing Ψ(x, t) = eiωt/2 ψ(x, t) Eq. (10.53a) becomes Ψ(x, t) =
∞ X
Cn e−inωt φn (x)
(10.55)
n=0
which is a Fourier series in t and is a periodic function of t. The strictly periodic behaviour of Ψ is due to the fact that En are equally spaced. For a discussion on fluid-like properties of probability density, interpretation of them for the one-dimensional Gaussian wave packet, stationary states of a particle in a box and hydrogen atom see refs.[42,43].
10.7 CONCLUDING REMARKS Suppose one shines a femtosecond optical pulse on a diatomic molecule that has enough broad bandwidth to cover several vibrational levels in the excited state. Then it is possible to create a nuclear wave packet that is spatially localized and no longer stationary. The wave packet thus created is a linear combination of several eigenstates. Laser pump-probe techniques in the femtosecond time domain have been applied to temporarily resolve the motion of vibrational wave packet molecules and study the femtochemistry of the transition state of several gas-phase reactions. It has been demonstrated that a wave packet, the coherent superposition of a set of vibrational eigenstates produced by an optical pulse the duration of which is short compared to the vibrational period, can be observed by detecting the molecular species itself as a dissociation product by
Wave Packet 257 fluorescence spectroscopy [44]. Since the temporal history of the wave packet bears the signature of the potential well in which it moves, the study on wave packet dynamics has been used to find the potential energy surface of diatomic molecules. All spectral information can be recovered by studying the temporal evolution of wave packet. Technological advances led to a significant amount of research being focused at investigating the time-dependent behaviour of electron wave packets. The ultimate theme is to generate wave packets that control electronic process in atoms and molecules. Several control schemes exist which use light to drive atoms or molecules along prescribed pathways. They exploit the coherence of laser light to manipulate the quantum phase relationships between various eigenstates of the system [45].
10.8 BIBLIOGRAPHY [1] A. Goldberg, H.M. Schey and J.L. Schwartz, Am. J. Phys. 35:177, 1967. [2] M.H. Bramhall and B.M. Casper, Am. J. Phys. 38:1136, 1970. [3] B. Diu, Eur. J. Phys. 1:231, 1980. [4] A. Edgar, Am. J. Phys. 63:136, 1995. [5] J.S. Boleman and S.B. Haley, Am. J. Phys. 43:270, 1975. [6] J.V. Greenman, Am. J. Phys. 40:1193, 1972. [7] C.U. Segre and J.D. Sullivan, Am. J. Phys. 44:729, 1976. [8] P.A. Deutchman, Am. J. Phys. 39:952, 1971. [9] E.A. Johnson and H.T. Williams, Am. J. Phys. 50:239, 1981. [10] J.C. Hamilton, J.L. Schwartz and W.A. Bowers, Am. J. Phys. 40:1657, 1972. [11] G. Friedman, W.A. Little, Am. J. Phys. 61:835, 1993. [12] G.W. Ford and R.F. O’Connell, Am. J. Phys. 70:319, 2002. [13] D.F. Styer, Am. J. Phys. 58:742, 1990. [14] L.S. Brown, Am. J. Phys. 41:525, 1972. [15] M. Nauenberg, Phys. Rev. A 40:1133, 1989. [16] M. Nauenberg, C. Stroud and J. Yeazell, Sci. Am. 270:44, 1994. [17] G. Alber and P. Zoller, Phys. Rep. 199:231, 1991.
258 Quantum Mechanics I: The Fundamentals [18] S. Brandt and H.D. Dahmen, Quantum Mechanics on the Personal Computer. Springer, New York, 1992. [19] M.A. Doncheski and R.W. Robinett, Am. J. Phys. 69:1084, 2001. [20] R.W. Robinett, Quantum Mechanics: Classical Results, Modern Systems and Visualized Examples. Oxford Univ. Press, New York, 1997, pp.102103. [21] J. Gea-Banacloche, Am. J. Phys. 67:776, 1999. [22] J.H. Eberly, N.B. Narozhny and J.J.S. Mondragon, Phys. Rev. Lett. 44:1323, 1980. [23] N.B. Narozhny, J.J.S. Mondragon and J.H. Eberly, Phys. Rev. A 23:236, 1981. [24] G. Rampe, H. Walther and N. Klein, Phys. Rev. Lett. 58:353, 1987. [25] J.A. Yeazell, M. Mallalieu and C.R. Stroud Jr, Phys. Rev. Lett. 64:2007, 1990. [26] T. Baumert, V. Engel, C. Rottgermann, W.T. Strunz and G. Gerber, Chem. Phys. Lett. 191:639, 1992. [27] R.W. Robinett and S. Heppelmann, Phys. Rev. A. 65:062103, 2002. [28] R. Bluhm, V.A. Kostelecky and J.A. Porter, Am. J. Phys. 64:944, 1996. [29] D.L. Aronstein and C.R. Stroud Jr, Phys. Rev. A 55:4526, 1997. [30] R.W. Robinett, Am. J. Phys. 68:410, 2000. [31] P. Stifter, E.W. Lamb Jr., W.P. Schleich, “The particle in a box revisited” in Frontiers of Quantum Optics and Laser Physics, Proceedings of the International Conference on Quantum Optics and Laser Physics. Springer, Singapore, 1997 pp.236-246. [32] K. Razi, Naqvi and S. Waldenstrom, Physica Scripta 62:12, 2000. [33] B.M. Garraway and K.A. Suominien, Contemporary Physics 43:97, 2002. [34] J. Parker and C.R. Stroud Jr, Phys. Rev. Lett. 56:716, 1986. [35] Z.D. Gaeta and C.R. Stroud Jr, Phys. Rev. A 42:6308, 1990. [36] I.Sh. Averbukh, Phys. Rev. A 46:R2205, 1992. [37] S.I. Vetchinkin and V.V. Eryomin, Chem. Phys. Lett. 222:394, 1994. [38] W.Y. Chen and G.J. Milburn, Phys. Rev. A 51:2328, 1995.
Wave Packet 259 [39] D.R. Meacher, P.E. Meyler, I.G. Hughes and P. Ewart, J. Phys. B 24:L63, 1991. [40] M.J.J. Vrakking, D.M. Villeneuve and A. Stolow, Phys. Rev. A 54:R37, 1996. [41] D. Mentrup and M. Luban, Am. J. Phys. 71:580, 2003. [42] A.S. Besicovitch, Almost Periodic Functions. Cambridge Univ. Press, Cambridge, 1932. [43] K. Mitra, Am. J. Phys. 71:894, 2003. [44] M. Dantus, M.H.M. Janssen and A.H. Zewail, Chem. Phys. Lett. 181:281, 1991. [45] J.R.R. Verlet and H.H. Fielding, Int. Rev. Phys. Chem. 20:283, 2001.
10.9 EXERCISES 10.1 Show that the wave packet of electromagnetic waves in free space travels undistorted with speed c. 10.2 What is the relation between the phase velocity vp of a free particle and the classical free particle velocity v. 10.3 Does the vacuum, dispersion for a quantum wave function? Why? √ 2 2 10.4 For the Gaussian wave packet ψ(x, 0) = [(1/ 2π)σ]1/2 e−(x−x0 ) /(4σ ) compute ∆x∆px . 10.5 Show that an even-parity wave packet of the infinite square-well potential manifests an exact revival at t = T2 /8 and an odd-parity wave packet at t = T2 /4 where T2 is defined for the infinite square-well potential by En = n2 π 2 ~2 /(2mL2 ) = 2πn2 /(~T2 ). 10.6 Determine the Gaussian wave packet for a quantum particle undergoing uniform acceleration. 10.7 Consider an initial wave packet of the form ψ(x, 0) = W (x)eiφ(x) where W and φ are real. Determine hpx i and hp2x i for this wave packet. 2
10.8 For the initial wave packet of the form ψ(x, 0) = (1/2π)1/4 e−x eiφ(x) of a free particle of unit mass find hpi, hp2 i and ψ(x, t) for φ(x) = x2 .
CHAPTER
11
Theory of Angular Momentum
11.1 INTRODUCTION A particle moving along a straight-line has a linear momentum mv. A revolving particle has angular momentum due to its mass and angular velocity. In classical mechanics, orbital angular momentum is the vector product of position (r) and linear momentum (p): L=r×p.
(11.1)
Let us imagine the electron as a small ball, revolving about some axis. It possesses angular momentum. Precisely, the angular momentum vector is a vector pointing along the axis of rotation with magnitude proportional to the speed of rotation. A vector which changes sign under space inversion (parity operation) is called a polar vector . r and p are polar vectors as they change sign. Since L = (−r) × (−p), L does not change sign under parity operation. Such vectors which do not change sign under parity operation are called axial vectors or psuedo-vectors. Classically, the angular momentum of the electron could have any magnitude. In quantum mechanics the orbital angular momentum operator is obtained by replacing p by −i~∇. The angular position of a particle is the angle at which it is located in two-dimensional polar coordinates. Why should one study about angular momentum? Let us point out some significances of angular momentum. Comet tails always point away from the sun because light carries linear momentum. In 1921 Einstein has shown that Planck’s black body law and the motion of molecules in a radiation fields could be explained if the linear momentum of a photon is ~k. In 1909 John Poynting noticed that polarized light has angular momentum − spin angular momentum associated with circular polarization. For a single photon it has a value of ±~. Gibson et al. [1] pioneered the use of orbital angular momentum to encode 261
262 Quantum Mechanics I: The Fundamentals information in free-space optical communication. Just like an intensity that is switched on or off can represent a value of zero or one (a bit), so different phase-front twists can represent different values. There are infinitely many different twists, and as a single photon can have these twists, photons can carry arbitrary amounts of information. Orbital angular momentum of light beams can be used for super-high density optical data storage [2]. For a particle rotating in a two-dimensional plane, we can choose the z-axis as the axis of rotation if the plane of rotation is the xy-plane. In this case the components Lx = Ly = 0 and Lz = L. For a central force problem, the force F acts in the direction of r. Hence, the torque is N = dL/dt = 0. Therefore, L is conserved in all central force problems. Since the potential is a spherically symmetric for the central force problem V (r) depends only on the magnitude of r. Because L is conserved classically for a central force, the motion takes place in a plane with L perpendicular to the plane. When the particle rotates in a three-dimensional space, the axis of rotation is perpendicular to the plane of rotation. However, the axis is free to take any direction. In this case L2 = L2x + L2y + L2z .
11.2 SCALAR WAVE FUNCTION UNDER ROTATIONS A rotation of r by an angle θ about an axis of unit vector n changes r to r′ . If the rotation operation in physical space is represented by R(n, θ), then r′ = R(n, θ)r. For example, if the z-axis is chosen as the rotation axis then we find cos θ − sin θ 0 R(k, θ) = (11.2) cos θ 0 sin θ . 0 0 1 Since rotation does not change the magnitude of the vector we have r′ ·r′ = r·r. Hence, RRT = RT R = I. The rotation matrix R is orthogonal. R(n, θ) gives the transformation matrix in physical space. We need to find the effect of such rotation in a wave function. As the wave function represents a vector in Hilbert space, the rotational transformation in physical space will lead to a unitary transformation UR (n, θ) in Hilbert space.
11.2.1
The Transformation Operator
Under a rotation through θ about n, ψ(r) → ψ ′ (r′ ) = ψ ′ (x′ , y ′ , z ′ ) where ψ ′ is the rotated function. We know that a rotated scalar function evaluated at the rotated point must be the same as the unrotated scalar function evaluated at the original point. In view of this because ψ is a scalar we have ψ ′ (r′ ) = ψ(r). The value of ψ(r) cannot change under rotation of the coordinate system even though our point of viewing is changed. From r′ = R(n, θ)r we write r = R(n, −θ)r′ by which ψ(r′ ) = ψ(R(n, −θ)r′ ). Dropping the prime on r′ ,
Theory of Angular Momentum 263 we get ψ ′ (r) = ψ(R(n, −θ)r). That is, the rotated wave function evaluated at r is the same as the unrotated function evaluated at the point R(n, −θ)r. If the transformation of ψ ′ (r) and ψ(r) is given by the operator UR (n, θ) then UR ψ(r) = ψ ′ (r) = ψ(R(n, −θ)r) . (11.3)
An expression for UR (n, θ) is obtained by choosing n along the z-axis and letting θ to be a first-order infinitesimal rotation angle ∆θ. With the choice n = k and θ → ∆θz , we obtain from Eq. (11.2) x + ∆θz y R(k, −∆θz )r ≈ (11.4) y − ∆θz x . z Hence, Eq. (11.3) becomes
UR (k, ∆θz )ψ(r) ≈ ψ(x + ∆θz y, y − ∆θz x, z) .
(11.5)
Using the Taylor series expansion ψ(x + ∆θz y, y − ∆θz x, z) =
∂ ∂ 1 − ∆θz x −y ψ(r) (11.6) ∂y ∂x
we get UR (k, ∆θz ) ≈ I − ∆θz
∂ ∂ x −y ∂y ∂x
=I−
i ∆θz k · L . ~
(11.7)
For a spinless particle, the infinitesimal generators of rotations about the zaxis is simply Lz . Generalizing Eq. (11.7) for an infinitesimal rotation ∆θ in the direction n, we get i UR (n, ∆θ) ≈ I − ∆θ n · L . ~
(11.8)
Suppose it is desired to find UR (n, θ) for some finite rotation θ. The effect of a small increase in the magnitude of θ from θ to θ + ∆θ is to follow the finite rotation UR (n, θ) by the infinitesimal rotation I − (i/~)∆θn · L to give i UR (n, θ + ∆θ) ≈ I − ∆θ n · L UR (n, θ) . (11.9) ~ Then UR (n, θ) satisfies the differential equation d i UR (n, θ) = − ∆θ n · L UR (n, θ) dθ ~
(11.10)
together with the boundary condition UR (0) = I. Its solution is UR (n, θ) = e−iθn·L/~ ,
ψ ′ (r) = e−iθn·L/~ψ(r) .
(11.11)
264 Quantum Mechanics I: The Fundamentals
11.2.2
Symmetry Groups
In classical mechanics the symmetry of a physical system leads to conservation laws. If the space is isotropic, it possesses rotational symmetry. Consequently, the angular momentum for such a system is conserved. Similarly, the homogeneity of space leads to translational symmetry and it conserves the linear momentum of the system. Invariance of a system under some symmetry operations results in some conservation laws and consequently we have a vast simplification of the mathematical theories of such systems. Associated with every symmetry is a mathematical object known as a symmetry group and the set of all the symmetry operations of a system satisfies the postulates of a group1 [3].
11.2.3
Lie Groups
The number of elements in a group is called its order . A group with finite number of elements is called a finite group whereas an infinite group contains infinite number of elements. An infinite group may further be either discrete or continuous. For example, a group of all integers including zero form a discrete infinite group under algebraic addition. If we consider a group of all real numbers x ∈ [−∞, ∞], they form an infinite continuous group under algebraic addition. The elements of a continuous group is characterized by a set of real, continuous, differentiable parameters (α1 , α2 , · · · , αr ). If r is finite then the group is said to be finite and r is called the order of the continuous group. For example, the two-dimensional rotation matrix cos θ − sin θ R(k, θ) = (11.12) sin θ cos θ forms a one-parameter infinite continuous group as θ can vary continuously between 0 to 2π. A continuously connected group in which the parameters of the product of two elements are continuous and differentiable functions of the parameters of the elements is called a Lie group. The dependence of the elements X1 , X2 , · · · of a Lie group on its r continuous parameters is written as Xi = Xi (α1 , α2 , · · · , αr ) ,
i = 1, 2, · · · .
(11.13)
It is convenient to choose the Cartesian parameters of a Lie group such that the image of the identity element is I = X(0, 0, · · · , 0). With this an element near the identity is written, due to the analytical properties of the Lie groups as X(0, 0, · · · , ǫj , · · · , 0) ≈ X(0, 0, · · · , 0) + iǫj Tj (0, 0, · · · , 0) 1 Group
theory is the mathematical tool to treat invariants and symmetries.
(11.14)
Theory of Angular Momentum 265 to first-order in ǫj . The operator Tj obtained from (11.14) is given by Tj
=
1 1 ∂X [X(0, · · · , ǫj , · · · , 0) − X(0, · · · , 0)] = . (11.15) ǫj →0 iǫj i ∂ǫj lim
Tj are called the generators of the group. If we write αj = N ǫj , where N is a large positive integer so that ǫj is a small quantity, then the element for any finite variation of the parameter αj is obtained from (11.13) as X (0, 0, · · · , αj , · · · , 0) = = =
lim [X(0, 0, · · · , ǫj , 0, · · · , 0)] N αj Tj lim 1 + i N →∞ N
N
N →∞
eiαj Tj .
(11.16)
We extend the above result to obtain r X X (α1 , α2 , · · · , αr ) = exp iαj Tj .
(11.17)
j=1
If Tj are known then all the elements of the Lie group corresponding to various αj values are found using Eq. (11.17). Hence, Tj are called the generators of the Lie group. We can prove that these generators of a group are linearly independent. Hence, they are not unique for a group. Any set of r-linear combination of these generators Tj can also be used as generators of the group. Since eiαk Tk and eiαl Tl are two elements of the group, their product must also belong to the group. So, the multiplication of two group elements will be given by the Baker–Campbell–Hausdorff formula [4] 1 1 A B e e = exp A + B + [A, B] + ([A, [A, B]] − [B, [A, B]]) + · · · . 2 12 (11.18) Therefore, it is clear that once the commutators of the generators are known, it is possible to evaluate the multiplication of any pair of group elements and the commutators P must be a linear combination of the generators Tj . We write [Tk , Tl ] = j fklj Tj , where fklj are called the structure constants of the group. Since [Tk , Tl ] = −[Tl , Tk ] we set fklj = −flkj . The group elements follow associative law of combination and the generators satisfy the Jacobi identity [Tk , [Tl , Tm ]] + [Tl , [Tm , Tk ]] + [Tm , [Tk , Tl ]] = 0 .
(11.19)
Moreover, we can prove that if the generators are Hermitian then the structure constants are purely imaginary.
266 Quantum Mechanics I: The Fundamentals
11.3 ORBITAL ANGULAR MOMENTUM In three-dimension with r = ix + jy + kz and p = ipx + jpy + kpz we write L = r × p as L = iLx + jLy + kLz where Lx = ypz − zpy ,
Ly = zpx − xpz ,
Lz = xpy − ypx .
(11.20)
The above are the classical components of L. The quantum mechanical operators Lx , Ly and Lz are obtained from Eq. (11.20) by replacing the variables by their corresponding operators. We obtain ∂ ∂ −z , (11.21a) Lx = −i~ y ∂z ∂y ∂ ∂ Ly = −i~ z −x , (11.21b) ∂x ∂z ∂ ∂ Lz = −i~ x −y . (11.21c) ∂y ∂x In general (11.22)
L = −i~r × ∇ .
Is the operator L a Hermitian? The operators x, y, z, −i~∂/∂x, −i~∂/∂y and −i~∂/∂z are Hermitian. Product of two operators is Hermitian only if their commutator bracket vanishes. Consider the operator Lx . Since y and pz , z and py are commuting operators their product is also a Hermitian. Hence Lx is Hermitian. Similarly, Ly and Lz are also Hermitian. Consequently, L is a Hermitian operator as well as L2 . Hence, L and L2 are observables. Let us transform Eq. (11.21) to a system of spherical polar coordinates x = r sin θ cos φ,
y = r sin θ sin φ,
z = r cos θ ,
(11.23)
where r2 = x2 + y 2 + z 2 ,
rdr = xdx + ydy + zdz
(11.24)
and ∂r ∂x ∂r ∂z ∂θ ∂x
= = =
tan φ =
x = sin θ cos φ , r
∂r y = = sin θ sin φ , ∂y r p 2 z x + y2 = cos θ , tan θ = , r z cos θ cos φ ∂θ cos θ sin φ ∂θ sin θ , = , =− , r ∂y r ∂z r y ∂φ sin φ ∂φ cos φ ∂φ , =− , = , =0. x ∂x r sin θ ∂y r sin θ ∂z
(11.25a) (11.25b) (11.25c) (11.25d)
Theory of Angular Momentum 267 Then ∂ ∂x
∂ ∂y ∂ ∂z
∂r ∂ ∂θ ∂ ∂φ ∂ + + ∂x ∂r ∂x ∂θ ∂x ∂φ ∂ cos θ cos φ ∂ sin φ ∂ = sin θ cos φ + − , ∂r r ∂θ r sin θ ∂φ ∂ cos θ sin φ ∂ cos φ ∂ = sin θ sin φ + + , ∂r r ∂θ r sin θ ∂φ ∂ sin θ ∂ = cos θ − . ∂r r ∂θ =
Substitution of Eqs. (11.23) and (11.26) in Eq. (11.21) gives ∂ ∂ Lx = −i~ − sin φ − cot θ cos φ , ∂θ ∂φ ∂ ∂ Ly = −i~ cos φ − cot θ sin φ , ∂θ ∂φ ∂ Lz = −i~ . ∂φ Using Eqs. (11.27) L2 is worked out as ∂ 1 ∂2 1 ∂ 2 2 L = −~ sin θ + . sin θ ∂θ ∂θ sin2 θ ∂φ2 Using the Eqs. (11.21) and (11.26) we get two new operators ∂ ∂ L± = Lx ± iLy = ~ e±iφ ∓ + i cot θ . ∂θ ∂φ
(11.26a) (11.26b) (11.26c)
(11.27a) (11.27b) (11.27c)
(11.28)
(11.29)
The components of L and L2 commute with r and ∂/∂r. Lz commutes with any operator and with ∇2 and with any spherically symmetric Hamiltonian, H, that is [L, H] = 0 and [L2 , H] = 0. This implies that (i) an eigenfunction of a spherically symmetric H is also an eigenfunction of Lx or Ly or Lz or of any linear combination of these operators and (ii) the law of conservation of angular momentum holds in both quantum and classical mechanics. Usually, only one of the three Cartesian components of angular momentum is specified in a quantum mechanical state. The remaining two are determined from the known third. Generally, the z-component of L is specified due to the simplicity of its operator form.
11.4 EIGENPAIRS OF L2 AND Lz The eigenvalue equation for L2 in spherical polar coordinates is 1 ∂ ∂ 1 ∂2 2 2 L ψ = −~ sin θ + ψ = λ~2 ψ , sin θ ∂θ ∂θ sin2 θ ∂ 2 φ
(11.30)
268 Quantum Mechanics I: The Fundamentals where λ~2 is the eigenvalue of L2 . Equation (11.30) is solved by the method of separation of variables. Equation (11.30) is same as the Eq. (4.157) which we considered earlier in the study of rigid rotator. The only difference is that instead of ~2 we have ~2 /(2I) in the left-side of Eq. (4.157) and instead of λ~2 we have E in the right-side of Eq. (4.157). Referring to our discussion on rigid rotator the eigenfunctions ψlm of L2 are written as ψlm
= = =
Θlm (θ)Φm (φ) 1/2 (2l + 1) (l − |m|)! |m| Pl (cos θ)eimφ 4π (l + |m|)! Ylm (θ, φ) .
(11.31)
The eigenvalues of L2 are λ~2 = l(l + 1)~2 . As L2 and Lz are commuting operators they have same set of eigenfunctions. Ylm are the eigenfunctions of L2 . Therefore, Ylm are the eigenfunctions of Lz also. Now, from the eigenvalue equation of Lz , Lz Ylm = −i~∂Ylm /∂φ we obtain Lz Ylm = m~Ylm . To summarize, L2 Ylm (θ, φ)
= l(l + 1)~2 Ylm (θ, φ) ,
l = 0, 1, 2, · · ·
(11.32a)
= m~Ylm (θ, φ), m = l, l − 1, · · · , −l. (11.32b) p The eigenvalues of L are l(l + 1)~. (Here l measures the magnitude of L). But as noted in sec. 1.4 the Bohr’s prediction is L = n~. The point is that we have to discard the planetary model of electrons moving about a nucleus. The angle φ is a measure of the rotation about the z-axis. The z-component of angular momentum Lz has eigenvalues m~. Since m takes only integer values we say that only certain orientations of L alone are possible. This is called space quantization in vector atom model2 . Figure 11.1 shows orientation of L for l = 2. Lz Ylm (θ, φ)
11.5 PROPERTIES OF COMPONENTS OF L AND L2 In the following we present some of the useful properties of the components of L and L2 . 2 When an atom is placed in a magnetic field applied along z-axis, the energy of the atom is changed by an amount proportional to the z-component of its magnetic moment. This change is related to Lz . For this reason m is called the magnetic quantum number .
Theory of Angular Momentum 269
2h
m=2 m=1
h 0
m=0
-h
m=-1 m=-2
-2 h
FIGURE 11.1
11.5.1
Orientations of L for l = 2.
Commutation Relations
It is easy to show that [Lx , x] = 0,
[Lx , y] = i~z,
[Lx , z] = −i~y ,
[Lx , px ] = 0, [Lx , py ] = i~pz , [Lx , pz ] = −i~py , Lx , p2 = 0, Ly , p2 = 0, Lz , p2 = 0,
(11.33a) (11.33b) (11.33c)
[Lx , Ly ] = i~Lz , [Ly , Lz ] = i~Lx , [Lz , Lx ] = i~Ly , (11.33d) Lx , L2 = L2 , Lx = 0 , Ly , L2 = L2 , Ly = 0 , (11.33e) Lz , L2 = L2 , Lz = 0, L × L = i~L . (11.33f)
We prove a few of the above results. Let us calculate [Lx , Ly ]. We obtain [Lx , Ly ]
= [ypz − zpy , zpx − xpz ] = [ypz , zpx ] − [ypz , xpz ] − [zpy , zpx ] + [zpy , xpz ]
= ypz (zpx ) − zpx (ypz ) − ypz (xpz ) + xpz (ypz ) − zpy (zpx ) +zpx (zpy ) + zpy (xpz ) − xpz (zpy )
= −i~ypx + yzpz px − yzpxpz − xypz pz + xypz pz − z 2 py px +z 2 px py + zxpy pz + i~xpy − xzpz py = i~ (xpy − ypx ) = i~Lz .
Next, compute [L2 , Lz ]. We get 2 L , Lz = L2x + L2y + L2z , Lz = L2x , Lz + L2y , Lz .
(11.34)
(11.35)
270 Quantum Mechanics I: The Fundamentals Using the relations [AB, C] = A[B, C] + [A, C]B we obtain 2 L , Lz = Lx [Lx , Lz ] + [Lx , Lz ] Lx + Ly [Ly , Lz ] + [Ly , Lz ] Ly = −i~Lx Ly − i~Ly Lx + i~Ly Lx + i~Lx Ly = 0 . (11.36) The nonzero values of [Lx , Ly ], [Ly , Lz ] and [Lz , Lx ] imply that Lx , Ly and Lz cannot be measured simultaneously with desired accuracy, that is, we cannot simultaneously have definite values of Lx , Ly and Lz . The operator L2 commutes with every component of L. However, the components of L are not commuting operators. Hence, it is impossible to form a simultaneous eigenfunction of two different components of L.
11.5.2
What is the Significance of Commutator of L2 and Lz = 0?
In order to understand the significance of the relation [L2 , Lz ] = 0 consider L2 and Lz . What is the condition for ψ to be a simultaneous eigenfunction of L2 and Lz ? If ψ is an eigenfunction of L2 and Lz with eigenvalues λ and λ′ respectively then L2 ψ Lz ψ
= =
λψ , ′
λψ.
(11.37a) (11.37b)
Operating −Lz from left on Eq. (11.37a) we have −Lz L2 ψ = −λλ′ ψ .
(11.38a)
Operating L2 from left on Eq. (11.37b) we get L2 Lz ψ = λλ′ ψ .
(11.38b)
Adding Eqs. (11.38a) and (11.38b) we obtain L2 Lz ψ−Lz L2 ψ = [L2 , Lz ]ψ = 0. That is, [L2 , Lz ] = 0. Thus, the above relation implies that if ψ is an eigenfunction of L2 then it is an eigenfunction of Lz also and vice-verse. Further, L2 and Lz can be measured simultaneously.
11.5.3
Ladder Operators L±
Earlier we defined the ladder operators L± as (Lx ± iLy ). Then [Lz , L± ]
= = = =
Solved Problem 1: Find the effect of L± on ψ.
[Lz , Lx ] ± i [Lz , Ly ]
i~Ly ± ~Lx ±~ (Lx ± iLy )
±~L± .
(11.39)
Theory of Angular Momentum 271 Consider Lz (L± ψ). We obtain Lz (L± ψ)
= (Lz L± ) ψ = (Lz L± − L± Lz + L± Lz ) ψ = ([Lz , L± ] + L± Lz ) ψ = (±~L± + L± Lz ) ψ
= (±~L± + m~L± ) ψ = (m ± 1) ~L± ψ .
(11.40)
The effect of Lz on ψ is to yield the eigenvalues m~. When L± is also operated then the eigenvalue is changed to (m ± 1)~ as seen from Eq. (11.40). Thus, the effect of L± is to change the eigenvalue by ±~. L+ increases the eigenvalue by +~ while L− decreases the eigenvalue by −~.
11.5.4
Expectation Values
Because of the symmetry properties of ψnlm in angular coordinates, we obtain (11.41)
hLx,y ilm = (ψnlm , Lx,y ψnlm ) = (Ylm , Lx,y Ylm ) = 0 . Further L± ψjm
(11.42)
= (λj ± ~) ψjm = ψjm±1
and hL2x ilm
= =
Then
1 hL2y ilm = Ylm , L2x Ylm = Ylm , L2 − L2z Ylm 2 1 2 2 ~ l(l + 1) − m . (11.43) 2
1/2 =~ ∆Lx = ∆Ly = hL2x ilm − hLx i2lm
1 l(l + 1) − m2 2
1/2
. (11.44)
p For Yll states ∆Lx = ∆Ly = ~ l/2. We note that hL2x i 6= 0. If hL2x i = 0 then all the Lx ’s are zero because L2x is positive. In this case the system is in a definite state for which Lx = 0 which means the uncertainty in Lx is zero. Similarly, for Ly also the uncertainty is zero. But they are noncommuting operators. Therefore, the uncertainty relation is (∆Lx )(∆Ly ) ≥ ~/2. Hence, hL2x i = 0 is not possible. We can find [5] Z Z ∗ 3 hLi = ψ Lψ d r = m r × J d3 r . (11.45) V
272 Quantum Mechanics I: The Fundamentals In terms of matrix representation hLi = hΨl |L|Ψl i. Expanding Ψl as Ψl =
l X
am ψlm
(11.46)
m=−l
we write hLi =
X
a∗m Lmm′ am′ ,
Lmm′ = hψlm |L|ψlm′ i .
m,m′
(11.47)
p ∆Lx and ∆Ly are minimum when m = l. In this case ∆Lx = ∆Ly = ~ l/2. This is the highest precision of a measurement on Lx and Ly in a state Ylm .
11.5.5
Lnz and eLz are Hermitians
In spherical polar coordinates Z 2π ψ ∗ Lz Φ dφ = 0
=
∂Φ dφ ψ ∗ −i~ ∂φ 0 Z 2π ∂ψ ∗ ∗ 2π dφ . [−i~ψ Φ]φ=0 + i~ Φ ∂φ 0
Z
2π
The first term in the above equation vanishes. Then ∗ Z 2π Z 2π ∂ψ ∗ ψ Lz Φ dφ = −i~ Φ dφ . ∂φ 0 0
(11.48)
(11.49)
Therefore, Lz is Hermitian. Since nth power of a Hermitian operator is also a Hermitian, Lnz are also Hermitians. Further, we write eL z =
∞ X 1 k L . k! z
(11.50)
k=0
Since Lkz , k = 0, 1, 2, · · · are Hermitians, eLz is also a Hermitian.
11.6 EIGENSPECTRA THROUGH COMMUTATION RELATIONS In sec. 11.4 we obtained the eigenvalues of L2 and Lz by solving their eigenvalue equations which are differential equations. We now determine the eigenvalues of L2 and Lz using commutation relations together with the ladder operators L± . The procedure is similar to the method which we used in sec. 4.4 for linear harmonic oscillator.
11.6.1
Determination of Eigenvalues
Let ψlm denote the state vector belonging to the eigenvalues λl ~2 of L2 and m~ of Lz . (This state can be represented as |l, mi or |lmi). Then Lz ψlm = m~ψlm ,
L2 ψlm = λl ~2 ψlm .
(11.51)
Theory of Angular Momentum 273 ± Introduce a function φ± lm through the equation L± ψlm = φlm . Since the effect of L± on ψlm is to change the eigenvalue by ±~ the function φ± lm is proportional to ψlm±1 . That is, φ± = N ~ψ where N is a constant to be determined. ± lm±1 ± lm We denote the minimum of the eigenvalue as m1 and the maximum as m2 . We determine their values. If L+ is operated on ψlm2 then the eigenvalue must be m2 + 1 which is not possible. Thus, L+ ψlm2 = 0. Similarly, L− ψlm1 = 0. Now, consider L− (L+ ψlm2 ) = 0 . (11.52)
Expanding the left-side of Eq. (11.52) we get 0 = = = The above equation gives
(Lx − iLy ) (Lx + iLy ) ψlm2 L2 − L2z − ~Lz ψlm2 ~2 λl − m22 − m2 ψlm2 .
(11.53) (11.54)
λl − m2 (m2 + 1) = 0 . Similarly, from L+ (L− ψlm1 ) = 0 we have
(11.55)
λl − m1 (m1 − 1) = 0 .
Comparison of Eqs. (11.54) and (11.55) gives m21 − m1 = m22 + m2 or (11.56)
(m2 + m1 ) (m2 − m1 + 1) = 0 .
From Eq. (11.56), m2 is −m1 or m2 = m1 − 1. Because m2 is the maximum value it cannot be m1 − 1. Therefore, we choose m2 = −m1 . Since the values of m are integers the number m2 − m1 must be an integer and we denote it as 2l. The possible values of m are m = l, l − 1, · · · , −l + 1, −l. Then the possible values of l are 0, 1/2, 1, · · · . Substitution of m2 = −m1 in the expression m2 − m1 = 2l gives m2 = l. Further, from Eq. (11.54) λl is obtained as (11.57)
λl = m2 (m2 + 1) = l(l + 1) . Next, we determine N± . Using Eq. (11.49), we write Z Z ± ∗ ± 2 2 ∗ ψlm±1 ψlm±1 dτ . φlm φlm dτ = |N± | ~
(11.58)
Since ψlm form a complete set, the integral in the right-side of the above ± equation is 1. Replacing φ± lm using L± ψlm = φlm , we have Z ∗ |N± |2 ~2 = (L± ψlm ) (L± ψlm ) dτ = ~2 [(l ∓ m)(l ± m + 1)] .
(11.59)
Now, the eigenvalue equations for Lz , L2 and L± are Lz ψlm 2
L ψlm L± ψlm
= = =
(11.60a)
m~ψlm , 2
(11.60b)
~ l(l + 1)ψlm , ~ [(l ∓ m)(l ± m + 1)]
1/2
ψlm±1 .
(11.60c)
274 Quantum Mechanics I: The Fundamentals
11.6.2
Properties of Eigenvalues
From (11.60b) we state the following features of eigenvalue spectra of L2 : 1. The eigenvalues of L2 are 6= (l~)2 but = l(l + 1)~2 and are greater than (l~)2 . The reason for this is as follows [6]. For a state ψlm we have hL2 i = λ~2 and hL2z i = m2 ~2 . Then hL2 i = hL2x i + hL2y i + hL2z i
(11.61)
which gives λ~2
=
hL2x i + hL2y i + (m~)2 .
(11.62)
The expectation values of L2x and L2y are nonnegative. This means λ ≥ m2 . If we want to have a state with some positive value of l for m then λ ≥ l2 . The eigenvalues of L2 are ≥ (l~)2 . To have equality sign we must have hL2x i = hL2y i = 0. If the projection of the angular momentum on the z-axis were the same length as the angular momentum vector then the projections on the x and y axes would be zero. This violates the uncertainty relation. This is because the components of angular momentum do not commute with each other as seen earlier. Thus, the eigenvalue of L2 is greater than (l~)2 . 2. For a given l the m can take (2l + 1) distinct values. Therefore, eigenvalues of L2 are (2l + 1)-fold degenerate. 3. What is the fundamental difference between the states with integer and half-integer values of l? Consider the equation e−iθLz ψlm = e−imθ ψlm . For integer values of l the states ψlm will not change under a rotation through θ = 2π and are single-valued. But the states will change sign for half-integer angular momentum and are hence double-valued under rotation. Next, consider the ψlm given by ψlm
= (−1)m Ylm (θ, φ) 1/2 (2l + 1)(l − |m|)! |m| = Pl (cos θ)eimφ . 4π(l + |m|)!
(11.63)
The above solution is valid only for integer values of l and m. For example, consider l = 1/2, m = ±1/2. We have √ 1 ψ 21 ± 12 = √ e±iφ/2 sin θ . 4π
(11.64)
Theory of Angular Momentum 275 Consider L± ψ 12 12 . We expect L+ ψ 12 12 = 0. But √ cos θ ∂ ∂ +i eiφ/2 sin θ L+ ψ 21 12 = eiφ ∂θ sin θ ∂φ √ 1 ∂ − cot θ sin θ = ei3φ/2 ∂θ 2 6= 0 .
(11.65a)
Similarly, L− ψ 21 − 12
= 6=
√ ∂ ∂ + i cot θ e−iφ/2 sin θ e−iφ − ∂θ ∂φ 0. (11.65b)
Therefore, the functions (11.63) corresponding to half-integer values of l are not admissible eigenfunctions of L. Half-integer values of l have to be discarded. 4. In sec. 11.4 the eigenfunctions and eigenvalues of L2 are obtained by solving the eigenvalue Eq. (11.30). In this case the eigenvalues are l(l + 1)~2 where l is a nonnegative integer, l = 0, 1, 2, · · · . In contrast to this, when the commutation relations are alone used as above, the eigenvalues are l(l +1)~2 where l = 0, 1/2, 1, · · · . In addition to integer l, half-integer values also exist, however, they do not correspond to orbital angular momentum. In working with the commutation relations instead of the quantum number l suppose we use a different quantum number j. Then the eigenvalues are j(j + 1)~ where j = 0, 1/2, 1, · · · . These j quantum numbers will be representing the total angular momentum J = L + S where S is spin angular momentum.
Solved Problem 2: Show that ψ± = N (x ± iy)f (r) is an eigenfunction both of L2 and Lz and state the corresponding eigenvalues. p We have x+iy = r sin θ(cos φ+i sin φ) = r sin θ eiφ = −r 8π/3 Y11 . The given ψ+ can be rewritten as p ψ+ = −N 8π/3 rf (r)Y11 . (11.66) Since L2 Ylm = l(l + 1)~2 Ylm and Lz Ylm = m~ we get L2 ψ+ = 2~2 ψ+ ,
Lz ψ+ = ~ψ+ .
(11.67)
The eigenvalues of L2 and Lz corresponding to ψ+ are 2~2 and ~p respectively. Next, we have x − iy = r sin θ(cos φ − i sin φ) = r sin θ e−iφ = r 8π/3 Y1−1 . Then p ψ− = N 8π/3 rf (r)Y1−1 . (11.68)
276 Quantum Mechanics I: The Fundamentals We obtain L2 ψ− = 2~2 ψ− ,
Lz ψ− = −~ψ− .
(11.69)
The eigenvalues of L2 and Lz corresponding to ψ− are 2~2 and −~ respectively.
11.7 MATRIX REPRESENTATION OF L2 , Lz AND L± Suppose the wave function ψ is a vector. For example, ψ of an atom is complex and we can treat it as a two-component vector function. If ψ is a regarded as a vector then the operator acting on it must be represented in matrix form. For illustrative purpose consider f (x) and g(x) such that Af (x) = g(x). Since the wave function has the completeness property, we write X X f (x) = am ψm , g(x) = bn ψn . (11.70) m
n
P
P Then Af (x) = g(x) is rewritten as m am Aψm = n bn ψn . am can be determined using normalization condition. To find bn , multiply the above expression by ψl∗ and integrate over entire space which gives Z X X X am Alm = am ψl∗ Aψm dτ = bn δln = bl , (11.71) n
where we have used Alm =
Z
ψl∗ Aψm dτ .
(11.72)
Alm form a matrix:
Alm
A11
A 21 = An1
A12 A22 An2
···
···
··· ···
This is called matrix representation.
11.7.1
A1n
A2n . Ann
(11.73)
Matrix Elements of L, Lz and L±
The matrix elements of an operator A in a given l state are given by Z ∗ Aψlm′ dτ, m, m′ = l, l − 1, · · · , −l (11.74) Amm′ = hlm|A|lm′ i = ψlm where m and m′ are the allowed range of eigenvalues of Lz . The matrices for any values of l are represented by rows labeled by m and columns by m′ . The
Theory of Angular Momentum 277 matrix form of A is written as hll|A|lli hll|A|ll − 1i ··· hll − 1|A|lli hll − 1|A|ll − 1i · · · A= ··· hl − l|A|lli hl − l|A|ll − 1i ··· The elements of L2 are 2
′
hlm|L |lm i =
Z
hll|A|l − li
hll − 1|A|l − li . hl − l|A|l − li
∗ ψlm L2 ψlm′ dτ = ~2 l(l + 1)δmm′ .
(11.75)
(11.76)
Similarly, hlm|Lz |lm′ i =
m~δmm′ ,
hlm|L± |lm i =
~ [(l ∓ m )(l ± m + 1)]
′
′
(11.77) ′
1/2
δmm′ ±1 .
(11.78)
For l = 0 the matrix elements of L2 , Lz , L± , Lx and Ly all are zero.
Solved Problem 3: Obtain the matrix representation of the components of angular momentum for j = 1/2 and l = 1. For j = 1/2, the values of m and m′ are −j, j, that is −1/2, 1/2. The eigenvalues of J2 are j(j + 1)~2 = 3~2 /4 whereas the eigenvalues of Jz are m~ = −~/2, ~/2. Now, hjj|J2 |jji hjj|J2 |j − ji J2 = 2 2 hj − j|J |jji hj − j|J |j − ji D 1 1 1 −1 E D 1 1 1 1 E 2 J2 J 22 22 2 2 22 . = (11.79) D 1 −1 2 1 1 E D 1 −1 2 1 −1 E J J 2 2 22 2 2 2 2
The matrix elements are determined from Eq. (11.76). They are D1 1 1 1E D 1 1 1 −1 E 3 2 2 2 = ~ , =0, J J 22 22 4 22 2 2 D 1 −1 E D 1 −1 E 3 2 1 1 2 1 −1 = 0, = ~2 . J J 2 2 22 2 2 2 2 4 Then
1 3 J2 = ~2 4 0
0
= 3 ~2 I . 4 1
(11.80a) (11.80b)
(11.81)
278 Quantum Mechanics I: The Fundamentals Similarly, we obtain Jz
=
J+
=
and Jx
=
Jy
=
0 ~ 1 = ~ σz , 2 2 0 −1 0 1 0 , J− = ~ ~ 0 0 1
(11.82a) 0 0
1 ~ 0 1 ~ (J+ + J− ) = = σx , 2 2 2 1 0 0 −i 1 ~ = ~ σy . (J+ − J− ) = 2i 2 2 i 0
(11.82b)
(11.82c)
(11.82d)
As already discussed l = 1/2 cannot represent orbital angular momentum, the above represents spin angular momentum with quantum number s = 1/2. For l = 1 we obtain 1 0 0 1 0 0 L2 = 2~2 (11.83a) 0 0 1 0 , Lz = ~ 0 0 , 0 0 1 0 0 −1 0 1 0 0 −1 0 i~ ~ , Ly = Lx = √ 1 0 1 1 0 −1 . (11.83b) 2 2 0 1 0 0 1 0
11.8 WHAT IS SPIN? So far, we have not introduced the term spin in quantum mechanics. In this section the notion of spin [7] is introduced and then a few examples of necessity of spin angular momentum are mentioned.
11.8.1
Physical Picture of Spin [7]
Soon after Bohr came up with his atom model, it was assumed that the electron revolving around the nucleus was also spinning about an axis. Later, it was realized that this assumption had no validity. Spin of particle is not a measure of how it is spinning. In fact, it is impossible to tell whether an electron is spinning. The word spin is a convenient way of representing the intrinsic angular momentum of a particle.
Theory of Angular Momentum 279 The spin is said to be a nonorbital, intrinsic or inherent angular momentum. Also, spin is thought of as due to an internal structure of the electron. Further, the spin arises in a natural way from Dirac equation. We can regard spin as due to a circulating flow of energy or a momentum density, in the electron wave field. Bellinfante established that this picture of spin is valid not only for electrons, but also for photons, vector mesons and gravitons. In all these cases the spin angular momentum is due to a circulating energy flow in the fields. The spin of the electron has a close classical analog. It is an angular momentum of the same kind as possessed by the fields of a circularly polarized electromagnetic wave. Gordon showed that the magnetic moment of the electron is due to the circulating flow of charge in the electron wave field. This means that spin and the magnetic moment are not due to the internal properties of the electron but with the structure of its wave field. The momentum density in the Dirac field is [7] G=
~ ∗ ~ [ψ ∇ψ − (∇ψ ∗ )ψ] + ∇ × (ψ ∗ σψ) , 2i 4
(11.84)
where σ1 = −iα2 α3 , σ2 = −iα3 α1 and σ3 = −iα1 α2 . Here σi ’s are Pauli matrices and αi ’s are Dirac matrices (see sec. 19.3.1). The first term in G is associated with the translational motion of the electron. The second term is associated with circulating flow of energy. For a Gaussian wave packet ψ=
2 2 1 e−r /2d , (πd2 )3/4
d≫
~ mc
(11.85)
the first term in Eq. (11.84) is zero and G=
~ 4d2
1 πd2
3/2
e−r
2
/d2
(−2y x ˆ + 2xˆ y) .
(11.86)
The flow lines for the energy are circular. Such a circulating flow of energy will give an angular momentum and is the spin. For historical continuity this property was called spin and not means that the electron is spinning. For an arbitrary wave packet, the net angular momentum is Z ˆ × G d3 x J = x Z ~ ˆ × [ψ ∗ ∇ψ − (∇ψ ∗ )ψ] d3 x = x 2i Z ~ ˆ × [∇ × (ψ ∗ σ)ψ] d3 x . + x (11.87) 4 Expanding the cross product in the second integral into two dot products and
280 Quantum Mechanics I: The Fundamentals then integrating both of these by parts we obtain Z Z ~ ~ ∗ ∗ 3 ˆ × [ψ ∇ψ − (∇ψ )ψ] d x + J = ψ ∗ σψ d3 x x 2i 2 ~ = L + hσi 2 = L+S.
(11.88)
The first term is the orbital angular momentum, and the second term is the spin S. The first term is independent of the spin-orbital state. The second term is dependent on the spin-orbital state. Because the spin S given by Eq. (11.88) is the expectation value of σ, the operator representing the spin must be Sop =
~ σ. 2
(11.89)
Thus, the spin operator obeys all the usual commutation relations. In particular, the eigenvalues of any component of the spin, say z-component, are ±~/2 (or ±1/2 in units of ~). Whatever we have said in the earlier sections of the present chapter for orbital angular momentum L is also true for spin angular momentum and also true for the total angular momentum J = L + S.
11.8.2
Components of Spin Operator
In matrix form the components of the spin operator are (the eigenvalues of components of spin are in units of ~) 1 0 0 1 ~ , S+ = ~ , Sz = (11.90a) 2 0 −1 0 0 0 0 0 1 , (11.90b) , Sx = 1 (S+ + S− ) = ~ S− = ~ 2 2 1 0 1 0 1 ~ 0 −i (S− − S+ ) = , (11.90c) Sy = 2i 2 i 0
Then
S2 = Sx2 + Sy2 + Sz2 =
1 0
3 2 . ~ 4 0 1
(11.91)
The eigenvalues of Sx , Sy , Sz and S2 are ±~/2, ±~/2, ±~/2 and 3~2 /4 respectively.
Theory of Angular Momentum 281 The components of S satisfy orbital angular momentum type commutation relations: [Sx , Sy ] = i~Sz ,
[Sy , Sz ] = i~Sx ,
[Sz , Sx ] = i~Sy .
(11.92)
Verify that S+ |αi = 0 ,
S− |αi = ~|βi ,
S+ |βi = ~|αi ,
S− |βi = 0 ,
(11.93)
where |αi and |βi are the eigenkets of Sz with eigenvalues ~/2 and −~/2 respectively. Using the above relations it is easy to prove that σx |αi
=
σy |βi
|βi ,
=
−i|αi ,
σx |βi = |αi ,
σz |αi = |αi ,
σy |αi = i|βi ,
σz |βi = −|βi .
(11.94a) (11.94b)
Because Sx = (S+ + S− )/2 we obtain Sx |αi =
1 ~ 1 (S+ + S− ) |αi = (S+ |αi + S− |αi) = |βi 2 2 2
(11.95a)
Sx |βi =
1 1 ~ (S+ + S− ) |βi = (S+ |βi + S− |βi) = |αi . 2 2 2
(11.95b)
and
Similarly, we obtain Sy |αi =
~ |βi , 2
Sy |βi = −
i~ |αi . 2
(11.95c)
Since |αi and |βi are eigenfunctions of Sz with eigenvalues ±~/2, we have Sz |αi =
~ |αi , 2
Sz |βi =
~ |βi . 2
(11.96)
Using the relation σx = (2/~)Sx we get σx |αi =
2 Sx |αi = |βi , ~
σx |βi = |αi .
(11.97a)
σy |βi = −i|αi , σz |βi = −|βi .
(11.97b) (11.97c)
In a similar manner we get σy |αi σz |αi
11.8.3
= i|βi , = |αi ,
Normalized Eigenfunctions of σx , σy and σz
We obtain the normalized eigenfunctions of σx , σy and σz . Since σx |αi = |βi, σx |βi = |αi we have σx (|αi + |βi) = |αi + |βi . (11.98)
282 Quantum Mechanics I: The Fundamentals Similarly, from Eqs. (11.97b) we obtain σx (|αi − |βi) σy (|αi + i|βi)
= =
−|αi + |βi , |αi + i|βi ,
(11.99a) (11.99b)
= −|αi + i|βi . (11.99c) √ The normalization gives the factor 1/ 2. What is the major difference between the forms of operators and eigenfunctions of orbital angular momentum and spin angular momentum? For a particle with spin ~/2 we have [5] Z Z ∗ hSi = ψ Sψ dτ = m r × Js d3 r . (11.100) σy (|αi − i|βi)
V
11.8.4
Some Examples of Necessity of Spin Angular Momentum
Without spin the quantum theory cannot explain many observed phenomena. Some examples are given below: 1. The energy levels of hydrogen atom are given by Eq. (12.35). Since En depends only on n and independent of l and m we say that the energy levels are degenerate with respect to l and m. The number of degeneracy is 2n2 (see Eq. (12.65)). To distinguish these states we use spin angular momentum. 2. The magnetic dipole moments and susceptibility of ferromagnets are explained with spin angular momentum. 3. The D-lines of sodium obtained at 5890A◦ and 5896A◦ show the existence of fine structure and is explained using spin. 4. A beam of atoms passing through an inhomogeneous magnetic field should split into (2l + 1) lines. In an experiment with l = 0 (ground state), the beam splits into two separate beams. Therefore, some additional angular momentum is required to explain the above observation.
Solved Problem 4: Show that S1 · S2 = S1z S2z + (S1+ S2− + S1− S2+ )/2. We have S1 · S2 S1x S1x S2x
S1y S2y
= S1x S2x + S1y S2y + S1z S2z , = (S1+ + S1− )/2 , S2x = (S2+ + S2− )/2 , S1+ + S1− S2+ + S2− = 2 2 = (S1+ S2+ + S1+ S2− + S1− S2+ + S1− S2− ) /4 , S1+ − S1− S2+ − S2− = 2i 2i = − (S1+ S2+ + S1+ S2− + S1− S2+ − S1− S2− ) /4 .(11.101)
Theory of Angular Momentum 283 Substituting these in the expression for S1 · S2 we get S1 · S2 = S1z S2z + (S1+ S2− + S1− S2+ )/2.
11.9 SPIN STATES OF AN ELECTRON Since the spin quantum number for electron is found to be 1/2 the components of S after setting ~ = 1 are Sx =
1 σx , 2
Sy =
1 σy , 2
Sz =
1 σz , 2
(11.102)
where the Pauli’s spin matrices σ’s are given by 0 1 0 −i 1 0 , σy = , σz = . (11.103) σx = 1 0 i 0 0 −1 We have
S2 = Sx2 + Sy2 + Sz2 =
3 1 0 = s(s + 1)I , 4 0 1
(11.104)
where s = 1/2. The operators S2 and Sz are diagonal implying that we can represent the spin-up and spin-down states of them by a two-component column vector called spinor . We write u 1 u . (11.105) Sz =± 2 v v That is, and get
u =± u v v −1
1
0
0
=± u . v −v u
(11.106)
(11.107)
For plus sign v = 0 and for minus sign u = 0. What are the eigenvectors of Sx and Sy ?
11.9.1
Eigenvectors of Sz
We have σz χ± = ±χ± . The following relations for Pauli spin matrices are easy to verify: σx2 = σy2 = σz2 = 1, σx σy = iσz , σy σz = iσx , σz σx = iσy , σx σy + σy σx = 0,
σy σz + σz σy = 0,
(11.108a)
σz σx + σx σz = 0. (11.108b)
284 Quantum Mechanics I: The Fundamentals and σx σy σz = i. We obtain χ+ =
1 , 0
χ− =
0 . 1
(11.109)
These two eigenstates are the spin-up and spin-down states respectively. They are the states where z-component of the spin is parallel (ms = 1/2) and antiparallel (ms = −1/2) to the z-axis. In terms of χ+ and χ− we write the arbitrary spinor as α+ 1 0 = α+ + α− , |α+ |2 + |α− |2 = 1 . (11.110) α− 0 1
11.9.2
Expectation Value of S
Let us calculate the expectation value of S. We write XX hα|S|αi = hα|iihi|S|jihj|αi i
We write the right-side of the above equation as α∗+ , α∗− S hSx i
= =
hSy i
=
1 ∗ ∗ 0 1 α+ α+ , α− 2 α− 1 0 1 ∗ α+ α− + α∗− α+ , 2
1 ∗ ∗ 0 −i α+ α+ , α− 2 α− i 0
= − hSz i
= =
(11.111)
j
i ∗ α+ α− − α∗− α+ , 2
1 ∗ ∗ α+ α ,α 2 + − −α− 1 |α+ |2 − |α− |2 . 2
α+ α−
. Then
(11.112a)
(11.112b)
(11.112c)
Solved Problem 5: ˆ where n ˆ Determine the eigenvalues and the eigenvectors of the operator σ · n is a unit vector in an arbitrary direction. We have
2
ˆ )2 = (σx nx + σy ny + σz nz ) = 1 . (σ · n
(11.113)
Theory of Angular Momentum 285 ˆ are thus ±1. Denote the eigenvectors as χn+ and χn− . The eigenvalues of σ · n To determine χn± , construct the vectors 1 1 (1 + σ · n ˆ )χ , (1 − σ · n ˆ )χ , (11.114) 2 2 where χ = a+ χ+ + a− χ− is an arbitrary spin vector. Operation of σ · n ˆ on each of these vectors gives 1 1 ˆ )χ = (1 + σ · n ˆ )χ , σ·n ˆ (1 + σ · n 2 2 1 1 σ·n ˆ (1 − σ · n ˆ )χ = − (1 − σ · n ˆ )χ . 2 2 For χ = χ+ these two equations become 1 (1 + σ · n ˆ )χ+ 2
= = =
1 (1 − σ · n ˆ )χ+ 2
=
(11.115a) (11.115b)
1 (1 + σx nx + σy ny + σz nz )χ+ 2 1 (χ+ + nx σx χ+ + ny σy χ+ + nz σz χ+ ) 2 1 [(1 + nz ) χ+ + (nx + iny ) χ− ] (11.116a) 2 1 [(1 − nz ) χ+ − (nx + iny ) χ− ] . (11.116b) 2
The square of norms of these two vectors are obtained as (1 + nz )/2 and (1 − nz )/2. Then the normalized eigenfunctions are χn+
=
χn−
=
1 p [(1 + nz )χ+ + (nx + iny )χ− ] , 2(1 + nz ) 1 p [(1 − nz )χ+ − (nx + iny )χ− ] . 2(1 − nz )
(11.117a) (11.117b)
In the state χn+ , a measurement of the z-component of the spin will give the values ±1/2 with the relative probabilities (1 + nz )2 : (n2x + n2y ) = cos2 (θ/2) : sin2 (θ/2) ,
(11.118)
where θ is the angle between n ˆ and z.
11.10 SPIN-ORBIT COUPLING There is a magnetic moment (µ) associated with the intrinsic rotation of the electron. It is proportional to the spin S. The magnetic moment interacts with the magnetic field created by the orbital motion of the electron. This is called spin-orbital interaction. The energy of this interaction is proportional to L · S. For an electron in a spherically symmetric potential the proportionality factor is shown to be (e/2m2 c2 r)dV /dr. That is, HSO =
e dV L·S. 2m2 c2 r dr
(11.119)
286 Quantum Mechanics I: The Fundamentals In the presence of any external electromagnetic field the perturbed Hamiltonian is 2 1 eA Horb = p− + V (r) + eφ . (11.120) 2m c If φ = ∇ · A = 0 then Horb =
1 2 e2 p − eA · p + V (r) + A2 . 2m 2mc2
(11.121)
For a homogeneous magnetic field HB we choose A = (HB × r)/2. Then −
e e e A·p= − (HB × r) · p = −HB · L = −HB · µL , (11.122) mc 2mc 2mc
where µL = eL/(2mc). If we include spin then we have the spin-magnetic moment interaction energy HSM given by HSM = −µS · HB ,
µS =
e e~ S= σ. mc 2mc
(11.123)
1 2 p +V , 2m
(11.124a)
Now, the total Hamiltonian is H = H0 + Hmag + HSO ,
H0 =
and Hmag = −
e HB · (L + 2S) . 2mc
(11.124b)
The spin-orbit interaction or coupling gives a shift in energy due to the electron possessing both orbital and spin angular momenta. Spin-orbit coupling is found to be important in the field of spintronics [8]. Spin-splitting in energy spectrum due to spin-orbit coupling in nanotubes and a model of spin-orbit interaction in graphene have been reported [9]. Spinorbit coupling induced spin-Hall effect [10], BE condensation [11,12] and new exotic phases of matter [13] were realized. The effect of spin-orbit interaction on band structure in quantum wires and graphene was investigated [14,15].
11.11 ROTATIONAL TRANSFORMATION For a scalar field ψ the operators Lx , Ly and Lz are the generators of the infinitesimal rotation about the three coordinate axes through the angles ∆θx , ∆θy and ∆θz respectively. A particle possessing internal degrees of freedom like spin is classified by considering the transformation properties of the wave function under rotation. For an infinitesimal rotation ∆θn = i∆θx + j∆θy + k∆θz
(11.125)
Theory of Angular Momentum 287 the relation r′ = R(n, θ)r up to the first-order in ∆θ is
r′ = r + ∆θn × r, R(n, ∆θ) ≈
1 ∆θz −∆θy
−∆θz
1
∆θx
∆θy
−∆θx . (11.126) 1
If ψ(r) transforms like a scalar then we have the Eqs. (11.3) and (11.8). For a particle described by a vector wave function Ψ(r) instead of a scalar wave function ψ(r), both r and Ψ(r) will transform like a vector due to a rotation. Hence, the transformation must be given as Ψ′ (r) = RΨ R−1 r = Ψ R−1 r + ∆θn × Ψ R−1 r = Ψ(R(n, −∆θ)r) + ∆θn × Ψ(r) i = Ψ(r) − (∆θn · L)Ψ(r) + ∆θn × Ψ(r) . ~
(11.127)
Using
i ∆θ n × Ψ(r) = − (∆θ n · S)Ψα (r) (11.128) ~ we get the components of S as the following 3 × 3 matrices 0 0 0 0 0 1 Sx = i~ 0 0 −1 , Sy = i~ 0 0 0 , 0 1 0 −1 0 0 0 −1 0 Sz = i~ 1 (11.129) 0 0 . 0 0 0
The transformation is given as
Ψ′ (r) = UR (n, ∆θ)ψ(r) .
(11.130)
From Eqs. (11.128)–(11.130) we get the transformation equation for the infinitesimal rotation of a vector wave function as UR (n, ∆θ) = I −
i ∆θ n · (L + S) . ~
(11.131)
If the wave function transforms like a vector then the three generators of infinitesimal rotations are Jx = L x + S x ,
Jy = Ly + Sy and Jz = Lz + Sz ,
(11.132)
288 Quantum Mechanics I: The Fundamentals where J = L + S is identified as the total angular momentum of the particle. Since Sx , Sy and Sz are independent of x, y and z, all the components of L commute with all the components of S. L acts only on the r dependence of Ψ(r) and the operator S rearranges the components of the vector wave function Ψ(r) without affecting its r dependence. In case where H contains terms coupling S and L to each other, L and S do not separately commutes with H but J commutes with it. We find from Eq. (11.129) that the total spin operator S2 = Sx2 +Sy2 +Sz2 has the eigenvalue 2~2 . Since the eigenvalue of the angular momentum operator is s(s + 1)~2 , we find s = 1. If the spin quantum number of a particle is 1 then its wave function transforms like a vector. For particles like electrons having a spin quantum number 1/2, the wave function transforms like a two-component spinor. So, we conclude that different transformation properties of the wave function lead to different values of the spin.
11.12 ADDITION OF ANGULAR MOMENTA Let us introduce Jx , Jy and Jz as the operators corresponding to the components of an angular momentum which may be an orbital angular momentum or a spin angular momentum or even an addition of two angular momenta. Suppose we consider a system of particles. The properties obeyed by them for a single particle are also satisfied by the respective operators for a system of particles. Let us consider the orbital angular momentum. The total angular momentum operator for a system of particles is defined as X J = J(i) , J2 = Jx2 + Jy2 + Jz2 , (11.133a) i
Jx
X
=
Jx (i) ,
Jy =
i
We obtain [Jx , Jy ] =
" X
Similarly,
Jy (i) ,
Jz =
i
Jx (i),
i
X
X
Jy (k) =
k
[Jy , Jz ] = 2 J , Jx =
#
Jz (i) . (11.133b)
i
XX i
X
[Jx (i), Jy (k)] = i~Jz .(11.134a)
k
i~Jx , [Jz , Jx ] = i~Jy , 2 2 J , Jy = 0 , J , Jz = 0 .
0,
(11.134b) (11.134c)
The eigenvalue equations for J2 , Jz and J± are J2 |jmi Jz |jmi
J± |jmi
= j(j + 1)~2 |jmi ,
(11.135a) (11.135b)
= m~|jmi ,
= [(j ∓ m) (j ± m + 1)]
1/2
|jm ± 1i .
(11.135c)
The eigenvalues of J2 can also be obtained based on the uncertainty relation for the components of J [6].
Theory of Angular Momentum 289
11.12.1
Clebsch–Gordan Coefficients
Now, we describe the combining of the angular momenta associated with various subsystems of a given system. For simplicity, consider the addition of two angular momenta J1 and J2 of two noninteracting systems 1 and 2. Here J1,2 is the sum of the orbital angular momenta L1,2 and spin angular momenta S1,2 . J1 and J2 are combined to form a total angular momentum J as J = J1 + J2 . The individual eigenstates of 1 and 2 satisfy J21 |j1 m1 i
J1z |j1 m1 i J22 |j2 m2 i J2z |j2 m2 i
= j1 (j1 + 1)~2 |j1 m1 i , = m1 ~|j1 m1 i , = j2 (j2 + 1)~2 |j2 m2 i ,
= m2 ~|j2 m2 i .
(11.136a) (11.136b) (11.136c) (11.136d)
For noninteracting systems J1 , J2 and J are fixed. The eigenstates |j1 m1 ; j2 m2 i separate into |j1 m1 i and |j2 m2 i. We write |j1 m1 ; j2 m2 i as a tensor product of |j1 m1 i and |j2 m2 i. We define the tensor product or basis vectors as |j1 m1 ; j2 m2 i = |j1 m1 i|j2 m2 i , (11.137)
where m1 = −j1 , · · · , j1 , m2 = −j2 , · · · , j2 . Let us denote the new basis set as |jmi with J2 |jmi = Jz |jmi =
j(j + 1)~2 |jmi , m~|jmi .
(11.138a) (11.138b)
Any |jmi is written as a linear combination of the states |j1 m1 ; j2 m2 i: XX |j1 m1 ; j2 m2 ihj1 m1 ; j2 m2 |jmi . (11.139) |jmi = m1 m2
The coefficients hj1 m1 ; j2 m2 |jmi are called Clebsch–Gordan3 (CG) coefficients or vector coupling coefficients. They are determined by a systematic procedure and are required to construct the states |jmi. For simplicity, we use the notation hm1 ; m2 | for hj1 m1 ; j2 m2 |. Because Jz = J1z + J2z from Eqs. (11.136b), (11.136d) and (11.138b) we get m = m1 + m2 . That is, hm1 ; m2 |jmi = 0 , unless m2 = m − m1 .
(11.140)
This restricts the summation in Eq. (11.140): X |jmi = |m1 ; m − m1 ihm1 ; m − m1 |jmi .
(11.141)
m1
3 Paul Albert Gordan was born on April 27, 1837 in Breslau, Germany (now Wroclaw, Poland) and died on December 21, 1912 in Erlangen, Germany. Rudolf Friedrich Alfred Clebsch was a German mathematician born on January 19, 1833 and died on November 7, 1872.
290 Quantum Mechanics I: The Fundamentals The maximum value of m is j1 + j2 since the maximum values of m1 and m2 are j1 and j2 respectively. j1 + j2 is also the maximum value of j. The minimum value of j is |j1 − j2 |. The possible values of j are j1 + j2 , j1 + j2 − 1, · · · , |j1 − j2 | .
(11.142)
We prove that if Eq. (11.142) holds then the dimensionality of the space spanned by |j1 m1 ; j2 m2 i is the same as that of the space spanned by |jmi. Since m1 takes 2j1 + 1 values and m2 takes 2j2 + 1 values, the dimensionality of the space spanned by |j1 m1 ; j2 m2 i is N = (2j1 + 1) (2j2 + 1) .
(11.143a)
To find the dimensionality of the space spanned by |jmi kets, we write (assuming j1 > j2 ) N
=
jX 1 +j2
(2j + 1)
j=j1 −j2
1 [2 (j1 − j2 ) + 1 + 2 (j1 + j2 ) + 1] (2j2 + 1) 2 = (2j1 + 1) (2j2 + 1) .
=
(11.143b)
Since both the space have the same dimensionality Eq. (11.139) holds.
11.12.2
Recursion Relations for the Clebsch–Gordan Coefficients
The CG coefficients form a unitary matrix. Further more, the matrix elements are taken to be real by convention. So, the unitary matrix becomes an orthogonal matrix. We can easily verify that the CG coefficients are real and orthogonal: hm1 ; m2 |jmi = hjm|m1 ; m2 i , XX hjm|m1 ; m2 ihm1 ; m2 |j ′ m′ i = δjj ′ δmm′ ,
(11.144a) (11.144b)
m1 m2
XX j
m
hm1 ; m2 |jmihjm|m′1 ; m′2 i = δm1 m′1 δm2 m′2 .
(11.144c)
Consider the special case: j1 = j, m′ = m = m1 + m2 . Then we obtain from (11.144b) XX |hm1 ; m2 |jmi|2 = 1 (11.144d) m1 m2
which is the normalization condition for |jmi. With j1 , j2 and j fixed, the coefficients with different m1 and m2 are related to each other by recursion relations. Using Eq. (11.138) we get XX J± |jmi = (J1± + J2± ) |m1 m2 ihm1 m2 |jmi . (11.144e) m1 m2
Theory of Angular Momentum 291 Using Eq. (11.135c), we obtain (with m1 → m′1 ; m2 → m′2 ) [j(j + 1) − m(m ± 1)]1/2 |jm ± 1i XX = [j1 (j1 + 1) − m1 (m1 ± 1)]1/2 |m′1 ± 1; m′2 i m′1 m′2
+ [j2 (j2 + 1) − m2 (m2 ± 1)]
1/2
|m′1 ; m′2 ± 1i.
