A Textbook on Modern Quantum Mechanics [1 ed.] 0367723441, 9780367723446

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A Textbook on Modern Quantum Mechanics

A Textbook on Modern Quantum Mechanics

A C Sharma

First edition published 2022 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2022 A C Sharma CRC Press is an imprint of Taylor & Francis Group, LLC 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, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. ISBN: [978-0-367-72344-6] (hbk) ISBN: [978-0-367-72347-7] (pbk) ISBN: [978-1-003-15445-7] (ebk) Typeset in Palatino by KnowledgeWorks Global Ltd. Access the Support Material: https://www.routledge.com/9780367723446

Contents Preface...............................................................................................................................................xi Acknowledgments....................................................................................................................... xiii About the Author...........................................................................................................................xv

1. Introduction to Quantum Mechanics..................................................................................1 1.1 Blackbody Radiation and Planck’s Hypothesis.........................................................2 1.2 Photoelectric Effect........................................................................................................3 1.3 Bohr’s Atomic Model and the Hydrogen Atom.........................................................4 1.4 Compton Scattering of Photons...................................................................................5 1.5 De Broglie Hypothesis..................................................................................................7 1.6 Pauli Exclusion Principle..............................................................................................7 1.7 Schrödinger Wave Equation.........................................................................................7 1.8 Born Interpretation of Wave Function........................................................................9 1.9 Heisenberg Uncertainty Principle............................................................................. 10 1.10 Davisson and Germer Wave Properties of Electrons............................................. 10 1.11 Bohr-Sommerfeld Quantization Condition............................................................. 11 1.12 Correspondence Principle........................................................................................... 12 1.13 Heisenberg Quantum Mechanics.............................................................................. 12 1.14 Dirac Theory of Quantum Mechanics...................................................................... 13 1.15 Important Quantum Mechanical Parameters in SI Units...................................... 13 1.16 Solved Examples.......................................................................................................... 14 1.17 Exercises........................................................................................................................ 19 2. Wave Mechanics and Its Simple Applications................................................................21 2.1 Schrödinger Equation.................................................................................................. 21 2.2 Bound States and Scattering States...........................................................................22 2.3 Probability Density, Probability Current, and Expectation Value........................ 23 2.4 Simple Applications of Time-Independent Schrödinger Equation...................... 25 2.4.1 Free Particle Motion....................................................................................... 25 2.4.2 Infinite Potential Well (Particle in a Box).................................................... 26 2.4.3 The Finite Potential Well................................................................................ 29 2.4.4 Step Potential...................................................................................................34 2.4.5 Finite Potential Barrier and Tunneling........................................................ 37 2.4.6 Relevance of Free Particle, Potential Wells, and Potential Barriers......... 40 2.5 Periodic Solids and their Band Structures............................................................... 41 2.5.1 The Kronig-Penney Model............................................................................42 2.5.2 Confined States in Quantum Wells, Wires, and Dots............................... 45 2.6 Solved Examples.......................................................................................................... 46 2.7 Exercises........................................................................................................................ 57 3. Matrix Formulation of Quantum Mechanics..................................................................59 3.1 Matrices and their Basic Algebra............................................................................... 60 3.2 Bra and Ket Notations................................................................................................. 62 v

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3.3

Vectors and Vector Space............................................................................................ 62 3.3.1 Linearly Independent Vectors....................................................................... 62 3.3.2 Orthogonal and Orthonormal Vectors........................................................63 3.3.3 Abstract Representation of a Vector.............................................................64 3.3.4 Outer Product of Vectors...............................................................................64 3.4 Gram-Schmidt Method for Orthogonalization of Vectors.....................................65 3.5 Schwarz Inequality...................................................................................................... 66 3.6 Linear Transformation of Vectors.............................................................................. 67 3.6.1 Eigenvalues and Eigenvectors of a Matrix.................................................. 67 3.6.2 Numerical Method to Find Eigenvalue and Eigenvector...................................................................................................... 69 3.7 Inverse Matrix.............................................................................................................. 70 3.8 Orthogonal Matrix....................................................................................................... 71 3.9 Hermitian Matrix......................................................................................................... 72 3.10 Unitary Matrix.............................................................................................................. 73 3.11 Diagonalization of a Matrix........................................................................................ 74 3.12 Cayley-Hamilton Theorem......................................................................................... 76 3.13 Bilinear, Quadratic, and Hermitian Forms..............................................................77 3.14 Change of Basis............................................................................................................ 78 3.14.1 Unitary Transformations............................................................................... 79 3.15 Infinite-dimensional Space.........................................................................................80 3.16 Hilbert Space................................................................................................................. 81 3.16.1 Basis Vectors in Hilbert Space....................................................................... 82 3.16.2 Quantum States and Operators in Hilbert Space.......................................83 3.16.3 Schrödinger Equation in Matrix Form........................................................84 3.17 Statement of Assumptions of Quantum Mechanics...............................................85 3.18 General Uncertainty Principle................................................................................... 86 3.19 One Dimensional Harmonic Oscillator.................................................................... 87 3.20 Solved Examples.......................................................................................................... 93 3.21 Exercises...................................................................................................................... 107 4. Transformations, Conservation Laws, and Symmetries.............................................109 4.1 Translation in Space................................................................................................... 109 4.2 Translation in Time.................................................................................................... 110 4.3 Rotation in Space........................................................................................................ 111 4.4 Quantum Generalization of the Rotation Operator.............................................. 114 4.5 Invariance and Conservation Laws......................................................................... 115 4.5.1 Infinitesimal Space Translation.................................................................. 116 4.5.2 Infinitesimal Time Translation................................................................... 117 4.5.3 Infinitesimal Rotation.................................................................................. 117 4.5.4 Conservation of Charge............................................................................... 118 4.6 Parity and Space Inversion....................................................................................... 118 4.6.1 Parity Operator.............................................................................................. 119 4.7 Time-Reversal Operator............................................................................................ 121 4.7.1 Properties of Antilinear Operator.............................................................. 123 4.7.2 Time Reversal Operator for Non-zero Spin Particles.............................. 126 4.8 Solved Examples........................................................................................................ 128 4.9 Exercises...................................................................................................................... 129

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5. Angular Momentum...........................................................................................................131 5.1 Orbital Angular Momentum.................................................................................... 131 5.2 Eigenvalues of Angular Momentum....................................................................... 134 5.3 Eigenfunctions of Orbital Angular Momentum................................................... 138 5.4 General Angular Momentum.................................................................................. 140 5.5 Spin Angular Momentum........................................................................................ 142 5.5.1 Pauli Theory of Spin One-half Systems..................................................... 143 5.6 Addition of Angular Momentum............................................................................ 146 5.6.1 Clebsch-Gordon Coefficients and their Properties.................................. 148 5.6.2 Recursion Relations for Clebsch-Gordon Coefficients............................ 149 5.6.3 Computation of Clebsch-Gordon Coefficients......................................... 150 5.7 Solved Examples........................................................................................................ 152 5.8 Exercises...................................................................................................................... 159 6. Schrödinger Equation for Central Potentials and 3D System.................................... 161 6.1 Motion in a Central Field.......................................................................................... 161 6.2 Energy Eigenvalues of the Hydrogen Atom.......................................................... 163 6.3 Wave Functions of the Hydrogen Atom................................................................. 165 6.4 Radial Probability Density....................................................................................... 168 6.5 Free Particle Motion................................................................................................... 169 6.6 Spherically Symmetric Potential Well..................................................................... 170 6.7 Electron Confined to a 3D Box................................................................................. 175 6.8 Solved Examples........................................................................................................ 176 6.9 Exercises...................................................................................................................... 186 7. Approximation Methods.................................................................................................... 187 7.1 Perturbation Theory.................................................................................................. 187 7.1.1 Perturbation Theory for Nondegenerate States........................................ 188 7.1.1.1 First Order Corrections to Energy and Wave Function Ket................................................................................... 189 7.1.1.2 Second Order Corrections to Energy and Wave Function Ket................................................................................... 190 7.1.1.3 kth Order Corrections to Energy and Wave Function Ket................................................................................... 192 7.1.1.4 Anharmonic Oscillator................................................................. 193 7.1.2 Perturbation Theory for Degenerate States............................................... 195 7.1.2.1 Effect of an Electric Field on the First Excited State in a Hydrogen Atom (Linear Stark Effect)................................. 197 7.2 Variation Method....................................................................................................... 200 7.2.1 The Ground State of the Helium Atom..................................................... 202 7.2.2 Rayleigh-Ritz Variational Method.............................................................. 206 7.2.3 The Hydrogen Molecule Ion....................................................................... 208 7.2.4 Variational Method for Excited States....................................................... 212 7.2.5 Application of Variational Method to the Excited State of a 1D Harmonic Oscillator........................................................................ 213 7.3 The W K B Approximations..................................................................................... 215 7.3.1 The Classical Region.................................................................................... 216 7.3.2 Alternative Derivation of the WKB Formula............................................ 218

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7.3.3 Non-classical or Tunneling Region............................................................ 219 7.3.4 Connecting Formulae................................................................................... 220 7.3.5 Quantum Condition for Bound State......................................................... 224 7.4 Solved Examples........................................................................................................ 227 7.5 Exercises...................................................................................................................... 242 8. Quantum Theory of Scattering........................................................................................245 8.1 Scattering Cross-Section and Frame of Reference................................................ 245 8.2 Asymptotic Expansion and Scattering Amplitude............................................... 246 8.3 Partial Wave Analysis............................................................................................... 248 8.3.1 Free Particle and Asymptotic Solutions.................................................... 249 8.3.2 Scattering Amplitude and Phase Shift...................................................... 251 8.3.3 Optical Theorem........................................................................................... 253 8.3.4 Scattering Length.......................................................................................... 253 8.3.5 Scattering by a Square Well Potential........................................................254 8.3.6 Scattering by a Hard Sphere Potential....................................................... 257 8.3.7 Interpretation of the Phase Shift................................................................. 258 8.4 Expression for Phase Shift........................................................................................ 260 8.5 Integral Equation........................................................................................................ 261 8.6 The Born Approximation.......................................................................................... 265 8.6.1 Scattering by Screened Coulomb Potential............................................... 266 8.6.2 Validity of the Born Approximation.......................................................... 267 8.7 Transformation from the Center of Mass Coordinate System to the Laboratory Coordinate System..................................................................... 269 8.8 Solved Examples........................................................................................................ 271 8.9 Exercises...................................................................................................................... 280 9. Quantum Theory of Many Particle Systems.................................................................281 9.1 System of Indistinguishable Particles..................................................................... 281 9.1.1 Non-interacting System of Particles........................................................... 283 9.1.2 Space and Spin Parts of Wave Function.................................................... 285 9.2 The Helium Atom...................................................................................................... 286 9.2.1 Ground State of Helium............................................................................... 289 9.2.2 Excited State of Helium................................................................................ 290 9.3 Systems of N-Electrons............................................................................................. 290 9.3.1 Hartree Approximation............................................................................... 291 9.3.2 Hartree-Fock Approximation..................................................................... 293 9.3.3 Thomas-Fermi Theory................................................................................. 296 9.3.4 Thomas-Fermi Model of Atom................................................................... 298 9.3.5 Density Functional Theory.......................................................................... 299 9.4 Solved Examples........................................................................................................ 301 9.5 Exercises...................................................................................................................... 313 10. Time-dependent Perturbations and Semi-classical Treatment of Interaction of Field with Matter............................................................................................................ 315 10.1 Time-dependent Potentials....................................................................................... 315 10.2 Exactly Solvable Time-dependent Two-state Systems.......................................... 317 10.3 Time-dependent Perturbation Theory.................................................................... 321 10.3.1 First Order Perturbation.............................................................................. 321

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10.4 Harmonic Perturbation............................................................................................. 322 10.4.1 Transition Probability................................................................................... 323 10.4.2 Fermi’s Golden Rule..................................................................................... 324 10.5 Constant Perturbation............................................................................................... 326 10.5.1 Fermi’s Golden Rule..................................................................................... 327 10.6 Semi-classical Treatment of Interaction of a Field with Matter................................................................................................................. 327 10.6.1 Absorption and Stimulated Emission........................................................ 327 10.6.2 Electric Dipole Approximation................................................................... 330 10.7 Spontaneous Emission and Einstein Coefficients................................................. 332 10.8 Dipole Selection Rules...............................................................................................334 10.9 Solved Examples........................................................................................................ 336 10.10 Exercises......................................................................................................................343 11. Relativistic Quantum Mechanics.....................................................................................345 11.1 The Klein-Gordon Equation.....................................................................................346 11.1.1 Probability Density and Probability Current...........................................346 11.1.2 The Klein-Gordon Equation in a Coulombic Field.................................................................................................................348 11.2 The Dirac Equation.................................................................................................... 350 11.2.1 Derivation of the Dirac Equation................................................................ 351 11.2.2 Covariant Form of the Dirac Equation......................................................354 11.2.3 Probability Density and Probability Current........................................... 355 11.3 Free Particle Solutions of the Dirac Equation........................................................ 357 11.3.1 Positive and Negative Energy Eigenvalues............................................... 360 11.4 The Dirac Equation and the Constants of Motion................................................ 361 11.5 Spin Magnetic Moment (the Dirac Electron in an Electromagnetic Field)............................................................................................... 365 11.6 Spin-Orbit Interaction Energy.................................................................................. 367 11.7 Solution of the Dirac Equation for Central Potential....................................................................................................................... 369 11.7.1 Hydrogen-like Atom.................................................................................... 374 11.8 Solved Examples........................................................................................................ 377 11.9 Exercises......................................................................................................................384 12. Quantization of Fields and Second Quantization........................................................385 12.1 Quantization of an Electromagnetic Field............................................................. 385 12.1.1 Field Operators.............................................................................................. 389 12.2 Second Quantization................................................................................................. 390 12.2.1 Second Quantization of the Schrödinger Equation for Bosons....................................................................................................... 391 12.2.2 Second Quantization of the Schrödinger Equation for Fermions................................................................................................... 394 12.2.3 Matrix Representation of Fermionic Operators....................................... 397 12.2.4 Number Operator......................................................................................... 397 12.3 System of Weakly Interacting Bosons..................................................................... 398 12.4 Free Electron System.................................................................................................400 12.5 Solved Examples........................................................................................................ 402 12.6 Exercises......................................................................................................................408

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Annexure A: Useful Formulae.................................................................................................409 I. Table of Integrals................................................................................................................ 409 II. Series and Expansions..................................................................................................... 409 III. Basic Functional Relations............................................................................................. 410 IV. Coordinate Systems........................................................................................................ 411 Annexure B: Dirac Delta Function..........................................................................................413 I. Properties of Delta Function............................................................................................ 414 II. Representation of Delta Function.................................................................................. 414 Answers to Exercises.................................................................................................................. 417 Chapter 1................................................................................................................................ 417 Chapter 2................................................................................................................................ 417 Chapter 3................................................................................................................................ 418 Chapter 5................................................................................................................................ 418 Chapter 6................................................................................................................................ 419 Chapter 7................................................................................................................................ 419 Chapter 8................................................................................................................................ 420 Chapter 9................................................................................................................................ 421 Chapter 10.............................................................................................................................. 421 Chapter 11..............................................................................................................................422 Chapter 12..............................................................................................................................422 Bibliography.................................................................................................................................423 Index..............................................................................................................................................425

Preface Quantum mechanics not only describes the microscopic world but is also essential to explain many macroscopic quantum systems such as semiconductors, superconductivity, and superfluidity. Concepts such as the particle properties of radiation, the wave properties of matter, quantized energy and momentum, and the idea that one can no longer know exactly where a single particle like an electron is at any instant of time are the foundations of modern sciences and are indispensable when explaining most of the experimental results in physical, chemical, material, and biological sciences. Consequently, advances in modern technologies are not possible without prior knowledge of modern quantum mechanics. I taught multiple courses on quantum mechanics during my 33-year teaching career at The Maharaja Sayajirao University of Baroda, Vadodara (India) and Jiwaji University, Gwalior (India), and as subject matter expert to several other universities. The purpose of writing this book has been to design a textbook that i. provides everything a student needs to know for succeeding at all levels of undergraduate and graduate studies, ii. meets international standards with detailed and elegant mathematical treatment and the up-to-date interpretation/presentation style, and iii. contains enough solved examples to illustrate the fundamental concepts. Over the past decades, quantum mechanics has continually evolved, and the tools of presentation and mathematical treatment of the subject have gone through significant changes. Therefore, I felt the need to write a textbook on modern quantum mechanics that has a rejuvenated approach to the subject matter. As a teacher, I believe that testing the understanding of a topic by problem-solving is vital for learning the concepts. The topics covered in this book are divided into 12 chapters. Several newly emerged topics from contemporary physics are incorporated either as text or as solved examples. There are over a hundred solved examples and an equal number of unsolved exercises in this book. Each chapter of this book consists of solved examples followed by unsolved exercises, with answers, to consolidate the readers’ understanding. Every effort has been made to make this book reader friendly. Chapter 1 starts with the historical development of quantum mechanics and explains the key concepts behind the fundamental discoveries, which made important contributions to the foundation of modern quantum mechanics. The emphasis of this chapter is to bring out clearly the concepts behind Schrödinger's wave mechanics, Heisenberg's matrix mechanics, and Dirac's relativistic quantum mechanics. Development of the Schrödinger wave equation and its 1D applications are covered in Chapter 2. The solution of Schrödinger's wave equation for free particles, quantum wells, quantum barriers, and potential steps along with its relevance and importance to contemporary physics and chemistry are highlighted and demonstrated through examples from condensed matter physics. Chapter 3 deals with matrix and vector formulation of quantum mechanics. Review of important properties of matrices and vector space which are relevant to quantum mechanics, the change of basis and irrelevance of choice of basis of a vector space, the numerical approach to finding eigenvalue and the inverse of a matrix, xi

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Preface

the equivalence of the Heisenberg and Schrödinger approaches to quantum mechanics, the general uncertainty principle, the 1D harmonic oscillator, and its matrix formulation are some of the main attractions of this chapter. A modern approach to explain the transformations, conservation laws, and symmetries is presented in Chapter 4. Reflection (parity operator) and time reversal are discussed at length in the chapter. Chapter 5 covers orbital, total, and spin angular momenta and the Clebsch-Gordon coefficients. The matrix approach to deriving eigenvalues of orbital angular momentum and the Pauli theory of spin half systems are special features of this chapter. Chapter 6 discusses the solution of Schrödinger’s equation for central potentials and 3D systems along with free particle motion and discretized energy states in finite volume. Chapter 7 deals with approximation methods: Perturbation theory for non-degenerate and degenerate systems, variational methods, and WKB approximations. The kth order corrections to energy and wave function for non-degenerate perturbation theory, two types of treatment for WKB approximation, and detailed derivation of connecting formulae, and application of the variational method to compute energy for excited states are some of the topics which are not generally found in textbooks on quantum mechanics, but are covered in this chapter. The quantum theory of scattering along with detailed derivations on partial wave analysis and integral equation are presented in Chapter 8. Chapter 9 details the quantum theory of many particle systems and includes explanations of space and spin parts of wave function, detailed derivations for direct and exchange integrals for the helium atom, and the Thomas-Fermi and density functional theories for systems of N-particles which are essential to teach at the graduate level but are not covered in most of the textbooks on quantum mechanics. Time-dependent perturbations and the semi-classical treatment of interaction of fields with matter are covered in Chapter 10 along with a discussion of the appropriate mathematical treatment of time-dependent potentials and exactly solvable time-dependent two-state systems. Chapter 11 deals with relativistic quantum mechanics which includes derivation of the Dirac equation, constants of motion, and a solution of the Dirac equation for central potentials. Essential contents on the quantization of fields and second quantization are presented in Chapter 12. A contemporary approach is used for presentation and discussions on field quantization and second quantization for systems of Bosons and Fermions. Access the Support Material at https://www.routledge.com/9780367723446.

Acknowledgments It is my pleasure to acknowledge the questions followed by intense discussions with my all students, on whom the matter of this book is well tested. I am indebted to two of my colleagues, Professor Mahesh Prakash, former head, School of Physics, Jiwaji University, Gwalior, and Professor J. P. Singh, former head, Physics Department, Faculty of Science, The Maharaja Sayajirao University of Baroda, Vadodara, for their critical reading of the first draft of this book and making several valuable suggestions. I would also like to acknowledge, with a sense of gratitude the valuable input and the assistance in drawing the figures from Dr. Digish K. Patel (one of my former graduate students), Assistant Professor, Ganpat University, Mehsana (Gujarat). Dr. Pushpendra Tripathi, Associate Professor, Aligarh Muslim University has also made many useful suggestions for the content of this book. With a deep sense of gratitude I thank my wife, Ragini Sharma, for her constant encouragement and wholehearted support throughout my professional career, especially when I was writing this book. Finally, I acknowledge with pleasure the encouragement and support received from our sons Sanshit and Mayank and our daughter-in-law Vrishti.

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About the Author

Dr. A C Sharma is a retired Professor of Physics from the Faculty of Science, The Maharaja Sayajirao University of Baroda, Vadodara (Gujarat) India. He has been head of the physics department, dean of the faculty of science for two terms, chairman of the board of studies in physics, chairman of the faculty board for two terms, and coordinator of many research programs awarded to physics department and to science faculty from the Department of Science and Technology, and from the University Grants Commission, New Delhi. He has distinguished educational records, obtained his Ph.D. from Roorkee University (now IIT Roorkee) and did his post-doctoral work at Cavendish Laboratory, University of Cambridge (U. K.). Professor Sharma has served as scientists pool officer at IIT Delhi, lecturer at Jiwaji University Gwalior, and then reader and professor at Maharaja Sayajirao University of Baroda. He has also been a visiting scientist to Chalmers University of Technology, Sweden and Indian Institute of Science, Bangalore. He has 33 years’ experience of teaching undergraduate, post-graduate, and graduate students and 40 years’ research experience in different areas of theoretical condensed matter physics. Semiconductors, surfaces and interfaces, superconductivity and nanostructures have been the areas of his research interest. Many-particle theory and numerical computations have been his expertized techniques to perform the research work. He has published 125 research papers. Some of his research papers are published in journals like Physical Review, Journal of Condensed Matter Physics, and Journal of Applied Physics and Solid State Communications. Professor Sharma has successfully completed 8 major research projects, delivered 41 invited talks at national and international conferences, organized 6 conferences, and supervised 12 students for their Ph.D. degrees. He has been awarded several scholarships, visiting fellowships and medals. Professor Sharma is a member of the Indian Physics Association, and a Fellow of Gujarat Science Academy. He has worked as a subject expert to evaluate the graduate thesis and research proposals for several universities/ institutions and referee to several search journals.

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1 Introduction to Quantum Mechanics During the late 1800s and early 1900s, it became clear that physics was due for major revision. Existing laws of physics at that time failed to explain adequately or even approximately a large number of phenomena and observations. The explanation of phenomena involving small systems like electrons and atoms and their interaction with electromagnetic fields faced serious problems when attempted with classical physics. The following clearly demonstrated the necessity for a departure from classical mechanics: 1. Classical physics laws suggest that an electron moving in an orbit of an atom must lose energy by emission of synchrotron radiation, and it must spiral gradually towards the nucleus of an atom, which was not observed experimentally. 2. The experiments that were performed for observing the interference of light, the photoelectric effect, and diffraction of electrons clearly demonstrated that under certain situations waves act like streams of particles, and streams of particles were found to act as if they were waves. The laws of classical physics were completely unable to explain this. 3. The laws of classical physics suggest that the energy density of an electromagnetic field in vacuum cannot be finite because of the divergence of energy carried by short wavelength modes, which was not observed in experiments, and the total energy density was found to be finite. Quantum mechanics, which was developed between 1900 and 1930 along with the general and special theory of relativity, completely revolutionized the field of physics, and it is now one of the pillars of modern sciences. The new concepts such as the wave properties of a particle and particle properties of radiation, quantized energy and momentum, and the idea that the position of a particle like an electron cannot be known exactly at any one instant of time had been found necessary to explain all the new experimental evidence that was available at that time. Classical mechanics and electromagnetism were replaced by quantum mechanics when a system was of a size comparable with that of an atom. Today, it is well understood that quantum mechanics is not only needed to describe systems at very small scales, but it is also required to explain macroscopic systems such as semiconductors, superconductivity, and superfluids. At present, several fields of physics and chemistry, like condensed matter physics, atomic/molecular physics, computational chemistry, particle physics and nuclear physics, are better understood within the mathematical framework of quantum mechanics. Since, everything is divisible into quantum-mechanical particles, the laws of quantum mechanics must reduce to the laws of classical physics in the appropriate limit known as the classical limit. Quantum mechanics was developed in both a non-relativistic and relativistic manner. Some of the most accurate theories of physics had been worked out within the framework of 1

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A Textbook on Modern Quantum Mechanics

relativistic quantum mechanics and quantum field theory. However, non-relativistic quantum mechanics is widely used, because of its simplicity for situations where relativistic effects are not so important. The research that contributed to the foundations of quantum mechanics was as follows:

1.1  Blackbody Radiation and Planck’s Hypothesis Heated up objects such as a hot iron generally glow to give off light, which displays a range of colors lying between red to yellow. On further increases in temperature, more light and a different spectrum of light result. Though different spectra are exhibited by different substances, the majority of these are approximated by an ideal known as blackbody radiation. An idealized body that completely absorbs all the energy of the radiation that is falling upon it and attains an equilibrium temperature and then reemits that energy as quickly as it is absorbed is termed a blackbody. The radiated energy has various wavelengths, and for every wavelength a maximum of energy occurs at some value that depends on the temperature of the body. The hotter body exhibits a shorter wavelength for maximum radiation. Calculations of the energy distribution for the radiation from a black body with the use of classical physics were unsuccessful. German theoretical physicist Max Planck suggested that the radiation energy is not emitted continuously but is emitted in discrete packets, called quanta. The energy E and frequency ν of a quanta are related by the relation E = hν . The quantity h, which is approximately equal to 6.62607 × 10 −34 joule∙second, is a universal constant. The energy of a spectrum was calculated and it was shown by Planck that the calculated energy spectrum agreed with observations made on the black body, for the entire wavelength range. It was assumed by Planck that the atoms in an oscillating state were the sources of radiation. The vibrational energy of each oscillator can take any number of discrete values, but no value in between two discrete values. He further assumed that when an oscillator changes its energy state from E1 to a state of lower energy E2 the discrete amount of energy ( E1 − E2 ) is equal to hν, which is determined from blackbody radiation data. Planck’s law for Eλ, radiated energy per unit volume of a black body in the wavelength interval of λ to λ + Δλ, is given in terms of h, speed of light c, the Boltzmann constant kB , and the absolute temperature T:

Eλ =

8 hc 1 (1.1) 5 hc λ   kBTλ − 1 e  

The sun releases its internal energy as light from the surface and does not reflect any incoming light and therefore it acts very similarly to a blackbody. Fig. 1.1 given below demonstrates what the shape of the curve for blackbody radiation looked like for objects at different temperatures, including the sun. The wavelength of maximum power emitted by a blackbody at temperature T is given by Wien’s displacement law; λ maxT = 2898  µm K.

Introduction to Quantum Mechanics

3

FIGURE 1.1 An illustrative diagram for blackbody radiation spectrum.

1.2  Photoelectric Effect Emission of electrons when light hits a metal surface is known as the photoelectric effect. The emitted electrons are called photoelectrons. The theory of photoelectric effect explains the following experimental observations on emission of electrons from an illuminated metal surface. 1. There exists a threshold frequency for every metal surface, which is a minimum frequency of incident light below which no photoelectrons are emitted. 2. Electrons are emitted above the threshold frequency. The maximum kinetic energy of each emitted photoelectron depends on the frequency, not on the intensity of the incident light, unless the intensity is too high. 3. The rate of ejected photoelectrons is directly proportional to the intensity of the incident light, for a given metal and frequency of incident radiation. For fixed frequency, an increase in the intensity of the incident beam increases the magnitude of the photoelectric current, when stopping voltage remains the same. 4. A small time lag, less than 10−9 second, between the incidence of radiation and the emission of a photoelectron is observed. 5. Direction of distribution of emitted electrons peaks in the direction of polarization (the direction of the electric field) of the incident light, for linearly polarized light. In 1905, Albert Einstein published a paper by borrowing Planck’s hypothesis about the quantized energy from his blackbody research, and he assumed that the incident radiation on a metal surface should be thought as quanta of energy hν . The quanta is known as a photon. In photoemission, one such photon is absorbed by one electron. Some energy will be lost in moving towards the surface when the electron is at some distance into the material of the cathode. There will always be some electrostatic cost as the electron leaves the

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A Textbook on Modern Quantum Mechanics

FIGURE 1.2 An illustrative diagram for ejection of photoelectrons.

surface, which is termed as work function φ . The electrons emitted from regions very close to the surface are the most energetic. These electrons leave the cathode with kinetic energy:

1 mv 2 = hν − φ (1.2) 2

There is a minimum energy for a given metal, for which the quantum of energy is equal to φ . Light below that energy, no matter how bright, will not eject electrons. Every photon with energy more than threshold energy excites only one electron. An increased intensity of light only increases the number of released electrons and not their kinetic energy. Since, a very small time is required for an atom to be heated to a critical temperature, release of the electron is nearly instantaneous upon absorption of the light.

1.3  Bohr’s Atomic Model and the Hydrogen Atom A model of an atom to explain how electrons can have stable orbits around the nucleus was proposed by Bohr in 1913. The size of the orbit determines the energy of an electron; it is lower for smaller orbits. When the electron jumps from a higher energy orbit to a lower energy orbit, radiation is released. Three postulates made by Bohr are: 1. The electron revolves in a stable orbit (stationary) around the nucleus without radiating any energy. These stationary orbits are attained at certain discrete distances from the nucleus. The electron is not allowed to have any energy in between the discrete energies of orbits. 2. The magnitude of angular momentum of a revolving electron in a stationary orbit is quantized and it attains the values which are integral multiples of  = h / 2 π :

mvr = n (1.3) where n is an integer, which takes values 1, 2, 3, 4,..... and is termed a principle quantum number. The smallest value of n is 1, which corresponds to an orbital radius equal to 0.0529 nm, known as the Bohr radius. Here, it is assumed that an electron of mass m and charge −e moves with velocity v, in an orbit of radius r.

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Introduction to Quantum Mechanics

3. Electrons can jump from one allowed orbit to another by gaining (absorption) or losing (emission) energy from electromagnetic radiation of frequency ν determined by the energy difference of the levels according to ∆E = E1 − E2 = hν . For the hydrogen atom, an attractive coulomb force provides the centripetal acceleration v2 to maintain orbital motion. The total force on the electron thus is: r F=

e2 ˆ v2 ˆ r m r = r 4πε 0 r 2

⇒ mv 2 =

e2 4πε 0 r

(1.4)

where ε 0 = 8.854 × 10−12  F / m (Farad per meter) is the permittivity of free space. The total energy of the electron in a hydrogen atom is given by:



E=

1 e2 mv 2 − 2 4πε 0 r

=

e2 e2 − 8πε 0 r 4πε 0 r

=−

(1.5)

e2 8πε 0 r

Also, Eqns. (1.3) and (1.4) can be combined to give the radius of the nth orbital of the hydrogen atom:  4πε 0  2  2   n = a0 n2 (1.6) rn =   me 2 



4πε 0  2 = a0 = 0.0529 nm is the radius of the first orbital of the hydrogen atom, termed me 2 the Bohr radius. On substituting Eqn. (1.6) into Eqn. (1.5), one gets energies of quantized levels:

Here,



En = −

me 4 e2 E =   − =   − 21 (1.7) 2 2 2 2 2 32 π ε 0  n 8πε 0 a0 n n

Here, E1 = −13.6 eV is the ground state energy of the hydrogen atom.

1.4  Compton Scattering of Photons During the early 20th century, research on interaction of X-rays with matter demonstrated that when X-rays of a known wavelength interact with atoms, the X-rays are scattered through an angle θ, and they emerge at a different wavelength that is related to θ . The

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A Textbook on Modern Quantum Mechanics

wavelength of the scattered rays was found longer (corresponding to lower energy) than the initial wavelength, in multiple experiments, whereas classical electromagnetism predicted that the wavelengths of scattered and initial X-rays should be equal. The observed X-ray shift was attributed to the particle-like momentum of light quanta, in the research work published by Compton in 1923. The energy and frequency of light quanta are interdependent. Compton assumed that each scattered X-ray photon interacted with only one electron, and derived the mathematical relationship between the shift in wavelength and the scattering angle:

λ′ − λ =

 ( 1 − cos θ ) (1.8) m0 c

where λ ′ and λ are the wavelengths after and before scattering; m0 is the rest mass of the  electron and θ is the scattering angle. The is known as the Compton wavelength of m0 c the electron, which is equal to 2.43 × 10−12 m. The wavelength shift, ∆λ = λ′ − λ, can take any value from zero to (for θ = 180°), twice the Compton wavelength of the electron (for θ = 0°). Compton also observed that no wavelength shift was experienced for some X-rays, despite being scattered through large angles. This happened because of the photon failing to eject an electron. The magnitude of the shift is related to the Compton wavelength of the entire atom (which could be 10,000 times smaller) not to the Compton wavelength of the electron. This is a different phenomenon known as coherent scattering off the entire atom, as the atom remains intact, gaining no internal excitation. In modern experiments, measurement of energies is more conventional than the measurec c ment of wavelengths of the scattered photons. By writing E = and E ′ = in the above λ′ λ relation we obtain:



E′ =

E (1.9)  E  1+  1 − cos θ ( )  m0 c 2 

FIGURE 1.3 An illustrative diagram of Compton scattering.

Introduction to Quantum Mechanics

7

1.5  De Broglie Hypothesis In 1924, de Broglie proposed that just as light has both wave-like and particle-like properties, electrons too have both particle-like and wave-like properties. One of the most important contributions to the development of quantum mechanics comes from matter waves. Just like a beam of light, all matter exhibits wave-like behavior, and a beam of electrons can be diffracted. This is known as the de Broglie hypothesis, and the matter waves are referred as de Broglie waves. The de Broglie wavelength λ, which is associated with a particle, relates to the momentum of particle p through the relation λ = h / p . George Paget Thomson’s thin metal diffraction experiment and the Davisson-Germer experiment independently demonstrated the wave-like behavior for electrons. It had also been confirmed for other elementary particles, neutral atoms and even molecules. All matter exhibits properties of both particles and waves, and hence the de Broglie relationship hold for all types of matter.

1.6  Pauli Exclusion Principle Pauli discovered the exclusion principle in 1925, which states that each electron must be in its own distinct quantum state. In other words, no two electrons in an atom or no two identical Fermions in more generalized terms, can simultaneously occupy the same quantum state, defined by a set of quantum numbers (n, l, m, ms ) , where n, l, m , and ms are termed the principal quantum number, azimuthal quantum number, magnetic quantum number and spin quantum number, respectively. If two electrons occupy the same orbital, then the values of n, l , and m are the same, and therefore their ms must be different, which means 1 1 that the electrons must have opposite half-integer spin projections of and − . Particles 2 2 with whole number spin (Bosons) are not subject to the Pauli exclusion principle. The exclusion principle was initially formulated by Wolfgang Pauli for electrons and later on it was extended to all Fermions. The Pauli exclusion principle and the requirement of wave function to be antisymmetric with respect to exchange are equivalent.

1.7  Schrödinger Wave Equation During the early 20th century, physicists began to acknowledge that matter possesses wavelike behavior. However, there existed no known equations that obeyed matter waves. In 1926, Austrian physicist Erwin Schrödinger, in continuation to de Broglie’s hypothesis, formulated a second-order differential equation to explain the wave nature of the matter and particle associated with the wave. The Schrödinger equation assumes that a particle behaves as a wave and yields a solution in terms of the function known as the wave function ψ and the energy E of the particle under consideration. Once the wave function is known, then everything about the particle can be deduced from the wave function. This makes ψ a most important quantity, which itself does not have any physical significance; however, the abso2 lute square of it, i.e. ψ gives the probability of finding the particle in a region of space at an

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A Textbook on Modern Quantum Mechanics

instant of time. The E of the particle, depending upon the potential energy V, and boundary conditions (constraints on the particle), can be continuous or quantized. One of the most remarkable features of quantum mechanics is the quantization of energy of the particle. The Schrödinger equation has been very popular, despite its initial criticism by the scientific community because of its limitations in regard to relativistic particles. The Schrödinger equation is used in two forms, one consisting of time, termed as the timedependent equation, and the other in which the time factor is eliminated, and hence named the time-independent equation. The Schrödinger equation, like Newton equations, cannot be derived but can be verified from experimental results. A simple way to get the Schrödinger equation could be as follows: Consider the motion of a free particle along the x-axis with momentum p. According to h de Broglie’s hypothesis, a wave of λ = is associated with the particle and hence we can p assume a wave function ψ( x , t) given by: ψ ( x , t ) = Ae i( kx −ωt ) (1.10)



The hypotheses of Planck and de Broglie suggest that E = hν = ω and p = i

ψ ( x , t ) = Ae 



( px − Et )

h = k . Therefore: λ

(1.11)

On partial differentiation of Eqn. (1.11) with respect to x and t , we obtain:

∂ψ ( x , t ) ip  ∂ ip  = ψ ( x , t ) ⇒  −  ψ ( x , t ) = 0 (1.12) ∂x   ∂x  



∂ψ ( x , t ) iE  ∂ iE  = − ψ ( x , t ) ⇒  +  ψ ( x , t ) = 0 (1.13) ∂t   ∂t  

∂ ∂ Equations (1.12) and (1.13) suggest that p = − i and E = i . For a non-relativistic free ∂ ∂ t x p2 particle E = and hence we get: 2m −



∂ψ ( x , t )  2 ∂2 ψ ( x, t ) (1.14) = i 2m ∂x2 ∂t

which is the Schrödinger equation for a free particle moving along the x-axis. It is to be noticed here: i. For a free particle moving along an arbitrary direction, Eqn. (1.14) can be generalized to:

−

∂ψ ( r , t )  2  ∂2 ∂2 ∂2   2 +   2 +   2  ψ ( r , t ) = i ∂y ∂z  ∂t 2m  ∂x

∂ψ ( r , t ) 2 2 ⇒− ∇ ψ ( r , t ) = i ∂t 2m

(1.15)

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Introduction to Quantum Mechanics

ii. For a particle moving under the influence of a field characterized by potential p2 energy, V (r ) , the total energy of the particle is E = + V (r ) and therefore Eqn. (1.15) 2m becomes:  2 2  ∂ψ ( r , t )  − 2 m ∇ + V ( r )  ψ ( r , t ) = i ∂t (1.16)  



iii. It is important to note that Eqn. (1.14) has been obtained by assuming the wave function given by Eqn. (1.10). However, several solutions like Eqn. (1.10) and their linear combination will also be the solutions of Eqn. (1.14). Therefore: ψ ( x, t ) =



∫ A(k)e

i( kx −ωt )

dk

(1.17)

will also be the solution of Eqn. (1.14). Equation (1.17) represents a group of waves having different k -values. The group of waves is known as a wave packet. Each k corresponds to a wave and in principle, k can take any value in between −∞ and ∞. dω ω However, for practical cases it varies over a certain range. The and have dimendk k h dω k sions of velocity. With the use of E = hν and λ = , we can write = = v, which is dk m p dω the velocity for a freely moving particle. Therefore, is termed group velocity and dk ω it is represented by v g , while , which is the velocity of an individual wave, is known k as phase velocity.

1.8  Born Interpretation of Wave Function As stated above, it became clear that matter must be considered to have wave-like properties to explain experimental data. In 1926, Max Born formulated a physical law of quantum mechanics, which gives the probability of getting the given results from a measurement on the quantum system. The Born law, which is one of the key principles of quantum mechanics, states that the probability density of finding a particle at a given point of time is proportional to the square of the magnitude of the wave function of the particle at that point. Thus, the wave function itself has no physical significance, but the square of its absolute magnitude has significance when evaluated at a point. The probability of finding a particle between a and b at time t is given by:

∫



b a

ψ ( x , t ) dx 2

(1.18)

where ψ ( x , t ) = ψ ( x , t )   ψ ( x , t ) is a complex square. At any given time: 2



*

∫ ψ ( x, t )

2

dx = 1

(1.19)

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A Textbook on Modern Quantum Mechanics

which would remain true at any other point of time, because the probability of finding a particle over a space cannot change with time. We need (i) the integral to be time-independent, because otherwise a probabilistic interpretation wouldn’t be possible, and (ii) the probability of finding the particle over the entire space to be 1.

1.9  Heisenberg Uncertainty Principle German physicist Werner Heisenberg introduced first in 1927 the uncertainty principle, which states that the more precisely the position of a particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating to standard deviations in one dimensional position ∆x and momentum ∆p was derived by Earle Hesse Kennard and by Hermann Wey in 1928:

∆x∆p ≥

 (1.20) 2

Two alternative explanations for the uncertainty principle are offered in quantum mechanics: (i) the wave mechanics picture and (ii) the matrix mechanics picture. The matrix mechanics approach formulates the uncertainty principle in a more generalized manner. In the wave mechanics description of quantum mechanics, the uncertainty relation in terms of position and momentum arises because the expressions of the wave function in the two corresponding orthonormal bases in Hilbert space are the Fourier transform of each other. A nonzero function and its Fourier transform both cannot be sharply localized, because position and momentum are conjugate variables. A similar relation arises between the variances of all pairs of Fourier conjugates. All pairs of non-commuting self-adjoint operators representing observables are subject to similar uncertainty relations in matrix formulations of quantum mechanics. One of the most fundamental concepts in quantum theory lies in the idea that xp ≠ px , which is not followed by pure numbers but by the operators and matrices. Our understanding of physical properties being represented by numbers is altered on being represented by operators and matrices. We know that matrices have the same non-commutativity as operators, and they too can be the mathematical formalism of quantum theory. In fact, Heisenberg’s original approach to quantum theory, which is called matrix mechanics, was to find matrices x and p, such that xp − px = i. An eigenstate of an observable represents the state of the wave function for a certain measurement of value (the eigenvalue). For example, if a measurement of an observable A is performed, then the system is in an eigenstate of that observable. However, the eigenstate of the observable A need not be an eigenstate of another observable B. In such a case, it does not have a uniquely associated measurement for it, as the system is not in an eigenstate of that observable.

1.10  Davisson and Germer Wave Properties of Electrons To test the de Broglie’s hypothesis that matter behaved like waves, Clinton Davisson and Lester Germer fired slow moving electrons at a crystalline nickel target, in 1927 at Bell Labs. In the experiment, they measured and determined the angular dependence of the

11

Introduction to Quantum Mechanics

FIGURE 1.4 Illustrative set up of Davisson-Germer experiment.

reflected electron intensity to verify the diffraction pattern that was predicted by Bragg for X-rays. The same effect by firing electrons through metal films to produce a diffraction pattern was independently demonstrated by George Paget Thomson at the same time. The de Broglie hypothesis that matter has wave-like behavior was confirmed by the Davisson-Germer experiment. The Compton effect and the Davisson-Germer experiment established the wave-particle duality hypothesis, which is a fundamental step in quantum theory. The Davisson-Germer experiment consisted of (i) firing an electron beam from a heated tungsten filament working as an electron gun, (ii) an electrostatic particle accelerator perpendicular to the surface of the nickel crystal, and (iii) measurement of how the number of reflected electrons varied with respect to variation in angle between the detector and the nickel surface. The electrons released from a heated tungsten filament were accelerated through an electric potential difference towards the nickel crystal to give them kinetic energy. The experiment was conducted in a vacuum chamber to avoid collisions of the electrons with other atoms on their way towards the surface. A Faraday cup electron detector was used to measure the number of electrons that were scattered at different angles. The detector accepted only elastically scattered electrons.

1.11  Bohr-Sommerfeld Quantization Condition One of the main tools of the old quantum theory is the Bohr-Sommerfeld quantization condition. The condition allows one to select out only certain states, no other states, of a classical system as allowed states in which the system can exist. The basic idea of BohrSommerfeld is a quantization condition that means that the motion in an atomic system is quantized, or discretized:

∫ p

i 

dqi = ni h (1.21)

where qi are generalized coordinates and the pi are corresponding canonical conjugate momenta of the system. The quantum numbers ni are integers and the integral is taken

12

A Textbook on Modern Quantum Mechanics

over one period of the motion at constant energy, described by the Hamiltonian. The integral is equal to an area in phase space, a quantity known as action, which is quantized in terms of Planck’s constant. The quantization condition, given by Eqn. (1.21) is often known as the Wilson-Sommerfeld rule, proposed independently by William Wilson and Arnold Sommerfeld. The Bohr theory was extended by Sommerfeld, and it turned out to be a powerful tool of atomic research at that time. The theory was adopted and further developed by physicists. Bohr originally used only one quantum number, while the Bohr-Sommerfeld theory is more generalized and describes the atom in terms of two quantum numbers. For a circular orbit, angular momentum is mvr, which is a constant of motion and hence for a circular orbit Eqn. (1.21) reduces to:

∫ mvr dθ = nh or 2πmvr = nh (1.22)

which is the Bohr quantization rule given by Eqn. (1.3).

1.12  Correspondence Principle The correspondence principle states that the behavior of systems described by quantum mechanical theory should reproduce that which is given by classical physics, in the limit of large quantum numbers. The conditions under which results from quantum mechanics agree with those from classical physics are called the correspondence limit. Quantum calculations must agree with classical calculations for large orbits and for large energies. In a more generalized manner, the correspondence principle states that a new scientific theory must reproduce the earlier existing scientific theory in appropriate circumstances, which means that a new theory explains all the phenomena under appropriate limits for which the preceding theory was valid. Two examples of this are the following: (i) Einstein’s theory of special relativity follows the correspondence principle, because it reduces to classical mechanics when velocities are small as compared to the speed of light. (ii) When the number of particles is large, statistical mechanics reproduces thermodynamics.

1.13  Heisenberg Quantum Mechanics The first formulation of quantum mechanics, which is conceptually autonomous and logically consistent, is Heisenberg’s matrix mechanics. The novel ideas published by Heisenberg in a research paper paved the way for a complete departure from the classical description of atomic physics, and advanced a new formulation of the laws of microphysics. According to the Heisenberg formulation, the spectral lines and their intensities are the observables, not the Bohr orbits. Heisenberg advanced two additional assumptions in regard to dynamics of motion: (1) he proposed that

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Introduction to Quantum Mechanics

the equations of motion should have the same mathematical form as the classical Newton’s second law, and (2) he modified the Bohr-Sommerfeld quantization condition. Heisenberg and Born also assumed that the Hamilton canonical equations of motion were to be preserved. The basis of Heisenberg quantum mechanics had been the fundamental commutation relation  p, q  = − i , where q and p are time-dependent position and momentum matrices.

1.14  Dirac Theory of Quantum Mechanics In 1928, Dirac proposed a formulation of quantum mechanics for an electron. In his theory spin emerges as a natural consequence of the relativistic treatment of quantum mechanics. The Dirac equation consists of both the principles of quantum mechanics and the theory of special relativity. It was the first quantum theory that accounts fully for special relativity and was validated to account for the fine details of the hydrogen spectrum in a completely rigorous manner. Also, the Dirac quantum theory implied the existence of antimatter, which was at that time unsuspected and unobserved. The existence of antimatter was experimentally confirmed several years later. In contrast to the Schrödinger equation that is described in terms of one complex value dependent wave function, the wave function in Dirac theory involves four complex values. A theoretical justification to introduce several component wave functions by Pauli in his spin theory is provided by the Dirac theory. The Dirac equation for freely moving particle is as follows: ∂  cα. p + βm0 c 2  ψ ( r , t ) = i ψ ( r , t ) (1.23) ∂t



The new elements in the Dirac equation are the 4 × 4 αk and β matrices, and the fourcomponent wave function, ψ ( r, t ) . m0 is the rest mass of the electron.

1.15  Important Quantum Mechanical Parameters in SI Units a. Bohr radius: a0 =

4πε 0  2 = 5.29177 × 10−11 meter . m0 e 2

b. Binding energy of hydrogen atom: E1 = c. Bohr magneton: µ B =

m0 e 4 = 13.6057 eV. 2 2 ( 4πε 0 )  2

e = 9.274 × 10−21 Joule/tesla. 2 m0

d. Compton wavelength of an electron: λ =

2 π = 2.43 × 10−12 meter. m0 c

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A Textbook on Modern Quantum Mechanics

e. Classical electron radius: re =

e2 = 2.82 × 10−15 meter. 4πε 0 m0 c 2

f. Electron rest mass energy: E0 = m0 c 2 = 0.511 MeV. g. Proton rest mass Energy: Ep = M p c 2 = 938 MeV. h. Fine structure constant:

e2 1 . ≈ 4πε 0 c 137

i. Planck’s constant:  = 1.05457 × 10−34 Joule second. j. Speed of light: c = 2.99792 × 108 meter/second. k. Rest mass of electron: m0 = 9.10939 × 10−31 kg. l. Charge on an electron: − e = −1.60218 × 10−19 Coulomb. m. Mass of proton: mp = 1.67262 × 10−27 kg. n. Boltzmann constant: kB = 1.38066 × 10−23 Joule/Kalvin.

1.16  Solved Examples 1. For the free electron described by ψ ( x , t ) = Ae i( kx −ωt ) , where k = 100 nm −1 , determine the de Broglie wavelength, momentum, kinetic energy, speed, and angular frequency. SOLUTION







λ=

2 π 2 × 3.1416 = = 6.2832 × 10−2 nm = 6.2832 × 10−11 m 100 k

Momentum, p =

h 6.626 × 10−34 =  = 1.054 × 10−23 kg m s −1 . λ 6.2832 × 10−11

Speed, v =

p = 1.157 × 107 m s −1 m

Kinetic energy, K =

p2 1   mv 2 = = ω 2 2m

K = 6.0975 × 10−17 J = 380.6  eV = 3.806 KeV Angular frequency,  ω =

K = vk = 1.157 × 1018 s−1 

2. A particle of mass m is described by ψ ( x ) = Ae −αx in a region where the energy of the particle is zero. Find the potential energy of the particle. 2

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Introduction to Quantum Mechanics

SOLUTION

−2 d2 ψ ( x) + The time-independent Schrödinger equation for a particle is: 2 m dx 2 V ( x)ψ ( x) = Eψ ( x) . In this case: 2 d2 ψ ( x) = V ( x)ψ ( x) 2 m dx 2 2 2 2 d2 ⇒ V ( x)Ae −αx = Ae −αx . 2 2 m dx 2α ⇒ V ( x) = ( 2αx − 1) m



2

3. Wave function ψ ( x) = Ae −αx describes the ground state of a quantum oscillator that satisfies the equation: −



 2 d 2 ψ ( x) 1 + mω 2 x 2 ψ ( x) = Eψ ( x) (1.24) 2 m dx 2 2

Calculate A, α and E and then show that it satisfies the Heisenberg uncertainty 2  2 relation ∆x∆p ≥ , where ∆x = x2 − x and ∆p = p2 − p . 2

(

(

)

)

SOLUTION

The oscillating particle can be found anywhere in the range of −∞ ≤ x ≤ ∞ . Therefore, from Eqn. (1.19), we have ∞

∫



∞

2  π  ψ ( x)ψ ( x)dx = 1 ⇒ A 2 e −2 αx dx = 1 ⇒ A 2 ×   2α 

∫

*

−∞

1/2

= 1 (1.25)

−∞

 2α  Therefore, A =   π 

1/4

. Substituting the value of ψ( x) in Eqn. (1.24), we get:

 2α 1 2α 2  2  2  − E +  mω 2 − x  ψ ( x) = 0 (1.26)  m   2  m



For the non-vanishing wave function, this equation can be satisfied for α = 2α 1  mω  = ω , which gives A =   π  2 m We next calculate:

and E =

1/4

mω 2

.

x = ∫ ∞−∞ ψ ( x ) xψ ( x ) dx   =   ∫ ∞−∞ A 2 x   e −2 αx dx   = 0, as the integrand is the odd function of x, *

2

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A Textbook on Modern Quantum Mechanics

and:

∫ = ∫

x2 =

∞

−∞ ∞

−∞

Hence, ∆x =

ψ ( x ) x 2 ψ ( x ) dx *

2

A 2 x 2   e −2 αx dx =

 2 mω

(1.27)

  . 2 mω ∞

dψ ( x )

∫ ψ ( x) dx dx de αx =   − i ∫ A  e dx dx = 2 iα ∫ A x e dx = 0

p = − i

*

−∞ ∞



−αx 2

2

−

2

−∞

∞

(1.28)

−2 αx 2

2

−∞

and: ∞

∫ = 2 α ∫

p2 = −2

2

ψ ( x)

*

−∞

∞

−∞

=

(

d2 ψ ( x ) dx dx 2

)

2 A 2 1 − 2αx 2 e −2 αx dx (1.29)

mω 2

mω . Therefore, ∆x∆p =  / 2 , which is the optimum value in the 2 uncertainty relation. Hence ∆p =

4. Consider two cases: (a) Zero point oscillation for a pendulum of mass, m = 1 kg and length, l = 1 m. (b) Motion of a tennis ball of mass m = 0.1 kg moving with speed v = 0.5 m/sec. Show that a quantum description is not required for these motions. SOLUTION

a. For pendulum, having zero-point oscillation, average potential energy is ˇ = 1 E = 1 mω 2 A 2 = 1 ω , where A is the root mean square amplitude of zeroV 4 2 4 point oscillation, and ω = g / l . Therefore, A =  / 2 mω , ω = 9.8 = 3.13 sec −1 .



A=

1.05 × 10−34 = 0.41 × 10−17 m. 2 × 3.13

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Introduction to Quantum Mechanics

b. For a tennis ball λ =

h h 6.626 × 10−34 = 1.32 × 10−32 m. = = p mv 0.1 × 0.5

Thus, it is concluded that the values of A for a pendulum and λ for a tennis ball are negligibly small to describe them quantum mechanically. 5. A metal surface when illuminated by the light sources of wavelengths λ and 0.5 λ ejects the photoelectrons with a maximum kinetic energy of 2.00 eV and 6.00 eV, respectively. What is the work function of the metal? SOLUTION

Let K1 and K 2 be the maximum kinetic energies of ejected photoelectrons and φ be the work function of metal. Then: K1 =

hc hc − φ and K 2 = − φ . Hence, φ = K 2 − 2 K1 = 2.0 eV. λ 0.5λ

6. The X-rays undergo Compton scattering with a target. The energy of X-rays is 300 keV and the scattered X-rays are detected at 30° relative to the incident X-rays. Calculate (a) the energy of the scattered X-ray, (b) the energy of the recoiling electron, and (c) Compton shift at this angle. SOLUTION

a. The energy of the scattered X-ray is given by Eqn. (1.9): E′ =



E ,  E  1+  1 − cos θ ( )  m0 c 2 



m0 c 2 = 9.11 × ( 2.9979 ) × 10−15 J =  511.8 keV,



E′ =

2

300 = 278.13 keV. 300   − 1+  1.0 0.866 ( )  511.08 

b. The energy of the recoil electron: E − E′ = 21.87 keV. c. Equation (1.8) gives the wavelength shift:

∆λ = λ ′ − λ =

 (1 − cos θ) m0 c

where λ′ and λ are the wavelengths after and before scattering, m0 is rest mass  of the electron, and θ is the scattering angle. The is known as the Compton m0 c wavelength of the electron, which is equal to 2.43 × 10−12 meter. Therefore:

∆λ = 2.43 × 10−12 ( 1.0 − 0.866 ) = 3.255 × 10−13 meter.

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7. Calculate the wavelength of maximum power emitted by a blackbody at 100   °C temperature. SOLUTION

The wavelength of maximum power emitted by a blackbody at temperature T is given by Wien’s displacement law, λ maxT = 2898  µm K. Therefore: λ max =



2898 = 7.77  µm. 373

8. Use the Bohr model for the hydrogen atom and calculate the energy change when the atom moves from energy level n = 5 to energy level n = 2. Is it radiation emission or absorption? Find the wavelength for this energy change. SOLUTION

The energy of nth level is given by Eqn. (1.7): En = −

where RH =

me 4 e2 RH 2 = − 2 2 2 2 = − n2 32 π ε 0  n 8πε 0 a0 n

e2 = 2.179 × 10−18 Joules  = 13.6 eV is a constant. 8πε 0 a0 1 1  ∆E = E2 − E5 = −13.6  −   4 25  = −2.856 eV



since this photon is moving down in energy levels 5 to 2 and the value of ∆E is negative. Radiation is emitted. To calculate the wavelength of emitted radiation, we use ∆E = hv = hc / λ , which implies λ = hc / ∆E . λ= 9. Show that ψ ( x) = Axe the wave function.

6.626 × 2.9979 × 10−26 =  4.34 × 10−7 m = 434 nm 2.856 × 1.602 × 10−19

− x2

 d2 2 is a wave function for the operator  − 2 + 4 x  . Normalize  dx 

SOLUTION

(

)

2  d2 d2 2 − x2 + 4 Ax 3e − x  − dx 2 + 4 x    ψ ( x ) = − dx 2 Axe   2 2 2 d = 4 Ax 3   e − x − A  e − x − 2 x 2 e − x    dx



2 2 2 2 = 4 Ax 3   e − x − A  −2 xe − x − 4xe − x + 4x 3   e − x    2

= 6 Axe − x = 6ψ (x)        

(1.30)

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Introduction to Quantum Mechanics

2  d2  which shows that ψ ( x) = Axe − x is the wave function of the operator  − 2 + 4 x 2  dx   with eigenvalue equal to 6. The normalization condition ψ ( x ) is:

∞

∫ ψ ( x)

2

dx = 1

−∞

∞

∫

2

⇒ A 2 x 2 e −2 x dx =



−∞

 2 ⇒ A = 2   π

A2 π / 2 . 4

1/4

 2 Therefore ψ ( x ) = 2    π



1/4

2

xe − x (1.31)

1.17 Exercises ∂ f ( x) d and then prove that xp − px = i , where p = − i ∂x dx and f ( x) is an arbitrary function x. 2. The helium ion can be treated as a hydrogen-like atom with ionization energy, Eion = 54.4eV in the ground state. Calculate the frequency ν and wavelength λ of electromagnetic radiation that will just ionize the helium ion. 3. On application of DC voltage V0 across the two superconducting layers separated  2 eV0  by a thin insulating layer, an oscillating current I = I 0 sin  t is produced    by the tunneling of paired electrons through the insulating layer. The phenomenon is known as the AC Josephson effect. Calculate the frequency of current for V0 = 2 × 10−6 Volts. 1. Show that f ( x)p − pf ( x) = i

4. Use the Bohr-Sommerfeld quantization condition and calculate the velocity of the electron in the first orbit of the hydrogen atom. Treat the first orbit as a circle of radius equal to the Bohr radius. 5. The uncertainty in the position of a moving particle is one-fourth of its de Broglie wavelength. Show that the maximum possible uncertainty in its velocity v is ∆v = . π 6. A free particle moving along the x-axis is represented by wave function, ψ ( x , t) = Ae i( kx−ωt ) . Use the time-dependent Schrödinger equation to find the relation between ω and k and then calculate the group velocity v g and phase velocity v ph .

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A Textbook on Modern Quantum Mechanics

7. The work function of aluminum is 4.26 eV. Its surface is irradiated by an ultraviolet beam of wavelength 220 nm. Calculate the maximum kinetic energy of electrons ejected from its surface.  2 2 e2  ψ (r ) = Eψ (r ) , ∇ + 8. Find for what value of E , ψ (r ) = Ne − r/a0 satisfies −  4πε 0 r   2m where N is constant. 9. Calculate N with the use of the condition of normalization for the wave function ψ (r ) = Ne − r/a0 .

2 Wave Mechanics and Its Simple Applications As stated in Chapter 1, Schrödinger developed quantum wave mechanics as a continuation of de Broglie’s hypothesis. He formulated a second-order differential equation to explain the wave nature of matter and the particle associated to the wave. The Schrödinger equation assumes that a particle behaves as a wave and yields a solution in terms of wave function and the energy of the particle under consideration. Once the wave function is known, then everything about the particle can be deduced from the wave function.

2.1  Schrödinger Equation This equation is used in two forms. In one of the forms, time explicitly appears to describe how the wave function of a particle evolves in time. This equation is referred to as the time-dependent Schrödinger equation. The other is the equation in which the time dependence has been dropped and the equation describes, among other things, what are allowed values of energies and hence it is known as the time-independent Schrödinger equation. However, these are not two independent equations. The timeindependent equation can be derived readily from the time-dependent equation (except if the potentials are time dependent). A simple derivation of the time-dependent Schrödinger equation is presented in Section 1.7 and the time-dependent Schrödinger equation, for a particle moving under the influence of a field defined by potential energy, V ( r ) , is given by Eqn. (1.16):

 2 2  ∂ψ (r , t)  − 2 m ∇ + V (r )  ψ (r , t) = i ∂t (2.1)  

2 2 ∇ + 2m V (r ) and other operators are time independent. The time independence of the Hamiltonian allows one to factorize the wave function into space- and time-dependent parts. Time dependence enters to wave function via a complex exponential factor, e − iEt/ , and hence the time-dependent wave function is written as: In the Schrödinger representation of quantum mechanics, the Hamiltonian, H = −



ψ ( r , t ) = ψ ( r ) e − iEt/ (2.2)

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Substitution of Eqn. (2.2) into Eqn. (2.1) yields:

 2 2   − 2 m ∇ + V ( r )  ψ ( r ) = Eψ ( r ) (2.3)  

which is the time-independent Schrödinger equation, also known as the eigenvalue equation. Solutions of Eqn. (2.1) describe the dynamical behavior of the particle, which is in some sense similar to the classical physics description obtained from Newton’s equation F = ma . But, there is an important difference between the two descriptions. By solving Newton’s equation, one can determine the position of a particle as a function of time, whereas solution of the Schrödinger equation gives a wave function ψ (r , t) which is related to the probability of finding the particle in some region where space varies as a function of time. E, termed as the energy of the particle in Eqn. (2.3), is a free parameter and in principle it can have all possible values without any restriction at any stage, suggesting that to determine the wave function for a particle moving under the influence of a potential with some specific value of E, we have to solve Eqn. (2.3) for the corresponding wave function. In doing so, we find different solutions ψ (r ) for different choices of E, for a given potential. We can emphasize this fact by writing ψ E ( r ) as the solution associated with a value of E. But all ψ E ( r ) are not acceptable. To be physically acceptable, ψ E ( r ) must satisfy two conditions; (i) it must be normalizable to fulfill the criteria of probability interpretation of the wave function, (ii) ψ E ( r ) and its derivative must be continuous. The first condition leads to a rather remarkable property of physical systems described by Eqn. (2.3), which is the quantization of energy:

∫ ψ (r , t )

2

d 3r =

∫ ψ (r )

2

d 3 r = 1 (2.4)

To have the finite (unity) value of the integral, we must have ψ E ( r ) = 0, for r → ±∞ . Thus, physically acceptable solutions are those which satisfy the normalization condition. Wave functions that do not satisfy the normalization condition and the corresponding value of the energy are not physically acceptable. A particle can never be observed to have any energy other than these values, referred to as allowed energies of the particle. We thus find that the probability interpretation of the wave function forces us to conclude that the allowed energies of a particle moving in a potential are restricted to certain discrete values, which are determined by the nature of the potential. The quantization of energy, a result of quantum mechanics, has enormous significance for determining the structure of atoms, and the properties of matter overall.

2.2  Bound States and Scattering States Equation (2.3) can be solved for two types of particles: (i) particles trapped by an attractive potential into what is known as a bound state, and (ii) the particles that are free to travel to infinity, also known as scattering states. A particle trapped in an infinitely deep potential well, an electron in an atom trapped by the attractive potential due to a positively charged atomic nucleus, and a nucleon trapped within a nucleus by attractive nuclear forces are examples of bound states. In all these cases, the probability of finding the particle at infinity is zero, because the wave function vanishes at infinity. Thus, when a particle is trapped

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Wave Mechanics and Its Simple Applications

or confined to a limited region of space by an attractive potential, we obtain wave functions that satisfy the above boundary condition Eqn. (2.4), and their energies are quantized. When a particle is not bound by any attractive potential or it is repelled by a repulsive potential, it is free to move as far as it likes in space. For such cases, we find that the wave function does not vanish at infinity, and its energy is not quantized. The problem then arises of how to reconcile this situation with the normalization condition, and the probability interpretation of the wave function. When wave function does not diverge or remains finite at infinity, a physical meaning can be assigned to such states as being idealized mathematical limiting cases which can still be dealt with in the same way as the bound state wave functions, provided some care is taken with the physical interpretation.