(11.144f)
Premultiplying with hm1 ; m2 | and using orthogonality condition which means nonvanishing contributions from the right-side are possible only with m1 = m′1 ± 1, m2 = m′2 for the first term and m1 = m′1 , m2 = m′2 ± 1 for the second term we obtain the desired recursion relations [j(j + 1) − m(m ± 1)]
1/2
hm1 ; m2 |jm ± 1i
= [j1 (j1 + 1) − m1 (m1 ∓ 1)]1/2 hm1 ± 1; m2 |jmi + [j2 (j2 + 1) − m2 (m2 ∓ 1)]
1/2
hm1 ; m2 ∓ 1|jmi. (11.145)
These recursion relations and the normalization condition determine all the CG coefficients.
11.12.3
General Formula for the Clebsch–Gordan Coefficients
The general formula for the CG coefficients is hm1 ; m2 |jmi = δm1 +m2 ;m α β ,
(11.146a)
where α
β
(s − 2j)! (s − 2j2 )! (s − 2j1 )! (j1 + m1 )! (j1 − m1 )! (s + 1)! ×(j2 + m2 )! (j2 − m2 )! (j + m)! (j − m)! , (11.146b) X (−1)ν = ν1 (j1 + j2 − j − ν)! (j1 − m1 − ν)! (j2 + m2 − ν)! ν 1 × . (11.146c) (j − j2 + m1 + ν)! (j − j2 − m2 + ν)!
= (2j + 1)
In Eqs. (11.146) s = j1 + j2 + j and the ν summation runs over all positive integer values for which all of the factorial arguments are ≥ 0. The above formula is difficult to work with, so we present three special cases. Case (i): m1 = j1 , m2 = j2 hm1 ; m2 |jmi = hj1 j1 ; j2 j2 |jji = 1 .
(11.147)
292 Quantum Mechanics I: The Fundamentals Case (ii): m1 = ±j1 or m2 = ±j2 and m = ±j hj1 − m1 ; j2 − m2 |j − ji 1/2 (2j + 1)! (j1 + j2 − j)! = (−1)j1 −m1 (j1 + j2 + j + 1)! (j + j1 − j2 )! (j − j1 + j2 )! 1/2 (j1 + m1 )! (j2 + m2 )! . (11.148) × (j1 − m1 )! (j2 − m2 )! Case (iii): j1 + j2 = j
(j + m)! (j − m)! (2j1 )! (2j2 )! hj1 m1 ; j2 m2 |jmi = (2j)! (j1 + m1 )! (j1 − m1 )! (j2 + m2 )! (j2 − m2 )!
11.12.4
1/2
.
(11.149)
Example
Consider the addition of spin-1/2 angular momenta. We have j1 = j2 = 1/2. The possible values of j are 1 and 0. The values of m1 and m2 are −1/2 and 1/2. The values of m are 1, 0 and −1. The states |jmi are |jmi = |11i, |10i, |1 − 1i, |00i .
(11.150)
Consider the state |11i. It is expanded as |11i =
1/2 X
1/2 X
m1 =−1/2 m2 =−1/2
m m ED m m E 1 2 1 2 ; ; 1 . 2 2 2 2
Since m1 = m2 = 1/2 alone gives m1 + m2 = 1 we have 1 1 1 1 ED 1 1 1 1 E |11i = ; ; 1 . 22 22 22 22
(11.151)
(11.152)
Here m1 = j1 , m2 = j2 and this belongs to the case (i). The CG coefficient D1 1 1 1 E ; 1 = 1. Then 22 22 1 1 1 1E |11i = ; . (11.153) 22 22 The state |10i is expanded as X X m1 m2 ED m1 m2 E |10i = ; ; (11.154) 0 . 2 2 2 2 m m 1
2
m = m1 +m2 = 0 is possible for (i) m1 = 1/2, m2 = −1/2 and (ii) m1 = −1/2, m2 = 1/2. So, 1 1 1 −1 ED 1 1 1 −1 E 1 −1 1 1 ED 1 −1 1 1 E ; ; ; |10i = ; 0 + 0 . (11.155) 22 2 2 22 2 2 2 2 22 2 2 22
Theory of Angular Momentum 293 Since j = j1 + j2 = 1 we use the result of the case (iii). We obtain r r D 1 1 −1 E 1! 1! 1! 1! 1 ; = =√ , (11.156a) 0 2 2 2 2! 1! 0! 0! 1! 2 r r D 1 −1 1 1 E 1! 1! 1! 1! 1 ; = =√ (11.156b) 0 2 2 22 2! 0! 1! 1! 0! 2 and
1 |10i = √ 2
E 1 −1 1 1 E 1 1 1 −1 + ; . ; 22 2 2 2 2 22
The state |00i is written as |00i = =
(11.156c)
X X m1 m2 ED m1 m2 E ; ; 0 2 2 2 2 m1 m2 D 1 1 1 −1 E D 1 −1 1 1 E ; ; 0 + 0 . 22 2 2 2 2 22
(11.157)
Since m = j from the case (iii) we get
r
r
We obtain
D 1 1 1 −1 E = ; 0 22 2 2 D 1 −1 1 1 E = ; 0 2 2 22
TABLE 11.1
The CG coefficients for j2 = 1/2 or 1.
(−1)
1 1 2−2
1! 1! 2! 0! 0!
1! 0! 1 = √ , 0! 1! 2
1 −√ . 2
(11.158a) (11.158b)
1 −1 1 −1 E |1 − 1i = ; . (11.159) 2 2 2 2 Tables 11.1 and 11.2 present the CG coefficients where j2 = 1/2 or 1.
m2 j j1 +
j1 −
s
1 2
1 2
−
s
1/2 j1 + m + 2j1 + 1
1 2
j1 − m + 2j1 + 1
1 2
s
s
−1/2 j1 − m + 2j1 + 1
1 2
j1 + m + 2j1 + 1
1 2
294 Quantum Mechanics I: The Fundamentals
TABLE 11.2
m + 1.
The CG coefficients hj1 m − m2 ; 1 m2 |jmi. Here α± = j1 ±
j j1 + 1 j1 j1 − 1
h
m2 = 1 (j1 +m)α+ (2j1 +1)(2j1 +2)
−
h
h
(j1 +m)α− 2j1 (j1 +1) (j1 −m)α− 2j1 (2j1 +1)
i1/2
h
i1/2 i1/2
−
h
m2 = 0 α− α+ (2j1 +1)(j1 +1)
h
m2 j1 (j1 +1)
(j1 −m)(j1 +m) j1 (2j1 +1)
i1/2
i1/2
i1/2
h
m2 = −1 (j1 −m)α− (2j1 +1)(2j1 +2)
h
h
(j1 −m)α+ 2j1 (j1 +1)
α+ (j1 +m) 2j1 (2j1 +1)
i1/2 i1/2 i1/2
The eigenvectors of L2 and Lz in coordinate representation are hθφ|lmi = Ylm (θ, φ). The coordinate representation for the composite state |l1 m1 ; l2 m2 i is given by the projection hθ1 φ1 ; θ2 φ2 |l1 m1 ; l2 m2 i = Yl1 m1 (θ1 , φ1 )Yl2 m2 (θ2 , φ2 ) . Therefore, hθφ|lmi = Ylm (θ, φ) gives XX hl1 m1 ; l2 m2 |lmiYl1 m1 (θ1 , φ1 )Yl2 m2 (θ2 , φ2 ) . Ylm (θ, φ) =
(11.160)
(11.161)
m1 m2
Thus, we notice that the CG expansion affords a way of obtaining the coordinate representation of the composite state Ylm (θ, φ).
11.13 ROTATIONAL PROPERTIES OF OPERATORS P A vector in classical physics transform like Vi → j Rij Vj , i, j = 1, 2, 3 where Rij = R(n, θ) is the 3 × 3 rotational matrix. It is reasonably assumed that hV i in quantum mechanics transform like a classical vector under rotation. Specifically, as the state ket |αi is changed under a finite rotation according to |αi → UR (n, θ)|αi the hV i is assumed to change as X hα|Vi |αi → hα|UR† (n, θ)Vi UR (n, θ)|αi = Rij hα|Vj |αi . (11.162) j
This is true for any arbitrary |αi. Therefore, UR† (n, θ)Vi UR (n, θ) =
X
Rij Vj
(11.163)
j
must hold as an operator equation. Let us consider a specific case, an infinitesimal rotation ∆θ in the direction n. When the rotation is infinitesimal,
Theory of Angular Momentum 295 we have UR (n, ∆θ) = I − i∆θ n · J/~ .
(11.164)
We now write (11.163) as Vi +
X ∆θ [Vi , n · J] = Rij (n, ∆θ)Vj . i~ j
In particular, for n along the z-axis, we have from 1 −∆θz R(n, ∆θz ) = ∆θz 1 0 0
So,
Eq. (11.2) 0 0 . 1
∆θ [Vx , Jz ] = Vx − ∆θz Vy . i~ ∆θ i = 2 : Vy + [Vy , Jz ] = Vy + ∆θz Vx . i~ ∆θ i = 3 : Vz + [Vz , Jz ] = Vz . i~
i = 1 : Vx +
(11.165)
(11.166)
(11.167a) (11.167b) (11.167c)
This means that the vector operator V must satisfy the commutation relations [Vi , Jj ] = iǫijk ~Vk where ǫijk , which is known as the totally antisymmetric unit tensor, is defined by ǫ123 = ǫ231 = ǫ312 = 1 and ǫ321 = ǫ213 = ǫ132 = −1 with all other components equal to zero. Clearly, the behaviour of any vector operator V under a finite rotation is completely determined by the commutation relations (11.167). We use (11.167) as the defining property of a vector operator. Notice that the angular momentum commutation relations (11.134a) and (11.134b) are a special case of (11.167) in which we let Vi → Ji , Vk → Jk . Other special cases Vi → r, Jk → Lk give [y, Lz ] = i~x, [x, Lz ] = −i~y, etc. Similarly, if Vi → pi , we get [px , Lz ] = −i~py , [py , Lz ] = i~pz , etc.
11.14 TENSOR OPERATORS In the previous sections we examined the rotational properties of scalar and vector operators. Operators with rotational properties of higher-order than scalars and vectors can also be constructed. A Cartesian tensor transforms like X Tijk··· = · · · Rii′ Rjj ′ Rkk′ · · · Ti′′ j ′ k′ (11.168) i′ j ′ k′
under a rotation where R is the 3 × 3 rotation matrix. The number of indices is called the rank of a tensor. For example, a Cartesian tensor of rank two
296 Quantum Mechanics I: The Fundamentals (called dyadic) is obtained as Tij = Ui Vj where Ui and Vj are the components of a Cartesian vector. Tij has nine components. But Cartesian tensors are not in a convenient form as they can be decomposed into objects that transform differently under rotation. In other words, Cartesian tensors are reducible. For example, the dyadic in Tij = Ui Vj can be decomposed as 1 1 1 1 Ui Vj = (U · V)δij + (Ui Vj − Uj Vi ) + (Ui Vj + Uj Vi ) − U · Vδij . 3 2 2 3 (11.169) The first term on the right-hand side, U · V, is a scalar product invariant under rotations. The second term is an antisymmetric tensor and is written as ǫijk (U× V)k . There are 3 independent components. The last term is a 3 × 3 symmetric traceless tensor with 5 (6 − 1, where 1 comes from the traceless condition) independent components, totally we get the required 9 independent components of Tij . We note that the spherical harmonics Ylm (θ, φ) has 1 independent component for l = 0, 3 independent components for l = 1 and 5 independent components for l = 2. This suggests that the dyadic Tij can be decomposed into tensor that transforms like spherical harmonics with l = 0, 1 and 2. In fact Eq. (11.169) illustrates an example of reducing a Cartesian tensor into irreducible spherical tensors as Ylm (θ, φ) is an example of spherical tensor. ˆ an operator representing some We have seen that |αi is any state, and O, dynamical variable of a quantum system, then rotation through an infinitesimal angle ∆θ about the direction n takes |αi → (I − i∆θn · J)|αi
(11.170)
and ˆ O
ˆ + i∆θn · J) ≈ O ˆ − i∆θn · [J, O] ˆ . (11.171) → (I − i∆θn · J)O(I
ˆ brought about by infinitesimal rotations, are Thus, the changes in |αi and O, ˆ respectively. The commutator directly expressed in terms of J|αi and [J, O] with J determines the rotation properties of an operator. Since Ylm (θφ) are tensor operators, let us study their commutation relation with J. Since Jz = −i~∂/∂φ we have Jz Ylm f (θ, φ) = f (θ, φ)Jz Ylm + Ylm Jz f (θ, φ)
(11.172)
[Jz Ylm − Ylm Jz ]f = m~Ylm f .
(11.173)
and Since f (θ, φ) is arbitrary, we obtain the commutation relation (11.174a)
[Jz , Ylm ] = m~Ylm . In a similar fashion 1/2
[J± , Ylm ] = [l(l + 1) − m(m ± 1)]
~Yl,m±1 .
(11.174b)
Theory of Angular Momentum 297 Now, define an irreducible tensor operator T (k), where k = 0, 1/2, 1, 3/2, · · · , to be a set of 2k + 1 operators T (k, q), where q = k, k − 1, · · · , −k that have commutations with J similar to those in Eqs. (11.174): [Jz , T (k, q)]
(11.175a)
= q~T (k, q) ,
[J± , T (k, q)]
= [k(k + 1) − q(q ± 1)]
1/2
~T (k, q ± 1) . (11.175b)
These relations can be considered as a definition of spherical tensors. Since the tensor operator T (k, q) and the spherical harmonics |jmi satisfy the same commutation relations (11.174) and (11.175), tensor operators can be combined in accordance with the same rule as angular momentum eigenstates. We have the basis |j1 m1 , j2 m2 i which is the tensor product of the basis |j1 m1 i and |j2 m2 i. We have seen in Eq. (11.139) that a new basis |jmi can be given as a linear coefficient of the tensor product basis |j1 m1 , j2 m2 i using the CG coefficients hj1 m1 , j2 m2 |jmi. Hence, an analog to Eq. (11.139) using the correspondence T1 (k1 , q1 ) → |j1 m1 i, T2 (k2 , q2 ) → |j2 , m2 i and T (k, q) → |jmi gives the product of tensors relation T (k, q) =
k1 X
k2 X
q1 =−k1 q2 =−k2
T1 (k1 , q1 )T2 (k2 , q2 )hq1 q2 |kqi ,
(11.176)
where the CG coefficients hq1 q2 |kqi corresponds to j1 = k1 and j2 = k2 . We can prove that if T1 (k1 , q1 ) and T2 (k2 , q2 ) are irreducible spherical tensors of rank k1 and k2 respectively then T (k, q) is also an irreducible tensor of rank k by verifying that T (k, q) satisfies Eq. (11.175). Using J = J1 + J2 and [J, T1 T2 ] = [J, T1 ] T2 + T1 [J, T2 ] we obtain XX {[Jz , T1 ] T2 + T1 [Jz , T2 ]} hq1 q2 |kqi . (11.177) [Jz , T (k, q)] = q1
q2
Since T1 and T2 satisfy Eq. (11.175), we get XX [Jz , T (k, q)] = (q1 + q2 )~T1 (k1 , q1 )T2 (k2 , q2 )hq1 q2 |kqi . q1
(11.178)
q2
Because hq1 q2 |kqi is zero unless q1 + q2 = q, we get from (11.177) [Jz , T (k, q)] = q~T (k, q) .
(11.179)
It is easy to show that [J± , T (k, q)] = [k(k + 1) − q(q ± 1)]
1/2
~T (k, q ± 1) .
(11.180)
From Eqs. (11.179) and (11.180) we find that the product T (k, q) is also an irreducible tensor operator. The construction of the operator T (k, q) is in exact analogy with that of the states |jmi from states |j1 m1 ; j2 m2 i.
298 Quantum Mechanics I: The Fundamentals
Solved Problem 6: Show that hjm, α′ |T (k1 , q1 )|α, j2 m2 i = 0 if m 6= q1 + m2 . The tensor operator T (k1 , q1 ) satisfies the relation [Tz , T (k1 , q1 )] = ~q1 T (k1 , q1 ) .
(11.181)
Then hjm, α′ | [Jz , T (k1 , q1 )] α, j2 m2 i = = =
hjm, α′ |Jz T (k1 , q1 )|α, j2 m2 i −hjm, α|T Jz |α, j2 m2 i
m~hjm, α′ |T |α, j2 m2 i
−m2 ~hjm, α′ |T |α, j2 m2 i (m − m2 )~hjm, α′ |T |α, j2 m2 i.
(11.182)
From Eqs. (11.181) and (11.182) we get (m − m2 )~hjm, α′ |T |α, j2 m2 i = ~q1 hjm, α′ |T |α, j2 m2 i . (11.183) That is, hjm, α′ |T (k1 , q1 )|α, j2 m2 i = 0 if m 6= q1 + m2 .
11.15 THE WIGNER–ECKART THEOREM In considering the interactions of an electromagnetic field with atoms and nuclei, it is often necessary to evaluate matrix elements of tensor operators with respect to angular momentum eigenstates |jmi. Since |jmi are also tensor operators, we write from Eq. (11.176) X |jmi = T1 (k1 , q1 )|j2 m2 i hq1 m2 |jmi . (11.184)
The series expansion of |jmi in terms of T1 (k1 , q1 )|j2 m2 i, given in Eq. (11.184) can be inverted by making use of the orthonormality of the CG coefficients: T1 (k1 , q1 )|j2 m2 i =
j X
kX 1 +j2
m=−j j=|k1 −j2 |
|jmihq1 m2 |jmi ,
(11.185)
where we have used hjm|q1 m2 i = hq1 m2 |jmi as CG coefficients are taken as real. If we include explicitly quantum numbers other than j and m for the state by α, then Eq. (11.185) is given as T1 (k1 , q1 )|αj2 m2 i =
j X
kX 1 +j2
m=−j j=|k1 −j2 |
|α, jmihq1 m2 |α, jmi ,
(11.186)
Theory of Angular Momentum 299 We may now multiply the above equation from left with hj ′ m′ , α′ | to get the matrix element hj ′ m′ , α′ |T1 (k1 , q1 )|α, j2 m2 i =
j X
kX 1 +j2
hj ′ m′ , α′ |α, jmihq1 m2 |jmi .
(11.187)
m=−j j=|k1 −j2 |
On making use of the orthogonality relation hj ′ m′ , α′ |α, jmi = 0 unless j = j ′ , m = m′
(11.188)
we obtain hj ′ m′ , α′ |T1 (k1 , q1 )|α, j2 m2 i = N hq1 m2 |j ′ m′ i ,
(11.189)
where N = hj ′ m′ , α|α′ , j ′ m′ i is the normalization factor. Dropping the primes and writing in the standard notation for N , the matrix element of tensor operators with respect to angular momentum eigenstates are given as hjm, α′ |T1 (k1 , q1 )|α, j2 m2 i = (−1)k1 −j2 +j (2j + 1)−1/2 hj, α′ ||T (k)||α, j2 ihq1 m2 |jmi. (11.190) The above equation is known as Wigner–Eckart theorem. This theorem states that the matrix element is the product of two terms. The physical properties of the matrix element are contained in hj, α′ ||T (k)||α, j2 i. The second term is a CG coefficient for adding j2 and k to get j. It depends only on the way the system is oriented with respect to the z-axis. The first factor depends on the dynamics, for instance, α may stand for the radial quantum number. On the other hand, it is independent of magnetic quantum numbers m1 , m2 and q1 which specify the orientation of the physical system. To evaluate the matrix element of the tensor operator with various combination of m1 , m2 and q1 , it is sufficient to know one of them; all others can be related geometrically because they are proportional to CG coefficients. The selection rule for tensor operator matrix element is read-off from the selection rules for adding angular momentum. Indeed, from the requirement that the CG coefficients can be nonvanishing we obtain the m-selection rule as m = q1 + m2 and the triangular relation |j2 − k1 | ≤ j ≤ j2 + k1 .
11.16 CONCLUDING REMARKS We note that angular position is restricted to values between −π to π whereas linear position can take any value. A consequence of this is that the uncertainty in the angular position is always finite. Another implication is that the angular momentum is quantized and can only have its eigenvalues whereas linear position can take any value. For a big fly-wheel or cycle-wheel one could think that by giving it various angular velocities one can make the fly-wheel
300 Quantum Mechanics I: The Fundamentals to have any angular momentum. But this is not true. Angular momentum can only be increased in discrete steps of ~. Because ~ = 1.0546 × 10−27 ergs-sec is very small that in the example of fly-wheel the step increased is hardly visible. Another point is that Y00 is an eigenfunction of the three operators Lx , Ly and Lz . It is a simultaneous eigenfunction of these three operators which are noncommuting. In this sense Y00 is a special eigenfunction. If a system is in Y00 state then we know the three components of angular momentum with certainty−they are all zero. We note that the effect of L± on a spherical eigenstate for any central potential is to change the value of m. The other quantum numbers l and n are unchanged. This is due to the symmetry of the central potential. Thus, L± are ladder operators for any central potential. According to Eq. (11.88), the mechanism underlying spin is not essentially different from that underlying orbital angular momentum. Both forms of angular momentum arise from the momentum density in the fields. The distinction between spin and orbital contributions to the total angular momentum merely results from the independence of these contributions. The position of the momentum density that gives rise to the spin can be reversed independently of the position that gives rise to the orbital angular momentum.
11.17 BIBLIOGRAPHY [1] G. Gibson, J. Courtial, M.J. Padgett, M. Vasnetsov, V. Pas’ko, S.M. Barnett and S. Franke-Arnold, Opt. Express 12:5448, 2004. [2] R.J. Voogd, M. Singh, S.F. Pereira, A.S. van de Nes and J.J.M. Braat, “The use of orbital angular momentum of light beams for super-high density optical data storage”, in Frontiers in Optics, OSA Technical Digest Series. Optical Society of America, 2004. [3] G. Burns and A.M. Glazer, Space Groups for Scientists and Engineers. Academic Press, Boston, 1990. [4] Yu.A. Bakhturin, “Campbell–Hausdorff formula”, in Encyclopaedia of Mathematics. Kluwer Academic Publishers, Dordrecht, 2000. [5] K.M. Mita, Am. J. Phys. 68:259, 2000. [6] D.H. Kobe, Eur. J. Phys. 19:119, 1998. [7] H.C. Ohanian, Am. J. Phys. 54:500, 1986. [8] J.D. Koralek, C.P. Weber, J. Orenstein, B.A. Bernevig, S.C. Zhang, S. Mack and D.D. Awschalom, Nature 458:610, 2009. [9] D.H. Hernando, F. Gulnea and A. Brataas, Phys. Rev. B 74:155426, 2006.
Theory of Angular Momentum 301 [10] J. Wunderllch, B. Kaestner, J. Sinova and T. Jungwirth, Phys. Rev. Lett. 94:047204, 2005. [11] W.C. Jun, I.M. Shen and Z.X. Fa, Chin. Phys. Lett. 28:097102, 2011. [12] Y.J. Lin, K.J. Garcia and I.B. Spielman, Nature 471:83, 2011. [13] L.W. Cheuk, A.T. Sommer, Z. Hadzibabic, T. Yefsah, W.S. Bakr and M.W. Zwieriein, Phys. Rev. Lett. 109:095302, 2012. [14] M. Governale and U. Zullcke, Phys. Rev. B 66:073311, 2002. [15] M. Gmitra, S. Konschuh, C. Ertier, C.A. Drazi and J. Fabean, Phys. Rev. B 80:235431, 2009.
11.18 EXERCISES
cos θ
11.1 Show that R(k, θ) = sin θ
0
− sin θ
cos θ 0
0
0 . 1
cos θ
11.2 Show that the rotation matrices R(k, θ) = sin θ
0
a continuous group.
− sin θ
cos θ 0
0
0 form 1
11.3 Show that p and −H are the generators of infinitesimal displacements in space and time respectively. 11.4 Show that a state of a system corresponding to finite angular momentum cannot be a simultaneous eigenstate of any two of the Cartesian components of L. 11.5 Show that if an operator O commutes with Lx , Ly and Lz then O2 commutes with L2 . 11.6 Find the values of [Lx , x] and [Lx , y]. 11.7 Show that if a particle has the normalized eigenfunction ψ = N eikz then the z component of its angular momentum is zero. 11.8 What are the expectation values of Lx and Ly for a system which is in an eigenstate of Lz ? 11.9 Show that L2 , r = −2i~(L × r − i~r).
302 Quantum Mechanics I: The Fundamentals 11.10 Simplify L × r + r × L. 11.11 Show that L · p = L · r = 0. 11.12 Determine the values of [Lx , r2 ], [Ly , r2 ] and [Lz , r2 ]. 11.13 Suppose an atom or a particle has the angular momentum L. If a magnetic field B is applied, then the magnetic moment is µ = γL where γ is the gyromagnetic ratio. Then its Hamiltonian operator H0 = −µ · B. ˙ = −γB × L. Prove that the equation of motion for the L operator is L 11.14 Let ψ(θ, φ) be the eigenfunctions of Lx operator corresponding to l = 1. Writing ψ = a1 Y11 + a2 Y10 + a3 Y1−1 , find the eigenvalues and eigenfunctions of Lx . 11.15 Show that |lli = Nl eilφ sinl θ is a solution of the differential equation L+ |lli = 0. 11.16 If J = J1 + J2 + J3 where all the Ji are for j = 1 and let J(J + 1) be the eigenvalues of J2 . (i) What are the possible values of J? (ii) How many linearly independent states are there for each of these values? 11.17 Show that the total number of eigenstate of J = J1 + J2 + · · · + JN are given by (2j1 + 1)(2j2 + 1) · · · (2jN + 1). Hence find the number of eigenstates for the coupling of one p, one d and one f electrons (excluding spin). 11.18 Find (i) the total number of eigenstate of J = J1 + J2 + · · · + JN and then (ii) the number of eigenstates for the coupling of one p, one d and one f electrons (excluding spin). 11.19 Find the transformation matrix between the Cartesian components of the position (x, y, z) and the spherical basis (Y11 , Y10 , Y1−1 ). 11.20 The rotational Hamiltonian for a symmetric top molecule is given by H = L2x + L2y /(2I1 ) + L2z /(2I2 ). Find its (i) eigenvalues and (ii) eigenfunctions. 11.21 Show that the rotation transformation operator for the spin angular momentum e−iS·nφ/~ = I cos(φ/2) − iσ · n sin(φ/2) where S = ~σ/2 and σ are the Pauli’s spin matrices. Note that (σ · n)k = 1 for k even and σ · n for k odd. √ 1 2 11.22 An electron is in the spin state |φi = √ . (i) What is the 3 1 probability that a measurement of Sz will give ~/2? (ii) What is the probability that a measurement of Sx will give −~/2?
Theory of Angular Momentum 303 11.23 Using the recursion relation [j(j + 1) − m(m ± 1)]
1/2
hm1 ; m2 |jm ± 1i
= [j1 (j1 + 1) − m1 (m1 ∓ 1)]1/2 hm1 ± 1; m2 |jmi + [j2 (j2 + 1) − m2 (m2 ∓ 1)]
show that
1/2
hm1 ; m2 ∓ 1|jmi
hj1 , j2 − 1|j1 + j2 , j1 + j2 − 1i = (j2 /(j1 + j2 ))1/2 and hj1 − 1, j2 |j1 + j2 , j1 + j2 − 1i = (j1 /(j1 + j2 ))1/2 .
11.24 Find the CG matrix for the case j1 = 1/2 and j2 = 1/2. 2 11.25 Show that the energy eigenvalue of a rigid rotator H l(l + 1)~ /(2I) follows from the Bohr–Sommerfeld quantization rule L dφ = (l + 1/2)h in the limit of large quantum number.
11.26 Obtain the value of h{L+ , L− }i = hL+ L− + L− L+ i where {} is used for anticommutator. 11.27 The Hamiltonian for a particle with orbital angular momentum unity (l = 1) is given by H = aL2z + b(L2x − L2y ), a 6= b. Determine the eigenvalues and eigenvectors of H. 11.28 A measurement Lxy = Lx + Ly is done on a system with l = 1. What are the possible results of such a measurement? 11.29 Construct the state with j = 3/2 and mj = −1/2 for a 2p-electron. 11.30 Given the normalized eigenvectors of Lx for l = 1 as √ √ ψ11 = (Y11 + 2Y10 + Y1−1 )/2, ψ10 = (Y11 − Y1−1 )/ 2, √ ψ1−1 = (Y11 − 2Y10 + Y1−1 )/2 find the matrix of Ly and Lz in this representation.
CHAPTER
12
Hydrogen Atom
12.1 INTRODUCTION The hydrogen atom is the simplest of all atoms consisting of proton of charge e and the electron of charge −e. It is a two particle system. A two-body problem can always be reduced to a one-body problem by choosing the center of mass coordinate and the relative coordinates. In the hydrogen atom the attractive force exerted by the proton on electron prevents it from escape. The potential energy V of the system is −e2 /r where r is the distance between the electron and the proton. Suppose the nucleus is held fixed at the origin of coordinates. V (r) is independent of θ and φ, that is, insensitive to the direction of r and depends only on the scalar magnitude r. The potential energy remains invariant under rotation and is said to be spherically symmetric. Figure 12.1 shows V (r) as a function of r. In this chapter the hydrogen atom problem in threedimension is solved and momentum eigenfunction is obtained. The problem in D-dimension where D may be noninteger is analysed. The magnetic field produced by a hydrogen atom is worked out. Finally, the system in parabolic coordinates is discussed.
12.2 HYDROGEN ATOM IN THREE-DIMENSION In the hydrogen atom an electron of mass me is coupled to a proton of mass mp . The force is the Coulomb force −e2 / |r1 − r2 |. The Schrödinger equation of hydrogen atom is ∂ψ ~2 2 ~2 2 e2 i~ = − ∇1 − ∇2 − ψ, (12.1) ∂t 2me 2mp |r1 − r2 | where ∇i , i = 1, 2 is the Laplacian operator and ψ = ψ(r1 , r2 , t) is the joint probability amplitude function of finding the electron at r1 and the proton at r2 at time t. For atoms which are isoelectronic with hydrogen atom such as He+ , Li++ , +++ Be , etc, the potential is −Ze2 /r where Z is 2, 3, 4, etc. respectively. For 305
306 Quantum Mechanics I: The Fundamentals 1
V (r)
0 -1 -2 -3 -4 -5
FIGURE 12.1
-1
0
1
2
r
3
4
5
Potential of hydrogen atom. The y-axis is V (r) = −1/r.
these atoms we have Eq. (12.1) with e2 /|r1 − r2 | replaced by −Ze2 /|r1 − r2 |. Introducing the relative displacement r = r1 − r2 , and the centre of mass displacement R = (me r1 + mp r2 )/(me + mp ), the Schrödinger Eq. (12.1) becomes ∂ψ ~2 2 ~2 2 e 2 i~ = − ∇R − ∇r − ψ, (12.2) ∂t 2M 2µ r where M = me + mp is the total mass, µ = me mp /(me + mp ) is the reduced mass and ψ = ψ(r, R, t). The advantage of transforming Eq. (12.1) to Eq. (12.2) is evident as the eigenfunctions of Eq. (12.1), ψ(r1 , r2 , t) cannot be written as a product ψ1 (r1 )ψ(r2 ) whereas the eigenfunctions of Eq. (12.2) can be written as the product of two eigenfunctions as Eq. (12.2) can be separated into two equations one depending on r and the other on R. We write the total energy of the system as Etot = E + Ecm where Ecm is the centre of mass energy and E is the energy of relative motion. Now, writing ψ(r, R, t) = ψE (r)ψcm (R) e−i(E+Ecm )t/~ ,
(12.3)
where ψcm (R) represents a free particle wave function for the centre of mass and ψE (r) is the probability amplitude function to find the electron and proton with a relative displacement r and relative energy E. The Schrödinger equation for the relative motion of the electron and proton takes the form ~2 2 e 2 − ∇r − ψE (r) = EψE (r) . (12.4) 2µ r Writing ∇r and ψE as simply ∇ and ψ respectively and defining V = −e2 /r, Eq. (12.4) is rewritten as 2 ~ − ∇2 + V ψ = Eψ . (12.5) 2µ
Hydrogen Atom 307 It is convenient to solve the Eq. (12.5) in spherical coordinates, because it is generally difficult to obtain the solution of it in Cartesian coordinates.
Solved Problem 1: Show that the time-independent Schrödinger equation ~2 2 e 2 ~2 2 ∇ − ∇ − ψ(r, R) = ET ψ(r, R) − 2m R 2µ r r can be separated into two independent equations for the variables r and R and find the solution corresponding to the R-part equation. Assume the solution of the given equation as ψ = ψE (r)ψCM (R). Substituting this in the given equation we get −
~2 ~2 e2 ψE ∇2R ψCM − ψCM ∇2r ψE − ψE ψCM = ET ψE ψCM . 2m 2µ r
(12.6)
Dividing throughout by ψE ψCM we get −
~2 1 ~2 1 2 e2 ∇2R ψCM = ET + ∇r ψE + . 2m ψCM 2µ ψE r
(12.7)
Since r and R are independent variables the above equation must be a constant and say that it is ECM . We write ~2 2 ∇ ψ = ECM ψCM 2m R CM
(12.8)
~2 2 e2 ∇r ψE − ψE = (ET − ECM ) ψE = EψE . 2µ r
(12.9)
− and −
Equation (12.8) gives the Schrödinger equation for the centre of mass coordinate R and Eq. (12.9) gives the Schrödinger equation for the relative coordinate r. The solution of Eq. (12.8) is ψCM = A e±iK·R ,
K2 =
2mECM ~2
(12.10)
which is a free particle eigenfunction.
12.2.1
Application of Variable Separable Method
Let us assume the solution of the form ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) = R(r)Y (θ, φ)
(12.11)
308 Quantum Mechanics I: The Fundamentals and obtain the differential equations for R, Θ and Φ. In spherical coordinates = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, (12.12a) ∂ 1 1 ∂ ∂ ∇ = rˆ + θˆ + φˆ , (12.12b) ∂r r ∂θ r sin θ ∂φ 1 ∂ 1 ∂ 1 ∂2 ∂ 2 ∂ ∇2 = r + sin θ + , 2 r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin θ ∂φ2 (12.12c) x
where rˆ, θˆ and φˆ are the usual unit vectors. Using Eqs. (12.11) and (12.12) in Eq. (12.5) the Schrödinger equation in spherical polar coordinates is 1 d 2µr2 2 dR r + 2 (E − V ) R dr dr ~ 1 1 ∂ ∂Y 1 ∂2Y =− sin θ + . (12.13) Y sin θ ∂θ ∂θ sin2 θ ∂φ2 The left-side of Eq. (12.13) is a function of r only while the right-side is a function of θ and φ only. Therefore, they must be equal to a same constant, say, λ. Hence, from Eq. (12.13) we write 1 d 2µ λ 2 dR r + 2 (E − V ) − 2 R = 0, (12.14) r2 dr dr ~ r ∂Y 1 ∂2Y 1 ∂ sin θ + + λY = 0 . (12.15) sin θ ∂θ ∂θ sin2 θ ∂φ2 For all central force problems with spherically symmetric potential the energy term occurs only in the radial equation. Equation (12.15) is solved without knowing the form of V as follows. Equation (12.15) is separated by substituting Y = Θ(θ)Φ(φ) which gives 1 d2 Φ sin θ d dΘ sin θ + λ sin2 θ . (12.16) − = Φ dφ2 Θ dθ dθ left-side of Eq. (12.16) is a function of φ only and the right-side is a function of θ only and hence they must be equal to same a constant which we set as m2 . Now, from Eq. (12.16) we write d2 Φ + m2 Φ = 0 , dφ2 1 d dΘ m2 sin θ + λ− Θ=0. sin θ dθ dθ sin2 θ
(12.17a) (12.17b)
Hydrogen Atom 309
12.2.1.1 Solution of Φ Equation The solution of Eq. (12.17a) is Φ = Aeimφ + Be−imφ . Applying the condition Φ(0) = Φ(2π), we observe that m = 0, ±1, ±2,· · · and thus Φ = Aeimφ .
(12.18)
The Φ given above is also an eigenfunction of Lz with the eigenvalue m~. Since the Hamiltonian for a magnetic field applied along the z-direction is HB = −eBLz /(2m), the first-order perturbation due to a magnetic field shifts energy levels proportional to m and hence m is some√times called a magnetic quantum number . Normalization of Φ gives A = 1/ 2π (verify).
12.2.1.2 Solution of Θ-Equation Introducing the change of variable as x = cos θ, rewrite the Eq. (12.17b) as d m2 2 dΘ (1 − x ) + λ− Θ=0. (12.19) dx dx 1 − x2 When λ = l(l + 1) with l = 0, 1, 2, · · · the above equation is the associated Legendre differential equation. To ensure a finite probability density for all x = cos θ, only polynomial solutions to Eq. (12.19) have to be considered. Then we get the equation d d m2 (1 − x2 ) Plm (x) + l(l + 1) − Plm (x) = 0 , (12.20a) dx dx 1 − x2 where Plm (x) is called associated Legendre polynomial which is given by the Rodrigues formula Plm (x) =
l+|m| 1 2 |m|/2 d (x2 − 1)l . (1 − x ) 2l l! dxl+|m|
(12.20b)
m The solution of Eq. (12.19) is written as Θ = BPlm (x) = BP θ) provided R πl (cos λ = l(l + 1) and |m| ≤ l. The normalization condition is 0 Θ∗ Θ sin θdθ = 1, R1 that is, −1 B 2 Plm (x)Plm (x)dx = 1. This gives
(2l + 1)(l − |m|)! B=ǫ 2(l + |m|)!
1/2
(12.21)
,
where ǫ = (−1)m for m > 0 and ǫ = 1 for m ≤ 0. Thus, Ylm (θ, φ) = ΘΦ = (−1)m
(2l + 1)(l − m)! 4π(l + m)!
1/2
eimφ Plm (cos θ),
(12.22)
where l = 0, 1, 2, · · · , m = l, l−1, · · · , −l+1, −l. l is called the orbital quantum number since Ylm (θ, φ) are the eigenfunctions of L2 operator with eigenvalues
310 Quantum Mechanics I: The Fundamentals l(l+1)~2 . Expression for some of the Ylm calculated using (12.22) and (12.20b) are given below (verify): 1/2 1/2 1/2 z 3 3 1 , Y10 = cos θ = , (12.23a) Y00 = 4π 4π 4π r 1/2 1/2 (x + iy) 3 3 iφ Y11 = − sin θ e = − , (12.23b) 8π 8π r 1/2 1/2 (x − iy) 3 3 −iφ Y1−1 = sin θ e = , (12.23c) 8π 8π r 1/2 1/2 2 5 3z 5 3 cos2 θ − 1 = Y20 = − 1 .(12.23d) 16π 16π r2
12.2.1.3 Solution of Radial Equation Having obtained the forms of Φ and Θ, next we solve the radial equation with V (r) = −e2 /r. The radial equation is d2 R 2 dR 2µ e2 l(l + 1)~2 + + 2 E+ − R=0. (12.24) dr2 r dr ~ r 2µr2 With the change of variable R = f (r)/r Eq. (12.24) becomes 2µ e2 l(l + 1)~2 d2 f + 2 E+ − f =0. dr2 ~ r 2µr2
(12.25)
Compare this equation with the one-dimensional Schrödinger equation. The effective potential energy is Veff (r) = −
e2 l(l + 1)~2 + . r 2µr2
(12.26)
Then Eq. (12.25) is the Schrödinger equation for a particle of mass µ moving in the effective potential Veff . From E = H = p2 /(2µ)+Veff we have E −Veff = 2p2 /(2µ) ≥ 0. The region with E − Veff ≥ 0 is the classically allowed region. E − Veff = 0 gives the turning points at which the velocity of the particle is zero. Figure 12.2 shows r versus Veff for a few values of l. We see from Fig. 12.2 that for l 6= 0, Veff (r → 0) → +∞. An infinite potential at some point means that the wave function must vanish there, that is, R(r) → 0 for l 6= 0. The turning points are the roots of the equation E+
e2 l(l + 1)~2 − =0. r 2µr2
The roots of the above equation are s e2 e4 l(l + 1)~2 r± = − ± + . 2 2E 4E 2µE
(12.27)
(12.28)
Hydrogen Atom 311 10
Veff
l=1 l=2 0
l=0 -10
0
1
r
2
The effective potential of hydrogen atom for a few values of l where we have set e2 = 1, ~ = 1 and µ = 1. FIGURE 12.2
If E > 0 then r+ ≥ 0 and r− < 0. Because r ∈ [0, ∞] r+ is the only turning point for E > 0 implying that the classical particle does not exhibit bounded motion. The two turning points are real and positive when −e4 µ/(l(l+1)~2) < E < 0 and the motion is bounded. Substituting the Bohr energy relation En = −e4 µ/(2n2 ~2 ) for E in Eq. (12.28) we obtain s n2 l(l + 1)~4 n4 ~4 n2 ~2 ± − r± = 2 4 2 e µ e ~ e 4 µ2 h i p ~2 = na0 n ± n2 − l(l + 1) , a0 = 2 (12.29) e µ where a0 is the Bohr radius. The classically allowed region is r− < r < r+ . With the change of variables ρ2 = −
8r2 µE , ~2
R=
f (ρ) p = −8Eµ/~2 f (ρ) r
Eq. (12.25) becomes 2√ e µ (2l + 1)2 − 1 1 ′′ √ f + − − f =0. 4ρ2 4 ~ρ −2E
(12.30)
(12.31)
The physically acceptable solution of Eq. (12.31) must vanish at ρ = 0 (that is at r = 0) since R = f (ρ)/r. When the electron is far from the nucleus we assume ρ be very large and neglect the terms containing 1/ρ and 1/ρ2 in Eq. (12.31). In this case f ′′ − f /4 = 0 whose solution is f (ρ) = e−ρ/2 . Note that f (ρ) → 0 as ρ → ∞ and is therefore a good approximation. For more general situation choose f (ρ) = F (ρ)e−ρ/2 where F (ρ) take care of small
312 Quantum Mechanics I: The Fundamentals values of ρ. The solution of the full Eq. (12.31) is obtained by comparing it with the equation [1] (2q + p + 1) (1 − p2 ) 1 ′′ y + + − y = 0. (12.32) 2ρ 4ρ2 4 The solution of the above equation is y = e−ρ/2 ρ(p+1)/2 L(p) q (ρ) ,
(12.33)
(p)
where Lq (ρ) is a generalized Laguerre polynomial. For a finite wave function q must be 0, 1, 2, · · · and we get a polynomial solution. We define q = n−l−1. Comparing Eq. (12.31) with Eq. (12.32) we obtain p = ±(2l + 1) and √ e2 µ 2q + p + 1 2(n − l − 1) + (2l + 1) + 1 √ = = =n (12.34) 2ρ 2ρ hρ −2E The result is
e4 µ , n = 1, 2, · · · (12.35) 2~2 n2 where n is called principal quantum number . En given by the above equation is in CGS system. In SI system 2 2 µ e En = − 2 2 , n = 1, 2, · · · . (12.36) 2~ n 4πǫ0 En = −
In atomic units (~ = m = e = 1) En = −1/(2n2). For hydrogen-like atoms the results are obtained by replacing e2 by Ze2 . Then En = −Z 2 e4 µ/(2~2 n2 ). For hydrogen atom p = (2l + 1) is physically acceptable and p = −(2l + 1) is unacceptable since R → ∞ as r → 0. Since q ≥ 0 we get from q = n − l − 1 the result l ≤ n − 1. Hence, the value of the orbital quantum number l cannot exceed n − 1 for a acceptable solution.
12.2.2
Orthogonality Relation and Radial Wave Function
From Eq. (12.32) we write d2 yq p + 1 (1 − p2 ) ρ ρ 2 + q+ + − yq = 0 dρ 2 4ρ 4 and ρ
d2 yq′ p + 1 (1 − p2 ) ρ ′ + q + + − yq ′ = 0 . dρ2 2 4ρ 4
(12.37)
(12.38)
For q 6= q ′ multiplication of Eq. (12.37) by yq and Eq. (12.38) by yq′ and subtracting one from another lead to d dyq dyq′ q′ − q yq ′ − yq = yq yq ′ . (12.39) dρ dρ dρ ρ
Hydrogen Atom 313 Integrating this equation gives ∞ Z ∞ yq yq ′ dyq dyq′ yq ′ − yq = (q ′ − q) dρ . dρ dρ 0 ρ 0 For ρ > −1, yq and yq′ vanish at ρ = 0 and ∞. So, Z ∞ yq yq ′ dρ = 0 . (q ′ − q) ρ 0 Substituting for yq and yq′ we find Z ∞ e−ρ ρp Lpq (ρ)Lpq′ (ρ) dρ = 0 , 0
(12.40)
(12.41)
(q 6= q ′ ) .
(12.42)
The polynomials Lpq (ρ) are orthogonal in (0, ∞) with respect to e−ρ ρp . The integral (12.42) for q = q ′ is evaluated with the help of the generating function ∞ X e−ρz/(1−z) = Lpq (ρ)z q . G(ρ, z) = (12.43) (1 − z)p+1 q=0 Then the integral in Eq. (12.42) becomes Z
∞
e
−ρ p
2
ρ G dρ =
0
∞ X ∞ Z X q=0 q′ =0
∞
′
e−ρ eρ Lpq Lpq′ z q z q dρ .
0
(12.44)
Using Eq. (12.42) the above equation becomes Z
∞
e−t ρp G2 dρ =
0
∞ Z X q=0
∞ 0
e−ρ eρ Lpq
2
z 2q dρ .
Substituting the generating function we obtain Z ∞ Z ∞ 1 tp e−ρ(1+z)/(1−z) dρ e−t ρp G2 dρ = (1 − z)2p+2 0 0 Γ(p + 1) = . (1 − z 2 )p+1 We have
∞
X Γ(q + p + 1) 1 = z 2q . 2 p+1 (1 − z ) Γ(q + 1) Γ(p + 1) q=0
(12.45)
(12.46)
(12.47)
Using (12.46) and (12.47) in (12.45) the result is ∞ X Γ(q + p + 1) q=0
Γ(q + 1)
z
2q
=
∞ Z X q=0
0
∞
e−ρ ρp Lpq
2
z 2q dρ .
(12.48)
314 Quantum Mechanics I: The Fundamentals That is,
Z
0
∞
e−ρ ρp Lpq
2
dρ =
Γ(q + p + 1) . Γ(q + 1)
(12.49)
Using the following recurrence relation of Laguerre polynomial ρLpq = (2q + p + 1)Lpq − (q + 1)Lpq+1 − (q + p)Lpq−1 we evaluate the integral is Z ∞ 2 e−ρ ρp+1 Lpq dρ 0
=
(12.50)
Z ∞ 2 (2q + p + 1) e−ρ ρp Lpq dρ 0 Z ∞ −(q + 1) e−ρ ρp Lpq Lpq+1 dρ 0 Z ∞ −(q + p) e−ρ ρp Lpq Lpq−1 dρ . (12.51) 0
The use of the orthogonality relation (12.42) and the Eq. (12.49) gives Z ∞ 2 Γ(p + q + 1) e−ρ ρp+1 Lpq dρ = (2q + p + 1) . (12.52) Γ(q + 1) 0 We have
y = e−ρ/2 ρ(p+1)/2 Lpq (ρ) .
(12.53)
Comparing Eq. (12.31) for the radial part of the hydrogen atom with that for y, we find that with q = n − l − 1 and p = 2l + 1 f ∼ e−ρ/2 ρl+1 L2l+1 n−l−1 (ρ)
(12.54)
and hence
f (ρ) = Nnl e−ρ/2 ρl L2l+1 (12.55) n−l−1 (ρ) , r with Nnl is the normalization constant and p 8µ|En | 2r ρ= r= , (12.56) ~ na0 R∞ 2 where a0 is the Bohr radius. Using the relation 0 Rnl (r)r2 dr = 1 the normalization condition takes the form Z ∞ na 3 2 0 2 1 = Nnl e−ρ ρ2l+2 L2l+1 dρ (12.57) n−l−1 2 0 Rnl (ρ) ∼
Defining p = 2l + 1, q = n − l − 1 we get Z ∞ 2 2n · (n + l)! e−ρ ρ2l+2 L2l+1 dρ = . n−l−1 (n − l − 1)! 0
(12.58)
Hydrogen Atom 315 Then Nnl =
"
2 na0
3
(n − l − 1)! 2n · (n + l)!
#1/2
.
(12.59)
Hence, Rnl (ρ) =
12.2.3
"
2 na0
3
(n − l − 1)! 2n · (n + l)!
#1/2
e−ρ/2 ρl L2l+1 n−l−1 .
(12.60)
Normalized Eigenfunctions
The normalized energy eigenfunctions for the hydrogen atom are 2r ψnlm (r, θ, φ) = Rnl Ylm (θ, φ) . na0
(12.61)
The radial eigenfunction Rnl (ρ) is evaluated using the Rodrigues formula Lpq (x) = ex x−p
dq e−x xp+q . q dx
(12.62)
For n = 1 and l = 0 with L10 (ρ) = 1 R10 = 2
1 a0
3/2
e−r/a0 .
(12.63a)
Similarly, we obtain R20
=
R21
=
R30
=
R31
=
R32
=
3/2 1 r 2− e−r/(2a0 ) , a0 a0 3/2 1 1 r −r/(2a0 ) √ e , a0 24 a0 3/2 " 2 # 2 1 r r √ 27 − 18 + 2 e−r/(3a0 ) a0 a0 81 3 a0 3/2 4 1 r r √ 6− e−r/(3a0 ) , a0 a0 81 6 a0 3/2 2 4 1 r √ e−r/(3a0 ) . a0 81 30 a0 1 √ 8
(12.63b) (12.63c) , (12.63d) (12.63e) (12.63f)
Figure 12.3 shows Rn0 (r) for a few values of n where a0 is set to unity. For hydrogen-like atoms the wave functions is obtained from Eq. (12.61) by replacing a0 by a0 /Z. Some Ylm are given in Eqs. (12.23).
316 Quantum Mechanics I: The Fundamentals
2
a0 Rn0
n=1 n=2
3/2
1
n=3
0 0
5
10
r/a0 FIGURE 12.3
Radial wave function Rnl with l = 0 and n = 1, 2 and 3.
The first few ψnlm are given by ψ100
12.2.4
= R10 Y00
ψ200
=
ψ210
=
ψ211
=
ψ21−1
=
1 = √ π
1 a0
3/2
e−r/a0 ,
(12.64a)
3/2 r R20 Y00 2− e−r/(2a0 ) , (12.64b) a0 3/2 1 1 r R21 Y10 = √ cos θ e−r/(2a0 ) , (12.64c) a0 32π a0 3/2 1 1 r R21 Y11 = − √ sin θ e−r/(2a0 ) eiφ , (12.64d) 8 π a0 a0 3/2 1 1 r sin θ e−r/(2a0 ) e−iφ . (12.64e) R21 Y1−1 = √ a0 8 π a0 1 = √ 32π
1 a0
Features of Energy Eigenstates
Let us enumerate the features of eigenstates of hydrogen atom in the following. 1. There are infinite number of energy levels as n takes values 1, 2, · · · 2. As n increases |En | decreases and En → 0 as n → ∞. The lowest energy corresponds to n = 1. An energy of 0 corresponds to ionization of the hydrogen atom, that is complete separation of the proton and the electron with both the particles at rest. Figure 12.4 shows first few energy levels of hydrogen atom along with the few energy levels of linear harmonic oscillation and particle in a box for comparison. 3. We write En = −E1 /n2 where E1 = µe4 /(2~2 ) ≈ −13.6eV.
Hydrogen Atom 317
Harmonic oscillator
Particle in a box
Hydrogen atom
First few energy levels of linear harmonic oscillator, particle in a box potential and the hydrogen atom. FIGURE 12.4
4. The energy difference between nth and (n+1)th levels is given by ∆En = (2n + 1)/(n2 (n + 1)2 ). As n increases ∆En decreases. 5. En depends on n and independent of l and m. Therefore, each of the functions RYlm (m = −l, · · · , l) is a solution of the Schrödinger equation of the hydrogen atom for the same value of E. The energy levels are thus degenerate. How many energy levels do exist with the same En ? We know that l takes the values 0, 1, · · · , n − 1 and for each l, m = l, l − 1, · · · , −l + 1, −l. There are (2l + 1) values of m labelled by n different values of l. Thus, the total number of degeneracy is given by n−1 X
(2l + 1) =
l=0
X
2l +
X
l=
2n(n − 1) + n = n2 . 2
(12.65)
Since electrons have two spin states, two electrons can be accommodated in a single state. The ground state (n = 1) is nondegenerate. All other states are degenerate states. The existence of degenerate energy eigenvalues implies that linear combination of the corresponding eigenfunctions are solutions of the wave equation with the same energy. Hence, the number of electrons Pn−1that can be placed in the shell corresponding to a particular n is l=0 2(2l + 1) = 2n2 .
6. In the s-state (l = 0) R is nonzero at the origin √ r = 0. On the other hand, for l 6= 0, Rnl (0) = 0. Since Y00 (θ, φ) = 1/ 4π we find that |ψn00 (0)|2 =
|Rn0 (0)|2 Z3 = . 4π πa30 n3
(12.66)
7. The probability distribution of the position of the electron is Pnl (r) = |Rnl |2 r2 . For n = 1, l = 0 (1s (ground)state) P = 4(r/a0 )2 e−2r/a0 /a0 . For n = 2, l = 0 (2s-state) P = (r/a0 )2 (1 − (r/2a0 ))2 e−r/a0 /(2a0 ). For n = 2, l = 1 (2p-state) P = (r/a0 )4 e−r/a0 /(24a0 ). Figure 12.5 depicts the variation of Pn0 for three values of n.
318 Quantum Mechanics I: The Fundamentals
0 .6
a0 Pn0
n=1 n=2
0 .4
n=3
0 .2 0
0
10
20
30
r/a0 Radial probability distribution of hydrogen atom for three values of n with l = 0. FIGURE 12.5
8. The probability density Pnlm
= =
2
2
2
|ψnlm | = |Rnl (r)| |Ylm (θ, φ)| 2 (2l + 1)(l − m)! |m| |Rnl (r)|2 Pl (cos θ) 4π(l + m)!
(12.67)
is independent of φ and function of r and θ only. |Rnl (r)|2 gives the electron density as a function of r along a given direction θ and the remaining term in Pnlm gives the angular distribution for a given r. 9. The eigenvalues of L2 are l(l + 1)~2 and is always greater than m2 ~2 . 10. The action of parity operator on the hydrogen-like wave function is P ψnlm = P [Rnl Ylm ] = Rnl (−1)l Ylm = (−1)l ψnlm .
(12.68)
The states ψnlm have the parity of l. An algebraic method for solving hydrogen-like atoms is described in ref.[2]. For the construction of necessary and sufficient conditions for the existence of bound states in central potentials one may refer to refs.[3, 4].
Solved Problem 2: What is the probability that an electron in the ground state of hydrogen atom be found in the classically forbidden region (that is, in the region where V (r) > E)? We have E1 = −µe4 /(2~2 ). The classical turning point can be found from −e2 /re = −µe4 /(2~2 ). We find re = 2~2 /(µe2 ) = 2a0 . The ground state
Hydrogen Atom 319 radial eigenfunction is R10 = 2(1/a0 )3/2 e−r/a0 . Hence, the probability for an electron to be found in the classically forbidden region is Z ∞ Z ∞ 4 2 2 R10 r dr = e−2r/a0 r2 dr a30 2a0 2a0 ∞ Z ∞ 4 e−2r/a0 r2 4 = e−2r/a0 r dr + a30 −2/a0 2a0 a20 2a0 ∞ Z ∞ 4 e−2r/a0 r 2 e−2r/a0 dr = 8e−4 + 2 + a0 −2/a0 2a0 a0 2a0 =
8e−4 + 4e−4 + e−4
=
13e−4 .
(12.69)
Solved Problem 3: Calculate the momentum eigenfunction for the ground state of hydrogen atom. From the Eqs. (12.64a) and (9.13b) we obtain Z 1 1 C(K) = e−r/a0 e−iK·r dr . (2π)3/2 (πa30 )1/2
(12.70)
Choosing the direction of K along z-axis we have K · r = kr cos θ and dr = r2 sin θ dr dθ dφ. Then Z ∞ Z π Z 2π 1 1 e−r/a0 e−ikr cos θ r2 sin θ dr dθ dφ C(K) = (2π)3/2 (πa30 )1/2 0 0 0 =
(2a0 )3/2 . π(1 + k 2 a20 )2
(12.71)
Next, consider the eigenfunction ψ200 given by Eq. (12.64b). We get (with √ 3/2 A = 1/( 32πa0 )) r 2− e−r/a0 e−ikr cos θ r2 sin θ dr dθ dφ a0 0 0 0 Z 2πA ∞ r2 = 2r − e−r/a0 eikr − 1 dr . (12.72) ik 0 a0 Z ∞ Using the integral formula xn e−µx dx = n!/µn+1 we obtain C(K)
= A
Z
∞
Z
π
Z
2π
0
ika30 a2 − 1 C(K) = i4πA 0 4 + a0 (1 − ika0 )3
.
(12.73)
320 Quantum Mechanics I: The Fundamentals
12.3 HYDROGEN ATOM IN D -DIMENSION In this section we are concerned with the bound states (E < 0) of the hydrogen atom in D-dimension where D may be noninteger [5]. We restrict ourselves to radial part of the problem. The Coulomb potential is taken as V = −e2 /r for all values of D. For D = 1, r is to be interpreted as |r|. For the three-dimensional case the radial Eq. (12.24) is rewritten as 2e2 µ 2µE 2 ∇ R+ + 2 R=0, (12.74a) ~2 ~ r where
d2 2 d l(l + 1) + − . (12.74b) dr2 r dr r2 For D-dimension the volume element is dτ = rD−1 dr and the Laplacian is ∇2 =
D−1 d l(l + D − 2) d2 + − . 2 dr r dr r2 Consequently, the Schrödinger equation in D-dimension is 2 2e µ 2µE l(l + D − 2) d2 R (D − 1) dR + + + 2 − R=0. dr2 r dr ~2 r ~ r2 ∇2 =
(12.75)
(12.76)
With the change of variables
ρ2 = −
8µEr2 , ~2
R=
f (ρ)
(12.77)
r−(D−1)/2
Eq. (12.76) becomes 2√ e µ 1 + (2l + D − 2)2 1 √ f ′′ + + − f =0. 4ρ2 4 ~ρ −2E
(12.78)
Defining q = n + l and comparing the above equation with Eq. (12.32) gives p = ±(2l + D − 2) ,
E=−
~2 (2n
2µe4 . + 2l + p + 1)2
(12.79)
The eigenfunction is R = e−ρ/2 ρ(p−D+2)/2 Lpq (ρ) .
(12.80)
For D = 1, the only value that l can take is 0. Then p = ±1. For p = 1 E=−
µe4 , + 1)2
2~2 (n
For p = −1
R = e−ρ/2 ρL1n (ρ) ,
ρ=
2µe2 r . + 1)
~2 (n
(12.81)
µe4 2µe2 r −ρ/2 −1 , R = e ρL (ρ) , ρ = . (12.82) n 2~2 n2 ~2 n The results obtained earlier for three-dimensional case is recovered from Eqs. (12.79) and (12.80) by the substitution D = 3. E=−
Hydrogen Atom 321
12.4 FIELD PRODUCED BY A HYDROGEN ATOM In this section we calculate the magnetic field in a hydrogen atom [6]. The contribution from both the orbital and spin motion of the electron are considered. The calculations are nonrelativistic and the energy eigenstates |n, l, ml , ms i and |n, l, j, mj i are alone considered. The calculation is tedious for a state of superposition of these eigenfunctions. This is because it is incorrect to add the fields due to each eigenstate.
12.4.1
General Considerations
For ml 6= 0 states, we treat the electron cloud as circulating round the z-axis. This orbital motion produces the magnetic field. The different parts of the cloud may be circulating in opposite directions. Now, the probability current density is given by ~ J= [ψ ∗ ∇ψ − (∇ψ ∗ )ψ] , (12.83) 2mi where m is mass of the electron. The electric current density j is j = −eJ = iµB [ψ ∗ ∇ψ − (∇ψ ∗ )ψ] ,
(12.84)
where µB = e~/(2m) is the Bohr magnetron . Assuming ψ = F (r, θ)eiml φ for an energy eigenstate |nlml i, Eq. (12.84) in spherical polar coordinates gives (the proof is left as an exercise to the reader) 2µB |F |2 ml φ1 . r sin θ
(12.85)
2µB ml ∗ 2µB |F |2 ml =− ψ ψ. r sin θ r sin θ
(12.86)
j = (0)r1 + (0)θ1 − Thus, jr = 0 ,
jθ = 0 ,
jφ = −
Using Eq. (12.86) the vector potential A and the field is calculated. Since A will be directed in the direction of the current density we have Ar = Aθ = 0 and ∂Aφ /∂φ = 0. Hence, r and θ components of ∇2 A vanish. The φcomponent of ∇2 A is ∇2 A = ∇2 Aφ −
1 2 ∂Vr 2 cos θ ∂Vθ Aφ + 2 + 2 2 . r sin θ ∂φ r2 sin2 θ r sin θ ∂φ
(12.87)
As Ar = Aφ = Aθ = 0 we get ∇2 A = φ1 ∇2 Aφ −
Aφ r2 sin θ
(12.88)
Since ∂Aφ /∂φ = 0, and 1 ∂ ∇ Aφ = 2 r ∂r 2
1 ∂ ∂Aφ 2 ∂Aφ r + 2 sin ∂r r sin θ ∂θ ∂θ
(12.89)
322 Quantum Mechanics I: The Fundamentals ∇2 A takes the form ∂Aφ 1 ∂ ∂Aφ Aφ φ1 ∂ r2 + sin θ − . ∇2 A = 2 r ∂r ∂r sin θ ∂θ ∂θ sin θ Then ∇2 A = −µ0 j becomes ∂ ∂Aφ 1 ∂ ∂Aφ Aφ r2 + sin θ − = −µ0 r2 jφ . ∂r ∂r sin θ ∂θ ∂θ sin θ
(12.90)
(12.91)
The field will involve the alternate multipoles, that is, magnetic dipole, octupole, · · · . The corresponding θ dependence of jφ are, respectively, sin θ, 4 cos2 θ sin θ − sin3 θ, · · · . Therefore, we write jφ = Rdip (r) sin θ + Roct (r)(4 cos2 θ sin θ − sin3 θ) + · · · .
12.4.2
(12.92)
Magnetic Dipole Contribution
Substitution of Eq. (12.92) with the dipole term alone in Eq. (12.91) gives ∂ 1 ∂ ∂Aφ Aφ ∂Aφ r2 + sin θ − = −µ0 r2 Rdip sin θ . (12.93) ∂r ∂r sin θ ∂θ ∂θ sin2 θ Assume a solution of this equation as Aφ = S(r) sin θ. The differential equation for S is obtained as r2 S ′′ + 2rS ′ − 2S = −µ0 Rdip r2 .
(12.94)
Solutions of the homogeneous part of Eq. (12.94) are r and 1/r2 . Therefore, Aφ → 0 as r → ∞, we choose S = T /r2 . Then Eq. (12.94) becomes d T′ = −µ0 Rdip . (12.95) dr r2 Integration of Eq. (12.95) once gives Z T′ = −µ Rdip dr + constant . 0 r2
12.4.3
(12.96)
Magnetic Octupole Contribution
Proceeding as before, we find Aφ =
V (r) 4 cos2 θ sin θ − sin3 θ , r4
(12.97a)
where V is obtained from
V′ = −µ0 r6
Z
Roct dr + constant . r2
(12.97b)
Hydrogen Atom 323
12.4.4
Application to Particular State
When ml = 0, Eq. (12.86) gives jφ = 0 and therefore the field vanishes. On the other hand, from Eq. (12.86) we note that if the sign of ml is changed, then j changes sign and hence so do A and B. Therefore, we consider only the states with ml > 0. For the state |2 1 1i with the eigenfunction ψ211 µB re−r/a0 sin θ . (12.98) jφ = − 32πa50 Comparison of Eq. (12.98) with Eq. (12.92) gives µB Rdip = − re−r/a0 . 32πa50
(12.99)
Substituting the above expression for Rdip in Eq. (12.96) we obtain T . Then Aφ is worked out to be µ0 µB sin θ −r/a0 r 8a0 8a20 k2 Aφ = e +4+ + 2 + k1 r + 2 , (12.100) 32πa20 a0 r r r where k1 and k2 are arbitrary constants. To determine k1 and k2 we apply the conditions A → 0 as r → ∞ and A →finite as r → 0. These gives k1 = 0 and k2 = −8a20 . Then r 1 a0 a20 µ0 µB sin θ −r/a0 a20 + + + 2 − 2 . (12.101a) A = φ1 e 4πa20 8a0 2 r r r We find Ar = Aθ = 0. Then the components of B are µ0 µB cos θ −r/a0 1 a0 a20 a30 a30 Br = e + + 2 + 3 − 3 , (12.101b) 2πa30 8 2r r r r 2 µ0 µB sin θ −r/a0 r 1 a0 a0 a30 a30 Bθ = + + + 2 + 3 − 3 . e 4πa30 8a0 4 2r r r r (12.101c) It is interesting to compare the Eqs. (12.101) with the fields due to a dipole moment Ml at the origin, pointing along the z-axis (θ = 0). These are µ0 Ml sin θ 2µ0 Ml cos θ µ0 Ml sin θ , Br = , Bθ = . (12.102) 2 3 4πr 4πr 4πr3 Only at large r the fields given by Eqs. (12.101) reduce to Eqs. (12.102) with Ml = −µB . At r < a0 the fields are different. As r → 0 A = φ1
A → 0,
Br → −
µ0 µB cos θ , 48πa30
Bθ →
µ0 µB sin θ . 48πa30
(12.103)
The above equation shows that at the origin, B is directed along the z-axis (θ = 0), as expected, with Z µ0 µB µ0 r×j , B= Bz (0) = − dτ . (12.104) 48πa30 4π r3
324 Quantum Mechanics I: The Fundamentals
12.4.5
Contribution of Electron Spin
The inclusion of spin involves an additional term in the probability current density. This term is J(s) = (~/2m)∇ × (ψ ∗ σψ) where σ = (σx , σy , σz ) are Pauli spin matrices. If ms = 1/2 then ψ = ψ
Similarly, if ml = −1/2 then ψ ∗ σψ = −ψ ∗ ψK. The spin current density is j(s)
= =
1 0
and ψ ∗ σψ = ψ ∗ ψK.
−eJ(s) = −2µB ms [∇(ψ ∗ ψ)] × K cos θ ∂ ∂ + (ψ ∗ ψ) . 2φ1 ms µB sin θ ∂r r ∂θ
(12.105)
The case ms = −1/2 gives the same fields of ms = 1/2 but with reversed sign. For the ground state the orbit contribution is zero for the magnetic field and (s)
jφ = −
2µB −2r/a0 e sin θ . πa40
(12.106)
(s)
Since, jφ has only dipole term we use Eq. (12.96) and obtain A(s)
=
Br(s)
=
(s)
=
Bθ
µ0 µB sin θ −2r/a0 a0 a20 a20 e 2 + 2 + − , (12.107a) 4πa20 r r2 r2 µ0 µB cos θ −2r/a0 2a0 2a20 a30 a30 e + + − , (12.107b) 2πa30 r r2 r3 r3 µ0 µB sin θ −2r/a0 2a0 2a20 a30 a30 e 4 + + + − . 4πa30 r r2 r3 r3 (12.107c)
φ1
At large r, A = −φµ0 µB sin θ/(4πr2 ) which implies that the dipole moment due to the spin is Ms = −µB as expected. At r = 0, B is in the z-direction (s) with Bz (0) = −2µ0 µB /(3πa50 ).