2.3  Probability Density, Probability Current, and Expectation Value Current density related to a wave function ψ(r , t) can be calculated, as ψ(r , t) evolves with time in accordance with the Schrödinger equation:

i

∂ψ (r , t) = Hψ (r , t) (2.5) ∂t

The complex conjugation of Eqn. (2.5) is: − i



∂ψ * (r , t) = H * ψ * (r , t) (2.6) ∂t

Probability density is defined as P = ψ ( r , t ) and then charge density is given by ρ = qP , where q is charge on the particle. We have: 2

2

∂P ∂ ψ ∂ * ∂ψ * ∂ψ = = ψψ = ψ + ψ* (2.7) ∂t ∂t ∂t ∂t ∂t



(

)

With the use of Eqns. (2.5) and (2.6), Eqn. (2.7) can be rewritten as: ∂P 1 =  ψ * ( Hψ ) − ( Hψ )* ψ  (2.8) ∂t i 

where H = −

2 2 ∇ + V ( r ) = H * , if V (r ) is real. We then get: 2m  ∂P 1  * 2 2 =  −ψ ∇ 2 ψ + ψ *V (r )ψ + ∇ 2 ψ * ψ − ψV (r )ψ *  ∂t i  2m 2m 

(



)

(

 2 1  * 2 2 −ψ ∇ ψ + ∇2ψ * ψ   i  2m 2m  i =− ∇.  ψ∇ψ * − ψ *∇ψ  2m  =

(

)

(

)

)

(2.9)

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which yields: ∂ P i + ∇.  ψ∇ψ * − ψ *∇ψ  = 0 ∂t 2 m  ∂P ⇒ + ∇.S = 0 ∂t



with: S=



i  ψ∇ψ * − ψ *∇ψ  (2.10) 2m 

S is called probability current, describing the flow of probability. This also implies: ∂ρ / ∂t + ∇.J = 0 (2.11)



Here, we have defined current density J = qS . Eqn. (2.11) is a well-known continuity equation. To understand more about Eqn. (2.10), let us consider a system completely confined to a volume V so that nothing is going in and out at the surface. On integrating both terms of the equation over the entire volume, we get:  ∂P



∫  ∂t + ∇.S d r = 0



⇒

∫

3

∂P 3 d r = − ∇.Sd 3 r =   − S. ds ∂t

∫

∫

(2.12a)

where, we used the Gauss theorem to convert volume integration to surface integration. ds is surface element defined outwards and integrated over the surface of the volume V. In this case, because we confined to system completely in the volume, there is nothing at the surface and therefore

∫ S. ds = 0 . On interchanging the order of the volume integral and

the time derivative over P, we write;

∂ ∂t



∫ Pd r = 0 (2.12b) 3

∫ Pd r = constant or a conserved quantity. If the volume does not confine the full system, the amount in Pd r is given by amount of flow out of volume, as is seen ∫ which states that

3

3

from Eqn. (2.12a). This means that S(r ) and P(r ) are the current and density, respectively, of a conserved quantity. The definition of probability density allows us to calculate expectation value of an observable (operator), A. When a large number of measurements on A of a particle are made in a particular state, the average of different measured values is the expectation 2 value. The ψ(r , t) represents the probability of measurement on A of a particle at position vector r and time t. The expectation value (average of measurements) of A is defined by

Wave Mechanics and Its Simple Applications

25

A = ∫ ψ (r , t)* Aψ (r , t)d 3 r for normalized wave function. If ψ(r , t) is not a normalized wavefunction, then A is defined as follows: A =



∫ ψ (r , t)* Aψ (r , t)d 3 r (2.13) ∫ ψ (r , t)* ψ (r , t)d 3 r

2.4  Simple Applications of Time-Independent Schrödinger Equation To illustrate how the time-independent Schrödinger equation can be solved in practice, and what are some of the characteristics of its solutions, we discuss here the some of its well-known solutions. 2.4.1  Free Particle Motion A particle that has no external forces acting on it (a constant potential energy V0 ) is called a free particle. The one-dimensional (1D) motion of a particle of mass m is described by:



−

2 d2ψ ( x ) + V0 ψ ( x ) = Eψ ( x ) (2.14) 2 m dx 2

d2ψ ( x ) 2 m ( E − V0 ) =− ψ ( x ) = − k 2 ψ ( x ) (2.15) dx 2 2

2 m ( E − V0 )  . k has real values when E > V0 and it has imaginary values for E < V0 .  For real values of k, a generalized solution of Eqn. (2.15) is given by: with k ±



ψ ( x ) = Ae ikx + Be − ikx (2.16)

2 k 2 which represents a traveling wave of kinetic energy, = ( E − V0 ) , classically allowed 2m values of kinetic energy. When E < V0 , values of k are imaginary, say k = iK , and the solution of Eqn. (2.15) is given as:

ψ ( x ) = Ce Kx + De − Kx (2.17)

2 k 2 2 K 2 =− = −(V0 − E) , which are classically forbidden values of kinetic energy. 2m 2m We thus find that in the case of E > V0 , kinetic energies of the particle are classically allowed, whereas in the situation in which E < V0 , kinetic energies are classically forbidden. This means that a particle, which is rolling on a potential surface described by V ( x ) , can exist in the region where V ( x ) < E and it must turn around as and when V ( x ) ≥ E . The points where V ( x ) = E (kinetic energies or velocities of the particle are zero) are known as the classical turning points. where

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For the traveling wave solutions given by Eqn. (2.16), probability density P ( x ) is given by:



iEt

P ( x ) = ψ ( x, t ) ψ ( x, t ) = ψ ( x ) e  ψ ( x ) e *

*

−

iEt 

   = A 2 + B2 + 2 AB cos ( 2 kx )

(2.18)

for real values of A and B. Equation (2.18) represents a standing wave. If we consider a wave is traveling along in only one direction, say the +x-direction, P ( x ) = A 2 , which is independent of position, implying that the particle is equally likely to be anywhere in space; it is completely delocalized. Solution of Eqn. (2.14) for E < V0 is given by Eqn. (2.17) and



iEt

P ( x ) = ψ ( x, t ) ψ ( x, t ) = ψ ( x ) e  ψ ( x ) e *

*

−

iEt 

  = A 2 e 2 Kx + B2 e −2 Kx + 2 AB cosh ( 2 Kx )

(2.19)

which suggests that the probability density of finding the particle in regions where V ( x ) > E is nonzero, implying that particle penetrates to classically forbidden region (barrier). Thus, quantum mechanical solutions yield a non-zero probability of finding the particle in the classically forbidden region. Quantum mechanics allows a particle to penetrate (tunnel through) a barrier (classically forbidden region) and emerge on the other side of the barrier. 2.4.2  Infinite Potential Well (Particle in a Box) Consider a single particle of mass m confined to within a region, − a / 2 < x < a / 2, of potential energy V ( x ) = V0 < E, bounded by infinitely high potential barriers, i.e. V ( x ) = ∞ at the a a walls at x = − and x = . Potential is written as: 2 2



  V0 < E  V ( x) =   ∞ 

a −a

2

for

The probability of finding the particle outside the well is zero; hence wave function  a must vanish at boundaries, i.e. ψ  ±  = 0 , otherwise V ( x ) ψx) will be infinite outside  2 the well, which has no meaning. Solutions of the 1D Schrödinger equation inside the well 2 k 2 ( − a / 2 < x < a / 2 ) are given by Eqn. (2.16) and 2m = E − V0 . At the boundaries we have:

 a ψ  −  = Ae − ika/2 + Be ika/2 = 0 (2.21)  2



 a ψ   = Ae ika/2 + Be − ika/2 = 0 (2.22)  2

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Wave Mechanics and Its Simple Applications

FIGURE 2.1 Infinite potential well.

The sum of Eqns. (2.21) and (2.22) yields:

( A + B) cos 

ka   = 0 (2.23) 2

while subtraction of Eqn. (2.21) from (2.22) gives:

 ka  i ( A − B ) sin   = 0 (2.24)  2

Two simple cases for which conditions Eqns. (2.23) and (2.24) can be satisfied are (i) both A and B are zero, and (ii) A = B or A = − B. The first case gives a trivial solution which is physically unacceptable. For the second case, we have:

 ka  cos   = 0, when  A = B (2.25)  2



 ka  sin   = 0, for  A = − B (2.26)  2

nπ satisfies Eqn. (2.25) for a odd values of n (= 1, 3, 5, 7,….) and it satisfies Eqn. (2.26) for even values of n (= 2, 4, 6……….). Note that energy E and momentum k of a particle take discrete values. Each of values of n corresponds to a quantum state of the system. For the nth quantum state, we have: Both the conditions (Eqns 2.25 and 2.26) must be satisfied. k =



En = V0 +

 2 ( nπ ) (2.27) 2 m  a 2 2

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and,



  nπx   for odd  n − values,  n = 1, 3, 5, 7 ……..  A  cos   a   ψ n ( x) =  (2.28) nπx   A  sin  − = …… n n 8 ..  for even  values,  2, 4, 6,  a  

or,

a  nπ  ψ n ( x) = sin   x +   for n = 1, 2, 3, 4, 5………. (2.29)  a  2 

En and ψ n ( x ) are called energy eigenvalue and eigenfunction, respectively. E1 , E2  and  E3 are energy values and ψ 1 , ψ 2 and ψ 3 are corresponding wave functions for the ground state, first exited state, and second excited state. The constant A can be determined by using the normalization condition on the wave function: a/2

∫ψ

n

( x )* ψ n ( x ) dx = 1

− a/2

a/2



⇒

 nπ 

a

∫ A   sin  a  x + 2   dx = 1 2

2

− a/2

(2.30)

⇒ A = 2/a It is to be noted that the normalization condition can be met for a range of complex amplitudes, A = e iθ 2 / a , in which the phase is arbitrary, which implies that the outcome of 2 a measurement, which is proportional to ψ ( x ) about the particle position, is invariant under a global phase factor. Also, note that a/2



∫ψ

m

( x )* ψ n ( x ) dx =  δ mn (2.31)

− a/2

where, δ mn is known as the Kronecker delta and it is defined as:

 1 δ mn =   0

when  m = n (2.32) otherwise

Wave functions that satisfy Eqn. (2.31) are called orthonormal wave functions. The infinite potential well is a valuable model because it shows in a simple way how energy quantization takes place in nature. This model describes very accurately the quantum character of such systems of electrons trapped in a block of metal, quantum wire, and nanoparticle or gas molecules contained in a bottle. The potential experienced by an electron as it approaches the edges of a block of metal, quantum wire, and nanoparticle or as experienced by a gas molecule as it approaches the walls of its container are effectively infinite, provided particles have sufficiently low kinetic energy as compared to the height of these potential barriers.

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Wave Mechanics and Its Simple Applications

2.4.3  The Finite Potential Well When a particle inside potential well has an energy comparable to the height of the potential barriers, use of the infinite well potential model is not justified, because chances of the particle escaping the well are there. The motion of the particle can be discussed both classically and quantum mechanically, though, as we expect, the quantum mechanical treatment is not necessarily consistent with our classical physics-based expectations. We describe here the quantum mechanical properties of a particle in a finite potential well depicted in Fig. 2.2. We choose a well in which the potential is symmetric about its origin, and we also choose zero potential energy at the bottom of the well. To find bound state solutions, E < V ( x ) , let us describe the potential:



 a a  0  for  − < x < 2 2  V ( x) =   V0 > E  for  x > a 2 

(2.33)

a a > x > , classical motion is forbidden because kinetic energy is negative. 2 2 However, quantum mechanically motion is allowed and it is described by the Schrödinger equation: For cases of −



d 2 ψ ( x) 2 m (V0 − E ) = ψ ( x) = α 2 ψ ( x) (2.34) dx 2 2

Where α 2 is a positive quantity. A general solution of Eqn. (2.34) is: ψ ( x) = Ae αx + Be −αx (2.35)



In the region of x < − a / 2 , inclusion of a term having e −αx leads to ψ ( x) → ∞ , for x → −∞ , which is an unacceptable solution. Hence,

ψ ( x ) = Ae αx

FIGURE 2.2 Schematic diagram for finite potential well.

for x < − a / 2 (2.36)

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Similarly, for x > a / 2, we cannot include e αx in our solution, and therefore, ψ ( x ) = Be −αx



for x > a / 2 (2.37)

Inside the well, − a / 2 < x < a / 2 , the Schrödinger equation is: d2ψ ( x ) = − k 2 ψ ( x ) , with  k = dx 2



2 mE   (2.38) 2

A general solution can be given by: ψ ( x ) = C   cos ( kx ) + D  sin ( kx ) (2.39)



It is to be noted that Eqns. (2.39) and (2.16) convey the same meaning. The allowed solutions of the Schrödinger equation are those which satisfy necessary boundary conditions: wave function and its first order derivatives must be continuous at all values of x. Thus, in order dψ to satisfy Eqns. (2.37), (2.38) and (2.39) everywhere, we require that ψ( x) and ψ ′( x) = be dx continuous at the walls of the well.

 a ψ−  :  2

 ka   ka  Ae −αa/2 = C cos   − D sin   (2.40)  2  2



 a ψ′  −  :  2

 ka   ka  α   Ae −αa/2 = kC sin   + kD cos   (2.41)  2  2



 a ψ  :  2

 ka   ka  Be −αa/2 = C cos   + D sin   (2.42)  2  2



 a ψ′   :  2

− α   Be

−

αa 2

 ka   ka  = − kC sin   + kD cos   (2.43)  2  2

Equations (2.40) to (2.43) form a matrix equation:



            

0

 ka  − cos    2

 ka  sin    2

α

0

 ka  − sin    2

 ka  − k cos    2

0

1

 ka  − cos    2

 ka  − sin    2

0

−α

 ka  sin    2

 ka  − k cos    2

1

            

 αa −  Ae 2  αa −  2 Be   C  D 

     =       

0 0 0 0

    (2.44)  

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Wave Mechanics and Its Simple Applications

The four equations can be solved to find the allowed discrete energy eigenvalues En of the states confined within the quantum well. A non-trivial solution of matrix Eqn. (2.44) is obtained when the determinant of the 4 × 4 matrix is zero:

1

0

 ka  − cos    2

 ka  sin    2

α

0

 ka  − sin    2

 ka  − k cos    2

0

1

 ka  − cos    2

 ka  − sin    2

0

−α

 ka  sin    2

 ka  − k cos    2



= 0 (2.45a)

Simplification gives that the determinant goes to zero when:

  ka   ka    k sin  2  − α cos  2    

  ka   ka    k cos  2  + α sin  2   = 0 (2.45b)  

which means, either

 ka  α = k  tan   (2.46a)  2

or

 ka  α = − k  cot   (2.46b)  2

Solutions presented by Eqns. (2.40) to (2.46) are found less frequently in quantum mechanics textbooks, but these are quite useful in applications like the double quantum well, the modeling of bonding in molecules, and in an infinite array of quantum wells that represents a periodic solid. As the potential is symmetric, an easier way to get Eqns. (2.46a) and (2.46b), which exists in textbooks on quantum mechanics, is as follows: Addition and subtraction of Eqns. (2.40) and (2.42) give:

( A + B) e



( A − B) e

−

−

αa 2

αa 2

 ka  = 2C   cos   (2.47a)  2  ka  = −2 D  sin   (2.47b)  2

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Addition and subtraction of Eqns. (2.41) and (2.43) yield:

α ( A − B) e

−

αa 2

 ka  = 2 kD  cos   (2.48a)  2



α ( A + B) e

−

αa 2

 ka  = 2 kC   sin   (2.48b)  2

Division of Eqn. (2.48b) by (2.47a) gives Eqn. (2.46a) provided A ≠ − B and C ≠ 0 . Similarly, we get Eqn. (2.46b) on dividing Eqn. (2.48a) by Eqn. (2.47b), when A ≠ B and D ≠ 0 . It is to be noted that Eqns. (2.46a) and (2.46b) cannot be satisfied simultaneously, because in that  ka  case tan 2   = −1, which cannot be satisfied for real values of k and a.  2 For the case of A = B and D = 0, allowed energy values within the well are given by Eqn. (2.46a) and the wave functions are:



 Ae αx   ψ ( x) =  C cos ( kx )   Be −αx  

when

x < −a / 2

a < x < a / 2 (2.49a) 2 a when x > 2

for −

Wave function in this case is symmetric with respect to x; ψ ( x ) = ψ ( − x ) . In other words, it is an even function. When A = − B and C = 0, allowed energies within the well are given by Eqn. (2.46b) and the wave functions are:



 Ae αx   ψ ( x) =  D sin ( kx )   Be −αx  

when

x < −a / 2

a < x < a / 2 (2.49b) 2 a when x > 2

for −

Wave function is antisymmetric, ψ ( − x ) = −ψ ( x ) in this case. It is an odd function. To find the energy levels that correspond to symmetric solutions, we solve numerically or graphically Eqn. (2.46a), while for energy levels that correspond to antisymmetric solutions, Eqn. (2.46b) is solved. A simple graphical solution method is as follows: define, λ = ka / 2 and µ = αa / 2 . Then, Eqns. (2.64a) and (2.46b) are:

µ =   λ  tan ( λ ) (2.50a)

and,

µ =   −λ  cot ( λ ) (2.50b)

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Wave Mechanics and Its Simple Applications

FIGURE 2.3 Graphical solution of Eqns. (2.50a) and (2.50b). Solid curves are µ = λ tan λ and dashed curves are µ = −λ cot λ .

with,

λ2 + µ2 =

a2 2 mV0 a 2 k + α2 = (2.51). 4 22

(

)

a mV0 / 2 in ( λ ,  µ ) -space. Equations (2.50a),  (2.50b) and (2.51) are plotted as a function of λ in Fig. 2.3. Intersections of the plots of µ =   λ  tan ( λ ) and λ 2 + µ 2 = ρ2 curves in ( µ , λ ) -plane for ­different

Equation (2.51) represents a circle of radius ρ =

 2 2   8 2   24 2  values of ρ ; ρ = 1   = V0 a 2  , ρ = 2 3    = V0 a 2  ,  ρ = 2  = V0 a 2  etc. provide even  m   m   m 

(symmetric) solutions, while, intersections of µ =   −λ cot ( λ ) and λ 2 + µ 2 = ρ2 for different values of ρ yield odd (antisymmetric) solutions. The number of bound states depends on the height and width of the well through V0 a 2 factor. As is seen from Fig. 2.3, there is π one bound state corresponding to the even (symmetric) wave function for 0 < ρ < , and 2 there are two bound states, one corresponding to the even (symmetric) wave function and π one corresponding to the odd (antisymmetric) solution, for < ρ < π . Similarly, there exist 2 3π ; two of these correspond to the symmetric solution and three bound states for π < ρ < 2 one corresponds to the antisymmetric solution. On increasing the values of ρ, the number of bound states energy levels increase successively. It is to be noted that there exists at least

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A Textbook on Modern Quantum Mechanics

one bound state for the given value of V0 , and there is no degeneracy in the solutions. When the intersection of the circle with Eqn. (2.46a) is represented by λ 1 , λ 2 , λ 3 …. , we have:

λ 2n =

ma 2 2 En ⇒ En = ( 2λ n / a )2 (2.52). 2 2 2m

The case of E > V0 does not provide any bound state. In this case, particle energy exceeds 2m the height of the well and k 2 = 2 ( E − V0 ) is positive everywhere giving the sinusoidal  solution of the Schrödinger equation in all regions, and probability density is distributed all over the space. A particle is incident on the potential well, and it would keep going as there is no probability of reflection from the potential well. This is the case of scattering of a particle from the potential well. 2.4.4  Step Potential We consider a case of a particle, initially traveling in a region of space of a constant potential, that suddenly moves into a region of different but again constant potential, as is shown in Fig. 2.4. The potential can be expressed as a piecewise function:



 V1 V ( x) =   V2

for

x V2 , it can continue to propagate arbitrarily far to the right, with an increased wavelength (decreased wave vector and momentum) across the step. But, if the particle has energy below the step E < V2 , it is classically forbidden to be found in the region of x > 0 . However, the quantum mechanical description allows the particle to tunnel into this region and thus have a non-zero probability to be found in the region of x > 0 . In both cases, there exists a finite reflectivity.

FIGURE 2.4 Potential step.

35

Wave Mechanics and Its Simple Applications

For the given energy of a particle E there are two cases of interest; (i) E is greater than both V1 and V2 , and (ii) E is greater than V1 but less than V2 .

Case-1: E is greater than both V1 and V2 The solution of the 1D Schrödinger equation in region-I ( x < 0 ), and in region-II ( x > 0 ), can be written as: ψ 1 ( x) = Ae ik1x + Be − ik1x (2.54a)

 2m  with k1 =  2 ( E − V1 )   

1/2

for x < 0 . And, ψ 2 ( x) = Ce ik2 x + De − ik2 x (2.54b)

1/2

 2m  where, k2 =  2 ( E − V2 ) , when x > 0. The coefficients ( A, B, C  and  D) are to be determined.    Both terms in ψ 1 ( x) are to be retained, as region-I can have both incident and reflected waves. However, we would not have a reflected wave in region-II, and hence, we take D = 0 for writing ψ 2 ( x). Equations (2.54a) and (2.54b) must satisfy the boundary conditions (the wave function and its first derivative must be continuous) at x = 0. Therefore:

ψ 1 (0) = ψ 2 (0) ⇒ A + B = C (2.55a)



ψ 1′ (0) = ψ ′2 (0) ⇒ ik1 A − ik1 B = ik2 C (2.55b).

Elimination of C from Eqns. (2.55a) and (2.55b) gives: k1 ( A − B ) = k 2 ( A + B )

B k1  B = 1−  − 1 (2.56)  A k2 A B 1 − k 2 / k1 ⇒ = A 1 + k 2 / k1 ⇒ 

with

k2 = k1

1 − V2 / E (2.57) 1 − V1 / E

B / A is the reflection coefficient of the barrier. The reflectivity of the barrier, which corresponds to the ratio of probabilities, is given by:

R=

2

2

1 − k 2 / k1 B = (2.58) 1 + k 2 / k1 A

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A Textbook on Modern Quantum Mechanics

Due to the conservation of the particle number, the transmissivity is given by: 2

T = 1− R = 1−



1 − k 2 / k1 (2.59) 1 + k 2 / k1

It is to be noted that on going from region-I to region-II, the de Broglie wavelength of the particle with energy E changes and becomes longer for an increased potential step. We see that in the quantum analogue of scattering, the energy of the particle is not quantized, and the wave function that describes the scattering of a particle of a given energy does not decrease as x → ±∞; therefore everything that leads to the quantization of energy for a bound particle does not apply here. Wave functions given by Eqns. (2.54a) and (2.54b) do not go to zero as x → ±∞ , implying that they do not satisfy the normalization condition – 2 the integral over ψ will always diverge. Then, how does one maintain the claim that the wave function must have a probability interpretation if one of the principal requirements, the normalization condition, does not hold true? A wave function that cannot be normalized to unity is inconsistent with the probability interpretation of the wave function, and hence it is not physically permitted. Nevertheless, it is possible to work with such wave functions 2 by reinterpreting the wave function and calling ψ( x , t) particle flux. We thus find that there can be quantum mechanical descriptions which agree with our classical intuition, but there also occur new kinds of phenomena that have no explanation within classical physics.

Case-2: E is greater than V1 but less than V2 In this case, the 1D Schrödinger equation in region-I ( x < 0) has a sinusoidal solution that can be given by Eqn. (2.54a). However, in region-II ( x > 0), the solution of the 1D Schrödinger equation cannot be given by Eqn. (2.54b). A general solution in this is: ψ 2 ( x) = Ce αx + De −αx (2.60)

1/2

 2m  where, α =  2 (V2 − E ) . Inclusion of term Ce αx in the solution in region-II would lead    to an unacceptable solution because the probability of finding the particle in this region will go infinity as x → ∞ . Therefore, we should set C = 0 . Applying again the boundary conditions at x = 0 , we obtain:

A + B = D (2.61a)

and,

ik ( A − B ) = αD (2.61b)

Elimination of D from Eqns. (2.61a) and (2.61b) gives:

B ik − α (2.62) = A ik + α

Wave Mechanics and Its Simple Applications

37

which is the reflection coefficient of the barrier. The reflectivity of the barrier is thus given by:

R=

2

B  ik − α   − ik − α  = = 1 (2.63)  ik + α   − ik + α  A

We thus find that though the particle has nonzero probability to penetrate the classically forbidden region, it is still being totally reflected away, and there is no transmission to the step potential. Note that the depth at which the particle penetrates into the classically forbidden region is given by the distance, ∆x from x   =  0 , at which the probability drops by 1/ e.

P ( ∆x ) = ψ 2 = D2 e −2 α∆x = D2 e −1 (2.64a) 2

which gives:

∆x =

1  = (2.64b) 2α 2 2 m (V2 − E )  

2.4.5  Finite Potential Barrier and Tunneling Another useful example of solving the 1D Schrödinger equation is the case of the finite potential barrier of height V0 and width a , shown in Fig. 2.5. The potential energy is defined by:

FIGURE 2.5 Finite potential barrier.

 V0 V ( x) =   0

for  − a / 2 < x < a / 2 when  x > a / 2

(2.65)

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A Textbook on Modern Quantum Mechanics

There can again be two types of behavior occurring. Case-1, where E > V0 , and case-2 where E < V0 . In case-1, the particle will have some reflection and some transmission as expected classically. Case-2 is of more interest, because classical physics predictions would be that the particle should be totally reflected and there should not be any transmission, which means that there should be zero probability to find the particle on the right-hand side of the barrier. However, quantum mechanics provide a finite (nonzero) probability of finding the particle on the right-hand side of the barrier, and it can continue to propagate to +∞ . This simple model is the precursor for discussing the decay of atomic nuclei as well as other quantum tunneling effects such as those associated with a scanning electron microscope. The particle is moving from −x to +x across the barrier centred on the origin. In region-I ( x < − a / 2) and in region-III ( x > a / 2) , V ( x ) is zero and hence the general solution of the 1D Schrödinger equation will be a sinusoidal solution like that given by Eqn. (2.54a). However, if the particle initially starts from the left-hand side of the barrier ( x < − a / 2) , then it is on the right side of the barrier ( x > a / 2) , and there is no way to have e − ikx in the solution, because this region cannot have a wave traveling to the −x direction. Therefore:

ψ 1 ( x) = Ae ikx + Be − ikx , for ( x < − a / 2) (2.66)



ψ 3 ( x) = Fe ikx , when ( x > a / 2) (2.67)

 2 mE  with k =  2    

1/2

. In region-II (− a / 2 < x < a / 2) , the solution of the 1D Schrödinger

equation is to be worked out for two possible cases: (i) E > V0 , and (ii) E < V0 .

Case-1: E > V0 The solution of the 1D Schrödinger equation will be similar to that given by Eqn. (2.66):

ψ 2 ( x) = Ce iKx + De − iKx (2.68)

with K = 2 m ( E − V0 ) /  . The wave associated with the particle will have reflection as well as transmission at both boundaries of the barrier.

Case-2: E < V0 The general solution of the 1D Schrödinger equation can be written as:

ψ 2 ( x) = Ce αx + De −αx (2.69)

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Wave Mechanics and Its Simple Applications

where α = 2 m (V0 − E ) /  . Application of boundary conditions on wave functions given by Eqns. (2.66), (2.67) and (2.88) provides:

 −a   −a  ψ1   = ψ2   :  2   2 

Ae − ika/2 + Be ika/2 = Ce −αa/2 + De αa/2 (2.70a)



 −a   −a  ψ 1′   = ψ ′2   :  2   2 

ika αa  − ika   − αa  k  Ae 2 − Be 2  = α  Ce 2 − De 2  (2.70b)    



 a  a ψ2   = ψ3   :  2  2

Ce αa/2 + De −αa/2 = Fe ika/2 (2.70c)



 a  a ψ ′2   = ψ ′3   :  2  2

αa  αa  − α  Ce 2 − De 2  = ikFe ika/2 (2.70d)  

Equations (2.70a) to (2.70d) are to be solved to find out reflection ( B / A ) and transmission ( F / A ) coefficients from the barrier. We first divide Eqns. (2.70c) and (2.70d) by A and then solve these for C / A and D / A in terms of F / A to get:

C ( α + ik ) F ika/2 −αa/2 = e e (2.71a) A 2αA



D ( α − ik ) F ika/2 αa/2 = e e (2.71b) A 2αA

Next, first dividing by A and then subtracting Eqn. (2.71b) from (2.71a), we obtain B / A in terms of C / A and D / A :

(α − ik ) C e − 2 e − B =− A 2 ikA αa



ika 2

+

( α + ik ) D e αa/2 e − ika/2 (2.72) 2 ikA

On substituting values of C / A and D / A from Eqns. (2.71a) and (2.71b) into Eqn. (2.72), we obtain:

(

)

2 2 B F α +k =     sinh ( αa ) (2.73) A A 2 ikα

Addition of Eqns. (2.70a) and (2.70b), after dividing by A , gives:

e − ika/2 =

(α + ik ) C e −αa/2 − ( α − ik ) D e αa/2 (2.74) 2 ikA

2 ikA

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A Textbook on Modern Quantum Mechanics

Substitution of C / A and D / A from Eqns. (2.71a) and (2.71b) into Eqn. (2.74) yields: e − ika/2 =

2 2 F  ( α + ik ) −αa ( α − ik ) αa  ika/2 e − e e  A  4ikα 4ikα 

F 4ikαe − ika ⇒ = A ( α + ik )2 e −αa − ( α − ik )2 e αa   

(2.75)

which gives: B 2(α 2 + k 2 )e − ika sinh(αa) = (2.76) A ( α + ik )2 e −αa − ( α − ik )2 e αa   



The reflectance R and transmittance T are the given by:

(

)

2

2 2 α 2 + k 2 sinh 2 (αa) B (2.77a) R= = A  α 2 + k 2 2 cosh(2αa) − α 4 + k 4 − 6 k 2 a 2   

(

)

F = A  α2 + k2 

)

(

)

(

)

and

T=

2

(

8k 2 α 2 2

cosh(2αa) − α 4 + k 4 − 6 k 2 a 2  

(2.77b)

In case of weak transmission ( αa  1) , the transmittance can be approximated by taking e 2 αa cosh ( 2αa ) ≈ and then on dropping the smaller terms in the denominator, we get: 2

T≈

( 4αk )2

(k

2

+α

)

2 2

e −2 αa (2.78)

2.4.6  Relevance of Free Particle, Potential Wells, and Potential Barriers Examples illustrated above may seem somewhat abstract and not directly related to any realistic problems that one finds in the real world. However, these are not useless simplified cases of the 1D Schrödinger equation, as one might think. Rather, these are first-order approximations to scenarios that are encountered in the real world. These simplified solutions can give us insight into the behavior (both qualitative and quantitative) of actual physical systems. For example, (i) several properties of some metals can be understood with the use of free particle motion, (ii) the quantum tunneling model through the step barrier gives reasonable approximations for quantum tunneling effects in a variety of physical phenomena, such as the scanning electron microscope, (iii) the bound states for the infinite potential well demonstrate the concept of bound states, quantized energy levels, and

Wave Mechanics and Its Simple Applications

41

temporal evolution of energy eigenstate superposition, which are linked to numerous real world scenarios such as the quantum wire, nanoparticle and hydrogen atom. Also, the propagation of a free particle, propagation of a particle across a potential step, and tunneling are all analogous to the propagation of light through various media with varying refractive indices. Particle-in-a-box solutions can be used to study the optical cavity.

2.5  Periodic Solids and their Band Structures For many solids, each electron effectively experiences a similar average potential as experienced by all other electrons. This allows one to use an independent electron approximation and study the electronic properties of a solid by employing the one electron Schrödinger equation instead of solving something like the 1022 body problem. The exact form of the average potential may not be known, but it is expected to relate closely with the isolated atomic potential of an atom, which is the part of the solid. Further, many of the solids are crystalline, with a periodic lattice. Because the ground state electronic structure must also be periodic, with the same charge distribution in each unit cell, the average potential is periodic too; V ( r ) = V ( r + R ) , where R is a vector joining the same points in two different unit cells. Individual electron wave functions must also reflect this periodicity by satisfying a condition referred to as Bloch’s theorem, which states that the wave functions of the p2 one electron Hamiltonian H = + V (r ) can be chosen to have the form of a plane wave 2m times a function with periodicity of the lattice:

ψ nk (r ) = e ik .R unk (r ) (2.79a)

with

unk ( r + R ) = unk ( r ) (2.79b)

where the subscript n refers to a state associated with wave vector k. Combining Eqns. (2.79a) and (2.79b), Bloch’s theorem is restated as:

ψ nk ( r + R ) = e ik. R ψ nk ( r ) (2.80)

The equation Hψ nk ( r ) = E nk ψ nk ( r ) is solved to find energy eigenvalues Enk . A plot of Enk versus k is referred as the band structure of a solid. For a 1D periodic structure with length L joining the same points in two different unit cells, Eqn. (2.79a) can be rewritten as:

ψ nk ( x ) = e iqx unk ( x ) (2.81a)

with

unk ( x + L ) = unk ( x ) (2.81b)

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2.5.1  The Kronig-Penney Model The Kronig-Penney model describes the electron in a 1D periodic potential. The possible eigenvalues that can be occupied by the electron and corresponding eigenstates are determined by solving a 1D Schrödinger equation. Many electronic properties of a periodic solid can be illustrated by using a periodic array of quantum wells, separated by barriers. A periodic structure having wells, each of width a and barriers each having width b and height V0 is shown in Fig. 2.6. Equations (2.81a) and (2.81b) describe the wave function in the 1D periodic potential, with L = nd , where d = a + b is period of potential. Within a well, V ( x ) = 0 . For the first well on the right-hand side to origin, 0 < x < a, and hence the solution of the 1D Schrödinger equation is: ψ 1 ( x) = Ae ikx + Be − ikx (2.82)

1/2

 2 mE  with k =  2  . The general solution of the 1D Schrödinger equation to the first barrier    on the left-hand side of the origin ( −b < x < 0 ) is: ψ 2 ( x) = Ce αx + De −αx (2.83)



Where α = 2 m (V0 − E ) /  because E < V0 inside the barrier. Two of the boundary conditions are applied at x = 0 and the other two at x = a . At x = 0 :

ψ 1 ( 0) = ψ 2 ( 0) : ψ 1′ ( 0 ) = ψ ′2 ( 0 ) :

FIGURE 2.6 The Kronig-Penney periodic potential.

A + B = C + D (2.84a) ik ( A − B ) = α ( C − D ) (2.84b)

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Wave Mechanics and Its Simple Applications

To apply boundary conditions at x = a, we have to make use of the Bloch theorem, Eqn. (2.80), for the 1D case, which becomes: ψ ( x + L ) = e iqL ψ ( x ) (2.85)



where q is the wave vector. x = a is related to x = −b via the Bloch Theorem, Eqn. (2.85). Therefore by choosing L = a + b, we have:

(

)

iq ( a + b )



ψ 1 ( a) = e iq( a + b) ψ 2 (−b) :    Ae ika + Be − ika = Ce −αb + De αb e e



ψ 1′ ( a) = e iq( a + b) ψ ′2 (−b) :     ik Ae ika − Be − ika = α Ce −αb − e αb e e

(

)

(

)

(2.86a)

iq ( a + b )

(2.86b).

Equations (2.84a), (2.84b), (2.86a), and (2.86b) are to be solved simultaneously to determine the energy eigenvalue E and the coefficients A,  B,  C , and D. These 4-equations are represented by the matrix equation:        

1 ik e ika ike ika

−1 −α

1 − ik e − ika − ike − ika

−1 α

− e −αb + iq( a + b)

− e αb + iq( a + b)

−αe −αb + iq( a + b)

αe αb + iq( a + b)

       

      

A B C D

      =      

0 0 0 0

    (2.87)   

A nontrivial solution of Eqn. (2.87) demands that the determinant of the 4 × 4 matrix must vanish. Hence:



1 ik e ika ike ika

−1 −α

1 − ik e − ika − ike − ika

−1 α

− e −αb + iq( a + b)

− e αb + iq( a + b)

−αe −αb + iq( a + b)

αe αb + iq( a + b)

= 0 (2.88)

Simplification of Eqn. (2.88) yields:

cos {q ( a + b )} = cos ( ka ) cosh ( αb ) +

1α k  −  sin ( ka ) sinh ( αb ) (2.89) 2  k α

Equation (2.89) is to be solved numerically or graphically to find energy eigenvalues for a 1D periodic solid. The left-hand side of the equation varies between −1 to +1. This suggests that the numerical values of the terms on right-hand side, which varies between −1 to +1, are acceptable (allowed) values. The values which are beyond the range of −1 to +1 are forbidden values. As is seen from Fig. 2.7, the plot of terms on the right-hand side: 1α k f ( y ) = cos ( ka ) cosh ( αb ) +  −  sin ( ka ) sinh ( αb ) as a function of y = E / V0 , is a continu2  k α ous curve having some portions of the curve (near the dip and peak) outside the range of −1 to +1.

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FIGURE 2.7 Graphical evaluation of energy gaps in the Kronig-Penney model.

Energies belonging to the portion of the curve outside the range of −1 to +1 are forbidden energy values. No solutions of the Schrödinger equation can be found for values of energy that lie beyond the range of −1 to +1. We thus find that the periodicity of potential along with the condition defined by Eqn. (2.81b) leads to the concept of energy gaps (forbidden energies). We deduce some important results from Eqn. (2.89): a. On choosing ka = nπ , we find that the right-hand side of the equation goes to ± cosh ( αb ) , since sin ( nπ ) = 0 and cos(nπ) = ±1 . The magnitude of cosh ( αb ) is always greater than 1 for states that are bound within the well, suggesting that there can be no solutions of Eqn. (2.89) for ka = nπ . This implies that no solutions 2  2 ( nπ ) , likewise for neighboring values of energy. The q − are possible for E = 2 2m a values that correspond to ka = nπ are forbidden values of the wave vector. b. For αb → ∞ , allowed energy levels from Eqn. (2.89) should reduce to those of the finite potential well. Equation (2.89) can be rewritten as: cos ( ka ) +



1α k  −  sin ( ka ) tanh ( αb ) = cos {q ( a + b )} / cosh ( αb ) (2.90) 2  k α

which for αb → ∞ ( tanh(αb) → 1 and cosh(αb) → ∞ ) reduces to: cos ( ka ) +



1α k  −  sin ( ka ) = 0 (2.91a) 2  k α

With the use of standard trigonometric identities, we find that Eqn. (2.45) reduces to Eqn. (2.91a). Equation (2.45) is expanded as:  ka   ka   ka   ka   ka   ka  k 2 sin   cos   − kα  cos 2   + kα  sin 2   − α 2 sin   cos   = 0  2  2  2  2  2  2 1 2 α − k 2  sin ( ka ) + kα cos ( ka ) = 0 2 1α k ⇒ cos ( ka ) +  −  sin ( ka ) = 0 2  k α ⇒

(

)

(2.91b)

Wave Mechanics and Its Simple Applications

45

2.5.2  Confined States in Quantum Wells, Wires, and Dots Quantum wells, wires, and dots offer the possibility of band structure engineering and wave function engineering, whereby composition and hence electronic and optoelectronic properties can be tailored to optimize for specific applications. Such band structure features are not possible in bulk semiconductors. The development of experimental techniques such as molecular beam epitaxy and metal-organic vapor phase epitaxy has made it possible to produce semiconductor quantum wells, wires, and dots structures. A layered structure of GaAs and AlGaAs, where few angstrom thick layers of GaAs and of AlGaAs are grown alternatively along one direction, say z-direction, exhibits a well–barrier structure in the conduction and valence band, as is seen in Fig. 2.8. Such a structure is periodic in the x − y plane and therefore Bloch’s theorem holds in the x − y plane. If we consider the states at the center of the 2D Brillouin zone ( k x = k y = 0 ), the band edge mismatch implies that electrons in the conduction band and holes in the valence band see a quantum well potential. The allowed energy levels for electrons can then be found by solving the 1D Schrödinger equation:

 2 d2   − 2 m* dz 2 + V ( z )  ψ ( z ) = Eψ ( z ) (2.92) e  

where me* is the electron effective mass ψ ( z ) is the electron envelope function, and V ( z ) is the band edge distribution potential. To calculate electron (hole) states in such a quantum well, we require that at the interface of the well and barrier layers, the wave functions of the well ψ w ( z) and of the barrier ψ b ( z) satisfy the conditions:

ψ w = ψ b (2.93a)

and

1 dψ w 1 dψ b = * (2.93b) me* dz mh dz

FIGURE 2.8 Simplified bottom of conduction band and top of valance band diagram for GaAs/AlGaAs heterostructure.

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A Textbook on Modern Quantum Mechanics

where, mh* is the effective mass of a hole. Equations (2.93) are the generalization of boundary conditions on wave function, to include the effect of mass change on crossing the boundary. Equation (2.93b) ensures the conservation of probability current density between different layers. Following the procedure laid down in Section 2.4.5, the confined energies are given by:  ka  k tan   = α   me* / mh* (2.94a)  2



for states corresponding to symmetric solutions, while for antisymmetric solutions, energies are given by:  ka  − k cot   = α   me* / mh* (2.94b)  2



2  me* E , and α 2 = 2  mh* ( 2 ∆Ec − E ) /  2 . The ∆Ec is the conduction 2 band energy offset (difference between the energies of the conduction band edges of AlGaAs and GaAS). The above analysis can easily be extended to structures of quantum wires and quantum dots. If we assume infinite confining potential (V = ∞ ) in the barrier for rectangular wells, wires, and dots, then the allowed energy states of confined electrons are given by: where a is well width, k 2 =



Er k x , k y =

2 r 2 2 + k x2 + k y2 ,   quantum well (2.95a) 8  me* L2z 2  me*



  Er , s ( k x ) =

 2  r 2 s2   2 k x2 + + ,   quantum wire (2.95b) 8  me*  L2z L2y  2  me*



Er , s ,t =

(

)

(

2 8  me*

)

 r 2 s2 t 2   L2 + L2 + L2  ,   quantum dot ( box ) (2.95c)  z y x 

where, Lx , Ly ,  and  Lz are confining lengths along x − ,  y − , and z − direction. r,s,t = 1, 2, 3…… are quantum numbers.

2.6  Solved Examples 1. For a particle in an infinite potential well constrained along the x − y axes, the 2 wave function is given by ψ ( x , y ) = sin( k x x)sin( k y y ) with k x = nx π / Lx and Lx Ly k y = ny π / Ly , where nx  and  ny take values 1,2,3…… Lx  and  Ly are confining lengths. Show that wave function is normalized.

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Wave Mechanics and Its Simple Applications

SOLUTION

The condition of normalization demands: Ly Lx

∫∫ψ (x.y)ψ(x, y)dxdy = 1 (2.96) *



0 0

We therefore have: Ly Lx

∫∫ψ (x.y)ψ(x, y)dxdy *

0 0

=



(

4 Lx Ly

 2 =  Lx

Lx

 1 =  Lx

Lx

Ly Lx

) ∫∫

∫ 0

(

( ))

dxdy sin ( k x x ) sin k y y

2

0 0

  2 Lx  sin ( k x x)dx       sin 2 ( k y y )dy    Ly  0 2

∫

   1 Ly {1 − cos ( 2 kx x )}  dx      L   1 − cos 2 ky y   dy    y 0  0

∫

∫{

(

)}

=1

2. A particle of mass m moves in a 1D infinite potential well of length L, having boundaries at x = 0 and x   =   L. V ( x )  =  0 for 0 < x < L , and V ( x ) = ∞, elsewhere. Show that 2 sin( kx) L with k = nπ / L , and energy eigenvalues are given by En =  2 n2 π 2 / 2 mL2 ; quantum number n takes the values n = 1, 2, 3……. . a. Calculate the probability to find the particle in the range of 0 < x ≤ L / 4 , when the particle is in the nth state. Show that probability depends on n. b. For what value of n, does the probability have the maximum value? c. Assume that ψ is a superposition of two eigenstates ψ = aψ n + bψ m at time t   =  0 . What is ψ at time t and what energy expectation value does ψ have at time t, and how does this relate to its value at t   =  0 ? d. What are the probabilities of measuring En and Em ? the normalized wave functions of the Hamiltonian are given by ψ ( x) =

SOLUTION

In this case: d2 ψ ( x ) + k 2 ψ ( x ) = 0 (2.97a) dx 2

which has the solution:

ψ ( x ) = Ae ikx + Be − ikx (2.97b)

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where k 2 = 2 mE /  2 . Conditions given by the Eqns. (2.21) and (2.22) are now modified to

ψ ( 0 ) = 0  :

A + B = 0 (2.98a)



ψ ( L ) = 0 :

Ae ikL + Be − ikL = 0 (2.98b)

Condition in Eqn. (2.98a) suggests that B = − A and therefore Eqn. (2.98b) yields: A  sin ( kL ) = 0 (2.98c)



For A ≠ 0, condition in Eqn. (2.98c) can only be satisfied for kL = nπ, where n = 1, 2, 3, 4……. Therefore, En =  2 n2 π 2 / 2 mL2 . Equation (2.97b) reduces to ψ ( x ) = 2 iA   sin ( kx ) . Normalization of wave function requires that:

∫

L

0



ψ ( x ) dx = 1 2

⇒ −4 A

∫

2

L

sin2 ( kx ) dx = 1 (2.99)

0

⇒ A = −i

1 2L

Therefore: 2  nπx   sin  (2.100)  L  L

 ψ n ( x) =



a. Probability density is defined by P ( x ) = ψ . Therefore, the probability to find a particle in the range of 0 < x ≤ L / 4 , when it is in the nth state is: 2

∫

L 4

0

ψ n ( x)

2

2 dx = L

∫

1 = L

L 4

0

∫

sin2 ( kx ) dx

L 4

0

1 − cos ( 2 kx )  dx

=

1 1  kL  sin   −  2 4 2 kL

=

1 1  nπ  − sin   ,  2  4 2 nπ

we thus find that probability depends on n. Its value will be ¼ for even values of n. 1  1 For odd values of n, probability will be given by  − for n = 1, 5, 9, 13… ,  4 2nπ  1  1 while it will be given by  + , for n = 3, 7, 11, 15…….  4 2nπ 

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Wave Mechanics and Its Simple Applications

b. The probability to find a particle in the range of 0 < x ≤ L / 4 will have maxi1 1 mum value for n = 3 . The maximum value is + ≈ 0.303 . 4 6π c. We are given: ψ ( x , 0 ) = aψ n ( x ) + bψ m ( x ) . The time-dependent wave function is:



ψ ( x, t ) = ψ ( x, 0) e

−

iHt 

= aψ n ( x ) e

−

iEnt 

+ bψ m ( x ) e

−

iEmt 

(2.101)

The expectation value of the Hamiltonian H at time t, can be given by: L

H=



∫ ψ ( x, t ) Hψ ( x, t ) dx (2.102) *

0

It is to be noted here that ψ n and ψ m are eigenstates of the Schrödinger equation. Hence, Hψ n = En ψ n  and  Hψ m = Em ψ m . Also, these are orthogonal wave functions: L

∫ψ



* m

( x)ψ n ( x)dx = δ mn (2.103)

0

Therefore: H=

=

=

∫

L

∫

L

∫

L

0

0

0 2

*

iE t iE t iE t iE t   − n − m  − n − m   + bψ e   + bψ e  a ψ e H a ψ e n m n m     dx     *

iE t iE t iE t iE t  − n − m   − n − m   + bψ e   + bHψ e  a ψ e aH ψ e n m n m     dx    

(2.104)

*

iE t iE t iE t iE t  − n − m   − n − m   + bψ e   + bE ψ e  a ψ e aE ψ e m m m  n   n n  dx    

= a En

∫

L

0

+ a* bEm

2

ψ *n (x)ψ n (x)dx + b Em

∫

L

0

ψ *n (x)ψ m (x)e

−

∫

i ( Em − En )t 

L

0

ψ *m (x)ψ m (x)dx

dx + ab * En

∫

L

0

ψ *m (x)ψ n (x)e

−

i ( En − Em )t 

dx

which simplifies to:

2

2

H = a En + b Em (2.105) The value of H at t = 0 is going to be same as that given by Eqn. (2.105), as the last two terms in Eqn. (2.104), which have time-dependent terms, do not contribute. Hence, the expectation value of the Hamiltonian is independent of ∂ time or H = 0. ∂t 2 2 d. The probabilities of measuring En and Em are; a and b .

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A Textbook on Modern Quantum Mechanics

3. For infinite well potential, define the above in problem-2, calculate x ,  x 2 ,  p ,

(

and p 2 , for the lowest quantum state. Use the definition ∆A   =   A   2   −   A

)

2 1/2

to define the uncertainty ∆A. Calculate ∆x and ∆p and verify the Heisenberg uncertainty relation that is ∆x∆p ≥  / 2. SOLUTION

The wave function for the lowest quantum state is given by ψ ( x ) = with k = π / L. Notice that ψ ( x )* = ψ ( x ) ; hence: L

x =

∫ 0



2 ψ ( x ) xψ ( x ) dx = L *

L

∫ x sin ( kx) dx 2

0

L

=

2  sin ( kx ) , L

(2.106a)

1 L x ( 1 − cos(2 kx)) dx = L 2

∫ 0

L

x

2

=

2 ψ ( x ) x ψ ( x ) dx =   L

∫

*

0

2

L

∫ x sin ( kx) dx 2

2

0

L

=



1 2 x ( 1 − cos(2 kx)) dx L

∫

(2.106b)

0

L2 L2 = − 2 3 2π Therefore:

(

)

2 1/2

∆x = x 2   −  x

1/2

 L2 L2 L2  = − 2 −  4  3 2π

1/2

 1 1  = L − 2   12 2 π  p =



∫

L 0

(2.107)

ψ ( x ) pψ ( x ) dx

= −i  

*

2 L

∫

L 0

sin ( kx )

d  (sin ( kx)) dx (2.108) dx

=0 p2 =



∫

L 0

ψ ( x ) p 2 ψ ( x ) dx *

∫ ∫

2 L d2   sin ( kx ) 2 {sin ( kx )} dx L 0 dx (2.109) 2 2 L L 2 k  = 2 k 2 sin2 ( kx ) dx = (1 − cos (2 kx)) dx 0 L 0 L =  2 k 2 =  2 π 2 / L2 = −   2

∫

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Wave Mechanics and Its Simple Applications

Hence: ∆p = =

(

p2  −   p

)

2 1/2

  2 π2 =  2 − 0   L

1/2

π L

Therefore: 1  π  1 ∆x∆p = L  −  12 2 π 2  L



= 0.58 >

 2

4. An electron is confined in the ground state of an infinite potential well of width 10−10 cm. Its energy in ground state is 40 eV. Calculate (a) the energies of the electron in its first, second, and third excited states; (b) the average force on the walls of the box when the electron is in the ground state. SOLUTION

a. From problem-2 above, we have En =  2 n2 π 2 / 2 mL2 . L = 10−10 m. For ground state (n = 1) and E1 =  2 π 2 / 2 mL2 = 40 eV . Therefore, the energy of the nth state in terms of the energy of the ground state is En = 40n2  eV . Then the energies of the first, second, and third excited states are E2 = 160 eV , E3 = 360 eV , and E4 = 640 eV . ∂H . Also, b. The average force on the walls of the well is given by F = −   ∂L Hψ = E ψ , which can be rewritten as: n

n

n

( H − En ) ψn = 0 (2.110)



On differentiating Eqn. (2.110) with L, we obtain: ∂ψ n  ∂ H ∂En  − = 0 (2.111)   ψ n + ( H − En ) ∂L ∂L  ∂L

Hence:

ψ n* ( En − H )



∂ψ n  ∂ H ∂En  = ψ n*  −  ψ n (2.112)  ∂L ∂L ∂L 

Since the Hamiltonian H is real and Hermitian, we can write:

∫ψ

n

*

( En − H )

∂ψ n dx = ∂L

∫

∂ψ n ( En − H )* ψ n* dx, ∂L

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A Textbook on Modern Quantum Mechanics

which is zero, since (( H − En ) ψ n ) = 0 . Therefore, integration of the right-hand side of Eqn. (2.112) gives: *

∫ψ



n

*

∂H ∂En  ∂ H ∂En  − − = 0 (2.113)  ψ n dx =  ∂L ∂L  ∂L ∂L

Hence,

−

∂En =  2 n2 π 2 / mL3 . ∂L

 2 π 2 2E1 , For the ground state (n = 1), force on the walls of the well is: F = = mL3 L which gives: F=



2 × 40 × 1.602 × 1019 = 1.28 × 10−7 Newton. 1010

5. Given an electron in a 1D potential well of width 0.8 nm and depth 6 eV, find the number of bound states present in the well. SOLUTION

a mV0 / 2 deter mines the number of the bound state in a finite potential well. In this case, we have:

As discussed in Section 2.4.3, the value of the parameter ρ =

a 8 × 10−10 mV0 / 2 =  1.0546 × 10−34 = 5.02

ρ=



( 9.11 × 10

−31

× 6 × 1.602 × 10−19

)

2

3π < ρ < 2 π , the number of bound states will be 5, 3 corresponding to 2 symmetric solutions and 2 corresponding to antisymmetric solutions. 6. A beam of 14 eV electrons is incident on a potential barrier of height 35 eV and width 0.06 nm. Calculate the transmittance with the use of both exact and approximate formulae and estimate the error in approximation. Since,

SOLUTION

We are given E = 14 × 1.602 × 10−19 Joules, V0 = 35 × 1.602 × 10−19 Joules and a = 6 × 10−11 m. Therefore:

ka = a 2 mE /  2 = 1.21,



αa = a 2 m (V0 − E ) /  2 = 1.48.

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Wave Mechanics and Its Simple Applications

Transmittance with the use of Eqn. (2.77b) is:

T=



T=

(k

2

+ α2

)

2(2 kα)2 2

(

cosh ( 2αa ) − k 4 + α 4 − 6 k 2 α 2

)

25.656 = 0.219. 13.355 × 9.675 − 12.3

With the use of Eqn. (2.78):

( 4k α ) T≈ (k + α ) 2



2

2 2 2 2

e −2 αa =

51.13 × 0.0518 = 0.196. 13.555

Error in approximation is around 10.5%. 7. Consider a stream of particles of mass m , each moving in the positive x − direction with kinetic energy E toward a potential jump located at x   =  0 . The potential is zero for x  0 . What fraction of the particles is reflected at x   =  0? SOLUTION

The Schrödinger equation for x ≤ 0 is: d2 ψ 1 + k12 ψ 1 = 0 (2.114a) dx 2



where k12 = 2 mE /  2 . For x   >  0 , we have: d2 ψ 2 + k22 ψ 2 = 0 (2.114b) dx 2



3E   2m  E −  2  4  mE k1 with k22 = = = . Hence, k2 = k1 / 2 . The solution of Eqn. (2.114) 2 22 4 is given by: ψ 1 ( x) = Ae ik1x + Be − ik1x (2.115a)



Since, there is no reflected wave in the region of x > 0 , we have: ψ 2 ( x) = Ce ik2 x (2.115b)



Application of boundary conditions at x = 0 gives:

A + B = C (2.116a)



ik1 ( A − B ) = ik2 C (2.116b) B 2 = 1/ 3 . Hence R = B / A = 1/ 9 . We thus find that one-ninth of A the particles are reflected at x   = 0. which give r =

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A Textbook on Modern Quantum Mechanics

8. Nearly Free electron model: With reference to the Kronig-Penney model described above, consider a periodic potential V ( x ) defined by:   V0 V ( x) =    0 



b b a

    

Show that bound state energies (E < V0 ) are given by the equation:  2 mE   E  tan    a = −  .  V0 − E    



SOLUTION

We are given:

d 2 ψ 1 2 mE + 2 ψ 1 = 0 for 0 ≤ x ≤ a (2.128a) dx 2 



d 2 ψ 2 2 m (V0 − E ) − ψ 2 = 0 for  x > a (2.128b) dx 2 2 The requirements that ψ 1 = 0 at x = 0 and ψ 2 → 0 for x → ∞ suggest that we can

2 m (V0 − E ) 2 mE take: ψ 1 = Asin( kx) and ψ 2 = Be −αx , with k 2 = 2 and α 2 = . Applying  2 boundary conditions at x = a , we get:

ψ 1 ( a) = ψ 2 ( a)  ⇒ Asin( ka) = Be −αa (2.129a)



ψ 1′ ( a ) = ψ ′2 ( a )  ⇒ Ak   cos( ka) = − Bαe −αa (2.129b) On dividing Eqn. (2.129a) by Eqn. (2.129b) we obtain:



tan ( ka ) = −

 2 mE   E  k   ⇒ tan    a = −  .  V0 − E  α   

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Wave Mechanics and Its Simple Applications

2.7 Exercises 1. A particle of mass 4.5 × 10−31 kg is confined to a one dimensional infinite square potential well of width 5 nm. It makes the transition from first excited state to the ground state. Calculate the wavelength of the emitted photon. 2. An electron confined to a one dimensional infinite square potential well makes the transition from n = 3 to n = 1 energy level. The frequency of the emitted photon is 2.5 × 1014 Hz . Find the width of the well. 3. A particle is in a state described by the wave function ψ (r , 0) = Ne −αr , where α is a real constant. (a) Calculate the normalization factor N. (b) Calculate ψ(r , t) and 2 show that the probability density ψ (r , t) is isotropic, independent of θ and φ . 4. Different states of a particle confined to a 1D infinite square potential well of width L 2  nπx  are given by: ψ n ( x) = sin  . Take L = 2 nm and then calculate the probability  L  L of finding the particle in between 0.2 nm to 0.4 nm, in the ground state (n = 1). 5. Show that ψ n ( x) belong to different levels (different values of n) in the above L

∫

exercise-4 are orthonormal ψ *m ( x)ψ n ( x)dx = δ mn . 0

6. A free particle is represented by ψ (r , t) = e i( k .r −ωt ) . Show its probability current S is dω equal to its group velocity v g = . dk e ikr 7. Calculate the probability current S when ψ (r ) = and compare your answer with r that obtained in exercise 6. 8. Show that for the 1D bound particle case 9. If ψ ( x) = Ae − x

2

/a2

d dt

∞

∫ ψ ( x , t)

2

dx = 0 .

−∞

is the wave function for the 1D Schrödinger equation,

 2 d2   − 2 m dx 2 + V ( x)  ψ ( x) = Eψ ( x) , what should be the values of V ( x) and E ?   10. Consider a particle whose normalized wave function is:

 2β 3/2 xe −βx for  x ≥ 0 ψ ( x) =   0 for x < 0 (a) Find for what value of x probability density is maximum. (b) Calculate expectation values x , x 2 , p and p 2 and show that they satisfy the Heisenberg uncertainty relation.  2  d 2 4x 2  11. The Hamiltonian and wave function of a particle are given by H = − − 4  2 m  dx 2 a  x − x 2 /a2 and ψ ( x) = e . Find the energy of the particle. a

3 Matrix Formulation of Quantum Mechanics In 1925, matrix mechanics was developed by Heisenberg while wave mechanics and the non-relativistic Schrödinger equation were invented by Schrödinger. While Schrödinger’s approach requires integral and differential calculus, matrices and vectors are the basis of Heisenberg’s representation of quantum mechanics. It was subsequently shown that the two approaches were equivalent. One of the most fundamental concepts in quantum theory lies in the idea xp   ≠   px, which is not followed by pure numbers but by operators and matrices. Here, x and p are the 1D position and the momentum. Our understanding of physical properties being represented by numbers is altered on being represented by operators and matrices. We know that matrices have the same non-commutativity as operators and they too can demonstrate the mathematical formalism of quantum theory. In fact, Heisenberg’s original approach to quantum theory, which is called matrix mechanics, was to find matrices representing x and p, such that xp − px = i. In the Schrödinger representation, wave function is time dependent and it evolves in accordance to the Schrödinger equation:



 2 2  ∂ψ (r , t)  − 2 m ∇ + V (r )  ψ (r , t) = i ∂t (3.1a)  

and an observable (operator) is treated independent of time. However, in the Heisenberg representation of quantum mechanics the vector representing a quantum state does not change with time, while an observable A satisfies:



dA 1  ∂A  = [ A, H ] +  (3.1b)  ∂t  classical dt i

where the second term on the right-hand side will be zero if A does not have explicit time dependence. The Heisenberg Eqn. (3.1b) becomes an equation in Hamiltonian mechanics (similar to classical physics) on replacing the commutator by the Poisson bracket. Also, the Lorentz invariance is manifested in the Heisenberg picture. Therefore, it can be inferred that the Heisenberg picture is more natural and fundamental than the Schrödinger picture.

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A Textbook on Modern Quantum Mechanics

3.1  Matrices and their Basic Algebra A two dimensional array of numbers (real or complex) or of functions is called the matrix:     A=    

a11 a21 a31 .... .... am 1

a12 a22 a32 .... .... am 2

.... .... ..... .... .... ....

.... .... .... .... .... ....

a1n a2 n a3 n .... .... amn

 f11 ( x)     f21 ( x)   ....   or F( x) =   ....   ....     fm1 ( x) 

f12 ( x)

....

....

f22 ( x)

....

....

.... .... .... f m 2 ( x)

..... .... .... ....

.... .... .... ....

f 1n ( x )   f 2 n ( x)  ....  (3.2) ....  ....   fmn ( x)  

where, the horizontal and vertical lines are called rows and columns, respectively, of the matrix. The dimensions of a matrix are defined with the number of rows first and then the number of columns. An entry of a matrix is represented by aij or fij ( x) with 1 ≤ i ≤ m and 1 ≤ j ≤ n. There is no universal convention to start indices i and j. Some of the programming languages start at zero. The elements, aii  or  fii are called diagonal elements, and the line on which they appear is termed the principal diagonal. A matrix with one row and n columns is called a row matrix, and the matrix of one column and m rows is called a column matrix. A matrix that consists of an equal number of rows and columns is termed a square matrix. A matrix whose each and every elements set is equal to zero is called a null matrix or zero matrix that satisfies the property 0 + A = A + 0 = A ; A − A = 0. A matrix that has all elements on the main diagonal set to 1 and all other elements set to 0 is known as a unit matrix. It satisfies MI n = M for any m × n matrix M; then I n is a unit matrix of order n. A matrix where elements on the main diagonal set equal to a constant value and all other elements are set to 0 is called a constant matrix. The algebra of matrices and operators differs from that of numbers. Matrices of the same order can only be added or subtracted. For given m × n matrices A and B, the sum A + B is a m × n matrix computed by adding corresponding elements ( A ± B )ij = aij ± bij . A scalar multiplication cA is computed by multiplying each matrix element by scalar c: c( cA )ij = caij. Two matrices can be multiplied if and only if the number of columns of the left matrix is the same as the number of rows of the right matrix. If A is an m × n matrix and then B is has to be an n × p matrix, to have the matrix product AB of order m ×  p:

( AB)ij =

∑a b

ik kj

= ai 1b1 j + ai 2 b2 j + ...... + ainbnj (3.3a)

k

for each pair (ij). Two matrices A and B are said to commute if AB = BA, for well-defined products AB and BA. However, commutativity does not generally hold, which means AB  ≠   BA. The transpose of a matrix A of order m × n is a matrix of order n × m denoted as Atr or AT  and it is formed by turning rows into columns and columns into rows: A  ij = A ji for all or A, T T T T T T indices i and j. Transposition follows: ( A + B ) = A + B and ( AB) = B A . A matrix formed by replacing each element by its complex conjugate is called a Complex Conjugation Matrix, represented by A*. The complex conjugation of the product: ( AB)* = B* A*. A matrix formed by performing transposition and complex conjugation operations, one

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after other, is called a Hermitian Conjugation Matrix or conjugate adjoint matrix, and it is represented by A †. It holds that ( A + B)† = A † + B†  and ( AB)† = B† A †. The sum of diagonal elements is termed as the trace of the matrix: tr( A) =



∑a (3.3b) ii

i

The trace holds the rules: (a) tr( kA) = k  tr( A) , where k is a real number, (b)  tr ( A + B ) = tr ( A ) + tr ( B ) , and tr ( AB ) = tr(BA) . Every square matrix reduces to single number or expression of scalars, which is called the determinant of the matrix. It is defined as follows:



A=

a11 a21 a31 .... .... an 1

a12 a22 a32 .... .... an 2

.... .... ..... .... .... ....