12.5 SYSTEM IN PARABOLIC COORDINATES Consider the hydrogen atom in the presence of an external field. Let the applied constant electric field E0 is along the z-axis so that E0 = E0 z. Then the electron and proton are stretched apart in the z-direction. Now the atom becomes elongated along the z-axis and the parabolic coordinates system is better suited than spherical polar coordinates to describe the system. The Schrödinger equation of the system is 2 ~ e2 − ∇2 − − er · E0 ψ = Eψ . (12.108) 2µ r
Hydrogen Atom 325 Dimensionless orthogonal parabolic coordinates (ξ, η, φ) are given by: p p a0 x = a0 ξη cos φ, y = a0 ξη sin φ, z = (ξ − η), (12.109a) 2 a0 2 2 2 2 r = (ξ + η), r = x + y + z , (12.109b) 2 where x,, y and z are the Cartesian coordinates, 0 ≤ ξ ≤ ∞, 0 ≤ η ≤ ∞, 0 ≤ φ ≤ 2π and a0 is the radius of first Bohr orbit. The differential elements of length ds and volume element dτ are written as ξ+η 2 2 2 2 2 ξ+η (dξ) + (dη) + ξη(dφ) , (12.110a) (ds) = a0 4ξ 4η a30 dτ = (ξ + η) dξ dη dφ . (12.110b) 4 The operator ∇2 is given by 1 4 ∂ ∂ ∂ ∂ 1 ∂2 ∇2 = 2 ξ + η + . a0 (ξ + η) ∂ξ ∂ξ ∂η ∂η ξη ∂φ2 Then the Schrödinger equation in parabolic coordinates is 4 ∂ ∂ ∂ ∂ 1 ∂2 − ξ + η − ξ + η ∂ξ ∂ξ ∂η ∂η ξη ∂φ2 4 − + E0 (ξ − η) ψ = ξ+η
(12.111)
2Eψ . (12.112)
Assume the solution of Eq. (12.112) as ψ = f1 (ξ)f2 (η)eimφ .
(12.113)
Multiplication of Eq. (12.112) by (ξ + η)/4 and substitution of Eq. (12.113) in the resulting equation decouples it into the equations with (β1 + β2 = 1) d 1 df1 m2 1 ξ + E0 ξ − − E0 ξ 2 + β1 f1 = 0 , (12.114a) dξ dξ 2 4ξ 4 2 df2 1 m 1 d η + E0 η − − E0 η 2 + β2 f2 = 0 , (12.114b) dη dη 2 4η 4
12.5.1
Solution for E0 = 0
For E0 = 0 the problem reduces to that of the hydrogen atom. As shown in sec. 12.2 the energy of a bound state is En = −1/(2n2 ). Now, introduce two new coordinates ρ1 = ξ/n, ρ2 = η/n and write nβi = αi + (|m|+ 1)/2, i = 1, 2. Then Eqs. (12.114) take the form d2 fi 1 dfi 1 1 1 + + − + αi + (|m| + 1) dρ2i ρi dρi 4 ρi 2 2 3 m n E0 ρi − 2− fi = 0 , i = 1, 2. (12.115) 4ρi 4
326 Quantum Mechanics I: The Fundamentals Solution of Eq. (12.115) is fi (ρi ) Wi (ρi )
|m|/2
= e−ρi /2 ρi Wi (ρi ) , |m|!αi ! L|m| (ρi ) = (|m| + αi )! αi
i = 1, 2
(12.116a) (12.116b)
|m|
with Lαi being an associated Laguerre polynomial. In Eq. (12.116b), α1 and α2 must take the integer values 0, 1, 2, · · · in order to have normalizable solutions. Because 1 1 1 1 α1 + (|m| + 1) + α2 + (|m| + 1) = 1 (12.117) β1 + β2 = n 2 n 2 n is obtained as n = α1 + α2 + |m| + 1, where 0 ≤ |m| ≤ n − 1, 0 ≤ αi ≤ (n − |m| − 1), m = ±|m|. Now, |m| |m| ψα1 ,α2 ,m = C(ξη)|m|/2 e−(ξ+η)/2n Lα (ξ/n)Lα (η/n) eimφ , 1 2
(12.118)
where C is a normalization constant and is determined from the normalization condition Z Z ∞ Z 2π 1 ∞ dξ dη dφ(ξ + η) |ψα1 ,α2 ,m |2 = 1 . (12.119) 4 0 0 0 C is worked out as C2 =
α1 ! α2 ! . + α1 )! (|m| + α2 )!
πn4 n2|m| (|m|
(12.120)
The bound state energy is En = −1/(2n2) = −1/[2(α1 + α2 + |m| + 1)2 ].
12.5.2
Solution for Large E0
For large enough external field, E0 , the motion is dominated by the states, α1 = 0, α2 > 0 and m = 0. In this case n = α2 + 1. The eigenfunction is r 1 1 −(ξ+η)/2n 0 ψ0,α2 ,0 = 2 e Lα2 (η/n) . (12.121) n π The most probable value of ξ is 0. Thus, x ≈ y ≈ 0 and z ≈ −η/2. Some of the dipole matrix elements are obtained as [7] hn; 0, α2 , 0|1/a0|n; 0, α2 , 0| = hn; 0, n − 1, 0|1/a0 |n ± 1; 0, n − 1 ± 1, 0i ≈ hn; 0, n − 1, 0|1/a0 |n ± 2; 0, n − 1 ± 2, 0i ≈
3 − n(n − 1),(12.122a) 2 0.32n2, (12.122b) 2 0.11n . (12.122c)
Hydrogen Atom 327
12.6 CONCLUDING REMARKS Since the potential for a hydrogen-like atom is given by V (r) = −Ze2 /r, and the nuclear mass is different, the energy eigenvalues and the position of maximum radial probability distribution will all be different. The hydrogenic ions He+ (Z = 2), Li2+ (Z = 3), Be3+ (Z = 4) etc are examples of such systems. The (neutral) isotopes of atomic hydrogen, deuterium and tritium also provide examples of hydrogenic systems. Since the mass of deuterium is md ≈ 2mp and that of tritium is mt ≈ 3mp , the reduced masses of them will be slightly different from that of hydrogen. These small differences in the values of µ give rise to isotopic shifts of the spectral lines. In addition to deuterium and tritium, there exists other hydrogenic systems in which the role of proton is played by another positive particle. For example, positronium is a bound hydrogenic system made of a positron e+ and an electron. Muonium is another example of such a system in which the proton is replaced by a positive muon µ+ , a particle which is very smaller to the positron, except that it has a mass mµ ≈ 207me and is unstable with a lifetime of about 2.2 × 10−6 sec. In all the above mentioned hydrogenic atoms the negative particle is an electron. Other negative particles could form a bound system with a positive nucleon. For example, a negative muon µ− can be captured by the Coulombic attraction of a nucleon of charge Ze, thus forming a muonic atom. The simplest example of muonic atom is the bound system (p+ µ− ) consisting of a proton p+ and a negative µ− . The radius a of muonic hydrogen (p+ µ− ) is nearly 186 times smaller than that of ordinary hydrogen atom. The reduced mass of muonic hydrogen is ≈ 186me .
12.7 BIBLIOGRAPHY [1] M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. Dover, New York, 1965 pp 773-778. [2] T.H. Cooke and J.L. Wood, Am. J. Phys. 70:945, 2002. [3] V. Glaser, H. Grosse, A. Martin and W. Thirring, Essays in Honour of Valentine Bargmann. Princeton University Press, Princeton, New Jersey, 1976. [4] F. Brau, J. Phys. A 37:6687, 2004. [5] R.E. Moss, Am. J. Phys. 55:397, 1987. [6] W. Gough, Eur. J. Phys. 17:208, 1996. [7] J.N. Bardsley and B. Sundaram, Phys. Rev. A 32:689, 1985.
328 Quantum Mechanics I: The Fundamentals
12.8 EXERCISES 12.1 Show that the equation ~2 2 ∂ψ ~2 2 e2 =− ψ ∇1 − ∇2 − ∂t 2me 2mp |r1 − r2 | reduces to ∂ψ ~2 2 ~2 2 e 2 i~ = − ∇R − ∇r − ψ, ∂t 2M 2µ r i~
where M = me + mp , µ = me mp /(me + mp ) under the centre of mass and relative coordinate transformation r = r1 − r2 and (me + mp ) R = me r1 + mp r2 .
12.2 The eigenfunction for 1s electron is ψ100 = (1/(πa30 ))1/2 e−r/a0 . Calculate the probability of finding the electron in a sphere of radium r = a0 . 12.3 Show that the total angular momentum is unchanged in the transformation (r1 , r2 ) → (R, r). 12.4 For the equation
d2 R 2 dR 2µE 2µe2 l(l + 1) + + + − R=0 dr2 r dr ~2 ~2 r r2 obtain the asymptotic solution of bound states for n → ∞.
12.5 Using the equation 2µ e2 l(l + 1)~2 f ′′ + 2 E + − f =0 ~ r 2µr2 show that near the origin r → 0, the physically acceptable solution behaves like f (r) ∼ rl+1 for l 6= 0. 12.6 What is the probability distribution of the radial position r? 12.7 At time t = 0, a hydrogen atom is described by the superposition state 1 1 1 ψT (r, 0) = √ ψ100 (r) − √ ψ210 (r) + √ ψ211 (r). 2 6 3 What is ψT (r, t) for t > 0? 12.8 Consider the wave function given in the problem 12.7. Determine the probability that at t > 0, a measurement would find (i) E = E2 , (ii) L2 = 2~2 and (iii) Lz = 0. 12.9 Determine whether the probability density associated with the wave function given in the problem 12.7 is periodic in time and if it is so find the period. 12.10 For a hydrogen atom in the ground state calculate the average distance of the electron from the nucleus.
Hydrogen Atom 329 12.11 If the interaction potential V is spherically symmetric and proportional to rk then the virial theorem states 2hT i = khV i. Prove the virial theorem for hydrogen atom. 12.12 According to Bohr’s correspondence principle the quantum mechanical results reduce to classical results at large quantum numbers. For large values of n find the location of the electron. 12.13 Show that the probability distribution of the radial function of the electron for the hydrogen atom Pnl (r) = |Rnl (r)|2 r2 has n − l maxima and for the lowest value of l = n − 1, there is just one maximum at the value given by the Bohr model. 12.14 Find the lowest energy and the classical turning radii of hydrogen atom in the l = 4 state. 12.15 Assuming the radius of the Li++ nucleus as 3fm calculate the probability of finding the 1s electron inside the nucleus. 12.16 Find the expectation value of the position vector of hydrogen atom. 12.17 Show that the transition between n = 2 to n = 1 levels of muonic hydrogen atom lies in the X-ray region. 12.18 A hydrogen atom is subjected to a potential a/r2 in addition to the potential −e2 /r. Assume the ground state eigenfunction as ψ(r) = rα e−βr . Find the ground state energy using the radial equation and the condition on the potential constant a for the admissible real energy value. 12.19 According to Bohr’s correspondence principle the quantum mechanical results reduce to classical results at large quantum numbers. Show that for large values of n, the electron has energy which is the same as that of a classical electron in a circular orbit. 12.20 Assume that the admissible wave function in the neighbourhood of the region for a system with a potential V (r) which is no more singular than r−2 as ψ(π, θ, 0) = rα f (θ, φ). Find the condition on α for the existence of hHi.
CHAPTER
13
Approximation Methods I: Time-Independent Perturbation Theory
13.1 INTRODUCTION The starting point in a quantum mechanical problem is the construction of a Schrödinger equation. We try to solve it exactly to obtain wave function. A Schrödinger equation is exactly solvable only for a certain number of simple potentials. In most of the cases the Schrödinger equation cannot be solved exactly. The solution of the Schrödinger equation for the real systems is usually complicated and often exact solutions are very difficult to find or solutions not exist. Examples include as anharmonic oscillator, hydrogen atom in an electric field, a plane rotator in the presence of an applied electric field and so on. Therefore, we need to rely on approximation methods. The simple and most interesting and powerful approximate method is the perturbation theory developed by Schrödinger in 1926 [1]. Earlier Lord Rayleigh [2] analysed harmonic vibrations of a string perturbed by small inhomogeneities. The work of Rayleigh was pointed out by Schrödinger. Consequently, the method is also called Rayleigh–Schrödinger perturbation theory. This method is powerful for determining the changes in the discrete energy values and the associated eigenfunctions of a system due to a weak perturbation, provided the eigenpairs of the undisturbed system are available. The approximations obtained from the perturbation theory are used to calculate energy level diagrams and to understand the splitting and assignments of atomic and molecular spectra [3– 5]. The perturbation theory has been proven to be useful in several fields of theoretical physics and also in chemistry. It provides considerable insight into various phenomena. The main disadvantage of this method is that in certain problems the perturbation series converges slowly or diverges.
331
332 Quantum Mechanics I: The Fundamentals In perturbation problems the Hamiltonian is split into two parts. One part called unperturbed part is large and the corresponding Schrödinger equation is assumed to be exactly solvable. The other part is treated as perturbation. The changes in the energies and the eigenfunctions are represented as a power series in the perturbation parameter. The approximation consists of neglecting terms in the series after the first few terms. Approximating the series to the first n terms gives the n-th order approximation. In this chapter we develop the perturbation theory for the timeindependent perturbation. We consider the theory for both degenerate and nondegenerate energy levels. We apply the theory to a few physically interesting problems. The problem of time-dependent perturbation will be discussed in the next chapter.
13.2 THEORY FOR NONDEGENERATE CASE Let H (0) be the unperturbed Hamiltonian operator and its energy eigenvalues are discrete and nondegenerate. As noted earlier in chapter 4, by nondegenerate we mean that there is one eigenfunction corresponding to each energy level. In case there are k-linearly independent eigenfunctions corresponding to one energy level the system is said to be k-fold degenerate. An example of a system with two-fold degeneracy is a free particle (see sec. 4.3). Let the total Hamiltonian of the system be H = H (0) + λH (1) ,
(13.1)
where H (0) is the Hamiltonian of the unperturbed system and H (1) is the perturbation. It is desired to solve Hφn = En φn , that is, H (0) + λH (1) φn = En φn . (13.2)
When λ = 0, Eq. (13.2) reduces to
(0) (0) H (0) φ(0) n = En φn .
(13.3) (0)
(0)
The system (13.3) is assumed to be exactly solvable so that φn and En are explicitly known. The goal is to express En and φn of the perturbed system as a power series in terms of λ as En φn (k)
(k)
= =
En(0) + λEn(1) + λ2 En(2) + · · · , φn(0)
+
λφ(1) n
+
λ2 φ(2) n
+ ··· .
(13.4) (13.5)
Here En and φn are the k-th order corrections. We restrict ourselves with (1) (1) (2) (2) the calculation of En , φn , En and φn . Substitution of Eqs. (13.4) and
Approximation Methods I: Time-Independent Perturbation Theory 333 (13.5) in Eq. (13.2) gives (1) 2 (2) H (0) + λH (1) φ(0) n + λφn + λ φn + · · · (1) 2 (2) = En(0) + λEn(1) + λ2 En(2) + · · · φ(0) n + λφn + λ φn + · · · .
(13.6)
For this equation to be satisfied for all values of λ the coefficients of various powers of λ must be equal to zero. The first few equations obtained in this way are (0) (0) Coeff. of λ0 : H (0) φ(0) n = En φn .
(13.7)
(1) (0) (1) (0) Coeff. of λ1 : H (0) φ(1) φn = En(0) φ(1) (13.8) n +H n + En φn . 2 (0) (2) (1) (1) (0) (2) (1) (1) (2) (0) Coeff. of λ : H φn + H φn = En φn + En φn + En φn .
(13.9) Equation (13.7) is assumed as already solved. We proceed to solve the (1) (1) Eqs. (13.8) and (13.9). First, we solve the Eq. (13.8) to find En and φn . (2) (2) Then we find En and φn .
13.2.1
(1)
(1)
First-Order Corrections En and φn (1)
To solve the Eq. (13.8) we write φn as X (0) φ(1) c(1) n = m φm .
(13.10)
m
(0)
The reason for the above form is that the various possible states φm are known and they form an orthonormal set and any arbitrary wave function can be written as a linear combination of these functions. Now Eq. (13.8) becomes X X (0) (0) (0) (0) (1) (0) c(1) φm + H (1) φ(0) c(1) (13.11) m H n = En m φm + En φn .
Using Eq. (13.3) the above equation is rewritten as X (0) (0) (1) (1) (0) − H φ(0) − E φ = E c(1) E n . n m n m m
(13.12)
(0)∗
Multiplication of Eq. (13.12) by φr and integrating over all space give (1) (0) (0) = En(1) δrn − Hrn , (13.13) c(1) E − E r r n
where
(1) Hrn =
and we used the result
R∞
−∞
Z
∞
−∞
(1) (0) φ(0) φn dτ r H
(0)∗ (0) φm dτ
φr
= δrm .
(13.14)
334 Quantum Mechanics I: The Fundamentals
13.2.2
(1)
Determination of En
When r = n Eq. (13.13) gives (1) En(1) = Hnn = hH (1) i .
(13.15)
The first-order correction to the nth energy eigenvalue is the expectation value of perturbation H (1) over the nth unperturbed state.
13.2.3
(1)
Determination of cm
For r 6= n, Eq. (13.13) gives (1)
c(1) r =
Hrn (0)
(0)
En − Er
,
r 6= n .
(13.16)
(s)
Further, note that any of the function φn can have an arbitrary multiple of (0) φn added to it without affecting the Eqs. (13.8) and (13.9) and hence without (s) affecting the determination of φn in terms of lower-order function. Hence the (s) (0) (s) multiple can be chosen in such a way that cn = hφn |φn i = 0 for all s > 0. (s) (s) Hence expand φn as (with cn = 0) X (0) φ(s) c(s) (13.17) n = m φm . m6=n
Now, φn
=
φ(0) n +λ
X
m6=n
En
=
(1)
Hmn (0)
(0)
En − Em
(1) En(0) + λHnn .
In order to use it we require (1) (0) , λHnn ≪ En(0) − Em
φ(0) m ,
for all n 6= m .
(13.18a) (13.18b)
(13.19)
(1)
Another criterion is that the change in energy λHnn should be small compared (0) to the spacing between En and the energies nearest to it. From Eq. (13.18a) we note that the first-order perturbation theory breaks down for degenerate (0) (0) energy levels since En − Em becomes zero.
Solved Problem 1: (2)
(2)
Determine the second-order corrections En and φn . We write φ(2) n =
X m
(0) c(2) m φm .
(13.20)
Approximation Methods I: Time-Independent Perturbation Theory 335 (1)
(2)
Substituting φn and φn in Eq. (13.9) and multiplying the resulting equation (0)∗ by φr and integrating over all space we get X X (2) (1) c(2) Er(0) − En(0) = En(1) c(1) c(1) r m δrm + En δrn − m Hrm . (13.21) m6=n
m6=n
When r = n from the above equation we obtain (1) 2 Hnm X X (1) . En(2) = c(1) m Hnm = (0) (0) m6=n m6=n En − Em When r 6= n from Eq. (13.21) we get (1)
c(2) r = −
(1)
En Hrn (0)
(0)
Er − En
2 −
X
m6=n
The normalization condition gives
c(2) n = − Now, φ(2) n
(1)
(13.22)
(1)
Hrm Hmn . (0) (0) (0) (0) Er − En En − Em
1 X (1) 2 cr . 2
(13.23)
(13.24)
r6=n
(1) (1) X X Hrm Hmn = (0) (0) (0) (0) En − Em r6=n m6=n En − Er 2 1 (1) (1) (1) H rn 2 En Hrn (0) (0) − . (13.25) 2 φr − 2 φn (0) (0) (0) (0) Er − En Er − En
We have found first- and second-order corrections to the unperturbed energy eigenvalues and eigenfunctions. If the Hamiltonian has both discrete and continuous states then we must include both the sum over all discrete states and the integral over all the continuous states. We notice several important points from Eqs. (13.22) and (13.25): 1. The ground state energy is always lowered in the second-order pertur(0) (0) bation because in Eq. (13.22) the quantity En − Em is negative. 2. Energies of two states are pushed apart in second-order perturbation. . (0) (0) 3. Due to the term 1 En − Em if the energy levels of unperturbed systems are close, the levels are more affected. 4. The theory breaks down if the energy levels are degenerate. We can proceed further to obtain higher-order correction. In practice we may stop the calculation after second-order correction.
336 Quantum Mechanics I: The Fundamentals (2)
13.2.4
Dalgarno and Lewis Method of Calculation of En (2)
The En given by Eq. (13.22) involves evaluation of infinite terms in the summation. In some cases computation of them is a difficult task. However, this is simplified by the method of Dalgarno and Lewis [6,7]. (2)
13.2.4.1 Formula for En
Suppose Fn is a state-dependent operator with the commutator relation h i (0) (1) Fn , H (0) φ(0) − H (0) Fn φ(0) − En(1) φ(0) (13.26) n = Fn H n = H n . (0)∗
Multiplying Eq. (13.26) by φm and integrating over all space we get E D E D h i E D m Fn , H (0) n = m Fn H (0) n − m H (0) Fn n E D = m H (1) − En(1) n . (13.27) When m = n
i E D h n Fn , H (0) n
E D n Fn H (0) − H (0) Fn n E D = n Fn En(0) − En(0) Fn n
=
= 0. That is,
E i E D D h n Fn , H (0) n = n H (1) − En(1) n = 0 .
(13.28)
(13.29)
When m 6= n we obtain from Eq. (13.27) E D E D E D E D m Fn H (0) n − m H (0) Fn n = m H (1) n − m En(1) n . (13.30)
That is,
E D (0) hm |Fn | ni En(0) − Em = m H (1) n .
(13.31)
Rewrite (13.31) as
(1)
hm |Fn | ni =
Hmn (0)
(0)
En − Em
,
m 6= n .
(13.32)
Substituting Eq. (13.32) in Eq. (13.22) we get En(2)
=
(1) (1) E X D H H nm mn = n H (1) m hm |Fn | ni . (13.33) (0) (0) m6=n En − Em m6=n
X
Approximation Methods I: Time-Independent Perturbation Theory 337 In Eq. (13.33) Z ∞denominator disappeared. Using the closure property X |mihm| + |kihk| dk = 1 we get m
−∞
E D En(2) = n H (1) Fn n − En(1) hn |Fn | ni .
(13.34)
The point is that instead of the infinite number of summations in Eq. (13.22), now we need to calculate only at most two integrals hn H (1) Fn ni, hn |Fn | ni (1) or only one if En is zero. Similarly, we obtain E D En(3) = n Fn H (1) Fn n − 2En(2) hn |Fn | ni − En(1) hn |Fn | ni . (13.35)
13.2.4.2 Expression for Fn
If H (1) has no differential operators then from Eq. (13.26) we obtain 1 2 1 2 (1) Fn px + V − px + V Fn φ(0) − En(1) φ(0) n = H n . (13.36) 2m 2m Substituting px = −i~d/dx, Eq. (13.36) becomes 2m ′′ ′ (0)′ φ(0) = 2 H (1) − En(1) φ(0) n . n Fn + 2Fn φn ~
After simple mathematics, from Eq. (13.37), we get x Z 2 2 2m x (1) (1) (0) ′ H − E φ dx . φ(0) (x) F (x) = n n n n ~2 a a
(13.37)
(13.38)
(0)
Suppose a is such that φn (a) = 0. Then Z x 2 2m H (1) − En(1) φ(0) Fn′ (x) = dx . 2 n (0) a ~2 φn
Integration of the above equation gives Z x Z 2 2m x (1) 1 (1) (0) Hξ − En φn (ξ) dξ dx Fn (x) = 2 ~2 a (0) φn
(13.39)
(13.40)
Solved Problem 2: Consider a particle in a box with the Hamiltonian H = H (0) + H (1) , H (0) = p2x /(2m) + V (x) where ( ( 0, for |x| ≤ π/2 αx, for |x| ≤ π/2 (1) V (x) = H = (13.41) ∞, for |x| > π/2 , 0, for |x| > π/2 .
338 Quantum Mechanics I: The Fundamentals (2)
Determine E0
of the particle.
The first few unperturbed eigenfunctions are p p (0) (0) φ0 = 2/π cos x , φ1 = 2/π sin 2x , p ~2 (0) 2/π sin 4x , En(0) = (n + 1)2 φ3 = . 2m
(13.42a) (13.42b)
(2)
Let us calculate E0 using Dalgarno–Lewis method. For the ground state, F0 is calculated from Eq. (13.40) as Z x Z 2mα x 1 2 ξ cos ξ dξ dx F0 (x) = ~2 cos2 x a 2 x Z αm x 2 ξ 1 = sec x + (2ξ sin 2ξ + cos 2ξ) dx ~2 2 4 a Z x αm 1 = x2 sec2 x + 2x tan x + 1 − sec2 x 2~2 2 1 2 2 − sec x a + a sin 2a + cos 2a dx . (13.43) 2 (0)
The condition φ0 (a) = 0 gives a = ±π/2. We identify the correct choice of a using Eq. (13.32). Both the values of a gives αm π2 2 F0 (x) = 2 x − tan x + x . (13.44) 2~ 4 Consider h0|F0 |1i. We get h0|F0 |1i =
h0|H (1) |1i (0)
(0)
E0 − E1
.
(13.45)
The right-side of the Eq. (13.45) gives h0|H (1) |1i
(0) E0
−
(0) E1
=−
32α . 27π
(13.46)
The correct choice of a is determined by computing h0|F0 |1i for a = π/2 and a = −π/2 and comparing them with Eq. (13.46). We find that both the choices lead to h0|F0 |1i = −32α/(27π). Then (2)
E0
=
h0|H (1) |1ih1|H (1) |0i (0) E0
−
(0) E1
+ ··· = −
29 α2 m + ··· . 3 5 π 2 ~2
(13.47)
Next, we have h0|F0 |2i = 0 and h0|F0 |3i = 26 αmπ/(33 53 ~2 ). Thus, (2)
E0
=
−
29 α2 m 211 α2 m 2.1097α2m − 5 5 2 2 − ··· ≈ − . 5 2 2 3 π ~ 3 5 π ~ ~2 π 2
(13.48)
Approximation Methods I: Time-Independent Perturbation Theory 339 We compare this result with the exact second-order correction given by Eq. (13.34), we get (2) E0
π2 2 = x x − tan x + x cos2 x dx 4 −π/2 π 2 α2 m = − 15 − π 2 . 24~2 π 2 α2 m ~2 π 2
Z
π/2
(13.49)
13.3 APPLICATIONS TO NONDEGENERATE LEVELS To demonstrate the use of perturbation theory let us apply it to some interesting systems. First, apply it to the linearly perturbed harmonic oscillator whose unperturbed and as well as the perturbed parts are exactly solvable.
13.3.1
Perturbed Linear Harmonic Oscillator
Consider the one-dimensional harmonic oscillator in an electric field. The potential and the Hamiltonian of the system are V (x) = mω 2 x2 /2 − eEx and H = H (0) + λH (1) ,
H (0) =
1 2 1 p + mω 2 x2 , 2m x 2
H (1) = −eEx . (13.50)
The system with λ = 0 is exactly solved and its energy levels and eigenfunctions are given by Eqs. (4.33) and (4.29) respectively. The eigenpairs of the system with λ 6= 0 are also explicitly known. Let us obtain En (exact) of the perturbed system. Introducing the change of variable x = y + α the Hamiltonian H becomes H=
1 2 1 px + mω 2 y 2 + 2αy + α2 − λeE (y + α) . 2m 2
(13.51)
We set the coefficient of y to zero. This gives α = λeE/(mω 2 ). Then λ2 e2 E 2 1 2 1 px + mω 2 y 2 − 2m 2 2mω 2
(13.52)
1 2 1 λ2 e2 E 2 = H′ = p + mω 2 y 2 . 2 2mω 2m x 2
(13.53)
H= which we rewrite as H+
H ′ is the Hamiltonian of the harmonic oscillator with the position operator y. Thus, for the perturbed oscillator 1 λ2 e2 E 2 En (exact) = n + ~ω − . (13.54) 2 2mω 2 The above equation shows that the change in energy is second-order in λ. In
340 Quantum Mechanics I: The Fundamentals the following we apply the perturbation theory to find the correction to the energy eigenvalues. We expect the first-order correction to energy as zero. The first-order correction to En is given by Z ∞ (1) (1) (1) (0) En = Hnn = φ(0)∗ φn dx . (13.55) n H −∞
The value of the above integral can be easily worked out if we replace the position operator x in terms of the raising and lowering operators a† and a introduced in sec. 4.4 and expressing the eigenstates and the integral in Eq. (13.55) in Dirac notation. p In terms of a and a† the operator x is expressed as x = ~/(2mω) a + a† . We denote the eigenstates as |ni. We have the relation √ √ a|ni = n|n − 1i , a† |ni = n + 1|n + 1i . (13.56) The orthogonality of two eigenstates |n′ i and |ni is written as hn′ |ni = δn′ n . (1) Now, En is obtained as En(1)
= = =
hn|H (1) |ni 1/2 √ √ ~ n hn|n − 1i + n + 1 hn|n + 1i −eE 2mω 0. (13.57)
(1)
(2)
Since En = 0 we find the second-order correction. En is calculated as X hn H (1) mi 2 (2) En = (0) (0) m6=n En − Em 2 √ √ e2 E 2 X mδn,m−1 + m + 1δn,m+1 = 2mω 2 n−m m6=n
2
=
−
2
e E 2mω 2
(13.58)
which is exactly the correction given in Eq. (13.54).
13.3.2
Hyperfine Splitting of the Ground State of Hydrogen [8]
The splitting of an energy level due to the interaction of the nuclear magnetic dipole and the magnetic field Be produced by the electron orbit current is called the fine structure. The term hyperfine structure is used to describe extremely small splitting of spectral lines. In the following we derive the hyperfine structure of hydrogen by employing first-order perturbation theory. We consider the hyperfine splitting of the ground state [8]. This splitting is associated with the interaction of the magnetic moments of the proton and the electron.
Approximation Methods I: Time-Independent Perturbation Theory 341 The electron and proton in atomic hydrogen constitute tiny magnetic dipoles. The magnetic field of a magnetic dipole kept at the centre of a sphere of radius R is given by B(r) =
µ0 2 [3 (m · r) r − m] + µ0 mδ 3 (r) , 4πr3 3
(13.59)
where the first term is only to the region outside an infinitesimal sphere about r = 0. Then the energy of a magnetic dipole m in the presence of a magnetic field B is H = −m · B. Now, the energy of one magnetic dipole (m1 ) in the field of another magnetic dipole (m2 ) is H=−
µ0 2 [3 (m1 · r1 ) (m2 · r2 ) − m1 · m2 ] − µ0 m1 · m2 δ 3 (r) , 4πr3 3
(13.60)
where r is their separation. According to the first-order perturbation theory, the change in the ground state energy is 1/2 Z ∞ 1 E′ = φ∗0 Hφ0 dτ , φ0 = e−r/a0 s , (13.61) 3 πa −∞ 0 ◦
with a0 = 0.52917706A is the Bohr radius and s denotes the spin of the electron. Because φ0 is spherically symmetrical, the θ integral is zero and hence 2 2 µ0 E ′ = − µ0 hm1 · m2 i|φ0 |2 = − hm1 · m2 i . (13.62) 3 3 πa3 Here m1 and m2 are the magnetic dipole moments of the proton and electron respectively. They are given by m1 = γp Sp and m2 = −γe Se where γ’s are the two gyromagnetic ratios. Thus, E′ =
2 µ0 γp γe hSe · Sp i . 3 πa3
(13.63)
The good quantum numbers of the system are the eigenvalues of the total 2 angular momentum J = Se + Sp . Now J2 = (Se + Sp ) = S2e + S2p + 2Se · Sp 2 2 2 so that Se · Sp = J − Se − Sp /2. Since the electrons and protons carry spin-1/2 the eigenvalues of S2e and S2p are 3~2 /4. The two spins form a spin-1 triplet (J2 = 2~2 ) and a spin-0 singlet (J2 = 0). Thus, ( ~2 /4 (triplet) hSe · Sp i = (13.64) −3~2 /4 (singlet) . The singlet level is lower because in this state the proton and electron spins are antiparallel (their magnetic moments are parallel). The energy gap between the triplet and singlet states is ∆Ehyd =
2µ0 ~2 γp γe = 5.884 × 10−6 eV , 3πa3
(13.65)
342 Quantum Mechanics I: The Fundamentals where we have substituted γe = ege /(2m), γp = egp /(2m), ge = 2.0023 and gp = 5.5857. The frequency of the photon emitted in a transition from the triplet state to the singlet state is γ = ∆E/h = 1422.8MHz while the experimental value is 1420.4MHz.
13.4 THEORY FOR DEGENERATE LEVELS The first-order perturbation theory developed for nondegenerate levels is not applicable when the energy level of the unperturbed system is degenerate. This (0) (0) is because, if the energy values of φm and φn are the same, say, E (0) then (0) (0) Em − En = 0 making the correction term in Eq. (13.18a) infinity. Therefore, we require a separate perturbation theory for degenerate levels. Note that degeneracy in any quantum mechanical system stems from symmetries inherent to the system. If any perturbation distorts the symmetry of the system, then it will tend to remove the related degeneracy. Suppose the eigenvalues of unperturbed Hamiltonian H (0) are m-fold degenerate. Then there are m linearly independent orthogonal eigenfunctions (0) (0) (0) (0) (0) (0) φ1 , φ2 ,· · · ,φm corresponding to E1 = E2 = · · · = Em = E (0) . Given (0) E the correct choice of the eigenfunction out of the above m eigenfunctions is not known. Therefore, choose the unperturbed eigenfunction of a degenerate level as a linear combination of m eigenfunctions and allow the perturbation to choose the appropriate one. So write φ(0) =
m X
(0)
ci φi
,
(13.66)
i=1
where ci ’s are constants. The perturbed eigenfunction and the corresponding energy are written as φ = E =
φ(0) + λφ(1) + λ2 φ(2) + · · · , E (0) + λE (1) + λ2 E (2) + · · · .
(13.67a) (13.67b)
Substitution of Eqs. (13.67) in the equation H (0) + λH (1) φ = Eφ gives H (0) + λH (1) φ(0) + λφ(1) + · · · = E (0) + λE (1) + · · · φ(0) + λφ(1) + · · · . (13.68)
Equating the coefficients of various powers of λ to zero we get λ0 1
λ
: H (0) φ(0) = E (0) φ(0) ,
(13.69a)
: H (0) φ(1) + H (1) φ(0) = E (0) φ(1) + E (1) φ(0) ,
(13.69b)
and so on. Next, write φ(1) =
X i
(0)
ai φi
.
(13.70)
Approximation Methods I: Time-Independent Perturbation Theory 343 Then Eq. (13.69b) becomes X X (0) (0) (0) ci φi . Ei − E (0) ai φi = E (1) − H (1)
(13.71)
i
i
(0)∗
Multiplication of Eq. (13.71) by φr
and integration over all space lead to
m X (1) Hri ci , ar Er(0) − E (0) = E (1) cr − i=1
(0)
For r ≤ m, Er
r = 0, 1, · · · .
(13.72)
= E (0) , therefore Eq. (13.72) gives m X (1) Hri − E (1) δri ci = 0 , i=1
(13.73)
r≤m.
Equation (13.73) generates m simultaneous equations for the ci ’s. For nontrivial solution the determinant of coefficient matrix of c must be zero: (1) (1) H − E (1) H (1) · · · H 11 12 1m (1) (1) (1) (1) H21 H22 − E ··· H2m (13.74) . =0. .. .. .. . . ··· (1) (1) (1) H H ··· Hmm − E (1) m1
m2
(1)
(1)
(1)
This is known as secular equation. The m real roots E1 , E2 , · · · , Em of the Eq. (13.74) are the required corrections to the eigenvalues in the first approx(1) imation. When all these Ei are distinct then the perturbed eigenvalues are distinct. The perturbation removed the degeneracy completely. On the other hand, if some of them are identical then the removal is partial. The first-order eigenfunction is determined by substituting cr = 0 for r > m in Eq. (13.72). This gives Pm (1) H ci ar = i=1 ri , (0) E (0) − Er
r>m.
(13.75)
In Eqs. (13.72), equations with r ≤ m are used for the calculation of E (1) . We are thus left with equations with r > m. Therefore, we set ar = 0 for r ≤ m. Then Ek
=
φk
=
(0)
(1)
Ek + λEk , X Pm H (1) ci (0) i=1 ri φ(0) φk + λ r . (0) − E (0) r>m E r
(13.76a) (13.76b)
344 Quantum Mechanics I: The Fundamentals In the next section we apply the degenerate perturbation theory to hydrogen atom in an applied electric field.
Solved Problem 3: Consider a two-dimensional harmonic oscillator with the Hamiltonian H (0) = (1/2m)p2x + (1/2m)p2y + (k/2) x2 + y 2 . Its eigenstates are denoted as |nx ny i. Analyse the effect of the perturbation H (1) = λxy on the two-fold degenerate states |10i and |01i. In terms of ladder operators H (0) is expressed as H (0) = ~ω a† a + b† b + 1 ,
where a b x px
1 1 d 1 1 d √ αx + , a† = √ αx − , α dx α dx 2 2 1 1 d 1 1 d = √ αy + , b† = √ αy − , α dy α dy 2 2 1 1 mω = √ , a + a† , y = √ b + b† , α = ~ 2α 2α i~α i~α = − √ (a − a† ), py = √ (b − b† ) . 2 2 =
(13.77)
(13.78a) (13.78b) (13.78c) (13.78d)
The eigenstates of H (0) are given by the product of the kets |nx i and |ny i: |nx i|ny i = |nx ny i. The corresponding energy is (13.79)
Enx ,ny = (nx + ny + 1) ~ω
which is (nx + ny + 1)-fold degenerate. The state |10i and |01i are degenerate both having the same energy eigenvalue 2~ω. For the case of H (1) = λxy, |10i and |01i the secular matrix is h10|λxy|10i − E (1) h10|λxy|01i =0. (13.80) h01|λxy|10i h01|λxy|01i − E (1)
We have h10|λxy|10i = λh1|x|1ih0|y|0i = 0 since h1|x|1i = h0|y|0i = 0. Next, h10|λxy|01i = λh1|x|0ih0|y|1i =
λ . 2α2
(13.81)
Then Eq. (13.80) gives E (1) = ±λ/(2α2 ). The degenerate states |10i and |01i with energy 2~ω splits into two levels with energies 2~ω0 + λ/(2α2 ) and 2~ω −
Approximation Methods I: Time-Independent Perturbation Theory 345 λ/(2α2 ). The corresponding new eigenfunctions are obtained by substituting E (1) = ±λ/(2α2 ) in the secular equation: −E (1) λ/2α2 u1 =0. (13.82) λ/2α2 −E (1) u2 For E (1) = λ/2α2 we get u1 = u2 so the normalized eigenfunction is
1 1 which gives φ1 = √1 (|10i + |01i). For E (1) = −λ/2α2 we get √ 2 2 1 1 u1 = −u2 and so φ2 = √ (|10i − |01i). 2
13.5 FIRST-ORDER STARK EFFECT IN HYDROGEN The splitting of energy levels of an atom by a uniform external field E is known as Stark effect . The effect was discovered by Stark in 1913 during the observation of Balmer series of hydrogen. Suppose the hydrogen atom is subjected to an electric field along zdirection. The first excited energy level is found to split into three energy levels. This is explained using perturbation theory. The perturbation is λH (1) = −eEr cos θ where E is the strength of the field. Choosing λ = eE we write H (1) = −r cos θ. In sec. 12.2 we solved the unperturbed hydrogen atom. The normalized eigenfunctions are designated as ψnlm and are given by Eq. (12.61). Applying the perturbation theory we calculate the change in the energy of the ground state and first excited state due to the applied field.
13.5.1
Ground State: n = 1
p The ground state eigenfunction is ψ100 = 1 πa30 e−r/a0 . According to the first-order perturbation theory for nondegenerate case the change in (ground state)energy E1 is given by (1)
E1
= = =
(1)
λhH100 i Z (1) ∗ H100 ψ100 dτ λ ψ100 Z ∞ Z π Z 2π eE − 3 r3 e−2r/a0 cos θ sin θ dr dθ dφ . (13.83) πa0 0 0 0
In the above integral, integration with respect to θ becomes zero. Therefore, (1) E1 = 0. Thus, there is no change in ground state energy according to firstorder perturbation theory. Since the first-order correction to E1 is zero we proceed to calculate the
346 Quantum Mechanics I: The Fundamentals second-order correction [9,10] by the Dalgarno and Lewis method described earlier. According to this method (2) E1 = − H (1) F = h1|H (1) F |1i , (13.84a) 11
where F is to be determined from
i h H (1) ψ100 = F, H (0) ψ100 .
Substituting for H (0) and H (1) we get −r cos θψ100 = − H (0) F − F H (0) ψ100 2 ~ e2 = ∇2 + F ψ100 + F H (0) ψ100 2m r ~2 = 2∇ψ100 · ∇F + ψ100 ∇2 F . 2m
Substituting for ψ100 in the above equation we get 2 ∂F 2m 2 ∇ F− + 2 r cos θ = 0 . a0 ∂r ~
Seeking a solution of the form F = f (r) cos θ we obtain 2f 2 df 2mr 1 d 2 df r − 2 − + 2 =0. 2 r dr dr r a0 dr ~ Its solution is f= Then
ma0 r a + r cos θ . 0 ~2 2
(13.84b)
(13.85)
(13.86)
(13.87)
(13.88)
Z ∞ Z π Z 2π r 4 ma0 −2r/a0 e a + r cos2 θ sin θ dr dθ dφ 0 ~2 πa30 0 2 0 0 9ma40 = − , (13.89) 4~2 R∞ (2) where we used 0 xn e−x dx = n!. Including λ = eE in E1 we write (2)
E1
=
−
(2)
E1
=−
9ma40 e2 E 2 9 = − a30 E 2 . 2 4~ 4
(13.90)
This shift in energy is interpreted in terms of the dipole moment p = αE acquired by the hydrogen atom where α is called electric polarizability of (2) hydrogen atom. Here p = −∂E1 /∂E and α is obtained as (2)
α=−
1 ∂E1 E ∂E
=
9 3 a . 2 0
(13.91)
Approximation Methods I: Time-Independent Perturbation Theory 347 The shift in the energy level, −9a30 E 2 /4, is due to the electric polarization of the atom. Before perturbation, that is, before applying the electric field, the (0) system has one ground state eigenfunction with an eigenvalue E1 . Now, the (0) (1) energy is changed to E1 + E1 by the applied field. There is no Stark effect in the ground state. Classically, an energy shift of −p · E will be realized if a system possesses an electric dipole moment p. Therefore, we say that the atom in its ground state has no permanent dipole moment. Generally, systems in nondegenerate states have no permanent dipole moments.
13.5.2
First Excited State: n = 2
Next, work out the effect of the applied field on the first excited state corresponding to n = 2. The possible values of (l, m) are (0, 0), (1, 0), (1, 1), (1, −1) (since l = 0, 1 and m = −l to l). There are four states given by ψnlm → ψ200 , ψ210 , ψ211 and ψ21−1 . The first excited state is thus four-fold degenerate. In this case the zeroth-order wave function is ψ (0) = c00 ψ200 + c10 ψ210 + c11 ψ211 + c1−1 ψ21−1 . The secular equation is written as (1) (1) (1) H H200,210 200,200 − E2 (1) (1) (1) H210,200 H210,210 − E2 (1) (1) H211,210 H211,200 (1) (1) H H21−1,210 21−1,200 where (1)
Hnlm,nl′ m′ =
Z
(1)
H200,21−1
(1)
H210,21−1
(1)
H200,211
(1)
H210,211 (1)
(1)
H211,211 − E2 (1)
H21−1,211
(13.92)
(1)
H211,21−1 (1)
(1)
H21−1,21−1 − E2
=0, ∗ ψnlm H (1) ψnl′ m′ dτ .
The eigenfunctions ψ200 , ψ210 , ψ211 , ψ21−1 are given by r ψ200 = N 2 − e−r/2a0 eimφ , m = 0 a0 r ψ210 = N e−r/2a0 cos θ eimφ , m = 1 a0 r e−r/2a0 sin θ eimφ , m = 1 ψ211 = N √ 2 a0 r ψ21−1 = N √ e−r/2a0 sin θ eimφ , m = −1 2 a0
(13.93) (13.94)
(13.95a) (13.95b) (13.95c) (13.95d)
348 Quantum Mechanics I: The Fundamentals p (1) where N = 1/ 32πa30 . The quantities H2lm,2l′ m′ are given by (1) H2lm,2l′ m′
Z
∞
Z
π
Z
2π
∗ ψ2lm H (1) ψ2l′ m′ r2 sin θ dr dθ dφ 0 0 Z ∞ Z π Z 2π ∗ − ψ2lm ψ2l′ m′ r3 cos θ sin θ dr dθ dφ. 0 0 0
= =
0
(13.96)
First, perform the integration with respect to φ. The result is ( Z 2π 2π, if m′ = m (1) i(m′ −m)φ dφ = H2lm,2l′ m′ (φ) = e 0, if m′ 6= m. 0
(13.97)
(1)
Hence the quantities H2lm,2l′ m′ with m′ 6= m are zero. Now, we need to evalu(1)
(1)
(1)
(1)
(1)
(1)
ate the quantities H200,200 , H200,210 , H210,200 , H210,210 , H211,211 , H21−1,21−1 . Integration with respect to θ makes some of these integrals vanish: Z π sin2 θ π (1) (13.98a) H200,200 (θ) = sin θ cos θ dθ = =0, 2 0 0 Z π cos4 θ π (1) H210,210 (θ) = cos3 θ sin θ dθ = − = 0 , (13.98b) 4 0 0 Z π sin4 θ π (1) H211,211 (θ) = sin3 θ cos θ dθ = (13.98c) =0, 4 0 0 Z π sin4 θ π (1) H21−1,21−1 (θ) = sin3 θ cos θ dθ = (13.98d) =0. 4 0 0 (1)
(1)
Thus, only the integrals H200,210 and H210,200 have to be evaluated. We obtain (1)
H200,210
= =
∗ (1) H210,200 Z ∞ Z π Z 2π 0
= =
0
1 − 24a40 3a0 .
Z
0
0 ∞
∗ ψ210 H (1) ψ200 r2 sin θ dr dθ dφ r 4 e−r/a0 dr r 2− a0
Now, the secular equation is written as (1) 3a0 0 −E2 (1) 3a0 −E2 0 (1) 0 0 −E2 0 0 0
0 0 =0. 0 (1) −E2
(13.99)
(13.100)
Approximation Methods I: Time-Independent Perturbation Theory 349 Expanding the determinant we obtain 2 2 (1) (1) 2 E2 E2 − 9a0 = 0 .
(13.101)
The roots of the above equation are (1)
E2
(13.102)
= 0, 0, 3a0 , −3a0 .
The energy E2 to the first-order correction is E2
(0)
(0)
(0)
(0)
= E2 , E2 , E2 + 3λa0 , E2 − 3λa0 .
(13.103)
The energy of the first excited state level in the presence of the applied field (0) (0) (0) (0) becomes E2 + 0, E2 + 0, E2 + 3λa0 , E2 − 3λa0 . That is, the level n = 2 is (0) splitted into three levels. The middle level E2 + 0 is still doubly degenerate. Since the energy of an electric dipole with moment p, oriented relative to the electric field is −p · λ, we note that the energy perturbations +3λa0 , −3λa0 , 0 are those of a dipole with moment 3λa0 oriented antiparallel, parallel and perpendicular to the electric field respectively. Next, determine the zeroth-order eigenfunction given by Eq. (13.92). We have to determine c00 , c10 , c11 and c1−1 . They are obtained by solving the set of equations (refer Eq. (13.73)) 4 X
(1)
(1)
Hri ci = E2 cr ,
r = 1, 2, 3, 4 .
(13.104)
i=1
(1)
In matrix form the above equation is written as, after using the values of Hri , 0 3a0 0 0 c00 c00 3a 0 0 0 0 c10 (1) c10 (13.105) = E2 . 0 c11 c11 0 0 0 0 0 0 0 c1−1 c1−1 (1)
For E2 = 0, from Eq. (13.105) we obtain 3a0 c10 = 0 and 3a0 c00 = 0 which imply c10 = 0, c00 = 0, c11 =arbitrary and c1−1 =arbitrary. The corresponding eigenfunctions are ψ211 and ψ21−1 where we have chosen c11 = c1−1 = 1 so that the eigenfunctions are normalized. These two eigenfunctions have E2 = (0) (1) E2 + 0. For E2 = ±3a0 Eq. (13.105) gives 3a0 c10 = ±3a0 c00 , 3a0 c00 = ±3a0 c10 , 3a0 c11 = 0, 3a0 c1−1 = 0 . (13.106) From √ Eq. (13.106) we √ have c10 = ±c00 , c11 = c1−1 = 0. We choose c00 = 1/ 2 and c10 = ±1/ 2 so that the eigenfunction ψ (0) = c00 ψ200 + c10 ψ210 is normalized.
350 Quantum Mechanics I: The Fundamentals (1)
The eigenfunctions for E2
= 3a0 and −3a0 are given by
1 (0) ψ+ = √ (ψ200 + ψ210 ) , 2
1 (0) ψ− = √ (ψ200 − ψ210 ) 2
(0)
(13.107)
(0)
respectively. The ψ+ state with the lowest energy E2 − 3a0 is the most stable state. Now, one may ask: What does happen in n = 3 and higher excited states? Particularly, we want to know whether the dipolar nature found in n = 2 states is preserved in the higher excited states. The dipolar nature is found to be preserved in the higher excited states also [10].
13.6 ALTERNATE PERTURBATION THEORIES Perturbation theory is one of the most widely used approximate methods in quantum mechanics due to its transparency which greatly facilitates physical insight. The calculation of sufficient perturbation corrections is not trivial in many cases of interest and for this reason several alternate implementations have been suggested. We have already seen in sec. 13.2.4 the Dalgarno and Lewis method. It consists of writing the nth perturbation correction to the eigenfunction φn as Fn φ0 and then solving the resulting equation for Fn . In this section we discuss another perturbation theory which has been successfully applied to a variety of problems.
13.6.1
Hypervirial Perturbation Theory
Direct application of Rayleigh–Schrödinger perturbation theory is tedious for the perturbation containing a polynomial function like the anharmonic potentials of a diatomic molecule. For such systems, analytic perturbation calculation of higher-order is straight-forward when we take into account the hypervirial and Hellmann–Feynman theorems. The method, commonly termed perturbation theory without eigenfunctions or the hypervirial perturbation method, has been applied to many quantum mechanical models [11].
13.6.2
Quantum Mechanical Hypervirial Theorem (QMHT)
Let H be the time-independent and self-adjoint operator with eigenfunction ψ. Then for any time-independent linear operator A on the space H, the QMHT states that the following expectation values vanishes: hψ, [A, H]ψi = 0 .
(13.108)
We illustrate the usefulness of this theorem with the one-dimensional Hamiltonian ~2 2 d H =− D + V (x) , Dx = . (13.109) 2m x dx
Approximation Methods I: Time-Independent Perturbation Theory 351 With A = xk Dx , k = 1, 2, · · · if we evaluate the commutator explicitly, then a set of recursion relations involving the expectation values, hxm i = hψ, xm ψi , m = 0, 1, 2, · · · hx0 i = 1 ,
(13.110)
are obtained. These relations are known as the hypervirial relations. In particular, the case k = 1 corresponds to the usual virial theorem. To evaluate the commutators, the following relations are useful: [Dx , H] =
xk , H
=
k x Dx , H =
dV , dx 1 ~2 kxk−1 Dx + k(k − 1)xk−2 , m 2 k k x [Dx , H] + x , H Dx .
(13.111a) (13.111b) (13.111c)
The identity Eq. (13.111c) is used to rewrite the commutator, while the appearance of Dx and Dx2 are eliminated using Eqs. (13.111a)–(13.111b) and the time-independent Schrödinger equation. Taking expectation values with respect to an eigenstate ψ with energy E yields the general hypervirial relations 2kEhxk−1 i = 2khxk−1 V i + hxk Dx V i −
~2 k(k − 1)(k − 2)hxk−3 i .(13.112) 4m
In the case of solvable Hamiltonians, like harmonic oscillator, (radial) hydrogen atom, these relations necessarily coincide with those obtained from the knowledge of the recursion relation between the orthogonal polynomials (Hermite and Laguerre respectively) which comprise the eigenfunctions. For one-dimensional harmonic oscillator, the Eq. (13.112) produces the following recursion relations: 2(k + 1)~ 1 k(k − 1)(k + 1)~ k−2 k+2 hx i= N+ hxk i + hx i, (13.113) (k + 2)mω 2 4(k + 2)mω where hx0 i = 1. Since hxi = 0, all the expectation values of the odd powers of x vanish.
13.6.3
Quantum Mechanical Hellmann–Feynman Theorem
Let Hλ be a time-independent, self-adjoint operator that depends explicitly on a scalar parameter λ and Hλ ψ = E(λ)ψ ,
hψ, ψi = 1
then the Hellmann–Feynman theorem states that D δH E δE λ = ψ, ψ . δλ δλ
(13.114a)
(13.114b)
352 Quantum Mechanics I: The Fundamentals
Solved Problem 4: If Hλ = H (0) + λH (1) , prove that
∂hEi D ∂Hλ E (1) . = ψ, ψ = H ∂λ ∂λ
We have Hλ ψ = E(λ)ψ, hψ, ψi = 1 and hEi = hψ, Hλ ψi. Then ∂hEi ∂λ
= = = = = =
13.6.4
D ∂ψ
E D ∂H E D ∂ψ E λ , Hλ ψ + ψ, ψ + ψ, Hλ ∂λ ∂λ ∂λ D ∂ψ E D ∂H E D ∂ψ E λ E , ψ + ψ, ψ + E ψ, ∂λ ∂λ ∂λ D E D D ∂H E ∂ψ E ∂ψ λ , ψ + ψ, + ψ, ψ E ∂λ ∂λ ∂λ D ∂H E ∂ λ E hψ, ψi + ψ, ψ ∂λ ∂λ
ψ, H (1) ψ
(1) H .
Perturbation Expansion
The quantum mechanical hypervirial and Hellmann-Feymann theorems are used to provide Rayleigh–Schrödinger perturbation expansions. The first step in the treatment of perturbed Hamiltonians is to obtain the hypervirial relations for hxk i and E for a particular perturbed eigenstate. Next, the following series expansions are assumed: E(λ) =
∞ X
n=0
E
(n) n
λ ,
k
hx i(λ) =
∞ X
(n)
Ck λn .
(13.115)
n=0
The unperturbed energy E (0) is known. As well, the normalized condition (n) hx0 i = 1 is imposed, implying that C0 = δ0n . Substitution of these into Eq. (13.113) and collecting like powers of λn yield a system of difference (i) equations involving Ck and E (j) . However, this system is not closed, since the relation between these two sets of coefficients has not yet been defined. The closure is provided by the Hellmann–Feynman relation ∂E/∂λ = hH (1) i. The series for E(λ), Eq. (13.115), is differentiated and matched term-wise in (i) powers of λ with the series for hH (1) i. The Ck array is generally calculated (i) column-wise. A knowledge of the nth column in Ck array permits calculation of E (n+1) in analogy with Rayleigh–Schrödinger perturbation theory using the wave function.
Approximation Methods I: Time-Independent Perturbation Theory 353
13.7 CONCLUDING REMARKS In this chapter we demonstrated the power of perturbation theory in calculating the small shifts in the spectral lines of some important systems due to the applied perturbation. Though the corrections to the energy eigenvalues given by the perturbation theory are not exact, the results are accurate as long as the perturbation parameter λ is very small. However, in certain cases the perturbation theory is undesirable, particularly, when the systems under considerations cannot be splitted into unperturbed and perturbed parts and also when the strength of the perturbation is very large. There are exactly solvable systems which when subjected to certain interactions posses new set of eigenstates and they cannot be treated as the unperturbed states + corrections. In such cases other approximation methods such as variational method and WKB approximations are useful. An alternative perturbation method on the application of a factorization method to a Riccati equation derived from the Schrödinger equation has been developed and studied [12,13].
13.8 BIBLIOGRAPHY [1] E. Schrödinger, Annalen der Physik 80:437, 1926. [2] J.W.S. Rayleigh, Theory of Sound. Macmillan, London, 1894. [3] E. O’Reilly, Quantum Theory of Solids. CRC Press, Florence, 2002. [4] V. Magnasco, Elementary Methods of Molecular Quantum Mechanics. Elsevier, Amsterdam, 2006. [5] M.A. Parker, Solid State and Quantum Theory for Optoelectronics. CRC Press, Boco Raton, 2009. [6] A. Dalgarno and J.T. Lewis, Proc. Royal Soc. London A 233:70, 1955. [7] H.A. Mavromatis, Am. J. Phys. 59:738, 1991. [8] D.J. Griffiths, Am. J. Phys. 50:698, 1982. [9] R.K. Prasad, Quantum Chemistry. New Age International Publications, New Delhi, 1997. [10] C. Barratt, Am. J. Phys. 51:610, 1983. [11] F.M. Fernandez and E.A. Castro, Hypervirial Theorems. Springer, Berlin, 1987. [12] C.N. Bessis and G. Bessis, J. Math. Phys. 38:5483,1997. [13] F.M. Fernandez, J. Phys. A 39:1683, 2006.
354 Quantum Mechanics I: The Fundamentals
13.9 EXERCISES (3)
13.1 Evaluate the third-order energy term En level.
for a nondegenerate energy
13.2 In nondegenerate time-independent perturbation theory, what is the probability of finding a system in a perturbed eigenstate φn if the corre(0) sponding unperturbed eigenstate is φn ? Solve this up to terms of order 2 λ . 13.3 A harmonic oscillator potential is perturbed by the term λbx2 . Calculate the first-order and second-order corrections to the energy eigenvalues. 13.4 Calculate the first and second-orders corrections to the energy eigenvalues of a linear harmonic oscillator with the cubic term −λµx3 added to the potential. Discuss the condition for the validity of the approximation. 13.5 A harmonic oscillator Hamiltonian is perturbed by a quartic term in x as λbx4 where b has units of energy/length4 . Calculate the first-order correction to the energy eigenvalues of the perturbed system treating the harmonic oscillator as the unperturbed system. 13.6 Consider the perturbed harmonic oscillator Hamiltonian H = p2x /(2m) p 2 + x /(2m) + λ~ m/~ x, −∞ ≤ x ≤ ∞. Applying Dalgarno–Lewis (2) method compute E0 . 13.7 Show that a constant perturbation H (1) = α to a particle in a box of one-dimension with walls at x = 0 and x = L shifts the energy levels without changing the eigenfunctions. (0)
13.8 Calculate the first-order correction to E3 for a particle in a onedimensional box with walls at x = 0 and x = L due to the perturbation H (1) = ax2 . 13.9 Consider a particle of mass m in a one-dimensional infinite square-well of length 2L with its centre at the origin. The perturbation is given by H (1) = V (x) = E0 Lδ(x), where E0 has the units of energy. Obtain the first-order corrections to energy eigenvalues and eigenfunctions [I.R. Lapidus, Am. J. Phys. 55:172, 1987]. 13.10 If the Hamiltonian of a particle in a box of length L is subjected to a uniform electric field given by H (1) = −eEx calculate the first-order correction to the energy eigenvalues. 13.11 A particle in a box potential of width L is perturbed by the term V0 for 0 < x < L/2 and zero otherwise. Compute the first-order correction to the energy eigenvalues.
Approximation Methods I: Time-Independent Perturbation Theory 355 (
−a/x, x > 0 is in its ground ∞, x≤0 state characterized by the eigenfunction φ0 = N xe−βx . The particle is subjected to an applied electric field Ee in the x-direction. Calculate the first-order correction to ground state energy eigenvalue.
13.12 An electron in the potential V (x) =
13.13 Assume that the perturbation added to the infinite height potential is ( 0, 0 < x < L/2 Vp (x) = V0 (x − L/2), L/2 < x < L. Calculate the first-order correction to the energy eigenvalues.
13.14 A particle in a box potential of width L is given by V (x) = 0 for 0 < x < L and 0 elsewhere. It is perturbed by the additional term V0 sin(πx/L) in the interval 0 to L. Applying the perturbation theory, compute the first-order correction to the energy levels. 13.15 Calculate the first-order correction to the energy eigenvalues if the infinite height potential of width L is subjected to the perturbation ( V0 , L/4 < x < 3L/4 Vp (x) = 0, otherwise. 13.16 A particle of mass m is in a two-dimensional quantum box whose infinite height sides are at x = y = 0, x = y = L. The particle is subjected to the perturbation λxy. Determine the first-order perturbed energy eigenvalues in the |nx , ny i state. 13.17 The exact eigenfunctions and energy eigenvalues of a plane rotator are √ given by φn = (1/ 2π)einφ , En = n2 ~2 /(8π 2 I) where I is the moment of inertia. The system is subjected to an electric field and −F µ cos φ represents the perturbed term where F is the strength of the field and µ is the electric moment. Determine the first- and second-order corrections to the energy levels. 13.18 A hydrogen atom Hamiltonian is perturbed by the potential r ≥ ρ0 0, V (x) = e2 r e2 − 3 + , r < ρ0 , 2ρ0 ρ0 r where ρ0 ≪ a0 and a0 is the Bohr radius. The perturbed term can be thought of as the change in the potential due to the finite value of a radius ρ0 occupied by the nucleus. Calculate the first-order Rb correction to the ground state energy. Assume that 0 rn e−αr dr = nRb (−∂/∂α) 0 e−αr dr and e−βρ0 ≈ 1 since ρ0 ≪ a0 .
356 Quantum Mechanics I: The Fundamentals 13.19 Applying the time-independent perturbation theory calculate the firstorder change in the ground state and first excited state energy levels of hydrogen atom due to the perturbation −eF (constant field). Compare the results with the exact values of energy eigenvalues. Can you observe Stark effect due to the application of the above fields? 13.20 Calculate the first-order change in the ground state and first excited state energy levels of hydrogen atom due to the perturbation −α/r. 13.21 Consider the hydrogen atom with the perturbation Vp (r) = −α/r2 . Calculate the correction to the ground state energy. Show that the perturbation partially removes the degeneracy of first excited state. (0) E1 λ∆ . 13.22 The Hamiltonian matrix of a two-state system is H = (0) λ∆ E2 The energy eigenfunctions unperturbed Hamiltonian (λ = 0) are for the 1 0 (0) (0) given by 1(0) = , 2(0) = with eigenvalues E1 and E2 0 1 respectively.
(a) Find the exact energy eigenvalues of H. (0) (0) (b) Assuming λ|∆| ≪ E1 − E2 and using the time-independent perturbation theory, evaluate the first-order and second-order energy corrections. 13.23 Assuming the degenerate eigenfunctions for a p-electron characterized by (0) (0) (0) |n, l = 1, m = ±1, 0i to be φx = xf (r), φy = yf (r) and φz = zf (r), show that the three-fold degeneracy is completely removed where it is subjected to a potential V = λ(x2 − y 2 ). You need not evaluate the energy shifts in detail. 13.24 Suppose the Hamiltonian of a rigid rotator in a magnetic field perpendicular to the axis is of the form AL2 + BLz + CLy if terms quadratic in the field are neglected. Assuming B ≫ C, use perturbation theory to lowest nonvanishing order to get approximate energy eigenvalues for l = 1 states. Use the following relations. p Ly = (L+ − L− )/2i, L+ |lm i = (l − m)(l + m + 1) ~|lm + 1i and p L− |lm i = (l + m)(l − m + 1) ~|lm − 1i. 13.25 Assume the nucleus of the hydrogen atom to be a sphere of radius ρ0 with the charge e uniformly distributed.
(a) Find the potential energy of the electron and hence the perturbation potential to a point nucleus.
Approximation Methods I: Time-Independent Perturbation Theory 357 (b) Evaluate the first-order correction to the state |nlmi assuming the radial part of the eigenfunction inside the nucleus as 2 Rnl (0) = δ . 3/2 l0 n3/2 a0
1
2ǫ
0
13.26 The Hamiltonian of a perturbed system is H = 2ǫ 2 + ǫ 3ǫ 0 3ǫ 3 + 2ǫ where ǫ ≪ 1. Work out the first-order eigenvalues and eigenvectors using the perturbation theory. 13.27 A hydrogen atom is in a constant uniform electric field E0 that points in the z-direction. If α is the polarizability of the hydrogen, then the change in the ground state energy is αE02 /2. Using the perturbation theory show that
(1)2 100 100 H 1 αE 2 < . (0) 0) 2 0 E −E 2
1
Hence, find the maximum value of the polarizability of the hydrogen atom implied by this inequality.
13.28 The eigenstates of a rotating dumb-bell with moment of inertia I, El = ~2 l(l + 1)/2I are (2l + 1)-fold degenerate (m = l, l − 1, · · · , −l). Its eigenfunctions are given by the spherical harmonics Nlm Ylm = (m) Nlm Pl (cos θ) eimφ . In the event that the dumb-bell is equally and oppositly charged at its ends, it becomes an electric dipole. The interaction energy between such a dipole and a constant uniform electric field E is given by H (1) = −d · E , H = H (0) − d · E where d is the dipole moment. Find the first-order effect. 13.29 Consider the dipole moment described in the previous problem. If both ends are equally charged, the rotating dipole constitutes a magnetic dipole. If the dipole has angular momentum L, the corresponding magnetic dipole moment is µ = eL/(2mc) where e is the net charge of the dipole. The interaction energy between the magnetic dipole and a constant magnetic field B is H (1) = −µ · B = −(e/(2mc))L · B. If B points in the z-direction, show that H (1) separate the (2l + 1)-fold degenerate El energies of the rotating dipole. 13.30 A symmetric two-dimensional oscillator in angular coordinates has 2 2 |m| eigenfunctions ψn,m (r, φ) = Nnm r|m| e−r /2λ Ln (r2 /λ2 )eimφ and energy eigenvalues En,m = (2n + |m| + 1) ~ω = (N + 1)~ω where n = 0, 1, · · · , m = 0, ±1, · · · and λ2 = ~/(µω). The two states with N = 1, n = 0 and m = ±1 are degenerate states with the eigenfunctions
358 Quantum Mechanics I: The Fundamentals √ ±iφ 2 2 e 2 ψ0,±1 (r, φ) = 2 re−r /2λ √ . What is the effect of the perturbation λ 2π H (1) = axy = ar2 sin φ cos φ on the degeneate states?
CHAPTER
14
Approximation Methods II: Time-Dependent Perturbation Theory
14.1 INTRODUCTION Approximation methods are developed for systems where the time dependence of the Hamiltonian is small, slowly varying or rapidly varying. These correspond to perturbation, adiabatic and sudden approximations respectively. In this chapter we consider the time-dependent perturbation theory applicable for small time-dependent perturbation. This is the case when we wish to investigate the response of systems to incident radiation. The incident electromagnetic wave makes the electric and magnetic potentials to oscillate and the perturbation is usually very small compared with the static Hamiltonian. The theory of time-dependent perturbation was developed by Dirac in 1927 [1]. In this chapter we discuss the theory for constant and harmonic perturbations applied during a finite time interval, adiabatic and sudden perturbations.
14.2 TRANSITION PROBABILITY The problem is to solve the time-dependent Schrödinger equation ∂ψ (0) i~ = H + λH (1) ψ = Hψ . ∂t
(14.1)
Let (En , φn ) are the stationary state eigenpairs of the unperturbed system. The time evolution of φn (x) is given by φn (x, t) = e−iEn t/~ φn (x) .