.... .... .... .... .... ....

a1n a2 n a3 n .... .... ann

=

∑ (−1) a σ

a a ....anin (3.4)

1i1 2 i2 3 i3

where the summation goes over all the permutations i1 , i2 , i3 ....in of indices 1, 2, 3,...., n and σ is the number of inversions in the permutation i1 , i2 , i3 ....in of the row indices. Some notable properties of the determinant are: (i) The determinants of a matrix and of its transpose are equal, i.e. A = AT . (ii) Multiplying all the elements of a row (column) of the determinant by a number is equivalent to the determinant multiplied by the same number. (iii) On interchanging two rows (columns) of the determinant, the determinant changes its sign. (iv) If two rows (columns) of the determinant are identical, then the determinant is equal to zero. (v) If each element in the row (column) of the determinant is a sum of two numbers, then the determinant expands into the sum of two determinants. (vi) The determinant will not change if a scalar multiple to a row (column) is added to another row (column). (vii) For arbitrary matrices A and B of order n it holds that AB = A B . The determinant theorem of expansion by cofactors is given as:

ai 1 Ak 1 + ai 2 Ak 2 + .... + ain Akn = A δ ik (3.5)



a1i A1k + a2 i A2 k + .... + ani Ank = A δ ik (3.6)

where the Kronecker delta δ ik is defined as:

 1 δ ik =   0

i=k (3.7) i≠k

The symbol Aik represent the product of the number (−1)i+k and the determinant of the (n − 1) × (n − 1) matrix obtained by deleting the i-th row and k-th column of the given matrix.

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3.2  Bra and Ket Notations P. A. M. Dirac created a powerful and concise formalism of quantum mechanics, which is referred to as the Dirac notation or bra-ket (bracket) notation. Two distinctly different computational approaches, wave mechanics and matrix mechanics, to quantum theory are equivalent, each with its strengths in certain applications. The quantum mechanical calculation can be set in Dirac’s notation in the first step. Then, depending on which method is most expedient computationally, one chooses one of the matrix mechanics and wave mechanics. Quantum states are preferred to describe the bra-ket notations. They are also used to denote abstract vectors and linear functionals in mathematics. The symbol is called ket and the conjugate adjoint of it is called bra, and it is denoted by . Bra and ket are derived from the word bracket.

3.3  Vectors and Vector Space A vector is characterized by a length and a direction, and a vector space is a set of vectors closed under addition and multiplication. Vectors are added and multiplied together, and are multiplied by constants, including complex numbers. A vector is represented by any of  the symbols A, A , A and A . A linear vector space is defined as an abstract set of vectors, which satisfy the following algebraic axioms: 1. For every pair of vectors A and B , there exists a unique vector C = A + B , and the vector addition is commutative as well as associative, that is A + B = B + A and ( A + B ) + C = A + ( B + C ). 2. The origin of vector space is zero or the null vector 0 , which is defined by A + 0 = A and associated with A ; there exists - A so that A − A = 0 . 3. There exists a vector called unit vector defined by 1 , which is of unit length. 4. Elements of vector space can be multiplied by real or complex numbers, and they follow the rule:   a ( A + B ) = a A + a B ; ( a + b ) A = a A + b A , where a and b are real or complex numbers. 5. Multiplication by numbers 0 and 1 is defined as: 0 A = 0 and 1 A = A . 3.3.1  Linearly Independent Vectors N-vectors forming a set satisfy:

{ xi } are said to be linearly independent of each other if they



c1 x1 + c2 x2 + c3 x3 + ......... + cN xN = 0 (3.8)

if and only if c1 = c2 = c3 ......... = c N = 0. And, if

c1 x1 + c2 x2 + c3 x3 + ......... + cN xN = 0 (3.9)

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for nonzero values of c1 , c2 , c3 ,......c N, vectors of set { xi } are called linearly dependent. Every linearly dependent set contains at least one vector that is a linear combination of others: xi = (c1 x1 + c2 x2 + c3 x3 + ........ + c N xN )/ ci (3.10)

for ci ≠ 0.

3.3.2  Orthogonal and Orthonormal Vectors For any two vectors there exists an associated complex number called the inner or scalar product X Y . The positive square root of the inner product of a vector with itself is called the norm of the vector: X = X X . Two vectors are called orthogonal vectors if their inner product is zero. X Y = 0 (3.11)



{ }

There exists a set y i of orthogonal vectors if y i y j = 0 for i ≠ j, for 1 ≤ (i, j) ≤ N . On replacing xi by y i for 1 ≤ i ≤ N in Eqn. (3.8) and then taking inner products with y1 , y 2 , y 3 …… y N ; with imposing the condition of orthogonality we find that Eqn. (3.8) can be satisfied if and only if c1 = c2 = c3 ......... = c N = 0. Thus, orthogonal vectors are linearly independent vectors, too. However, the converse is not true; linearly independent vectors need not necessarily be orthogonal. The N-vectors belonging to the set { ei }, 1 ≤ i ≤ N , are said to be orthonormal to each other if they satisfy: ei e j = δ ij. The norm of each is unity. As is obvious orthonormal vectors are orthogonal, too. On replacing xi by ei for 1 ≤ i ≤ N in Eqn. (3.8) and then taking inner products with e1 , e2 , e3 …… eN and then applying ei e j = δ ij, we find that Eqn. (3.8) can be satisfied if and only if c1 = c2 = c3 ......... = c N = 0, suggesting that orthonormal vectors are also linearly independent. Linearly independent, orthogonal, and orthonormal vectors have the property that every vector that belongs to a linear vector space can be expressed as (i) a linear combination of linearly independent vectors, (ii) a linear combination of orthogonal vectors, and (iii) a linear combination of orthonormal vectors. The set of linearly independent vectors { xi } or orthogonal vectors y i or orthonormal vectors { ei } can be used as a coordinate system or basis of vector space. The representation of any vector in terms of vec-

{ }

{ }

tors of set { xi } or y i , or { ei

} is unique. Suppose

N

A =

N

∑

ai ei and also A =

i=1

∑a′ e i

i

.

i=1

N

Then 0 =

∑(a − a′) e i

i

i

, which can be true if and only if ai = ai′. The numbers ai, which

i=1

are uniquely defined, can be real or complex. They are called components or expansion coefficients of A relative to the basis vectors { ei }. The number of vectors N in a set { ei } can be finite or infinite. Accordingly, vector space is called finite dimensional or infi∞

nite-dimensional space. For an infinite-dimensional space: A =

∑a

i

ei . The ai = ei A ,

i=1

whether A belongs to finite N -dimensional or infinite-dimensional vector space. Every vector, including a basis vector, has components equal to the dimensionality of the vector space. For example:

ei = 0 e1 + 0 e2 + 0 e3 + ....... + 1 ei + ... + 0 eN (3.12)

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3.3.3  Abstract Representation of a Vector In an N -dimensional space, there is one-to-one correspondence between the vectors in the set { ei } and the expansion coefficients ( a1 , a2 , a3 ......aN ). Also, the linear combination and inner product of two vectors A and B involve only the expansion coefficients with respect to basis { ei }:

A + B = ( a1 + b1 ) e1 + ( a2 + b2 ) e2 + ........ + ( aN + bN ) e N (3.13)



A B = ( a1*b1 + a2* b2 + a3* b3 + ....... + aN* bN ) (3.14)

Every relation among vectors has its exact copy in relation among representative expansion coefficients of vectors with respect to basis { ei }. Because of this it is possible to develop an entire theory of vectors in terms of representative expansion coefficients, without making further reference to basis vectors. Thus, intrinsic properties of a vector space, which are independent of basis, are physically or mathematically important. Physically significant properties of a vector space in quantum mechanics are independent of the choice of coordinate system or basis vectors. Equation (3.14) can equivalently be written in terms of row and column matrices as follows: A B = ( a1*b1 + a2* b2 + a3* b3 + ....... + aN* bN )



= [ a1*

a2*

   aN* ]     

a3* ....

b1 b2 b3  bN

       

(3.15)

= A† B 3.3.4  Outer Product of Vectors The product of bra and ket vectors is called the inner product, while the product of ket and bra vectors A B is termed the outer product. The inner product is a scaler quantity, while the outer product is a matrix as is shown below:



   A B =        =    

a1 a2 a3  aN

     b1*    

b2*

b3*

.....

a1b1*

a1b2*

a1b3*



a2 b1*

a2 b2*

a2 b3*



* 3 1

* 3 2

* 3 3

  

ab  an b1*

ab  an b2*

ab  an b3*

bn* 

a1bn*   a2 bn*   a3 bn*     an bn* 

(3.16)

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Since, each column of A B represents a vector, the outer product is an analogue of the vector product of two vectors. The orthonormal vectors, which can be chosen as basis vectors for an N-dimensional vector space, are:



   e1 =    

1 0 0 . . 0

  0   0   0           1   0   0   , e =  0  , e =  1 ………. e =  0  2 3 N   .   .   .  (3.17)          .   .   .    0   0   1 

Depending upon the requirements for the solution of a problem, we can choose any of the sets of linearly independent, orthogonal, or orthonormal vectors as the basis vectors. A set of linearly independent vectors can be converted to a set of orthogonal vectors, which then can be converted to a set of orthonormal vectors.

3.4  Gram-Schmidt Method for Orthogonalization of Vectors A set of orthogonal vectors is obtainable from a set of linearly independent vectors. The Gram-Schmidt method is a procedure that takes a set { U i } of linearly independent vectors and constructs a set { Vi } of orthogonal vectors. We take V1 = U 1 and then express the next orthogonal vector V2 as a leaner combination of U 2 and V1 :

V2 = U 2 + a21 V1 (3.18)

On taking the inner product of V2 with V1 , we get:

V1 V2 = V1 U 2 + a21 V1 V1 (3.19)

which gives:

2

a21 = − V1 U 2 / V1 (3.20)

Similarly define:

V3 = U 3 + a32 V2 + a31 V1 (3.21)

Then take inner products of V3 with V2 and V1 to obtain: 2



a32 = − V2 U 3 / V2 (3.22)



a31 = − V1 U 3 / V1 (3.23)

2

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By looking at Eqns. (3.20), (3.22) and (3.23), we can generalize to:

Vi = U i + ai , i − 1 Vi − 1 + ai , i − 2 Vi − 2 + ...... + ai 1 V1 (3.24)

with: 2

aij = − Vj U i / Vj (3.25)



where 2 ≤ i ≤ N and 1 ≤ j ≤ N − 1. A set of orthonormal vectors { Wi } can then be formed from orthogonal vectors by dividing each orthogonal vector by its norm: Wi = Vi / Vi .

3.5  Schwarz Inequality The Schwarz inequality also known as the Cauchy-Schwarz inequality is a useful inequality. The norm of the scalar product of two vectors is X Y cos θ ≤ X Y , because cos θ ≤ 1, the generalization of which leads to the Schwarz inequality:

XY

2

≤ XX YY

X Y ≤ X Y (3.26)

or

The two sides are equal if and only if X and Y are linearly independent or in the geometrical sense they are parallel. To prove the Schwarz inequality, let us define Z = X − λY , where λ can be taken as a complex number. We then have: 2



0 ≤ Z = X − λY X − λY ⇒ 0 ≤ X X − λ X Y − λ* Y X + λ

2

YY

(3.27)

Taking partial differentiation with λ * , we get:

2

λ= Y X /Y ,

2

λ * = X Y / Y (3.28)

and

Substitution of values of λ and λ * yields:

2

2

0≤ X − X Y

2

/ Y (3.29)

which is true if:

XY

2

2

2

≤ X Y (3.30)

or equivalently:

X Y ≤ X Y (3.31)

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For the case of the Euclidean space, Eqn. (3.30) is:   



2

  ( xi y i ) ≤    i=1 n

∑

 xi2    i=1 n

∑

 y i2  (3.32)  i=1 n

∑

In terms of square integrable complex-valued functions, the Schwarz inequality is expressed as:



∫

2

f ( x) g( x) dx ≤  





∫ f (x) dx  ∫ g(x) dx (3.33) 2

2

3.6  Linear Transformation of Vectors For an m × n matrix A and a vector X having n-components, there exists a uniquely defined vector Z = A X that has m-components. The transforming matrix has the number of columns equal to the number of components in X and the number of rows equal to the number of components in vector Z . The transformation is linear if: A( a X + b Y ) = aA X + bA Y for the complex numbers a and b. Transformation with use of the unit matrix leaves a vector unchanged, while transformation by the null matrix yields a zero vector: I X = X and 0 X = 0. 3.6.1  Eigenvalues and Eigenvectors of a Matrix If for a square matrix A of order-n A X = λ X , where λ is a real or complex scalar quantity, then matrix A is said to satisfy an eigenvalue equation. The λ and X are called the eigenvalue and eigenvector of matrix A, respectively. The equation A X = λ X can also be written as:

( A − λI ) X = 0 (3.34)

I being the unit matrix of order n. Note that the null vector is an eigenvector for every square matrix. In the following, we want to find the non-trivial eigenvectors. The Eqn. (3.34) presents a system of homogenous linear algebraic equations, which has a non-trivial solution if the matrix ( A − λI ) of the system is singular.

( A − λI ) = 0 (3.35)

The Eqn. (3.35) is known as the characteristic equation of the matrix, and the polynomial:

ρ ( λ ) = ( A − λI ) (3.36)

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is called the characteristic polynomial. The Eqn. (3.35) is an algebraic equation of order n in terms of λ.



a11 − λ a21 a31 .... an 1

a12 a22 − λ a32 .... an 2

a13 a23 a33 − λ .... an 3

.... .... .... .... ....

a1n a2 n a3 n .... ann − λ

= 0 (3.37)

which can also be expressed as:

ρ ( λ ) = λ n + C1 λ n− 1 + C2 λ n− 2 + C3 λ n− 3 + ……….. + Cn (3.38)

The solution of Eqn. (3.38) yields n-values of λ, and hence matrix A has n-eigenvalues. The set of all eigenvalues, λ 1 , λ 2 , λ 3 .........λ n is called the spectrum of eigenvalues of the matrix. Thus, there are always n, not necessarily distinct, eigenvalues. Equal eigenvalues that correspond to multiple roots are called degenerate. Each of n-eigenvalues corresponds to an eigenvector: The different eigenvectors that belong to the same eigenvalue are called degenerate eigenvectors. Some important properties of eigenvalues and eigenvectors are as follows: 1. If λ 1 , λ 2 , λ 3 , λ 4 .....λ n are the eigenvalues of the matrix A of order n, then λ 1k , λ 2k ,.....λ nk too are the eigenvalues of the matrix A k . Proof: Multiply both sides of A X i = λ i X i by A k − 1 to get: Ak Xi = λ i Ak −1 Xi = λ i Ak −2 (A Xi )

= λ 2i A k − 2 X i = λ 3i A k − 3 X i

(3.39)

 = λ ik X i 2. The eigenvectors of the matrix A, which correspond to different eigenvalues, are linearly independent. Proof: Let X 1 , X 2 , X 3 ,....... X n be the eigenvectors of the matrix corresponding to the different eigenvalues λ 1 , λ 2 , λ 3 , λ 4 .....λ n . It is to be shown that the eigenvectors are linearly independent. Let us take:

X = c1 X 1 + c2 X 2 + c3 X 3 + ......... + c N X N (3.40) On operating each term of Eqn. (3.40) by matrix A, we obtain:



A X = A ( c1 X 1 + c2 X 2 + c3 X 3 + ......... + c N X N

)

= c1 λ 1 X 1 + c2 λ 2 X 2 + c3 λ 3 X 3 + ......... + c N λ N X N

(3.41)

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Multiply each term of (3.40) by λ 1 and subtract the resulting equation from Eqn. (3.41) to get:

A X − λ 1 X = c1 ( λ 1 − λ 1 ) X 1 + c 2 ( λ 2 − λ 1 ) X 2 + c 3 ( λ 3 − λ 1 ) X 3 + ...... + c j (λ j − λ 1 ) X j + ... + cN (λ N − λ 1 ) X N

(3.42)

Since λ 1 is one of the eigenvalues of matrix A, we should have:

A X − λ 1 X = 0 (3.43) λ 1 , λ 2 , λ 3 , λ 4 .....λ n are distinct eigenvalues. Hence to satisfy Eqn. (3.43), we must have c2 = c3 ......... = c N = 0. Similarly, on multiplying each term of Eqn. (3.40) by λ 2 and then subtracting from Eqn. (3.42), it can be proved that c1 = c3 ......... = c N = 0 , and so on. It therefore proves that eigenvectors X 1 , X 2 , X 3 ,....... X n are linearly independent. 3. If the eigenvalues of the matrices A and B are nondegenerate and the matrices A and B are commutative, then they have common eigenvectors. Proof: Take A X = λ X and multiply both sides from left with matrix B to obtain:



BA X = B(λ X ) = λ( BX ) (3.44) Since matrices A and B commute, BA = AB, therefore:



BA X = AB X = λ( BX ) (3.45) Hence, if X is an eigenvector of the matrix A corresponding to eigenvalue λ, then B X is also an eigenvector of the matrix A corresponding to the same eigenvalue, and then the vectors X and B X are collinear: B X = µ X . Therefore, the eigenvector X of A corresponding to the eigenvalue λ is also the eigenvector of matrix B corresponding to eigenvalue µ. In a similar manner, it can be shown that each eigenvector of the matrix B is an eigenvector of the matrix A.

3.6.2  Numerical Method to Find Eigenvalue and Eigenvector The methods of diagonalization of a matrix are used to find the eigenvalues of a matrix. A simple iterative method can be used to find the largest eigenvalue and its corresponding eigenvector of a matrix. Consider a matrix A of order-n and vector



   X =    

x1 x2 x3  xn

    (3.46a)    

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Start with x1 = 1, x2 = 0, x3 = 0,......., xn = 0 and calculate:    AX = Y =    



where, zi =

y1   1   y2   z2 y 3  = y 1  z3         zn yn  

    (3.46b)   

yi for 2 ≤ i ≤ n. Redefine X as: y1



 1   z2 X =  z3     zn

    (3.46c)   

and then calculate A X . Repeat the process until the value of X is the same, within the allowed error, in two successive iterations. At that stage y1 is the eigenvalue and X is the eigenvector of matrix A.

3.7  Inverse Matrix A n  ×   n matrix A is invertible if and only if there exists a matrix B such that:

AB  =   I n  (   =   BA ) . (3.47a)

The matrix B is known as the inverse matrix of A, commonly represented by A −1. Invertible matrices are precisely those matrices whose determinant is nonzero. The elements of A −1 α ij are given by ( A −1 )ij = , where α ij is termed the cofactor of element aij of matrix A, and it A is defined by: α ij = (−1)i + j × (determinant formed after removing ith row and jth column of A ). For two matrices C and D of the same order:

( CD)−1 = D−1C −1 (3.47b)

If X is an eigenvector of a matrix A corresponding to the eigenvalue λ, then the same eigenvector X is also the eigenvector of the inverse matrix A −1 corresponding to the eigenvalue 1/ λ . This can be seen on multiplying both the sides of the A X = λ X from the left by A −1; A −1 A X = λA −1 X ⇒ A −1 X = (1/ λ) X .

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An inverse matrix can also be obtained numerically. One of the numerical methods for matrix inversion is the Gauss-Jordan elimination method. To find the inverse matrix with the use of the Gauss-Jordan elimination method, one operates on the matrix:



 A 

   I  ≡    

a11 a21  an 1

a1n a2 n  ann

   

1 0  0

0 1  0

   

0 0  1

    (3.48)   

where I is the unit matrix. By applying the Gaussian elimination method, Eqn. (3.48) is transformed to:



 I 

   B  ≡    

1 0  0

0 1  0

   

b11 b21  bn1

0 0  1

   

b1n b2 n  bnn

    (3.49)   

The matrix:



 b11   b21     bn1

   

b1n b2 n  bnn

   (3.50)   

is then A −1. Reduction of matrix A to a diagonal matrix can be obtained by row-operation: a  aij ⇒ aij −  ik  akj where i, j and k vary between 1 and n. The corresponding operation  akk  a  applied to matrix I is: bij ⇒ δ ij −  ik  δ kj and then divide bij by aii to get the resultant matrix B.  akk  One finds that the procedure is numerically unstable unless pivoting (exchanging rows and columns as appropriate) is used. A good choice is picking the largest available element as the pivot.

3.8  Orthogonal Matrix A square matrix Q whose transposition and inverse are equal is known as an orthogonal matrix:

Q = Q −1

 = I (3.51) or QQ

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Linear transformation of a vector by an orthogonal matrix preserves the vector length: VQ QV = V V , where Q can represent a linear translation, rotations, reflections, and the combinations of rotation and reflection in a finite dimensional space. Orthogonal matrices are important for a number of theoretical and practical reasons, and these are widely used  1 0   1 0  in physics. For example,   is used for identity transformation, while    0 1   0 −1  gives reflection across the x-axis.  = Q Q = 1 ⇒ Q = ±1 The determinant of an orthogonal matrix is either +1 or −1: QQ because Q = Q . The eigenvectors of an orthogonal matrix are orthogonal and an eigenvalue is either +1 or −1.

3.9  Hermitian Matrix  = A or aij = a ji is known as a real symmetric matrix. While a matrix A matrix that satisfies; A † that obeys A = A or aij = a*ji is termed as a Hermitian or self-adjoint matrix. Real symmetric and Hermitian matrices are the commonest matrices in quantum mechanics as most of the Hamiltonians can be represented this way. The elements of a Hermitian matrix are symmetric about the principal diagonal. Hermitian matrices have an important application in quantum physics in the form of representations of measurement in Heisenberg’s quantum mechanics. There is a Hermitian matrix that corresponds to each observable parameter of a physical system. The eigenvalues of a matrix are the possible values that can result from a measurement of that parameter. The eigenvectors of a matrix are the corresponding states of the system for a measurement. Each measurement yields precisely one real value and leaves the system in only one of a set of mutually orthogonal states. Therefore, the measured values must correspond to the eigenvalues of the Hermitian matrix, and the resulting states should be the eigenstate. A matrix in which elements about the diagonal are such that aij = − a*ji or A † = − A is known as a skew Hermitian or anti-Hermitian matrix. For a Hermitian matrix, eigenvalues are real and the eigenvectors are orthogonal. If λ and X are the eigenvalue and eigenvector of a Hermitian matrix H , then we write:

H X = λ X (3.52)

Taking the Hermitian conjugate of both sides, then noting that multiplication by a scalar is commutative and H † = H, we get:

X H = λ * X (3.53)

Multiply Eqn. (3.52) from the right by X and Eqn. (3.53) from the left with X to get:

X A X = λ X X (3.54a)



X A X = λ * X X (3.54b)

(

)

The left-hand sides of Eqns. (3.54a) and (3.54b) are equal; therefore λ − λ * X X = 0. For a nonzero vector, X X ≠ 0, and hence λ = λ * which is possible if λ is real. However, its

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converse may not be true; a matrix with real eigenvalues is not necessarily Hermitian. For  a + iα b + iβ   has real eigenvalues but it is not a Hermitian matrix. example, matrix  d + iδ   c + iγ  Next consider the case where λ 1 and λ 2 are two distinct eigenvalues, which correspond to eigenvectors X 1  and  X 2 , respectively, of Hermitian matrix H :

A X 1 = λ 1 X 1 (3.55a)



A X 2 = λ 2 X 2 (3.55b)

Take the Hermitian conjugate of Eqn. (3.55a) and then multiply both sides by X 2 : X 1 A X 2 = λ 1 X 1 X 2 (3.56a)



where we used λ *1 = λ 1. Multiply both sides of Eqn. (3.54b) by X 1 to get: X 1 A X 2 = λ 2 X 1 X 2 (3.56b)



On subtracting Eqn. (3.56b) from Eqn. (3.56a), we obtain: X 1 X 2 (λ 1 − λ 2 ) = 0 (3.57)



Since eigenvalues are distinct ( λ 1 ≠ λ 2 ), X 1 X 2 must vanish implying that X 1 and X 2 are orthogonal.

3.10  Unitary Matrix A square matrix of order n whose product with its conjugate transpose is a unit matrix U †U = UU † = I n is known as a unitary matrix. Another definition of a unitary matrix is given as follows: a matrix whose columns are the eigenvectors of a Hermitian matrix is a unitary matrix. Also, a matrix whose conjugate transpose is equal to its inverse is a unitary matrix. There is no difference between a unitary matrix and an orthogonal matrix when elements are real. For a unitary matrix, the inner product X Y = X U †U Y = UX UY for all complex vectors X and Y . We thus find that the transformation of a vector by a unitary matrix does not change its norm. For a unitary matrix U of order n, the following holds: 1. U is unitary then U † is unitary. 2. If the columns of U form an orthogonal basis of a vector space, then the rows of U too form an orthonormal basis with respect to this inner product. 3. The vectors represented by the columns of U are orthonormal vectors. 4. Eigenvalues of U are complex numbers of absolute value 1. Let us say that U satisfies the eigenvalue equation:

U X = λ X (3.58)

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The conjugate transposition of it is: X U † = λ * X (3.59)

which gives:

X U †U X = λ * λ X X



⇒ XX =λ

2

(3.60)

XX

2

Therefore, λ = 1, which suggests that the eigenvalues of a unitary matrix are complex numbers of absolute value 1. 5. The determinant of a unitary matrix is ±1. If U = d then UU † = U U † = d 2 = 1, which gives d = ±1.

3.11  Diagonalization of a Matrix Diagonalization of a matrix plays an important role in the determination of eigenvalues and eigenvectors, which are key aspects of quantum mechanics. A matrix of eigenvalues is a diagonal matrix, and hence, finding the eigenvalues is equivalent to diagonalization of a matrix. A triangular matrix precedes to a diagonal matrix. When all the matrix elements below the main diagonal are zero, aij = 0 for i > j, the matrix is called an upper triangular matrix, while a lower triangular matrix is the other way around, aij = 0 for i < j. The eigenvalues of both upper triangular and lower triangular matrices are also the elements of the main diagonal. The characteristic equation of an upper triangular matrix A is:



a11 − λ 0  0

a12 a22 − λ  0

   

a1n a2 n  ann − λ

= 0 (3.61)

Expansion of the determinant along the columns gives:

( a11 − λ)( a22 − λ)( a33 − λ).......( ann − λ) = 0. (3.62)

A matrix with all off-diagonal elements equal to zero is called a diagonal matrix.



   D=   

d11 0  0

0 d22  0

…   

0 0 0 dnn

    (3.63)   

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To have a non-singular diagonal matrix all diagonal elements must be different from zero. The inverse of a diagonal matrix can be found by replacing all diagonal elements by their respective reciprocals and the multiplication of two diagonal matrices results in a diagonal matrix: D1 D2 = D2 D1. Important propositions related to diagonalization of a matrix are as follows: 1. If a matrix A has n linearly independent eigenvectors, X 1 , X 2 , X 3 ,....... X n corresponding to the eigenvalues λ 1 , λ 2 , λ 3 , λ 4 .....λ n then the matrix can be expressed in the form A = SΛS−1, where Λ is diagonal matrix having λ 1 , λ 2 , λ 3 , λ 4 .....λ n along the main diagonal and the matrix S is the matrix whose columns 1,2,3….n represent vectors X 1 , X 2 , X 3 ,....... X n , respectively:  λ1   0 Λ =    ...   0

0 λ2 ... 0

.... ... ... ...

0 0 ... λn

  x11    , and S =  x21   ...    xn1 

x12 x22 ... xn 2

... ... ... ...

x1n x2 n ... xnn

 x1i      ;  X =  x2 i i   ...     xni

   (3.64)   

To prove this, we multiply both sides of A = SΛS−1 by S from the right to obtain AS = SΛ. Then: AS = A  X 1 .... X i .... X n 

(3.65)

=  A X 1 .... A X i .... A X n  =  λ 1 X 1 ....λ 2 X i ....λ n X n  and



 λ1  0 SΛ =  X 1 .... X i .... X n    ...   0

0 λ2 ... 0

... ... ... ...

0 0 ... λn

    (3.66).  

=  λ 1 X 1 ....λ 2 X i ....λ n X n  Therefore, A = SΛS−1. This also proves that Λ = S−1 AS. Thus, a matrix A is diagonalizable if there exists a non-singular matrix P and a diagonal matrix Λ such that Λ = P −1 AP. Note that an n × n matrix may fail to be diagonalizable if (i) all roots of its characteristic equation are not real numbers, and (ii) it does not have n-linearly independent eigenvectors. 2. A Hermitian matrix H can be diagonalized using a unitary matrix U. Which means that in Λ = P −1 AP, if A is Hermitian then P is a unitary matrix. To prove this, let us take Λ = P −1 HP , where Λ is the diagonal matrix of the eigenvalues of

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H = H †, which are real. Therefore, Λ too is a Hermitian matrix: Λ = Λ †. On taking conjugate transposition of both sides we get: Λ † = ( P −1 HP)†

⇒ Λ = P † H ( P −1 )† (3.67) or   P −1 HP = P † H ( P −1 )† This implies P −1 = P † and ( P −1 )† = P, which is the property of a unitary matrix: U † = U −1 . Therefore, Λ = U † HU, where Λ is a diagonal matrix having eigenvalues of H along its principle diagonal. 3. Two commuting Hermitian matrices can be diagonalized by the same unitary transformation. Let us say that a unitary transformation brings two commuting matrices A and B to diagonal matrices Λ and Λ ′:



Λ = U † AU (3.68a)



Λ ′ = U † BU (3.68b) Then ΛΛ ′ − Λ ′Λ = U † ( AB − BA)U = 0, because of AB − BA = 0, a commuting property.

3.12  Cayley-Hamilton Theorem If A is a matrix of order n and ρ(λ) = A − λI then ρ ( A ) = 0, meaning that every matrix satisfies its own characteristic equation. Proof: Putting λ = A in Eqn. (3.38), we get:

ρ( A) = A n + c1 A n− 1 + c2 λA n− 2 + c3 A n− 3 + ...... + cn = 0 (3.69) Let B(λ) be the matrix of the cofactors of determinant A − λI :



 b11 (λ)   b21 (λ)  ... B(λ) =  ...   ...  b m 1 (λ ) 

b12 (λ)

...

...

b22 (λ) ... ... ... bm 2 (λ)

... ... ... ... ...

... ... ... ... ...

b1n (λ)   b2 n (λ)   ...  (3.70a) ...   ...  bnn (λ)  

where bij (λ) is the cofactor of the (i, j) element of matrix ( A − λI ). The bij (λ) is the polynomial of order (n − 1). Therefore:

B(λ) = B0 + B1 λ + B2 λ 2 + B3 λ 3 + ......Bn− 1 λ n− 1 (3.70b)

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Each of the matrices B0, B1, B2, ....Bn− 1, is a matrix of order n. The inverse of matrix ( A − λI ) is defined as: ( A − λI )−1 = B (λ)/ A − λI



(3.71a)

or ( A − λI )

−1

A − λI = B (λ) = B 0 + B 1 λ + B 2 λ + B 3 λ + ......B n− 1 λ 2

3

n− 1

where B (λ) is a transposition of B(λ). Multiply both sides of Eqn. (3.71a) from the left by: ( A − λI ): A − λI = ( A − λI )(B 0 + B 1 λ + B 2 λ 2 + B 3 λ 3 + ......B n− 1 λ n− 1 ) (3.71b)

Hence,

ρ(λ) = AB 0 + λ( AB 1 − B 0 ) + λ 2 ( AB 2 − B 1 ) + ...... + λ n− 1 ( AB n− 1 − B n− 2 ) + λ n B n− 1 (3.71c)

Hence, ρ( A) = 0.

3.13  Bilinear, Quadratic, and Hermitian Forms Consider vectors X and Y in an N-dimensional vector space:  y1   y2 Y =    y n



     

and

 x1  x2 X =     xn

   (3.72)   

The expression Y A X where A is a square matrix of order n, is known as a bilinear form.