(14.2)
Assume that the perturbation is switched-on at t = 0 and H is very close to H (0) so that the set {φn (x, t)} is also a complete set of eigenstates of the total 359
360 Quantum Mechanics I: The Fundamentals Hamiltonian. We write the solution of Eq. (14.1) as X ψ(x, t) = an (t)e−iEn t/~ φn (x),
(14.3)
n
where an (t) to be determined. Here an (t) is the probability amplitude of finding the system in the state φn (x, t) at time t. If the perturbation is absent then an (t) = an (0). To determine an (t) we substitute Eq. (14.3) in Eq. (14.1) and obtain X X i~ a˙ n e−iEn t/~ φn = λ an H (1) e−iEn t/~ φn . (14.4) Multiplication of Eq. (14.4) by φ∗f and integrating overall space lead to i~a˙ f = λ
X
an eiωf n t Hf n ,
(14.5a)
n
where ωf n = (Ef − En ) /~ ,
Hf n =
Z
∞
−∞
φ∗f H (1) (t)φn dτ .
(14.5b)
Next, we expand af (t) as (0)
(1)
(2)
af (t) = af + λaf + λ2 af + · · · .
(14.6)
The obvious criterion for the applicability of the above expansion of af (t) is (1) (0) af ≪ af . Substituting the above series in Eq. (14.5) and collecting equal powers of λ and equating them to zero individually we obtain (0)
Coeff. of λ0 : i~a˙ f = 0 . X (1) iωf n t Coeff. of λ1 : i~a˙ f = a(0) Hf n . n e
(14.7a) (14.7b)
n
(2)
Coeff. of λ2 : i~a˙ f = Coeff. of λs+1 :
X
iωf n t a(1) Hf n . n e
n (s+1) i~a˙ f =
X
iωf n t a(s) Hf n . n e
(14.7c) (14.7d)
n
We note that to zeroth-order (no perturbation), the probability amplitudes (0) (0) (0) af are constants in time: af (t) = af (0) and specify the state of the system before the perturbation is applied. Now, we analyse the first-order effect on the system. Equation (14.7) is a set of coupled integro-differential equations. If the system has m eigenstates then Eq. (14.7) generates m coupled equations and are difficult to solve. Therefore, instead of solving Eq. (14.7) we slightly modify the problem. Suppose the perturbation is switched-on at t = 0 and switchedoff at t = T . Perturbation is applied during a finite time interval T . Assume
Approximation Methods II: Time-Dependent Perturbation Theory 361 that the system is initially in a state i with the eigenfunction φi . Then at t = 0 the probability of finding the system in φi is unity and the probability (0) of finding the system in other states φn (n 6= i) is zero: an = δni . Due to the applied perturbation transition from the state i to an another state after the time T can happen. Once the perturbation is switched-off the system settles down to a stationary state and denote this final state as f . What is the probability for the transition from φi to φf state? It is simply the probability of finding the system in φf : Pf i = a∗f af . Now, Eq. (14.7b) is 1 iωf i t e Hf i . i~
(1)
a˙ f =
(14.8)
Integrating the above equation from 0 to T we obtain 1 = i~
(1) af
Z
T
eiωf i t Hf i (t) dt .
(14.9)
0
The transition probability for ith state to f th state when f 6= i is given by (1) Pf i = af 2 . We consider the cases of H (1) (t) = H (1) (t) =
W (x) , W (x) e−iωt ,
(14.10) (14.11)
H (1) (t) = H (1) (t) =
W (x) sin ωt , W (x) cos ωt
(14.12) (14.13)
separately in the following sections.
14.3 CONSTANT PERTURBATION For the constant perturbation, H (1) (t) = W (x), Eq. (14.9) becomes Z ∞ 1 (1) a˙ f = eiωf i t Hf i , Hf i = φ∗f W (x)φi dτ . i~ −∞
(14.14)
Its solution is obtained as (1) af
1 = Hf i i~
Z
T
eiωf i t dt =
0
Hf i 1 − eiωf i T . ~ωf i
(14.15)
Then the first-order transition probability is Pf i
(1)∗ (1) af
= af
=
4|Hf i |2 sin2 (ωf i T /2) . ~2 ωf2 i
(14.16)
4 sin2(ωf iT /2)/ωf2 i
362 Quantum Mechanics I: The Fundamentals
1 0 .5 0 -20
FIGURE 14.1
14.3.1
-10
0
ωf i
10
20
Plot of [4 sin2 (ωf i T /2)]/ωf2 i as a function of ωf i for T = 1.
Features of First-Order Transition Probability
Some of the features of Eq. (14.16) are summarized below. 1. The Pf i given by Eq. (14.16) is the probability for the system to undergo a transition from a state i to f . This transition is due to the applied perturbation. Pf i depends on Hf i , T and ωf i (that is, Ef and Ei0 ). 2. Selection rule: The transition from ith state to f th state is forbidden if Hf i = 0. The selection rule is thus Hf i 6= 0.
3. Figure 14.1 shows the plot of s = [4 sin2 (ωf i T /2)]/ωf2 i versus ωf i for fixed T . The function s has a sharp peak centered at ωf i = 0. For small ωf i , s = T 2 . The height of the main peak is T 2 . In this case Pf i ∝ T 2 . Hence, the probability per unit time or transition rate Γf i is ∝ T . 4. For large ωf i , Pf i → 0.
5. Pf i is zero when sin2 (ωf i T /2) = 0, that is, ωf i ≈ 2nπ/T . 6. A striking feature of Pf i from Fig. 14.1 is that it does not change monotonically with ωf i . It neither increases nor decreases continuously with ωf i but oscillates with rapidly decreasing amplitude. 7. Pf i for closely spaced levels: Equation (14.16) represents the probability for transition to one particular state φf . In real systems we wish to know PRf i for transitions to a group of closely spaced states. In this case ∞ Pf i = −∞ |af |2 ρf (Ef ) dEf , where ρf dE represents number of states in the range dE. Assume that ρ is almost constant in the range dE. Then Z ∞ 2π Pf i = ρf |af |2 dEf = ρf |Hf i |2 T , (14.17) ~ −∞ R∞ where we have used the result −∞ (sin2 ξ)/ξ 2 dξ = π. Thus, the total transition probability is proportional to T .
Approximation Methods II: Time-Dependent Perturbation Theory 363 8. The transition rate per unit time is 2π dPf i = Γf i = ρf |Hf i |2 dT ~
(14.18)
which is independent of T . The general form of the above formula is applicable to atomic transitions, nuclear decay and many physical transitions and hence Fermi [2] called it the (Fermi) golden rule. This rule is often used in spectroscopy to describe the transition rate between states. Calculations of intensities of spectral lines and many of cross-sections for various processes are based on the use of this rule. 9. For large T the area under the central peak is large compared to the other peaks. Therefore, the transitions to the states within this peak occur mostly. In this case |ωf i | ≤ 2π/T . Suppose, we wish to measure ωf i by measuring Pf i . Then the uncertainty ∆ω ∼ 2π/T or ∆E ∼ 2π~/T or ∆E ∼ h/T . The energy change ∆E lasts for a time ∆t = T . Therefore, ∆E∆T ≈ h. This is known as time-energy uncertainty. If T is small, we have a broader peak in Fig. 14.1 and as a result the energy is not conserved. On the other hand, if the perturbation is applied for a very long time, we have a very narrow peak, and approximate energy conservation is required for a transition with appreciable probability. 10. From Eq. (14.16) we note that the transition probability varies for various states. In experiments, how do we account for this? If we look at a spectrum we will notice that the intensity of spectral lines varies − some are very dark while some are very faint and so on. If the intensity of a line is large then that line corresponds to a transition which takes place very often, that is, this corresponds to large Pij .
Solved Problem 1: A linear harmonic oscillator is subjected to an applied electric field during the time t = 0 to T . Assume that the initial state of the system is φi and the perturbation to the Hamiltonian is H (1) = −eEx. Find the probability for the transition to the state φf in time T . The transition probability is given by Eq. (14.16). We obtain Hf i
= −eEhf |x|ii 1/2 ~ = −eE hf |a + a† |ii 2mω 1/2 h i √ √ ~ = −eE i + 1 δf,i+1 + i δf,i−1 . 2mω
(14.19)
According to first-order perturbation theory, the possible transitions are (i) i → i, (ii) i → i + 1 and (iii) i → i − 1. In the first case the probability is 1
364 Quantum Mechanics I: The Fundamentals (0)
(1)
because ai = 1 and ai = 0. However, we are interested in the transition from ith state to f with f 6= i. Therefore, the selection rule is f = i ± 1. For the transition i to i + 1 (absorption process) ~e2 (i + 1)E 2 (1) 2 . (14.20) Hi+1,i = 2mω
For i to i − 1 (emission process) (1) 2 ~e2 iE 2 . Hi−1,i = 2mω
(14.21)
Solved Problem 2:
The motion of a particle of mass m and electric charge e confined to an interval of length L is subjected to a uniform electric field E from t = 0 to T . Assume that the particle is initially in the state i. Calculate the probability that after a time T the system is in the state f where f > i. The Hamiltonian of the unperturbed system is H (0) = p2x /(2m) + V , V = 0 for 0 < x < L and V = ∞ elsewhere. The unperturbed eigenstates are r nπx n2 π 2 ~2 2 sin , En = . (14.22) φn = L L 2mL2 The perturbation due to the applied field is H (1) = −eEx. The transition probability Pf i is given by (1) 2 4 Hf i π 2 ~2 (f 2 − i2 ) 2 Pf i = sin (ω T /2) , ω = . (14.23) f i f i ~2 ωf2 i 2mL2 (1)
We obtain Hf i = hφf |H (1) |φi i as (1)
Hf i
= −
We have the results cos(f + i)π cos(f + i)π
eEL (f + i)2 cos(f − i)π − 1) π 2 (f 2 − i2 )2 −(f − i)2 (cos(f + i)π − 1) .
(14.24)
for f + i = even
(14.25a)
= cos(f − i)π = −1, for f + i = odd .
(14.25b)
= cos(f − i)π = 1, (1)
Therefore, for f + i = even Hf i = 0. For f + i = odd (1)
Hf i
=
8eELf i . π 2 (f 2 − i2 )2
(14.26)
Thus, for f + i = even we have Pf i = 0 and for f + i = odd it is given by (1) Eq. (14.23) with Hf i given by Eq. (14.26).
Approximation Methods II: Time-Dependent Perturbation Theory 365
14.4 HARMONIC PERTURBATION Suppose the perturbation H (1) is a slowly varying harmonic function of time: H (1) (t) = W (x) e−iωt where ω is the frequency of the applied force. Assume that the perturbation is switched on at t = 0 and switched-off at t = T . Following the treatment of previous section we have Z ∞ 1 φ∗f W (x)φi dτ . (14.27) a˙ f = ei(ωf i −ω)t Hf i , Hf i = i~ −∞ Integration of the above equation from t = 0 to T gives Hf i af (T ) = 1 − ei(ωf i −ω)T . ~(ωf i − ω)
(14.28)
Then
Pf i =
a∗f af
2 4 Hf i = 2 sin2 [(ωf i − ω)T /2] . ~ (ωf i − ω)2
(14.29)
This expression of Pf i differs from that of the constant perturbation by the term (ωf i − ω). With this replacement the entire discussion on Pf i of constant perturbation is valid for the harmonic perturbation. We now list some of the differences between constant and harmonic perturbations: 1. For constant perturbation Pf i ∝ ωf i whereas for harmonic perturbation it is ∝ ωf i − ω. 2. For constant perturbation Pf i is large when ωf i ≈ 0; that is, Ef ≈ Ei . For harmonic perturbation Pf i is large when ωf i − ω ≈ 0, that is, Ef − Ei ≈ ~ω. Therefore, most of the transitions take place between the states for which Ef − Ei ≈ ~ω. 3. For constant perturbation Pf i is proportional to sin2 (ωf i T /2). For harmonic perturbation we have all frequencies ω. Integration over all frequencies gives Pf i proportional to T .
14.4.1
Perturbation with Whole Spectrum of Frequencies
In the above we considered the perturbation with a single frequency ω. Suppose the perturbation contains a whole spectrum of frequencies ω and the phases of the frequency components are unrelated. Then the total transition probability is Z Pf i = |a2f | ρ(ω) dω 2 Z H (ω) sin2 [(ωf i − ω)T /2] fi 4 = ρ(ω) dω . (14.30) ~2 (ωf i − ω)2
366 Quantum Mechanics I: The Fundamentals Then Γf i =
2 2π Hf i (ωf i ) ρ(ωf i ) . 2 ~
(14.31)
The upward transition rate is (i = n, f = m, Em > En ) Γmn =
2 2π Hmn (ωmn ) ρ(ωmn ) . 2 ~
(14.32)
The transition rate for downward transition is (i = m, f = n, Em > En ) Γnm
= =
2 2π ρ(ωnm ) H (ω ) nm nm ~2 2 2π ρ(−ωmn ) . H (−ω ) nm mn ~2
(14.33)
Because of ρ(−ω) = ρ(ω) and Hf i (−ω) = (Hif (ω))† we have Γnm = Γmn . That is, the transition rate for upward and downward transitions are the same. Suppose H (1) = W (x) sin ωt. Then a˙ f =
h i 1 (1) iωf i t 1 Hf i e = Hf i ei(ωf i −ω)t − ei(ωf i +ω)t . i~ 2~
Integrating the above equation we obtain 1 − ei(ωf i −ω)T iHf i 1 − ei(ωf i +ω)T − af = − 2~ ωf i + ω ωf i − ω
(14.34)
(14.35)
and 2 Hf i 2 1 − ei(ωf i +ω)T 1 − ei(ωf i −ω)T Pf i (T ) = − . 4~2 ωf i + ω ωf i − ω
Similarly, for H (1) = W (x) cos ωt we have Hf i 1 − ei(ωf i +ω)T 1 − ei(ωf i −ω)T af = + , 2~ ωf i + ω ωf i − ω 2 Hf i 2 1 − ei(ωf i +ω)T 1 − ei(ωf i −ω)T Pf i (T ) = + . 4~2 ωf i + ω ωf i − ω
(14.36)
(14.37a) (14.37b)
For the harmonic perturbations sin ωt and cos ωt, the term | · |2 in Eqs. (14.36) and (14.37c) has an appreciable amplitude only if the denominator of one of the two terms in it is ≈ 0: ωf i = ±ω or Ef = Ei ± ~ω. Thus, according to first-order perturbation theory for the harmonic perturbation the system must either absorb or emit a quanta of energy ~ωf i from the perturbation.
Approximation Methods II: Time-Dependent Perturbation Theory 367 Suppose the system is in the eigenstate φi at t = a (for example at t = −∞ or 0). The perturbation is switched-on at t = a and left forever. It is desired to know the state at t = b (for example t = ∞ or T ). In this case Rb af = (1/i~) a Hf i eiωf i t dt and the probability of finding the particle in the state f at t = b is Z 2 1 b 2 iωf i t |af | = 2 Hf i e dt . (14.38) ~ a
Solved Problem 3:
A linear harmonic oscillator is subjected to the time varying field described by H (1) = −eExe−iΩt during the time t = 0 to T . If the system is initially in its ground state calculate the probability of finding the system in the firstexcited state for t > T . Find the probability if (i) H (1) = −eEx sin Ωt and (ii) H (1) = −eEx cos Ωt. The transition probability is given by 2 4 H10 P10 = 2 sin2 [(ω10 − Ω)T /2] . ~ (ω10 − ω)2 3 1 We obtain ω10 = ~ω − ~ω /~ = ω and 2 2 H10
Then P10 =
= −eEh1|x|0i 1/2 ~ = −eE h1|a + a† |0i 2mω 1/2 ~ . = −eE 2mω 2e2 E 2 sin2 [(ω − Ω)T /2] . ~mω(ω − Ω)2
(14.39)
(14.40)
(14.41)
In the limit Ω → ω, P10 = e2 E 2 T 2 /(2m~ω). For H (1) = −eEx sin Ωt P10
e2 E 2 = 8~mω
2 1 − ei(ω+Ω)T 1 − ei(ω−Ω)T − (ω + Ω) (ω − Ω)
and for H (1) = −eEx cos ωt 2 e2 E 2 1 − ei(ω+Ω)T 1 − ei(ω−Ω)T P10 = + . 8~mω (ω + Ω) (ω − Ω)
(14.42a)
(14.42b)
368 Quantum Mechanics I: The Fundamentals
Solved Problem 4: Assume that a hydrogen atom with an electron and an infinitely heavy proton is subjected to a spatially uniform and a harmonically time-varying electric field E(t) = 2kξ0 cos ωt, t > t0 . If the perturbation H (1) (t) is taken as (14.43) H (1) (t) = W (x) eiωt + e−iωt calculate the differential ionization probability per unit solid angle.
The time-independent part of the hydrogen atom is given by W (x) = eE0 z = eE0 r cos p θ. The initial ground state of the hydrogen atom is given by ψ100 = (1/ πa30 )e−r/a0 . The final states should correspond to a positive energy electron in Coulomb field of the proton. Since the eigenfunction for such a state is quite complicated, a simple eigenfunction for a free particle confined in a box of side L can be assumed. So the final state k is given (1) as ψk = L−3/2 eik·r . The quantities |Hf i |2 in (14.36) will depend on the direction of k. Hence, the range of k is assumed to be in a solid angle. Since kx = 2πnx /L, ky = 2πny /L and kz = 2πnz /L, the number of states in the range dkx dky dkz becomes (L/2π)2 dkx dky dkz which expressed in spherical polar coordinates gives
L 2π
3
k 2 sin θ dkdθdφ ,
(14.44)
where θ and φ are the polar angles of k with respect to the electric field E. We have L3 2 ρ(k) dEk = k dk sin θdθdφ . (14.45) 2π Since Ek = ~2 k 2 /(2me ) we get dEk = ~2 k/me . Then ρ(k) =
me L3 k sin θ dθ dφ . 8π 3 ~2
(14.46)
The initial state is the state of hydrogen atom ρ(ωf i ) = ρ(k) and i32πeE0 ka50 cos θ
(1)
Hf i = −
1/3
(πa30 L3 )
3
(1 + k 2 a20 )
.
(14.47)
Then the probability per unit time that the electron of the hydrogen atom is ejected into the solid angle sin θ dθ dφ is Γf i =
256me E02 k 3 e2 a70 π~3
(1 +
6 k 2 a20 )
cos2 θ sin θ dθ dφ .
(14.48)
The differential ionization probability per unit solid angle Γf i /(sin θ dθ dφ) is ∝ cos2 θ as the driving force in the direction of ejection is equal to eE0 cos θ.
Approximation Methods II: Time-Dependent Perturbation Theory 369
14.5 ADIABATIC PERTURBATION What will happen if a potential changes very slowly? For each differential change in the potential, a differential change say ǫ is caused in the wave function. But, if the changes take place at different times, the differential amplitudes contribute P 2 with uncorrelated phases. Then the total change in the probability is |ǫ| = 0. Hence, the probability associated with each state remains the same. This is known as adiabatic approximation 1 . For example, consider a pendulum exhibiting oscillation in a vertical plane. Suppose we move the support. The mode of oscillation of the system change. What will happen if the support is moved very slowly? The motion of the pendulum relative to the support will be unchanged. That is, when the perturbation is sufficiently slow then the state of the system remains the same. This is known as adiabatic process.
14.5.1
Transition Probability
Suppose the perturbation applied to a quantum mechanical system varies slowly with time and is written as H (1) (t) = W (x)V (t). An example is V (t) = eηt where 0 < η ≪ 1. The perturbation is switched-on at t = −∞ and the initial state of the system is φi . The perturbation H (1) (t) is assumed small in comparison with E/τ , where E and τ are energy and time typically characteristic of the system. We wish to find the probability of finding the system in the state φf at time t. From Eq. (14.8) the evolution equation for transition amplitude is written as a˙ f (t) =
1 (1) H V (t) eiωf i t . i~ f i
(14.49)
Integrating the above equation with respect to time from t = −∞ we get (1)
af (t) =
Hf i i~
Z
t
V (t) eiωf i t dt .
(14.50)
−∞
Since V (t) is a slowly varying function of t we assume that it is almost constant during a time interval dt and obtain (1)
af (t) = −
Hf i V (t)eiωf i t ~ωf i
.
(14.51)
(1)
We require |af | ≪ 1. That is, |Hf i | ≪ ~|ωf i |. Next, (1) 2 2 H V fi Pf i = . ~2 ωf2 i
(14.52)
1 According to Max Born and Vladimir Fock [3] a physical system remains in its instantaneous eigenstate if a given perturbation acting on it is slowly enough and if there is a gap between the eigenvalue and the rest of the Hamiltonian’s spectrum.
370 Quantum Mechanics I: The Fundamentals When V (t) = eηt from Eq. (14.52) Pf i =
(1) 2 2ηt H e fi
(14.53)
.
~2 ωf2 i
On the other hand, from Eq. (14.50) we obtain (1)
af (t) = − Now, Pf (t) =
Hf i ei(ωf i −iη)t ~(ωf i − iη) (1) 2 2ηt H e fi
~2 (ωf2 i + η 2 )
.
.
(14.54)
(14.55)
Since 0 < η ≪ 1 we can neglect η 2 in the above equation. Then Eq. (14.55) becomes the Eq. (14.53).
Solved Problem 5: A harmonic oscillator in the state φi is subjected to the perturbation Vp (x, t) = −αxeηt , 0 < η ≪ 1 from t = −∞. Calculate the transition probability to a state φf in time t. We apply the adiabatic perturbation theory since the perturbation Vp (x, t) is a slowly varying function of t. From Eq. (14.55) we obtain Pf (t) =
(1) 2 2ηt H e fi
~2 (ωf2 i + η 2 )
,
(14.56a)
where ωf i
1 Ef − Ei = = ~ ~
1 1 f+ ~ω − i + ~ω = (f − i)ω (14.56b) 2 2
and (1)
Hf i
= hφf |αx|φi i 1/2 ~ = α hφf |a + a† |φi i 2mω 1/2 h i √ √ ~ = α i + 1 δf,i+1 + i δf,i−1 . 2mω
(14.56c)
For the upward transition i to i + 1 (ωf i = ω) 2 (1) 2 H = α ~(i + 1) , fi 2mω
ω = ωf i
(14.57)
Approximation Methods II: Time-Dependent Perturbation Theory 371 and Pf (t) =
α2 (i + 1)e2ηt . 2mω~(ω 2 + η 2 )
(14.58)
For the downward transition i to i − 1 (ωf i = −ω) 2 (1) 2 H = α ~i , fi 2mω
and
Pf (t) =
ω = −ωf i
α2 i e2ηt . 2mω~(ω 2 + η 2 )
(14.59)
(14.60)
14.6 SUDDEN APPROXIMATION In the previous sections we assumed that the perturbation is applied during a time interval T . Suppose a time-independent perturbation is switched-on suddenly at t = t′ and is left on forever. This is known as sudden or impulse approximation. In the sudden approximation the Hamiltonian of the system is H = H (0) for t < t′ and H = H (0) + λH (1) for t > t′ . Also, for such problems it is preassumed that the eigenstates of both Hamiltonians H (0) and H = H (0) + λH (1) are known. Sudden approximation theory has been used in the study of single and multiple electron losses of high and intermediate energy projectiles in their collisions with neutral targets, pressure broadening of rotational transitions, core hole ionization in CO, multielectron transition in complex atoms, collisions of fast multicharged ions with atoms and in the collisions of charged particles with highly-excited atoms. What is ψ(t) in the sudden approximation? For t < t′ to get ψ(t) we use (0) H and for t > t′ we use the Hamiltonian H (0) + λH (1) . Usually t′ is chosen as −∞ or 0 for simplicity. Let H (0) φ< n Hφ> n
= En< φ< n , =
En> φ> n
,
t 0.
(14.61b)
For t < 0 and t > 0 the wave functions of the system are X X > < t/~ < −iEn t/~ > −iEm , ψ> = a> , ψ< = a< m φm e n φn e n
(14.62)
m
> respectively, where φ> m and Em are the eigenfunctions and eigenvalues of the < perturbed system. How do we determine a> m ? We use the continuity of ψ > > < > and ψ at t = 0 to calculate am . At t = 0, ψ = ψ , that is, X X < > a< a> (14.63) n φn = m φm . n
m
372 Quantum Mechanics I: The Fundamentals Multiplying both sides by φ>∗ f and integrating over all space we get X n
That is, a> f = a< n
P
n
> < a< n hφf |φn i =
X m
> > a> m hφf |φm i =
X
a> m δf m .
(14.64)
m
> < a< n hφf |φn i.
If we assume that for t < 0 the system is in the state i then a< i = 1 and = 0 for n 6= i. Then X < > < a> δni hφ> (14.65) f = f |φn i = hφf |φi i . n
>∗ > To calculate a> f we require both initial and final eigenstates. af af gives the probability for the transition from the ith state to the f th state of the perturbed system. An interesting consequence of sudden approximation is that the system can undergo an adiabatic change due to the sudden approximation. By adiabatic change we mean a change such that the perturbed system remains in an energy eigenstate with the same quantum number. That is, for example, a system may remain in the nth state of the system for t < 0 and t > 0 also. However, for t < 0 it is in the nth state (φ< n ) of the unperturbed Hamiltonian while for t > 0 it is in the nth state (φ> ) of the perturbed system. n > φ< and φ states have the same quantum number but with different energies n n En< and En> respectively. This is often illustrated with the following example. Consider the infinite height potential confining a particle to the region 0 < x < L. Suppose the right-wall is suddenly moved to the position x = 2L at t = 0. If the system is in the ground state for t < 0, what is the probability of it being in the ground state of the new well? What is the probability of it being in the nth state of the new well? For t < 0 the energy eigenfunctions and eigenvalues are given by r 2 nπx n2 π 2 ~2 < φn = sin , En< = . (14.66) L L 2mL2
The expanded well is also exactly solvable. Its eigenstates are obtained from Eq. (14.66) by simply replacing L by 2L: r 1 nπx n2 π 2 ~2 > φn = sin , En> = . (14.67) L 2L 8mL2 The transition amplitude for finding the expanded system in the ground state is Z 2L
< < φ = φ>∗ (14.68) a1 = φ> 1 φ1 dx . 1 1 0
Since
φ< 1
= 0 for x > L we have √ Z L √ Z L πx πx 4 2 2 >∗ < a1 = φ1 φ1 dx = sin sin dx = . L 0 2L L 3π 0
(14.69)
Approximation Methods II: Time-Dependent Perturbation Theory 373 The transition probability is then P1 = a∗1 a1 = 32/(9π 2). For the second question,
< an = φ> n φ1 =
√ π/2 2 sin(n − 2)y sin(n + 2)y − . π n−2 n+2 0
(14.70)
Since sin(mπ/2) = (−1)(m+3)/2 for odd m and 0 for even m, we obtain an = 0 for even n and for odd n √ √ 2 (−1)(n+1)/2 (−1)(n+1)/2 4 2 (−1)(n+1)/2 an = − = . (14.71) π n−2 n+2 π(n2 − 4) Then Pn = a∗n an = 32/(π 2 (n2 − 4)2 ) for odd n and Pn = 0 for even n.
14.7 THE SEMICLASSICAL THEORY OF RADIATION Based on the considerations of matter in equilibrium with radiation Einstein postulated that the ratio of the probabilities for spontaneous emission, A, and stimulated emission, B, is a constant: A ~ω 3 = 2 3 , B π c
(14.72)
where ω is the frequency of the radiation. Equilibrium is possible and the laws of thermodynamics are satisfied only for the relation (14.72). In 1920, the coefficients A and B were calculated.. Consider two energy levels of an atomic system corresponding to the energies E1 and E2 , E2 > E1 . Let N1 and N2 represent the number of atoms per unit volume in levels 1 and 2 respectively. An atom in the lower energy level is excited to E2 by absorbing radiation. This process can occur only in the presence of radiation. Such a process is known as stimulated absorption or simply absorption. The rate of absorption depends on the density ρ(ω), associated with the radiation field corresponding to the frequency ω = (E2 − E1 )/~. The rate of absorption is proportional to N1 and also to ρ(ω). Thus, the number of absorption per unit volume and per unit time is written as Nab = CN1 ρ(ω) where C is the coefficient of proportionality. On the other hand, when the atom is in an excited state, it can make a transition to a lower energy state by emitting electromagnetic radiation. There are two different types of emission process: spontaneous emission and stimulated emission. In the spontaneous emission an atom in the excited state emits radiation in the absence of incident radiation. It is not stimulated by incident radiation but occurs spontaneously. The rate of spontaneous emission is proportional to the number of atoms in the excited state. Then the number of spontaneous emissions per unit volume per unit time is given by AN2 . In the stimulated emission an incident radiation of appropriate frequency induces an atom in an excited state to emit radiation. For example, a photon
374 Quantum Mechanics I: The Fundamentals of appropriate frequency can cause emission of a photon with the same energy. This is responsible for the working of a laser. The rate of transition to the lower energy level is proportional to N2 and ρ(ω). Thus, the number of stimulated emissions per unit volume and per unit time is BN2 ρ(ω). In recognition of Einstein’s insight the quantities A, B and C are called Einstein coefficients. The coefficients A and B received important attention in the field of lasers. At thermal equilibrium, the number of upward transitions must be equal to the number of downward transitions and so A, B and C are related as CN1 ρ(ω) = AN2 + BN2 ρ(ω)
(14.73)
From this equation we write ρ(ω) = B
A CN1 BN2
−1
.
(14.74)
In statistical mechanics for an ensemble of systems in thermal equilibrium at temperature T , the average number of systems occupying a state of energy E is proportional to the Boltzmann factor e−E/(kB T ) where kB is the Boltzmann constant. We have N1 ∝ e−E1 /(kB T ) and N2 ∝ e−E2 /(kB T ) and N1 = e(E2 −E1 )/(kB T ) = e~ω/(kB T ) . N2
(14.75)
Then Eq. (14.74) becomes ρ(ω) =
A . C ~ω/(kB T ) B e −1 B
(14.76)
This distribution must agree with the Planck distribution ρ(ω) =
~ω 3
π 2 c3 e~ω/(kB T ) − 1
Comparison of Eqs. (14.76) and (14.77) gives B=C ,
A ~ω 3 = 2 3 . B π c
.
(14.77)
(14.78)
Knowing A, B and C we are able to calculate spontaneous emission, stimulated emission and spontaneous absorption. But the Eqs. (14.78) are insufficient to calculate these constants because we have only two equations for three unknowns. The relation B = C indicates that stimulated emission is as strong as absorption. That is, In the presence of incident radiation the probabilities of pumping of electrons into upper states and stimulating the already excited electron to undergo the downward transition are same. Thus, population inversion cannot occur in a two level system. The second equation in (14.78) gives the relation between stimulated and spontaneous emissions.
Approximation Methods II: Time-Dependent Perturbation Theory 375
14.8 CALCULATION OF EINSTEIN COEFFICIENTS In order to calculate the Einstein coefficients we consider an atom in a oscillating electric field E(t) = eˆE0 cos ωt. The field is switched-on at t = 0 and switched-off at t = T . eˆ represents the unit vector along the direction of the applied field. The interaction energy of the electron with the applied field is given by H (1) (r, t) = qE ·r = qE0 (ˆ e ·r) cos ωt where q represents the magnitude of the electronic charge. Let H (0) represents the Hamiltonian of the atomic system and φn denote the eigenfunctions of H (0) belonging to the energy En = ~ωn . Writing X ψ(r, t) = an (t)e−iωn t φn (r) (14.79) we obtain
i~a˙ f =
X
(1)
an (t)Hf n (t)eiωf n t ,
(14.80a)
n
where (1)
Hf n (t) =
Z
φ∗f H (1) φn dr = qE0 eˆ ·
We define Df n = q
R
1 iωt e + e−iωt 2
Z
φ∗f rφn dr . (14.80b)
φ∗f rφn dr and Df n = eˆ · Df n . Then (1)
Hf n (t) =
1 E0 eiωt + e−iωt Df n . 2
(14.81)
R The quantity q φ∗f rφn dr = ρf n is called the dipole matrix element. Equation (14.80a) is rewritten as i~a˙ f =
h i 1 X E0 Df n an (t) ei(ωf n +ω)t + ei(ωf n −ω)t . 2 n
(14.82)
We assume that at t = 0 the state of the atom is φi and hence ai (0) = 1, an (0) = 0 for n 6= i. Then we have i~a˙ f =
h i 1 E0 Df i ei(ωf i +ω)t + ei(ωf i −ω)t . 2
(14.83)
Solving this equation we get
T iE0 Df i ei(ωf i +ω)t ei(ωf i −ω)t af (t) = − + . 2~ (ωf i + ω) (ωf i − ω) 0
(14.84)
The transition may be emission or absorption. In the absorption process Ei < Ef and ωf i is positive. The radiation field may contain any range of frequencies. However, since ωf i − ω < ω < ωf i + ω, the term ωf i − ω would vanish or become very small. ωf i − ω ≈ 0 means (Ef − Ei )/~ ≈ ω or Ef =
376 Quantum Mechanics I: The Fundamentals Ei + ~ω. Then the second term in (14.84) is very large compared to the first term. That is, the second term is the resonance term. We neglect the first term, the nonresonant term. The neglect of it is called rotating wave approximation, often noticed in magnetic resonance and laser theory. In the emission process Ei > Ef and ωf i is negative. In this case the first term in Eq. (14.84) is dominant. Let us consider the emission of radiation and assume that at t = 0 the atom is in state 2. Also, assume that ω is close to ω21 = (E2 − E1 )/~. The transition amplitude is a2 (t) = −
iE0 D21 ei(ω21 −ω)t . 2~ (ω21 − ω)
(14.85)
The transition probability is now calculated as Γ21 = |af |2 =
2
π |D21 | ρ(ω21 )T . 3E0 ~2
(14.86)
The transition rate is then given by W21 =
dΓ21 π |D21 |2 ρ(ω21 ) ≈ . dT 3E0 ~2
(14.87)
If there are N2 atoms per unit volume in state 2 then the number of stimulated emission per unit volume is W 21 = N2 W21 =
N2 π |D21 |2 ρ . 3E0 ~2
(14.88)
and is equal to BN2 ρ. Hence, B= Then A is given by
Z 2 2 4π 2 q 2 π |D21 | ∗ . φ = rφ dr 2 1 2 2 3E0 ~ 3~ 4πE0
A=
Z 2 ω2 ~ω 2 B φ∗1 rφ2 dr . = 2 3 3 π C 3πC E0 ~
(14.89)
(14.90)
The corresponding expression for stimulated absorption is obtained by starting with the first term on the right-side of Eq. (14.84) and continuing in a similar way. The final expression is identical to (14.89) except for an interchange of indices 1 and 2.
14.9 CONCLUDING REMARKS In the previous chapter we were concerned with the theoretical treatment of time-independent weak perturbation. Because the perturbation is timeindependent, the quantum state of the system is time-independent. As the
Approximation Methods II: Time-Dependent Perturbation Theory 377 perturbation is weak its effect is only to shift the energy eigenvalues by a fraction. Therefore, we developed a theory to compute corrections to energy eigenvalues and eigenfunctions. In the case of time-dependent perturbations the system undergoes transition from one state to another during the action of the perturbation. Further, once the perturbation is switched-off the system settles to one of the stationary states. Therefore, we are interested in knowing the probability for the transition from an initial state to another state. One can numerically solve the equation of the transition coefficient and compute transition probability when it is difficult to compute it theoretically. This is described in the last chapter. In the case of adiabatic perturbation where the potential changes slowly with time we obtained a formula for the transition probability for a transition between the initial and final states. When a timeindependent perturbation is applied to a system suddenly at an instant of time, say, t0 the Hamiltonian of the perturbed system is time-independent. For t > t0 the perturbed system will be in one of the stationary states of it. E of nth state of the perturbed and unperturbed systems are different. Consequently, we are interested in finding the transition probability for the system to make a transition from an initial state i (state of the unperturbed system for t < t0 ) to jth state of the perturbed system for t > t0 .
14.10 BIBLIOGRAPHY [1] P.A.M. Dirac, Proc. Roy. Soc. London A 114:243, 1927. [2] E. Fermi, Nuclear Physics. Univ. of Chicago Press, Chicago, 1950. [3] M. Born and V.A. Fock, Zeitschrift für Physik a Hadrons and Nuclei 51:165, 1928; http://en.wikipedia.org/wiki/Adiabatic−approximation.
14.11 EXERCISES 14.1 Evaluate the transition amplitude up to the second-order for the con( 0, t T . 14.12 A plane rotator is subjected to a constant electric field F ′ during 0 < t < T and then the field is switched-off. √ The normalized eigenfunctions of the unperturbed system are Φn = (1/ 2π)eimφ . Find the probability for the transition from the state l to f .
Approximation Methods II: Time-Dependent Perturbation Theory 379 14.13 A plane rotator is subjected to an electric field −F µ cos φ during t ∈ [0, T ]. The system is in the lth state at t √ = 0. The normalized eigenfunctions and energy eigenvalues are ψ = (1/ 2π) eimφ , Em = ~2 m2 /(8π 2 I). Find the probability for the transition from the state l to f . Obtain the selection rule for the transition. 14.14 A linear harmonic oscillator is subjected to the time-varying electric field described by H (1) = −eF xe−iΩt during t = 0 to T . If the system is initially in its ground state calculate the probability for finding the system in the first excited state for time t > T . Find the probability if (i) H (1) = −eF x sin Ωt and (ii) H (1) = −eF x cos Ωt. 14.15 Assume that an adiabatic perturbation of the form H (1) = W (x)eαt is turned on slowly from t = −∞. Obtain the expression for second-order transition amplitude. Also write the time-independent wave function up to second-order correction. 14.16 Assume that a harmonic oscillator in the ground state is subjected to the perturbation Vp (x, t) = −αxe−µt from t = 0. Calculate the transition probability to a state φn in the limit t → ∞. 14.17 Suppose that H0 , H = H1 , Hi ,
the Hamiltonian of a system changes as shown below: for t < 0 for t > t0 for 0 < t < t0 .
H0 , H1 and Hi have complete set of energy eigenfunctions. Show that the error in the sudden approximation is proportional to t0 for small t0 .
14.18 A harmonic oscillator is in its ground state at t = −∞. Its potential is 2 perturbed by the term −αx2 e−µt between t = −∞ and ∞. Find the first-order transition probability to a state En . 14.19 A one-dimensional harmonic oscillator consisting of a mass m suspended from a spring with spring constant k is initially in its ground state. At t = 0 the upper end of the spring is suddenly raised to a distance d. This disturbance exists during the interval 0 < t < T and at t = T the spring is suddenly brought to its original position. Obtain the transition amplitude for the system to be found in its first excited state for t > T . 14.20 A harmonic oscillator Hamiltonian is perturbed by the term mω 2 (b2 + 2b)x2 /2 suddenly at t = 0. For t < 0 the system is in the ground state. Obtain the energy eigenfunctions and eigenvalues of the perturbed system. Compute the probability of finding the system for t > 0 in the ground state of the perturbed system. 14.21 A one-dimensional harmonic oscillator has its spring constant k suddenly reduced by a factor of 1/2. The oscillator is initially in its ground state.
380 Quantum Mechanics I: The Fundamentals Find the probability for the oscillator to remain in the ground state after the perturbation. 14.22 A one-dimensional harmonic oscillator has its equilibrium point shifted suddenly from x = 0 to x = a. The oscillator is initially in its ground state. Find the probability for the oscillator to remain in the ground state after the perturbation. 14.23 At t = 0 a linear harmonic oscillator potential is suddenly subjected to the time-independent electric field so that the change in the Hamiltonian is H (1) = −eEx. At t = 0 the system is in its ground state. Find the energy eigenvalues and eigenfunctions of the perturbed system by a suitable change of variable. Then calculate the probability of finding the system in the ground state of the new Hamiltonian. 14.24 For t < 0 a hydrogen-like atom with atomic number Z is in its ground state. At t = 0 its potential is suddenly perturbed by the additional term Vp (r) = −1/r. Determine the eigenstates of the perturbed system. Applying the sudden approximation theory compute the probability for finding the electron in the ground state of the perturbed system and 14.25 An electron which was initially in the hydrogen atom ground state (ψ100 , Z = 1) has undergone transition to the He atom (Z = 2) as a result of beta decay of the nucleus (Z = 1 → Z = 2). Assume that the change in the Hamiltonian is fast compared to the characteristic atomic time so that sudden approximation theory is applied. Find the probability for the transition to the ground state of He.
CHAPTER
15
Approximation Methods III: WKB and Asymptotic Methods
15.1 INTRODUCTION The WKB method is due to Gregor Wentzel, Hendrik Anthony Kramers and Leon Nicolas Brillouin [1-3]. The method is applicable for slowly varying potentials, that is, the potentials that change not more than a few wavelengths. This corresponds to states of the systems with large kinetic energy and hence high quantum number. In other words ~ is small. Therefore, the approximation is known as semiclassical approximation. The method is useful for estimating eigenfunctions and tunnelling probabilities for smooth potentials or for potentials with only a few discontinuities. The WKB approximation is a physically intuitive method and makes many connections with intuition about waves and classical mechanics as well as with results from the early days of quantum theory as it leads to the Bohr quantization condition. The results obtained by the WKB technique are also obtained employing two other different methods, namely, the semiclassical path integral and the instanton procedure [4,5]. In the WKB method, solutions on either side of a turning point are matched with a third solution valid near the turning point. This difficulty is avoided in an asymptotic method of Keller [6]. We discuss the WKB and the asymptotic method in this chapter.
15.2 PRINCIPLE OF WKB METHOD The method essentially consists of three steps: 1. The eigenfunction is written as a series expansion in the powers of ~. The approximation breaks down at turning points (E − V (x) = 0). 381
382 Quantum Mechanics I: The Fundamentals 2. A solution valid near turning points is constructed. 3. The third step is the application of boundary conditions.
15.2.1
Asymptotic Solution
We assume the solution of the Schrödinger equation ~2 ψ ′′ + 2m(E − V (x))ψ = 0
(15.1)
ψ(x) = A(x) eiS(x)/~ ,
(15.2)
as where A and S are real functions. Substitution of Eq. (15.2) in Eq. (15.1) gives ~2 A′′ + 2i~A′ S ′ + i~AS ′′ − AS ′2 + 2m(E − V )A = 0 . (15.3) Equating the real and imaginary parts of the above equation separately to zero we obtain ~2 A′′ − AS ′2 + 2m(E − V )A = 0 , ′
′
′′
2
(15.4a)
′ ′
2A S + AS = (A S ) = 0 . (15.4b) √ The solution of (15.4b) is A = C/ S ′ where C is an arbitrary constant. Since Eq. (15.4a) is difficult solve exactly, assume a solution in series as A = A0 + ~2 A1 + · · · ,
S = S0 + ~ 2 S1 + · · · .
(15.5)
Using Eq. (15.5) in Eq. (15.4a) and equating the coefficient of ~0 to zero we get S0′ = ± [2m(E − V )]1/2 . (15.6) p ′ A0 = C/ S0 . Next, setting the coefficient of ~2 to zero in Eq. (15.4a) gives A′′0 − 2S0′ S1′ A0 = 0 .
(15.7)
Defining ~p(x) = [2m(ER − V )]1/2 Eq. (15.6) is written as S0′ = ±p(x). The x solution of it is S0 = ± p(x′ ) dx′ . Also C
S1′
, A0 = p p(x)
′ −1/2 ′′ (S0 ) p = . 2 S0′
(15.8)
In the lowest-order approximation we consider only A0 and S0 . Then C1 C2 ψ(x) = A0 (x) eiS0 /~ = √ eiy1 + √ e−iy1 , p p where
1 y1 = ~
Z
0
x
p(x′ ) dx′ .
(15.9)
(15.10)
Approximation Methods III: WKB and Asymptotic Methods 383 p Since p(x) = 2m(E − V ) , y1 is real if V < E and imaginary if V > E. So, the solution of Eq. (15.9) is oscillating for the region V < E while for V > E it is an exponential form. The above approximation is valid only if the series is convergent. That is, ~2 S1 < S0 or ~2 S1 /S0 < 1. If ~2 S1 /S0 is small then ~2 S1′ /S0′ will also be small. Therefore, the condition for good approximation is ~2 S1′ /S0′ ≪ 1. Using S0′ = ~/p = h/λ the above condition becomes ′2 λ − 2λλ′′ ≪ 32π 2 , (15.11)
where λ′ = (dλ/dx)/λ is the fractional change in A over one wavelength. This condition is also written as λ|dp/dx|/p ≪ 1 where λ = ~/p is the de Broglie wavelength of the particle. Thus, the condition for the applicability of the WKB approximation is that the change of momentum over a de Broglie wavelength of the particle be small. The WKB approximation is valid in the region of x where the condition (15.11) is satisfied. For the region nearer to x0 where E − V (x0 ) = 0 the p(x0 ) is zero. Thus, the wavelength λ = h2π/p(x0 ) is infinity and hence λ′ and λ′′ also become infinity. The point is that the condition (15.11) is not satisfied near the turning point x0 and the solution (15.9) is not valid near x0 . They are said to be valid asymptotically provided they are used several wavelengths from the nearest turning points if the wavelength is slowly varying. At one side of the turning points (15.9) become oscillatory and on the other side, it would be exponential for a potential well. The exponential eigenfunction on one side must be connected to the oscillatory eigenfunction on the other side using connecting formulas.
15.2.2
Solution near a Turning Point
The solutions of the Schrödinger Eq. (15.1) are regular and analytic. To obtain a solution like (15.9), the Eq. (15.1) is modified slightly so that an exact solution can be constructed. Assume that x = 0 is the turning point. Writing the Taylor series expansion of E − V or p2 about x0 = 0 as p2 (x) = ρxn 1 + αx + βx2 + · · · , n > 0 (15.12)
and considering the first term alone the wave equation becomes ψ ′′ + ρxn ψ = 0 .
(15.13)
Its solution is p p ψ(x) = A y/p Jm (y) + B y/p J−m (y) , (15.14) Rx where m = 1/(n + 2), y = 0 p(x′ ) dx′ and Jm (y) is the Bessel function of order m.
384 Quantum Mechanics I: The Fundamentals
15.2.3
Connection at the Turning Point
The turning point x = 0 separates V (x) into two region. Region-I is with x < 0 and E > V (x). Region-II is with x > 0 where E < V (x). Suppose a particle of energy E moves in the potential V (x). Let it approach x = 0 from x1 < 0. Classically at x = 0, E is V (x = 0) and the kinetic energy is zero. Hence, the velocity is zero and particle begins to turn back at x = 0. But in quantum mechanics because of tunnelling there is a possibility of a particle to be found in the region-II. For this reason, the region-I is called classical region and the region-II is called nonclassical region. The equations relating the WKB eigenfunctions in these two regions are called connection formulas. In the regions-I and II we write Z 1 0 y1 = p(x′ ) dx′ , (real) (15.15a) ~ x Z x 1 p(x′ ) dx′ , (imaginary) (15.15b) y2 = ~ 0 respectively so that y1 and y2 increase as x moves away from the turning point. If we assume the potential near x > 0 is linear then the solutions are p p ψ1 (x) = A1 y1 /p J1/3 (y1 ) + B1 y1 /p J−1/3 (y1 ), x < 0 (15.16a) p p ψ2 (x) = A2 y2 /p I1/3 (y2 ) + B2 y2 /p I−1/3 (y2 ), x > 0 (15.16b)
where I is the Bessel function of imaginary argument. To match ψ1 and ψ2 at x = 0 we write the solution near x = 0. For x ≈ 0, p2 is ρx. Then y1 ≈ y2 ≈ (2/3)ρ1/2 |x|3/2 . This condition together with ψ1 ≈ ψ2 at x = 0 leads to A2 = −A1 , B1 = B2 .
15.2.4
Asymptotic Connection Formulas
The behaviours of ψ1 and ψ2 far from the turning points, that is at ±∞, is obtained by using the asymptotic forms of Bessel function. These are given by p α ψ1osc (x → −∞) ∼ √ sin(y1 + π/4) , α = 2/π (15.17a) p α ψ2osc (x → −∞) ∼ √ cos(y1 + π/4) , (15.17b) p α p e−y2 , ψ1exp (x → ∞) ∼ (15.17c) 2 |p| α ψ2exp (x → ∞) ∼ p ey2 . (15.17d) |p| Here ψiexp and ψiosc are approximations to the same eigenfunction. ψiexp is the continuation into the nonclassical region of ψiosc . Thus, the connection between
Approximation Methods III: WKB and Asymptotic Methods 385
III
V(x)
I II
x-
x
x+
A one-dimensional potential with two turning points at x− and x+ . The regions I and III are classically forbidden regions and the region II is the classically allowed region. FIGURE 15.1
the approximate forms of the eigenfunctions in the classical and nonclassical regions are given by Classical(osc) α √ sin(y1 + π/4) p α √ cos(y1 + π/4) p
Nonclassical(exp) α p e−y2 2 |p| α p ey 2 . |p|
15.3 APPLICATIONS OF WKB METHOD
In the following we discuss a few applications of WKB method.
15.3.1
Energy Levels of a One-Dimensional Potential Well
For any energy level E of a one-dimensional potential well shown in Fig. 15.1, there are two turning points x− and x+ such that E−V (x− ) = E−V (x+ ) = 0. Region-II (x− < x < x+ ) is classically allowed where the eigenfunction is oscillatory. Regions-I and III are nonclassical with exponentially decreasing eigenfunctions. The regions-I, II and III where the WKB approximation is valid are defined by x ≤ x− , x− ≤ x < x+ and x ≥ x+ respectively. The eigenfunctions in the regions-I and III are ψI (x)
≈
ψIII (x)
≈
R x− A1 p e−(1/~) x |p(x)| dx , 2 |p| R x− A p3 e−(1/~) x+ |p(x)| dx , 2 |p|
(15.18a) (15.18b)
386 Quantum Mechanics I: The Fundamentals p where p(x) = 2m(E − V (x)) . Using the connection formula the eigenfunction in the region-II is obtained as " Z # A1 π 1 x ψII (x) ≈ √ sin p(x) dx + , (15.19a) p ~ x− 4 Z x+ A3 1 π ≈ √ sin . (15.19b) p(x) dx + p ~ x 4 Equation (15.19a) is the solution to the right of x− and (15.19b) gives the solution to the left of x+ . These two solutions must join smoothly in the interior region-II. So " Z # Z x+ π 1 x π A3 1 A1 p(x) dx + p(x) dx + = √ sin . (15.20) √ sin p ~ x− 4 p ~ x 4 We rewrite this equation as " Z Z x+ # 1 x+ π π 1 A1 sin p(x) dx + − p(x) dx + ~ x− 2 ~ x 4 Z x+ π 1 p(x) dx + = A3 sin . ~ x 4
(15.21)
Comparison of Eq. (15.21) with sin(mπ − θ) = (−1)m−1 sin θ, m = 1, 2, · · · gives A3 /A1 = (−1)m and Z π 1 x+ p(x) dx + = mπ = (n + 1)π , n = 0, 1, · · · ~ x− 2 or
Z 1 1 x+ p(x) dx . n+ π= 2 ~ x−
φ=
(15.22)
From (15.22) we write Z
2
x+
x−
1 p(x) dx = 2 n + ~π . 2
(15.23)
Now, 2
Z
x+
x−
p(x) dx =
Z
x+
x−
p(x) dx −
Z
x−
x+
p(x) dx =
I
p(x) dx ,
(15.24)
where φ represents integration over a complete period of the particle. Substitution of Eq. (15.24) in Eq. (15.23) gives I 1 p(x) dx = n + h, (15.25) 2
Approximation Methods III: WKB and Asymptotic Methods 387 p where p(x) = 2m(E − V (x)) . Integration of the above equation gives E in terms of n and the Eq. (15.25) corresponds to Bohr–Sommerfeld quantum rule. Using Eq. (15.22) we obtain the quantized energy levels of certain systems.
Solved Problem 1: Determine the energy levels of one-dimensional harmonic oscillator. For the harmonic oscillator potential V (x) = kx2 /2 the p turning points are 2m(E − V (x)) . We obtained by equating p(x) = 0 in the equation p(x) = p p get the turning points as x− = − 2E/k and x+ = 2E/k . Then φ is obtained as Z 1 x+ p(x) dx φ = ~ x− s Z 1 2 1 x+ 2m E − kx dx = ~ x− 2 √ Z x+ r 2mE kx2 1− = dx ~ 2E x− r Z 2E m 1 p 1 − y 2 dy = ~ k −1 Z 2E π/2 = cos2 z dz ~ω −π/2 Eπ = . (15.26) ~ω Equating φ = Eπ/(~ω) to (n + 1/2)π (refer Eq. (15.22)) we get E = (n + 1/2) ~ω.
Solved Problem 2: Applying the WKB method obtain the energy eigenvalues of hydrogen atom. For a spherically symmetrical potential V (r) the radial equation for R(r) when ψ(r, θ, φ) = is
1 R(r) Ylm (θ, φ) r
(15.27)
2m ~2 l(l + 1) R + 2 E − V (r) − R=0 (15.28) ~ 2mr2 which has the general form as Eq. (15.1) but the boundary conditions are not at r = ±∞ but at r = 0, ∞. However, the change of variables r = es , R(r) = es/2 F (s) change the boundary conditions into r = −∞ to ∞. Then Eq. (15.28) becomes " 2 # 2m ~2 1 ′′ 2s F + 2 (E − V )e − l+ F =0. (15.29) ~ 2m 2 ′′
388 Quantum Mechanics I: The Fundamentals The bound state condition is F = (n + 1/2)π where 1 F = ~
Z
s+ s−
"
s
2m(E − V (e ))e
2s
2 #1/2 1 ds . −~ l+ 2 2
(15.30)
Substituting s = ln r the above equation is rewritten as 1 F = ~
Z
b
a
"
~2 l + 2m(E − V (r)) − r2
1 2 2
#1/2
dr ,
(15.31)
where a and b are the turning points. We write F
= =
2 #1/2 Z " ~2 l + 21 1 b e2 2m E + − dr ~ a r r2 p Z 2m|E| b 1 1/2 [−(r − a)(r − b)] dr ~ r a
(15.32)
2 2 2 where and ab = ~ (l + 1/2) /(2m|E|). The value of the integral a + b = e√/|E| is π a + b − 2 ab /2, 0 < a < b. Then
√ π m e2 1 F = p π. − l+ 2 ~ 2|E|
(15.33)
Equating this expression to (n + 1/2)π we obtain E=−
me4 . 2~2 (n + l + 1)2
(15.34)
This is the exact expression for the hydrogen energy eigenvalues provided the principle quantum number is now n + l + 1. We note that here n gives the number of nodes in the WKB (radial) eigenfunction while the principle quantum number counts both radial and spherical harmonic nodes and we start from 1 instead of 0.
15.3.2
Penetration Probability of a Barrier
The Schrödinger equation is solved to find the penetration of a square potential barrier. For a barrier of complicated shape, the Schrödinger equation cannot usually be solved exactly and the WKB approximation is often suitable for the problem. The WKB method is applied to obtain a transmission coefficient for penetration through a barrier. Consider a slowly varying barrier, for example, as shown in Fig. 15.2 and assume that the wave is incident from left and E < V . The region-III in Fig. 15.2 has transmitted wave alone: Therefore, the solution of the Schrödinger equation is
V(x)
Approximation Methods III: WKB and Asymptotic Methods 389
I
II
III
x FIGURE 15.2
A slowly varying potential barrier. A A ψIII = √ eiy3 = √ (cos y3 + i sin y3 ) , p2 p2
where y3 = (1/~) ψII
p p2 (x) dx and p2 (x) = 2m(E − V ) . Similarly, Rb A 1 −(1/~) R b p1 dx x = √ e − ie(1/~) x p1 dx p1 2 A 1 −(1/~) R b p1 dx (1/~) R x p1 dx a a e e = √ p1 2 i Rb Rx −ie(1/~) a p1 dx e−(1/~) a p1 dx ,
(15.35)
Rx b
p where p1 (x) = 2m(V − E) . In the region-I Z a Rb 1 A 1 ψI = √ − e−(1/~) a p1 dx sin p2 dx p2 2 ~ x Z a R 2i (1/~) b p1 dx 1 a −√ e cos p2 dx p2 ~ x = ψI inc + ψI ref ,
(15.36)
(15.37)
where ψI inc
=
ψI ref
=
iA (i/~) R a p2 dx (1/~) R b p1 dx 1 −(1/~) R b p1 dx x a a −√ e e − e (15.38) p2 4 Ra Rb Rb iA 1 − √ e−(i/~) x p2 dx e(1/~) a p1 dx + e−(1/~) a p1 dx . p2 4 (15.39)
The transmission coefficient is given by T = transmitted flux/incident flux. The transmitted flux from Eq. (15.35a) is transmitted flux =
|A|2 . p2
(15.40a)
390 Quantum Mechanics I: The Fundamentals From Eq. (15.38) the incident flux is ! ! 2 Z Z |A|2 1 1 b 1 b incident flux = p1 dx − exp − p1 dx . exp p2 ~ a 4 ~ a
(15.40b)
Then
"
T = exp
1 ~
Z
!
b
p1 dx
a
1 1 − exp − 4 ~
Z
b
!#−2
p1 dx a
.
(15.41)
Rb The WKB approximation is applicable if (1/~) a p1 dx ≫ 1. Then we neglect the second exponential term in Eq. (15.41). The resulting quantity is called barrier penetration factor and is ! ! Z Z 2 b 2 bp τ = exp − 2m(E − V (x)) dx . (15.42) p1 dx = exp − ~ a ~ a
In the next two subsections we apply the WKB method to α-decay and cold emission of electrons from metals.
15.3.3
Theory of α-Decay
The stability of a nucleus is explained by the short-range attractive force existing between the nucleons inside the nucleus. Inside the nuccleus two neutrons and protons combine and form the α-particle. Since α-particle is a highly stable configuration, it spends an appreciable fraction of time inside the nucleus before being emitted. How they escape from the forces binding the nucleus together was explained by Gamow and others as an example of quantum mechanical penetration of a potential barrier. The existence of such a barrier can be seen if we look at it from either side. An α-particle incident from outside will be repelled by the Coulomb field of the nucleus and on the other hand, it is attracted by a strong nuclear force inside the nucleus. The potential outside the nucleus is s due to the Coulomb force between the daughter nuclei with atomic number Z plus the centrifugal force. The form of the potential inside the nucleus cannot be given in detail. The energy of the α-particle E, is always less than the barrier height. So the α-particle emission takes place only due to quantum mechanical tunnelling effect. The inner turning point is assumed as the radius of the daughter nucleus and the outer turning point r0 is obtained from E=
2Ze2 l(l + 1)~2 + . r0 2µr02
The radial part of the Schrödinger equation with R = u(r)/r is d2 u 2µ 2Ze2 l(l + 1) + =− 2 −E u. dr2 ~ r 2µr2
(15.43)
(15.44)
Approximation Methods III: WKB and Asymptotic Methods 391 From (15.42) the barrier penetration factor is given by ( 1/2 ) Z 2 r0 l(l + 1)~2 2Ze2 T = exp − + −E dr . 2µ ~ R r 2µr2
(15.45)
For l = 0, from (15.43), we get r0 = 2Ze2 /E. Then the integral in (15.45) is Z
2Ze2 /E
R
2µ
2Ze2 −E r
1/2
Z √ dr = 2mE
2Ze2 /E
R
2Ze2 −1 Er
1/2
dr .
(15.46) Making the substitution r = 2Ze2 cos2 θ/E we get r 1/2 Z 2Ze2 /E i p 2Ze2 2µ h −1 2µ −E dr = 2Ze2 cos s − s 1 − s2 , r E R (15.47) p where s = RE/(2Ze2) . Hence ( ) r i p 4Ze2 2µ h −1 T = exp − . (15.48) cos s − s 1 − s2 ~ E If the speed of the α-particle inside the nucleus is v then the rate of hitting the barrier is v/R. Then the probability of escape per second (which is the inverse of mean-life time τ ) is given by 1 v ≈ T . τ R
(15.49)
Since, v ∼ 109 cm/sec and R ∼ 10−12 cm, we have v/R ∼ 1021 sec−1 . Further, for low energy α-particle, R ≪ 2Ze2 /E. In the limit RE/(2Ze2 ) → 0, we get cos−1 s = π/2. Therefore, we get from (15.48) 2Ze2 p T ≈ exp − π 2µ/E . (15.50) ~ Substituting (15.50) in (15.49), we write the decay constant as 1 v 2Ze2 p λ= = exp − π 2µ/E . τ R ~
(15.51)
Taking logarithm on both sides, we get the Geiger and Nuttal empirical law Z ln λ = A − B √ . E
(15.52)
√ Experimentally, it has been found that a plot of ln λ against 1/ E gives fairly a straight-line. The effect of introducing the term l(l +1)~2 /(2mr2 ) is not very great because even for l = 5 the value of τ changes by a factor not more than 10.
392 Quantum Mechanics I: The Fundamentals
V V0
V=V0-eEz W
EF z z1=0 z2=W/eE
FIGURE 15.3
Potential for electron in a metal with the external electric
field. 15.3.4
Cold Emission of Electrons from Metals
When a strong electric field is applied metals emit electrons even at room temperature. This phenomenon is called cold emission. In a metal, the electrons are filled up to the Fermi level as shown in Fig. 15.3. This is the required energy to remove an electron from the Fermi level. If an external field E is applied then the potential will vary as a function of the distance from the surface of the metal as shown in Fig. 15.3. The two classical turning points are z1 = 0 and z2 = W/(eE). We have V (z) = V0 − eEz and E − V (z) = V0 − eEz − EF = W − eEz . The transmission probability for cold emission is given by Z 2 x2 p T = exp − 2m(V (z) − E) dz ~ x " √ 1 # 4 2mW 3 = exp − . 3~eE
(15.53)
(15.54)
The electron penetrates the barrier due to the applied electric field.
15.4 WKB QUANTIZATION WITH PERTURBATION Earlier we have seen the usefulness of a perturbation theory in obtaining correction to energy eigenvalues and eigenfunctions for perturbed systems whose unperturbed part is exactly solvable. In this section we examine how the quantized energies, En , arising from the WKB quantization condition change when the exactly solvable potential is modified slightly. Starting from the WKB phase integral we obtain a formula for first-order correction to energy. We apply the method to some simple potentials and compare the result with perturbation theory.
Approximation Methods III: WKB and Asymptotic Methods 393
15.4.1
Formula for First-Order Shift in Energy
Let us obtain a formula for first-order shift in energy [7]. We write the WKB quantization condition as √ Z 2m x+ p En − V (x) dx = (n + cl + cr ) π , (15.55) ~ x−
where cl = cr = 1/4 for linear functions, cl = cr = 1/2 for infinite walls and other choices for radial and Coulomb-type problems. Let us write V (x) → V (x) + λV ′ (x), where V ′ (x) is the perturbation part of the potential and the change in the quantized energies up to first-order correction as En → (0) (1) En + λEn . Now, the integral in Eq. (15.55) takes the form Z x+ (λ) r (0)
(1)
En − V (x) + λ En − V ′ (x) dx .
In (λ) =
x− (λ)
(15.56)
(1)
Since En − V ′ (x) is assumed to be small we write the Taylor series expansion of the right-side of Eq. (15.56). We obtain dIn In (λ) = In (0) + λ + ··· , (15.57a) dλ λ=0 where
In (0) =
Z
x+ (0)
x− (0)
dIn dλ λ=0
=
1 2
Z
q (0) En − V (x) dx ,
x+ (0)
x− (0)
(1) En − V ′ (x) q dx . (0) En − V (x)
(15.57b)
(15.57c)
Even though we have expanded about λ = 0, for a given value of n, In must be independent of λ since Eq. (15.55) is satisfied always. Therefore, any change in V (x) and in the x± must be accompanied by a corresponding change in En so as to make In (λ) = In (0). This implies that dIn /dλ|λ=0 = 0. Then the substitution r . (0)
v(x) =
2 En − V (x)
m
(15.58)
in Eq. (15.57c) gives
En(1) Since have
R x+ (0) x− (0)
Z
x+ (0) x− (0)
1 dx = v(x)
Z
x+ (0)
x− (0)
V ′ (x) dx . v(x)
(15.59)
1/v(x) dx = τ /2 where τ is the period of the classical motion we En(1) =
2 τ
Z
x+ (0) x− (0)
V ′ (x) dx . v(x)
(15.60)
394 Quantum Mechanics I: The Fundamentals (0)
Further, PCM (x) = 2/(τ v(x)). Therefore, En(1)
=
Z
x+ (0)
x− (0)
(0)
PCM (x)V ′ (x) dx = hV ′ (x)i .
(15.61)
Solved Problem 3: Find the first-order correction to the energy eigenvalues of the linear harmonic oscillator perturbed by λbx2 . For the unperturbed harmonic oscillator with V (x) = kx2 /2 = mω 2 x2 /2 the unperturbed energy eigenvalues obtained by solving the Schrödinger equation (0) are En = (n + 1/2)~ω. The same result is obtained by WKB method. Now, we consider the effect of the additional quadratic potential V ′ (x) r = bx2 . The 1 2bλ perturbed potential is V (x) = mω 2 x2 + λbx2 . By defining ω ¯ =ω 1+ 2 mω 2 we write V (x) = m¯ ω 2 x2 /2 so En =
1/2 1 2bλ λb (0) n+ ~ 1+ = E + · · · 1 + n 2 mω 2 mω 2
(15.62)
which is an exact result. For the unperturbed p harmonic oscillator PCM (x) is given by (refer Eq. (4.36)) PCM (x) = 1/(π A2n − x2 ). Then En(1)
λb = π
Z
An
−An
x2 p dx . A2n − x2
(15.63)
Introducing the change of variable x = An sin θ the above integral is easily (1) evaluated. We obtain En = bλA2n /2. The turning points of the unperturbed system are (obtained by equating p = 0 and H = (n + 1/2)~ω in the unper1/2 turbed Hamiltonian) are An = ± [2~(n + 1/2)/(mω)] . Therefore, En(1) =
λb~ mω
(0) 1 λbEn n+ = 2 mω 2
(15.64)
which is same as the first-order in λ in Eq. (15.62). For the perturbation of the form V ′ (x) = −eEx it is easy to check that the WKB first-order correction to En is 0. The same result is obtained earlier using time-independent perturbation theory.
15.5 AN ASYMPTOTIC METHOD The WKB method has an intrinsic defect making it inconvenient to use in many situations. In this method the different expressions for the solution must be used on each side of a turning point and further they have to be matched
Approximation Methods III: WKB and Asymptotic Methods 395 to a third expression which is valid near the turning point. Consequently, in a problem with two turning points, five different expressions for the solution must be used and matched to one another. This defect can be overcome by using one spatially uniform asymptotic representation of the solution instead of the WKB five-part nonuniform representation. In this section we present the treatment of the asymptotic method [6] for the harmonic oscillator potential which can be generalized to a general potential. Let us consider the Schrödinger equation in the form ψxx + (E ′ − V ′ (x))ψ = 0 ,
(15.65)
where E ′ = 2mE/~2 and V ′ (x) = 2mV (x)/~2 and V (x) is the potential of the system. For simplicity we set ~ = 1, m = 1 and ω = 1. For the harmonic oscillator V ′ (x) = x2 /4 and the Schrödinger equation is x2 ψ=0. (15.66) ψxx + E ′ − 4 The above equation admits bound state solution only for E ′ = (n + 1/2). In this case 2 ψ(x) = e−x /4 Hn (x) . (15.67) For a general potential we write the solution of Eq. (15.65) as 2
ψ(x) = φ−1/2 e−φx /4 Hn (φn (x)) . x
(15.68)
We substitute Eq. (15.68) in Eq. (15.65) and use the result that Eq. (15.67) obeys Eq. (15.66) with E ′ = n + 1/2. Then Eq. (15.68) is found to satisfy Eq. (15.65) provided 1 φ2 −1/2 φ2x n + − = E ′ − V ′ + φ1/2 φ . (15.69) x x 2 4 xx √ We assume that V ′ (x) changes very slowly in the distance 2π E ′ , which is the wavelength. Then the last term in Eq. (15.69) is negligible compared to the other terms. Now, 1 φ2 φ2x n + − = E′ − V ′ . (15.70) 2 4 Suppose E ′ − V ′ = 0 at x− and x+ with xp − < x+ . Then Eq. (15.70) gives φ2 −4(n+1/2) = 0. Its roots are φ(x± ) = ±2 n + 1/2. One of these conditions can be viewed as an initial condition and the other to determine E ′ . Integration of Eq. (15.70) from x− to x gives 1/2 Z x √ 1 φ2 φx n + − dx = E ′ − V ′ dx 2 4 x− x−
Z
x
396 Quantum Mechanics I: The Fundamentals or 1 π φ 1 φ2 1 n+ + n+ − + n+ sin−1 2 2 4 2 4 2 Z x √ = E ′ − V ′ dx .
φ
p 2 n + 1/2
!