Y A X =  y1* 

y 2*

n

=

∑y a x * i ij

( i , j )= 1

j

y 3*

....

y n*

        

a11 a21 a31  an 1

a12 a22 a32  an 2

a13 a23 a33  an 3

.... .... ....  ....

a1n a2 n a3 n  ann

       

x1 x2 x3  xn

       (3.73)  

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On replacing Y by X in bilinear form Y A X , we get: n



∑ x a x (3.74) * i ij

X AX =

j

( i , j )= 1

X A X is termed a quadratic form. The quadratic expression is analogous to the expectation value of an operator A, and to the scalar quadratic expression αx 2 . If the matrix A in Eqn. (3.74) is a Hermitian matrix, then it is called a Hermitian form:

†

X H X = X H X (3.75)

3.14  Change of Basis Choice of basis vectors for representing a quantum state is purely arbitrary, as physical measurements do not depend on our choice of basis vectors. If { ei } is set of orthonormal basis vectors in a N-dimensional vector space, then another set { ei′ } of orthonormal vectors can also be the basis vectors for the same vector space. Orthonormal vectors of set { ei′ } can be constructed by linear combinations of vectors of set { ei }. Each vector of set

{ ei′ } can be expressed as a linear combination of vectors of set { ei } as follows: n

ei′ =



∑u

* ij

e j (3.76)

j=1

in which uij are suitably chosen complex numbers. Since vectors of new and old basis sets are orthonormal, we have ei e j = δ ij and ei′ e ′j = δ ij. δ ij = ei′ e ′j =

=

∑u u k ,l

∑

* jl

ik

uik u*jl δ kl =

k ,l

e k el (3.77a)

∑

uik u*jk

k

Also: δ ij = ei e j = ei UU † e j

∑e Ue =∑ e U e = ∑u u =



i

k

ek U † e j

i

k

e j U ek

k =1

k =1

ik

k =1

* jk

†

(3.77b)

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In other words, matrix uik and ukj* are (i, j) elements of matrices U and U †, respectively, and UU † = I. It is also apparent that a necessary and sufficient condition is that the vectors of set { ei′ } are orthonormal and U is a unitary matrix. 3.14.1  Unitary Transformations Let us say that vector X is expressed with respect to basis { ei X =



∑x

i

} and { ei′ } as follows:

ei (3.78a)

i

and X =



∑x′ e′ (3.78b). i

i

i

We then have:

∑x

i

∑x′ e′ = ∑x ′u e

ei =

i

i

i

i

* i ij



j

(3.79)

i, j

∑x′u

=

j

* ji

ei

i, j

Hence:

∑ i



  xi − 

⇒ xi =

 x ′j u*ji  ei = 0 

∑ j

∑

u*ji x ' j

(3.80)

j

which provides a connection between components xi and x 'i . Assigning 1 to n to both i and j we obtain:



 x1   x2     xn

*   u11   *  =  u12      *   u1n

* u21



* 22

  

u  u2* n

un* 1   x1′  un* 2   x2′      *   xn′ unn 

   (3.81)   

or

X = U† X′ ⇒ U X = UU † X ′ = X ′

(3.82)

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Next consider the case of vector transformation in the vector space of basis { ei }: Y = A X (3.83a)



The corresponding transformation in the vector space of basis { ei′

} can be represented as:

Y ′ = A′ X ′ (3.83b)



where A′ is the transforming matrix in the vector space of basis { ei′ }. The objective is to find a relation between A and A′. From Eqn. (3.82):

Y ′ = U Y and X ′ = U X (3.84)

Then from Eqn. (3.83b): U Y = A ′U X

⇒ UA X = A′U X

(3.85a)

†

⇒ A = U A ′U which implies: A′ = UAU † (3.85b)

Also:

X ′ Y ′ = X U †U Y = X Y (3.86)

which shows that the inner product of two vectors, representing two eigenstates, is independent of the choice of basis vectors. For commuting matrices A and B, we have:



UABU † = UBAU † ⇒ UAU †UBU † = UBU †UAU † (3.87) ⇒ A ′B ′ = B ′ A ′

3.15  Infinite-dimensional Space The theory of N-dimensional vector space can be extended to infinite-dimensional space by taking N → ∞ limit. Each vector of infinite-dimensional space then has an infinite number of components:



 x1   x2 X =  x3      

    (3.88)    

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The inner product of two vectors is then defined by:

X Y = x1* y1 + x2* y 2 + x3* y 3 + .... + xi* yi + ...... (3.89)

The properties of an infinite-dimensional space can be inferred by generalization of the definitions and theorems of N-dimensional space. However, arguments such as the diagonalization of a matrix, which depends upon the assumption that N is finite must be re-examined. The assumption or theorem that N linearly-independent vectors span a space of N dimensions loses its meaning for N → ∞ . A set of finitely many orthonormal vectors { ei } is said to be complete if every vector X can be expressed as a convergent sum: ∞

X =



∑x

i

ei (3.90)

i=1

which implies: ∞



XX =

∑x

i

2

, with xi = ei X (3.91)

i=1

This is known as the completeness relation for a set { ei }. This relation is true for every vector in the vector space.

3.16  Hilbert Space The functions f ( x) and g( x) are square-integrable in the interval ( a, b) if the norms: b



(f, f)=

∫ f ( x)

2

dx (3.92a)

2

dx (3.92b)

a

and b



( g , g) =

∫ g( x) a

2

exist. The Schwarz-inequality for functions ( f , g ) ≤ ( f , f )( g , g ) suggests that the inner product of two functions ( f , g ) exists for square-integrable functions, and hence a linear combination h( x) = af ( x) + bg( x) is too square-integrable. The functions behave like vectors, and vector space defined this way is called Hilbert space. The continuous variable a ≤ x ≤ b in this space takes the place of discrete variable i of the linear vector space, defined

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earlier. The vector f has infinitely uncountable components, because f ( x) corresponds to uncountable values of x. A linear transformation that corresponds to Y = A X in the Hilbert space is: b



∫

g( x) = A( x , x ′) f ( x ′)dx ′ (3.93) a

The continuous matrix A( x , x ′) is a function of two variables. An identity transformation like X = 1 X is defined by: b



∫

f ( x) = δ( x − x ′) f ( x ′)dx ′ (3.94) a

The δ( x − x ′) is the Dirac delta function that is discussed in Annexure B. We thus find that every definition of N-dimensional space has an analogue in Hilbert space. A function is called a normalized function if its norm is equal to unity. Two normalized functions whose inner product is equal to zero are known as orthonormal functions. A Hilbert space can consist of a set of N-orthonormal functions {φi ( x)}. They are said to fulfill the criteria of completeness and closure if: b



∫

(φi , φ j ) = φ*i ( x)φ j ( x)dx = δ ij (3.95) a



∑φ (x)φ (x′) = δ(x − x′) (3.96) * i

i

i

3.16.1  Basis Vectors in Hilbert Space We can define a set of N-orthonormal vectors { φi ( x) }, where each vector is a continuous function of x. The functions φ1 ( x), φ2 ( x), φ3 ( x),........φn ( x) satisfy Eqn. (3.96). The ket vectors φ1 , φ2 , φ3 ,....... φn can then be used as basis vectors to define a vector space, known as a Hilbert vector space. A quantum mechanical state in this vector space is then defined by:

ψ =

∑c

i

φi (3.97a)

i



with ci = φi ψ . (3.97b)

Therefore,

ψ =

∑φ

i

ψ φi (3.98)

i

where, c1 , c2 ,.....cn are complex or real components of ψ along the basis vectors. The quantum state of a system is represented by the wave function ψ(r ) in coordinate space, and by ψ(p) in momentum space. The above definitions, Eqns. (3.97) and (3.98), of vector

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Hilbert space are generalized on replacing x either by r or by p for coordinate (momentum) 3 * 3 * space. An important property of vector Hilbert space is: φ ϕ = ∫ d rφ (r )ϕ(r ) = ∫ d pφ (p)ϕ(p). 3.16.2  Quantum States and Operators in Hilbert Space The wave functions defining the quantum states are represented as vectors in linear vector Hilbert space. The wave function ψ a ( r ) , where a is representative of all quantum numbers, is denoted by ψ a or simply by a as the vector. The ψ a can be expressed in terms of basis kets as follows: ψa =



∑c

i

φi

a =

or

i

∑c

i

i (3.99)

i

which in abstract representation becomes:    a =    



    (3.100)    

c1 c2 c3 ... cn

An operator A takes the wave function ψ(r ) to a new function ξ ( r ) : ξ = A ψ (3.101)



Expressing ξ and ψ in terms of basis kets φ1 , φ2 , φ3 ,..... φn , we get:

∑b φ i

i

i



∑c φ = ∑c A φ =A

i

i

i

i

(3.102)

i

i

Taking the inner product with φ j :

∑b



i

φ j φi =

i

∑c

i

φ j A φi (3.103)

i

Since, φ j φi = δ ij , Eqn. (3.103) simplifies to: bj =



∑c A (3.104) i

ji

i

where, A ji = φ j A φi for 1 ≤ (i, j) ≤ n. Thus:



       

b1 b2 b3  bn

      =        

A11 A21 A31  An1

A12 A22 A32  An2

A13 A23 A33  An3

... ... ...  ...

A1n A2 n A3 n  Ann

       

c1 c2 c3  cn

    (3.105)    

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Equations (3.101) and (3.105) are equivalent. In Eqn. (3.105), a quantum state is represented by a vector while the operator is represented by a matrix. If A is a Hermitian operator, * then the corresponding matrix too is Hermitian and hence Aij = A ji . If φi , for 1 ≤ i ≤ n, are eigenvectors of operator A, then: A φi = α i φi (3.106)



or φ j A φi = α i δ ji and the resulting matrix is a diagonal matrix. Thus, the matrix representation of an operator with respect to its own eigenvectors is a diagonal matrix and the matrix elements are the eigenvalues of the operator. 3.16.3  Schrödinger Equation in Matrix Form The time-dependent Schrödinger equation is:

i

∂ ψ = H ψ (3.107) ∂t

As discussed in Section 2.1, the time independence of the Schrödinger Hamiltonian allows us to factorize ψ (r , t) into time-dependent and space dependable parts. We therefore c j (t)φ j (r ). Substituting ψ = c j φ j in Eqn. (3.107) and then taking can write: ψ (r , t) =

∑

∑

j

j

the inner product with φi , we obtain: dci (t) −i =− dt 



∑c (t)H (3.108a) j

ij

i

or:        



c1 c2 c3  cn

      =        

H 11 H 21 H 31  H n1

H 12 H 22 H 32  H n2

... ... ...  ...

H 13 H 23 H 33  H n3

H 1n H2n H 3n  H nn

       

c1 c2 c3  cn

    (3.108b)    

where H ij = φi H φ j . The time-independent Schrödinger equation Hψ (r ) = Eψ (r ), with the use of ψ (r ) = ai φi (r ), can be written in matrix form as follows:

∑ i

a jE =



∑a H (3.109a) i

ji

i

or:



 ( H 11 − E)  H 21   H 31     H n1 

H 13

...

( H 22 − E)

H 23

...

H 32

( H 33 − E)

...

 H n2

 H n3

 ...

H 12

  H2n   H 3n    ( H nn − E)    H 1n

a1 a2 a3  an

    = 0 (3.109b)    

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3.17  Statement of Assumptions of Quantum Mechanics 1. Every physical system corresponds to a linear vector space, where each vector of the space represents a possible quantum state of the system. The principle of superposition is implied by this assumption. 2. Each physically measurable property of an observable of the system corresponds to a linear operator, and the eigenvalues of the operator are the possible results of measurement on the observable. Thus, the normalized eigenvectors of a Hermitian operator that corresponds to an observable span the entire statespace, and they can be used as basis vectors. Consequently, any ψ can be expressed as a linear combination of eigenvectors of the Hermitian operator. If u1 , u2 ...... un are eigenvectors and α 1 , α 2 ,.......α n are corresponding eigenvalues of the operator, then: A ui = α i ui (3.110)

and:

ψ =



∑c

i

ui (3.111)

i

If an observation of a quantity that corresponds to an operator A is made in the state ψ of the system then the relative probability of obtaining the result α i is: 2

2

P(α i ) = ci = ui ψ (3.112)



The expectation value of operator A in the state of ψ is given by:

A =



ψ Aψ ψψ

∑α c = ∑c i

i

i

i

2

2

(3.113)

i

if ui are orthonormal vectors. 3. The ψ can be a simultaneous eigenvector of two operators A and B if they commute. This means that simultaneous precise measurements of two observables A and B are possible in the state of ψ , if they commute. When they do not commute, the uncertainties in simultaneous measurements (∆A and ∆B) are related as follows: ∆A ∆B ≥



1 C (3.114a) 2

where:

[ A, B] = iC (3.114b) Equation (3.114a) is known as the general principle of uncertainty.

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3.18  General Uncertainty Principle The quantities corresponding to two non-commuting operators cannot be well defined in the same quantum state. If for two Hermitian operators A and B:

[ A, B] = iC (3.115)



with C too as a nonzero Hermitian operator, then eigenvectors of A are not eigenvectors of B. A linear combination of several eigenvectors of A is the eigenvectors of B. The results of measurements performed on B and on A define the probability distributions of finite widths--∆A for A and ∆B for B, as follows:

∆A =

{ (A −

A )2

}

1/2

and ∆B =

{ (B − B ) } 2

1/2

(3.116)

For any state ψ , we then have: ∆A ∆B ≥



1 C (3.117) 2

where: C = ψ C ψ . Equation (3.117) is the generalized statement of the uncertainty principle. Proof: Let us define A = A − A and B = B − B . Evaluate the commutator:



 A, B  = ( A − A )( B − B ) − ( B − B )( A − A ) = AB − A B − A B + A B − BA + B A + B A − B A = AB − BA = [ A, B ]

(3.118)

Note that A is scalar while A is an operator. Therefore, A and B too are Hermitian. Construct an operator R = A + iλB, where λ is a real scalar quantity. Let us say ξ = R ψ . Since the norm square of a vector can never be negative, we have:

ξ ξ = ψ R † R ψ ≥ 0 (3.119)

Therefore:

(

) ( A + iλB ) ψ ≥ 0 ψ ( A − iλB )( A + iλB ) ψ ≥ 0

ψ A + iλB

⇒

†

⇒ ψ ( A2 + λ 2 B 2 + iλ[ A, B]) ψ ≥ 0 ⇒ A2 + λ 2 B 2 − λ C ≥ 0

(3.120)

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We have made use of Eqn. (3.118) in driving Eqn. (3.120). Find the minimum value of λ by taking partial differentiation of Eqn. (3.120):

∂ ∂λ

(A

2

)

+ λ 2 B 2 − λ C = 0 (3.121a)

which gives: λ=



C (3.121b) 2 B2

substitution of which in Eqn. (3.120), yields:

(A −

A

)2 ( B −

B

)2

≥

1 2 C (3.122a) 4

or:

( ∆A)( ∆B) =

1 C (3.122b) 2

3.19  One Dimensional Harmonic Oscillator The problem of the harmonic oscillator is a prominent problem of physics. Many of the physical phenomena are represented by the harmonic oscillator. For example, the Maxwell field can be represented as a linear combination of harmonic oscillators of different frequencies. The electromagnetic fields can then be represented by the Hamiltonian describing a set of harmonic oscillators. As discussed in Section 3.16, one can always construct a Hilbert space that corresponds to the given system, which consists of different oscillators. All possible eigenstates of the system then are represented by the different vectors of that Hilbert space. We here consider the case of a single harmonic oscillator, whose Hamiltonian is given by:

H=

p2 1 + mω 2 x 2 (3.123) 2m 2

where ω is the natural frequency of the oscillator. The eigenvalue equation for a single oscillator is: H ψ = E ψ (3.124)



where E is the energy eigenvalue and ψ is the state vector. Let us define new operators in terms of operators x and p as follows:

a=

i ( p − imωx) (3.125a) 2 mω

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and: −i ( p + imωx) (3.125b) 2 mω

a† =

which gives:

a − a † = ip

2 , and a + a † = mω

2 mω x (3.125c) 

With the use of  x , p  = i, we obtain: 1  p 2 + m2 ω 2 x 2 − imω( xp − px)  2 mω  (3.126a) 1  p 2 + m2 ω 2 x 2 + mω  = 2 mω

aa † =

and: 1  p 2 + m2 ω 2 x 2 + imω( xp − px)  2 mω  (3.126b) 1  p 2 + m2 ω 2 x 2 − mω  = 2 mω

a† a =

Therefore:

aa † − a † a =  a, a †  = 1 (3.127)

and:

( aa † + a † a) =

1  p 2 + m2 ω 2 x 2  (3.128a) mω 

Thus: 1 † ( aa + a † a)ω 2 (3.128b) 1  † =  a a +  ω  2

H=

Let us evaluate the commutators [ a, H ] and  a † , H :

[ a, H ] = ω(aa† a − a† aa) = ω(aa† − a† a)aω = aω (3.129a)



 a † , H  = ω( a † a † a − a † aa † ) = ωa † ( a † a − aa † ) = − a † ω (3.129b)

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 † H is Hermitian; therefore a a =  H/ω −  culate its norm square:

1  too is Hermitian. We define φ = a ψ and cal2

φ φ = ψ a † a ψ = ψ ( H/ω − 1/2) ψ = (E/ω − 1/2) ψ ψ (3.130)

The norm of any vector can never be negative, therefore: φ φ ≥ 0 (3.131)



Thus, if ψ is not a null vector, ψ ψ ≠ 0 then the minimum value that can be assigned to 1 E is E0 = ω , which corresponds to the minimum value of φ φ = 0 in Eqn. (3.131). It can 2 then be inferred that the E0 corresponds to an eigenvector ψ 0 and no state exists below the state ψ 0 . Hence:

a ψ 0 = 0 (3.132)

Further: H φ = Ha ψ

= ( aH − aω ) ψ

(3.133)

= ( E − ω ) a ψ which means that a ψ is an eigenvector of H , and corresponds to the eigenvalue E − ω . In a similar manner: Ha 2 ψ = Ha( a ψ ) = ( aH − aω )( a ψ )

(

)

= aH a ψ − ω a 2 ψ

(3.134)

= a ( E − ω ) a ψ − ω a 2 ψ = ( E − 2 ω ) a 2 ψ Equations (3.133) and (3.134) can be generalized to:

Ha n ψ = (E − nω )a n ψ (3.135)

We thus find that eigenvalues E, ( E − ω ), ( E − 2ω ) and …... ( E − nω ) correspond to eigenvectors ψ , a ψ , a 2 ψ and ………… a n ψ , respectively. In a similar manner:

Ha † ψ = ( a † H + a † ω ) ψ = (E + ω ) ψ

(3.136a)

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and by successive operation with a † on a † ψ , we can construct ( a † )n ψ which on operating by H gives:

H ( a † )n ψ = (E + nω )( a † )n ψ (3.136b)

We again notice that eigenvalues E, ( E + ω ), ( E + 2ω ) and …... ( E + nω ) correspond to eigenvectors ψ , a † ψ , ( a † )2 ψ and ………… ( a † )n ψ , respectively. We thus infer that operation of a on ψ , which corresponds to energy eigenvalue E, takes it to a lower energy eigenvalue ( E − ω )  state, while operation by a † on ψ takes it to a higher energy eigenvalue ( E + ω ) state. The operators a and a † are therefore termed as lowering (annihilation) and raising (creation) operators. 1 As stated before, E − nω in Eqn. (3.135) cannot be smaller than ω. We therefore have 2 1 E − nω = ω . Thus, the energy eigenvalue of the nth state of the harmonic oscillator is: 2 1  En =  n +  ω (3.137)  2



1 ω is ψ 0 . 2 Further higher eigenvalue states can then be constructed by successive operation of a † on ψ0 . Let us say that the quantum state that corresponds to En is represented by eigenvector ψ n and we have: where n = 0, 1, 2......... The state that corresponds to the lowest eigenvalue

H ψ n = En ψ n (3.138)



If we assume that eigenvectors are normalized then ψ m ψ n = m n = δ mn, here m n is an abbreviated form of ψ m ψ n . From Eqns. (3.133) and (3.136a), we notice that a ψ n and a † ψ n correspond to eigenvalues ( En − ω ) and ( En + ω ), respectively. Also, as per Eqns. (3.137) and (3.138), (En − ω ) and ( En + ω ) belong to ψ n− 1 and ψ n+ 1 , respectively. It can therefore be concluded that vectors a ψ n and ψ n− 1 , which have the same eigenvalue, are collinear, and hence we can write:

a ψ n = α n ψ n− 1

or

a n = α n n − 1 (3.139a)

a † ψ n = β n ψ n+ 1

or

a † n = β n n + 1 (3.139b)

Similarly, we can write:

where α n and β n are scalars and can be determined as follows: 2

αn = αn

2

n−1 n−1

= (n − 1) α *nα n (n − 1) = n a† a n =n

(3.140a)

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Here we have made use of n − 1 n − 1 = n n = 1 and the Hermitian conjugation of Eqn. (3.139a). Similarly, we write: 2

2

βn = βn

n+1 n+1

= n + 1 β*nβ n n + 1



= n aa † n

(3.140b)

= (n + 1) We therefore obtain:

α n = n , and β n = n + 1 (3.141)

which gives:

a ψ n = n ψ n− 1 (3.142a)



a † ψ n = n + 1 ψ n+ 1 (3.142b)

As is obvious, Eqns. (3.142) are not the eigenvalue equations. Equation (3.132a) with the use of Eqn. (3.125a) gives:

 d mω  + x  ψ 0 ( x) = 0 (3.143a)  dx  

which is a first order differential equation, whose solution can be given by: ψ 0 ( x) = Ne − mωx



2

/2 

(3.143b)

where N is a constant, which is determined by applying the normalization condition: ∞

∫ ψ ( x)

2

0

dx = 1

−∞

∞



⇒ 2N

2

∫e

− mωx 2 /

dx = 1 (3.143c)

0

 mω  ⇒N=  π 

1/4

Equation (3.142b) along with Eqn. (3.125b) is then used to determine: ψ 1 = a† ψ 0 =

  d mω  − mωx2 /2  + x  Ne −   2 mω  dx

2 mω − mωx2 /2  xe =N 

(3.143d)

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Also from Eqn. (3.125c), we can write:

2 mω x ψ n = 0 (3.144) 

a ψ n + a† ψ n −

which with the use of Eqn. (3.142) give:

n + 1 ψ n+ 1 =

2 mω x ψ n − n ψ n− 1 (3.145) 

Equation (3.145) is the recurrence relation, which can be used to derive ψ 2 after knowing ψ 0 and ψ 1 . Thus, all eigenstates of the 1D harmonic oscillator can be generated with the use of Eqn. (3.145). The ψ n (ρ) is then given by:

ψ n (ρ) =

 mω    n (2 n !)  π  1

1/4

2

e −ρ

/2

H n (ρ) (3.146)

mω x . H n (ρ) is the Hermite polynomial of degree n. The first few Hermite poly nomials are: H 0 (ρ) = 1, H 1 (ρ) = 2ρ, H 2 (ρ) = 4ρ2 − 2 . The matrix representation of various operators used to describe the 1D harmonic oscillator can be given by employing the following essential properties of operators a, a †, and a † a: where ρ =



amn = m a n = nδ m , n− 1 (3.147a)



† amn = m a † n = n + 1δ m , n+ 1 (3.147b)

and:

( a a) †



mn

= m a † a n = nδ m , n (3.147c)

The matrices representing a and a † are then given by:



   a =    

0

1

0

0

0

0

2

0

0

0

0

3

0 

0 

0 

0 

  ...   (3.148a) ...  4    ...

and:



   a† =     

0

0

0

0

1

0

0

0

0

2

0

0

0 

0 

3 

0 

...   ...  ...  (3.148b) ...    

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The operators x and p are represented by the matrices:

x=





p=

     2 mω    

0

1

0

0

0

1

0

2

0

0

0

2

0

3

0

0 

0 

3 

0 

4 

 0   i 1 mω  2  0  0   

... ... ... ... 

−i 1

0

0

0

0

−i 2

0

0

i 2

0

−i 3

0

0 

i 3 

0 

−i 4 

     (3.149a)    ... ... ... ... 

     (3.149b)   

The Hamiltonian is represented by the diagonal matrix:       H = ω      



1 2

0

0

0

0

3 2

0

0

0

0

5 2

0

0

0

0







7 2 

 ...    ...   ...  (3.150)   ....    

Note that ψ 0 , ψ 1 , ψ 2 ..... ψ n ....... are orthonormal vectors, which can be used as basis vectors for a Hilbert space that corresponds to a system of large number of harmonic oscillators.

3.20  Solved Examples 1. Show that the vectors:

 1 A1 =  0   −1

 0    and , A =   2  1   0  

 1  A3 =  0  ,    1 

are (a) linearly independent, (b) orthogonal, (c) but not orthonormal.

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SOLUTION

a. The condition for linear independence is that c1 A1 + c2 A2 + c3 A3 = 0 can only be satisfied for c1 = c2 = c3 = 0. We have:  c1   0  − c1 



  0    +  c2    0

  c3   + 0   c3

  0      =  0  , which gives:   0  



c1 + c3 = 0, (3.151a)



c2 = 0, (3.151b)



c3 − c1 = 0. (3.151c) Addition and subtraction of Eqns. (3.151a) and (3.151c) give c1 = c2 = c3 = 0.

b. The orthogonality condition demands that Ai A j = 0 for 1 ≤ ( i,  j ) ≤ 3 :

A1 A2 =  1

0

 0    −1   1  = 0,  0 

similarly:



 0 

1

 1    0   0  = 0  1 

 1    and  1 0 1   0  = 0.  −1  Hence, it is proved that vectors A1 , A2  and  A3 are orthogonal. c. To satisfy the orthonormal condition, the requirement in addition to the orthogonality condition is that the norm of each vector should be unity. A1 = 12 + 02 + (−1)2 = 2 , A2 = 1 , and A3 = 2 ; therefore A1 , A2 and A3 are not orthonormal. Division of A1 and A3 by 2 will provide a set of orthonormal vectors. 2. Show that the vectors



 i  1     A =  0 , B =  0  − i  i 

  , and  

 i   C =  1   0 

are linearly independent but not orthogonal. Orthogonalize them using the GramSchmidt method.

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Matrix Formulation of Quantum Mechanics

SOLUTION

We have:  i  1    a  0  + b  0  − i  i 



  i   0  + c  (3.152a)     1 = 0    0   0 

which gives:

a + ib + ic = 0 (3.152b)



c = 0 (3.152c)



ia − ib = 0 (3.152d) Equation (3.152d) gives a = b. Substitution of c = 0  and a = b in Eqn. (3.152b) gives

( 1 + i ) b = 0, which means a = 0  = b. Hence, vectors are linearly independent. The inner products of A , B and C are:

A B =  1



B C =  − i

C A =  − i

0

 i  − i   0  − i

  = i − 1 ≠ 0,  

0

 i    i   1  = 1 ≠ 0, and  (3.153)  0 

1

 1     0   0  = −i ≠ 0  i 

Hence A , B and C are not orthogonal. To orthogonalize these, we choose a set of three vectors U , V and W , and then use the Gram-Schmidt method of orthogonalization. We take U = A , and V = B + a U . On imposing the condition that V 2 and U are orthogonal U V = 0, we obtain a = − U B / U , which gives:



 1+ i  1  a = (1 − i)/2, and  V = 0 . 2  1 − i  We next take W = C + b V + c U and then we use U W = 0 and V W = 0 to 2 2 determine b and c. This yields b = − V C / V and c = − U C / U , which provides b = − ( 1 + i )/2 and c = − i/2 . Therefore:



 i   1+ i   1   0     (1 + i)   i   W =  1 − 0  −  0  =  1  (3.154)  4 2  0   1 − i   i   0 

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Three orthogonalized vectors are:  1+ i   1  1   U =  0  , V =  0  and 2  1 − i   i 



 0   W =  1  (3.155)  0 

As is seen, vectors V and W are orthonormal vectors, while U is not. To orthonormalize U , divide it by its norm 2 . 3. Construct a 3 × 3 matrix A that transforms the vectors:  1 X1 =  0   −1



 0    , X =  2  1  and   0  

 1  X3 =  0     1 

to the vectors:  2   1 Y1 =  0  , Y2 =  0   0   1



  and  

 1 Y3 =  0   −1

   

SOLUTION

The transforming equation is Yi = A X i for 1 ≤ i ≤ 3. Therefore:







 2   a11     0  =  a21  0   a31  1   a11  = a  0   21  1   a31  1   0  −1

  a11 = a   21   a31

a12 a22 a32

a13 a23 a33

a12 a22 a32 a12 a22 a32

a13 a23 a33 a13 a23 a33

 1   0   −1 

  2 = a11 − a13 ⇒ 0= a −a 21 23    0 = a31 − a33  

  0   1 = a12      1  ⇒  0 = a22   0   1 = a32  

   for i = 1 (3.156a)  

   when i = 2 (3.156b)  

  1   1 = a11 + a13      0  ⇒  0 = a21 + a23   1   −1 = a31 + a33  

   for i = 3 (3.156c)  

the solution of which yields: a11 =



3 1 , a13 = a31 = a33 = − ,  a22 = a21 = a23 = 0,  a12 = a32 = 1. The matrix is: 2 2  3   2 A= 0  −1   2

1 0 1

−1 2 0 −1 2

    (3.157)   

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4. A vector space is represented by basis vectors:  1   0       0  ,  1  and  0   0 



 0     0 .  1 

Construct a matrix that transforms these to another set of basis vectors:  1/ 2  0   i/ 2 



  (1 + i)/ 2   0 ,   (1 − i)/ 2  

   , and  

 0     1 .  0 

Is the transforming matrix unitary? SOLUTION

We have:



 1   a11 1    0  =  a21  2  i   a31 

a12 a22 a32

a13 a23 a33

 1 = a11  1   2      0  ⇒  0 = a21   0   i  = a31  2 

    (3.158a)   



 1 + i   a11 1   0  =  a21  2  1 − i   a31 

a12 a22 a32

a13 a23 a33

 1+ i   0   2 = a12      1  ⇒  0 = a22   0   1 − i  = a32   2

    (3.158b)   

 0   a11     1  =  a21  0   a31 



a12 a22 a32

a13 a23 a33

  0   0 = a13      0  ⇒  1 = a23   1   0 = a33  

   (3.158c).  

The transforming matrix is:

and:



 1/ 2  0 A=  / 2 i   1/ 2  A =  (1 − i)/2  0  †

(1 + i)/ 2 0 (1 − i)/ 2

0 0 1

0   1  (3.159a)  0  

− i/ 2 (1 + i)/2 0

   , (3.159b)  

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which give:  1 AA † =  0  0



0 1 0

   (3.159c) 

0 0 1

This shows that the transforming matrix is a unitary matrix. 5. Find the eigenvalues and eigenvectors of the matrix:  1  A= 1  1



0 2 0

  . 

0 3 3

SOLUTION

The characteristic equation of the matrix is: 1− λ 1 1



0 2−λ 0

0 3 3−λ

= 0 (3.160)

Simplification of the determinant gives (1 − λ)(2 − λ)(3 − λ) = 0, whose solution yields three eigenvalues: λ 1 = 1, λ 2 = 2 and  λ 3 = 3. The eigenvector corresponding to the eigenvalue λ 1 = 1 is obtained from:  0   1  1



0 1 0

0 3 2

  x11    x21   x31 

  0      =  0  (3.161a)   0  

which provides two independent equations: x11 + x21 + 3 x31 = 0 (3.161b) x11 + 2 x31 = 0



Here the number of degrees of freedom is 1, and therefore by choosing x31 = 1, we get the eigenvector corresponding to the eigenvalue λ 1 = 1:  −2    X 1 =  −1  (3.161c)  1 



To obtain the eigenvector that belongs to λ 2 = 2 we have:



 −1   1  1

0 0 0

0 3 1

  x12  x   22   x32

  0      =  0  ⇒ − x12 = 0 ,   x12 + 3 x32 = 0, and  x12 + x32 = 0.   0  

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This suggests that x12  and  x32 are zero and x22 can be assigned an arbitrary value. We therefore can take:  0    X 2 =  1  (3.161d)  0 



The eigenvector corresponding to λ 3 = 3 is obtained from:



 −2   1  1

0 −1 0

0 3 0

  x13    x23   x33 

  0      =  0    ⇒  x13 = 0,  x13 − x23 + 3 x33 = 0  (3.161e)   0  

which yields x23 = 3 x33. By choosing x33 = 1, we get:  0    X 3 =  3  (3.161f)  1 

6. For the matrix:   A=   



4 0 0 0

2 2 0 4

0 −1 3 0

4 0 3 7

     

if λ 1 = 4, λ 2 = 1, and λ 3 = 6 are three eigenvalues of A. Find the fourth eigenvalue of the matrix and its determinant. SOLUTION

Since the trace of any matrix is equal to the sum of all eigenvalues, we have: 4 + 2 + 3 + 7 = 4 + 1 + 6 + λ 4 ⇒ λ 4 = 5. The determinant of a matrix is equal to the n multiple of all eigenvalues and ( −1) . Therefore A = λ 1 λ 2 λ 3 λ 4 = 4 × 1 × 6 × 5 = 120. 7. The eigenvalues and corresponding eigenvectors of a 3 × 3 matrix are given by:

 −1  λ 1 = 3, X 1 = −1   1

 0   2     and λ = 5, X =   , λ = −3, X =  2 3 3  0 ,  1   2  1   0  

so, find the matrix. SOLUTION

We know that a matrix A can be written as A = SΛS−1, where:

 3 Λ =  0  0

0 −3 0

0 0 5

  −1  and S =    −1   1

2 1 0

0 0 1

 .  