(15.71)
x−
Setting x = x+ in the above equation we get Z x+ √ 1 n+ π= E ′ − V ′ dx . 2 x−
(15.72)
The above result is same as the one given by the WKB method. But the present derivation is simpler. Further, the representation (15.68) for the wave function is valid for all x.
15.6 CONCLUDING REMARKS A novel WKB approach for calculating the life-time of quasi-stationary states in the potential wells of the form V (x) = P (x) − µQ(x) is suggested in ref.[8]. An alternative formulation of WKB approximation is proposed by Vandyck [9]. An exact quantization rule of the form Z xb Z xb K ′ (x) [φ(x)/φ′ (x)] dx , (15.73) K(x) dx = (n + 1)π + xa
xa
p where K(x) = 2m(EV (x))/~, xa and xb are two turning points determined by E = V (x) and φ(x) = (1/ψ)dψ/dx has been proposed [10]. Here n + 1 is the number of the nodes of φ(x) in the region E ≥ V (x). The second term in (15.73) is found to be independent of the number of nodes of the wave function. Thus, it is sufficient to make use Rof the ground Rstate in computing x x it. A proper quantization rule of the form xab K(x) dx − xab K0 (x) dx = nπ where n is the number of nodes of ψ(x) has been proposed [11]. Using this rule all the bound state eigenvalues can be calculated from the ground state energy only. Evaluation of WKB integral for the special cases (i) n (radial quantum number) → ∞ and l (azimuthal quantum number)-fixed and (ii) l → ∞ and n-fixed have been discussed in refs.[12,13].
15.7 BIBLIOGRAPHY [1] G. Wentzel, Z. Phys. 38:518, 1926. [2] H.A. Kramers, Z. Phys. 39:828, 1926. [3] L. Brillouin, Comptes Rendus 183:24, 1926.
Approximation Methods III: WKB and Asymptotic Methods 397 [4] C. Itzykson and J. Zuber, Quantum Field Theory. McGraw–Hill, New York, 1980. [5] B.R. Holstein, Am. J. Phys. 56:338, 1988. [6] J.B. Keller, Am. J. Phys. 54:546, 1986. [7] R.W. Robinett, Am. J. Phys. 65:320, 1997. [8] J. Zamastil, J. Cizek and L. Skala, Phys. Rev. Lett. 84: 5683, 2000. [9] M.A. Vandyck, Eur. J. Phys. 12:112, 1991. [10] Z.Q. Ma and B.W. Xu, Europhys. Lett. 69:685, 2005. [11] W.C. Qiang and S.H. Dong, Europhys. Lett. 89:10003, 2010. [12] G. Paiano and M. Pellicoro, Nuovo Cimento B 119:239, 2004. [13] G. Paiano, Europhys. Lett. 83:30013, 2008.
15.8 EXERCISES 15.1 Show that the gradient of the phase S in ψ(X) = A(X)eiS(X)/~ characterizes the probability flux. 15.2 What is the equation we get by applying the semiclassical approximations ~ → 0 and ~ ∇2 S ≪ (∇S)2 to the time-dependent Schrödinger equation?
15.3 Using the Hamilton–Jacobi equation 1 ∂S |∇S(x, t)|2 + V (x) + =0 2m ∂t show that the semiclassical approximation ∇2 S ≪ |∇S|2 for a stationary state is valid if the de Broglie wavelength is very small compared with the characteristic distance over which the potential varies appreciably (Hint: For stationary states, the Hamilton’s principal function S and Hamilton’s characteristic function W are related by S = W (x) − Et). Rx 15.4 Given φ = (1/~) x12 p(x) dx find ∂φ/∂E. 15.5 Using the WKB quantization rule find the eigenvalues of the anharmonic oscillator with the Hamiltonian H = −(~2 /2m)d2 /dx2 + λx4 .
15.6 Use the WKB approximation to determine the bound state energies of V 0 |x| ≤ a |x|, a 2 V (x) = 1 h V0 = , |x| ≥ a. m a
398 Quantum Mechanics I: The Fundamentals 15.7 Use the WKB-based perturbation method to find the first-order correction to the energy of a harmonic oscillator perturbed by a small potential V ′ = λx4 . Compare it with the result of the perturbation theory. 15.8 The nonrelativistic expansion of the energy of a free particle in onedimension can be written in the form p 1 1 2 E = p2x c2 + m2 c4 ≈ mc2 + p − p4 . 2m x 8m3 c2 x Treating −p4x /(8m3 c2 ) as a perturbation to the harmonic oscillator, estimate the correction to the energy using WKB perturbation method. 15.9 A neutron falling in a gravitational field and bouncing-off a horizontal mirror exhibits quantized energy levels. The potential of the problem is V (z) = mgz for z > 0 and ∞ for z < 0. Applying the WKB method obtain the energy levels of the system. 15.10 For a particle in a box potential, V (x) = 0 for 0 < x < L and ∞ otherwise, determine the energy eigenvalues by WKB method.
CHAPTER
16
Approximation Methods IV: Variational Approach
16.1 INTRODUCTION In chapters 13 and 14 we focused on perturbation theories for time-independent and time-dependent perturbations. For time-independent perturbations we obtained first-order and second-order corrections to energy eigenvalues and eigenfunctions. For time-dependent perturbations we obtained the probability for a system to make a transition from an ith initial state to f th final state. We have also discussed how to study the effect of adiabatic and sudden perturbations. In the present chapter we consider the variational method. The variational method can be used to construct an approximate ground state eigenfunction. In some cases the method can be used for finding approximately the energy of first few discrete states of a bound system. The perturbation theory uses a set of unperturbed states and the perturbation Hamiltonian H (1) to construct the eigenfunctions and eigenvalues through highly formalized approximations. In contrast, the variational method is based on a guess for the state—a trial state which is used to determine the corresponding energy. Any well behaved trial state introduced will yield an appropriate energy, although some trial states will be better than other, but whatever the guess, we can always make systematic improvements. We can construct a reasonably accurate eigenfunction often by a physical insight and ingenuity.
16.2 CALCULATION OF GROUND STATE ENERGY Let us briefly describe the basic idea of the variational method of calculating ground state energy.
399
400 Quantum Mechanics I: The Fundamentals
16.2.1
Basic Principle
Let φn are the eigenfunctions of the eigenvalue equation n = 0, 1, . . . . (16.1) P An arbitrary R ∞ wave function is written as ψ = n an φn . Now, consider the integral −∞ ψ ∗ Hψ dτ . We obtain Z ∞ Z ∞X X ψ ∗ Hψ dτ = a∗m φ∗m H an φn dτ Hφn = En φn ,
−∞
−∞ m
=
X n
≥
X n
n
2
|an | En
|an |2 E0 ,
(16.2)
where E0 is the ground state energy. For normalized ψ the requirement is P 2 n |an | = 1. Then Z ∞ E0 ≤ ψ ∗ Hψ dτ . (16.3) −∞
Equation (16.3) is known as variational theorem. For normalized ψ Z ∞ .Z ∞ ∗ ψ ∗ ψ dτ . ψ Hψ dτ E0 ≤ −∞
(16.4)
−∞
Equation (16.3) states that whatever be the R ∞form of wave function ψ the upper bound for the ground state energy E0 is −∞ ψ ∗ Hψ dτ . Thus, we can choose a suitable trial function ψ = φ(ci ) which depends on the parameters ci and determine the R ∞ best upper bound for ground state energy by minimizing the expression −∞ φ∗ Hφdτ with respect to the parameters ci . R∞ The conditions for E(ci ) = −∞ φ∗ (ci )Hφ(ci )dτ to be a mininum are ∂E ∂2E >0. = 0 and ∂ci ∂c2i
(16.5)
The above conditions are used to determine the minimum value of E.
Solved Problem 1: Determine the error in the ground state energy given by the variational method. Suppose φ is a trial ground state eigenfunction while φ0 is the true ground state eigenfunction. The difference between them is φ − φ0 = λ∆φ where λ is a small constant and ∆φ is orthogonal to φ0 . Now, Eφ is given by Z ∞ (φ0 + λ∆φ)∗ H(φ0 + λ∆φ) dτ −∞ Z ∞ Eφ = . (16.6) ∗ (φ0 + λ∆φ) (φ0 + λ∆φ) dτ −∞
Approximation Methods IV: Variational Approach 401 We rewrite it as
Eφ
=
Z
Z
∞
−∞ ∞
−∞
2
E0 + λ =
Z
1 + λ2 = =
(16.7)
.
(φ∗0 φ0 + λφ∗0 ∆φ + λ(∆φ)∗ φ0 + λ2 ∆φ∗ ∆φ) dτ
That is,
Eφ
φ∗0 H(φ0 + λ∆φ) + λ∆φ∗ H(φ0 + λ∆φ) dτ
∞
Z−∞ ∞
E0 + λ2
∆φ∗ H∆φ dτ ∆φ∗ ∆φ dτ
−∞ Z ∞
∗
∆φ H∆φ dτ
−∞ 2
E0 + O(λ ) ,
2
1 − λ E0
Z
∞
∗
∆φ ∆φ dτ
−∞
(16.8)
where we have neglected terms containing λ4 and higher powers of λ. Thus, the error of the variational energy is of the order of λ2 .
16.2.2
Applications
In the following we apply the variational method to some interesting quantum systems.
16.2.3
Harmonic Oscillator
We calculate the ground state energy of the harmonic oscillator with H=−
~2 d2 1 + mω 2 x2 . 2m dx2 2
(16.9)
2
We choose the trial eigenfunction as φ = Ae−αx . First, we normalize the φ: Z ∞ Z ∞ Z ∞ 2 2 1= φ∗ φ dx = A2 e−2αx dx = 2A2 e−2αx dx . (16.10) −∞
−∞
R∞
0
2
p dx = π/(4β) we obtain A = (2α/π)1/4 .
Using the integral relation 0 e−βx Next, Z ∞ E = φ∗ Hφ dx −∞ Z ∞ 2 ~2 d2 1 2 −αx2 2 2 = A e − + mω x e−αx dx 2 2m dx 2 −∞ Z 2 2~2 A2 α ∞ = (1 − 2x2 )e−2αx dx m 0 Z ∞ 2 2 2 +mω A x2 e−2αx dx . 0
(16.11)
402 Quantum Mechanics I: The Fundamentals Using the result Z
∞
2n −βx2
x e
0
p (2n − 1)!! π/β dx = , 2n+1 β n
(16.12)
where double factorial of n is 1 if n = 0 or 1 and n × (n − 2)!! if n ≥ 2. Using the identity (2n − 1)!! = (2n)!/(2n n!) we get ~2 α mω 2 + . 2m 8α2
(16.13)
∂E ~2 mω 2 = − ∂α 2m 8α2
(16.14)
E= The minimization condition 0=
gives α = mω/(2~). Further, we note that ∂ 2 E/∂α2 = mω 2 /(4α3 ) > 0. Substitution of α = mω/(2~) in Eq. (16.13) gives E = ~ω/2 which is the well known expression for zero-point energy of harmonic oscillator. The obtained value of ground state energy is exact because the trial eigenfunction chosen is the correct functional form with the adjustable parameter α.
16.2.4
Hydrogen Atom
The Hamiltonian of the hydrogen atom is H = −(~2 /2m)∇2 −e2 /r. We choose the trial eigenfunction as φ = Ae−ar . The normalization of φ gives Z ∞ Z π Z 2π Z ∞ ∗ 2 2 1= φ φ r sin θ dr dθ dφ = 4πA r2 e−2ar dr . (16.15) 0
0
0
0
p Using the relation 0 x e R dx = n!/β we get A = a3 /π . Next, we calculate E, φ∗ Hφ dτ . Since φ is a function of r only, it is sufficient to consider the radial component of H which is given by ~2 1 ∂ e2 2 ∂ Hrad = − r − = H1 + H2 . (16.16) 2m r2 ∂r ∂r r R We write E = φ∗ (H1 + H2 )φ dτ = E1 + E2 . We obtain Z ∞ Z π Z 2π E1 = A2 e−ar H1 e−ar r2 sin θ dr dθ dφ 0 0 0 Z 2π~2 A2 a2 ∞ −ar = e 2re−ar − ar2 e−ar dr m 0 2 2 π~ A (16.17a) = 2ma R∞
n −βx
n+1
and E2 =
Z
∗
2 2
φ H2 φ dτ = −4πA e
Z
0
∞
re−2ar dr = −
πA2 e2 . a2
(16.17b)
Approximation Methods IV: Variational Approach 403 Then E = E1 +E2 = ~2 a2 /(2m)−ae2. The minimization condition ∂E/∂a = 0 yields a = e2 m/~2 . Substituting this value for a in the above expression for E we get E = −me4 /(2~2 ) which is the exact ground state energy.
16.2.5
Helium Atom
The helium atom consists of a nucleus of charge 2e and two electrons each of charge −e at distances r1 and r2 from the nucleus. The Hamiltonian of the system is 2e2 2e2 ~2 e2 H=− ∇21 + ∇22 − − + . (16.18) 2m r1 r2 r12 If e2 /r12 is neglected then Eq. (16.18) simply represents Hamiltonian of two hydrogen atoms whose ground state energy is E1 = −Z 2 β where Z = 1, β = me4 /2~2 . Therefore, we choose the trial eigenfunction as φ = Ae−Z The normalization condition Z 1 = φ∗ φ dτ Z Z Z = A2 r1 ,r2
θ1 ,θ2
′
(r1 +r2 )/a0
e−2Z φ1 ,φ2
′
(16.19)
.
(r1 +r2 )/a0 2 r1
sin θ1 dr1 dθ1 dφ1
×r22 sin θ2 dr2 dθ2 dφ2
(16.20)
gives A = Z ′3 /(πa30 ). We proceed to calculate E. The eigenfunction φ given by Eq. (16.19) is the ground state energy of hydrogen-like atom with charge Z ′ e. The corresponding ground state energy is E1 = −Z ′2 β. So, we have ~2 2 Z ′2 e2 ∇1 − φ = E1 φ = −Z ′2 βφ . (16.21) − 2m r1 That is, −
Z ′ e2 ~2 2 ∇1 = −Z ′2 β + . 2m r1
(16.22a)
−
~2 2 Z ′ e2 ∇2 = −Z ′2 β + . 2m r2
(16.22b)
Similarly,
Then
~2 − ∇21 + ∇22 = −2Z ′2 β + Z ′ e2 2m
1 1 + r1 r2
.
(16.23)
404 Quantum Mechanics I: The Fundamentals The Hamiltonian, Eq. (16.18), is rewritten as H = H1 + H2 + H3 , where e2 1 1 H1 = −2Z ′2 β, H2 = e2 (Z ′ − 2) , H3 = + . (16.24) r1 r2 r12 The energy E is given by
We obtain Z Z
E=
Z
φ∗ H1 φ dτ
=
−2Z ′2 β ,
φ∗ H3 φ dτ
=
5 ′ Zβ. 4
Then
φ∗ (H1 + H2 + H3 )φ dτ .
Z
(16.25)
φ∗ H2 φ dτ = 4Z ′ (Z ′ − 2)β , (16.26a) (16.26b)
27 E = 2β Z ′2 − Z ′ . 8
(16.27)
The minimization condition ∂E/∂Z ′ = 0 gives Z ′ = 27/16. With this value of Z ′ the ground state energy is obtained as 2 2 me4 27 27 =− 2 , (16.28) E = −2β 16 ~ 16 whereas the exact value of energy is Eexact = −22 me4 /~2 .
16.3 TRIAL EIGENFUNCTIONS FOR EXCITED STATES Now, we consider the excited states.
16.3.1
Trial Eigenfunctions Orthogonal to All Lower Levels
The variational method can also be applied to compute the excited state energy values by choosing the trial eigenfunctions to be orthogonal to all the lower state eigenfunctions. For example, if a trial eigenfunction Z ψt = ψ − φ0 φ∗ ψ dτ (16.29) is chosen then
Z
Z
φ∗0 ψt dτ =
Z
φ∗0 φ0 dτ = 1 we get
φ∗0 ψ dτ − R
Z
φ∗0 φ0 dτ
Z
φ∗0 ψ dτ .
(16.30)
φ∗0 ψt dτ = 0. ψt is orthogonal to the ground P state eigenfunctions. Hence, the expansion ψt = n=1 an φn will not contain Since
Approximation Methods IV: Variational Approach 405 a0 . We obtain Z
hEt i =
ψt∗ Hψt
dτ =
∞ X
n=1
Therefore, hEt i
∞ X
≥
n=1
|an |2 En .
(16.31)
|an |2 E1 (16.32)
≥ E1 .
So, the trial wave function ψt gives an upper limit for the first excited state. Extending the above, we can prove that if ψt is orthogonal to φi , i = 0, 1, · · · , n then ψt will give an upper limit of the (n + 1)th energy eigenvalue.
16.3.2
Finite Basis Set Expansion
A simple procedure for obtaining approximate excited state energies is to expand the trial wave function ψ using states taken from a known complete set {φn }. The trial wave function is chosen as linear combination of known functions as r X ψ= ai φi , (16.33) i=1
where ai ’s are variational parameters. The φi may not be normalized or mutually orthogonal. The energy is Z ∞ .Z ∞ ∗ E= ψ Hψ dτ ψ ∗ ψ dτ . (16.34) −∞
−∞
Substitution of (16.33) in (16.34) yields Pr Pr E
=
Pi=1 Pj=1 r r i=1
where Hij =
Z
∞ −∞
φ∗i Hφj dτ ,
j=1
a∗i aj Hij a∗i aj Sij Z
Sij =
∞
,
(16.35a)
φ∗i φj dτ .
(16.35b)
a∗i aj Hij = 0 .
(16.36)
−∞
From the above equation we write r X r X i=1 j=1
Ea∗i aj Sij −
r X r X i=1 j=1
Differentiating the above equation with respect to a∗i and substituting ∂E/∂a∗i = 0 we obtain r X j=1
(Hij − ESij ) aj = 0,
i = 1, 2, · · · , r .
(16.37)
406 Quantum Mechanics I: The Fundamentals Differentiation of Eq. (16.36) with respect to ai and using ∂E/∂ai = 0 gives the complex conjugate of Eq. (16.37). Equations (16.37) form a set of r homogeneous linear equations for the r unknowns ai . For nontrivial solution, the determinant of the coefficients matrix of Eq. (16.37) must vanish. That is, H11 − ES11 H12 − ES12 · · · H1r − ES1r H21 − ES21 H22 − ES22 · · · H2r − ES2r (16.38) . =0. .. .. . Hr1 − ESr1 Hr2 − ESr2 · · · Hrr − ESrr
This will give an algebraic equation of degree r in E. If the r roots of E are arranged in increasing order E0 , E1 , · · · , Er−1 then E0 will be the ground state energy, E1 will be the first excited state energy and so on. Substituting these roots in Eq. (16.37) and solving it we are able determine aj corresponding to each roots. These aj when substituted in Eq. (16.33) will give the corresponding energy eigenfunction. If the basis chosen as an orthonormal basis then Sij = δij . Then Eq. (16.38) will become H11 − E H12 · · · H1r H21 H22 − E · · · H2r (16.39) =0. . .. .. . Hr1 Hr2 · · · Hrr − E
The accuracy and convergence of the energy values are determined by including one more function φr+1 to the basis and solving the resulting (r + 1) equations. If |En (r) − En (r + 1)| is very small then the convergence is very good. The basis set is expanded till good convergence is realized. Although any well-behaved set of basis states may be used to perform the diagonalization of H, one set is noteworthy. If H is partitioned as H0 + H1 , where H1 need not be weak, then the eigenfunctions of H0 can be chosen as the basis if they are known.
Solved Problem 2: If the first n − 1 eigenfunctions of a particular Hamiltonian are known, write a formal expression for a variational method trial function that could be used to get the upper limit on the nth energy level. Let u1 , u2 , · · · , un−1 are known eigenfunctions. If φ is any arbitrary trial function then using the Schmidt orthogonalization process, a trial eigenfunction that could be used to get the upper limit on the nth energy level can be expressed as n−1 X Z ∞ ψ =φ− ui u∗i φ dτ . (16.40) i=1
−∞
Approximation Methods IV: Variational Approach 407
r12
2
r
r 2B
1A
1
2A
r
r
1B
R
A FIGURE 16.1
B
Coordinate system of hydrogen molecule.
Now, Z
u∗j ψ dτ
= = =
Z
Z
Z
= 0.
u∗j φ dτ
−
u∗j φ dτ − u∗j φ dτ −
n−1 XZ
u∗j ui
i=1
n−1 X i=1
Z
δji
Z
dτ
Z
u∗i φ dτ ,
j ≤n−1
u∗i φ dτ
u∗j φ dτ (16.41)
Since ψ is orthogonal to the first n − 1 eigenfunctions, it can be used as a trial function for the n energy level.
16.4 APPLICATION TO HYDROGEN MOLECULE As an example for the application of the linear combination method we shall estimate the ground state energy of the hydrogen molecule. Consider the coordinate system of the hydrogen molecule as shown in Fig. 16.1. We assume that the nuclei of the two hydrogen atoms A and B are fixed in space at a distance R apart. The position of the first electron is given by the position vector r1A with respect to A and that of the second electron is given by r2B with respect to the second nuclei B. The Hamiltonian for the hydrogen molecule is given by H = H (0) + H (1) ,
(16.42a)
408 Quantum Mechanics I: The Fundamentals where H (0)
=
H (1)
=
2 2 e2 ~ e2 ~ + − ∇22 − , − ∇21 − 2µ r1A 2µ r2B e2 e2 e2 e2 + − − . R r12 r1B r2A
(16.42b) (16.42c)
If the interaction between the two atoms (H (1) ) is ignored then the ground state eigenfunction can be either φ1 = uA (1)uB (2) or φ2 = uA (2)uB (1) when the second electron is considered as belonging to the nuclei A and the first electron is belonging to the nuclei B where uA (1) =
1 πa30
1/2
e−r1A /a0
(16.43)
H (0) φ2 = 2E0 φ2 ,
(16.44)
and so on. We have H (0) φ1 = 2E0 φ1 ,
where E0 = −e2 /(2a0 ). Construct a trial eigenfunction as a linear combination of φ1 and φ2 as ψ = a1 φ1 + a2 φ2 . (16.45) From Eq. (16.45) the determinantal equation is H11 − ES11 H12 − ES12 H21 − ES21 H22 − ES22 where
Hij
=
Sij
=
Z Z Z Z
φ∗i Hφj dτ1 dτ2 , φ∗i φj dτ1 dτ2 ,
=0, i, j = 1, 2
i, j = 1, 2.
(16.46)
(16.47a) (16.47b)
It can be proved that H11 = H22 ,
H12 = H21 ,
S11 = S22 = 1,
Using (16.48) in (16.46) we get H11 − E H12 − ES12 The roots of E are
ES =
H11 + H12 , 1 + S12
S12 = S21 .
H12 − ES12 =0. H11 − E EA =
H11 − H12 . 1 − S12
(16.48)
(16.49)
(16.50)
Approximation Methods IV: Variational Approach 409 From the homogeneous equation H11 − E H12 − ES12 a1 =0 H12 − ES12 H22 − E a2
(16.51)
we get the normalized eigenvectors as 1 1 1 1 √ and √ 2 2 1 −1
(16.52)
1 ψS = √ (φ1 + φ2 ) , 2
(16.53)
for ES and EA respectively. Then
1 ψA = √ (φ1 − φ2 ) . 2
The exchange of the electrons changes φ1 → φ2 and φ2 → φ1 and hence ψS → ψS and ψA → −ψA . ψS is a symmetric eigenfunction and ψA is antisymmetric. We have not considered the spin part of the eigenfunction as H does not have spin operators. Since the total eigenfunction of fermions must be antisymmetric, ψS must be multiplied by the antisymmetric spin function (singlet state) and ψA must be multiplied by the symmetric spin functions (triplet-state). We obtain Z Z ∗ S12 = uA (1)uB (1) dτ1 u∗B (2)uA (2) dτ2 = S 2 , (16.54) R
u∗A (r)uB (r) dτ . Next, Z Z H11 = φ∗1 H (0) + H (1) φ1 dτ1 dτ2 = 2E0 + k ,
where S =
(16.55a)
where k=
Z Z
2
2
e |uA (1)| |uB (2)|
2
−
1 r2A
−
1 r1B
1 1 + + R r12
dτ1 dτ2 , (16.55b)
and H12 = 2E0 S 2 + J ,
(16.56a)
where J
=
Z Z
e2 (uA (1)uB (1)) (uA (2)uB (2)) 1 1 1 1 × − − + + dτ1 dτ2 . r2A r1B R r12
(16.56b)
410 Quantum Mechanics I: The Fundamentals
EA Req
2E0
R ES
Energy of the hydrogen molecule as a function of the interatomic distance. FIGURE 16.2
Using (16.54a), (16.55a) and (16.56a) we get ES = 2E0 +
K +J , 1 + S2
EA = 2E0 +
K −J . 1 − S2
(16.57)
The integrals K, J and S can be evaluated. The variational of ES and EA with respect to the interaction distance R is shown in Fig. 16.2. From this figure we note that the ground state is a symmetric and bound state. The antisymmetric state has energy greater than that of two separated hydrogen atoms (2E0 ) and hence it is not bound. The theoretical equilibrium bond length Req is found to be 0.77A◦ . The difference between EA and ES arises since J and S are not zero. K is the energy of the classical Coulomb electrostatic interaction between the electronic charge distribution. For this reason K is called the Coulombic integral . J and S are due to the fact that ψA and ψS have the terms φ1 and φ2 wherein the position of the electrons are interchanged. For this reason J is called the exchange integral and S is called the overlap integral . Though the exchange interaction arises because of the basic Coulomb interaction, it is purely a quantum mechanical phenomenon which cannot be explained classically. This exchange interaction is mainly responsible for the binding of atoms into molecules. It is also responsible for the ordering of spins in ferromagnetic materials.
16.5 HYDROGEN MOLECULE ION Consider the coordinate system of a H+ 2 ion shown in Fig. 16.3. The Hamiltonian of H+ ion neglecting the nucleon motion and spin-orbit interaction is 2 given as H =−
~2 2 e 2 e2 e2 ∇ − − + , 2m rA rB R
(16.58a)
rB = |r − RB | ,
(16.58b)
where rA = |r − RA | ,
R = |RA − RB | ,
Approximation Methods IV: Variational Approach 411
e
A
r
rB
A FIGURE 16.3
R
B
The coordinate system of a hydrogen ion.
with r being the position of the electron, RA and RB being the positions of the two protons. If R is sufficiently large then depending upon whether the electron is attached to the nucleon A or B, it will be described by uA or uB where uA and uB are the hydrogen ground state eigenfunctions centred at A and B respectively. This suggests that we use a linear combination of atomic orbitals (LCAO) to construct a molecular orbitals (MO) as (16.59a)
ψ = CA uA + CB uB , where uA =
1 πa30
1/2
e−rA /a0 ,
uB =
1 πa30
1/2
e−rB /a0 .
To find the energy levels, we solve the secular determinant HAA − ESAA HAB − ESAB =0, HBA − ESBA HBB − ESBB
(16.59b)
(16.60)
where
SAA = SBB
= =
SAB = SBA
=
HAA = HBB
=
HAB = HBA
=
Z ∞ Z π Z 2π 1 2 e−2rA /a0 rA sin θA drA dθA dφA πa30 0 0 0 1, (16.61a) Z 1 e−rA /a0 − e−rB /a0 dτ , (16.61b) πa30 Z 1 e−rA /a0 He−rA /a0 dτ , (16.61c) πa30 Z 1 e−rA /a0 He−rB /a0 dτ . (16.61d) πa30
The evaluation of the integrals (16.61b-d) is not simple as they are two-centre integrals, since they involve the distance between the point of integration
412 Quantum Mechanics I: The Fundamentals (x, y, z) and two different fixed points A and B. They are evaluated by a change of coordinate system from the usual Cartesian or polar coordinates to a new system in which rA and rB appear explicitly. We choose the new coordinates rA − rB rA + rB =ξ, = η and φ . (16.62) R R We find the ranges of ξ, η and φ as 1≤ξ L,
using the trial eigenfunction φ = |L|α − |x|α with α the trial parameter and compare it with the exact energy value.
16.3 Using the calculation of variation, show that the time-independent Schrödinger equation Ris the Euler–Lagrange equation that minimizes ∞ the functional I(c) = −∞ φ∗α (x, c)H(x)φα (x, c) dx with respect to variR∞ ation of φα (x) and φ∗α (x) with −∞ φ∗α (x)φα (x) dx = 1 and the Lagrange multiplier being the energy eigenvalue. 16.4 The Hamiltonian of a quartic oscillator is H = −(~2 /2m)d2 /dx2 + λx4 . 2 2 Choosing the normalized trial eigenfunction as φ0 = (α2 /π)1/4 e−α x /2 with α as theR variational parameter estimate the ground state energy. m ∞ Assume that 0 e−(ax) xn dx = Γ [(n + 1)/m] /man+1 .
414 Quantum Mechanics I: The Fundamentals 16.5 Consider the triangular potential V (x) = F x, if x > 0 and ∞, if x < 0. This is used as a model for an electron trapped on the surface of liquid helium by an electric field due to two capacitor plates bracketing the helium and vacuum above it or for the MOSFET. Applying the variational method calculate ground state energy. Use the trial eigenfunction as φ = xe−ax . 16.6 Estimate the ground state energy of the three-dimensional symmetric harmonic oscillator having the Hamiltonian H = −(~2 /2m)∇2 + mω 2 r2 /2 by taking the 1s hydrogenic normalized orbital ψ(r, β) = (2/β 3/2 ) e−r/β Y00 as the trial eigenfunction with β as the variation parameter. Compare the result with the exact ground state energy and find the relative error. 16.7 A hydrogen atom is subjected to an electric field E. Estimate the ground state energy of the atom by taking ψ = N (1 + qEz)u100 as the trial eigenfunction with q as the variational parameter. Find its polarizability by assuming E is small. 16.8 A particle of mass m is acted on by the three-dimensional potential V (r) = −V0 e−r/a where ~2 /(V0 a2 m) = 3/4. Use the trial function e−r/β to obtain a bound on the energy. 16.9 Let the total Hamiltonian of a system is of the form H 0 + H ′ with H ′ small compared to H 0 . By choosing the eigenfunction of H 0 as the trial eigenfunction, show that the first-order perturbation is always in excess of the exact value. 16.10 The nonrelativistic hydrogen has the Hamiltonian H = −(~2 /2m)∇2 − e2 /r. Estimate its ground state energy by taking the normalized trial eigenfunction 1/4 3/2 2 2 2 2 ψ(r, β) = e−r /β Y00 π β with β as the variational parameter. Compare it with the exact ground state energy and find the relative error.
CHAPTER
17
Scattering Theory
17.1 INTRODUCTION Consider a pond in which water is very still. Suppose a pole is stuck in the pond and is projecting out of the surface of water. When a small stone is dropped into the pond we notice ripples. When the ripples reach the pole, secondary waves are generated. We say that the pole scatters the incident waves. Experimentally, scattering is a process where a particle (or a system) with a certain energy is sent toward another system (target) and the final motion of the particle is studied after it interact with the system. When the kinetic energy is conserved then the scattering is said to be elastic, otherwise inelastic. Why are we interested in scattering processes? One of the best ways to explore the structure of particles and the forces between them is to allow them to scatter. This is practically true at the microscopic level where the systems cannot be seen in the literal sense and must be probed by indirect ways. A general scattering event is of the form A1 (α1 ) + A2 (α2 ) + · · · → B1 (β1 ) + B2 (β2 ) · · · , where {A1 , A2 , · · · B1 , B2 , · · · } are names of the particle and {α1 , β1 , · · · } are the variables specifying their states such as spin, momentum, etc. Many of the most important breakthroughs in physics have been made by means of scattering. There are two different ways of developing the theory of scattering processes in quantum mechanics. They give the same results but they proceed from different formulations. The first way is to consider the scattering of a particle as a transition out of an initial state of given momentum into a final state of different momentum but with the same energy. Let the time-dependent free (0) particle wave function representing the initial state be ψi and that of one of (0) the final states be ψf . Then using the time-dependent perturbation theory (0)
the probability of finding an incident particle scattered out of the state ψi
415
416 Quantum Mechanics I: The Fundamentals (0)
into ψf
is obtained from the Fermi formula (see sec. 14.2) Γ=
2π |hf |H ′ |ii|2 ρf (E) , T
(17.1)
where ρf (E) is the density of states in the neighbourhood of the energy E. The second approach is to regard the whole process as a scattering of waves by a potential and setup a differential equation. Its solution will give the scattered wave function. The first approach is more general while the second is more convenient and we will follow the second approach in this book. If the incident particle and the scattered particle have same energy but different direction of the wave vector k, then it is called elastic scattering. On the other hand, if the incident particle excites the target nuclei of atoms then the energy of the scattered particle will be different from that of the incident particle. Such a process is called inelastic scattering. In inelastic scattering, one has to find the reaction cross-section in addition to scattering cross-section to obtain the total cross-section. The general idea of scattering is to shoot projectiles at the system (target) under study and see how these projectiles bounce-off from the target. Historically, scattering has been used as an important tool for exploring the structure of matter. The search for fundamental particles is done entirely using scattering experiments. We note that Rutherford discovered the nucleus of the atom via scattering experiments. Some importance of scattering experiments are as follows: 1. Scattering of nucleons of various energies from a nucleus provides an idea of the strength and range of nuclear forces and structure of nucleus. 2. Scattering of high energy electrons reveals details about charge distribution in nucleus. 3. Scattering of neutrons from a nucleus gives magnetic properties of the nucleus. What is the aim of quantum scattering theory? Our interest in scattering process is to calculate the probability of certain final states, given the initial state and the Hamiltonian of the system. The aim of the scattering theory in quantum mechanics is to deduce details about the force or the interaction responsible for the scattering. The theory achieves this by establishing a relationship between the cross-section and the wave function of the system. The wave function is obtained by solving the Schrödinger equation. Thus, in a bound state problem the emphasis is on the energy eigenvalues whereas in the scattering problem the focus is on the wave function. In this chapter, first we review the classical definition of scattering crosssection and derive the relation between the scattering cross-sections of centre of mass and laboratory frame of reference. We establish the relationship between scattering cross-section and the wave function and introduce scattering
Scattering Theory 417
dA
(a)
(b)
dΩ θ Incident
FIGURE 17.1
x θ
φ
Target
Target
y z
Representation of a scattering event.
amplitude. Then we get a Green’s function for scattering amplitude and discuss Born’s approximation for scattering amplitude. We present partial wave analysis of scattering for spherically symmetric systems. Also, we consider the collisions between identical particles and the case of inelastic scattering.
17.2 CLASSICAL SCATTERING CROSS-SECTION Consider a scattering process. Suppose a parallel beam of particles of certain energy or momentum is sent towards a target (often a thin metal foil). The target deflects or scatters the particles in various directions. The scattered particles diverge. We choose the position of the target as the origin of the coordinate system (laboratory system). Only a fraction of the incident particles are scattered and others are transmitted and reflected. The direction of the scattered particles is described by the polar angles θ and φ as shown in Fig. 17.1. Suppose call the number of incident particles crossing unit area of crosssection per unit time as the incident flux Ni . Let ∆Ns is the number of particles scattered into a small solid angle dΩ. Evidently, ∆Ns ∝ Ni dΩ. We write ∆Ns =
dσ Ni dΩ , dΩ
σ =⇒ σ(θ, φ)
(17.2)
where dσ/dΩ is the proportionality constant and is called differential scattering cross-section. dσ/dΩ has units of area/steradian (SI units of solid angle): ∆σ = (dσ/dΩ)/dΩ is the area an incident particle must strike per target nucleus, in order to scatter into the solid angle dΩ(= 2π sin θ dθ).
418 Quantum Mechanics I: The Fundamentals From (17.2) we write Number of particles scattered through dΩ per unit time ∆Ns dσ dΩ = = . dΩ Ni Number of particles incident per unit time
(17.3)
per unit area of cross − section It gives nothing but the probability of a particle to be scattered into the angle dΩ. The total number of particles scattered per unit time is then calculated by integrating Eq. (17.2) over the range of Ω: Z Z Z dσ ∆Ns dΩ = Ni dΩ dΩ = Ni dσ dΩ . (17.4) dΩ That is, Ns = Ni σ, where σ is called total scattering cross-section. We write σ = Ns /Ni . The cross-sectional area of the nucleus is πR2 = 7.07 × 10−26A2/3 where R = 1.5 × 10−15 m is radius of the nucleus and A is the mass number. For Al27 the cross-sectional area is 0.588 × 10−28 m2 and for Au197 it is 2.4 × 10−28 m2 . The nuclear cross-sections are typically around 10−28 m2 . Crosssections are often measured in barns (b) where 1 barn = 10−28 m2 . Crosssections run from 0 to several 105 barns, but a few barns are typical.
17.3 CENTRE OF MASS AND LABORATORY COORDINATES SYSTEMS Often the scattering cross-section is experimentally measured and compared with the cross-section derived using theory. There is generally a correction which must be made before the results of the theory and experiment is compared. This correction arises from the necessity of taking into account the recoil of target particle. Let us recall the scattering process. In a laboratory experiment the target is usually a thin metal foil and a narrow beam of particles is directed to fall on it. The particles are scattered by nuclei in the target and their number is counted by suitable detectors kept at various angles. This is shown in Fig. 17.2a. The target is assumed to be stationary after interaction with the incident particle. However, in reality the target particles will not remain stationary, unless they are very massive than the incident particles. It is convenient to work out the theory of scattering in a frame of reference in which the centre of mass of the two particles is regarded as the origin with respect to which the coordinates of the incident particles are defined. Note that the experimental cross-sections are measured with respect to the laboratory coordinates. So, we transform the theoretically calculated quantities to the laboratory frame of reference for comparing the results of the theory with the experimental observation. When the target particle is in motion after interaction then the scattering
Scattering Theory 419
(a)
(b) θC
θ θL
FIGURE 17.2 Scattering angle in the laboratory system and centre of mass system.
(b)
(a) θC θL
rC
θC
V= µ v m2 0 R
θL
rL
θL
(c) vC
B θC
A E
θL
v θL vL
H
F G
Scattering process to an observer in the centre of mass system. Here µ = m1 m2 /(m1 + m2 ). FIGURE 17.3
angle is defined as the angle between the initial and the final relative vectors of the positions of the two particles as depicted in Fig. 17.2b. θL measured in the laboratory frame is different from the actual scattering angle θC . We must therefore find out the relation between θL and θC and only then we can transform the expression for the cross-section to the laboratory frame. The relationship between θL and θC are determined by examining the way in which scattering takes place in a coordinate system moving with the centre of mass of both particles. An advantage of centre of mass coordinate system is that here the total linear momentum is zero and two particles always move with equal and opposite momenta. Figure 17.3a depicts scattering process to an observer in the centre of the mass system. Here the particles execute opposite courses and the angle between the initial and final directions of the relative vector is the scattering angle. It is the scattering angle for both the particles. We define the following: • rL and vL are the position and velocity after scattering of the incident particle in the laboratory system.
420 Quantum Mechanics I: The Fundamentals • rC and vC are the position and velocity after scattering of the incident particle in the centre of the mass system. • R and V are the position and velocity of the centre of mass in the laboratory system. • v0 is the velocity of the incident particle. At any time rL = R + rC and vL = V + vC as shown in Fig. 17.3b. Since the target is at rest in the laboratory system the relative velocity of the particle in the laboratory system is v0 (before scattering). According to conservation of linear momentum we have m1 v0 = m1 V + m2 V which gives m1 v0 . m1 + m2
(17.5)
sin θL HG/AH BE = = . cos θL AG/AH AE + BH
(17.6)
V= From Fig. 17.3c we have tan θL =
In ABE triangle BE = vC sin θC and AE = vC cos θC . Therefore, tan θL =
vC sin θC sin θC = , vC cos θC + V cos θC + ρ
(17.7)
where ρ = V/vC . Also, vL cos θC = vC cos θL + V. In the centre of mass system rC =
m2 r, m1 + m2
vC =
m2 m2 v0 = V. m1 + m2 m1
(17.8)
That is, V/vC = m1 /m2 . comparison of this with ρ = V/vC gives ρ = m1 /m2 . Further, 2 vL2 = vC + V2 + 2vL · V cos θL . (17.9) Now, cos θC =
cos θL + ρ 1/2
(1 + 2ρ cos θL + ρ2 )
(17.10)
.
Then we obtain 1 + 2ρ cos θL + ρ2 d(cos θL ) = d(cos θC ) 1 + ρ cos θL
3/2
.
(17.11)
We notice that in general the values of cos θL and cos θC are different. The number of particles scattered per unit time into a given solid angle measured in terms of (i) cos θL and (ii) cos θC must be the same. Then dσθ dσθ 2π sin θL dθL = 2π sin θC dθC (17.12) dΩ L dΩ C
Scattering Theory 421 from which we get
dσθ dΩ
=
C
dσθ dΩ
L
d(cos θL ) . d(cos θC )
(17.13)
Using (17.11) in (17.13) we obtain
dσθ dΩ
=
dσθ dΩ L
C
1 + 2ρ cos θL + ρ2 1 + ρ cos θL
3/2
(17.14)
.
When ρ = 1 ( m1 = m2 ) we find θC = θL /2. Therefore, dσθ dσθ = 4 cos θL . dΩ C dΩ L
(17.15)
These formulas are nonrelativistic and therefore of no use in particle physics. In relativity, the centre of mass is not a useful concept as one would have to know the simultaneous position of two particles with different motions. We can also express the differential cross-section relations between the two frames in terms of cos θC . From (17.13) we obtain
dσθ dΩ
=
L
dσθ dΩ
C
1 + 2ρ cos θC + ρ2 1 + ρ cos θC
3/2
.
(17.16)
Though the differential cross-sections are different in both laboratory and centre of mass systems the total cross-section is same in both systems. Why?
Solved Problem 1: Show that for proton-proton scattering θL = θC /2. We have sin θC tan θL = m1 . + cos θC m2
(17.17)
For proton-proton scattering m1 = m2 = mp then tan θL
= = =
That is, θL = θC /2.
sin θC 1 + cos θC 2 sin(θC /2) cos(θC /2) 2 cos2 (θC /2) tan(θC /2) .
(17.18)
422 Quantum Mechanics I: The Fundamentals
ikz
e
ikr f (θ) e − r r
z − direction
r r
r
Incident plane wave Scattered wave FIGURE 17.4
Spherical wavefront formed by the scattered particles.
17.4 SCATTERING AMPLITUDE The total scattering cross-section σ = Ns /Ni is a classical result. Now we ask: What is the corresponding situation in quantum mechanics? In other words, what is the relation between σ and the wave function ψ? Suppose the particles involved in a scattering event are quantum mechanical particles. They must we described by ψ. The Schrödinger equation of the scattering problem is ~2 2 − ∇ ψ + V ψ = Eψ . (17.19) 2m We seek solution to (17.19) subjected to the following conditions: 1. Far before scattering ψ represents a homogeneous parallel beam of particles moving in the positive z-direction. Therefore, ψinc ∼ eikz . Generally, ψ is a function of r, θ, φ. Since particles move along z-direction, ψ is independent of φ. 2. Far after scattering ψ represents particles moving in various directions from the target. We consider elastic scattering where the kinetic energy (mv 2 /2) is conserved. This means that all particles move with the same constant v. Therefore, at any time the scattered particles form a spherical wave front. This is depicted in Fig. 17.4. The area of the surface of the sphere increases with r. Consequently, the probability of finding a particle in a fixed area dA decreases with increase in r. So, we write ψsca ∼ f (θ, φ)
eik·r , r
r→∞
(17.20)
where f (θ, φ) is called scattering amplitude. It is the amplitude of the
Scattering Theory 423 outgoing spherical wave. The general solution of (17.19) is written as ψr→∞ ∼ eikz + f
eik·r . r
(17.21)
To determine scattering cross-section σ = Ns /Ni we need to find number of particles scattered through dΩ and number of particles incident per unit time per unit area of cross-section. These are determined as follows. For the uniform monochromatic beam of particles of given energy, the incident density Jinc in the z-direction is evaluated from ψinc ∼ eikz as |Jinc | =
~k . m
(17.22)
Let the scattered current density be Jsca along the r direction. Consider a solid angle dΩ in the direction of r, subtended by an area dA. Then the number of particles per unit time scattered through dA is Jsca · dA. The number of particles passing through dA per second is ∆Ns = Jsca ·dA = r2 Jsca ·dΩ. Since the number of such scattered particles will be proportional to the incident current density Jinc , we write ∆Ns = |Jinc |dσ. From the above two expressions for ∆Ns we get r2 Jsca · dΩ = |Jinc |dσ and Jsca · dΩ = Jsca,r dσ ,
(17.23)
where Jsca,r is the r-component of Jsca . From (17.23) we get dσ Jsca,r = r2 . dΩ |Jinc |
(17.24)
The r-component of the current density Jsca is ~ ∂ ∗ ∗ ∂ Jsca,r = ψsca ψsca − ψsca ψsca . 2mi ∂r ∂r
(17.25)
Substituting for ψsca from (17.20) we get Jsca,r =
~k |f (θ, φ)|2 . mr2
(17.26)
Substituting (17.26) and (17.22) in (17.24) we obtain dσ = |f (θ, φ)|2 . dΩ The total scattering cross-section is then given by σ = is independent of φ then Z π σ = 2π |f |2 sin θ dθ . 0
(17.27) R
|f |2 sin θ dθ dφ. If f (17.28)
424 Quantum Mechanics I: The Fundamentals In classical mechanics, collisions of two particles are entirely determined by their velocities and impact parameter. In quantum mechanics, we cannot consider a path for a particle. So, we are concerned with the probability of the particle scattered through any given angle after collision.
Solved Problem 2: The wave function of a scattering process is given by ψ = eikz + Calculate total scattering cross-section.
cos θ eik·r . 5 r
The scattering amplitude is f = (cos θ)/5. Then Z π Z 2π π σ = 2π |f |2 sin θ dθ = cos2 θ sin θ dθ. 25 0 0 Carrying out the integration we get π 2π cos3 θ 2π 4π σ= − = (1 + 1) = . 25 3 75 75 0
17.5 GREEN’S FUNCTION APPROACH It is possible to find a solution for the scattering amplitude by transforming the Schrödinger equation into an integral equation employing Green’s function method. Suppose L be a linear differential operator. If Ly = f (x) then its solution can be obtained by Green’s function approach. Green’s function is the solution of Ly = δ(x − x′ ). Once the Green’s function is known then the solution of Ly = f (x) in the interval x ∈ [a, b] is given by y(x) =
Z
b
G(x, x′ )f (x′ ) dx′ .
(17.29)
a
Further, if y0 (x) is a solution of Ly = 0 then y(x) = y0 (x) +
Z
b
G(x, x′ )f (x′ ) dx′ .
(17.30)
a
This can be easily verified by substituting (17.30) in Ly = f (x). In a quantum mechanical scattering we wish to solve the Schrödinger equation ~2 2 ~2 k 2 − ∇ + V ψ = Eψ = ψ (17.31) 2m 2m which is rewritten as
2mV ∇2 + k 2 ψ = ψ. ~2
(17.32)
Scattering Theory 425 We are justified in writing E = ~2 k 2 /(2m) in (17.31) as the interaction V produces no energy shift in elastic scattering processes. Defining U (r) = 2mV (r)/~2 we rewrite the above equation as (17.33) ∇2 + k 2 ψ = U (r)ψ(r) = F (r)
which is of the form Ly = f (x) with L = ∇2 + k 2 . The Green’s function of Eq. (17.33) satisfies the equation (17.34) ∇2 + k 2 G(r, r′ ) = δ(r − r′ ) .
Let us assume that
′
G∝ Then the left-side of Eq. (17.33) is ∇2 + k 2 G =
e±ik|r−r | . |r − r′ |
(17.35)
′
∇2 + k 2 2e
e±ik|r−r | |r − r′ |
±ik|r−r′ |
′
±ik|r−r′ |
=
−k
=
−4πδ(r − r′ )e±ik|r−r | ,
+e
|r − r′ |
′
±ik|r−r | 1 2e ∇ + k |r − r′ | |r − r′ | 2
(17.36)
where we have used ∇2 (1/|r − r′ |) = −4πδ(r − r′ ). From Eq. (17.36) we write ′
∇2 + k 2
e±ik|r−r | ′ = δ(r − r′ )e±ik|r−r | . ′ −4π|r − r |
(17.37)
Comparison of Eqs. (17.34) and (17.37) gives
′
G± (r, r′ ) =
e±ik|r−r | −4π|r − r′ |
and
(17.38)
′
δ(r − r′ ) = δ(r − r′ )e±ik|r−r | . ′
(17.39)
′
Equation (17.39) is true for r = r and r 6= r . Thus, G given by Eq. (17.38) is a solution of Eq. (17.34). Moreover, the solution of (∇2 + k 2 )ψ = 0 is ψ0 (r) = eikz . Hence, ψ(r)
= eikz −
1 4π
Z
′
eik|r−r | U (r′ )ψ(r′ ) dτ ′ , |r − r′ |
(17.40)
where we have used only G+ as r → ∞, it must give Eq. (17.21) as the solution. The above equation indicates that the scattered wave is made of spherical waves arising at each point r′ of space. The amplitude of each contribution is proportional to U (r′ )ψ(r′ ), proportional jointly to the strength of the interaction and the amplitude of ψ at r′ . All these waves are combined at r and
426 Quantum Mechanics I: The Fundamentals form the total scattered wave. This is then combined to the incident wave to produce the total wave function at r. In this sense the Green’s function G+ is thought of as a propagator carrying the waves from their last point of scattering to the next. In the limit r → ∞ we write Z ′ 1 eik·r ikz ψ(r) = e − e−ik·r U (r′ )ψ(r′ ) dτ ′ . (17.41) 4π r Comparing this with the equation ψ(r) = eikz + f (θ, φ)
eik·r r
(17.42)
we get 1 f (θ, φ) = − 4π
Z
e−ik·r U (r)ψ(r) dτ .
(17.43)
The above equation is not a straight-forward formula to calculate f . Because ψ(r) which appears in the integral is still not known. However, it provides a very useful approximation called the Born approximation [1] to be discussed in the next section. We can use the process of interaction to evaluate ψ(r). If r is replaced by r′ in Eq. (17.40), we get ′
ψ(r ) = e
ikz ′
1 − 4π
Z
′
′′
eik|r −r | U (r′′ )ψ(r′′ ) dτ ′′ . |r′ − r′′ |
(17.44)
Substituting the above ψ(r′ ) in (17.40) we obtain ψ(r)
Z
′
′ eik|r−r | U (r′ ) ek·r dτ ′ ′ |r − r | 2 Z Z ik|r−r′ | ik|r′ −r′′ | −1 e ′ e − U (r ) ′ 4π |r − r′ | |r − r′′ | ′′ ′′ ′ ′′ ×U (r )ψ(r ) dτ dτ .
= eikz −
1 4π
(17.45)
The above equation gives the first iterated solution. This process is repeated indefinitely, resulting an infinite series called Neumann series. If the series converges, then it represents the solution to the Schrödinger equation. The first term in Eq. (17.45) is the incident beam. The second term represents single scattering by the interacted potential V (r′ ) in the volume element dτ ′ . The third term represents scattering of the first scattered wave at dτ ′ by interacted potential V (r′′ ) at volume dτ ′′ . The total effect of such double scattering is obtained by integrating with respect to τ ′ and τ ′′ . The Neumann series will converge rapidly if the interaction is weak.
k
Scattering Theory 427
s
θ k0 FIGURE 17.5
Vectors s, k0 and k and the scattering angle θ.
17.6 BORN APPROXIMATION We have seen that ψ can be given in an infinite series. The equation obtained by cutting-off the series at the nth term is called the nth Born approximation.
17.6.1
First Born Approximation for Scattering Amplitude
Suppose ψ is approximated by a plane wave function eikz . In doing this we assume that the deviation of ψ from eikz is infinitely small: g(r) = ψ − eikz ≪ eikz = 1 . (17.46)
Then the first Born approximation for the scattering amplitude is obtained by setting ψ = eik0 z where k0 is the magnitude of the vector k0 in the incident direction. We obtain Z Z 1 1 f (θ, φ) = − e−ik·r U (r)eik0 ·r dτ = − e−is·r U dτ , (17.47) 4π 4π where s = k − k0 and U = U (r). Thus, the scattering amplitude in the Born approximation is the Fourier transform of the potential apart from the constant factor. This is the first Born approximation. From Fig. 17.5 we find s = 2k sin(θ/2). In spherical polar coordinates s · r = sr cos θ. Then Z ∞ Z π Z 2π ′ ′ 1 e−isr cos θ U (r′ )r′2 sin θ′ dr′ dθ′ dφ′ f = − 4π 0 0 0 Z 1 ∞ = − r sin sr U (r) dr . (17.48) s 0 We note the following important features of Eq. (17.48): 1. f is a function of θ only (since s is a function of θ). 2. f is always real.
428 Quantum Mechanics I: The Fundamentals 3. ~s = ~(k − k0 ) is the momentum transfer. f depends on momentum transfer and not on incident momentum. R∞ 4. For small k, s is small, then f = − 0 r2 u(r) dr. f is independent of θ.
5. f is small for large s.
Solved Problem 3: Obtain the validity condition for the Born approximation. The Born approximation is valid only if (17.46) is satisfied. For the ψ given by Eq. (17.40) the validity condition (17.46) takes the form 1 g=− 4π
Z
′
eik|r−r | U (r′ )ψ(r′ ) dτ ′ . |r − r′ |
(17.49)
g takes the largest value near the origin. We wish to have, at the origin Z ′ 1 eik|−r | ′ ′ ′ U (r )ψ(r ) dτ (17.50) ≪1. 4π | − r′ | That is,
Z 1 eikr 2mV ≪1. ψ dτ 2 4π r ~
(17.51)
Replacing ψ by eikr cos θ , dτ by r2 sin θ dr dθ dφ and integrating with respect to θ and φ we get Z m ∞ 2ikr V e − 1 dr ≪ 1 (17.52) k~2 0
if V is spherically symmetric. For a given potential the validity condition can be obtained by evaluating the above integral.
Born approximation gives good result only when the second integral in the infinite series is negligibly small. If the energy is sufficiently high, the wave function ψ is a rapidly oscillating function within the region of interaction, and it can be expected that the integral will be small because of cancellation of contributions from different parts of the region of interaction. Hence, although the interaction may be quite strong, the scattered wave is small and the Born approximation is valid provided the energy is sufficiently high.
17.6.2
Application
Now, we apply the Born approximation to square-well potential and Coulomb potential systems.
Scattering Theory 429
17.6.2.1 Scattering from a Square-Well Potential An attractive square-well potential is represented by −V , for r ≤ a 0 V (r) = 0, for r > a .
For this potential f (θ) given by Eq. (17.48) is Z 2mV0 2mV0 a r sin sr dr = 2 3 (sin sa − as cos sa). f (θ) = ~s ~ s 0
(17.53)
(17.54)
Now, the differential scattering cross-section is dσ 4m2 V02 a6 g(sa) = |f 2 | = , dΩ h4
g(sa) =
1 (sin sa − sa cos sa)2 . (17.55) (sa)6
When sa ≪ 1 (low energy case) lim g(sa) =
sa→0
=
2 (sa)3 (sa)2 1 lim sa − + · · · − sa 1 − + ··· sa→0 (sa)6 3! 2! 1 . (17.56) 9
Then
dσ 4m2 V02 a6 = . dΩ 9~4 Next, the total scattering cross-section is Z π Z 2π 16πm2 V02 a6 dσ sin θ dθ dφ = . σ= dΩ 9~4 0 0
(17.57)
(17.58)
For the square-well potential the validity condition (17.52) becomes Z mV0 a 2ikr ≪1. e − 1 dr (17.59) ~2 k 0
Evaluation of the integral in the above equation gives mV0 2ika e − 1 − 2ika ≪ 1 . 2 2 2~ k
(17.60)
Defining 2ka = A and after simple mathematics we obtain mV0 (2 + A2 − 2 cos A − 2A sin A)1/2 ≪ 1 . 2~2 k 2 When A ≪ 1, that is, 2ka ≪ 1 (low-energy limit) we get mV0 a/(~2 k) ≪ 1.
(17.61)
(17.62)
430 Quantum Mechanics I: The Fundamentals Therefore, a and V0 should be small. In other words, the potential must be weak. Alternatively, if the potential is not weak but the denominator is very large then the above condition is satisfied and the Born approximation is still applicable. Because ~k is the momentum, we require the particle momentum should be large. Since momentum is proportional to velocity this is equivalent to saying that the velocity of the incident particle should be large. Therefore, the Born approximation is valid if either the potential is weak or the velocity or thepmomentum of the incident particle is sufficiently large. Substituting k = 2mE/~2 the condition (17.62) becomes E ≫ 2~2 /(ma2 V02 ) . Since the particle energy must be high, Born approximation is a high energy approximation.
17.6.2.2 Scattering from a Screened Coulomb Potential If the nucleus is considered as a point charge and orbital electrons are absent then the potential energy of an electron at a distance r from the nucleus is V (r) = −Ze2 /r. The effect of an orbital electron is to screen the nucleus from projectiles (electrons). In this case V (r) = −Ze2 e−r/a /r where a is the atomic radius. The scattering amplitude is Z imZe2 ∞ −r/a isr 2mZe2 a2 f =− 2 e . (17.63) e − e−isr dr = 2 2 2 ~ s ~ (a s + 1) 0
Then
dσ 4m2 Z 2 e4 a4 = f2 = 4 2 2 . dΩ ~ (a s + 1)2
(17.64)
The total scattering cross-section is Z π Z 2π dσ σ = sin θ dθ dφ dΩ 0 0 Z 8πm2 Z 2 e4 a4 π 2 sin(θ/2) cos(θ/2) = dθ 2 2 2 2 ~4 0 [1 + 4k a sin (θ/2)] =
16πm2 Z 2 e4 a4 . ~4 (1 + 4k 2 a2 )
(17.65)
For the validity condition of Born approximation see the problem 10 at the end of the chapter. If we set a → ∞, then the screened Coulomb potential V = −Ze2 e−r/a /r becomes the Coulomb potential. The differential scattering cross-section for Coulombic potential can be obtained from (17.64) in the limit a → ∞. We obtain dσ 4m2 Z 2 e4 m2 Z 2 e 4 = = 4 4 4 (17.66) 4 4 dΩ Coulomb ~ s 4~ k sin (θ/2) which is the famous Rutherford scattering formula for Coulomb scattering.
Scattering Theory 431
17.7 PARTIAL WAVE ANALYSIS When the Born approximation is not valid and the potential is spherically symmetric the partial wave analysis is much useful. In this approximation, a series solution for ψ is written. Each term in the solution is called partial wave characterized by a particular angular momentum.
17.7.1
Scattered Wave and Phase Shift
For a spherically symmetric potential ∇2 in Eq. (17.33) becomes ∂ 1 ∂ 1 1 ∂ ∂ ∂2 r2 + 2 sin θ + 2 2 ∇2 = 2 . r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 The square of the angular momentum L2 is given by ∂ 1 ∂2 1 ∂ 2 2 L = −~ sin θ + . sin θ ∂θ ∂θ sin2 θ ∂φ2
(17.67)
(17.68)
Then 1 ∂ ∇ = 2 r ∂r 2
L2 2 ∂ r − 2 2 . ∂r ~ r
Using Eq. (17.69) in (17.33) we get 1 ∂ L2 2 ∂ 2 r − 2 2 +k −U ψ =0. r2 ∂r ∂r ~ r
(17.69)
(17.70)
We assume the solution of the form ψ(X) = Rl (r)Yl (θ, φ) .
(17.71)
If the polar axis (z-axis) is chosen parallel to the direction of incident beam then ψ is independent of φ. Since Eq. (17.70) is a linear equation we write the general solution as ∞ X ψ= Rl (r)Yl (θ) . (17.72) l=0
The solution corresponding to a particular l is called l-th partial wave. Rl and Yl are determined by substituting (17.71) in (17.70) and then solving the corresponding differential equations for them. We get 1 ∂ 1 2 2 ∂ r Rl + k 2 − U r2 − L Yl = 0 . (17.73) Rl ∂r ∂r Yl ~2 The first and second terms are functions of r only while the last term is
432 Quantum Mechanics I: The Fundamentals function of θ only. The above equation is thus valid only if the first two terms and the last term are equal to a same constant. Therefore, we write 1 2 L Yl = l(l + 1) . ~2 Yl
(17.74)
This is a Legendre differential equation with argument cos θ. Its solution is Pl (cos θ). The radial equation is 2 dRl l(l + 1) d2 Rl 2 + + k −U − Rl = 0 . (17.75) dr2 r dr r2 Substituting Rl = Xl /r we get for scattering d2 Xl,s l(l + 1) 2 + k − U − Xl,s = 0 . dr2 r2
(17.76)
In the limit r → ∞ the above equation becomes d2 Xl,s + k 2 Xl,s = 0 . dr2
(17.77)
Xl,s (r) = Cl sin(kr + ∆l ) ,
(17.78)
Its solution is where Cl and ∆l are constants. For the incident wave the equation for Xl,in is d2 Xl,in l(l + 1) 2 + k − Xl,in = 0 . dr2 r2 It is a spherical Bessel equation with the solution lπ Xl,in (r) = sin kr − . 2
(17.79)
(17.80)
The difference between (17.78) and (17.80) is in the phase of the waves. The effect of the potential is thus to change the phase of the wave from −lπ/2 to ∆l . The difference in the phase is called phase shift δl : lπ lπ δl = kr + ∆l − kr − = ∆l + . (17.81) 2 2 δl is called the phase shift in the lth partial wave and it depends on l and k. It measures the amount by which Xl,s (r) is displaced in r relative to the Xl,in (r). If V = 0 then (17.78) and (17.80) are identical. The entire effect of scattering force is therefore described in terms of the set of phase shifts. Figure 17.6 shows an example of shift in the phase of the scattered wave by repulsive and attractive potentials [2].
Scattering Theory 433
Attractive poten t ial Repulsive poten t ial Zer o poten t ial
FIGURE 17.6
An example of phase shifts induced by scattering poten-
tials. In elastic scattering the number of particles per second entering the scattering region must be equal to the number of particles per second leaving that region. Therefore, the moduli of the amplitudes of the incoming and outgoing waves must be the same. As a result, the effect of the potential is only to produce a phase difference between the two waves. Scattering of quantum particles and the underlying phase shifts are important in many areas of atomic physics, including Bose–Einstein condensation, degenerate Fermi gases, frequency shifts in atomic clocks and magnetically tuned Feshbach resonances [3].
17.7.2
Scattering Amplitude in Terms of Phase Shift
From (17.72) and (17.78) we write ψ=
X Cl r
Pl sin(kr + ∆l ) .
(17.82)
Replacing Cl by Cl′ /k (so that everywhere the dependence is kr) and dropping the prime we get X Cl lπ ψ= Pl sin kr − + δl . (17.83) kr 2
We wish to compare this wave function with the wave function given by Eq. (17.21). For this purpose we express eikz in series form as eikz = eikr cos θ =
∞ X
il (2l + 1)Pl (cos θ)jl (kr) .
(17.84)
l=0
Since we are interested in the wave function far away from the scattering centre, we consider only the asymptotic value of r. Because limr→∞ jl (kr) ≈
434 Quantum Mechanics I: The Fundamentals sin(kr − lπ/2)/(kr) we get eikz =
∞ X l=0
lπ sin kr − 2 il (2l + 1)Pl (cos θ) . kr
(17.85)
Now, comparing (17.83) and (17.21) we obtain X Cl X lπ Pl lπ Pl sin kr − + δl = il (2l + 1) sin kr − kr 2 kr 2 +f
eikr . r
(17.86)
We note that the plane wave sin(kr − lπ/2)/(kr) is a superposition of the outgoing spherical waves ei(kr−lπ/2) /(kr) and the incoming spherical waves e−i(kr−lπ/2) /(kr) converging on origin. Using the relation sin x = (eix − e−ix )/(2i) Eq. (17.86) is rewritten as hX i X eikr Cl Pl e−ilπ/2 eiδl − il (2l + 1)Pl e−ilπ/2 − i2kf h X i X +e−ikr − Cl Pl eilπ/2 e−iδl + il (2l + 1)Pl eilπ/2 (17.87)
=0.
Since eikr and e−ikr are two linearly independent solutions the coefficients of e±ikr must be identically equal to zero. Equating the coefficients of e−ikr to zero in Eq. (17.87) gives Cl = il (2l + 1)eiδl . Then from the coefficients of eikr we get i X 1 hX Cl Pl i−l eiδl − (2l + 1)Pl f = 2ik 1X = (2l + 1)Pl eiδl sin δl . k
(17.88)
(17.89)
We call the fl ’s given by
fl
=
1 (2l + 1)Pl eiδl sin δl k
(17.90)
as partial scattering amplitudes. l = 0 gives f of the wave with L2 = 0. The significance of the phase shifts δl (k) can be seen in a different way by rewriting Eq. (17.83) using (17.88) as " # lπ ∞ −i(kr− lπ X 2 ) Sl (k)ei(kr− 2 ) l (2l + 1) e ψ = − i − , (17.91) 2ik r r l=0
Scattering Theory 435 where Sl (k) = e2iδl (k) and is called scattering coefficient of lth partial wave. In the square-bracket of (17.91) the first and the second terms represent incoming and outgoing spherical waves respectively. The effect of the potential appears in the outgoing spherical wave as scattering coefficient Sl (k) in each partial wave. The scattering potential can only alter the outgoing wave. We express fl 1 (2l + 1)(Sl (k) − 1). We see from Sl (k) = e2iδl (k) in terms of Sl (k) as fl = 2ik that the quantity Sl (k) can be generalized to describe inelastic scattering processes (see sec. 17.10). Conservation of probability is then expressed by requiring |Sl |2 = 1.
17.7.3
Total Scattering Cross-Section
We obtain σ
= 2π
Z
π
0
= = =
|f |2 sin θ dθ
Z π 2π X X ′ iδl −iδl′ (2l + 1)(2l + 1) sin δl sin δl′ e e Pl Pl′ sin θ dθ k2 0 l l′ 2δll′ 2π X X (2l + 1)(2l′ + 1) sin δl sin δl′ eiδl e−iδl′ ′ k2 (2l + 1) l l′ 4π X (2l + 1) sin2 δl . (17.92) k2 l
We write σl =
4π (2l + 1) sin2 δl k2
(17.93)
P so that σ = l σl . σl is zero for δl = 0 or π and is maximum when δl = ±π/2, ±3π/2. It is possible to express σ in terms of f also. From Eq. (17.89) the imaginary part of f when θ = 0 is fim (0) =
1X (2l + 1) sin2 δl . k
(17.94)
l
Then Eq. (17.92) gives σ=
4π fim (0) . k
(17.95)
This equation is known as optical theorem and is the statement of conservation of probability. It relates the imaginary part of scattering amplitude at θ = 0 to total scattering cross-section. Does the optical theorem applicable to Born approximation? Why?
436 Quantum Mechanics I: The Fundamentals
17.7.4
Relation Between δl , V (r) and Xl
To obtain an equation for δl consider the Eqs. (17.76) and (17.79). Multiplying Eq. (17.76) by Xl,in and Eq. (17.79) by Xl,s and subtracting one from another and then replacing Xl by rRl result in rRl,in
d2 d2 (rR ) − rR (rRl,in ) = r2 U Rl,in Rl,s . l,s l,s dr2 dr2
Integrating the above equation with respect to r once gives ∞ Z d d (rRl,in ) = r2 U Rl,in Rl,s dr . rRl,in (rRl,s ) − rRl,s dr dr 0 Substituting the solutions Rl,in and Rl,s we obtain Z 1 sin δl = − r2 U (r)Rl,in Rl,s dr . k
(17.96)
(17.97)
(17.98)
An interesting feature is that the phase shifts are smooth function of energy. Therefore, the dependence of δl on energy and V (r) can be easily analysed. We can use Born approximation to Eq. (17.98) for weak potentials V (r). Such a weak potential causes small or less scattering and therefore approximate Rl,s by Rl,in in Eq. (17.98) so that Z 1 2 sin δl = − r2 U (r) |Rl,in | dr . (17.99) k From Eq. (17.99), we conclude that if the potential is predominantly attractive, then sin δl is positive and if V (r) is repulsive then sin δl is negative. The nucleon-nucleon scattering system exemplifies this quite nicely. From the analysis of nucleon-nucleon scattering experiments, we are able to determine the phase shifts. For laboratory energies of less than 300MeV, a particular n-p phase shift is positive implying that the potential experienced by the n-p system is attractive, but around 300MeV this phase shift goes to zero and then for larger energies it becomes negative. Interpretation of this observation leads to the conclusion that for energies greater than 300MeV, n-p scattering probe the short range behaviour of the potential and the negative value of phase shift means that there is a repulsive core to the n-p interaction. Hence, for the relevant total angular momentum, the short range nucleon potential is attractive until a very small core radius rc is reached. For r < rc , the potential changes from attractive to repulsive. The form of the Eq. (17.92) for total cross-section suggests that, for a given k, there should be a maximum value of l, say lmax (k) such that δl (k) = 0,
l > lmax (k) .
(17.100)
If Eq. (17.100) is not true and if all the δl were nonzero, then Eq. (17.92)
Scattering Theory 437 would diverge and become infinity. Therefore, σ≈ and f (θ) ≈
lmax 4π X (2l + 1) sin2 δl k2
(17.101)
l=0
lmax 1X (2l + 1)Pl (cos θ)eiδl sin δl . k
(17.102)
l=0
Even if lmax is as small as 3 the angular distribution obtained from Eq. (17.102) will be complicated. We can get a rough estimate of lmax using the classical argument. If p s is the impact parameter then the pangular momentum is ~ks. So, we write lmax (lmax + 1) ~ ≈ ~ks. That is, lmax (lmax + 1) ≈ ks. For a given range of potential, lmax increases with k. In general, for low energy (small k), s-wave (l = 0) scattering dominates (isotropic scattering). In n-p scattering with energies below 10MeV, the scattering is essentially due to neutrons with l = 0 and the angular distribution of the scattered neutrons is isotropic in the centre of mass system, and this has been verified experimentally at low energies.
17.7.5
Scattering Length
We have noticed that for low energy scattering only s-state is important. In this case, from Eq. (17.89) we get 1 iδ0 4π e sin δ0 , σ = 2 sin2 δ0 . (17.103) k k The scattering is the result of the modification of the incident wave produced by the potential energy inside the range of the force. For s-wave, the behaviour of the wave function in this region is governed by E − V (r) in the Schrödinger equation. Consequently, if E is small compared to V (r), that is, to the depth of the potential well, the scattering is practically independent of E. The limiting value of f as E approaches zero is called the scattering length ls : f (θ) =
lim f (θ) = −ls .
E→0
(17.104)
Then σ = 4πls2 . If the interaction is weak and the phase shift is small, then it follows from the Eqs. (17.103) and (17.104) that limk→0 δ0 /k = −ls . Hence, for small energy δ0 = −kls .
17.8 SCATTERING FROM A SQUARE-WELL SYSTEM As an example of partial wave analysis we analyse the case of the attractive square-well potential ( −V0 , r < a V (r) = (17.105) 0, r > a.
438 Quantum Mechanics I: The Fundamentals The radial equation with l = 0 is given by ¨ + (k 2 − U )X = 0 , X
U=
2mV0 = k02 . ~2
(17.106)
We write the Eq. (17.106) separately for r < a and r > a: ¨ > + k 2 X> X ¨ < + k12 X< X
=
0,
r>a
(17.107a)
=
0,
r = C sin(kr + δ0 ) .
The constants C and δ0 must be such that X< , X> , X˙ < and X˙ > are continuous at r = a. Then X< = X> and X˙ < = X˙ > at r = a give sin k1 a = C sin(ka + δ0 ) ,
k1 cos k1 a = Ck cos(ka + δ0 ) .
(17.109)
From the Eqs. (17.109) we obtain −1
δ0 = tan
For δ0 ≪ k0 a we get sin δ0 = ka
k tan k1 a − ka . k1
tan k0 a −1 k0 a
(17.110)
(17.111)
.
For l = 0 the total scattering cross-section is 4π σ = 2 sin2 δ0 = 4πa2 k
tan k0 a −1 k0 a
2
.
(17.112)
Figure 17.7 shows the plot σ/a2 versus k0 a. We notice that for low bombarding energies whenever k0 a ≈ (2n + 1)π/2, n-integer or V0 a2 =
π 2 ~2 9π 2 ~2 25π 2 ~2 , , , ··· 8m 8m 8m
(17.113)
the scattering cross-section is very large. These V0 a2 are the values for which new energy levels with l = 0 appear. We say that an incident particle that has nearly the correct energy to be bound by the potential tends to concentrate there and give a large distortion in the wave function resulting in a large scattering cross-section. Further, σ becomes ∞ for k0 a = π/2, 3π/2,· · · . Since it was derived from the assumption that δ0 ≪ k0 a, the infinite cross-section is inadmissible. A detailed analysis gives σmax = 4π/k 2 . Such maxima are called resonances.
Scattering Theory 439 50
σ / a2
40 30 20 10 0
0
π 2
3π 2
5π 2
k0a
7π 2
Variation of σ/a2 with k0 a for the square-well potential given by Eq. (17.105). The dashed lines are k0 a = π/2, 3π/2, 5π/2 and 7π/2. FIGURE 17.7
When tan k0 a ≈ k0 a (that is k0 a ≪ 1) we have σ ≈ 0. If the values of V0 and a are so chosen that σ is close to zero or minimum for a nonzero potential then no scattering occurs. This was first observed by Carl Ramsauer in 1921 and is called Ramsauer effect. (Similar effect was observed a year later by John Sealy and Victor Albert Bailey. This effect is also known as Ramsauer– Townsend effect.) The Ramsauer effect is a very strong evidence for the wave nature of the electron. It was found in the scattering of electrons from atoms of noble (rare) gases, such as neon and argon. Rare gas atoms have a closed shell electronic configuration so that the orbital electrons are fully packed in a well defined spherical region. The combined electrostatic forces due to the nucleus and the orbital electrons are almost zero beyond the radius and hence an incident electron realizes a strong attractive potential-well of finite range. Can Ramsauer effect occur with a repulsive potential? Why? If the sign of V in (17.105) is changed, we get the low-energy scattering cross-section. Solution for the corresponding scattering problem is obtained by simply replacing k0 in Eq. (17.106) to iκ in all the relations. Then we obtain σ = 4πa2
tanh κa −1 κa
2
.
(17.114)
Solved Problem 4: Find the expression for the scattering length for the square-well potential in terms of the potential and its range. For a square-well potential we have tan k0 α sin δ0 = kα −1 , k0 α
k02 = U .
(17.115)
440 Quantum Mechanics I: The Fundamentals For low energy δ0 → 0 and sin δ0 → δ0 we get δ0 tan k0 α =α −1 . k k0 α
(17.116)
Then the scattering length ls is given by √ ! δ0 tan k0 α tan α U √ . ls = − = α 1 − =α 1− k k0 α α U
(17.117)
17.9 PHASE-SHIFT OF ONE-DIMENSIONAL CASE The partial wave analysis developed for spherically symmetric system has been formulated for one-dimensional systems also [4,5]. In the following we outline this. Far from scattering centre we write ψscat± (x) = eikx + f (ε)eiεkx ,
(17.118)
where ε = x/|x| = ±1. The conventional form of the ψ± with the particle incident on the potential from left is ( ψ− (x) = eikx + Be−ikx , x → −∞ (17.119) ψ= ψ+ (x) = Ceikx , x → ∞. We write ψ+ (x) = eikx + (C − 1)eikx
(17.120)
and identify f (+) = C − 1 and f (−) = B. As an analytic function can be uniquely expressed as a sum of even and odd parts, the general solution of the one-dimensional Schrödinger equation is expanded as X 1 ψ(x) = εl Bl cos εkx + lπ + δl , (17.121) 2 l=0,1
where Bl and δl are determined by solving the Schrödinger equation and the phases lπ/2 are a matter of convention. Comparison of the coefficients of incoming and outgoing waves in (17.120) and (17.121) we get X f (ε) = εl fl , (17.122) l=0,1
where fl = (eiδl − 1)/2 and Bl = (−1)l eiδl . Then the total scattering crosssection is given by X σtot = |f (+)|2 + |f (−)|2 = 2 sin2 δl . (17.123) l=0,1
Scattering Theory 441
17.10 INELASTIC SCATTERING In this section we obtain total scattering cross-section for inelastic scattering. A plane wave can be treated as a partial wave expansion of incoming and outgoing spherical waves e±ikr /kr. Due to the scattering object the outgoing spherical waves eikr/r becomes Sl eikr /r. where Sl = e2iδl and |Sl | = 1. For inelastic scattering we write Sl = ηl (k)e2iδl where 0 < ηl < 1. There is a loss of flux between the incoming spherical waves and the outgoing spherical waves due to particle decay, particle capture and the excitation of internal states of the target. The integrated radial flux of the incoming part in Eq. (17.91) can be calculated. The incoming part is written as f eikr /r where ∞ 1 X (2l + 1)Pl (cos θ) . 2ik
f=
(17.124)
l=0
For a spherical wave of the form f eikr /r the radial flux is given by [~k|f |2 /(mr2 )]b r . Then the total integrated flux over all direction is Jinc =
~k m
Z
|f |2 dΩ =
∞ ~k π X (2l + 1) . m k2
(17.125)
l=0
For inelastic scattering |Sl |2 6= 1 and Jout =
∞ ~k π X (2l + 1)|Sl (k)|2 . m k2
(17.126)
l=0
Then the inelastic cross-section is given by σinel =
∞ Jinc − Jout π X = 2 1 − |Sl (k)|2 . ~k/m k
(17.127)
l=0
The total cross-section is σtotal
=
σel + σinel =
∞ 2π X (2l + 1)(1 − ReSl (k)) . k2
(17.128)
l=0
17.11 CONCLUDING REMARKS Several advancements have been made both in the theory and experiments of quantum scattering. For example, using supercomputers a complete solution of the ionization of the hydrogen atom by collision with an electron was obtained. This interaction is responsible for the glow of fluorescent lights and for the ion beams that engrave silicon chips [6]. A quantum scattering theory based on the impact approximation is applied to the level shift and the spectral line broadening in plasmas [7]. A self-contained discussion of scattering is presented
442 Quantum Mechanics I: The Fundamentals for the case of central potentials in one space dimension, which facilitates the understanding of the more complex scattering theory in two and three dimensions [8]. Detection of quantum scattering phase shifts of individual atoms in a quantum scattering experiment using a novel atom interferometer has been reported. By performing an atomic clock measurement using only the scattered part of wave function of each atom the difference of the s-wave phase shifts for the two clock states has been measured [3]. An expression for the cross-section in the case of non-plane-wave scattering was obtained [9]. For more details on quantum scattering one may refer to refs.[10-13].
17.12 BIBLIOGRAPHY [1] M. Born, Z. Phys. 38:803, 1926. [2] M.R.C. Hunt: Online lecture notes at http://www.physics.gla.ac.uk/ donnelly/files/xsection/index.html. [3] R.A. Hart, X. Xu, R. Legere and K. Gibble, Nature 446:892, 2007 and references therein. [4] J.H. Eberly, Am. J. Phys. 33:771, 1965. [5] B. Stec and C. Jedrzejek, Eur. J. Phys. 11:75, 1990. [6] T.N. Rescigno, M. Baertschy, W.A. Isaacs and C.W. McCurdy, Science 286:2474, 1999. [7] K. Yamamoto and H. Narumi, J. Phys. Soc. Jpn. 47:1383, 1979. [8] V.E. Barlette, M.M. Leite and S.K. Adhikari, Eur. J. Phys. 21:435, 2000. [9] G. Gouesbet, J. Math. Phys. 50:112302, 2009. [10] L.D. Faddeev and S.P. Merkuriev, Quantum Scattering Theory For Several Particle Systems. Kluwer, Dorolrect, 1993. [11] D.G. Truhlar and B. Simon, Multi-Particle Quantum Scattering with Applications to Nuclear, Atomic and Molecular Physics. Springer, Berlin, 1997. [12] D. Belkic, Principles of Quantum Scattering Theory. IOP, Bristol, 2004. [13] J. Derezinski and C. Gerard, Scattering Theory of Classical and Quantum N-Particle Systems. Springer, Berlin, 1997.
Scattering Theory 443
17.13 EXERCISES 17.1 Consider a particle of mass m1 approaching a target particle of mass m2 with a velocity v0 . Using the energy conservation theorem show that the kinetic energies in the centre of of mass (EC ) and in the laboratory coordinate system (EL ) are related as EC = [m2 /(m1 + m2 )]EL where EL = m1 v 2 /2, EC = µv 2 /2 and µ = m1 m2 /(m1 + m2 ). 17.2 Find the largest laboratory scattering angle θL for (i) m2 < m1 and (ii) m2 ≥ m1 . 17.3 Show that
d(cos θL ) 1 + ρ cos θC . = d(cos θC ) (1 + 2ρ cos θC + ρ2 )3/2
17.4 Assume that the differential cross-section dσ/dΩ for a given interaction potential is isotropic in the centre of mass frame. For mass ratio ρ ≪ 1, what is the ratio of the differential cross-section in the forward direction to that in the θ = π/2 direction in the laboratory frame? 17.5 Using ψsca ∼ f (θ, φ)eik·r /r show that Jsca,r = ~k|f |2 /(mr2 ). 17.6 Find the value of ∇2r
1 . |r − r′ |
17.7 For the screened Coulomb potential V (r) = −Ae−αr /r obtain the validity condition for Born approximation and discuss it for both lower and upper velocity limits. 17.8 Calculate the cross-section for a central Gaussian potential √ differential 2 2 V (r) = (V0 / 4π)e−r /4a under Born approximation. 17.9 For the Yukawa potential V = V0 e−r/a obtain the total scattering crosssection by Born approximation. 17.10 For a delta potential V0 δ(r−a) obtain scattering cross-section for sa ≪ 1 by Born approximation. 17.11 The scattering amplitude for the Coulombic interaction of two identical bosons of charge Ze is given by γ f (θ) = e−i2γ ln[sin(θ/2)]+iπ+2iη0 2k sin2 (θ/2) where Z 2 e2 γ= , η0 = argΓ(1 + iγ). 4πǫ0 ~v Obtain the differential cross-section for the Coulomb scattering (Mott’s formula) of two identical spinless bosons.
444 Quantum Mechanics I: The Fundamentals 17.12 A perfectly rigid sphere is represented by a potential V (r) = ∞ for r < a and 0 for r > a. Particles cannot penetrate into the region r < a because V (r) = ∞ in this region. Therefore, the wave function ψ(r) is zero for r < a. Because of infinite potential within the radius a, Born approximation is not appropriate. Applying the partial wave analysis calculate the scattering cross-section. 17.13 For the hard sphere considered in the previous problem find the total scattering cross-section for lmax = kα. 17.14 Show that in an elastic scattering between two spinless particles, the centre of mass differential cross-section may be represented by an expression of the type dσ/dΩ = A + BP1 (cos θ) + · · · . Express the coefficients A and B in terms of the phase shifts δl . Use the orthogonality relation Z π 2 Pl (cos θ)Pl′ (cos θ) sin θ dθ = δll′ 2l +1 0 and the recurrence relation (l + 1)Pl+1 (cos θ) = (2l + 1) cos θPl (cos θ) − lPl−1 (cos θ).
CHAPTER
18
Identical Particles
18.1 INTRODUCTION Two particles are said to be identical when it is not possible to distinguish one particle from the other particle with their inherent physical properties such as mass, charge, spin, etc. Study of identical particles is important for two reasons: 1. Many-body systems contain usually identical particles and 2. they give rise quantum mechanical phenomena that have no classical analogues. In classical mechanics, we can able to know the trajectory of any particle given its initial position, momentum and the forces acting on it. When we consider a system with two or more identical particles, in principle, we can distinguish them at any time as each one of the particles follows a definite predictable path. This is because classical mechanics is a deterministic theory. For example, when two identical particles collide, we can determine the paths of these particles after collision as their paths are known before the collision. So, in classical mechanics, it is possible to distinguish two identical particles. Also, any particle can be observed without affecting its behaviour. In quantum mechanics the situation is different because the uncertainty principle does not allow us to know the initial positions and momenta of the particles. The individual particle is described by a wave packet. Since the wave packet is not localized and spreading with time, it is impossible to distinguish two identical particles as their wave functions always overlap at any point in space. Unless the two particles are well separated with no overlap of their wave functions, quantum identical particles cannot be distinguishable. In quantum mechanics the finite extent of the wave functions associated with each identical particle may result in overlapping of these wave functions. This makes impossible to state the wave function associated with each particle. Thus, a quantum mechanical description of a system of identical particles 445
446 Quantum Mechanics I: The Fundamentals must be formulated in such a way that the indistinguishability of identical particles is explicitly taken into account [1-3].
18.2 PERMUTATION SYMMETRY Quantum theory describes any object at energies below its excitation threshold with four degrees of freedom. For example, these degrees of freedom could be the coordinate vector r and its spin σ or the momentum vector p and the helicity. Let us treat the variables corresponding to an N -particles system as (1, 2, · · · , N ). Here 1 corresponds to r1 and σ1 of particle 1 and correspondingly for the other particles. So, the Hamiltonian for a system of N identical particles is H = H(1, 2, · · · , N ) and the wave function of the system is ψ = ψ(1, 2, · · · , N ). The Hamiltonian H(1, 2, · · · , N ) of the N indistinguishable particles is invariant with respect to all their interchanges or permutations. Let Pij be the operator that permutes i and j and its eigenvalue be ω. Pij is called the permutation operator. The effect of the Pij , which interchanges i and j, on an N -particle wave function is Pij ψ(1, 2, · · · , i, · · · , j, · · · , N ) = ψ(1, 2, · · · , j, · · · , i, · · · , N ) .
(18.1)
We have Pij ψ(1, 2, · · · , i, · · · , j, · · · , N ) =
=
ψ(1, 2, · · · , j, · · · , i, · · · , N )
ωψ(1, 2, · · · , i, · · · , j, · · · , N ) (18.2)
and Pij2 ψ(1, 2, · · · , i, · · · , j, · · · , N ) =
=
ωPij ψ(1, 2, · · · , j, · · · , i, · · · , N )
ω 2 ψ(1, 2, · · · , i, · · · , j, · · · , N ) (18.3)
where ω 2 = 1, that is, ω = ±1. So, the eigenvalues of the permutation operator are ±1. Since the eigenvalues are real, Pij is a Hermitian operator. Because Pij2 = I we have Pij = Pij−1 . Further, Pij is unitary (see problem 1 at the end of this chapter). An operator O changes to O′ under the unitary transformation as U OU −1 = O′ . Since the Hamiltonian H is symmetrical in its arguments (1, 2, · · · , N ), because of the indistinguishability of identical particles, we write P HP −1 = H. Therefore, for every permutation operator P H = HP . In fact, for any symmetric operator O(1, 2, · · · , N ) we have [P, O] = 0. It is to be noted that all observables O must be permutation invariant. Why? The permutation P has the following properties: 1. If ψ(1, 2, · · · , N ) is an eigenfunction of H with eigenvalue E then P ψ is also an eigenstate with the same eigenvalue. This type of degeneracy is called exchange degeneracy.
Identical Particles 447 2. For every permutation hφ|ψi = hP φ|P ψi. 3. The matrix elements of all observables are same in ψ and P ψ. The principle of indistinguishability with respect to permutation requires at least one simultaneous eigenfunction of all the N ! exchange operators and H. Let ψ denotes such a function. Then Pij ψ = ωij ψ,
(18.4)
Pij Pik = Pjk Pij = Pik Pjk
(18.5)
Hψ = Eψ, where ωij = ±1. It can be proved that
for any three different values of i, j, and k. Then from Eq. (18.5) we get ωij ωik = ωjk ωij = ωik ωjk
(18.6)
which gives ωij = ωik and ωjk = ωik or ωij = ωik = ωjk . That is, the common eigenfunction ψ belongs to the same eigenvalue of all the exchange operators. If ωij = 1 then we say that ψ is symmetric under the interchange of any pair of particles and denote it by ψs (1, 2, · · · , N ). If ωij = −1, ψ is antisymmetric and denote it by ψa (1, 2, · · · , N ). We get Pij ψs (1, 2, · · · , i, · · · , j, · · · , N ) = =
ψs (1, 2, · · · , j, · · · , i, · · · , N ) ψs (1, 2, · · · , i, · · · , j, · · · , N ) (18.7)
and Pij ψa (1, 2, · · · , i, · · · , j, · · · , N ) = ψa (1, 2, · · · , j, · · · , i, · · · , N )
= −ψa (1, 2, · · · , i, · · · , j, · · · , N ). (18.8)
For any arbitrary permutation operator P , ψs and ψa satisfy P ψs
= ψs ,
P ψa
= (−1)P ψa ,
(−1)P =
(
(18.9a) 1, for even permutation (18.9b) −1, for odd permutation.
For ψs , every P of the N ! elements are assigned the number 1 and for ψa , every even (odd) element is assigned the number 1 (−1). The two-particle case is trivial. There is only one permutation operator, P12 (apart from the identity operator I). For three or more particles, the Pij do not commute with one another, and in addition to ψs and ψa , there are states for which not all Pij are diagonal. These states are bases functions of higher-dimensional representation of the permutation group. It is an experimental fact that these states are not realized in nature. Only symmetric and antisymmetric states are realized in nature.
448 Quantum Mechanics I: The Fundamentals In consequence, the Hilbert space of N identical particles breaksup into subspaces Hs and Ha spanned by symmetric and antisymmetric states and a remainder unphysical (N ! − 2) states Hr which can be further broken in terms of multi-dimensional irreducible representations. Since P commutes with H, the symmetric character of a wave function remains the same in time and cannot be changed by any perturbation because of the principle of indistinguishability. As a consequence, Hs and Ha are two disconnected subspaces, because there would be no observables. If a nonstationary state is initially in one subspace, it will stay forever in that subspace. It is an experimental fact that both subspaces exist in nature. The wave functions of all the states of a given system of identical particles must have the same symmetry; otherwise, the wave function of a state formally a superposition of states of different symmetry would be neither symmetrical nor antisymmetrical. The symmetrical or antisymmetrical property of wave functions depends on the nature of the particle. This dependence arises from the spin of the particle is given by the spin-statistics theorem: The wave functions for any system of N identical particles having halfodd-integer spin must be antisymmetric under interchange of the labels of any pair of the N particles. No others are allowed. Correspondingly, the wave functions of any system of N identical, zero-spin or integer-spin particles must be symmetric under the interchange of the labels of any pair of the N particles. This theorem gives the connection between spin and statistics. Due to the principle of indistinguishability, the quantum statistics is different from the classical Maxwell–Boltzmann statistics. According to the above theorem, there are two types of particles, with symmetric and antisymmetric wave functions depending upon their spin. So, we have two quantum statistics. The statistics of particles with integer spin is known as Bose–Einstein statistics and that of particles with half-integer spin is known as Fermi–Dirac statistics. Particles obeying Bose–Einstein statistics are called bosons while those obeying Fermi– Dirac statistics are referred to as fermions. This correlation between spin and statistics applies to elementary particles and also to composite particles (such as atoms and nuclei). This is quite clear since a composite particle composed of fermions will have integral or half-integral spin as the number of fermions is even or odd. For example, He3 nucleus is fermion as it has two proton and one neutron. α-particle is a boson because it contains two protons and two neutrons. Examples of fermions are electrons, protons, neutrons, quarks, neutrinos etc. and examples of bosons are photons, π mesons, 12 C nuclei etc.
Solved Problem 1: Prove that Pij Pik = Pjk Pij . We obtain Pij Pik ψ(1, 2, · · · , i, · · · , j, · · · , k, · · · , N )
= ψ(1, 2, · · · , k, · · · , i, · · · , j, · · · , N ) .
(18.10)
Identical Particles 449 Also Pij ψ(1, 2, · · · , i, · · · , j, · · · , k, · · · , N )
= ψ(1, 2, · · · , j, · · · , i, · · · , k, · · · , N ) Pjk Pij ψ(1, 2, · · · , i, · · · , j, · · · , k, · · · , N )
= ψ(1, 2, · · · , k, · · · , i, · · · , j, · · · , N ) .
(18.11a) (18.11b)
From Eqs. (18.10) and (18.11) we find Pij Pik = Pjk Pij . Similarly, it can be proved that Pik Pij = Pik Pjk .
18.3 SYMMETRIC AND ANTISYMMETRIC WAVE FUNCTIONS The many-particle Schrödinger equation i~
∂ ψ(1, 2, · · · , N, t) = H(1, 2, · · · , N )ψ(1, 2, · · · , N, t) ∂t
(18.12)
gives the wave function ψ(1, 2, · · · , N, t) which is a function of time, the spin and the spatial coordinates of the N particles. The solution of Eq. (18.12) may or may not have a well-defined symmetry. Out of all possible solutions, we must choose antisymmetric functions for fermion systems and symmetric functions for boson systems. We have to construct symmetric or antisymmetric functions from such unsymmetrized functions. We know that there are N ! solutions to Eq. (18.12) corresponding to same energy eigenvalue as P ψ is also a solution to Eq. (18.12) if ψ is a solution. The sum of all these N ! functions gives the unnormalized symmetric wave function ψs , since the interchange of any pair of particles changes any one of the component functions into another of them as N ! permutation operators form a group. An antisymmetric unnormalized wave function ψa is setup by adding all the permuted functions arising from the original solution by means of an even number of interchanges of pairs of particles and subtracting the sum of all the permuted functions arising by means of an odd number of interchanges of pairs of particles in the original solution. Let us consider, for example, a system of two identical particles. If ψ(1, 2) is a solution of the Schrödinger Eq. (18.12) then P12 ψ(1, 2) = ψ(2, 1) is also a solution. Then apart from the normalizing factors, the symmetric and antisymmetric wave functions are of the form ψa = B(ψ(1, 2) − ψ(2, 1)) . (18.13) √ If ψ(1, 2) and ψ(2, 1) are normalized then A = B = 1/ 2! . For a N -particle system, the symmetrizing and antisymmetrizing operators Ss and Sa are 1 X P ψs = Ss ψ(1, · · · , N, t) = √ (1) P ψ(1, · · · , N, t), (18.14a) N! P 1 X ψa = Sa ψ(1, · · · , N, t) = √ (−1)P P ψ(1, · · · , N, t). (18.14b) N! P ψs = A(ψ(1, 2) + ψ(2, 1)) ,
450 Quantum Mechanics I: The Fundamentals
Solved Problem 2: For a three particle systems write the symmetric and antisymmetric wave functions. For a three-particle system ψs
ψa
=
=
1 √ [ψ(1, 2, 3) + ψ(1, 3, 2) + ψ(2, 1, 3) 6 +ψ(2, 3, 1) + ψ(3, 2, 1) + ψ(3, 1, 2)] , 1 √ [ψ(1, 2, 3) + ψ(2, 3, 1) + ψ(3, 1, 2) 6 −ψ(2, 1, 3) − ψ(1, 3, 2) − ψ(3, 2, 1)] .
(18.15a)
(18.15b)
The exchange of two particles in ψs does not change the function, whereas the exchange of two particles in ψa changes the sign of ψa . Hence, for both the symmetric and antisymmetric total eigenfunctions, ψs∗ ψs and ψa∗ ψa are unchanged under an exchange of the particle. So, any measurable quantity obtainable from the total eigenfunctions of a system of identical particles is unaffected by an exchange of the particle labels. The labels 1 and 2 appear in the expressions for ψs and ψa , but this labelling does not violate the requirements of indistinguishability because the value of any observable obtained from the eigenfunctions is independent of the assignment of the labels. For the wave functions given by Eq. (18.13) we have |ψs,a |2 =
1 |ψ(1, 2)|2 + |ψ(2, 1)|2 ± R.P. [ψ(1, 2)ψ ∗ (2, 1)] . 2
(18.16)
The first term is the joint probability distribution for two noninteracting distinguishable particles. The second term results from the interference between ψs and ψa and is due to the indistinguishability of identical particles. It should be noted that this interference term vanishes in the regimes where the two oneparticle wave functions do not overlap. That is, the wave function ψ(1, 2) is different from zero only when the coordinate 1 is in some region R1 , the coordinate 2 is in a region R2 , and R1 and R2 have no common domain. Under such a situation, ψ(1, 2) vanishes whenever 1 is not in R1 and 2 is not in R2 . Now, R1 and R2 do not overlap and the interference term is zero everywhere. So, the two identical particles can be distinguished as the particles are well separated. Hence, for a system of identical particles, at high energies (E ≫ kB T ) and low concentration, probable number of particles per quantum state will be very small. Wave functions of the particles are well separated. The particles can now be distinguished and we can apply the classical statistics instead of quantum statistics.
Identical Particles 451
18.4 THE EXCLUSION PRINCIPLE From the analysis of data concerning the energy levels of atom, Pauli in 1925, discovered his famous exclusion principle: In a multielectron atom there can never be more than one electron in the same quantum state. This principle is applicable not for electron system alone. All fermions follow this principle. An exact solution of a many-body problem in quantum mechanics cannot be found directly. It is usually found by using the method of successive approximations. In this method in the zeroth-order approximation the particles are noninteractive while in the higher-order approximations the interaction is included through perturbation theory. So, the Hamiltonian in zeroth-order approximation is given as sum of the Hamiltonians of the separate particles: H0 (1, 2, · · · , N ) = H0 (1) + H0 (2) + · · · + H0 (N ) .
(18.17a)
The zeroth-order eigenfunction is the product of one-particle eigenfunctions: ψ(1, 2, · · · , N ) = φα1 (1)φα2 (2) · · · φαN (N ) , E0 H0 (j)φαj (j)
= Eα1 + Eα2 + · · · + EαN , = Eαj φαj (j) .
(18.17b) (18.17c) (18.17d)
φαj (j) are the energy eigenfunctions of the jth particle and αj are the quantum numbers specifying its state. The eigenfunctions corresponding to E0 will then be linear combination of the functions ψ(1, 2, · · · , N ). ψs (1, 2, · · · , N, t) and ψa (1, 2, · · · , N, t) given by Eq. (18.16) are the zeroth-order wave functions of bosons and fermions respectively. We write ψa in the form of a determinant (Slater determinant) 1 ψa = √ N !
φα1 (1)
φα1 (2)
φα2 (1) .. .
φα2 (2) .. .
φαN (1)
φαN (2)
···
··· ··· ···
φα1 (N ) φα2 (N ) . .. . φα (N )
(18.18)
N
The change in the sign of the function (18.18) under a permutation of any two particles follows immediately from the change of sign of a determinant when two of its columns are interchanged. From (18.18), the Pauli exclusion principle follows. According to the Pauli exclusion principle, a system of identical fermions cannot be in a state described by a wave function (18.18) with two single-particle states which are the same. If any of the two single-particle states among α1 , α2 , · · · , αN are the same, the determinant vanishes. Thus, on a system of identical fermions, twoor more-particles cannot be in the same state. For a boson system (described
452 Quantum Mechanics I: The Fundamentals by symmetric wave function) any number of particles can occupy a single state. So, the occupation number of bosons in a state n can be 0, 1, 2, · · · whereas it is either 0 or 1 for fermions. Hence, the statistics for bosons and fermions are different and we have two different quantum statistics: Bose–Einstein statistics for bosons and Fermi–Dirac statistics for fermions.
18.5 SPIN EIGENFUNCTIONS OF TWO ELECTRONS The Hamiltonian of a system of identical particles N
H=
1 X 2 p + V (r1 , r2 , · · · , rN ) 2m i=1 i
(18.19)
does not contain the spin operators of the particles in the nonrelativistic limit if no external applied magnetic field and the spin-orbit interaction is neglected. In this case, we solve the Schrödinger equation and obtain Φ(r1 , r2 , · · · , rN ). This is the spatial part of the total wave function. In order to take into account the spin degrees of freedom, the total wave function is written as a product of Φ and the spin eigenfunction χ(r1 , r2 , · · · , rN ) as ψ(1, 2, · · · , N ) = Φ(r1 , r2 , · · · , rN )χ(r1 , r2 , · · · , rN )
(18.20)
or as a linear combination of such products. The above wave function is the first approximation when dealing with a system with a Hamiltonian containing spin-orbit interaction. The symmetry requirements of the wave function with permutations of particles is applicable to the total wave function ψ(1, 2, · · · , N ) since a permutation of particles corresponds to a permutation of both the spatial and the spin variables. The quantum numbers αj , corresponding to the wave function of jth particle, contains three space quantum numbers and one spin quantum number for a three-dimensional case. We can construct symmetric and antisymmetric wave functions for both the spatial and spin parts of the total wave function. If both the spatial and spin part of the wave functions are symmetric or antisymmetric then the total wave function will be symmetric. If they differ in their symmetries then the total wave function will be antisymmetric. But the total wave function for a system of bosons must be symmetric and that for a system of fermions must be antisymmetric. For two systems with angular momentum J1 with eigenstates |j1 m1 i and J2 with eigenstates |j2 m2 i, the combined system with total angular momentum J = J1 + J2 is found from the (2j1 + 1)(2j2 + 1) eigenstates |j1 m1 i|j2 m2 i using the Clebsch–Gordan (CG) transformation matrix. This also holds for the addition of spin angular momenta of two electrons. 1 The 1 1number E spin quantum E 1 of the electron s = 1/2 gives two eigenfunction = α and − = β. 22 2 2 For the case of two electrons, we get four eigenfunctions α(1)α(2), β(1)β(2), α(1)β(2) and β(1)α(2).
Identical Particles 453 In the coupled representation, we can construct eigenstates of the four commuting operators S2 = (S1 +S2 )2 = |S1 |2 +|S2 |2 +2S1 ·S2 , Sz = S1z +S2z , S12 and S22 . The simultaneous eigenstates of these operators are written as |sms i where s takes the values 1 and 0. ms takes the values 1, 0, −1 for s = 1 and 0 for s = 0. If s = 0 then the spins are antiparallel and for s = 1 the two spins are parallel. So, there are four eigenkets |sms i in the coupled representation. We have found in sec. 11.12, the CG transformation matrix between these two representations for the case j1 = 1/2 and j2 = 1/2. If we consider the two spin degrees of freedom alone then s1 = j1 = 1/2 and s2 = j2 = 1/2. So, using the CG matrix we obtain the spin wave function for two electrons in the coupled representation as χ(1) s χs(0) χ(−1) s χa
= |11i = α(1)α(2) , 1 = |10i = √ [α(1)β(2) + β(1)α(2)] , 2 = |1 − 1i = β(1)β(2) , 1 = |00i = √ [α(1)β(2) − β(1)α(2)] . 2
(18.21a) (18.21b) (18.21c) (18.21d)
Since S 2 |sms i = s(s + 1)~2|sms i and Sz |sms i = ms ~|sms i we can easily prove that Eqs. (18.21b) and (18.21d) are eigenstates of S 2 with eigenvalues 2~2 and 0 corresponding to s values of 1 and 0 respectively. The eigenvalues of Sz for the three states given by Eqs. (18.21a-c) are found to be ~, 0 and −~ respectively. The eigenvalues of Sz for the state given by Eq. (18.21d) is 0. The triplet states corresponding to s = 1 are all symmetric functions and the singlet state corresponding to s = 0 is antisymmetric.
Solved Problem 3: Show that the spin matrices for s = 1/2 can be defined by ~ i~ Sx = [(|αihβ|) + (|βihα|)], Sy = [(|βihα|) − (|αihβ|)], 2 2 ~ Sz = [(|αihα|) − (|βihβ|)]. 2 1 0 We have |αi = , |βi = . Then 0 1 0 ~ 1 Sx = 0 1 + 1 2 0 1
0 1 ~ . (18.22) 0 = 2 1 0
454 Quantum Mechanics I: The Fundamentals Next, Sy
=
=
Sz
=
=
1 i~ 0 1 0 − 0 1 2 1 0 ~ 0 −i , 2 i 0 0 ~ 1 1 0 − 0 1 2 0 1 0 ~ 1 . 2 0 −1
(18.23)
(18.24)
18.6 EXCHANGE INTERACTION The nonrelativistic Schrödinger equation does not take into account the spin of the particle as in the nonrelativistic approximation the electrical interaction of the particle does not depend on their spins. However, there is a peculiar dependence of the energy of the system on its total spin, due to the principle of indistinguishability of identical particles. Let us consider a system with two identical particles and with no external interaction. The inter-particle interaction is spin-independent. The relevant Hamiltonian for such a system will be H(r1 , r2 ) = K1 (r1 ) + K2 (r2 ) + V (|r1 − r2 |) .
(18.25)
In the centre of mass coordinates R and r = r1 − r2 , Eq. (18.25) becomes H(r, R) = Hrel (r) + Kcm (R) ,
(18.26)
where Hrel =
1 2 P (r) + V (r), 2µ rel
Kcm (R) =
1 2 P (R) . 2M cm
(18.27)
Since P12 commutes with H(r1 , r2 ) we have [P12 , Kcm ] = 0 = [P12 , Hrel ]. Hence, the eigenstates of Kcm and Hrel can be either symmetric or antisymmetric under the exchange of the two particles. The eigenstates of Kcm = eiKcm ·R = eiKcm ·(r1 +r2 ) , are already symmetric under the interchange of the labels 1 and 2 and thus the symmetry character of the overall eigenstate of H(r1 , r2 ) is determined by the eigenstates of Hrel . The eigenstates of Hrel (r) are the products of the spatial and spin states as given by Eq. (18.20). Suppose that the particles have zero spin. So, we have only the spatial
Identical Particles 455 wave function Φ(r1 , r2 ) as the spin wave function χ(s1 , s2 ) is absent for such particles. Since spin zero particle is a boson, the total wave function Φ(r1 , r2 ) must be symmetric. The spatial eigenstates of Hrel are the usual ψnlm (r, θ, φ). The interchange of two similar particles is equivalent to the inversion operation with the origin as the centre of the line joining the two particles. We know that the effect of inversion is to multiply the wave function ψnlm (r, θ, φ) by (−1)l . Therefore, for even values of l, the function ψnlm (r, θ, φ) is symmetric with respect to interchange of the two particles and antisymmetric for odd values of l. The conclusion is that a system of two identical particles with zero spin can have only an even orbital angular momentum. Next, consider a system consisting of two particles of spin- 21 (say, two electrons), Then the complete wave function must be antisymmetric. As given by Eqs. (18.21b) and (18.21d), the spin wave functions χ(s1 , s2 ) can be either symmetric (the triplet state with s = 1) or antisymmetric (the singlet state with s = 0) respectively. Since the total wave function must be antisymmetric, the triplet state ψnlm (r, θ, φ) must be antisymmetric and symmetric for the singlet state. Hence, for the triplet state l must be odd and even for the singlet state. So, the antisymmetry requirement leads to l + s = even. That is, l = even, s = 0 (the singlet state) and l = odd, s = 1 (the triplet state). This is a very important conclusion which will be applied for the case of He-atom in the next section. If φα1 and φα2 are the spatial eigenfunctions in the zeroth-order approximation then the symmetric spacial eigenfunction is given by 1 Φs (1, 2) = √ [φα1 (1)φα2 (2) + φα2 (1)φα1 (2)] 2
(18.28)
and the antisymmetric spatial function is 1 Φa (1, 2) = √ [φα1 (1)φα2 (2) − φα2 (1)φα1 (2)] . 2
(18.29)
So, for the singlet state of two electrons, the spatial wave function is given by Eq. (18.28). Equation (18.29) gives the spatial wave function for the triplet state. Let us consider a case in which the space variable of the two electrons have almost the same value. Then φα1 (1) ≈ φα1 (2) and φα2 (1) ≈ φα2 (2). Then we find from Eq. (18.29), Φa (1, 2) ≈ 0. This shows that when the triplet state electrons are close together then the probability density will be very small . Since there is almost zero probability of the two electrons coming closer, the triplet electrons act as if they repel each other. This type of repulsion does not arise due to their charges. It is due to the spin-orientation of the two electrons and the requirement of antisymmetric wave functions for fermions system. Symmetric space eigenfunction given by Eq. (18.28) has inverse property √ as Φs ≈ 2φα1 (1)φα2 (2) when φα1 (1) ≈ φα1 (2) and φα2 (1) ≈ φα2 (2). This is twice the average value of over all space of the probability density of the symmetric wave function Φs (1, 2). Hence, there is a particularly large chance
456 Quantum Mechanics I: The Fundamentals of finding the two noninteracting electrons close together if their space eigenfunction is symmetric. Thus, if the spins of the two electrons are antiparallel and the spin eigenfunction is the antisymmetric singlet (s = 0) then the two electrons act as if they attract each other since there is a large chance of finding them close together. The requirement that the fermions systems must have antisymmetric wave functions gives rise a coupling between their spin and space variables. They act as if they move under the influence of a force whose spin depends on their relative orientation. In other words, the possible values of the energy of a system of electrons depend on the total spin. This peculiar interaction arising due to the symmetry requirement is known as exchange interaction. It is a purely quantum effect which vanishes in the limit of classical mechanics. Exchange force arises because of the quantum mechanical indistinguishability of identical particles.
18.7 EXCITED STATES OF THE HELIUM ATOM In the zeroth-order approximation, the two electrons are in the ground state with the electronic configuration 1s2 . Both electrons are in the hydrogenic state ψ100 . We neglect the interaction between the two electrons and write its spatial wave function as 3 1 Z ψ0 (1, 2) = ψ100 (1)ψ100 (2) = e−Z(r1 +r2 )/a0 (18.30) π a0 with Z = 2 for helium atom. The above wave function is symmetric under permutation of the spatial coordinates of the two particles. To obtain an antisymmetric wave function, we multiply (18.30) by the spin antisymmetric wave function χa given by Eq. (18.21d). So, the ground states of helium-like atoms correspond to states with total spin zero. We have treated the electronelectron interaction as a perturbation in sec. 16.2 and found the ground state energy of the helium-like atoms. In this section, we shall consider the excited states of helium-like atoms. The space part of the wave function for the case of one of the electrons is in the ground state (1s) and the other is in an excited state (characterized by nlm) is given by 1 ψ(1, 2) = √ [ψ100 (1)ψnlm (2) ± ψnlm (1)ψ100 (2)] , 2
(18.31)
where + sign gives symmetric spatial wave function and − sign gives the antisymmetric spatial wave function. Since the total wave function must be antisymmetric, the symmetric wave function is associated with the singlet state with the antisymmetric spin wave function χa . The antisymmetric wave function corresponds to the triplet state with the spin wave functions χs with electrons having parallel spins. States with antiparallel spins (singlet states) are called para-states and those with parallel spins (triplet states) are termed as ortho-states.
Identical Particles 457 The first excited state of helium atom corresponds to the electron configuration 1s1 2s1 . The wave functions for the para- and ortho-states are given respectively as Φs (1, 2) = Φa (1, 2) =
1 √ [ψ100 (1)ψ200 (2) + ψ200 (1)ψ100 (2)] , 2 1 √ [ψ100 (1)ψ200 (2) − ψ200 (1)ψ100 (2)] . 2
(18.32) (18.33)
The Hamiltonian for the He-atom is given by ~2 1 e2 1 2 2 2 ∇1 + ∇2 − Ze + + H(1, 2) = − 2µ r1 r2 |r1 − r2 | = H0 + H ′ . (18.34) The electron-electron interaction H ′ = e2 /|r1 − r2 | can be considered as a perturbation to the unperturbed part of the Hamiltonian H0 which has exact solutions. To find the energies of the ortho- and para-states in the first-order perturbation theory, it is sufficient to determine hH(1, 2)i in these states. Since ψ100 and ψ200 are hydrogen-like functions corresponding to energies E1s and E2s of the hydrogen-like atom with n = 1 and n = 2 we get the energy of the para-state as Z Es = Φ∗s HΦs dτ = E1s + E2s + I + J (18.35) and the energy of the ortho-state as Z Es = Φ∗a HΦa dτ = E1s + E2s + I − J ,
(18.36)
where I J
Z
e2 d3 r1 d3 r2 (18.37) |r1 − r2 | Z e2 ψ100 (2)ψ200 (1)d3 r1 d3 r2 . (18.38) = [ψ100 (1)]∗ [ψ200 (2)]∗ |r1 − r2 |
=
[ψ100 (1)]2 [ψ200 (2)]2
The integral I is usually called the Coulomb integral and the integral J is called as exchange integral . It can be shown that both I and J are positive. So, we find that helium in spin-singlet state (para-state) lies higher than the helium in spin-triplet state (ortho-state). As already discussed, in the singlet case the space function is symmetric and the electrons have a tendency to come close to each other. The effect of the electrostatic repulsion is more significant, hence, a higher energy results. In the triplet case, the space function is antisymmetric and the electrons tend to avoid each other. Each configuration splits into the para-state and the ortho-state with the para-state lying higher. For the
458 Quantum Mechanics I: The Fundamentals
1s12p1
para 1P1 ortho 3P210
1s12s1
para 1S0 ortho 3S1
1s2
FIGURE 18.1
1S 0
Energy levels for low-lying configuration of the helium
atom. ground state, only para-helium is possible. Similar to 1s1 1s2 configuration, other configurations like 1s1 2p1 can be considered for the excited state. Then also there will be para- and ortho-states, giving two different energy levels as shown in Fig. 18.1. Though the spin orientations of the two electrons lead to two different energy levels for the excited states of helium, it must be remembered that this difference of energy between the two states occurs only due to Coulombic interactions as the Hamiltonian (18.34) contains only the Coulombic terms and no spin dependent term. This type of exchange interaction is called Heisenberg exchange interaction which is considered to be responsible for spin-ordering in ferromagnets. Para-helium atoms have no magnetic moment and form a diamagnetic gas. Ortho-helium atoms have a magnetic moment and form a paramagnetic gas.
18.8 COLLISIONS BETWEEN IDENTICAL PARTICLES Collisions between identical particles are particularly interesting as a direct illustration of the fundamental differences between classical and quantum mechanics. Consider the collision of two identical particles in the centre of mass frame as shown in Fig. 18.2. When the two particles are identical, it is impossible to decide from the observation of a particle reaching a detector whether it is the incident particle 1 or the recoiling target particle 2. A detector counts the particles scattered into the direction characterized by the polar angles (θ, φ). Since the particles 1 and 2 are identical there is no way of deciding whether a particle recorded by the detector results from a collision event in which particle 1 is scattered in the direction (θ, φ) (Fig. 18.2a) or from a collision process in which particle 2 is scattered in that direction, so that particle 1 is scattered in the opposite direction (π − θ, φ + π) (Fig. 18.2b). In classical mechanics, the differential cross-section for scattering in the direction (θ, φ) is the sum of the differential cross-sections for observation of
Identical Particles 459 Detector
Detector
1 1
2 1
θ
θ π− θ
2 2
FIGURE 18.2
1
(a)
2
(b)
The scattering of identical particles in the centre of mass
frame. particle 1 and particle 2 as the detection does not distinguish the incident and target particles since they are identical. So, the classical differential scattering cross-section for two identical particles is given by dσCM = |f (θ, φ)|2 + |f (π − θ, φ + π)|2 , dΩ
(18.39)
where the scattering amplitude is defined by the scattered wave function ψr→∞ → eik·r + f (θ, φ)
eik·r . r
(18.40)
We note that the total wave function (product of the spatial and spin part of the wave function) must be symmetric for bosons and antisymmetric for fermions. As the wave function (18.40) is asymmetric it must be properly symmetrized with respect to permutations of identical particles. For example, in the case of two identical spinless bosons, the wave function must be symmetric under the interchange of the spatial coordinates of two particles. Now, the interchange of r1 → r2 corresponds to replacing the relative position vector r = r1 − r2 by −r, which in polar coordinates corresponds to (r, θ, φ) being replaced by (r, π − θ, φ + π). So, in place of Eq. (18.40) we use ′
ψS (r → ∞) = eik·r + e−ik·r + [f (θ, φ) + f (π − θ, φ + π)]eik·r .
(18.41)
This results in a differential cross-section for spinless boson particles dσQM dΩ
= |f (θ, φ) + f (π − θ, φ + π)|2 = |f (θ, φ)|2 + |f (π − θ, φ + π)|2 +2Re [f (θ, φ)f ∗ (π − θ, φ + π)] .
(18.42)
It is important to note that this expression differs from the classical expression (18.39) by the presence of the third term on the right-side. The third
460 Quantum Mechanics I: The Fundamentals term represents a truly quantum mechanical effect, which arises due to the interference between the amplitude for scattering of the projectile and target in the direction (θ, φ). If the interaction potential is central then f (θ, φ) is independent of φ. So, we get from Eq. (18.42), dσQM = |f (θ)|2 + |f (π − θ)|2 + 2Re [f (θ)f ∗ (π − θ)] dΩ
(18.43)
so that the scattering is symmetric about the angle θ = π/2 in the centre of mass system, since the partial wave expansion of the symmetrized scattered amplitude fs (θ) = f (θ) + f (π − θ) contains only even values of l. Moreover, at θ = π/2, dσQM dσCM = 4|f (θ = π/2)|2 and = 2|f (θ = π/2)|2 . dΩ dΩ
(18.44)
Hence, the quantum mechanical cross-section is twice as large as the classical result. If we consider scattering by two spin-1/2 identical fermions, then the total wave function must be antisymmetric. Since the spin wave function is either symmetric (triplet state) or antisymmetric (singlet state), the spatial part of the singlet state must be symmetric and that for the triplet state must be antisymmetric. If we assume that the interaction between the two fermions is spin independent and central, then for the singlet state the scattering amplitude is fs = f (θ) + f (π − θ) and for the triplet state the scattering amplitude is fA = f (θ) − f (π − θ). Therefore, the quantum mechanical scattering crosssections are given by dσQM = |f (θ)|2 + |f (π − θ)|2 + 2Re [f (θ)f ∗ (π − θ)] (18.45) dΩ singlet and If spins three given
dσQM dΩ
triplet
= |f (θ)|2 + |f (π − θ)|2 − 2Re [f (θ)f ∗ (π − θ)] .
(18.46)
the incident and target particles are unpolarized (UP) (that is, their are randomly oriented), the probability of obtaining triplet states is times that of a singlet state, so that the differential cross-section is by dσQM 1 dσQM 3 dσQM = + . (18.47) dΩ UP 4 dΩ singlet 4 dΩ triplet
From (18.45) and (18.46) we get dσQM = |f (θ)|2 + |f (π − θ)|2 − Re [f (θ)f ∗ (π − θ)] . dΩ UP
(18.48)
Identical Particles 461 At θ = π/2, Eq. (18.48) gives dσQM = |f (θ = π/2)|2 . dΩ (θ=π/2)
(18.49)
Comparing it with Eq. (18.44), we note that at θ = π/2, the quantum mechanical differential cross-section is equal to one-half of the classical result.
18.9 CONCLUDING REMARKS The symmetry of the Hamiltonian with respect to permutations of its arguments implies the existence of a group of transformations on the state function. The permutation group SN which consists of all permutations of N identical objects has N ! elements. Since the states ψ and P ψ are experimentally indistinguishable, the question arises whether all these N ! states are realized in nature. It can be proved that the exchange operators do not commute among themselves. We can prove that, in general [Pij , Pkl ] 6= 0 and they will commute if the labels (i, j) both differ from the labels (k, l) and both are identical. Hence, only a limited set of permutation operators can be diagonalized simultaneously. Due to noncommutivity of Pij operators, a complete set of common eigenfunctions of all Pij cannot exist. With reference to Eqs. (18.17) it must be noted that one can speak about the states of separate particles like φα1 , φα2 , · · · only if the particles do not interact among themselves or weakly interact among themselves. So, Pauli’s exclusion principle can be formulated only for weakly interacting system of identical fermions. Further, the function (18.18) characterizes a state of the system in which different particles are in single-particle states, α1 , α2 , · · · , αN , it is impossible to say just which particle is in each of these states due to the indistinguishability of identical particles.
18.10 BIBLIOGRAPHY [1] A.S. Davydov, Quantum Mechanics. Pergamon Press, Oxford, 1965. [2] L.I. Schiff, Quantum Mechanics. McGraw Hill, New York, 1968. [3] J.J. Sakurai, Modern Quantum Mechanics. Addison–Wesley, New York, 1994.
18.11 EXERCISES 18.1 Prove the following properties of the permutation operator P . (a) If ψ(1, 2, · · · , N ) is an eigenfunction of H with an eigenvalue E then P ψ is also an eigenstate of H with the same eigenvalue. (b) hφ|ψi = hP φ|P ψi. (c) hP ψi |O|P ψj i = hψi |O|ψj i.
462 Quantum Mechanics I: The Fundamentals 18.2 Does the operator Pij defined through Pij ψ(1, 2, · · · , i, · · · , j, · · · , N ) = ψ(1, 2, · · · , j, · · · , i, · · · , N )
a unitary operator?
18.3 Show that the symmetric characters of the wave function does not change with time. 18.4 Find whether [Pij , Pkl ] is zero or not if any of the labels (i, j) do not differ from the labels (k, l). 18.5 The S3 permutation group contains six permutation operators. Find the two-dimensional representation of S3 in the basis |ψ1 i =
|ψ2 i =
1 √ [2φα1 (1)φα2 (2)φα3 (3) + 2φα2 (1)φα1 (2)φα3 (3) 12 −φα1 (1)φα3 (2)φα2 (3) − φα3 (1)φα2 (2)φα1 (3) −φα3 (1)φα1 (2)φα2 (3) − φα2 (1)φα3 (2)φα1 (3)] 1 √ [φα3 (1)φα2 (2)φα1 (3) + φα3 (1)φα1 (2)φα2 (3) 2 −φα1 (1)φα3 (2)φα2 (3) − φα2 (1)φα3 (2)φα1 (3)] .
18.6 Determine the nature of the wave function obtained by applying (i) the symmetrising and (ii) the antisymmetrising operators to an arbitrary linear combination of ψ(1, 2) and ψ(2, 1). 18.7 Show that the exchange operator for the spin can be given as 1 4 P12 = I + 2 S1 · S2 . 2 ~ 18.8 The spatial eigenfunctions and eigenvalues of a particle confined in a one-dimensional box located between x = −a and x = a are given by 2 2 2 1 (2n + 1)πx 1 ~ π , En(e) = n + , n = 0, 1, 2, · · · ψn(e) = √ cos a 2a 2 2ma2 1 nπx n2 ~2 π 2 ψn(o) = √ sin , En(o) = , n = 1, 2, · · · a a 2ma2 Determine the lowest three energy levels and the corresponding eigenfunctions if there are two identical particles in the box of (i) spin-0 bosons and (ii) spin-1/2 fermions each in spin-up state α.
CHAPTER
19
Relativistic Quantum Theory
19.1 INTRODUCTION Most of the applications of quantum mechanics are in the realm of the three phases of matters: gases, liquids and solids. In these cases a nonrelativistic quantum theory is sufficient. However, in principle, a strictly correct treatment should include relativity. For relativistic particles, that is, for particles travelling at speeds close to the speed of light, the nonrelativistic Schrödinger equation is not applicable. Oskar Klein and Walter Gordon1 proposed a relativistic wave equation for a free particle. Their equation fell into dispute because the probability density function is not positive definite, thereby preventing the probability interpretation. In 1928 Paul Adrian Maurice Dirac (shared Nobel Prize with Erwin Schrödinger in 1933) introduced a wave equation for a relativistic electron. His equation is considered as one of the highest achievements of 20th century physics. The Dirac equation led to a positive definite probability density function, thereby allowing a probability interpretation as in the nonrelativistic Schrödinger theory. It accounted the spin of the electron. That is, a relativistic electron has the spin property introduced by Goudsmit and Uhlenbeck [1] in order to explain anomalous Zeeman effect. Further, the Dirac equation accounted gyromagnetic ratio, magnetic moment and provided satisfactory explanation for the fine structure of the hydrogen atom. However, the equation predicted the existence of negative energy states which implied that matter was unstable. There are three unusual features of the negative energy states: 1. The hydrogen atom spectrum obtained from the Dirac equation has no ground state. 1 Walter Gordon was born on August 13, 1893 Apolda, Germany and died on December 24, 1939 Stockholm, Sweden.
463
464 Quantum Mechanics I: The Fundamentals 2. Zitterbewegung (trembling motion): The effect is that the expectation value of the position of a free particle displays rapid oscillatory motion, that is, the velocity is not a constant. 3. Klein paradox : For a certain range of incident energy the amplitude of the reflected wave from a step potential is larger than the incident wave. In this chapter we introduce the relativistic Klein–Gordon and Dirac equations and present their various features in detail.
19.2 KLEIN–GORDON EQUATION For simplicity, we start with a relativistic free particle.
19.2.1
Construction of Klein–Gordon Equation
The nonrelativistic Hamiltonian H = p2 /(2m) is not valid for relativistic case. The above H is generalized by considering E 2 = p2 c2 + m2 c4 where E is the total energy, m is the rest mass and p is the momentum. Then p H = E = p2 c 2 + m 2 c 4 . (19.1)
This is a classical result. The relativistic wave equation is obtained by replacing H and p by their corresponding operators in Hψ = Eψ. This gives i~
1/2 ∂ψ = −~2 c2 ∇2 + m2 c4 ψ. ∂t
(19.2)
Right-side of Eq. (19.2) contains a square-root of an operator. We can avoid the square-root by expanding it in a series which will then contain number of terms of even powers of ∇ making the equation difficult to solve. So, consider the squared Hamiltonian and the equation H 2 ψ = E 2 ψ. That is, −~2
∂2ψ = −~2 c2 ∇2 ψ + m2 c4 ψ . ∂t2
The above equation is rewritten as m2 c2 2 − 2 ψ=0, ~
2
= ∇2 −
1 ∂2 , c2 ∂t2
(19.3)
(19.4)
2
where is known as D’Alembertian operator. Equation (19.4) was proposed by Oskar Klein and Walter Gordon [2] and is known as Klein–Gordon (KG) equation. For a particle in a potential V (x) the KG equation is d2 ψ 1 + 2 2 (E − V (x))2 − m2 c4 ψ = 0 . dx2 c ~
(19.5a)
Relativistic Quantum Theory 465 In the Schrödinger equation form the above equation is written as d2 ψ + (Eeff − Veff ) ψ = 0 , dx2
(19.5b)
where Eeff = (E 2 − m2 c4 )/(c2 ~2 ), Veff = (2EV (x) − V 2 (x))/(c2 ~2 ). Since 2 , the KG Eq. (19.4) in covariant form takes the form ∂ µ ∂µ = m2 c2 µ ∂ ∂µ − 2 ψ=0. (19.6) ~ Only four-component potentials can be added to the KG equation as it will keep the equation in covariant form. What is the probability density P in relativistic case? We calculate it in the next subsection. The KG equation without mass corresponds to elastic stretching of an isolated vortex filament in the ideal fluid. The wave function has the meaning of the curve’s position vector. The mass part of the KG equation describes the rotation of the helical curve about the screw axis due to the hydrodynamic self-inductor of the bent vortex filament [3].
19.2.2
Continuity Equation – Probability and Current Densities
Multiplication of Eq. (19.4) by ψ ∗ from left gives m2 c2 2 ψ∗ − 2 ψ=0. ~
(19.7)
Multiplication of complex conjugate of Eq. (19.4) by ψ from right gives
2
m2 c2 ψ∗ ψ − 2 ψ∗ ψ = 0 . ~
Subtracting Eq. (19.8) from Eq. (19.7) we obtain 2 1 ∂ 2 ψ∗ ∗∂ ψ ψ ∗ ∇2 ψ − ∇2 ψ ∗ ψ + 2 ψ − ψ =0. c ∂t2 ∂t2
(19.8)
(19.9)
Using ∇ · (ψ ∗ ∇ψ) − ∇ · (ψ∇ψ ∗ ) = ψ ∗ ∇2 ψ − ψ∇2 ψ ∗ Eq. (19.9) is written as ∗ ∂ ~ ∂ψ ~ ∗ ∂ψ ∇· (ψ ∗ ∇ψ − ψ∇ψ ∗ ) + ψ − ψ = 0 , (19.10) 2mi ∂t 2mc2 i ∂t ∂t where the term ~/(2mi) is introduced for convenience. Defining ~ ∂ψ ∗ 1 ∗ ∂ψ ρ= ψ − ψ = Eψ ∗ ψ 2mc2 i ∂t ∂t mc2
(19.11a)
and J=
~ [ψ ∗ ∇ψ − ψ∇ψ ∗ ] 2mi
(19.11b)
466 Quantum Mechanics I: The Fundamentals Eq. (19.10) becomes the continuity equation ∂ρ +∇·J= 0 . ∂t
(19.12)
Relativistic J coincides with the nonrelativistic J given by, Eq. (2.38). However, ρ is different. ρ of nonrelativistic particle is positive definite at all points for all times. But ρ given by the Eq. (19.11a) is negative if E < 0. Since the values of ψ and ∂ψ/∂t at a given time can be arbitrary, ρ defined by Eq. (19.11a) can be positive, negative or zero. It is thus impossible to interpret ρ as the probability density of finding the particles at a point. In the relativistic quantum mechanics, a particle cannot be localized in space within a volume of dimension less than ~/(4mc), since otherwise the energy of the particle p2 /(2m) will by virtue of the Heisenberg uncertainty principle be larger than 2mc2 , which would suffice for the production of a pair of particles. So, in relativistic quantum mechanics, we cannot speak of a single particle theory. The idea of a single particle cannot be retained in relativistic quantum mechanics if there are interactions which lead to creation and annihilation of pair of particles. Therefore, the number of particles is not conserved in the relativistic theory. It is thus impossible to trace at high energies (relativistic region) the motion of a single particle. The total charge is however conserved. So, ρ given in Eq. (19.11a) is multiplied by the electric charge e of the particle and is interpreted as volume charge current density. Multiplying (19.11b) with e will give the electric current density. Then Eq. (19.12) will represent the continuity equation for charge R density. In this case Eq. (19.12) gives the conservation of total charge as ρ d3 r = constant.
Solved Problem 1:
Discuss the nonrelativistic limit of KG equation. We go over from the relativistic KG Eq. (19.4) to the nonrelativistic Schrödinger equation through the unitary transformation Ψ(r, t) = ψ(r, t)e−imc
2
t/~
.
(19.13)
In the nonrelativistic case the total energy is nearly equal to mc2 . Hence E = E ′ + mc2 where E ′ ≪ mc2 is the nonrelativistic energy of the particle. Therefore, E ′ ψ = |i~∂ψ/∂t| ≪ mc2 ψ. We can thus write 2 ∂Ψ ∂ψ imc2 imc2 −imc2 t/~ = − ψ e−imc t/~ ≈ − ψe (19.14) ∂t ∂t ~ ~ and ∂2Ψ ∂t2
≈
−
2 2imc2 ∂ψ m2 c4 + 2 ψ e−imc t/~ . ~ ∂t ~
(19.15)
Using (19.13) and (19.14) in Eq. (19.4) we get the the nonrelativistic
Relativistic Quantum Theory 467 Schrödinger equation for a free particle. Substituting Eq. (19.14) in (19.11a) we get ρ = ψ ∗ ψ which is the probability density corresponding to the nonrelativistic Schrödinger equation. Substituting (19.14) in (19.11b) gives the nonrelativistic current density.
19.2.3
Spin of Klein–Gordon Equation
To know the spin of the particles described by KG equation we must consider the transformation laws P for the wavePfunction under the original coordinate transformation x′µ = ν Λµν xν and µ Λµν Λµν ′ = δνν ′ with µ, ν = 1, 2, 3, 4. This transformation does not change the length of a four-vector and corresponds to a rotation in three-dimensional space, a proper Lorentz transformation, or an inversion. According to the special theory of relativity, the relativistic wave equation must retain its form under the above transformation. Hence, from the covariant form of the KG equation it follows that the wave function can be multiplied only by a factor with modulus unity. The wave function of Eq. (19.6) thus transform as ψ(x) → ψ ′ (x′ ) = λψ(x) with |λ| = 1. λ = 1 if the transformation corresponds to proper rotations (continuous transformations). Application of r → r′ = −r, t′ = t twice will give identical transformation. Hence, λ2 = 1, λ = ±1. For λ = 1, ψ(r′ , t′ ) = ψ ′ (−r, t) = ψ(r, t) and ψ transforms like a scalar. For λ = −1, ψ(r′ , t′ ) = ψ ′ (−r, t) = −ψ(r, t) and ψ transforms like a pseudo-scalar. So, the KG wave functions transform like a scalar or a pseudo-scalar under orthogonal transformation or proper Lorentz transformation. The spin of a particle is determined by the transformation characteristics of wave functions under coordinate transformations. If a wave function transforms like a scalar then it describes a particle of spin zero. So, the KG equation describes particles of zero spin like pions and kaons.
19.2.4
Plane Wave Solution
Assume the solution of Eq. (19.4) as ψ(r, t) = f (t)eik·r . Substituting it in Eq. (19.4) leads to the following differential equation for f : f ′′ + ω 2 f = 0,
ω 2 = k 2 c2 + (m2 c4 /~2 ) .
(19.16)
Using E = ~ω, ~k = p, in the second equation of (19.16) we obtain E = ± p2 c 2 + m 2 c 4
1/2
.
(19.17)
Note that, E can be negative. Solution of Eq. (19.16) is f (t) = e±iωt . Hence ψ(r, t) = ei(k·r±ωt) . From this solution we write the wave functions of two kinds for well-defined values of momentum as ψ± = ei(k·r±ωt) = ei(p·r±Et)/~ .
(19.18)
468 Quantum Mechanics I: The Fundamentals Substituting (19.18) in the charge density expression we get ∗ ∂ψ± ie~ eE ∗ ∗ ∗ ∂ψ± ρ± = ψ± − ψ± = ψ ψ . 2 2mc ∂t ∂t mc2 ± ±
(19.19)
Thus, the solutions of the type ψ+ correspond to free motion of a particle with momentum p and charge e while solutions of the type ψ− correspond to the free motion of a particle with the opposite sign of the charge.
19.2.5
Inadequacies of KG Equation
When the KG equation was reported many physicists thought that the problem of relativistic quantum theory of electron was solved. But Dirac did not think so and very much worried with the negative probability. Some of the inadequacies of the KG equation are: 1. There appears negative energy states. For a free particle E = K.E + P.E = K.E = 12 mv 2 > 0; E cannot be negative. If E < 0 then P = Eψ ∗ ψ/(mc2 ) < 0 but we require 0 ≤ P ≤ 1. 2. Equation (19.4) is second-order in time. This equation can be solved provided if ψ and ∂ψ/∂t are known at t = 0. In nonrelativistic case ψ at t = 0 is sufficient to solve the time-dependent Schrödinger equation.
19.3 DIRAC EQUATION FOR A FREE PARTICLE Dirac set out in seeking for an equation of first-order in t, x, y, z.
19.3.1
Dirac Hamiltonian and Dirac Matrices
Dirac in 1928 proposed the following Hamiltonian for a free particle [4]: E = H = c α · p + βmc2 ,
(19.20)
where α and β are to be determined. The corresponding wave equation is i~
∂ψ = −i~c (α · ∇) ψ + βmc2 ψ ∂t
(19.21)
which is first-order in space and time. In connection with the Hamiltonian given by Eq. (19.20), Dirac made the following two assumptions: 1. For a free particle H should be independent of space and time coordinates; otherwise such terms indicate the presence of forces. Therefore, α and β should not depend on time. 2. H is linear in p. This implies that α and β must be independent of p. They should also be independent of space and time coordinates in order to describe a free particle.