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Computation of the inverse of S gives:  1  S = 1  −1

−2 −1 2

−1



   

0 0 1

Therefore:  −1 A =  −1  1





 −1 A =  −1  1

2 1 0 2 1 0

8. For the matrix,

0 0 1 0 0 1

 3   0   0

0 −3 0

0 0 5

 1  1    −1

 3    −3   −5

−6 3 10

0 0 5

  −9 =   −6   −2

 −4 A=  3



−2 −1 2 12 9 4

   

0 0 1 0 0 5

   

−6  . 5 

find the non-singular matrices S and S−1 to compute Λ = S−1 AS. Find Λ 4 and then show that A 4 = SΛ 4 S−1. Compute A 4. SOLUTION

We first need to find the eigenvalues and eigenvectors of A. The characteristic equation of A is:

λI − A =

λ+4 −3

6 λ−5

= ( λ − 2 )( λ + 1) = 0 (3.162a)

giving λ 1 = 2 and λ 2 = −1. The eigenvector that corresponds to λ 1 = 2 is:



 −6 A X 1 = 2 X 1     ⇒   3 ⇒ ( x11 + x21 ) = 0

−6   x11  3   x21

 =0 (3.162b) 

 −1  which can be satisfied for   X 1 = r  , where r is an arbitrary constant.  1  When λ 2 = −1 we have:



 −3 A X 2 = − X 2     ⇒      3 ⇒ (  x12 + 2 x22 ) = 0  

−6   x12  6   x22

  = 0   (3.162c) 

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 −2  whose solution can be X 2 = t  , where t, too, is an arbitrary constant. Two  1   −1   −2  linearly independent eigenvectors of matrix are   and  . We thus 1    1  have:  2 −2  , Λ =  1   0

 −1 S=   1



 1 0  −1  and S =  −1   −1

2   (3.163a) −1 

and: −2   −4  1  3

 −1 S−1 AS =   1



−6   1  5   −1

2   2 = −1   0

0   (3.163b) −1 

Direct multiplication yields:  24 Λ4 =   0 



0

( −1)

4

 .  

Also, we find that:

(

)(

)(

)(

)

 −1 A4 =   1

 4 −2   2  1  0 

0

( −1)

 −1 =  1

−2   16  1   −1

32   −1 

Λ 4 = S−1 AS S−1 AS S−1 AS S−1 AS = S−1 A 4 S (3.164a)



which implies A 4 = SΛ 4 S−1:



4

  1   −1 

2   −1  (3.164b)

 −14 −30  =  31   15                It is to be noted here that for any n × n diagonalizable matrix A, Λ = S−1 AS, then A k = SΛ k S−1 ,   k = 1, 2,…

where:



 λk  1  0 Λk =     0 

0



0

λ 2k  0

  

0  λ nk

    . (3.164c)   

9. Show that scalar product of two vectors is invariant under unitary transformation.

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SOLUTION

Let us say that after unitary transformation two vectors A and B are:

A′ = U A (3.165a)



B′ = U B (3.165b) Notice that B′ = B U † , and then calculate the inner product of A′ and B′ : B′ A′ = B U †U A = B A (3.165c)



Hence, we find that the scalar (inner) product of two vectors is invariant under unitary transformation. 10. Show that every Hermitian matrix can give rise to a unitary matrix. SOLUTION

For Hermitian matrix H † = H, take the matrix: U = ( H − iI )( H + iI ) (3.166b) −1



where I is a unit matrix. We then have:

(

U † = ( H + iI )



) ( H − iI )

−1 †

†

= ( H + iI )( H − iI ) (3.166b) −1

which yields UU † = I. This proves that every Hermitian matrix can give rise to a unitary matrix. 11. Consider the case of a 1D harmonic oscillator, where for ground state ψ 0 ( x) =

α

(π)

1/4

e −α

2 2

x /2

 mω  , with α =    

1/2

. Calculate ψ 1 ( x), ψ 2 ( x) and then generalize

it for ψ n ( x). SOLUTION:

We know that operation by a † on a state vector takes it to the next eigenvalue state: ψ 1 ( x) = a † ψ 0 ( x), where:

a† =

1 1 d −1   ( p + imωx) = + imωx  =  − i  dx α 2 i 2 mω i 2 mω

 d  − α 2 x  (3.167a)   dx

Thus: ψ 1 ( x) = −

=−

1  d  − α 2 x  ψ 0 ( x)   α 2  dx 1  d  22 − α 2 x  e −α x /2 1/4   2α ( π )  dx

 1 =  2α π

 2 −α 2 x2 /2  α  1/2 2 2 H 1 (αx)e −α x /2 =  2α xe  2 π 

(3.167b)

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With the use of Eqn. (3.145), we get: 1 ψ0 2

ψ 2 = αx ψ 1 −

  3 2 −α 2 x2 /2 1 1 = − 1/4  2α x e ( π )1/4  2α ( π ) 

α −α 2 x2 /2 e 2

2 2 α 2α 2 x 2 − 1 e −α x /2 2

=

1 π ( )1/4

=

1 1/4 2 (π)

(

)

(3.168)

2 2 α H 2 (αx)e −α x /2 2

 α 1  = 2  π 2 (2!) 

1/2

H 2 (αx)e −α

2 2

x /2

,

Generalization of ψ 2 ( x) and ψ 1 ( x) yields:  α 1  ψ n ( x) =  n  π 2 ( n !) 



1/2

H n (αx)e −α

2 2

x /2

(3.169)

12. Show that operators a and a † are not Hermitian while operator a † a is Hermitian. SOLUTION

For a Hermitian matrix A, A = A † ⇒ Aij = A*ji . The matrix elements of a and a † are: amn = m a n = n m n − 1 = nδ m , n− 1 (3.170a)

and:

( anm )* =



*

nam =

(

m n m−1

)

*

= mδ n, m− 1 (3.170b)

which yield amn ≠ ( anm ) , and hence a is not a Hermitian operator. Further: *

(a ) †



mn

= m a † n = n + 1 m n + 1 = n + 1δ m , n+ 1 , (3.171)

and:

(a )

* † nm

*

= n a† m = m + 1 n m + 1 † = m + 1δ n, m+ 1 ≠ anm

(3.172)

We thus find that a † is not a Hermitian operator. The matrix elements of operator a † a are:

( a a) †

mn

= m a † a n = nδ mn (3.173)

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and:

{( a a) } = n a a m *

†



†

*

nm

( )

= mδ nm (3.174)

( )

Since, m and n are integer number, a † a = a † a . Hence a † a is a Hermitian nm mn operator. 13. Use the Heisenberg equation and show that the time dependence of annihilation and creation operators for a 1D harmonic oscillator can be expressed as: a(t) = a(0)e − iωt and a † (t) = a † (0)e iωt. SOLUTION

From Eqn. (3.125), we have:

a=

i ( p − imωx) and 2 mω

a† =

−i ( p + imωx) (3.175) 2 mω

The Heisenberg equation for time-independent operators gives:  dp 1 1  p2 1 =  p , H  =  p , + mω 2 x 2  dt i i  2 m 2  =

 p3 1 1  p3 1 + mω 2 px 2 − − mω 2 x 2 p   i  2 m 2 2m 2 

mω 2 px 2 − x 2 p 2 i mω 2 = px 2 − xpx + xpx − x 2 p 2 i mω 2 = ( −2ix ) 2 i =



(

)

(

which implies

(3.176a)

)

dp = − mω 2 x . And: dt  dx 1 1  p2 1 = [ x, H ] =  x, + mω 2 x 2  dt i i  2 m 2  =



 p2 x 1 1  xp 2 1 + mω 2 x 3 − − mω 2 x 3   i  2 m 2 2m 2 

1 xp 2 − p 2 x 2 im 1 = xp 2 − pxp + pxp − p 2 x 2 im 1 = ( 2ip ) 2 im =

( (

)

(3.176b)

)

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which means

dx p = . Therefore: dt m dx   dp − imω   dt dt 

da = dt

i 2 mω i = 2 mω = − iωa



( −mω x − iωp ) 2

(3.177a)

and: da † = dt =



−i 2 mω −i 2 mω

dx   dp + imω   dt dt 

( −mω x + iωp ) 2

(3.177b)

= iωa † The solution of Eqns. (3.177a) and (3.177b) can always be written as a(t) = a(0)e − iωt and a † (t) = a † (0)e iωt. Note that though a and a † are time dependent, the Hamiltonian 1  H =  a † a +  ω is independent of time.  2 14. Consider the matrix:



    A=     

2

2 3

0

2 3

2

1 3

0

1 3

2

    .     

(a) Is A a Hermitian matrix? If yes, find the Hermitian form X A X . (b) Take the unitary matrix:



    U=    

1 3 1 3 1 3

1 2 0 −

1 2

1 6 2 − 6 1 6

        

and then compute Λ = UAU † to show that a Hermitian matrix is diagonalized with the use of a unitary matrix. Verify that elements along the diagonal are eigenvalues of A.

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SOLUTION

a. As can be seen A † = A; hence it is a Hermitian matrix satisfying: 3

†

X A X = X A X . We have X A X =

3

3

j

i ij

i=1 j=1

that xi* = xi. Therefore:

3

∑∑x a x = ∑∑x a x , which implies * i ij

* j

i=1 j=1

X A X = x1 ( a11 x1 + a12 x2 + a13 x3 ) + x2 ( a21 x1 + a22 x2 + a23 x3 ) + x3 ( a31 x1 + a32 x2 + a33 x3 )

(

)

= 2 x12 + x22 + x32 + 2

(3.178a)

2 1 x1 x2 + 2 x2 x3 3 3

b. To compute Λ = UAU †, let us first calculate:     † AU =     



1 3 1 2 1 6

1 3

1 3 1 − 2 1 6

0 −

2 6

    (3.178b)     

Next calculate:     † Λ = UAU =       3 =  0  0

1 3 1 3 1 3

1 2 0 − 0 2 0

1 2 0 0 1

         

1 6 2 − 6 1 6

3

2 3

3 2

0

3 2

−

4 6

1 3 1 − 2 1 6

        (3.179a) 

   

The eigenvalues of A are given by:



A − λI = 0

2−λ

2 3

0

2 3

2−λ

1 3

1 3

0

{

= 0 

(3.179b)

2−λ

}

⇒ (2 − λ) (2 − λ) − 1 = 0 2

Solution of this gives three values of λ: λ 1 = 3,  λ 2 = 2 and  λ 3 = 1, which are the diagonal elements of matrix Λ.

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3.21 Exercises 1. If λ 1 , λ 2 , λ 3 , λ 4 .....λ n are the eigenvalues of matrix A, then show that the determinant A = (−1)n λ 1 λ 2 λ 3 .....λ n. 2. Show that if λ 1 , λ 2 , λ 3 , λ 4 .....λ n are eigenvalues of the matrix A, then λ 1 ± α , λ 2 ± α........λ n ± α are eigenvalues of the matrix A ± αI. 3. Show that if X is an eigenvector of a matrix A corresponding to the eigenvalue λ, then the same vector X is also an eigenvector of the matrix A2 corresponding to the eigenvalue λ 2. 4. Consider the following three vectors:



 25   5     A1 =  64  , A2 =  8  and  144   12  Are these linearly independent? 5. Consider the matrix and vector:   A=   

0 1 1 0

1 0 0 1

1 0 0 1

   and   

0 1 1 0

 1   A3 = 1  .    1 

 −2    1  . X =  −1   0   

(a) Find the vector Y = A X . (b) Find a vector X such that A X = 2 X . 6. Consider the matrices:



 0 σ1 =   1

 1 1   , and σ 3 =  0   0

 0 σ2 =   i

−i   0 

0  . −1 

(a) Show that they satisfy the relations:

σ 12 = σ 22 = σ 23 = 1, σ 1 σ 2 = iσ 3 , σ 2 σ 3 = iσ 1 , σ 3 σ 1 = iσ 2 , and σ 1 σ 2 + σ 2 σ 1 = 0. (b) Find eigenvalues of σ 1, σ 2 , and σ 3 . 7. Show that matrix



    A=    

1 3 1 3 1 3

1 2 0 −

1 2

1 6 2 − 6 1 6

is an orthogonal matrix. Calculate its determinant.

        

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8. Show that the matrix

    A=     



2

2 3

0

2 3

2

1 3

0

1 3

2

         

is a Hermitian matrix. Calculate its eigenvalues and eigenvectors and show that eigenvectors are orthogonal vectors. 9. X and Y are eigenkets of Hermitian matrix H . Under what condition will X + Y also be an eigenket of H. 10. An observable is represented by the matrix:     A=    



0

1 2

0

1 2

0

1 2

0

1 2

0

    .    

Calculate the eigenvalues and the normalized eigenvectors of this observable. Is there degeneracy? 11. Show that the matrices



 1 A =  0  0

0 −1 0

0 0 −1

  2  and B =    0   0

0 0 2i

0 −2 i 0

 .  

commute. Find eigenvalues of the matrices. Is there degeneracy? 12. Show that the vectors

 1  X =  0  and  i 

 1+ i  1 Y =  0  2  1 − i 

satisfy the Schwarz inequality. 13. Show that matrix U =

1  1  2 i

1   is a unitary matrix. Calculate Y = U X −i 

 2  where X =   and then show that Y Y = X X .  1  14. Show that the Hamiltonian of the 1D harmonic oscillator can be given by:

1 1  H =  aa † −  ω , and H = a † a + aa † ω .  2 2

(

)

4 Transformations, Conservation Laws, and Symmetries There would be no laws of physics without invariance principles. We rely on experimental results remaining the same from day to day and place to place. An invariance principle is intimately related to a conservation law and it reflects a basic symmetry. Analysis of various transformations makes the form of the laws involved invariant. For example, (i) the invariance of physical systems with respect to spatial translation (in other word that the laws of physics do not change with locations in space) gives the law of conservation of linear momentum. (ii) Invariance with respect to rotation gives the law of conservation of angular momentum. (iii) Invariance with respect to time translation gives the well-known law of conservation of energy. In quantum field theory, the gauge invariance of the electric potential and vector potential yields conservation of electric charge.

4.1  Translation in Space In the case of translation in space, it is important to notice whether the system is being translated by vector a or the coordinate axes are being translated by a, which would result in the opposite change in the wave function. Results for translating the wave function by +a would be the same as for translating the coordinate axes along with the apparatus and any external fields by −a relative to the wave function. These two equivalent operations are analogous to the time development in the Schrödinger picture and in the Heisenberg picture. In the Schrödinger picture, bras and kets change in time while operators do not. In the Heisenberg picture, operators develop with time while the bras and kets do not change. Therefore, there are two possible ways to deal with space translations: transform the bras and kets or transform the operators. We here consider the situation where bras and kets are transformed while operators are left unchanged. Let us consider a 1D case, where the translational is made by an operator T ( a) such that: T ( a)ψ ( x) = ψ ( x − a) (4.1)



Thus, if ψ( x) is a wave function centered at the origin, ψ ( x − a) is a wave function centered at the point a. To get the form of the operator T ( a), we make the Taylor series expansion of ψ ( x − a) : ψ ( x − a) = ψ ( x ) − a

dψ ( x) a 2 d 2 ψ ( x) + − ........ dx 2! dx 2

  d a2 d2 = 1 − a + − .....  ψ ( x) 2 dx 2! dx   =e

−a

d dx

(4.2)

ψ ( x) 109

110

Since p = − i

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d , the translation operator is: dx T ( a) = e − iap/h (4.3)



For two successive translations in space: T ( a ′)T ( a)ψ ( x) = ψ ( x − a − a ′) , which would be the same if we take T ( a)T ( a ′)ψ ( x) . Therefore, T ( a ′)T ( a) = T ( a)T ( a ′) , meaning that the space translation operators are commutative. In the case of an infinitesimal translation by ε: T (ε) = e − iεp/ ≈ 1 −



iεp (4.4) 

The linear momentum p is said to be the generator of the translation. Once we have established that the momentum operator is the generator of spatial translations, its generalization to three dimensions is trivial: T ( a ) = e − ia. p/ (4.5)



4.2  Translation in Time The time translation operator T (τ) is an operator that it when acts on ψ(t) gives the new wave function ψ (t + τ) : T (τ)ψ (t) = ψ (t + τ) (4.6)



The Taylor series expansion of ψ (t + τ) gives: T (τ)ψ (t) = ψ (t + τ) = ψ (t) + τ

=e =e

τ

∂ ∂t

∂ τ 2 ∂2 ψ (t) + ψ (t) + ....... ∂t 2! ∂t 2 (4.7)

ψ (t)

− iHτ 

ψ (t)

Thus T (τ) = e − iHτ/ , where we have made use of:

i

∂ ψ (r , t) = Hψ (r , t) (4.8) ∂t

For infinitesimal time translation, in a manner like that for space translation, we have:

T (ξ ) = e

− iHξ 

 1−

− iHξ (4.9) 

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Transformations, Conservation Laws, and Symmetries

4.3  Rotation in Space In contrast to translations in space and time, rotations do not commute even for a classical system. Rotating a book through π/2, first about the z-axis and then about the x-axis, leaves it in a different orientation from that obtained by rotating from the same starting position, first π/2 about the x-axis and then π/2 about the z -axis. Though the commutator is of second order, still operations for smallest rotations do not commute. The R -operators that represent rotations reflect this noncommutativity structure. This is seen by considering ordinary classical rotations of a real vector in three dimensional space. Consider that a vector V with components V1 , V2 and V3 is rotated by an angle θ about an arbitrary direction so that components become V′1 , V′2 and V′3 ; V ′ = R(θ) V . The old and new components are related via a 3 × 3 orthogonal matrix: R(θ) :



 V1′   R11     V2′  =  R21  V3′   R31   

R12 R22 R32

    

R13 R23 R33

 V1   V2  V3 

    , with RR = I (4.10)  

The matrix for rotation of a vector by an angle θ about the x -axis is:



 1 Rx (θ) =  0   0

0 cos θ sin θ

0 − sin θ cos θ

  (4.11a)  

Similarly, the matrices representing the rotation of a vector by an angle θ about y and z axes, respectively are:



 cos θ Ry (θ) =  0   − sin θ



 cos θ Rz (θ) =  sin θ  0 

0 1 0

sin θ 0 cos θ

− sin θ cos θ 0

0 0 1

  (4.11b)     (4.11c)  

Let us consider the case of rotation by an infinitesimal angle ε. On ignoring the terms having ε 3 and higher powers, we obtain:



 0   0 Rx ( ε ) = I + ε    0 

0 ε − 2 1

  −1   (4.12a) ε  −  2  0

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 ε  −2  Ry ( ε ) = I + ε  0   −1 



   0  (4.12b) ε  −  2 

0

1

0 0

and:  ε  −2  Rz ( ε ) = I + ε   1   0



−1 ε 2 0

−

 0   (4.12c) 0   0 

The elementary matrix multiplication yields:  ε  − 2  Rx (ε)Ry (ε) = I + ε   ε   −1

0 ε 2 1

−

  ε   − 2    , and Ry (ε)Rx (ε) = I + ε  −1   0   −ε   −1 1

ε ε 2 1

−

    (4.13) −1   −ε  1

which exhibits that Rx (ε)Ry (ε) ≠ Ry (ε)Rx (ε) , meaning that the infinitesimal rotations about different axes do not commute and we find that:



 0  Rx (ε), Ry (ε)  = ε 2  1   0

−1 0 0

0 0 0

  (4.14a)  

Ignoring the terms having ε 3 and ε 4 , we can write:



 0 Rz ( ε ) = I + ε  1   0 2

2

−1 0 0

0 0 0

  (4.14b)  

Multiplying Ry (ε)Rx (ε) with Rz (ε 2 ) from the left and then leaving the terms having ε 3 and ε 4 , we get:

Rx (ε)Ry (ε) = Rz (ε2 )Ry (ε)Rx (ε) (4.15a)

which suggest that the application of rotation operators on quantum mechanical kets must follow the pattern:

 Rx (ε)Ry (ε) − Rz (ε 2 )Ry (ε)Rx (ε)  ψ = 0 (4.15b)

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FIGURE 4.1 Schematic diagram for rotation of vector.

This in fact leads to very important results in quantum mechanics, as would be shown in forthcoming discussions. Let us consider the rotation by an infinitesimal angle δθ about an axis through the origin. After rotation, the system that is initially at r shifts to r + δr , where δr = δθ × r . We transform a wave function under this small rotation. Like the linear translation case R(δθ)ψ (r ) = ψ (r − δr ), where R ( δθ ) is an operator causing the rotation. The minus sign can again be understood from the discussions on linear translations in Section 4.1. The procedure is similar to that has been used to obtain Eqn. (4.5) gives us:

R(δθ)ψ (r ) = ψ (r − δr ) = e

− iδr .p 

ψ (r ) (4.16a)

Using δr = δθ × r and L = r × p, we get:



R(δθ)ψ (r ) = e =e

− iδr .p 

ψ (r )

− iδθ× r .p/

(

ψ (r )

(4.16b)

)

= e − iδθ.L/ ψ (r ) which yields:

R ( δθ ) = e

− iδθ . L 

(4.17)

Note that a finite rotation by an angle θ can be represented as multiplication together of a large number of δθ operators, which is equivalent to replacing δθ by θ in the exponential. For the case of infinitesimal rotation, Eqn. (4.16b) reduces to:

i   R(δθ)ψ (r )   1 − δθ.L ψ (r ) (4.18)   

Equation (4.17) establishes that the orbital angular momentum operator L is the generator of spatial rotations. The appropriately transformed wave function is generated by the action of R(δθ) on the original wave function on rotating our apparatus and the wave function with it. To clarify it further, we consider the case of rotating the system, and

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therefore the wave function, through an infinitesimal angle δθ z about the z-axis. Let us denote the rotated wave function by ψrot( x , y ). We then have: i ψ rot ( x , y ) = 1 − (δθ z )Lz  ψ ( x , y )   

  d d  = 1 − (δθ z )  x − y   ψ ( x , y ) (4.19) dx    dy   ψ ( x + (δθ z )y , y − (δθ z )x )

Thus, the value of the new wave function at ( x , y ) is the value of the old wave function at the point { x + (δθ z )y , y − (δθ z )x} .

4.4  Quantum Generalization of the Rotation Operator It is well known that in quantum mechanics orbital angular momentum is not the total angular momentum. It is found experimentally that particles like the electron exhibit an internal angular momentum, called spin angular momentum. The total angular momentum, called J , is the sum of the orbital angular momentum L and spin angular momentum S; J = L + S. The operator J is defined as the generator of rotations on any wave function, including possible spin components. Therefore, Eqn. (4.16b) is generalized to:

(

)

R(δθ)ψ (r ) = e − iδθ. J/ ψ (r ) (4.20)

which is similar to the equation for L , whose components are written as differentials. Up to this point, we considered ψ(r) a complex valued function of position. But the wave function at a point can have several components in some vector space. The rotation operator will operate in that space, and it is a differential operator too with respect to position. Then, the state vector ψ is a vector at each point, and the rotation of the system rotates this vector as well as moving it to a different value of r . The ψ , in general, has n-components at each point in space; R(δθ) is then a n × n matrix in the component space, and Eqn. (4.20) is the definition of J in that component space. This definition is used to study the properi i ties of J . For an infinitesimal angle ε, we can write: Rx (ε)  1 − εJ x ; Ry (ε)  1 − εJ y and   i Rz (ε 2 )  1 − ε 2 J x , which on substituting into Eqn. (4.15b) yield: 

i i i 2  i i        1 −  εJ x   1 −  εJ y  −  1 −  ε J z   1 −  εJ y   1 −  εJ x   ψ = 0 (4.21)  

All the zeroth and first-order terms in ε cancel, and the second-order term gives:

 J x , J y  = iJ z (4.22a)

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which is generalized to:  J i , J j  = iε ijk J k (4.22b)



where the symbol ε ijk is equal to +1, if ijk take values in cyclic order (123, 231, 312) and it is equal to −1 when ijk take values that are not in cyclic order. It equals to zero if any two of ijk are assigned the same value, for example, 112. As is seen, the use of Eqns. (4.15b) and (4.18) would give us:  Li , L j  = iε ijk Lk (4.22c)



for orbital angular momentum. Equations (4.22b) and (4.22c) constitute important formulas in quantum mechanics.

4.5  Invariance and Conservation Laws Any of the three transformations (space translation, time translation, and rotation in space) changes a state vector φ , which is imagined to be tied to a system, to a new state vector φ ′ . Thus, the general form of a transformation is: φ ′ = S φ (4.23)



− iap

− iτH

where S is a transforming operator that is equal to e  for space translation, e  for time − iδθ.J translation, and e  for rotation. We thus are considering two systems of coordinates, one before transformation (without prime) and the other after transformation (with prime). In the first system of coordinates, suppose the application of operator F on φ changes it to χ : χ = F φ (4.24a)



The corresponding procedure in the second system of coordinates, let us say, is represented by primed symbols to give: χ ′ = F ′ φ ′ (4.24b)



which, with the use of Eqn. (4.23), implies: S χ = F ′S φ (4.25a)

The use of Eqn. (4.24a) then yields:

SF φ = F ′S φ ⇒ SF = F ′S

(4.25b)

or:

F ′ = SFS−1 and F = S−1 F ′S (4.26)

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Transformations described by Eqn. (4.26) are applicable to all operators including the Hamiltonian operator H . We therefore write: H ′ = SHS−1 (4.27)



where H is Hamiltonian in the system of coordinates before transformation, while H ′ is the Hamiltonian in the system of coordinates after transformation. The invariance of the Hamiltonian under transformations requires that H ′ = H , meaning that a Hamiltonian has the same form whether expressed in terms of original or transformed coordinates of the system. In the following we consider translation and rotation operations. 4.5.1  Infinitesimal Space Translation For space translation S = e − iεp/ , which in the case of an infinitesimal translation is given by: − iεp S = 1− . We apply Eqn. (4.26) to cases of 3-operators: (a) F = x , (b) F = p , and (c) F = H .  Here p is linear momentum component along x-axis. a. x ′ = T −1 (ε)xT (ε) = e − iεp/ xe iεp/ iεp  iεp   iε   1+  x  1 −   x + ( px − xp)  x + ε (4.28a)    



The new position coordinate after translation is x + ε . Thus, the position vector is not conserved during linear motion. b.

p ′ = T −1 (ε)pT (ε)



 − i εp   i εp  =  e   p e      

(4.28b)

εp  εp     1− i  p1+ i       p

where we ignored terms having ε 2 or higher powers of ε. This means that linear momentum remains unchanged during translation in space. In other words, momentum is conserved during the linear motion of the system. Equation (4.28b) can easily be generalized for a system of n-particles by saying that p = p1 + p2 + p3 + p4 + ............ + pn . c.

H ′ = T −1 (ε)HT (ε)  − i ε.p   i ε.p  = e   He       ε.p   ε.p    1− i  H1+ i        H−

i ε.[ p, H ] 



(4.28c)

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where the term having ε 2 has been ignored. The requirement H ′ = H demands dp dp that [ p, H ] = 0 . Use of the Heisenberg equation [ p, H ] = i implies = 0 or p dt dt is the constant of motion. Thus, the invariance of the Hamiltonian –shifting of the system and equipment (as a whole) from one place to another in space should not affect the measurements – demands that linear momentum be conserved under the linear motion of system. Note that Eqns. (4.28b) and (4.28c) convey the same message. 4.5.2  Infinitesimal Time Translation Invariance of the measurements with time requires that H ′ = H . From Eqn. (4.27), we get: iτH   iτH   H′ =  1 −  H1+        (4.29) H



on ignoring terms having τ 2 . This suggests that the total energy of the system is conserved under time translation or there is no change in total energy with time. 4.5.3  Infinitesimal Rotation For rotational motion S = e − iδθ. J/ . We here again consider two cases: a.

(

)(

J′ = e − iδθ. J/ J e iδθ. J/

)

i i       1 − δθ.J J  1 + δθ.J       J

 (4.30a)

where we ignored terms having ( δθ ) . This suggests that the total angular momentum of a system does not change during the rotation motion of the system. b. We next take the transformation: 2



 − iδθ.J   iδθ.J  H′ =  e  He  , (4.30b)       which for the infinitesimal value of δθ reduces to:



iδθ.J   iδθ.J   H′   1 −   H1+       (4.30c) i  H − δθ.[ J, H ]  The invariance of the Hamiltonian under rotational motion (rotating system and measuring equipment as a whole) demands that [ J, H ] should be zero or J and H should commute. Therefore, angular momentum J is conserved or it is a constant of motion under rotational motion.

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4.5.4  Conservation of Charge For the invariance of physical systems under a translation in the electrostatic potential, we require that the total electric charge on a system be a conserved quantity. The physics of a system only depends on potential differences. Quantum mechanically, we can define a charge operator Qˆ which on operating upon a wave function ket ψ q returns to an eigenvalue of q . The ψ q describes a system of total charge q . Qˆ ψq = q ψq (4.31a)



If q is conserved, then Qˆ and H must commute: Qˆ , H  = 0 . It is assured by the invariance   under a global phase (or gauge) transformation like: ˆ ψ q ′ = e iεQ ψ q (4.31b)



where ε is an arbitrary real parameter. The above continuous transformations led to additive conservation laws: The sum of all charges is conserved. 