Relativistic Quantum Theory 469 The E given by Eq. (19.20) must satisfy the relativistic energy-momentum relationship E 2 = p2 c2 + m2 c4 . Denoting α1 , α2 and α3 for αx , αy and αz respectively, Eq. (19.20) takes the form E=c
3 X
αi pi + βmc2 .
(19.22)
i=1
Squaring Eq. (19.22) we get X XX E 2 = c2 α2i p2i + c2 (αi αj + αj αi ) pi pj i
+mc
3
X
i,i6=j
j
(αi β + βαi ) pi + m2 c4 β 2 .
(19.23)
i
Comparison of E 2 = p2 c2 + m2 c4 and (19.23) gives αi αj + αj αi = 0, i 6= j,
αi β + βαi = 0,
α2i = β 2 = 1 .
(19.24)
If αi and αj are nonzero numbers then αi αj + αj αi = 0, i 6= j is not true as numbers do not anticommute. Similarly, from αi β + βαi = 0 it is clear that β is not a number. However, they can be matrices. Since α2i = β 2 = I, the eigenvalues of these matrices are ±1. We write αi = αi I = αi β 2 = αi ββ. Using the relation αi β + βαi = 0 we have αi = −βαi β. Taking trace on both sides of αi = −βαi β we get trαi = −trβαi β. Since the trace of ABC is same as that of BCA, we get trαi = trαi β 2 = −trαi which implies trαi = 0. Then because the trace is the sum of all the eigenvalues, the numbers +1 and −1 must occur same number of times. Hence, the dimension of the matrices αi and β must be even. If n = 2 then the number of possible anticommuting matrices are found to be only 3. They are the Pauli matrices: 0 1 0 −i 1 0 , σ2 = , σ3 = . σ1 = (19.25) 1 0 i 0 0 −1
But we require four matrices α1 , α2 , α3 and β. When n = 4 there are four matrices and are Hermitian) satisfying the relations (19.24). They are 0 σ1 0 σ2 , α2 = , α1 = σ1 0 σ2 0 0 σ3 I 0 , β = . α3 = (19.26) σ3 0 0 −I
470 Quantum Mechanics I: The Fundamentals
19.3.2
Dirac Equation in Covariant Form
Rewrite the Eq. (19.21) as 3
X ∂ψ ∂ψ + i~c αk − βmc2 ψ = 0 ∂t ∂xk
(19.27)
X ∂ψ ∂ψ mc − iβαk ψ=0. + ∂(ict) ∂xk ~
(19.28)
i~
k=1
or β
Define x4 = ict, γ4 = β, γk = −iβαk , k = 1, 2, 3. Then Eq. (19.28) becomes 3
γ4
X ∂ψ ∂ψ mc + γk + ψ=0. ∂x4 ∂xk ~ k=1
That is, 4 X
k=1
γk
∂ψ mc + ψ=0. ∂xk ~
(19.29)
The above equation is called covariant form of Dirac equation. In Einstein summation conversion form the above equation is written as ∂ mc γk + ψ=0. (19.30) ∂xk ~ We can easily verify that γ matrices are Hermitian, γ 2 = I, γi γj + γj γi = 2δij and Trγi = 0.
Solved Problem 2: Let U be the transformation matrix that transforms the wave function ψ in the S frame to the wave function ψ ′ in the S ′ frame. Then show that the condition for the relativistic invariance of the Dirac equation is U −1 γk U Lkµ = γµ where Lkµ is the Lorentz transformation matrix. Consider the Dirac Eq. (19.30). We write ψ ′ = U ψ. Since the γ matrices are c-numbers, they remain unchanged in this transformation. We can write ∂ ∂ ∂xµ ∂ = = Lkµ . ∂x′k ∂xµ ∂x′k ∂xµ We have the Dirac equation in S ′ frame as ∂ mc γk ′ + ψ′ = 0 . ∂xk ~ Substituting ψ ′ = U ψ and (19.31) in (19.32) we get ∂ mc γk Lkµ + Uψ = 0 . ∂xµ ~
(19.31)
(19.32)
(19.33)
Relativistic Quantum Theory 471 Multiplying this equation from the left with U −1 , we get ∂ mc U −1 γk U Lkµ ψ=0. + ∂xµ ~
(19.34)
The Lorentz invariance requires that Eq. (19.34) must be same as Eq. (19.30). This is true if U −1 γk U Lkµ = γµ .
19.3.3
Probability and Current Densities
Consider the Dirac equation i~
∂ψ = −i~c α · ∇ψ + mc2 βψ . ∂t
(19.35)
Multiplication of Eq. (19.35) by ψ † (ψ is a column vector) from left gives i~ψ †
∂ψ = −i~c ψ † α · ∇ψ + mc2 ψ † βψ . ∂t
(19.36)
Taking the Hermitian adjoint of Eq. (19.35) and then multiplying it by ψ from right gives ∂ψ † ψ = i~c ∇ψ † · αψ + mc2 ψ † βψ . (19.37) −i~ ∂t Subtraction of Eq. (19.37) from Eq. (19.36) yields ∂ ψ † ψ + c ∇ · ψ † αψ = 0 . ∂t
(19.38)
Defining ρ = ψ † ψ and J = c ψ † αψ, Eq. (19.38) takes the form ∂ρ +∇·J =0 ∂t
(19.39)
which is the continuity equation. Since ρ is nonnegativeR the Dirac equation has a positive definite probability density. We note that ψ † ψ d3 r is conserved. As the Dirac equation is covariant and the derivatives in the Eq. (19.39) form a four-vector, the probability and current densities form a four-vector: J k = (J, icρ). Then the covariant form of Eq. (19.39) is ∂k J k = 0.
19.3.4
Plane Wave Solution
Consider the Dirac wave equation for a free particle in the form c α · p + βmc2 ψ = Eψ .
(19.40)
Since α and β are 4 × 4 matrices, the Eq. (19.40) has meaning only when ψ is a four-component function or represented by a 4 × 1 matrix. Let us seek a plane wave solution of the form ψ(r, t) = N u(p)ei(p·r−Et)/~ ,
(19.41)
472 Quantum Mechanics I: The Fundamentals where N is the normalization constant and u is linear in p. Substitution of Eq. (19.41) in Eq. (19.40) gives (19.42) c α · p + βmc2 u(p) = Eu(p) .
Since α and β are 4 × 4 matrices u will have four components and is called spinor Therefore, we write u1 u 2 u= (19.43) u3 u4 and partition it into u=
v w
,
v=
u1 u2
,
w=
Using Eqs. (19.26) the Eq. (19.42) becomes v I 0 v 0 − mc2 = c E w 0 −I w σ·p From the above equation we obtain E − mc2 v E + mc2 w
= c (σ · p) w ,
= c (σ · p) v .
u3 u4
.
σ·p 0
(19.44)
v w
.
(19.45) (19.46a) (19.46b)
When p is replaced by −i~∇ Eqs. (19.46) become a set of coupled first-order differential equations for the unknowns v and w. In the following we obtain the energy spectrum and eigenfunctions u(p) from Eqs. (19.46).
19.3.5
Energy Spectrum
Multiplication of Eq. (19.46a) by (E + mc2 ) and replacing (E + mc2 )w by the right-side of Eq. (19.46b) we have 2 E 2 − m2 c4 v = c2 (σ · p) v . (19.47)
2 2 Using the identity (σ · A)(σ · B) = A · B + iσ · (A × B), we get (σ · p) = p 2 2 4 2 2 the above equation becomes E − m c − c p v = 0, which gives p (19.48) E± = ± p2 c2 + m2 c4 , E− = −E+ .
When |p| = 0 we have E− = −mc2 and E+ = mc2 . As |p| increases from zero, E+ starts from mc2 and extends to ∞. On the other hand, E− begins from
Relativistic Quantum Theory 473 −mc2 and goes to −∞ as |p| → ∞. Thus, there are positive and negative energy regions separated by an energy gap 2mc2 . E− to E+ is a forbidden energy gap of width 2mc2 . Since the energy spectra extends to −∞ there is no ground state.
19.3.6
Determination of Eigenfunctions u(p)
Equations (19.46) are two-coupled linear equations and their solutions are v and w. To solve them, first, decide whether E+ or E− is to be used in Eqs. (19.46). When E is E+ = mc2 then left-side of Eq. (19.46a) becomes zero. Therefore, choose E = E− for Eq. (19.46a). Similarly, the choice E = E+ for Eq. (19.46b) is useful. From Eqs. (19.46) for negative energy states cσ · p w, E+ + mc2
(19.49a)
cσ · p v. E+ + mc2
(19.49b)
v=− and for positive energy states w=
Earlier, we assumed that u is linear in p. In Eq. (19.49b) if v is a function of p then w will contain nonlinear terms (that is, power or degree of p is greater than 1) and consequently u will also have nonlinear terms in p. Therefore, v’s are treated independent of p and can be a constant. Similarly, in Eq. (19.49a) w’s are constant. We choose the constants as 0, 1 taking note of the presence of the normalization constant N . There are four possible v’s and are 0 0 1 1 , , , . 0 1 0 1
Since
as
0 1
0 0
leads to trivial solution and
and
1 0
1 1
is a linear combination of
we consider the remaining two choices. σ · p is computed
σ · p = σx px + σy py + σz pz =
pz
p−
p+
−pz
where p± = px ± ipy . Now, Eq. (19.49b) becomes pz p− c v . w= E+ + mc2 p+ −pz
,
(19.50)
(19.51)
474 Quantum Mechanics I: The Fundamentals
Using v = v
v
0 1
and
1 0
the following two solutions are obtained:
p− v c , w = , u = .(19.52a) = E+ + mc2 1 −pz w pz v 1 c , u = . (19.52b) , w = = E+ + mc2 p+ w 0 0
Similarly, for the choices w = w
=
w
=
0 1
and
1 0
we get two more solutions
p− v c , v = − , u = .(19.52c) E+ + mc2 1 −pz w pz v 1 c , u = . (19.52d) , v = − E+ + mc2 p+ w 0 0
The solutions (19.52a) and (19.52b) are positive energy solutions while (19.52c) and (19.52d) are negative energy solutions. Totally, there are four linearly independent solutions. We rewrite the Eq. (19.52a) as u1 = 0, u2 = 1, u3 =
c c p− , u 4 = − pz . E+ + mc2 E+ + mc2
(19.53)
Similarly, we express the Eqs. (19.52b-d) in terms of u1 , u2 , u3 and u4 . The fours solutions given by Eqs. (19.52) are normalized using the condition ψ † ψ = 1 (why is this form? ). For the solution given by Eq. (19.53) we obtain 1
= |N |2 [u∗1 u1 + u∗2 u2 + u∗3 u3 + u∗4 u4 ] c 2 p2 = |N |2 1 + . (E+ + mc2 )2
Therefore, the normalization constant is
c 2 p2 N = 1+ (E+ + mc2 )2
−1/2
.
(19.54)
Remember that the four solutions are solutions of Dirac equation for a free particle. We may ask: How are the negative energy solutions are to be understood and what is the reason for the existence of two positive (or negative)
Relativistic Quantum Theory 475 energy solutions? The answer to the second question has been provided by showing that the Dirac equation has spin-1/2 (such as the electron) so that there are two independent spin orientations for given momentum and energy. The wave function ψ of Dirac equation has four components while that of the KG equation has only one component. That is, for Dirac equation more degrees of freedom are available than the required degrees of freedom to describe the translatory motion. Since the additional components are absent for the KG equation, we say that it describes particles of spin zero like π-mesons. For a free particle we found the energy states which are nondegenerate and there is a forbidden energy range including the value E = 0. Now, for a particle in a potential V we ask: What is the situation for a particle in a potential V ? Does degeneracy possible in nonzero potential? Does E = 0 allowed for a scalar (that is time-independent) potential? Can we observe energy level crossing? We answer these important questions in latter sections.
Solved Problem 3: Construct the plane wave solution of Dirac’s free electron in one-dimension. In one-dimension the Dirac equation Eψ = (cαx px + βmc2 )ψ is rewritten as 0 σx I 0 ∂ψ ∂ψ ψ = 0 . i~ + i~c − mc2 (19.55) ∂t ∂x σx 0 0 −I Writing ψ in four-components form we have ∂ψ1 ∂t ∂ψ2 i~ ∂t ∂ψ3 i~ ∂t ∂ψ4 i~ ∂t i~
∂ψ4 ∂x ∂ψ3 + i~c ∂x ∂ψ2 + i~c ∂x ∂ψ1 + i~c ∂x + i~c
− mc2 ψ1
= 0,
(19.56a)
− mc2 ψ2
= 0,
(19.56b)
+ mc2 ψ3
= 0,
(19.56c)
+ mc2 ψ4
= 0.
(19.56d)
An interesting observation is that the equations for the four components of ψ are coupled in pairs only; ψ1 with ψ4 ; ψ2 with ψ3 . This suggests that we
can introduce the two component spinor
ψu
where ψu = ψ1 or ψ2 and ψl ψl = ψ4 or ψ3 with u and l stand for ‘upper’ and ‘lower’ respectively. For stationary states Eqs. (19.56) become (19.57a) i~c ψ1′ + E + mc2 ψ4 = 0, i~c ψ2′ + E + mc2 ψ3 = 0, ′ 2 ′ 2 i~c ψ3 + E − mc ψ2 = 0, i~c ψ4 + E − mc ψ1 = 0 , (19.57b)
476 Quantum Mechanics I: The Fundamentals where we have written the last equation of (19.56) as first equation and so on. Equations (19.57a) and (19.57b) are rewritten in single equation as (19.58a) i~c ψu′ + E + mc2 ψl = 0 . Similarly, Eqs. (19.57c) and (19.57d) are written as i~c ψl′ + E − mc2 ψu = 0 .
(19.58b)
For simplicity we write ψu as v and ψl as w so that Eqs. (19.58) read as (19.59a) i~c v ′ + E + mc2 w = 0 , ′ 2 i~c w + E − mc v = 0 . (19.59b) Differentiating (19.59a) with respect to x and using Eq. (19.59b) we get p2 v=0, ~2
1 E 2 − m2 c4 . (19.60) 2 c Then w is given by w = −(i~c)v ′ / E + mc2 . Now, we need to determine v only. The general solution of Eq. (19.60) is v(x) = Aeipx/~ + Be−ipx/~ . v ′′ +
p2 =
19.4 NEGATIVE ENERGY STATES We have seen that the Dirac’s theory admits both positive and negative energy solutions. The energy eigenvalues of the plane wave solutions are 1/2 E+ = + c2 p2 + m2 c4 ≥ mc2 , (19.61a) 1/2 E− = − c2 p2 + m2 c4 ≤ −mc2 . (19.61b)
Thus, the energy of a free particle in the Dirac’s theory is continuous and ranges from −∞ to ∞ except for a forbidden gap of width 2mc2 . An energy gap of width 2mc2 exists in the Dirac’s theory of electron. We cannot ignore the negative energy solutions without giving a physical interpretation. As electron can undergo a transition from higher energy states to lower energy states, we must consider the possibility of transitions from E+ to E− energy states. Hence, the negative energy states cannot be regarded as independent of positive energy states. An excited state makes a transition to a lower level by emitting a photon. When it is in its ground state, it cannot make any transition to lower states as there are no further lower energy states. But in Dirac’s theory there is no lower bound as the allowed negative energies range from −mc2 to −∞. So, an electron with a positive energy E+ can make a transition onto the negative energy state E− emitting a γ-ray. Furthermore, once it reaches a negative energy state, it will continuously lower its energy indefinitely by emitting photons since there is no lower bound to the negative energy spectrum. Ultimately there will be no electron in positive energy states at all.
Relativistic Quantum Theory 477 To resolve this difficulty Dirac proposed that all the negative energy states are filled under normal conditions. The catastrophic transitions mentioned above are then disallowed because of the Pauli’s exclusion principle. So, the normal state of the vacuum contains of an infinite density of negative energy electrons. When a negative energy electron from this Fermi sea is excited into the positive energy states, a hole is left in the sea. This hole allows another negative-energy electron to make a transition to it, leaving a hole elsewhere in the sea. In semiconductor physics this hole acts as a positive charged particle. Dirac considered first the particle as proton. But Julius Robert Oppenheimer and Hermann Klaus Hugo Weyl theoretically proved that the particle must have the same mass as electrons. In 1932 Carl David Anderson discovered in a Wilson cloud chamber of cosmic rays a track occurring due to a particle of charge +e and mass equal to that of electron. This was identified as the hole in the Dirac’s theory and is named positron. The absorption of a photon in the Fermi sea of energy in excess of 2mc2 leads to the creation of an electron-positron pair. To ensure conservation of energy and momentum, this event must take place in the presence of a third body carrying away the momentum. A positive energy electron can fall to this hole and fill the Fermi sea. The decay energy of this electron-positron annihilation is carried away in two photons. Note that physically observable positron has positive energy values. For the conservation of momentum and spin, the positive energy positron will have opposite momentum and spin of the negative energy electron states. But the helicity σ ′ · p will not change, since the velocity v = c2 p/E as −E → +E, −p → p will be same for the positive energy positron and negative energy electron. With an infinite sea of electrons, Dirac’s theory is no longer the one-particle theory. A many particle theory of Dirac equation can be obtained using the formulation of quantized field. Even though the energy spectrum of free KG particle is identical to the free Dirac particle, it is not possible to construct a sensible hole theory out of KG particles which obey Bose–Einstein statistics.
19.5 JITTERY MOTION OF A FREE PARTICLE The velocity of a free particle is dX 1 c = [X, H] = − α · pX = c α . dt i~ i~
(19.62)
The time evolution of α is found to be dα 1 1 = [α, H] = c p + αβmc2 − Hα . dt i~ i~
(19.63)
478 Quantum Mechanics I: The Fundamentals Adding and subtracting Hα to the above equation we get dα dt
= = =
1 c p + αβmc2 − Hα + Hα − Hα i~ 1 c p + αβmc2 + c p + βαmc2 − 2Hα i~ 2 (c p − Hα) . i~
(19.64)
Since H and p are time-independent the Eq. (19.64) is written as dα 2iH c p = − +α . dt ~ H
(19.65)
dX c2 p c p = + c e2iHt/~ α(0) − . dt H H
(19.67)
Defining α′ = α − c p/H the above equation becomes dα′ /dt = (2iH/~)α′ whose solution is α′ (t) = α′ (0)e2iHt/~ . Then α(t) is obtained as cp c p α(t) = + e2iHt/~ α(0) − . (19.66) H H Now, Eq. (19.62) gives
Although the momentum p is a constant of motion, that is independent of time, the velocity is not. The solution of Eq. (19.67) is c2 p t ~ 2iHt/~ c2 p X(t) = X(0) + + e −1 c α(0) − . (19.68) H 2iH H By analogy with classical free particle we expect the term linear in t. In Eq. (19.68) the first two terms of the right-side describe the uniform motion and it corresponds to the classical dynamics of a free relativistic particle. The last term is a feature of relativistic quantum mechanics. This term oscillates extremely rapidly with high frequency, ω ∼ mc2 /~, (or period ∼ ~/(mc2 )) and with amplitude ~/(mc) (how?). This high frequency oscillatory or jitterly or trembling motion is called zitterbewegung. The frequency ω is so high that the deviation from the classical mechanics term c2 pt/H ≈ vt is undetectable. This paradox arises from the coexistence of particles and antiparticles in Dirac’s theory and that only a quantum field theory description predicting pair creation and annihilation could cure this problem. The zitterbewegung has fascinated many physicists [5-7] ever since, because the spin of the electron can be associated with the orbital angular momentum of the internal helical motion. The zitterbewegung amplitude X is equal to the electron Compton wavelength [8] X = c/ω0 = ~/(mc) = 3.8 × 10−13 m and s = XE0 /c = ~/2 and E0 = sω0 = mc2 /2. The last property implies V0 = mc2 /2. Further, since the electron is charged, the zitterbewegung motion generates a magnetic moment
Relativistic Quantum Theory 479 of magnitude µ = ecπX 2 /(2πXc) = eX/2. For more discussion on zitterbewegung one may refer to ref.[8]. Now, we may ask: Does the zitterbewegung occur only for spin-1/2 particles? Can spin-0 particles exhibit such a phenomenon? What is the cause of zitterbewegung? When does it vanish? In order to get the answers for these questions zitterbewegung has been examined for spin-0 particles using the Hamiltonian form of the KG equation [9]. For a spin-0 particle with the KG Hamiltonian p2 H = (σ3 + iσ2 ) + mσ3 (19.69) 2m X is obtained as e2iHt − 1 pt p X(t) = X(0) + + v(0) − . (19.70) H 2iH H
The last term describes the zitterbewegung. Next, we evaluate hX(t)i with a wave packet, which is a superposition of eigenstates of p and H, given by ψkσ (X) =
ωσ eik·X (2π)3/2
(19.71a)
with σ = ±1,
p ω = k 2 + m2 ,
1 m + σω ωσ = √ . 2 mω m − σω
(19.71b)
The states with σ = 1 and −1 havePenergies ω and −ω respectively. Writing R the wave packet in the form |ψi = σ ψkσ i aσk d3 k we obtain X Z k 2 aσ t d3 k hX(t)i = hX(0)i + k ω σ Z k (−)∗ (+)∗ −iωt +2 sin ωt Re a a e d3 k . (19.72) k k ω2
The last term describes oscillatory motion. If the wave packet contains only positive or only negative energy waves then the last term vanishes and hence the absence of zitterbewegung. Thus, the zitterbewegung is caused by the interference between the positive and negative energy states in the wave packet. The zitterbewegung effect has not been observed experimentally for electrons because the corresponding frequency of the free electron is ≈ 1020 Hz. It was recognized that the zitterbewegung effect can appear in solids where the electron at the boundary between two interacting bands obeys the Dirac equation [10-12]. This effect has been studied through simulations in ultra cold atoms [13], superconductivity [14], a single trapped ion [15,16] and graphene [17,18]. An optical analog of the effect was found in a two-dimensional photonic crystal composed by a triangular lattice of the dielectric cylinders where
480 Quantum Mechanics I: The Fundamentals two bands touch each other forming a double-cone structure with a linear dispersion [19,20]. The optical zitterbewegung effect was demonstrated by solving Maxwell’s equations for the electromagnetic pulses propagating through a negative-zero-positive index metamaterial [21].
Solved Problem 4: Show that the zitterbewegung effect can be removed by projecting the standard position operator to the particle and antiparticle subspaces of Hilbert space respectively. We introduce the projection operators P0± =
1 IH + |H|−1 H , 2
|H| = (H ∗ H)1/2
(19.73)
with P0+ P0− = 0 and P0+ + P0− = IH . The time evolution of the projected ± position operators x ˆ± ˆP0± are given by 0 = P0 x 2 −1 x ˆ± ˆ± tP0± . 0 (t) = x 0 (0) + c pH
(19.74)
It exactly corresponds to the classical expression. This is valid for particles and as well as for antiparticles. The interpretation is that particles and antiparticles show different time evolutions and the zitterbewegung represents the interference term P0+ x ˆP0− + P0− xˆP0+ =
i~ −1 ˆ H F 2
(19.75)
which is absent in a classical case.
19.6 SPIN OF A DIRAC PARTICLE In this section we show that for a Dirac particle orbital angular momentum is not conserved but the sum of orbital angular momentum and spin angular momentum is conserved. In the Heisenberg picture the time evolution of an operator is described by the Eq. (7.27). For a Dirac particle with the Hamiltonian (19.20) the time evolution of orbital angular momentum L is given by −i~
d (iL1 + jL2 + kL3 ) = [H, iL1 + jL2 + kL3 ] , dt
(19.76)
where L1 , L2 and L3 represent Lx , Ly and Lz respectively. Consider the component L3 . We write " 3 # X d −i~ L3 = [H, L3 ] = c αk pk + βmc2 , x1 p2 − x2 p1 . (19.77) dt k=1
Relativistic Quantum Theory 481 Since β is a matrix and x1 p2 and x2 p1 are scalars the quantity [βmc2 , x1 p2 − x2 p1 ] = 0. Therefore, −i~
d L3 dt
i hX i c αk pk , x1 p2 − c αk pk , x2 p1 X X = −i~c αk δk1 p2 + i~c αk δk2 p1
=
hX
= −i~c (α1 p2 − α2 p1 ) = −i~c (α × p)3 .
That is, dL3 /dt = c(α × p)3 . Similarly, dL2 /dt = c(α × p)2 and dL1 /dt = c(α × p)1 . Therefore, d L = c(α × p) (19.78) dt which is not zero. Hence L is not a constant of motion or a conserved quantity. However, it implies that L + A where A is such that dA/dt = −c(α × p) ′ is a conserved quantity. In the following we show that A = ~σ /2 where
σ′ =
σ
0
. 0 σ The equation of motion of σ3′ is simplified into −i~
dσ3′ dt
= [H, σ3′ ] hX i = c αk pk + βmc2 , σ3′
(19.79)
= 2ic (α1 p2 − p1 α2 ) ,
where we have used the result [βmc2 , σ3′ ] = 0, [α3 , σ3′ ] = 0, σ1′ σ3′ = −σ3′ σ1′ and σ1′ σ3′ = −iσ2′ . That is, (~/2)dσ3′ /dt = −c(α × p)3 . Therefore, we write d ~ ′ σ = −c(α × p) . dt 2
(19.80)
Combining Eqs. (19.78) and (19.80) we obtain d ~ ′ L+ σ =0. dt 2
(19.81)
~σ ′ /2 is called spin angular momentum and J = L + (~/2)σ ′ is the total angular momentum. Thus, J is a conserved quantity. Hence, we conclude that the Dirac particle is endowed with spin ~σ ′ /2.
Solved Problem 5:
Show that ~k = β(σ ′ · L + ~) = β(σ ′ · J − ~/2) where σ ′ =
σ
0
0 σ
.
482 Quantum Mechanics I: The Fundamentals We have ~k = β(σ ′ · L + ~), L = J − ~σ/2′ . Hence, ~ ′2 ~ ′ ′ ′ ~k = β σ · J − σ + ~ = β σ · J − |σ | + ~ 2 2 ~ = β σ′ · J − 2 2
2
(19.82)
2
since |σ ′ |2 = σx′ + σy′ + σz′ = 3.
19.7 PARTICLE IN A POTENTIAL In the following we construct the solution of Dirac equation for a particle in a potential V [22]. We consider the Dirac Hamiltonian H
= c α · p + βmc2 + V 0 d 0 −i 1 mc2 + V , = −i~c + dz i 0 0 −1
(19.83)
where α is chosen as α3 . Now, the equation Hψ = Eψ is written as 0 −i 1 0 v v d + mc2 −i~c + V =E . dz i 0 0 −1 w w (19.84) From the above equation we obtain −~c w′ + v mc2 + V = Ev, ~c v ′ − w mc2 + V = Ew, (19.85a) where prime denotes differentiation with respect to z. Suppose there are two bound state solutions v1 v2 , ψ2 = ψ1 = (19.86) w1 w2
with eigenvalues E1 and E2 respectively. Now, the Eqs. (19.85) are −~c w1′ + mc2 + V v1 = E1 w1 , (19.87a) 2 ′ (19.87b) ~c v1 − mc + V w1 = E1 v1 , ′ 2 −~c w2 + mc + V v2 = E2 v2 , (19.87c) ′ 2 ~c v2 − mc + V w2 = E2 w2 . (19.87d)
Multiplying Eq. (19.87a) by v2 and Eq. (19.87c) by v1 and subtracting one from another we get ~c (v1 w2′ − w1′ v2 ) = (E1 − E2 ) v1 v2 .
(19.88)
Relativistic Quantum Theory 483 Similarly, from Eqs. (19.87b) and (19.87d) we have ~c (v1′ w2 − w1 v2′ ) = (E1 − E2 ) w1 w2 .
(19.89)
Combining of the above two equations gives ~c
d (v1 w2 − w1 v2 ) = dz
(E1 − E2 ) ψ1∗ ψ2 .
(19.90)
If E1 = E2 , that is the eigenstates are degenerate, then d (v1 w2 − w1 v2 ) = 0 or v1 w2 − w1 v2 = constant . dz
(19.91)
Because v and w → 0 as |x| → ∞, this constant must be zero. Hence, v1 /w1 = v2 /w2 from which we obtain E + mc2 + V w v′ = . (19.92) v ~c v That is, E + mc2 + V w1 E + mc2 + V w2 v1′ v′ = = = 2 . v1 ~c v1 ~c v2 v2
(19.93)
Equations v1 /w1 = v2 /w2 and (19.92) imply that ψ1 ∝ ψ2 . Thus, they represent the same state. Therefore, the energy states are nondegenerate irrespective of the potential. Further, if E1 6= E2 [22], Z ∞ 1 (v1 w2 − w1 v2 ) |∞ hψ1 |ψ2 i = ψ1∗ ψ2 dz = (19.94) −∞ = 0 . (E − E ) 1 2 −∞ That is, ψ1 and ψ2 are orthogonal. This condition with the nondegeneracy of energy implies that level crossing cannot occur. When the potential is varied smoothly, for example, by varying a parameter in it then the energies and wave functions will also vary smoothly. Here by level crossing we mean that two energies E1 and E2 cross each other. This requires a singular behaviour of hψ1 |ψ2 i and is zero for E1 6= E2 and suddenly jumps to a finite value at E1 = E2 . This is not possible when ψ varies smoothly.
19.8 KLEIN PARADOX The step potential in the relativistic Dirac equation gives rise to the peculiar situation called Klein paradox [23,24,25].
19.8.1
Scattering by Step Potential
Let us consider the step potential, Eq. (5.34) and Fig. 5.7. For x < 0 the Dirac equation is the free particle equation given by Eq. (19.60). For x > 0
484 Quantum Mechanics I: The Fundamentals the corresponding equation is obtained by replacing E in Eq. (19.60) (p is a function of E) by E − V0 . From the solution for x < 0 the solution for x > 0 is obtained by replacing E by E − V0 since V (x) in x > 0 region is a constant. Let us obtain the solution [24]. The solution of Eq. (19.60) is v< (x) = A eipx/~ + Re−ipx/~ . Then (19.95) w< (x) = A aeipx/~ − aRe−ipx/~ , where a = cp/(E + mc2 ). Now, v< 1 1 = A eipx/~ + R e−ipx/~ ψ< = w< a −a h i = A u+ eipx/~ + Ru− e−ipx/~ , (19.96)
where u± =
1 ±a
. For x > 0 there is no reflected wave. Hence
ψ> =
where u ¯ = b
=
1 b
,
v> w>
¯ = Du ¯ eipx/~ ,
p¯ =
(19.97a)
1/2 1 (E − V0 )2 − m2 c4 , c
c p¯ . (E − V0 + mc2 )
(19.97b)
To determine R and D we use the boundary conditions: ψ< (0) = ψ> (0) and ′ ′ ψ< (0) = ψ> (0). These conditions lead to A(u+ + Ru− ) = Du¯. That is, 1 1 1 + R = D . A (19.98) a −a b
We get A(1 + R) = D, Aa(1 − R) = bD. Solving them results in R=
a−b , a+b
T =
2a . a+b
(19.99)
The behaviour of the wave functions given by Eqs. (19.96) and (19.97) depends on V0 . We consider the following cases: 1. E > V0 + mc2 .
Relativistic Quantum Theory 485 2. V0 − mc2 < E < V0 + mc2 . 3. E < V0 − mc2 .
Case 1: E > V0 + mc2 In this case E 2 > m2 c4 + V02 + 2mc2 V0 . Then E 2 − m2 c4 > 0. That is, p2 c2 = E 2 − m2 c4 is positive and hence p is real. Similarly, E − V0 > mc2 or (E − V0 )2 > m2 c4 or p¯ = [(E − V0 )2 − m2 c4 ]1/2 > 0. So p¯ is real. Then in Eq. (19.96) eipx/~ and e−ipx/~ represent incident wave moving towards positive x-direction and reflected wave moving towards negative x-direction ¯ respectively. eipx/~ in Eq. (19.97) represents transmitted wave, moving in the positive x-direction in region-II (x > 0). There is no reflected wave in regionII. This behaviour is similar to the nonrelativistic Schrödinger equation for E > V0 .
Case 2: V0 − mc2 < E < V0 + mc2
In this case (E − V0 )2 < m2 c4 . Hence, p¯ in Eq. (19.97b) is imaginary. Then ¯ ψ> (x) = AT u ¯ + e−px/~ ,
i1/2 h p¯ = (E − V0 )2 − m2 c4 .
(19.100)
The wave in the region-II is an exponentially decaying wave. In the region-I, we have the right-moving incoming and the left-moving reflected wave. This situation is also similar to the nonrelativistic case for 0 < E < V0 .
Case 3: E < V0 − mc2
We have E − V0 < −mc2 and E − V0 negative. Further, (E − V0 )2 > m2 c4 and hence p¯ is real leading to an oscillatory solution in region-II (the solution in the region-I is oscillatory). In the nonrelativistic case, for E < V0 , no such solution is obtained. Furthermore , E − V0 + mc2 is negative and thus b is negative. We write a − b a + |b| |R| = (19.101) = >1. a + b a − |b|
The paradoxical result is that |R| > 1. The amplitude of the reflected wave is larger than the incident wave! So, more particles are reflected by the step than the incident on it. This result was first reported by Oskar Klein. The above situation is called the Klein paradox . In the nonrelativistic case, for E < V0 , there is no oscillatory solution in the potential region (x > 0) and R is unity. In contrast, in the relativistic case, even when E is much smaller than V0 we have oscillatory waves in region-II. The occurrence of Klein paradox has been verified in a numerical simulation [26].
19.8.2
Interpretation of Klein Paradox in Terms of the Hole Theory
The resolution of the problem is associated with pair production [27]. The potential is strong enough to generate particle-antiparticle pairs from vacuum.
486 Quantum Mechanics I: The Fundamentals
V0 + mc2 V0 V0 - mc2
I E mc2 0 -mc2
FIGURE 19.1
II
Positive and negative energy in a step potential.
That is, multiparticle phenomenon is interpreted using a simple single particle wave function. As shown in Fig. 19.1 if E + mc2 < V0 then the positive energy continuum in region-I contacts with the negative energy continuum in regionII at the boundary. At the boundary, incident electrons stimulate the vacuum in the region-II, that is, filled by the negative energy electrons. As a result, negative energy electrons are ejected into region-I (being observed as positive energy electrons). Holes remaining in region-II are interpreted as positrons. The incident electrons induce pair creation. Here the reflected current becomes larger than the incident current because it consists of reflected electrons and created ones. The current in the region-II is regarded as the flow of positrons. The Klein paradox is treated as the simplest model of the decay of the charge vacuum in a field creating electron-positron pairs at the potential barrier [28]. Discussion on Klein paradox using Feynman’s picture of antiparticles as negative energy solutions travelling backward in time is presented in ref.[24]. Klein paradox was not found for bosons [29].
19.8.3
Further Works on Klein Paradox
Sauter [30] calculated the transmission coefficient for the potential 0 a the reflection and transmission coefficients are obtained as [31] R = T where k2 =
=
(1 − k)2 sin2 2pa , 4k 2 + (1 − k 2 )2 sin2 2pa 4k 2 , 4k 2 + (1 − k 2 )2 sin2 2pa
V − E + mc2 E + mc2 . (V − E − mc2 ) (E − mc2 )
(19.103a) (19.103b)
(19.103c)
R = 0, T = 1 when 2pa = nπ where n is an integer. The electron passes through the barrier without reflection and is called a transmission resonance. As a becomes very large for fixed mc2 , E and V the quantity pa becomes very large and sin pa oscillates very rapidly. In this case we average over the phase angle pa using sin2 pa = cos2 pa = 1/2 and obtain R∞ =
(1 − k 2 )2 , + (1 − k 2 )
8k 2
T∞ =
8k 2
8k 2 . + (1 − k 2 )2
(19.104)
For a study on relativistic quantum scattering by locally periodic potentials see ref.[32]. Scattering of a low momentum Dirac particle by an arbitrary potential of finite range is also studied [33]. It has been shown that the amplitude for reflection is 1 and the transmission amplitude is 0 in general. If the potential supports a half-bound state at momentum k = 0, this result does not hold. In the case of an asymmetric potential the transmission coefficient is found to be nonzero and is 1 for a symmetric potential. For discussions on relativistic linear harmonic oscillator one may refer to refs.[34-37]. It has been shown that the Klein paradoxial behaviour can be tested experimentally using electrostatic barriers in single- and bi-layer graphene [38]. For a quantum simulation of Klein paradox one may refer to the ref.[39].
19.9 RELATIVISTIC PARTICLE IN A BOX In sec. 4.6 the problem of a nonrelativistic particle in a box potential is solved exactly. For the box of width L described by the equation 0, 0 0 + N (a, −b, −c) + N (a, b, −c) .
(20.17)
We note that N (a, −b, −c) + N (a, b, −c) = N (a, −c) ,
N (a, −b, c) + N (a, −b, −c) = N (a, −b) , N (−a, b, −c) + N (a, b, −c) = N (b, −c) .
(20.18a) (20.18b) (20.18c)
Using (20.18) in (20.17) we observe N (a, −b) + N (b, −c) ≥ N (a, −c) .
(20.19)
For presentations of Bell’s theorem without inequalities see refs.[24,25]. It has been shown that the prediction of quantum mechanics violates this inequality. We present another version of Bell’s inequality which we use in the next section for a quantum prediction. Suppose we consider an experiment in which object 1 is picked randomly and a second object 2 is picked with each property opposite of object 1. P (a, −b) for object 1 is equal to the probability that object 1 has property a and the object 2 has the property b (that is, opposite to −b). We designate it as P (1a, 2b). Then Eq. (20.19) becomes P (1a, 2b) + P (1b, 2c) ≥ P (1a, 2c) .
(20.20)
20.7 VIOLATION OF BELL’S THEOREM Let us examine the Bell’s theorem, the inequality given by Eq. (20.20), in quantum mechanics [3]. 4 Bell’s theorem was called as the most profound discovery of science by Stapp [H.P. Stapp, Nuovo Cimento 29B:271, 1975][3].
Mysteries in Quantum Mechanics 511
z θ1
y
θ2
x
' P2
2
S
P'1
1
FIGURE 20.4 Measuring the z-component of spin of two atoms. The polarization of the two polarizers is the same. Each polarizer Pi′ has been rotated about the y-axis by an angle θi . (Reproduced with permission from D. Harrison, Am. J. Phys. 50:811, 1982. Copyright 1982, American Association of Physics Teachers).
20.7.1
Spin Measurement Experiment
We consider Fig. 20.4. The two Stern–Gerlach analyzers P1′ and P2′ are rotated about the y-axis through the angles θ1 and θ2 respectively. Now, we have P ′ φ′± = ±φ′± .
(20.21)
For Fig. 20.4 we have φ1+ φ1−
1
1
= cos(θ1 /2)φ′ + − sin(θ1 /2)φ′ − ,
(20.22a)
=
(20.22b)
1 cos(θ1 /2)φ′ −
+
1 sin(θ1 /2)φ′ +
and similarly for atom-2. The 1/2 in θ1 /2 is because we considered spin-1/2 atoms. Substituting the obtained φ’s in Eq. (20.3) we have ψ
=
1 h 1 2 1 2 √ cos ((θ1 − θ2 ) /2) φ′ + φ′ − − φ′ − φ′ + 2 i 1 2 1 2 − sin ((θ1 − θ2 ) /2) φ′ + φ′ + + φ′ − φ′ − .
(20.23)
For θ1 = θ2 Eq. (20.23) is identical to Eq. (20.3).
Solved Problem 2: What are the probabilities for (i) P1′ measuring atom-1 in spin-up and simultaneously P2′ measuring atom-2 also in spin-up and (ii) P1′ measuring atom-1 in spin-up and simultaneously P2′ measuring atom-2 in spin-down? (i) Let us denote P++ be the probability for P1′ measuring atom-1 in spin-up
512 Quantum Mechanics I: The Fundamentals and simultaneously P2′ measuring atom-2 also in spin-up. P++ is given by the ′2 square of the coefficient of φ′1 + φ− in Eq. (20.23): 1 1 1 P++ = − √ sin((θ1 − θ2 )/2) × − √ sin((θ1 − θ2 )/2) = sin2 ((θ1 − θ2 )/2) . 2 2 2 (20.24) For θ1 = θ2 we have P++ = 0 as expected. (ii) In this case the probability P+− is simply the square of the coefficient of ′2 φ′1 + φ− in Eq. (20.23). We obtain P+− =
1 cos2 ((θ1 − θ2 )/2) . 2
(20.25)
It becomes maximum (P+− = 1/2) when θ1 = θ2 . We use the result in the above solved problem to verify the Bell’s theorem. To test the Bell’s inequality, first write its quantum version. Assume that pair of the atom emitted by a source have a total spin 0 and they travel in opposite directions. If the right-side atom is spin-up then its companion atom is spindown provided the Stern–Gerlach filters have the same orientation. We set a, b and c as in table 20.3 and assume the following: 1. The polarizers measure the spins at locations separated by a space interval. Then the polarizers cannot influence each other. That is, we assume that no influence propagates faster than the speed of light. Otherwise, the measurement by one polarizer introduce uncontrollable phase factors into the wave function. This is a kind of uncertainty principle. 2. The objects have properties (such as spin) even if they are not measured. Suppose the Stern–Gerlach filters allow only spin-up atoms to pass through. We consider the application of Eq. (20.20) to spin measurements. Remember that Eq. (20.20) is for the case of each with properties opposite to the other. In a spin measurement also we consider the same. This is realized if the two polarizers have the same orientation implying that the atoms have opposite spins. We determine whether the right-side atom is spin-up, 0o , not spin-up 45o by measuring its spin at 0o and its companion’s spin at 45o . The Bell’s inequality, Eq. (20.20), becomes P (1up 0o , 2up 45o ) + P (1up 45o , 2up 90o ) ≥ P (1up 0o , 2up 90o ) .
(20.26)
Evaluate each term in Eq. (20.26) using Eq. (20.24). We have P (1up 0o , 2up 45o ) = P (1up 45o , 2up 90o ) = P (1up 0o , 2up 90o ) =
1 sin2 (−45o /2) , 2 1 sin2 (−45o /2) , 2 1 sin2 (90o /2) . 2
(20.27a) (20.27b) (20.27c)
Mysteries in Quantum Mechanics 513
Examples of three double-valued atomic spins. (Reproduced with permission from D. Harrison, Am. J. Phys. 50:811, 1982. Copyright 1982, American Association of Physics Teachers.) TABLE 20.3
a = spin-up θ = 0o
b = spin-up θ = 45o
c = spin-up θ = 90o
−a = spin-down θ = 0o
−b = spin-down θ = 45o
−c = spin-down θ = 90o
Thus, Eq. (20.26) becomes 1 1 1 sin2 (45o /2) + sin2 (45o /2) ≥ sin2 (90o /2) . 2 2 2
(20.28)
That is, 0.0732 + 0.0732 = 0.1464 ≥ 0.25 which is false. The local hidden variables model with the particles having a definite spin orientation before they are detected is wrong. Thus, the prediction of quantum mechanics violates the Bell’s inequality. The existence of local hidden variables is inconsistent with the predictions of quantum theory. Thus, Bell discovered that assumption of local realism is not in accordance with quantum physics.
Solved Problem 3: Verify that in table 20.3 instead of the values θ = 0o , 45o and 90o if the angles are 0o , 90o and 180o respectively then there is no conflict between the predictions of quantum mechanics and the Bell’s inequality. The Bell’s inequality P (1a, 2b) + P (2b, 2c) ≥ P (1a, 2c) becomes P (1up 0o , 2up 90o ) + P (1up 90o , 2up 180o) ≥ P (1up 0o , 2up 180o) . (20.29) Since P (1up 0o , 2up 90o ) = P (1up 90o , 2up 180o ) = P (1up 0o , 2up 180o ) =
1 1 sin2 (−90o /2) = , 2 4 1 1 2 o sin (−90 /2) = , 2 4 1 1 2 o sin (−180 /2) = 2 2
(20.30a) (20.30b) (20.30c)
we get the Bell’s inequality as 1/4+1/4 = 1/2. So, there is no conflict between the predictions of quantum mechanics and the Bell’s inequality.
514 Quantum Mechanics I: The Fundamentals
20.7.2
Experimental Tests of Bell’s Theorem
What is the outcome of real experiments? The result is that the inequality is violated. The predictions of quantum mechanics and Bell’s theorem were tested experimentally for photons and protons. These experiments helped us to understand the basics of quantum mechanics. Moreover, gave birth to fascinating new topics such as quantum information, quantum teleportation, quantum computation and quantum cryptography. The first experiment was carried out by Clauser and Freedman at Berkely [25]. The most famous experiment was performed by Aspect, Dalibard and Roger in Paris [26]. These experiments are quite difficult. The most striking point is that all the experiments agreed with the quantum mechanical results and Bell’s inequality was violated. Most of the experiments were based on correlations between the polarizations of pairs of photons. Correlated photons can be produced in many ways. An example is annihilation of positronium into two photons. Here initial and final angular momenta are zero. Alternatively, an atom in an excited state with angular momentum zero decays by J = 0 → J = 1 → J = 0 cascade emitting two photons. In a typical experiment a source emits polarization-conserved pairs of photons. Each photon is passed through a polarizer. Behind each polarizer, the transmitted photons are registered. Conservation of polarization implies that if the photon-1 is measured and found in horizontal (H) polarization then photon-2 will be vertical (V ) and vice-versa. Before measurement the total wave function is 1 |ψi = √ (|Hi1 |V i2 − |V i1 |Hi2 ) . (20.31) 2 A version of Bell’s inequality is |E(a, b) − E(−a, b)| + |E(a, −b) + E(−a, −b)| ≤ 2 ,
(20.32)
where E(a, b) =
1 [C++ (a, b) + C−− (a, b) − C+− (a, b) − C−+ (a, b)] . N
(20.33)
Here + and − are the outputs of a two-channel polarizer. Then C++ (a, b) is the number of coincidences between the + output port of the polarizer measuring photon-1 along a and + output port of the polarizer measuring √ photon-2 along b. For a = 0, b = 22.5o Eq. (20.32) is found to be 2 2 < 2 a clear violation of Bell’s inequality. One may wonder about the contradiction between Bell’s theorem and quantum mechanics. A possible reason is the difference between ideal experiment and measurement. But in real experiments we can consider reasonable assumptions about the working of the apparatus, noise, etc. Moreover, in the Bell’s theorem the polarizers are assumed to act independently. If this is incorrect then signal (passing faster than light) connecting the two polarizers
Mysteries in Quantum Mechanics 515 is possible. Next, the experiments have always involved fixing the angles of polarizers and then taking data. Thus, one may argue that the polarizers have much time to know orientation of each other. Greenberger et al. [27] have demonstrated Bell’s theorem in a new way by analyzing a system consisting of 3 or more correlated spin-1/2 particles. Here again Bell’s theorem is violated by quantum mechanical predictions. In 1993, Hardy proposed a situation where nonlocality can be inferred without using inequalities [28].
20.8 RESOLVING EPR PARADOX The EPR’s nonlocality puzzle can be resolved [29] if the correct physical input is used for the calculation of the quantum correlations. Suppose the result of spin measurement is two-valued. Let A and B denote the outcomes ±1. a and b denote the settings of the apparatus for the first particle and the second particle respectively. The statement of locality is A(a) = ±1 , B(b) = ±1. The existence of reality is that some hidden variables decide the outcomes even before the measurement. This is stated as A(a, h1 ) = ±1 , B(b, h2 ) = ±1 where h1 and h2 are hidden variables associated with the outcomes. The experimenter Pcan compute a correlation function of the outcomes defined by P (a, b) = N1 (Ai Bi ). This is the classical function obtained by averaging the products of the form (++ = +), (−− = +), (+− = −) and (−+ = −). The joint events (++) and (−−) are coincidences and the events (+−) and (−+) are anticoincidences. P (a, b) denotes the average The Bell correlation is then R of the quantity (number of anticoincidences). R dhP (h)A(a, h1 )B(b, h2 ) where dhP (h) = 1. Assume that a measurement on one particle does not alter either the magnitude or phase of the complex amplitude of the other particle. In particular, measurement on one particle does not make the other particle to possess a definite state. At each location a measurement yield two-valued denoted by + and − for each particle. The local amplitudes for the events + and − for the two particles are denoted by the complex functions C1+ , C1− , C2+ and C2− . Amplitudes C1+ and C2− are mutually orthogonal. Then the statement of locality is C1± = C1± (a, φ1 ) and C2± = C2± (b, φ2 ) where φ1 and φ2 are the hidden variables. The correlation function is now of the form U (a, b) = Re(N Ci Cj∗ ), where √ N is a normalization factor. Let as C1+√= (1/ 2)eis(θ1 −φ1 ) for the first particle at the first polarizer and C2+ = (1/ 2)eis(θ2 −φ2 ) for the second particle at the second polarizer. θ1 and θ2 are the directions of the two polarizers and s is the spin of the particle. We have C1 C1∗ = C2 C2∗ = 1/2. The correlation function for an outcome of either (++) or (−−) of two particles is U (θ1 , θ2 , φ0 ) = 2Re(C1 C2∗ ) = cos [s(θ1 − θ2 ) + sφ0 ] .
(20.34)
U 2 is the probability for a coincidence detection (++ or −−) and 1 − U 2 is
516 Quantum Mechanics I: The Fundamentals the probability for an anticoincidence (events of the type +− and −+). The average of the quantity (number of coincidences−number of anticoincidences) is U 2 − (1 − U 2 ) = 2U 2 − 1. Then P (a, b) = 2U 2 − 1 = 2 cos2 [s(θ1 − θ2 ) + sφ0 ] − 1 .
(20.35)
For the singlet state breaking up into two spin-1/2 particles propagating in opposite directions with total spin 0 the quantities U and P are obtained as π 1 = − sin((θ1 − θ2 )/2) (20.36) U = cos (θ1 − θ2 ) + 2 2 and P (a, b) = 2 sin2 ((θ1 − θ2 )/2) − 1 = − cos(θ1 − θ2 ). This is identical to the quantum mechanical prediction obtained from the singlet entangled state and the Pauli spin operator.
20.9 CONCLUDING REMARKS Quantum theory does not give exact outcome of a measurement but it gives various possible outcomes. Even if the event has occurred the outcome is known (according to quantum theory) only a measurement or observation is made. With out observation, the value of a dynamical variable cannot be given any meaning in quantum mechanics. All quantum systems are supposed to exist in some sort of state of probabilistic uncertainty until an observation or measurement is made and the wave function collapses. A measurement process not only makes the wave function collapse into one of the possibilities but in some experiments it is responsible for the wave or particle character of the system. For example, the wave behaviour of light in double-slit experiment and particle-like character in photoelectric experiment are not a property of light but are the property of our interaction with light through measuring process. Schrödinger equation does not tell us which event is going to happen at various times and at various places and how an event will occur at a place at a given time. Its goal is not to predict the actual events going to take place at various times and places but to predict the possible events. An important point is that given the value of ψ at t = t0 and assuming fixed boundary conditions the Schrödinger equation determines the value of ψ at t > t0 . In this sense and because there is no stochastic term in the equation the quantum description of state of a system is deterministic. If the events occur purely by chance then how we are able to find the probabilities of the events. This is because probabilities are given by probability function, and probability function is obtained from wave function. Wave function is the solution of the Schrödinger equation − which is completely deterministic.
20.10 BIBLIOGRAPHY [1] R. Deltete and R. Guy, Am. J. Phys. 58:673, 1990.
Mysteries in Quantum Mechanics 517 [2] A. Einstein, B. Podolsky and N. Rosen, Phys. Rev. 47:777, 1935. [3] D. Harrison, Am. J. Phys. 50:811, 1982. [4] B.M. Terhal, M.M. Wolf and A.C. Doherty, Physics Today, April pp.46, 2003. [5] T. Leggett, Physics World , December pp.73, 1999. [6] D.M. Greenberger, M.A. Horne and A. Zeilinger, Physics Today, August, pp.22, 1993. [7] C. Monroe, D.M. Meekhof, B.E. King and D.J. Wineland, Science 272:1131, 1996. [8] S. Wallentowitz, R.L. de Matos Filho and W. Vogel, Phys. Rev. A 56:1205, 1997. [9] D. Bouwmeester, J.W. Pan, M. Daniells, H. Weinfurter and A. Zeilinger, Phys. Rev. Lett. 82:1345, 1999; Nature 403:515, 2000. [10] C.A. Sackett, D. Kielpinski, B.E. King, C. Langer, V. Meyer, C.J. Myatt, M. Rowe, Q.A. Turchette, W.M. Itano, D.J. Wineland and C. Monroe, Nature 404:256, 2000. [11] S. Ghosh, T.F. Rosenbaum, G. Aeppli and S.N. Coppersmith, Nature 425:48, 2003. [12] C.P. Williams and S.H. Clearwater, Explorations in Quantum Computing. Springer, New York, 1998. [13] H.K. Lo, S. Popescu and T. Spiller (Eds.), Introduction to Quantum Computation and Information. World Scientific, Singapore, 1998. [14] S.F. Huelga, C. Macchiavello, T. Pellizzari, A.K. Ekert, M.B. Plenio and J.I. Cirac, Phys. Rev. Lett. 79:3865, 1997. [15] L.K. Grover, Phys. Rev. Lett. 78:325, 1997. [16] J.I. Cirac, A.K. Ekert, S.F. Huelga and C. Macchiavello, Phys. Rev. A 59:4249, 1999. [17] S. Bose, V. Vedral and P.L. Knight, Phys. Rev. A 57:822, 1998. [18] M. Murao, M.B. Plenio, S. Popescu, V. Vedral and P.L. Knight, Phys. Rev. A 57:R4075, 1998. [19] R. Penrose, The Emperor’s New Mind. Vintage, New York, 1990. [20] A.J. Leggett, Contemporary Physics 25:583, 1984. [21] N.D. Mermin, Physics Today April pp.38, 1985.
518 Quantum Mechanics I: The Fundamentals [22] J.S. Bell, Physics 1:195, 1964. [23] d’Espagnat, Am. J. Phys. 50:811, 1982. [24] N.D. Mermin, Am. J. Phys. 58:731, 1990. [25] J.J. Freedman and J.S. Clauser, Phys. Rev. Lett. 28:938, 1972. [26] A. Aspect, J. Dalibard and G. Roger, Phys. Rev. Lett. 49:1804, 1982. [27] D.M. Greenberger, M.A. Horne, A. Shimony and A. Zeilinger, Am. J. Phys. 58:1131, 1990. [28] L. Hardy, Phys. Rev. Lett. 71:1665, 1993. [29] C.S. Unnikrishnan, Current Science 79:195, 2000.
20.11 EXERCISES 20.1 List the assumptions used to derive the Bell’s inequality. 20.2 When do you get equality in Eq. (20.15)? 20.3 If you make measurements of a physically observable A on a system in rapid succession, do you expect the outcomes to be the same every time? 20.4 For a pair of spin-1/2 particles in the singlet state determine the expectation value of the measurements of the correlated state in the different directions of the unit vectors a and b, coplanar with σ. 20.5 Assume that at time t = 0 a fair coin was tossed by one of the two observers A and B. At t = t1 the coin landed on the ground and shown an outcome. At t = t2 , A looked at the coin and observed the outcome as head. B saw the coin at t = t3 > t2 . Assume that the above event is quantum one. Write the wave function of the state of the coin (a) for the two observers at t1 , t2 and t3 . (b) What would be the outcome of the coin tossing for the observer A during t1 < t < t2 knowing the outcome at t = t2 (as head) if the event is (i) classical and (ii) quantum. 20.6 What are the eigenvalues of P1′ and P2′ in Eq. (20.21)? 20.7 Evaluate the quantum correlation function CQM = P++ + P−− − P+− − P−+ corresponding to the wave function (20.23) associated with the measuring the z-component of spin of two atoms in the setup shown in Fig.20.4. 20.8 Compute the expectation value of P1′ P2′ for the setup shown in Fig.20.4 using the wave function given by Eq. (20.23). 20.9 Consider the table 20.3. Suppose the angles 45o and 90o are interchanged. Test the Bell’s inequality.
CHAPTER
21
Numerical Methods for Quantum Mechanics
21.1 INTRODUCTION In this book so far we are concerned with the various fundamentals of quantum mechanics. In the latter part of the century the subject has started growing on multi direction. Newer quantum phenomena were identified. Quantum version of certain classical phenomena and concepts have been formulated. New topics in quantum physics have born. Application of quantum physics to other branches of science and technology has started. For example, a great deal of research has been focused on the topics such as supersymmetry, Berry’s phase, Aharonov–Bohm effect, Wigner distribution function, coherent and squeezed states, quantum computing, quantum cryptography, quantum imaging and quantum lithography to name a few. Theoretical and experimental progress have been made in these topics. The fundamentals and the progresses made in these advanced topics are covered in the volume-II of this book. Before we move on to these advanced topics we wish to mention that in classical physics several fascinating newer phenomena were first realized in numerical simulation and then necessary theory and mathematical tools were developed. Thus, it is important to develop numerical schemes to study quantum systems. We have seen that solving the Schrödinger equation is the central problem of nonrelativistic quantum mechanics. Accurate solutions of the onedimensional Schrödinger equation with an arbitrary potential are important for understanding many phenomena in physics. But exactly solvable Schrödinger equations are very limited. Most of the Schrödinger equations cannot be solved analytically. However, number of simple and efficient numerical techniques are developed to study bound and scattering states of quantum mechanical systems. Particularly, there are techniques which allow us to compute eigenvalues and eigenfunctions of stationary states and transmission and reflection probabilities in a very accurate way [1-24]. Such numerical methods are important for the study of complex systems, design and analysis of quan519
520 Quantum Mechanics I: The Fundamentals tum effect devices [25,26]. With the advent of modern computer technology, robust numerical study of various quantum phenomena are now common. The present chapter is concerned with the study of numerical computation of 1. bound state eigenvalues, 2. bound state eigenfunctions, 3. ground state eigenvalue and eigenfunction, 4. transmission and reflection probabilities, 5. transition probabilities and 6. electronic distribution in atoms. There is a simple method developed by van der Maelen Uria, GarciaGranda and Menendez-Velazquez [14] to compute stationary states eigenvalues and eigenfunctions for arbitrary analytical or numerical potentials. In their method, the stationary state equation is transformed into a set of linear finite-difference equations. When these equations are expressed in a matrix form, (A − λI)ψ = 0, the problem becomes an algebraic eigenvalue problem. The eigenvalues and eigenfunctions of A can be determined numerically and are related to the energy eigenvalues and eigenfunctions of the original quantum mechanical problem. Chen [15] extended the matrix method to timedependent Schrödinger equation which is useful to analyse time evolution of an initial wave function. The final finite-difference formula derived by Chen is equivalent to the one obtained by Goldberg et al. [16]. A finite-difference time domain method has been developed by Sudiarta and Geldart [20] for finding stationary state eigenvalues and eigenfunctions. Their method is efficient for the calculation of ground state eigenvalue and it uses random wave function as an initial wave function. For a scattering potential Pang [17] described a numerical method which uses numerical integration of Schrödinger equation and function minimization technique to obtain the solution in the barrier region satisfying appropriate boundary conditions. Wang and Champagne [21] described a method called coupled channel method for numerical calculation of transition probabilities for systems with time-dependent perturbations switched-on at t = 0 and switched-off at t = τ . A simple Monte Carlo simulation for easy visualization of electronic distribution in atoms is demonstrated by de Aquino et al. [22]. In this chapter, the above methods are presented in detail. The methods are applied to a number of physically interesting potentials. Numerical results are compared with exact results wherever possible.
Numerical Methods for Quantum Mechanics 521
21.2 MATRIX METHOD FOR COMPUTING STATIONARY STATE SOLUTIONS Let us rewrite the time-independent Schrödinger equation into the dimensionless form ψ ′′ (x) +
2m [E − V (x)]ψ(x) = 0 , x ∈ [xmin , xmax ] . ~2
(21.1)
A Schrödinger equation containing a first derivative term can be transformed into the form (21.1) by a change of variable. In the numerical simulation we choose the boundary condition ψ → 0 as x → ±∞ as ψ(xmin ) = ψ(xmax ) = 0 and we set ~ = 1 and m = 1. ψ is zero outside the interval [xmin , xmax ]. |xmin | and |xmax | are sufficiently large. In the following we illustrate the method of van der Maelen Uria et al. [14] for finding eigenvalues and eigenfunctions of (21.1).
21.2.1
Discrete Equivalent of Schrödinger Equation
We discretize the space variable x in the interval [a, b] by choosing the grid of points x0 , x1 , · · · , xN with xi+1 − xi = h where h is a constant. The eigenfunctions will take values only at these grid of points. That is, ψ will be a set of numbers ψ0 , ψ1 , · · · , ψN where ψi = ψ(xi ). Usually ψ0 = ψ(xmin ) = 0 and ψN = ψ(xmax ) = 0. Now, Eq. (21.1) becomes the following discrete equation ψi′′ + 2 [E − V (xi )] ψi = 0,
i = 1, 2, · · · N − 1.
(21.2)
We convert the above equation into a difference equation by replacing ψi′′ by a central-difference formula. A three-point central difference formula for ψi′′ is ψ ′′ |x=xi =
1 [ψi+1 − 2ψi + ψi−1 ] + O(h2 ) . h2
(21.3)
Neglecting the error term and substituting Eq. (21.3) in Eq. (21.2) we obtain ψi−1 + 2 h2 E − h2 V (xi ) − 1 ψi + ψi+1 = 0, i = 1, 2, · · · , N − 1. (21.4) Let us rewrite Eq. (21.4) as −ψi−1 + 2 h2 V (xi ) + 1 − h2 E ψi − ψi+1 = 0,
i = 1, 2, · · · , N − 1. (21.5)
Suppose say N = 4. Then we have the following set −ψ0 + 2 h2 V1 + 1 − h2 E ψ1 − ψ2 −ψ1 + 2 h2 V2 + 1 − h2 E ψ2 − ψ3 −ψ2 + 2 h2 V3 + 1 − h2 E ψ3 − ψ4
of equations = 0,
(21.6a)
= 0,
(21.6b)
= 0.
(21.6c)
522 Quantum Mechanics I: The Fundamentals In the above set of equations ψ0 = ψ4 = 0. In matrix form (21.6) is written as
d11 − λ
−1
0
ψ1
−1 ψ2 = 0 , −1 d33 − λ ψ3
−1 d22 − λ 0
(21.7)
where dii = 2 1 + h2 V (xi ) and λ = 2h2 E. This is of the form (A − λI)ψ = 0 with A being symmetric tridiagonal. An N × N matrix A is said to symmetric tridiagonal if dij di(i+1)
= 0,
j 6= i − 1, i, i + 1,
= d(i+1)i ,
i = 1, 2, · · · , N
(21.8)
i = 1, 2, · · · , N − 1.
For arbitrary value of N the N − 1 coupled linear finite-difference Eqs. (21.4) are expressed in a compact matrix form (A − λI)ψ = 0 ,
(21.9a)
with the elements of A and λ are given by λ = 2h2 E, dii = 2 1 + h2 V (xi ) , i = 1, 2, · · · , N − 1 di(i+1) = d(i+1)i = −1, i = 1, 2, · · · , N − 2 dij
=
0,
j 6= i − 1, i, i + 1,
i = 1, 2, · · · , N − 1
(21.9b)
and I is the unit matrix. That is, the original differential eigenvalue problem now becomes an algebraic eigenvalue problem. Note that the calculation of E and ψ of the stationary states of a given quantum mechanical system is reduced to the calculation of λ and ψ of the matrix A. For nontrivial solution of (A − λI)ψ = 0 we require |A − λI| = 0 from which we can compute the eigenvalues of A and then the energy eigenvalues E. Then we can compute ψ from Aψ = λψ.
21.2.2
Computation of Eigenvalues of the Matrix A
How do we compute the eigenvalues of the matrix A? Interestingly, the matrix A is a symmetric tridiagonal. For a symmetric tridiagonal matrix efficient algorithms are available to calculate λ. An iterative method is the Jacobi method, but it is very slow for matrices of order greater than, say, ten. An alternate and more efficient technique is QL algorithm [27, 28]. This algorithm is presented in appendix F. We note that the truncation error in the finite-difference approximation of the second derivative term, Eq. (21.3), is proportional to h2 . The error can be made proportional to h4 using Richardson’s extrapolation. This technique (1) (2) is the following. Suppose Ei (h) and Ei (αh) are two approximate values
Numerical Methods for Quantum Mechanics 523 of Ei obtained by a method of order p with step size h and αh respectively. Then we can write (1)
Ei (h) = (2) Ei (αh)
=
Ei (0) + chp + O(h2p ) ,
(21.10a)
Ei (0) + cαp hp + O(h2p ) .
(21.10b)
Multiplying of Eq. (21.10a) by αp and subtracting it from (21.10b) gives (1)
Ei (0) =
(2)
αp Ei − Ei αp − 1
+ O(h2p ) .
(21.11)
The above formula is known as Richardson’s extrapolation. For h = h1 , αh = h2 , α = h2 /h1 and p = 2 Eq. (21.11) becomes 1 2 (1) 2 (2) Ei (more accurate) = 2 h E − h E . (21.12) 2 i 1 i h2 − h21 The eigenvalues of the given quantum system may be infinite in number. Since A is (N − 1) × (N − 1) matrix in the numerical computation we get only N − 1 eigenvalues, a subset of the possible eigenvalues.
21.2.3
Computation of the Eigenfunctions of the Matrix A
An eigenfunction of the matrix A for an eigenvalue is obtained by solving the algebraic eigenvalue problem, Eq. (21.9), where λ (= 2h2 E) is a given eigenvalue. We write the Eq. (21.9a) (after redefining N − 1 as N ) and Ai = dii − λ as A1 −1 0 0 ··· 0 0 0 ψ1 −1 A2 −1 0 ··· 0 0 0 ψ2 0 −1 A3 −1 · · · ψ3 0 0 0 . = 0. (21.13) . .. .. 0 0 0 0 · · · −1 AN −1 −1 ψN −1 0 0 0 0 ··· 0 −1 AN ψN We have to normalize any admissible eigenfunction. Let Cψ is the normalized eigenfunction where C is the normalization constant. C can be determined from the normalization condition Z b C2 ψ ∗ ψ dx = 1 . (21.14) a
Therefore, keeping this in mind, in the numerical scheme we choose ψ1 = 1 and compute other ψi ’s from (21.13). The first row in Eq. (21.13) gives ψ2 = A1 ψ1 (from row 1) .
(21.15)
524 Quantum Mechanics I: The Fundamentals That is, use row 1 to compute ψ2 , use row 2 to compute ψ3 and so on. From rows 2 to N − 2 we have ψ3 ψ4
ψN −1
= −ψ1 + A2 ψ2 (from row 2) , = −ψ2 + A3 ψ3 (from row 3) , .. . = −ψN −3 + AN −2 ψN −2 (from row N − 2) .
(21.16)
Note that (N − 2)th row gives ψN −1 . Since, (N − 1)th and N th rows are not yet used and only ψN to be determined, we add these last two rows and obtain a single new equation. From this new equation we compute ψN : ψN =
1 [−ψN −2 − (1 − AN −1 ) ψN −1 ] . 1 − AN
(21.17)
To summarize, with ψ1 = 1, ψ2 = A1 ψ1 , the expressions for ψ3 , · · · , ψN −1 are ψi = −ψi−2 + Ai−1 ψi−1 ,
i = 3, 4, · · · , N − 1
(21.18)
and ψN is given by the Eq. (21.17). The computed eigenfunction is unnormalized. To normalize it we calculate the normalization constant C from Eq. (21.14). We can evaluate the integral in Eq. (21.14), for example, using the composite trapezoidal rule: Z b 1 2 1 1 2 2 2 2 2 |ψ| dx = h = ψ ψ + ψ . (21.19) + ψ + · · · + ψ + 2 3 N −1 C2 2 1 2 N a
21.2.4
Application
We apply the method to three quantum mechanical systems and compare the numerically computed eigenvalues and eigenfunctions with the exact values.
21.2.4.1 Harmonic Oscillator For the linear harmonic oscillator potential V (x) = kx2 /2 with k = 1 the exact eigenvalues are Ee = (n + 1/2), n = 0, 1, · · · . It is easy to develop a program to compute the eigenvalues by the QL method (see the problem 21.1 at the end of the present chapter). In order to get a more accurate result, the eigenvalues are computed for two values of h, h1 and h2 . Then using the Richardson’s extrapolation more accurate values of E are obtained. Table 21.1 presents first few eigenvalues computed with h1 = 0.1 and h2 = 0.05, by extrapolation and exact values. Here xmin = −10 and xmax = 10. Accuracy can be improved by reducing the values of h1 and h2 and choosing large values of xmin and xmax . 2 The exact ground state eigenfunction is given by ψ(x) = (1/π)1/4 e−x /2 . Table 21.2 gives the numerically computed ground state eigenfunction values at some selected grid points for three values of h where xmin = −5 and
Numerical Methods for Quantum Mechanics 525
Numerically computed first few eigenvalues (En ) of the linear harmonic oscillator potential V (x) = kx2 /2. Here xmin = −10 and xmax = 10 and k = 1.
TABLE 21.1
En (h = 0.1) 9.44309 8.45444 7.46452 6.47333 5.48087 4.48715 3.49217 2.49593 1.49844 0.49969
En (h = 0.05) 9.48584 8.48866 7.49116 6.49335 5.49523 4.49679 3.49805 2.49898 1.49961 0.49992
Extrapolated En 9.50009 8.50007 7.50004 6.50002 5.50002 4.50000 3.50001 2.50000 1.50000 0.50000
Ee 9.5 8.5 7.5 6.5 5.5 4.5 3.5 2.5 1.5 0.5
Numerically computed ground state energy eigenfunction values at some grid points for three values of h. The last column is the exact result. TABLE 21.2
Numerical ψ x
h = 0.020 h = 0.010 h = 0.001 Exact ψ
±4.0 ±3.0 ±2.5 ±2.0 ±1.5 ±1.0 ±0.5 0.0
0.00024 0.00803 0.03175 0.09779 0.23459 0.43828 0.63770 0.72261
0.00025 0.00832 0.03292 0.10138 0.24321 0.45438 0.66112 0.74915
0.00025 0.00834 0.03300 0.10165 0.24385 0.45558 0.66287 0.75113
0.00025 0.00833 0.03300 0.10165 0.24385 0.45558 0.66287 0.75113
xmax = 5 (for a complete program to compute a normalized eigenfunction see the problem 21.3 at the end of the present chapter). ψ(x) approaches the exact result as h decreases. Very good agreement with exact eigenfunction is observed for other values of E also. The numerical solution is symmetric with respect to the origin. This solution is called an even-parity solution. An odd-parity solution can be obtained by choosing ψ(xmin ) = −1. Figure 21.1
526 Quantum Mechanics I: The Fundamentals
ψ
1 E=0.5
E=1.5
(b)
(c)
E=3.5
(d)
0
-1 1 E=2.5 ψ
(a)
0
-1 -5
0
x
5 -5
0
x
5
Numerically computed eigenfunctions of the harmonic oscillator (continuous curve) for first few values of E. The exact values of the eigenfunctions for some values of x are represented by solid circles.
FIGURE 21.1
shows the numerically computed eigenfunctions for E = 0.5, 1.5, 2.5 and 3.5. For E = 0.5 and 2.5 the value of ψ(xmin ) is set as 1 whereas for E = 1.5 and 3.5 the value of ψ(xmin ) is set as −1.
21.2.4.2 Square-Well Potential As a second example consider the square-well potential (5.16) which admits only a finite number of bound states. Its exact energy eigenvalues and eigenfunctions are determined in sec. 5.3. In the numerical computation we fix V0 as 35 and a = 1 for which only 6 bound states exist. The numerical (exact) energy eigenvalues are computed as 0.98336 (0.98336), 3.92084 (3.92084), 8.77009 (8.77009), 15.43900 (15.43901), 23.71532 (23.71539) and 32.84380 (32.84389). In the numerical simulation we get eigenvalues above 35 which are closely spaced and we can treat them as forming continuous spectrum. Figure 21.2 shows the numerically computed and exact energy eigenfunctions ψ0 , ψ1 and ψ5 corresponding to the energy eigenvalues E0 = 0.98336, 3.92084 and 32.8438 respectively.
21.2.4.3 Hydrogen Atom The radial Schrödinger equation for hydrogen-like atom is 1 d 2m l(l + 1) 2 dR r + (E − V (r)) − R=0. r2 dr dr ~2 r2
(21.20)
Numerical Methods for Quantum Mechanics 527
1
0 0
4
x
(b)
0
-1 -4
1
ψ 5
-1 -4
1
(a) ψ
ψ
0
1
0
x
4
(c)
0
-1 -4
0
x
4
Numerically computed eigenfunctions of the square-well potential (continuous curve) for a few values of E. The exact values of the eigenfunctions for some values of x are represented by solid circles. FIGURE 21.2
Introducing the change of variable ρ(r) = rR(r) and using the Hartree’s atomic units, ~ = 1 and m = 1, Eq. (21.20) becomes l(l + 1) d2 ρ + 2(E − V (r)) − ρ=0. (21.21) dr2 r2 The Coulomb potential is given by V (r) = −Z/r. Eigenvalues and eigenfunctions of Eq. (21.21) can be computed numerically. First, we consider the case of Z = 1 and l = 0. The exact energy eigenvalues are −1/(2n2). Table 21.3 gives the first few energy values calculated numerically with h1 = 0.1, h2 = 0.05 and by extrapolation. The last column is the exact E. Here, r range is fixed as r ∈ (0, 100). E1 to E5 are identical to the exact values within 5 decimal accuracy. Other values deviate from the exact energy values. As mentioned earlier, accuracy can be improved by choosing large value for maximum value of r and small values for h1 and h2 . Table 21.4 illustrates the influence of b = rmax on the energy values E1 to E5 . As expected these values converge to the exact values as the value of b increases. Moreover, the number of negative values of E increases with the increase in the values of b. For b = 100, 200, 300, 400 and 500 the number of negative values of E are 8, 12, 15, 17 and 19 respectively. Figure 21.3a shows the numerically computed radial eigenfunction for l = 0 and for first three energy eigenvalues.
528 Quantum Mechanics I: The Fundamentals
Numerically computed first few energy eigenvalues (En ) of the hydrogen atom. Here, r ∈ [0, 100]. TABLE 21.3
En (h = 0.1)
En (h = 0.05)
Extrapolated E
Ee
−0.00960 −0.01387 −0.02000 −0.03125 −0.05554 −0.12492 −0.49876
−0.00960 −0.01387 −0.02000 −0.03125 −0.05555 −0.12498 −0.49969
−0.00960 −0.01387 −0.02000 −0.03125 −0.05555 −0.12500 −0.50000
−0.01020 −0.01389 −0.02000 −0.03125 −0.05555 −0.12500 −0.50000
Numerically computed first few eigenvalues of the hydrogen atom as a function of b = rmax . The last row gives exact values. The numerical eigenvalues are obtained from the Richardson’s extrapolation of the eigenvalues for h1 = 0.1 and h2 = 0.05. TABLE 21.4
rmax
E1
E2
E3
E4
E5
10 20 30 40 50 100
−0.50000 −0.50000 −0.50000 −0.50000 −0.50000 −0.50000
−0.11305 −0.12499 −0.12500 −0.12500 −0.12500 −0.12500
0.08974 −0.04999 −0.05542 −0.05555 −0.05555 −0.05555
0.40122 0.01640 −0.02469 −0.03056 −0.03120 −0.03125
0.81953 0.11220 0.01317 −0.01110 −0.01788 −0.02000
∞
−0.50000
−0.12500
−0.05555
−0.03125
−0.02000
Suppose we choose l 6= 0. The possible values of l are 0, 1, · · · , n − 1. If we fix l = 0 then n must assume the values 1, 2, · · · . On the other hand, for l = 1 and 2 the quantum number n starts from 2 and 3 respectively. That is, if l is fixed as 1 then the numerically computed eigenvalue spectrum should not contain the energy value E1 . This is also realized in the numerical computation of E. Figure 21.3b shows the numerically computed radial eigenfunctions ρ21 and ρ31 .
Numerical Methods for Quantum Mechanics 529
1
(a)
nl = 10
ρ
nl = 30 0
nl = 20 -1
0
10
20
r
30
0 .5
40
(b)
nl = 21
ρ
0
nl = 31 - 0. 5
0
10
20
r
30
40
Numerically computed eigenfunctions (represented by continuous curves) and exact eigenfunctions (marked by solid circles at selected values of r) of hydrogen atom for (a) nl = 10, 20 and 30 and (b) nl = 21 and 31. For E1 , E2 and E3 the range of values of r used are [0, 10], [0, 20] and [0, 40] respectively. FIGURE 21.3
21.3 FINITE-DIFFERENCE TIME-DOMAIN METHOD There is another method called finite-difference time-domain (FDTD) method introduced by Sudiarta and Geldart [20] for computation of energy eigenvalues and eigenfunctions of time-independent Schrödinger equation. Their method can be easily extended for higher-dimensional systems. In the FDTD method the time-dependent Schrödinger equation is discretized using a centraldifference formula in both time and space. Using different initial wave function we can obtain various eigenvalues and the corresponding eigenfunctions. The method with random initial wave function is efficient in obtaining ground state eigenvalue and eigenfunction.
530 Quantum Mechanics I: The Fundamentals
21.3.1
Basic Idea
We consider the three-dimensional time-dependent Schrödinger equation ∂ ~2 2 i~ ψ(x, y, z, t) = − ∇ + V (x, y, z) ψ(x, y, z, t) . (21.22) ∂t 2m The solution of (21.22) in terms of stationary state eigenfunctions is written as ∞ X ψ(x, y, z, t) = Cn ψn (x, y, z)e−iEn t/~ , (21.23) n=0
where ψn ’s are bound state eigenfunctions of the equation ∇2 ψ(x, y, z) +
2m [E − V (x, y, z)]ψ(x, y, z) = 0 . ~2
(21.24)
The substitution τ = it (called Wick’s rotational transformation), ~ = 1 and m = 1 in Eqs. (21.22) and (21.23) gives 1 ∂ ψ(x, y, z, τ ) = ∇2 ψ(x, y, z, τ ) − V (x, y, z)ψ(x, y, z, τ ) ∂τ 2 and ψ(x, y, z, τ ) =
∞ X
Cn ψn (x, y, z)e−En τ .
(21.25)
(21.26)
n=0
In the long time limit limτ →∞ ψ ≈ C0 ψ0 e−E0 τ . Thus, in an initial wave function ψ(x, y, z, 0) if C0 6= 0 then after a sufficiently long time, the wave function ψ(x, y, z, τ ) → ψ0 (x, y, z). The obtained ψ0 is generally unnormalized. The corresponding energy E0 is obtained from Z ∞ ψ0∗ Hψ0 d3 X E0 = Z−∞ , d3 X = dx dy dz . (21.27) ∞ ∗ 3 ψ0 ψ0 d X −∞
21.3.2
Finite-Difference Algorithm [20]
Because the bound state wave functions are localized, in the numerical simulation we consider finite region of space. Let x ∈ [xmin , xmax ] and similarly for the y and z variables. We divide the x, y and z ranges into discrete lattice sites with spacing between the sites given by ∆x = (xmax − xmin )/Nx where Nx + 1 is the number of lattice points along the x-direction. The lattice points xi are given by xi = xmin + i∆x, i = 0, 1, · · · , Nx . Like-wise, define ∆y, ∆z, yi and zi . We denote the temporal increment as ∆τ . The wave function ψ(x, y, z, τ ) = ψ(xi , yj , zk , τn ) or ψ(i∆x, j∆y, k∆z, n∆τ ) is denoted as ψ n (i, j, k).
Numerical Methods for Quantum Mechanics 531 To discretize the time and space derivative in Eq. (21.25) we use the forward- and central-difference schemes respectively given by 1 n+1 ψ (i, j, k) − ψ n (i, j, k) , ∆τ 1 ≈ [ψ n (i + 1, j, k) − 2ψ n (i, j, k) + ψ n (i − 1, j, k)] , (∆x)2 1 ≈ [ψ n (i, j + 1, k) − 2ψ n (i, j, k) + ψ n (i, j − 1, k)] , (∆y)2 1 ≈ [ψ n (i, j, k + 1) − 2ψ n (i, j, k) + ψ n (i, j, k − 1)] . (∆z)2 (21.28) Further, ψ(x, y, z) ≈ ψ n+1 (i, j, k) + ψ n (i, j, k) /2. Substitution of (21.28) in (21.25) leads to ∂ψ ∂τ ∂2ψ ∂x2 ∂2ψ ∂y 2 ∂2ψ ∂z 2
≈
ψ n+1 (i, j, k) =
αψ n (i, j, k) +βγ1 [ψ n (i + 1, j, k) − 2ψ n (i, j, k) + ψ n (i − 1, j, k)] +βγ2 [ψ n (i, j + 1, k) − 2ψ n (i, j, k) + ψ n (i, j − 1, k)]
+βγ3 [ψ n (i, j, k + 1) − 2ψ n (i, j, k) + ψ n (i, j, k − 1)] , (21.29a)
where
α =
γ1
=
∆τ V (i, j, k) 1 2 , β= , ∆τ ∆τ 1+ V (i, j, k) 1+ V (i, j, k) 2 2 ∆τ ∆τ ∆τ , γ2 = , γ3 = . 2 2 2(∆x) 2(∆y) 2(∆z)2 1−
(21.29b)
(21.29c)
At each iteration the energy can be computed using (21.27). The discrete version of (21.27) is E0
=
P
Xh 1 V (i, j, k)ψ(i, j, k)2 2 ψ(i, j, k) i,j,k i,j,k
ψ(i, j, k) [ψ(i + 1, j, k) − 2ψ(i, j, k) + ψ(i − 1, j, k)] 2(∆x)2 ψ(i, j, k) − [ψ(i, j + 1, k) − 2ψ(i, j, k) + ψ(i, j − 1, k)] 2(∆y)2 i ψ(i, j, k) − [ψ(i, j, k + 1) − 2ψ(i, j, k) + ψ(i, j, k − 1)] . 2 2(∆z) (21.30) −
532 Quantum Mechanics I: The Fundamentals The boundary condition is ψ(x, y, z)boundary = 0. We note that since the final wave function ψ0 (x, y, z, τ ) is proportional to e−E0 τ , the magnitude of the final wave function may generally be quite small. For this reason we normalize the initial wave function and the wave function after each iteration. If ψ is a solution of the Schrödinger equation then −ψ is also a solution. The normalized ψ(x, y, z, τ = large) can be either +ψS or −ψS where ψS is a stationary state wave function.
Solved Problem 1: Obtain the stability condition for the FDTD scheme. The stability of the method, that is, the condition for the error occurred at a stage of computation to decay, can be analysed by the Fourier method (also called von Neumann analysis) [29]. To determine the stability condition consider the FDTD algorithm for the one-dimensional Schrödinger equation. In this case (21.29) becomes n n ψin+1 = αψin + β ψi+1 − 2ψin + ψi−1 , (21.31a)
where
α =
1 − (∆τ /2)Vi , 1 + ∆τ /2)Vi
β=
2∆x2
∆τ . (1 + (∆τ /2)Vi )
(21.31b)
Now, we assume the numerical solution as ψ = ψe + δψ, where δψ = eiax ebτ where ψe is the exact solution. Substituting x = i∆x, τ = n∆τ we get ψ = ψe + eiai∆x ebn∆τ . Define ξ = eb∆τ . Then ψ = ψe + eiai∆x ξ n . Substituting this in the finite-difference formula (21.31a) we get ψen+1 (i) + eiai∆x ξ n+1
That is,
= αψen (i) + αeiai∆x ξ n +β [ψen (i + 1) − 2ψen (i) + ψen (i − 1)] h i +β eia(i+1)∆x − 2eiai∆x + eia(i−1)∆x ξ n . (21.32)
i h eiai∆x ξ n+1 = αeiai∆x ξ n + β eia(i+1)∆x − 2eiai∆x + eia(i−1)∆x ξ n . (21.33)
The above equation gives ξ = α + β eia∆x − 2 + e−ia∆x = α − 4β sin2 (a∆x/2) .
(21.34)
For bound states the necessary and sufficient condition for stability is |ξ| ≤ 1. −1 Thus, for stability we require ∆τ ≤ 1/(∆x)2 . For the Eq. (21.29) the stability condition becomes −1 1 1 1 + + . (21.35) ∆τ ≤ (∆x)2 (∆y)2 (∆z)2
Numerical Methods for Quantum Mechanics 533
21.3.3
Application
In the following we apply the FDTD method to certain one- and twodimensional systems.
21.3.3.1 One-Dimensional Linear Harmonic Oscillator The ground and first excited states eigenvalues and eigenfunctions of the timeindependent Schrödinger equation of the linear harmonic oscillator with the potential V (x) = x2 /2 are E0 = 1/2, E1 = 3/2 and ψ0 =
1 −x2 /2 e , π 1/4
ψ1 =
2 2 x e−x /2 . (4π)1/4
(21.36)
We choose x ∈ [−5, 5], ∆x = 0.01, ∆τ = (∆x)2 /5 and the initial wave function as a random wave function, that is ψ(xi , 0) = ξi where ξi ’s are uniform random numbers in the interval [−0.5, 0.5] generated using Park–Miller method (see appendix G). The boundary condition is ψ(−5, 0) = ψ(5, 0) = 0. ψ(x, τ ) is normalized after every iteration. Figure 21.4a shows energy E versus τ . As τ increases E approaches the ground state energy value E0 = 0.5. In Fig.21.4b an arbitrary wave function evolves to the ground state wave function (−ψ0 ). Figures 21.4c and 21.4d show E versus τ and ψ(x, τ ) versus τ respectively where the initial wave function is chosen as ψ(x, τ = 0) = N sin 6x, N being the normalization constant. E → E1 = 1.5 and ψ(x, τ ) → −ψ1 . Other excited energy eigenvalues and eigenfunctions can be obtained with different initial wave function. The values of E0 and E1 obtained are 0.4999969 and 1.499984 respectively.
21.3.3.2 One-Dimensional Particle in a Box As a second example, consider a particle in a box potential V (x) = 0 for 0 ≤ x ≤ L and ∞ elsewhere. Its stationary state p eigenvalues and eigenfunctions are En = n2 π 2 ~2 /(2mL2 ) and ψn (x) = 2/L sin(nπx/L). We fix L = 1, ∆x = 0.01, ∆τ = (∆x)2 /5, ψ(x, τ = 0) as a random function and use the boundary condition ψ(0, τ ) = ψ(L, τ ) = 0. In Fig. 21.5a the numerically calculated energy is 4.934396 while the exact value of the ground state energy is 4.934802. The numerically calculated ψ at τ = 1 and the exact ground state ψ are plotted in Fig. 21.5b.
21.3.3.3 One-Dimensional Anharmonic Oscillator Another important quantum system is the anharmonic oscillator with the potential V (x) = (x2 − 2)2 . It is a double-well system and is used to describe structural phase transitions, tunneling of protons in hydrogen bonded systems and many other systems. Exact eigenvalues and eigenfunctions of the anharmonic oscillator is not known. Its ground state energy obtained by an another method [30] is 1.80081. The ground state energy eigenvalue obtained by the
534 Quantum Mechanics I: The Fundamentals 40
(a)
0.8
20 10 0 0
1
40
τ
2
3
1
0
-0.8 -5
0.5 0
x
(c)
30 20
50
0.8
10 0 0
1
τ
2
3
τ
(d) ψ(x, τ )
E
(b) ψ(x, τ )
E
30
0
-0.8 -5
0
x
5
3 2 τ 1 0
Variation of (a) ground state energy and (c) first excited state energy of linear harmonic oscillator with τ . The horizontal dashed line indicates the exact energy value. (b) Evolution of initial random wave function into the ground state wave function. The final wave function is ψ(x, τ = 1) = ψ0 . (d) Evolution of ψ(x, τ = 0) = N sin 6x into the first excited wave function. FIGURE 21.4
FDTD method is 1.80078 where we used ∆x = 0.01, ∆τ = (∆x)2 /5 and x ∈ [−5, 5]. Figures 21.5c and 21.5d present the results of the FDTD method.
21.3.3.4 Two-Dimensional Systems Next, we consider the two-dimensional harmonic oscillator with the potential V (x, y) = (x2 + y 2 )/2. Its exact ground state energy is 1. In the numerical simulation we choose x, y ∈ [−5, 5], ∆x = ∆y = 0.05, ∆τ = (∆x)2 /5 and ψ(x, y, τ = 0) as a random function. In Fig. 21.6a for large τ , E = 0.9998437. As expected the ground state wave function obtained (see Fig. 21.6b) is the two-dimensional Gaussian function. Figures 21.6c and 21.6d present the results for the two-dimensional box potential V (x, y) = 0 for 0 ≤ x, y ≤ 1 and ∞ otherwise. The exact eigenvalues and eigenfunctions are E(nx , ny ) = n2x + n2y π 2 /2, nx , ny = 1, 2, · · · and ψnx ,ny (x, y) = 2 sin(nx πx) sin(ny πy) for 0 ≤ x, y ≤ 1 otherwise 0. We fix ∆x = ∆y = 0.02, ∆τ = (∆x)2 /5. The numerically computed ground state energy is 9.866358 while the exact value
Numerical Methods for Quantum Mechanics 535 50
30
(b)
-ψ
E
2
(a) Box Potential
40
1
20 10 0
0
50
1
τ
0
0 1
(c) Anharmonic oscillator
40
0 .5
x
1
(d) Anharmonic oscillator
30
-ψ
E
0 .5
0 .5
20 10 0
0
0 .5
τ
1
0
-5
0
x
5
FIGURE 21.5 (a) Variation of ground state energy with τ and (b) the final wave function computed by the FDTD method (continuous line) and the exact wave function (solid circles) of a particle in a box potential. The horizontal dashed line indicates the exact energy value. The results for the anharmonic oscillator are presented in (c) and (d).
is 9.869604. For ∆x = 0.01 the obtained energy value is 9.868793. That is, accuracy can be improved by decreasing ∆x, ∆y and ∆τ .
21.4 TIME-DEPENDENT SCHRÖDINGER EQUATION In sec. 21.2 we described the finite-difference matrix method for computing energy eigenvalues and eigenfunctions of bound states of time-independent Schrödinger equation in one-dimension. In the present section we consider the problem of solving the time-dependent Schrödinger equation. Interestingly, the finite-difference scheme can be extended to the study of time development of certain initial wave profile [15]. We note that solving the time-independent Schrödinger equation is a boundary-value problem while solving the timedependent Schrödinger equation is essentially an initial-value problem. Setting ~ = 1, m = 1 the time-dependent Schrödinger equation becomes i
∂ψ 1 ∂2ψ =− + V (x)ψ . ∂t 2 ∂x2
(21.37)
536 Quantum Mechanics I: The Fundamentals
FIGURE 21.6 (a) Variation of ground state energy with τ and (b) the final wave function of two-dimensional linear harmonic oscillator computed by the FDTD method. The horizontal dashed line indicates the exact energy value. The results for a particle in a two-dimensional box potential are presented in (c) and (d).
We assume the boundary conditions ψ(x, 0) = φ(x) and ψ(±∞, t) = 0. As usual we discretize the space variable x ∈ [xmin , xmax ] as x0 , x1 , · · · , xN with xi+1 − xi = ∆x and t ∈ [0, T ] as 0, t1 , t2 , · · · , tm with tj+1 − tj = ∆t. The solution ψ(x, t) at the grid point (i, j) is denoted as ψi,j . We approximate ψ, ∂ψ/∂t, ∂ 2 ψ/∂x2 as ψ(x, t) ∂ ψ(x, t) ∂t ∂2 ψ(x, t) ∂x2
= ψ(xi , tj ) = = =
1 (ψi,j + ψi,j−1 ) , 2
1 (ψi,j − ψi,j−1 ) , ∆t 1 [(ψi+1,j − 2ψi,j + ψi−1,j ) 2(∆x)2 + (ψi+1,j−1 − 2ψi,j−1 + ψi−1,j−1 )] .
(21.38a) (21.38b)
(21.38c)
Using Eqs. (21.38) the time-dependent Schrödinger equation is rewritten as aψi−1,j + dψi,j + cψi+1,j = bi .
(21.39)
Numerical Methods for Quantum Mechanics 537 where a bi
1 1 2i , di = − − Vi , = 2(∆x)2 ∆t (∆x)2 1 2i = + + Vi ψi,j−1 ∆t (∆x)2 1 (ψi+1,j−1 + ψi−1,j−1 ) . − 2(∆x)2
(21.40a)
(21.40b)
With ψ0,j = 0 and ψN,j = 0 Eq. (21.39) is written in matrix form as d1 a1 ψ1,j b1 a d a ψ b 2 2 1 2,j 2 b3 a2 d3 ψ3,j . .. .. = . . . . dN −3 aN −3 ψN −3,j bN −3 aN −3 dN −2 aN −2 ψN −2,j bN −2 aN −2
dN −1
ψN −1,j
.
bN −1 (21.41) The above equation is of the form Aψ = B. An interesting observation on the matrix A is that it is a symmetric tridiagonal form and can be solved easily. The system given by Eq. (21.41) can be solved by the QL algorithm described in appendix F.
Solved Problem 2:
√ 2 The initial wave function of a free particle is ψ(x, 0) = (1/ 2π ) e−x . Numerically compute ψ(x, t) and state its behaviour. For numerical computation we choose ∆x = 0.05, ∆t = 0.01, x ∈ [−5, 5]. Since the Schrödinger equation is a complex equation, ψ(x, t) is complex even if ψ(x, 0) is a pure real. Therefore, ψ(x, t) is declared as a complex variable in the numerical algorithm of solving the Eq. (21.41). Real and imaginary parts of complex ψ are obtained as ψR = [ψ + conjg(ψ)]/2,
ψI = −i[ψ − conjg(ψ)]/2,
(21.42)
where conjg(ψ) denotes complex conjugate of ψ, that is ψ ∗ . |ψ|2 = ψ ∗ ψ is obtained through |ψ|2 = ψ × conjg(ψ) . (21.43) The wave is centered at x = 0. As expected the numerically computed ψ(x, t) spreads over the space variable. This is clear from Fig. 21.7.
538 Quantum Mechanics I: The Fundamentals
0.8
|ψ|2
ψ(x, 0) = N exp(−x2)
0.4 40
3
t
FIGURE 21.7
ticle. 21.4.1
2
1
0 -4
-2
0
x
2
4
Numerically computed time evolution of |ψ|2 of free par-
Application
In the following we solve the system (21.41) for certain quantum mechanical systems and capture the time evolution of an initial wave function.
21.4.2
Harmonic Oscillator
For the harmonic oscillator with the potential V (x) = kx2 /2 the stationary state energy eigenvalues and eigenfunctions are given by 2 2 1 En = n + ~ω, ψn (x) = Nn Hn (αx) e−α x /2 , (21.44) 2 p √ 1/2 where α = mω/~, Hn are Hermite polynomials and Nn = (α/( π 2n n!)) . The exact time evolution of ψn (x) is given by ψn (x, t) = ψn (x) e−iEn t/~ .
(21.45)
We solve the Schrödinger equation with ψ0 (x, 0) as the initial wave function. We choose ∆x = ∆t = 0.05 and x ∈ [−5, 5]. Further, we fix m = 1, ω = 1 and ~ = 1. Figure 21.8 shows the numerically computed real and imaginary parts of ψ0 and |ψ0 |2 . ψ0 (x, t) → 0 as x → ±∞. ψR and ψI evolve periodically in time while |ψ(x, t)|2 is independent of time. The period of the wave function is 4π. Figure 21.9 depicts the numerically computed time evolution of the following two initial states √ √ ψ(x, 0) = 0.1ψ0 + 0.9ψ1 , (21.46a) √ √ √ ψ(x, 0) = 0.35ψ0 + 0.15ψ1 + 0.5ψ2 . (21.46b)
Numerical Methods for Quantum Mechanics 539
0.8
5
ψR 0 -0.8 40
0 30
t
20
10
0.6
|ψ|2
-5 0
0 40 30
t
0.8
ψI
x
5
0 -0.8 40
0 30
t 20
10
20
5
10 0 -5
0x
x
-5 0
Numerically computed time evolution of real and imaginary parts of ground state eigenfunction of harmonic oscillator. |ψ(x, t)|2 is also plotted.
FIGURE 21.8
21.4.3
Particle in the Presence of a Step Potential
Next, we consider a step potential V (x) = 0 for x ≤ x0 and V0 for x > x0 which has no bound states. The incident wave function undergoes reflection and re√ 1/2 −x2 ik x fraction. Suppose the initial wave function is ψ(x, 0) = 1/ 2π e e 0 . In the numerical simulation we fix x0 = 5, k0 = 5, ∆x = ∆t = 0.05 and x ∈ [−10, 20]. Figure 21.10 shows ψR and |ψ|2 for V0 = 5, 10 and 50. In Figs. 21.10a and 21.10b for V0 = 5 and 10 we note that part of the incident wave is reflected while the remaining part is refracted. The amplitude of the reflected wave is relatively larger for V0 = 10. We expect complete reflection for sufficiently large value of V0 . This is the case shown in Figs. 21.10c and Figs. 21.11c for V0 = 50.
21.4.4
Tunneling
Now, we show that tunneling phenomenon can also be captured by numerically solving the time-dependent Schrödinger equation by the finite-difference method. We consider the rectangular barrier potential with height V0 and width 0.1. That is, V (x) = V0 for x ∈ [5, 5.1] and 0 elsewhere. We choose the 2 initial wave function as e−x +i5x and fix ∆x = ∆t = 0.05 and x ∈ [−10, 20]. The time evolution of ψR is shown in Fig. 21.12 for V0 = 5, 20 and 500. For small values of V0 , ψR is nonzero in the classically forbidden region x > 5.1.
540 Quantum Mechanics I: The Fundamentals
ψ=
√
0.1ψ0 +
√
0.9ψ1
|ψ|2
0.6 0 9 6
t
3 0 √
0.35ψ0 +
5
x
√ √ 0.15ψ1 + 0.5ψ2
|ψ|2
ψ= 0.6 0 9
-5
0
t
6 3 0
FIGURE 21.9
(21.46).
-5
0
5
x
Numerically computed evolution of |ψ(x, 0)|2 given by
The amplitude of the wave in this region decreases with increase in V0 . For V0 = 500 (see Fig. 21.12c) ψR is almost negligible in magnitude.
21.4.5
Wave Packet Revivals
In the final example, we consider a particle bouncing on a perfectly reflecting surface under the influence of gravity. The potential of the particle is v(x) = mgx for x > 0 and ∞ for x < 0. We illustrated the wave packet revival phenomenon in this system in sec. 10.3. Now, we show this feature can also be reproduced through numerical simulation. Figure 21.13 shows the plot of
ψR
Numerical Methods for Quantum Mechanics 541
1 -1 2
10
ψR
t 1 (a) V0 = 5
5 0
0
x
1 -1 2
10
t 1 (b) V0 = 10
5 0
0
x
ψR
1 -1 2 10
1
t (c) V0 = 50
5 0
0
x
Time evolution of real part of ψ of a particle in the presence of a step potential.
FIGURE 21.10
|ψ|2
542 Quantum Mechanics I: The Fundamentals
1 0 2
t
10
1
5 0
0
x
|ψ|2
(a) V0 = 5
1 0
2
10
t
1 0
0
x
|ψ|2
(b) V0 = 10
5
2 0 2
t
10
1
(c) V0 = 50
5 0
0
x
Time evolution of |ψ|2 of a particle in the presence of a step potential.
FIGURE 21.11
Numerical Methods for Quantum Mechanics 543
ψR
1 -1 3 2
t
10 5
1
(a) V0 = 5
0 -5
0
x
ψR
1 -1 3 2
t
10 5
1
(b) V0 = 20
0 -5
0
x
ψR
1 -1 3 2
t
10 1
(c) V0 = 500 0 -5
5 0
x
Numerically computed time evolution of real part of wave function ψ in the presence of a rectangular barrier of height V0 and width 0.1 at x = 5. The effect of V0 is clearly seen. FIGURE 21.12
544 Quantum Mechanics I: The Fundamentals
|ψ|2
0.6 0.3 0 10
8
t
6
4
2
0 0
5
10
15
20
x
Numerically computed time evolution of |ψ|2 of the Gaussian wave function given by Eq. (21.47) in the presence of the potential V (x) = mgx for x > 0 and ∞ for x < 0. FIGURE 21.13
|ψ|2 versus t of the Gaussian wave function
1/4 −(x−x0 )2 /σ2 ψ(x, 0) = 2/(πσ 2 ) e
(21.47)
at selected values of t where x0 = 10, σ = 2, g = 1 and m = 1. The localized wave packed solution spreads in time but at a later time T it reforms and appears as its initial state (the last continuous curve). (The dashed curve is the form of the initial wave.) This is called wave packet revival .
21.5 QUANTUM SCATTERING Having studied how to compute eigenvalues and eigenfunctions of stationary states of quantum mechanical systems we turn our attention to scattering potentials. A systematic numerical procedure for computing transmission coefficient was described by Pang [17]. We show how transmission probability of quantum mechanical barriers can be computed numerically. Quantum mechanical scattering and tunneling problems in one-dimension are described by the Schrödinger equation d2 ψ(x) + 2[E − V (x)]ψ(x) = 0 , dx2
(21.48)
where ~2 and m are set unity. Let us rewrite the above equation as dy1 = y2 , dx
dy2 = 2[V (x) − E]y1 , dx
(21.49)
where y1 = ψ. Assume that V (x) is nonzero only in the region x ∈ [0, a] and
Numerical Methods for Quantum Mechanics 545 the incident particle comes from the left with a unit flux. The general solution of Eq. (21.48) is written as eikx + Be−ikx , x < 0 ψ(x) = (21.50) Ceik(x−a) , x>a
where the parameters B and p C are generally √ complex and are to be determined. The wave vector k is 2mE/~2 = 2E. The solution ψ(x) in the region x ∈ [0, a] is to be obtained numerically. The complex solution and its first derivative satisfy the continuity conditions at x = 0 and a. These conditions are ψ(0) = ′
ψ (0) =
1 + B,
ψ(a) = C,
ik(1 − B),
′
ψ (a) = ikC.
(21.51a) (21.51b)
Once the constants B and C are known the transmission and reflections probabilities are obtained from the relations T = |C|2 , R = |B|2 and T + R = 1.
21.5.1
Numerical Solution in the Region 0 ≤ x ≤ a
Now, the intricate problem is to determine the solution of Eq. (21.48) in the region x ∈ [0, a] obeying the continuity conditions, Eq. (21.51). The correct solution is obtained using the numerical procedure outlined below. The procedure consists of four main steps. 1. For a given potential barrier and energy value make an initial guess for the parameter B = Br + iBi with |Br |, |Bi | ≤ 1. We can choose Br = α, √ Bi = 1 − α2 , |α| ≤ 1. 2. Integrate Eq. (21.49) from x = 0 to a using, say, the fourth-order Runge– Kutta method with the initial condition y1 (0) = 1 + B and y2 (0) = ik(1 − B). This is only a trial solution and not the solution satisfying the conditions, Eqs. (21.51). 3. From Eqs. (21.51) we have two expressions for C: C = C1 = y1 (a),
i C = C2 = − y2 (a) . k
(21.52)
These two values depend on the values of Br and Bi . Compute C1 and C2 from Eq. (21.52). 4. We require C1 = C2 . This will happen when the trial solution is exact. Introduce the function F = |C1 − C2 |2 . (21.53) Minimize the function F by varying y1 (a) and y2 (a) so that F ≈ 0. When this is achieved C1 and C2 are identical.
546 Quantum Mechanics I: The Fundamentals
21.5.2
Minimization of F
A simple method of minimization of a multivariable continuous function is the Nelder–Mead simplex method (NMS) [28]. The method is simple, effective and computationally compact. For the basic idea of the method with an example see the problem 21.12 at the end of this chapter. The method starts with (N + 1) initial guess for an N -dimension function. For the function F given by Eq. (21.53), C1 = C1R +iC1I , C2 = C2R +iC2I . If real and imaginary components are treated as separate variables then F is a function of 4 variables, that is N = 4. Therefore, 5 initial guesses on B p are needed. Br and Bi are chosen as BR = 0.2, 0.4, 0.6, 0.8, 1 and BI = 2 . The variables B, C, y and y are complex functions. Therefore, the 1 − BR 1 2 Eqs. (21.49), (21.51), (21.52) and (21.53) are rewritten as follows by treating real and imaginary parts separately so that the variables involved are all real: dy1R dx dy2R dx
dy1I = y2I , dx
=
y2R ,
=
2 [V (x) − E] y1R ,
(21.54a) dy2I = 2 [V (x) − E] y1I , (21.54b) dx
with the boundary conditions y1R (0) = y2R (0) =
1 + BR , y1I (0) = BI , kBI , y2I (0) = k(1 − BR ),
(21.55a) (21.55b)
and C1R
=
C2R
=
F
=
y1R (a), C1I = y1I (a), 1 1 y2I (a), C2I = − y2R (a) k k y2I 2 y2R 2 y1R − + y1I + . k k
(21.56a) (21.56b) (21.56c)
For the chosen 5 set of values of (BR , BI ) the corresponding y’s at x = a are obtained by solving the Schrödinger equation by the fourth-order Runge– Kutta method. After this step, NMS method is used to find correct y1 (a) and y2 (a) for which F ≈ 0. Minimization of F is performed actually by varying y1 (a) and y2 (a) and not by directly varying BR and BI . The transmission probability T and the reflection probability R are then computed from the 2 2 expressions T = |C|2 = C1R + C1I and R = 1 − T .
Solved Problem 3: A standard example for a quantum tunneling is the square-barrier potential V (x) = V0 for 0 ≤ x ≤ a and 0 otherwise. Applying the NMS method compute numerically the transmission probability T and compare it with the exact T .
Numerical Methods for Quantum Mechanics 547
Tnum
1
0 .5
0
0
1
2
3
4
5
E Numerically computed transmission probability Tnum (continuous line) as a function of energy for the square-barrier potential with V0 = 1 and a = 5. The exact values of T for a few values of E are marked by solid circles.
FIGURE 21.14
The Schrödinger equation for this potential is exactly solvable. The analytic expression for Texa is given by 4µ (1 + µ)2 sin2 ka + 4µ cos2 ka−1 , E > V0 Texa = (21.57a) −1 2 2 ′ 2 ′ ′ ′ 4µ (1 − µ) sinh k a + 4µ cosh k a , E < V0
where
µ = k
=
(E − V0 )/E, µ′ = E/(V0 − E), p p 2(E − V0 ) , k ′ = 2(V0 − E) .
(21.57b) (21.57c)
Suppose a = 5 and V0 = 1. Step size in the Runge–Kutta method is h = 0.01. Tolerance in the NMS is fixed as 10−6 . Developing a program to compute T as a function of energy is given as a problem (see the problem 21.13 at the end of this chapter). Figure 21.14 shows numerically computed T , Tnum , as a function of E. Table 21.5 presents the values of Tnum , Texa , Texa − Tnum and percentage of error, 100 × (Tnum − Texa )/Texa for several values of E. Very good agreement between numerical and analytical T is observed. The accuracy can be improved by choosing smaller values of step size h in the Runge–Kutta method and the tolerance value in the NMS method.
Solved Problem 4: T for a locally periodic potential with the unit cell being Dirac delta-function exhibits a band structure character. This property is found to occur for the
548 Quantum Mechanics I: The Fundamentals TABLE 21.5
Results for the square-barrier potential with V0 = 1 and
a = 5. E
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
Transmission probability Texa − Tnum Tnum
Texa
0.000180 0.074227 0.766294 0.940954 0.968169 0.988881 0.973517 0.997733 0.994056 0.988368
0.000182 0.074074 0.765397 0.940901 0.969047 0.987819 0.973134 0.997956 0.994032 0.987655
0.1240E-05 −0.1528E-03 −0.8973E-03 −0.5265E-04 0.8779E-03 −0.1063E-02 −0.3829E-03 0.2231E-03 −0.2387E-04 −0.7132E-03
% of error 0.683 0.206 0.117 0.006 0.091 0.108 0.039 0.022 0.002 0.072
potential consisting of finite number of identical square-barriers. The potential is given by V , (i − 1)2b ≤ x ≤ (2i − 1)b 0 V (x) = (21.58) 0, otherwise
where i = 1, 2, · · · , N . Numerically compute T for a few values of N and point out the features of T .
For the numerical simulation we fix b = 1 and V0 = 10. Figure 21.15 shows Tnum versus E for N = 1, 2, 4 and 10. One of the most striking features of this figure is the appearance of energy band structure similar to the Dirac comb. The number of maxima with Tnum = 1 increases with increase in N . For N = 1, 2 and 4 Tnum is always greater than zero. But for large N , for example N = 10 (Fig. 21.15d), Tnum is zero or ≈ 0 for a range of E. Thus, for large N , by suitably choosing E, V0 and a we can have almost complete transmission or complete reflection.
21.6 ELECTRONIC DISTRIBUTION OF HYDROGEN ATOM The Monte Carlo technique can be used to represent electronic distribution of hydrogen atom [22], particles distribution in harmonic oscillator, tunneling through a potential barrier and scattering of particles from a potential. In this section the electronic distribution of hydrogen atom is considered. Let us consider the ground state of hydrogen atom. It is given by ψnlm = √ ψ100 = (1/ π )e−r . The probability density for the electron is P100 (r) = |ψ100 |2 = e−2r /π. If we consider Cartesian coordinate system then r2 = x2 +
Numerical Methods for Quantum Mechanics 549
Tnum
1
0 .5
0 10
N =1 20
30
40
(a) 50
E
Tnum
1
0 .5
0 10
N =2 20
30
40
(b) 50
E
Tnum
1
0 .5
0 10
N =4 20
30
40
(c) 50
E
Tnum
1
0 .5
0 10
N = 10 20
30
40
(d) 50
E Numerically computed transmission probability as a function of energy E for a locally periodic potential with square-barrier as a unit cell. (a) For one barrier, (b) for two barriers, (c) for four barriers and (d) for ten barriers. Here V0 = 10 and b = 1. FIGURE 21.15
550 Quantum Mechanics I: The Fundamentals y 2 + z 2 . For the present discussion the multiplicative factor 1/π in P100 is dropped so that P100 = e−2r . Suppose we wish to obtain electronic distribution in x−y −z plane. Monte Carlo technique can be employed to this problem. For convenience let us choose the plane x − z with y = 0. Essentially, a sequence of N pair of points (x, z) with the distribution P100 = e−2r represent the electronic distribution in ground state. Thus, the problem is to generate a set of numbers {(x, z)} with the probability distribution P100 . How do we generate {(x, z)} with P100 = e−2r ? A sequence of numbers with a desired distribution can be obtained using the simple and efficient random sampling technique called accept-reject or simply rejection method. This method is described in appendix G for a function f (x). This method can be easily extended for f (x, z).
21.6.1
Rejection Technique to Generate Random Numbers with the Dis2 2 tribution e−(x +z )
Let f (x, z) be the desired distribution defined over the interval x ∈ [ax , bx ], z ∈ [az , bz ]. Select a suitable bounding function cg(x, z) with maximum of 2 2 cg(x, z) = cgm is ≥ maximum of f (x, z). For the exponential function e−(x +z ) we may choose x ∈ [−10, 10], z ∈ [−10, 10] and maximum of cg(x, z) as 2 2 cgm = 1 since the maximum of e−(x +z ) is 1 (cgm must be ≥ 1). Now, treat x and z are as uniformly distributed random numbers in the range [−10, 10] and cg(x, z) as uniformly distributed random numbers in the range [0, 1] since f (x, z) ∈ [0, 1]. Then the scheme is the following: 1. Choose a value for x and a value for z. 2. Choose a value for cg(x, z). 3. Compute f (x, z). 4. If f (x, z) ≤ cg(x, z) accept the pair (x, z). Otherwise reject it. Repeating the above algorithm we can generate a desired number of (x, z). The obtained pairs of numbers obey the chosen distribution f (x, z). Using the above method we can generate any number of pairs of random numbers (or in general n-dimensional random numbers) satisfying a given distribution.
21.6.2
Electronic Distribution in Various States
A program can be developed to generate a sequence of, say, N = 2000 pairs of points (x, z) with f (x, z) = P (x, y = 0, z) = e−2r (see the problem 21.17 at the end of this chapter). This corresponds to ground state nlm = 100. The values of a, b and cgm are chosen as −10, 10 and 1 respectively. The plot of x versus z represents electronic distribution of hydrogen atom in ground
Numerical Methods for Quantum Mechanics 551 state. To obtain the distribution in other states the corresponding expression for P (x, y, z) is to be used. Further, we have to choose appropriate values for a, b and cgm . These values for some of the states are mentioned in the program. Figure 21.16 displays the electronic distribution in several states. The distributions corresponding to nlm = 100, 200 and 300 exhibit spherical symmetry around their centre. The distributions for nlm = 210, 211, 310 and 311 show rotational symmetry around the z-axis. The influence of applied uniform electric field along z-axis on the probability distribution, called Stark effect can be visualized by the Monte Carlo method.
21.7 SCHRÖDINGER EQUATION WITH AN EXTERNAL FIELD In this section we present the coupled channel method (CCM) [21] of solving the time-dependent Schrödinger equation with an external field. For i~
∂ψ = Hψ , ∂t
where H = H (0) + λH (1) we write the solution as X ψ(x, t) = an (t)φn (x)e−iEn t/~ .
(21.59)
(21.60)
n
In the above solution φn (x) and En are the eigenfunctions and energy eigenvalues respectively of the unperturbed Hamiltonian H (0) and an (t) are the 2 expansion coefficients. Pn (t) = |aP n (t)| is the probability of finding the system in the state n at time t and n |an (t)|2 = 1. To determine an ’s substitute (21.60) in Eq. (21.59) and obtain i~
X dan n
dt
φn e−iEn t/~ = λ
X
an H (1) φn e−iEn t/~ .
(21.61)
n
Multiplying Eq. (21.61) by φ∗m and integrating over all space we get dam dt
=
(1) Hmn (t)
=
λ X (1) an Hmn (t)eiωmn t , i~ n Z ∞ φ∗m (x)H (1) (x, t)φn (x) dx
(21.62) (21.63)
−∞
and ωmn = (Em − En )/~. The above equation is exact. This equation is essentially a system of coupled integro-differential equation which is difficult to solve analytically. Therefore, we pose the problem in a slightly different manner. Suppose the perturbation is switched-on at t = 0 and switched-off at t = T . The system is, say, initially in the nth state and after the time T it settles on to the state m. In numerical simulation we choose a small number of states N , for example, 1 < N < 5. In this case Eq. (21.62) becomes a closed system of N coupled ordinary differential equations which we can solve by the
552 Quantum Mechanics I: The Fundamentals 10
20
nlm = 200 z
z
nlm = 100 0
-10 -10
0
10
x
0
-20 -20
nlm = 211 z
z
nlm = 210 0
0
0
x
0
-20 -20
0
-
-
0
0
x
20
nlm = 310 z
z
nlm = 300
-
20
20
0
- 0 - 0
0
x
x
0
-
0
x
z
nlm = 311
-
0
-
0
x
Electronic distribution of the hydrogen atom in some of its states produced by 2000 points in x − z plane with y = 0. FIGURE 21.16
Numerical Methods for Quantum Mechanics 553 fourth-order Runge–Kutta method by treating an (t) as complex or writing an (t) = Rn (t) + iIn (t). In the latter case we have dRm (t) dt
=
dIm (t) dt
=
1X (1) [Rn sin ωmn t + In cos ωmn t] Hmn (t) , ~ n 1X (1) [In sin ωmn t − Rn cos ωmn t] Hmn (t) . ~ n
(21.64a) (21.64b)
For a given system with an interaction term H (1) (x, t) consider first, say, five states, that is N = 5 (if n = 0 is not a quantum number, otherwise (1) N = 4). Hmn (t) given by Eq. (21.63) can be calculated either analytically or numerically. With an initial condition an (0) = Rn (0) + iIn (0), the 2N coupled Eqs. (21.64) can be solved for 0 ≤ t ≤ τ . Then, |an (t)|2 = [Rn (t)]2 + [In (t)]2 .
Solved Problem 5: Consider the linear harmonic oscillator with H (0) = p2x /(2M ) + M ω 2 x2 /2 interacting with a laser field. M denotes mass of the particle. Suppose the particle is subjected to a laser pulse of electric field F (t) given by F (t) = f0 sin2 (πt/τ ) cos ωL t ,
0≤t 0; a is the multiplier with 0 < a < m; c is the increment with 0 ≤ c ≤ m; I1 is the starting value with 0 ≤ I1 < m. We start with an integer I1 and calculate aI1 + c which is also an integer. We divide it by m, obtain the remainder and denote it as I2 . Repeating the above 575
576 Quantum Mechanics I: The Fundamentals procedure we can generate a sequence of integers {I1 , I2 , · · · }. The sequence obtained with m = 1000, I1 = 10, a = c = 89 is 979, 220, 669, 630, 159, · · · . Such a sequence generated from Eq. (G.1) will repeat itself with a period less than m. If we divide the obtained numbers by m we get a sequence in the interval [0, 1]. We can convert it into any suitable range [α, β] through xi → (β − α)xi + α. Park and Miller proposed a generator based on the choices [1] a = 75 , c = 0, m = 231 − 1 = 214783647.
(G.2)
When the value of aIj is greater than the maximum value of, say, a 32-bit integer then we cannot implement the Eqs. (G.1)-(G.2) in a computer. To overcome this problem the following suggestion of Schrage is very helpful. From the approximate factorization of m given by m = aq + r ,
q = int.[m/a], r = m (mod a),
(G.3)
where int denotes integer part. For r < q and 0 < I < m − 1 both aI(mod q) and r.int.[I/q] lie in the range [0, m − 1]. We can write ( aI (mod q) − r.int.[I/q], if it is > 0 (G.4) aI (mod m) = aI (mod q) − r.int.[I/q] + m, otherwise. With P2 = int.[I/q]
(G.5)
aP1 − rP2 , if it is > 0 aP1 − rP2 + m, otherwise.
(G.6)
P1 = I (mod q), we rewrite the Eq. (G.4) as aI (mod a) =
(
Some useful choices of a, q and r are given below: (i) a = 75 , q = 127773, r = 2836, (ii) a = 48271, q = 44488, r = 3399 and (iii) a = 69621, q = 30845, r = 23702. The period of the numbers generated using the above procedure is 231 − 2 = 2147483646. The above procedure generates numbers in the interval [0, m]. Note that the random numbers are now integers. The numbers generated can be converted into the range [0, 1] by dividing them by m. The following steps generate N random numbers in the interval [0, 1]. M = 2147483647 A = 7**5 R = 2836 Q = 127773 I=3 DO J = 1, N P1 = MOD(I,Q)
Random Number Generators for Desired 1
Distributions 577
(a)
0 .5
0
0
25
50
75
0.03
100
(b)
0.02 0.01 0
0
0.25
0 .5
0.75
1
(a) Plot of first 100 uniform random numbers (x) generated. (b) Computed probability distribution of 5×104 numbers. The interval [0, 1] is divided into 50 subintervals. P (x) is almost constant. FIGURE G.1
P2 = INT(I/Q) I = A*P1 - R*P2 IF(I.LE.0) I = I + M RAN = FLOAT(I)/FLOAT(M) END DO Figure G.1a depicts first 100 random numbers of the sequence generated. The mean and variance obtained with 104 numbers are 0.5017 and 0.0833 respectively whereas the exact values are 0.5 and 0.0833 · · · respectively.
B. GENERATION OF RANDOM NUMBERS WITH A DESIRED DISTRIBUTION
Using uniform random numbers we can generate random numbers of other types of distribution like Gaussian, exponential, binomial, etc. This can be achieved by employing the methods such as direct inversion, rejection method and Metropolis. We describe the rejection technique [2]. Suppose f (x) be the desired distribution defined over the interval [α, β]. We select a suitable bounding function g(x) such that cg(x) ≥ f (x) for all 2 values of α ≤ x ≤ β. For the exponential function f (x) = e−x for α ≤ x ≤ β and 0 otherwise we can choose the range [α, β] as [−5, 5]. Since f (x) ≤ 1 for
578 Quantum Mechanics I: The Fundamentals 0.6
P (x)
f (x)
1
0.5
0 -5
(b)
(a)
cg(x)
0
x
5
0.4 0.2 0 -5
0
5
x
(a) Description of rejection technique. The x-component of a point falling below the curve f (x) is a number with the chosen exponential distribution. (b) Probability distribution (unnormalized) of 5 × 105 random numbers generated. FIGURE G.2
x ∈ [α, β] we have cg(x) ≥ 1. We set c = 1 so that g(x) ∈ [0, 1]. We choose g(x) and x as uniformly distributed random numbers in the intervals [0, 1] and [−5, 5] respectively. The scheme is the following. 1. Choose a value of x randomly in the interval [−5, 5] and calculate f (x). 2. Choose a value of cg(x), the value of g(x) is between 0 and 1, and call it y. 3. If y > f (x) then go to step (1) otherwise x is a member of the chosen distribution number. We can obtain a sequence of random numbers with the distribution f (x) by repeating the above steps. The procedure is simply the following. Consider the Fig. G.2a. The dots are the points with coordinate values (xi , yi ) with xi ∈ [−5, 5] and yi ∈ [0, cg] (= [0, 1]). Discard the points falling outside the curve f (x). The x values of the points lying within the curve form random 2 numbers with the distribution e−x . Here cg(x) not necessarily defined in [0, 1]. It can be [0, 1.5] or [0, 2] or [0, > 1]. The number of points discarded to generate N numbers with a required distribution depends on the choice of cg(x). The number of discarded points will be minimum if the maximum value of cg(x) is chosen as the maximum value of f (x). Figure G.2b shows the probability distribution function P (x) of generated random numbers.
BIBLIOGRAPHY [1] W.H. Press, S.A. Teukolsky, W.T. Vetterling and B.P. Flannery, Numerical Receipes in Fortran. Foundation Books, New Delhi, 1998. [2] K.P.N. Murthy, Monte Carlo Basics. Indian Society for Radiation Physics, Kalpakkam, 2000.
Solutions to Selected Exercises ◦
1.2
3.106A .
1.4
1674 nm.
1.6
2.7756 × 10−19 J.
1.8
24847.5 V.
1.10
130797 × 1018 photons.
1.12
λ′ = 0.76224A and E = 2.60784 × 10−15 J.
1.14
3.
◦
2.2
i~∂ψ/∂t = −(~2 /2m)∇2 ψ − (qE(t) · X)ψ.
2.6
Not equal to E1 + E2 .
2.8
V (X, t) = V ∗ (X, −t),
2.14
ux − u2 − λ + g = 0, λ = 2mE/~2 and g = 2mV /~2 .
2.22
√ 1/2 1/ (σ π ) . p 2(1 + α2 )/α3 .
2.26
1 − e−b b2 + 2b + 2 /4. 0.
2.28
3~ω/2.
2.30
1/2.
2.34
~π/(E2 − E1 ).
2.38
5~2 /(mL2 ). √ 3 1 φ1 + φ2 . 2 2
2.18 2.20
2.44
579
580 Solutions to Selected Exercises 2.48 2.50
P
n
|Cn |2 En .
n2 ~2 π 2 /(2ma2 ).
3.2
x6 + 10x4 + 25x2 .
3.6
Not linear.
3.8
To shift t by t + τ .
3.12
y = A sin nx and λ = n2 , n = 1, 2, · · · .
3.16
ψ(x) = N eikx , E = V0 + (h2 k 2 /(2m)), k 2 = 2m (E − V0 ) /~2 .
3.18
(i) f (0). (ii) Iˆ2 .
3.22
[xn , px ] = i~nxn−1 and [x, pnx ] = i~npn−1 . x
3.26
(i) i~. (ii) 0.
3.30
[H, x] = −i~px /m and [H, px ] = 0.
3.34
0.
3.38
No.
3.46
To change A to A′ = T AT −1 . ′
4.2
ψ(x) = Ae±ik x and E = a + (~2 k ′2 /(2m)).
4.6
hxi = hpx i = 0.
4.8
~ω.
4.10
0.888.
4.14
En =
4.18 4.22 4.26
1 e2 E 2 n+ ~ω − . 2 2mω 2 √ a = −1/(2b), b = mk/~, N = (4b5 /π)1/4 and E = 5~ω/2.
1.50574 × 10−20 Js.
hxi = hpx i = 0, hx2 i = a2 n2 π 2 − 6 /(3n2 π 2 ), hp2x i = ~2 n2 π 2 /(4a2 ).
4.30
hEi = 5~2 /(4ma2 ).
4.32
c/(2a). p N = 15/(16a5) .
4.34 4.36 4.38
1099.9608nm.
−V0 + (~2 π 2 /(4ma)).
Solutions to Selected Exercises 581 1 i mω tan(ωt + α0 ). β(t) = − ln [cos(ωt + α0 )] + β0 . 2~ 2
4.42
α=
4.46
ψ(x) = 2β 3/2 x e−βx , β = ma/~2 and E = −ma2 /(2~2 ).
5.2 5.4 5.6 5.8
1−
V02 . (2E − V0 )2
−1 1 + (7/3)sinh2 (3/2) . √
[16/(4 + 3.2)] e−160 20/~ . √ √ (1 + 2)2 V0 /(4 2 ) eV.
5.10
E = 9V0 /8.
5.12
(i) 0.83 × 10−18 m. (ii) 0.7809 × 10−10 m. P P ∗ i j Di Cj hi|A|ji.
6.2 6.4 6.6
Normalized. 1 1 1 1 5 (b) √ . (c) 2|v1 i − 3|v2 i and − √ |v1′ i + √ |v2′ i. 2 1 −1 2 2
6.8
∂U/∂t = (1/i~)HU .
6.10
[C].
6.12
A = A† = A−1 .
6.14
λ∗ B|ψ2 ihψ1 |A† .
6.16
|ci |2 . eθ 0
6.18
6.20
3 2
0 e−θ
~ mω
.
2 1 n2 + n + . 2
7.2
˙ f˙A + f A.
7.4
(a) hxpx + px xi/m. (b) hp2x i/m + hxF i.
7.6
(i) Px + x. (ii) −Px − X.
7.8
x˙ = f + px and p˙x = −g(t).
7.10
e−iHt/~
dAH iHt/~ e . dt
582 Solutions to Selected Exercises 7.12
d 1 d xint = pxint and pxint = 0. dt m dt
7.14
da/dt = −iωa and da† /dt = iωa† .
7.16
(a) px /m, (b) qEx and (c) qE.
7.18
x(0) cos ωt + (1/mω)px (0) sin ωt.
7.20
3 1 |0ih0| + |1ih1|. 4 4
8.2
5.7748 × 10−12 m.
8.4
≈ 0.01157536 m.
8.6
v/(4π).
8.8
~.
8.10
~2 /(me2 ).
8.12
0.5 × 1011 Hz.
8.14
6.59 × 10−9 eV. r ~π 1 2 − . 2a 3 π 2
8.16 8.18
hx(0)i = hpx (0)i = 0, hx2 (0)i = σ 2 and hp2x (0)i = ~2 (1 + c2 )/(4σ 2 ).
8.20
≈ −e2 /(8πǫ0 a0 ).
9.2
It is diagonal with the continuum eigenvalues.
9.4
0.
9.6
√ A 2 sin ka √ . π k
9.8
1 sin(k − k0 )L √ . πL (k − k0 )
9.10 9.12 9.14 9.16
(2r0 )3/2 r0 , r0 = ~/∆p. π(r2 + r02 )2 √ 2 2 (σ/ π)1/2 e−σ k /2 . n Z ∞ ∂ φ∗ (p, t) i~ φ(p, t) dp. ∂p −∞ i~C(p).
Solutions to Selected Exercises 583 9.18
Ae−ixp/~ .
10.2
vp = 0.5v.
10.4
~/2.
10.6
( ) 2 1 p 0 t F t2 2 |ψ(x, t)| = √ exp − x − (x0 + + ) σ (t) . m 2m πσ(t)
10.8
hpi = 0 and hp2 i = 17~2 /4.
11.6
0 and i~z.
11.8
0.
2
11.10
2i~r.
11.12
0.
11.14
0, 1 and −1.
11.16
(i) 0, 1, 2 and 3. (ii) 27.
11.18
(i) (2j1 + 1)(2j2 + 1) · · · (2jN + 1). (ii) 105. 1 1 l(l + 1)~2 + − (i) Elm = m2 ~2 . (ii) Ylm . 2I1 2I2 2I1 √ √ (i) 2/3. (ii) ( 2 + 1)/ 6 )2 .
11.20 11.22 11.26 11.28
2~2 [l(l + 1) − m2 ]. √ √ 0 or 2 ~ or − 2 ~.
12.2
0.3233.
12.4
e−αr .
12.6
|Rnl |2 r2 .
12.8
(i) 1/2, (ii) 1/2 and (iii) 2/3.
12.10
3a0 /2.
12.12
The electron is located near the surface of a sphere of radius n2 a0 . √ E5 = −µe4 /(50~2 ) and r± = [25~2 ± 5 5 ~]/(µe2 ).
12.14 12.16
hri = 0.
12.18
E=
~2
me4 2ma √ , b = 2 . ~ 1 + 2b + 1 + 4b
584 Solutions to Selected Exercises 12.20 13.2 13.4
α > −1/2.
2 2 (1) 2 (0) P 1 − (1/2) Hrn / Er − En(0) . r6=n
Both µ2 ~/(m3 ω 5 ) and (30n2 + 30n + 11)/4(2n + 1) to be small.
13.6
−~/2.
13.8
aL2 (6π 2 − 1)/(18π 2 ).
13.10
eEL/2.
13.12
3eEe ~2 /(2ma).
13.14
8V0 n2 /(π(4n2 − 1)).
13.16
λL2 /4.
13.18
−3e2 (a0 − ρ0 )/(2a0 ρ0 ).
13.20
E1
(1)
(1)
= −α/a0 and E2 "
(0) E1
= −α/4a0 . r (0) E2
(0) E1
(0) E2
2
4λ2 ∆2
13.22
(a) E1,2 = (1/2)
13.24
Elm = 0 and E10 = 0, E11 = −E1−1 = C 2 ~2 /(2B).
13.26
E1
13.28
En = 0.
13.30
E (1) = ±aλ2 /2.
(1)
(1)
+
(2)
(1)
= 0, E2
±
(2)
(1)
= ǫ and E2
−
+
(2)
= 2ǫ.
(1)
14.2
The selection rule is f = i ± 2.
14.4
The selection rules are f = i ± 2, i ± 4.
14.6
[F02 π 2 /(2m~ω 3 )] e−2ωτ .
14.8
2 [256V02 /(225~2 π 2 ω31 )] sin2 (ω31 T /2), ω31 = 4π 2 ~/(mL2 ).
14.10
[m2 L6 V02 /(16π 8 ~4 )] sin2 (ω31 T /2), ω31 = 4π 2 ~/(mL2 ).
14.12
0.
14.14
[e2 E02 /(2mω0 ~)]δn1 .
14.16
[α2 /(2mω~(µ2 + ω 2 ))]. h √ i2 √ 2 2 α πe−ω /µ /( 2 mωµ) .
14.18
#
.
Solutions to Selected Exercises 585 14.20 15.2 15.4 15.6 15.8
8(1 + b)/(2 + b)4 . 1 ∂S |∇S|2 + V + = 0. 2m ∂t 1/(2~ν), ν-classical frequency of the motion of the particle. √ E0 = [3/(16 2)]2/3 V0 , E1 = 32/3 E0 , E2 = 52/3 E0 and E3 = 72/3 E0 . −[3/(16m3c2 )](~mω)2 (n + 1/2)2 .
15.10
n2 ~2 π 2 /(2mL2 ).
16.2
(α + 1)(2α + 1) 2α − 1
16.4 16.6 16.8 16.10
~2 , α ≈ 1.72. 4mL2
1.082(~2/(2m))2/3 λ1/3 . √ 3 ~ω. −V0 /32. −me4 /(2~2 ).
17.2
(i) θLmax = sin−1 (m2 /m1 ). (ii) θL can take any angle.
17.4
≈ 1 + 2ρ.
17.6
−4πδ(|r − r′ |).
17.8
(4m2 V02 a6 /~4 ) e−2s
2 2
17.10
16πm2 a4 V02 /~4 .
17.12
σ = 4πa2 .
a
.
18.2
Pij is a unitary operator.
18.4
[Pij , Pil ] 6= 0.
18.6
(i) Symmetric wave function. (ii) Antisymmetric wave function.
18.8
(i) 5ǫ/2 and (ii) 13ǫ/4 where ǫ = ~2 π 2 /(2ma2 ).
19.2
~ω = (c2 ~2 k 2 + m2 c4 )1/2 .
19.4
cα.
19.6
∂ µ Jµ = 0 where Jµ is the four-vector.
19.8
0.
19.10
dX/dt = cα and dp/dt = e∇φ − e∇(α · A).
586 Solutions to Selected Exercises 19.12
8 components column vector.
19.14
2mc2 .
19.16
β(t) = (mc2 /H) + β ′ (0)e2iHt/~ and hβ(t)i = (1 − (c2 p2 /E 2 ))1/2 .
19.18
[H, ~k] = 0.
19.20
The Dirac equation is unchanged.
20.2
If there are no objects in certain category.
20.4
− cos(θ1 − θ2 ).
20.6
±1.
20.8
− cos(θ1 − θ2 ).
21.2
E = −0.12500, −0.05556, −0.03125, −0.02000 and −0.01389.
21.4
E = 4.92822, 19.71288, 44.35400 and 78.85155.
21.6
(a) E = 1.5, 2.5, 3.5, 4.50001 and 5.50000.
21.8
E = 1.018 and 2.33765 for φ(x, τ = 0) = cos x and sin x respectively.
21.14
T = 1 for large E.
Physics
“… these two books by Rajasekar and Velusamy will definitely tell you how to do quantum mechanics.” —Dr. K.P.N. Murthy, Professor, School of Physics, and Director, Centre for Integrated Studies, University of Hyderabad Quantum Mechanics I: The Fundamentals provides a graduate-level account of the behavior of matter and energy at the molecular, atomic, nuclear, and sub-nuclear levels. It covers basic concepts, mathematical formalism, and applications to physically important systems. The text addresses many topics not typically found in books at this level, including:
A collection of problems at the end of each chapter develops readers’ understanding of both basic concepts and the application of theory to various physically important systems. This book, along with the authors’ follow-up Quantum Mechanics II: Advanced Topics, provides readers with a broad, up-to-date introduction to quantum mechanics.
• Access online or download to your smartphone, tablet or PC/Mac • Search the full text of this and other titles you own • Make and share notes and highlights • Copy and paste text and figures for use in your own documents • Customize your view by changing font size and layout
K24364
Quantum Mechanics I The Fundamentals
Quantum Mechanics I
Bound state solutions of quantum pendulum Pöschl–Teller potential Solutions of classical counterpart of quantum mechanical systems A criterion for bound state Scattering from a locally periodic potential and reflection-less potential Modified Heisenberg relation Wave packet revival and its dynamics Hydrogen atom in D-dimension Alternate perturbation theories An asymptotic method for slowly varying potentials Klein paradox, Einstein-Podolsky-Rosen (EPR) paradox, and Bell’s theorem Numerical methods for quantum systems
The Fundamentals
• • • • • • • • • • • •
WITH VITALSOURCE ® EBOOK
Rajasekar • Velusamy
“Excellent, up-to-date… Quantum Mechanics I: The Fundamentals covers the canonical basics and Quantum Mechanics II: Advanced Topics covers a range of modern developments…I recommend this set highly.” Dr. Jonathan P. Dowling, Hearne Professor of Theoretical Physics, Louisiana State University
S. Rajasekar R. Velusamy
w w w. c rc p r e s s . c o m
K24364_cover.indd 1
11/5/14 1:23 PM