4.6  Parity and Space Inversion Parity in physics refers to the relationship between an object or process and the mirror image of it. For example, any right-handed object produces a mirror image that is identical to it in every way except that the mirror image is left handed. A moving particle that spins in a clockwise manner will have a mirror image particle that is identical to it in every manner except that it spins anticlockwise. The law of conservation of parity states that for every real object or process there exists a mirror image that obeys the same physical laws. This concept played a very important role in quantum physics. Because of this law, it was inferred that all elementary particles and their interactions possessed mirror image counterparts, and therefore they exist. However, it has also been said and argued that parity may not conserve in weak interactions. It is conserved in the strong nuclear interactions and in the electromagnetic interactions. A parity operator P is an operator that describes the behavior of the wave function of any system of particles when the spatial coordinates ( x,  y ,  z ) of the wave function are reflected through the origin to ( − x,  − y ,  − z ) . The simultaneous flip in the sign of all spatial coordinates is therefore referred as a parity transformation or parity inversion:



 x   −x P :  y  ⇒  − y  z   − z

  (4.32)  

A reflection, mirror image, can be obtained by an inversion followed by rotation. Conversely, an inversion is same as three successive reflections with respect to three axes. Unlike translation and rotation, which can proceed continuously in terms of infinitesimal displacements, it is not possible to transform a system into a mirror image without

Transformations, Conservation Laws, and Symmetries

119

distorting it into a final configuration that is very different physically from the original system. It is therefore not at all clear how quantities like electric charge, which is unrelated to coordinate displacements, should be treated in reflection if symmetry is to be preserved. The equation F   =  m a that equates two vectors for constant mass is invariant under parity in classical mechanics. The law of gravity too involves only vectors and hence it is invariant under parity. Angular momentum L = r × P, which is an axial vector, is also invariant under parity because of P ( L ) = P ( r × P ) = ( − r ) × ( − P ) = L . Classical variables that do not change under parity are the time, the energy of the particle, power, total angular momentum of the particle, the electric potential, the magnetic induction, the magnetization, energy density of the electromagnetic field, the Maxwell stress tensor, all masses, charges, coupling constants, and other physical constants. Classical variables which are negated on spatial inversion include the position, velocity, acceleration and linear momentum of a particle, the force on the particles, the electric field, the electric displacement vector, electric polarization, electromagnetic vector potential, and Poynting vector. 4.6.1  Parity Operator If we ignore the spin motion, then the parity operator P is an inversion operator. The operation of P on the wave function of the system gives:

Pψ (r ) = ψ (−r ) (4.33a)

In the case of a system of many particles, wave function depends on coordinates of n particles, and we write:

Pψ (r1 , r2 , r3 ,......, rn ) = ψ (−r1 , −r2 , −r3 ,......, −rn ) (4.33b)

Also notice that:

P 2 ψ (r ) = Pψ (−r ) = ψ (r ) (4.34)

Let us say that P satisfies the eigenvalue equation:

Pψ (r ) = λψ (r ) (4.35a)

where λ and ψ(r ) are the eigenvalue and eigenfunction of the parity operator. The parity operator and the Hamiltonian can have simultaneous eigenfunctions, if they commute. On operating with P, Eqn. (4.35a) goes to:

P 2 ψ (r ) = λ 2 ψ (r ) = ψ (r ) ⇒ (λ 2 − 1)ψ (r ) = 0

(4.35b)

The nontrivial solution demands that λ = ±1. It is to be noted here that one can also have 2 2 P 2 ψ (r ) = e iϕ ψ (r ) , since an overall phase is unobservable because of e iϕ ψ (r ) = ψ (r ) . For the eigenstate of P that corresponds to λ = 1:

Pψ e (r ) = λψ e (r )  ≡   ψ e (−r ) = ψ e (r ) (4.36)

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Similarly, for the eigenstate of P that corresponds to λ = − 1: Pψ o (r ) = λψ o (r )  ≡   ψ o (−r ) = −ψ o (r ) (4.37)



We thus find that all wave functions, which remain unchanged on replacing r by −r , correspond to λ = 1, while all wave functions that change sign on replacing r by −r correspond to λ = −1. Some of the important properties of parity operators are: (i) It is a Hermitian operator. (ii) It is a unitary operator, and its eigenstates are orthogonal. To prove that it is a Hermitian operator, we require that elements of the matrix representing the parity operator should follow: Pij = Pji*



⇒ ψi P ψ j = ψ j P ψi

*

(4.38)

for all eigenvectors ψ i and ψ j . We take:

Pji* = ψ j P ψ i

*

(

*

∞

−∞

∞

−∞

∞

−∞

∫ f (− x)dx = − ∫ f (x′)dx′ = ∫ f (x′)dx′ , we get:

Letting r ′ = −r and then noting that

Pji* =

)

= ∫ ψ *j (r )Pψ i (r )d 3 r = ∫ ψ *i (−r )ψ j (r )d 3 r (4.39)

∫ ψ (−r)ψ (r)d r * i

j

3

−∞ −∞ −∞



=−

∫ ∫ ∫ ψ (x′, y ′, z′)ψ (− x′, − y ′, − z′)dx′dy ′dz′ * i

j

(4.40)

∞ ∞ ∞

∞ ∞ ∞

=

∫ ∫ ∫ ψ (x′, y ′, z′)Pψ (x′, y ′, z′)dx′dy′dz′ = P * i

j

ij

−∞ −∞ −∞

To prove that P is a unitary operator and its eigenstates ψ e and ψ o are orthogonal, take:

P ψe = ψe ⇒ ψ e P† = ψ e

(4.41a)

and,

P ψ o = − ψ o (4.41b)

Therefore:

ψ e P† P ψ e = ψ e ψ e ⇒ P† P = I

(4.42a)

Transformations, Conservation Laws, and Symmetries

121

and:

ψ e P † P ψ o = − ψ e ψ o (4.42b)



⇒ ψ e ψ o = − ψ e ψ o (4.42c)

which implies that ψ e ψ o = 0, proving that wave functions are orthogonal. As is noticed from Eqns. (4.5), (4.7), and (4.20), the operators which cause linear translation, time translation, rotation, and reflection (parity operator) are unitary operators, because each of these satisfies the condition AA † = 1. Also, these are linear operators. We next discuss the operator that is not a unitary and a linear operator.

4.7  Time-Reversal Operator Let us first consider time reversal in classical mechanics. For example, motion of a charged particle with charge q and mass m in an electric E is described by:

m

d2r = qE(r ) (4.43) dt 2

If r(t) is a solution of Eqn. (4.43), then so is r(−t), as can be seen from the fact that the Eqn. (4.43) is second order in time, so that the two changes of sign coming from t → −t cancel out. However, this does not hold for magnetic forces, where the equations of motion include first order time derivatives:

m

d 2 r q dr = × B(r ) (4.44) dt 2 c dt

To restore the invariance under time reversal, we have to replace B by −B along with t → −t in Eqn. (4.44), which means that the time reversed motion is physically allowed in the reversed magnetic field. We find that when a system interacting with external fields does not possess time reversal invariance by itself, inclusion of the charges and currents producing external fields in our definition of the system may restore the time reversal invariance. If the motions of all the charges in a closed system are reversed, then charge densities and currents transform according to ρ → ρ and J → − J , while the electric and magnetic fields produced by these charges and currents transform according to E → E and B → −B. Thus, electromagnetic effects are invariant under time reversal when these basic transformations along with Eqns. (4.43) and (4.44) are considered. Though, we have shown this only at the classical level, it is true at the quantum level as well. A quantum mechanical equation that is analogous to the equations of motion of the charged particle under an electromagnetic field is the Schrödinger equation:

i

 ∂ψ α (r , t)  −  2 2 = ∇ + V (r )  ψ α (r , t) (4.45) 2 ∂t m  

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Equation (4.45) describes the motion of a system in the quantum state α , at position r and time t. Here, α compositely represents all quantum numbers. As is seen, ψ α (r , t) and ψ *α (r , −t) are solutions but ψ α (r , −t) is not the solution of Eqn. (4.45). It is apparent that the complex conjugate of ψ α (r , t) is necessary, because without it, the left-hand side of Eqn. (4.45) would change sign under t → −t. In the Schrödinger representation, one can write ψ n (r , t) = un (r )e − iEnt/ and ψ *n (r , −t) = un* (r )e − iEnt/ , when system is in the state of energy En . We thus conjecture that the time reversal has something to do with complex conjugation. As was shown in Sections 4.5 and 4.6, under symmetry operations such as translation, rotation, and parity, the inner product of two vectors is preserved, which means that if a symmetry operation changes α to α ′ and β to β ′ , then: β α ′ = β α (4.46)



where, we have abbreviated ψ β ψ α by β α . Eqn. (4.46) states that if α is rotated and β is also rotated in same manner, then β α remains unchanged during rotation. Equation (4.46) arises from the fact that the translation, rotation, and parity operators are unitary operators: α ′ = U α and β ′ = β U † : β α ′ = β U †U α = β α (4.47)



For a time reversed state ψ α (r , −t) = θψ α (r , t) or in short α ′ = θ α , θ being the time reversal operator, we cannot fulfill the requirement Eqn. (4.46). However, a more generalized requirement: β α ′ = β α (4.48)



fulfills the Eqn. (4.46), and at the same time it works equally well for: * β α ′ = β α = α β (4.49)



Let us pursue the possibility shown in Eqn. (4.49), because we have said that time reversal has something to do with complex conjugation, in the case of the Schrödinger equation. Transformations α ′ = θ α and β ′ = θ β are said to be antiunitary if they fulfill criteria Eqn. (4.49). And the operator θ is called an antilinear operator if:

θ(c1 α + c2 β ) = c1*θ α + c2* θ β (4.50)

because for a linear operator, A :

A(c1 α + c2 β ) = c1 A α + c2 A β (4.51)

Thus, an antilinear operator does not commute with a constant because θc = c * θ. We thus infer that the product of two antilinear operators is linear, and the product of a linear with an antilinear operator is antilinear. In a more generalized manner, we say that a product of operators is linear if the number of antilinear factors is even and it is antilinear when the number of antilinear factors is odd.

Transformations, Conservation Laws, and Symmetries

123

The reason that the only quantities which are physically measurable are the absolute squares of scalar products allows us the relaxing of requirement Eqn. (4.46) to Eqn. (4.48). These are the probabilities, which are experimentally measurable. Its relevance for the discussion of symmetries in quantum mechanics is that a symmetry operation presumably must preserve the probabilities of all experimental outcomes. This implies that transformations in quantum mechanics must be implemented by means of either unitary or antiunitary operators. For example, rotations are implemented by means of unitary operators, as are most other symmetries commonly encountered in quantum mechanics. But time reversal is an exception, and it is implemented by means of antiunitary operators. If the time reversed operation transforms an operator A to A′ then A′ = θAθ−1 . (r0 , p0 ) → (r0 , − p0 ) is the initial condition of a motion transform under time reversal in classical mechanics. This allows postulating that the time reversal operator in quantum mechanics satisfies the conjugation relations:

r ′ = θrθ−1 = r (4.52a)



p′ = θpθ−1 = −p (4.52b)

which results in:

L ′ = θLθ−1 = −L. (4.53a).

Further, Eqn. (4.53a) can be generalized to be applicable to all kinds of angular momentum, orbital as well as spin:

J′ = θJθ−1 = − J (4.53b)

which is reasonable, because if we think of a simple model of a charged spinning particle as a charged–rotating sphere, then we see that reversing the motion will reverse both the angular momentum as well as the magnetic field produced by the spin. We now take the canonical commutation relations  xi , p j  = iδ ij and conjugate with θ :

θ  xi , p j  θ−1 =  xi , − p j  = − iδ ij (4.54a)



θ(i)θ−1 = − i ⇒ θ(i) = (− i)θ (4.54b)

Equation (4.54b), which is in conformity with Eqn. (4.50), forces us to conclude that the time reversal operator θ is antilinear, so that i on the right-hand side changes into −i when θ is pulled through it. 4.7.1  Properties of Antilinear Operator In the case of a linear operator A , ( α A ) ( β ) is the same as ( α ) ( A β ) , and we conclude that the positioning of the parentheses is irrelevant; we can write it simply as α A β . In other words, we can think of A as acting either to the right or to left. However, this is not true for an antilinear operator. A complex-valued linear operator on kets and the value of a bra acting on a ket are usual scalar products, and we write α β for bra α . To understand the act of the antilinear operator on bra, let us assume that the action of a given antilinear

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operator θ on kets is known, and we want to define its action on bras. For example, if α is a bra, we wish to define α θ. The antilinear operators cannot be treated like linear operators. α θ, a complex-valued operator acting on kets, is an antilinear operator. However, bras are supposed to be linear operators. Let us introduce a complex conjugation to make α θ a linear operator on kets. Set:

( α θ) ( β ) = ( α ) (θ β ) * (4.55)

The rule says that for an antilinear operator, it does matter whether the operator acts to the right or to the left in a matrix element. On changing the direction in which the operator acts, one must complex conjugate the matrix element. Therefore, in the case of antilinear operators, parentheses are necessary to indicate in which direction the operator acts. We next, consider the Hermitian conjugation for linear and antilinear operators. In the case of † linear operators, the Hermitian conjugate is ( A β ) = β A † for all kets β or equivalently by: †

φα A † ψ β = ψ β A φα (4.56)



A similar definition can also work for antilinear operators, and we can set ( θ β ) = β θ† . Note that θ† is an antilinear operator if θ is antilinear. Now, however, when we try to write an equation analogous to Eqn. (4.56), we must be careful about the parentheses. Thus, we have: †

(

)

α θ† β = ( β θ ) α  (4.57a)



†

or:

(

)

α θ† β = β ( θ α ) (4.57b)



Equations (4.55), (4.57a), and (4.57b) summarize the principal rules for antilinear operators, which differ from those of linear operators. We next consider antiunitary operators. An operator is antiunitary if it is antilinear and it satisfies: θθ† = θ† θ = 1 as has been said earlier. Similar to the case of unitary operators, antiunitary operators too preserve the absolute values of scalar products. Let us say that an antiunitary operator can be written as θ = UK , where U is a unitary operator and K is a complex-conjugation operator, which forms complex conjugation of any coefficient that multiplies a ket from right: Kc α = c * K α (4.58)



We examine the action of K on ψ α , when it is expended in terms of base ket: φα ′ = K φα

∑c u =∑ u φ =K

n

n

n

α

∑u φ u K u =∑ φ u

=K

n

n

n

α

n

(4.59)

n

∗

α

n

n

n

un

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Transformations, Conservation Laws, and Symmetries

∗

Here, we made use of cn = un φα , un φα

= φα un and have taken base ket:

   un =    



    (4.60)   

0  1  0

which is real and does not change under the action of the complex–conjugation operator. We then take: ϕ α ′ = θ ϕ α = UK

∑u =∑ ϕ =



∑u

ϕ α um

m

m

*

m

ϕ α UK um

α

um U um =

(4.61)

m

m

∑U u

m

*

um ϕ α

m

Also, consider: ψ β ′ = UK ψ β = UK =

∑

∑u

ψ β un

n

n

∗

un ψ β UK un =

n

∑

(4.62)

∗

un ψ β U un

n

The Hermitian adjoint of Eqn. (4.62) gives:

( ψ )′ = ∑ u



β

n

ψ β un U † (4.63)

n

Equations (4.61) and (4.63) give:

∑∑ u = ∑∑ u =∑ u u

ψ β ϕα ′ =

n



n

n

ψ β un U †U um um φα

n

ψ β um ϕ α δ nm =

m

m

n

n

∗

∗

∑u

n

ψ β ϕ α un

(4.64)

n

ϕα ψ β = ϕα ψ β = ψ β ϕα

∗

n

which is exactly the same as stated in Eqn. (4.49), and hence, it is proved that an antiunitary operator can be expressed as a product of a unitary operator and a complex-conjugation operator.

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Spinless particles are described in terms of position r , linear momentum p, and orbital angular momentum L . Therefore, requirements given by Eqns. (4.52a) to (4.53b) are satisfied by taking U = 1 and then θ = K : θrθ−1 = KrK −1 = r θpθ−1 = K (− i∇)K −1 = (i∇)KK −1 = −p



θLθ−1 = K (r × p)K −1 = −L

(4.65)

θJθ−1 = θ(L + S)θ−1 = −L − S The Hamiltonian is invariant under time reversal, and hence, the time reversal operator commutes with the Hamiltonian. However, it cannot be said that θ is the constant of motion, dA because the equation i = [ AH − HA] holds only for linear operators. dt

4.7.2  Time Reversal Operator for Non-zero Spin Particles To satisfy Eqns. (4.52a) to (4.53b), we require that spin operator S should change sign under time reversal operation θSθ−1 = −S . The explicit form of θ depends on chosen coordinate  representation. We take standard representation S = σ , where σ represents the Pauli 2 matrices:

 0 σx =   1

 0 1  , σy =  0   i

 1 −i  , and σ z =  0   0

0  (4.66) −1 

As is seen, r , S x and S z are real and p and S y are imaginary, which means that KSx = Sx K , KSy = −Sy K , and KSz = Sz K . On using θ = UK , we obtain:

θSx θ−1 = UKSx K −1U −1 = USx KK −1U † = USxU −1 ⇒ −Sx = USxU †

(4.67a)

and:



θSy θ−1 = UKSy K −1U −1 = −USy KK −1U −1 = −USyU −1 ⇒ Sy = USyU †

(4.67b)

similarly:

−Sz = USzU † (4.67c)

We thus find that transformation with the unitary operator U leaves Sy invariant and changes the sign of Sx and Sz . Therefore, U should represent the rotation of the coordinate

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system around the y-axis through an angle of π . Such a rotation can be represented by iπSy 1 iπS / e y and hence θ = e  K . For a spin particle: 2

iπSy

e



(4.68) = iσ y iπSy

iπSy

iπSy

because iSy or iσ y is real, K commutes with e  or Ke  = e  K . Two successive operations of K on a scalar c give K 2 c = Kc * = c, which means K 2 = 1. Also, operation of U 2 yields a rotation of 2 π through the y-axis. And, the 2 π -rotation leaves the wave function of a system of integer spin (2s is even) unchanged, while the wave function of a system of particles of half-integer spin (2s is odd) changes. This implies that θ2 = 1 for particles of integer spin and θ2 = −1 for a system of a half-integer spin. For the half-integer spin particle θ2 = (iσ y K )(iσ y K ) = −σ 2y = −1 . We have θn = iσ 1y iσ 2 y iσ 3 y iσ 4 y ..... iσ ny K for a system of n particles, and therefore θ2n = (−1)n , which is 1 for the even value of n and −1 for the odd value of n. Again, θ commutes with H and therefore the eigenket ψ α (r , t) is also the eigenket of θn and we can write:

(

)(

)(

)(

) (

)

θn ψ α (r , t) = c ψ α (r , t) (4.69a)



where c is the eigenvalue of θn . Operating on both sides of Eqn. (4.69a) with θn , we get: θ2n ψ α (r , t) = θn c ψ α (r , t)



2

= c ∗θ n ψ α ( r , t ) = c ∗ c ψ α ( r , t) = c ψ α ( r , t ) 2

2

(4.69b)

2

which yields c = 1 , when θ2n = 1 and c = −1 for θ2n = −1 . c = 1 yields eigenvalue c = ±1 2 that could be a possible solution if the system is non-degenerate, for c = −1 results to no possible eigenvalue that can satisfy an eigenvalue equation, which suggests there cannot be degeneracy in the system. The two eigenkets ψ β and θ ψ β are orthogonal in the case of a system of half-integer spin ∗ particles. To prove it, we take β α ′ = β α = α β , where β ′ = θ β and α ′ = θ α . Then:

βα′= αβ ⇒ θβ θα = α β

(4.70a)

Use α = θ β on left of Eqn. (4.70a) to get:

θβ θ2β = θβ β (4.70b)

Since θ2 = −1 for the half-integer spin particle, the left-hand side of Eqn. (4.70b) is equal to − θβ β and hence:

− θβ β = θβ β ⇒ θβ β = 0 (4.70c)

Equation (4.70c) proves that ψ β and θ ψ β are orthogonal.

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4.8  Solved Examples p2 1 2 + kx is invariant 2m 2 under space inversion and its wave function kets can be divided into two categories.

1. Show that the 1D harmonic oscillator Hamiltonian H =

SOLUTION

p2 1 2 + kx remains unchanged on 2m 2 replacing x by −x, as it has x 2 and hence H (− x) = H ( x) . We take H ( x)ψ ( x) = Eψ ( x). On applying space inversion and using H (− x) = H ( x) , we get: Under space inversion, x is changed to −x. The H =

H (− x) ψ (− x) = Eψ (− x)



⇒ H ( x) ψ (− x) = Eψ (− x)

(4.71a)

We thus find that both ψ( x) and ψ (− x) are wave function kets of H ( x) and hence these should be either parallel vectors or a scalar multiple of each other. Therefore, we can write: ψ (− x) = λ ψ ( x) (4.71b)



On replacing x by −x in Eqn. (4.71b), we get: ψ ( x) = λ ψ (− x) = λ 2 ψ ( x) (4.71c)



which yields λ 2 = 1 or λ = ±1. We can thus say that the wave function kets of a 1D harmonic oscillator can be divided into two groups; one of those belongs to λ = 1 and the other corresponds to λ = −1. 2. Show that spherical harmonics Ylm (θ, φ) are eigenfunctions of the parity operator.  SOLUTION

In spherical polar coordinates, the parity operator P changes: r → r , θ → π − θ and φ → π + φ. The operation of P on Ylm (θ, φ) = CPlm (cos θ)e imφ , where C is constant, gives: PYlm (θ, φ) = Ylm (π − θ, π + φ) = CPlm (cos(π − θ))e im( π+φ)

= CPlm (− cos θ)e imφ e imπ l + m imφ

= CPl (cos θ)(−1) m

= (−1) CPl (cos θ)e l

m

e

(4.72a) (−1)

m

imφ

where we used (−1)2m = 1 and Plm (− x) = (−1)l + m Plm ( x) . Therefore,

PYlm (θ, φ) = λYlm (θ, φ) = (−1)l Ylm (θ, φ) (4.72b) Hence λ = (−1)l , which could be +1 or −1 depending on whether l is even or odd.

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Transformations, Conservation Laws, and Symmetries

3. An operator is represented by matrices A and A′ before and after rotation, respectively. If A is Hermitian then show that A′ too is Hermitian. SOLUTION

From Eqn. (4.26), we have: A′ = SAS−1 with S = e − iδθ. j/ (4.73)



and A † = A . The Hermitian conjugate of A′ is:

(

A′ † = SAS−1



) = (S ) †

−1 †

A † S† (4.74)

Since SS−1 = I = SS† , S is a unitary matrix that satisfies S−1 = S† . Hence:

( )

A′ † = S−1

†

A † S†

= SAS−1 = A′



4. Show that the parity operator commutes with the orbital angular momentum operator. SOLUTION

The orbital angular momentum operator is L = r × p. Let us consider a state represented by ψ(r ) . Then: PLψ (r ) = P(r × p)ψ (r ) = ( − r × − p ) ψ (− r )



= LPψ (r ) which implies ( PL − LP ) ψ (r ) = 0 or [ P, L ] = 0 .

4.9 Exercises 1. A particle is described by the wave function ψ (r , t) = Ae i(k.r −ωt ) . Show that ψ(r , t) and ψ * (r , −t) are acceptable solutions of the Schrödinger equation, while ψ (r , −t) is not an acceptable solution.   0 −i  2. Calculate the eigenvalues and eigenvectors of the matrix sy =   and 0  2 i show that its eigenvectors are orthogonal. 3. Show that the time reversal operator commutes with the Hamiltonian. 2

4. Show that probability density P = ψ(r , t) remains invariant under space inversion, rotation, reflection, and time reversal operations.

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5. Matrices representing the rotation by an infinitesimal angle ε around the y-axis and z-axis are as follows:



 ε2  1− 2  Ry ( ε ) =  0   −ε 

0

ε

1

0 ε2 1− 2

0

  ε2   1− 2    and Rz (ε) =  ε       0

Show that Ry (ε) and Rz (ε) do not commute.

−ε ε2 2 0

1−

 0   . 0   1 

5 Angular Momentum Angular momentum plays a central role, both classically as well as quantum mechanically, in understanding the structure of atoms and other systems that involve rotational symmetry centered around the angular momentum. Like other observable quantities, angular momentum, too, is described by an operator in quantum mechanics, and it is a vector operator, like the linear momentum operator. In this chapter, we discuss various properties of angular momentum as a vector operator.

5.1  Orbital Angular Momentum The classical definition of the orbital angular momentum of a particle with position vector r and conjugate linear vector momentum p about the origin is L = r × p. The 3-components of L are Lx = ( ypz − zpy ) , Ly = ( zpx − xpz ) , Lz = ( xpy − ypx ) . Components of the position and linear momentum, as operators, satisfy the fundamental commutation relations: [ pi , p j ] = 0 , [ xi , x j ] = 0 , [ xi , p j ] = iδ ij , where i and j stand for x , y, and z . The commutation relation for Lx and Ly is: [Lx , Ly ] = Lx Ly − Ly Lx

(

)

(

= ypz − zpy ( zpx − xpz ) − ( zpx − xpz ) ypz − zpy

)

= ypz (zpx ) − yxpz2 − z 2 py px + zxpy pz

− zypx pz + z 2 px py + xypz2 − xpz (zpy ) = ypz (zpx ) − xpz (zpy ) + zxpy pz − zypx pz

(5.1)

= − iypx + yzpz px + ixpy − xzpz py + zxpy pz − zypx pz = i(xpy − ypx ) = iLz The commutators  Ly , Lz  and [ Lz , Lx ] can be found by changing x , y  and z in cyclic order to have:

[Lx , Ly ] = iLz , [Ly , Lz ] = iLx , and [Lz , Lx ] = iLy (5.2)

In quantum mechanics, the foundation of the theory of angular momentum is based on three commutation relations given by Eqn. (5.2). Other dynamical variables which are 131

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represented by the above three commuting relations are found to have identical properties to those of the components of an angular momentum. Next take:

L × L = iˆ(Ly Lz − Lz Ly ) + ˆj(Lz Lx − Lx Lz ) + kˆ(Lx Ly − Ly Lx ) ˆ ) = iL ˆ + ˆjL + kL = i(iL x y z

(5.3)

For a system of N-particles with orbital angular momentum vectors Li (where, i runs from 1 to N), each of these vectors satisfies: Li × Li = iLi (5.4)



The orbital angular momentum operators of different particles represent the different degrees of freedom of the system, and hence we expect them to commute with each other. We therefore write: Li × L j + L j × Li = 0 for i ≠ j where i and j refer to different particles. N

For total orbital angular momentum of the system L =

∑L : i

i=1

N

N

L×L=

∑ ∑L Li ×

i=1

j

j=1

N

N



∑

∑

1 Li × Li + (L i × L j + L j × L i ) (5.5) = 2 (i ≠ j) = 1 i=1 N

= i

∑L = iL i

i=1

We thus find that the sum of two or more angular momentum vectors and the total orbital angular momentum of the system satisfies the same commutation relation as a primitive orbital angular momentum vector. An important conclusion drawn from the commutation relations is that the three components of an angular momentum operator cannot be specified (or measured) simultaneously. Once we specify one component, the values of the other two components become uncertain. It has been conventional to specify the z − component Lz . The magnitude squared of the orbital angular momentum L2 and Lz commutes one with the other:

L2 , Lz  =  L2x , Lz  +  L2y , Lz  +  L2z , Lz  (5.6)

To prove this, take:  L2x , Lz  = L2x Lz − Lz L2x

= L2x Lz − Lx Lz Lx + Lx Lz Lx − Lz L2x = Lx ( Lx Lz − Lz Lx ) − ( Lz Lx − Lx Lz ) Lx = − i(Lx Ly + Ly Lx )

(5.7a)

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Similarly:  L2y , Lz  = L2y Lz − Lz L2y

= L2y Lz − Ly Lz Ly + Ly Lz Ly − Lz L2y

(

) (

)

= Ly Ly Lz − Lz Ly + Ly Lz − Lz Ly Ly

(5.7b)

= i(Ly Lx + Lx Ly ) and:

 L2x , Lx  = L2x Lx − Lx L2x =0

(5.7c)

Therefore:

L2 , Lz  =  L2x , Lz  +  L2y , Lz  +  L2z , Lz  (5.8) =0

Similarly we can prove that L2 commutes with L x  and L y , because there is nothing special about the z -axis. This suggests that one can specify the magnitude of an angular momentum vector along with one of its components (by convention, the z -component). The operators L+ = Lx + iLy and L− = Lx − iLy are also used in quantum mechanics. They commute with L2 and they satisfy the relations:



 L+ , Lz  = L+ Lz − Lz L+ = Lx Lz + iLy Lz − Lz Lx − iLz Ly = − iLy − Lx

(5.9a)

= − L+



 L− , Lz  = L− Lz − Lz L− = Lx Lz − iLy Lz − Lz Lx + iLz Ly = − iLy + Lx

(5.9b)

= L− and:  L+ , L−  = L+ L− − L− L+ = (Lx + iLy )(Lx − iLy ) − (Lx − iLy )(Lx + iLy )

= − iLx Ly + iLy Lx − iLx Ly + iLy Lx = −2 i(Lx Ly − Ly Lx ) = 2 Lz

(5.10a)

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From Eqn. (5.8), we then have:

 L2 , L+  =  L2 , L−  = 0 (5.10b)

5.2  Eigenvalues of Angular Momentum Though the procedure presented in this section is used to find eigenvalues for L2  and L z , it is equally applicable to compute the eigenvalues for J2  and J z as well as eigenvalues for S2  and S z . The J  and  S are total angular momentum and spin angular momentum, respectively. We start with the assumption that the simultaneous eigenkets of L2 and Lz are specified by two quantum numbers, l and m , and the wavefunction kets are denoted by l, m . Then the quantum number m is defined by:

Lz l, m = m l, m (5.11)

Thus, m is the eigenvalue of Lz . Since the dimensions of  and Lz are same, it is possible to write such an equation. Because Lz is a Hermitian operator, m has to be a real number. We take:

L2 l, m = λ lm  2 l, m (5.12)

without loss of generality, where λ lm is some real dimensionless function of l and m . We next take:

l, m (L2 − L2z ) l, m = (λ lm  2 − m2  2 ) l, m l, m = λ lm  2 − m2  2

(5.13)

where we have assumed that l, m is a normalized ket. Also

l, m (L2 − L2z ) l, m = l, m (L2x + L2y ) l, m = l, m L2x l, m + l, m L2y l, m

(5.14)

We know that for any general ket, φ and a Hermitian operator A , ξ = A φ and ξ ξ = φ A † A φ = φ A 2 φ ≥ 0 , since the norm square of a vector can never be negative. Therefore, from Eqn. (5.13), we conclude that m2 ≤ λ lm . We next consider the effect of the L+ on l, m . L2 , L+  = 0 implies that L2 L+ = L+ L2 . Therefore:

L2 L+ l, m = L+ L2 l, m =  2 λ lm L+ l, m

(5.15)

From Eqns. (5.12) and (5.15) we note that both l, m and L+ l, m belong to the same eigenvalue of L2 . Thus, the action of the shift operator L+ on a wave function ket does not affect the magnitude of the angular momentum. Further:

Lz L+ l, m = (Lz Lx + iLz Ly ) l, m (5.16)

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Angular Momentum

Equation (5.2) gives Lz Lx = Lx Lz + iLy and Lz Ly = Ly Lz − iLx . Therefore, Eqn. (5.16) reduces to: Lz L+ l , m = (Lx Lz + iLy + iLy Lz + Lx ) l , m = (L+ Lz + L+ ) l, m



= L+ m l, m + L+ l , m

(5.17a)

= (m + 1)L+ l, m We thus find that L+ l, m is an eigenket of operator Lz having the eigenvalue (m + 1) . Further note that from Eqn. (5.11), we get: Lz l, m + 1 = (m + 1) l, m + 1 (5.17b)



Equations (5.17a) and (5.17b) imply that L+ l, m and l, m + 1 are collinear vectors and hence we write: + L+ l, m = clm  l, m + 1 (5.18)



+ where clm is a number. We thus infer from the above discussion that when the operator + L acts on a simultaneous eigenstate of L2 and Lz , the eigenvalue of L2 is not changed, while the eigenvalue of Lz is increased by  . For this reason, L+ is termed a raising operator. Adopting a procedure similar to that used above, we find: − L− l, m = clm  l, m − 1 (5.19)



Hence, L− is called a lowering operator. The shift operators step the value of m up and down by unity each time, operating on one of the simultaneous eigenstates of L2 and Lz . At first sight, it appears that any value of m can be obtained by applying the shift operators for enough numbers of times. However, as per m2 ≤ λ lm , there is a definite upper bound to the values that can be taken by m2 . This upper bound is determined by the eigenvalue of L2 . It follows that there is a maximum and a minimum possible value which can be assigned to m . Let us say that mmax and mmin represent upper and lower cut-offs. Now, there is no state with m > mmax ; any attempt to raise the value of m above its maximum value mmax will result to zero and hence: L+ l, mmax = 0 (5.20a)

This implies that:

L− L+ l, mmax = 0 (5.20b)



(

) (

)

But L− L+ = (Lx − iLy )(Lx + iLy ) = L2x + L2y + i(Lx Ly − Ly Lx ) = L2x + L2y − Lz , which means that:

(L

2 x

)

+ L2y − Lz l, mmax = 0 (5.20c)

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We write Eqn. (5.20c) as follows:

(L

2 x



)

+ L2y + L2z l, mmax = (L2z + Lz ) l, mmax

⇒ L2 l, mmax = (L2z + Lz ) l, mmax = mmax (mmax + 1) 2 l, mmax

(5.21)

⇒ L2 l, mmax = mmax (mmax + 1) 2 l, mmax Comparison between Eqns. (5.12) and (5.21) gives λ lmmax = mmax (mmax + 1) . We, however, note that the successive operation of L− on a wave function ket q , mmax generates q , mmax − 1 , q , mmax − 2 , etc. Application of a lowering operator leaves unchanged the eigenvalue of L2 ; hence all these states must correspond to the same value of λ lm , namely mmax (mmax + 1) . Thus, L2 l, m = mmax (mmax + 1) 2 l, m (5.22)



Therefore, an unknown quantum number l can be equated to mmax , without any loss of generality and we write: L2 l, m = l(l + 1) 2 l, m (5.23)

and then:

{

}

l, m L− L+ l, m = l, m L2 − (L2z + Lz ) l, m



=  2 {l(l + 1) − m(m + 1)}

(5.24)

However, we also have: + l, m L− L+ l, m = l, m L− clm l, m + 1 + − = l, m clm L l, m + 1



+ − cl , m + 1 l , m l , m =  2 clm

(5.25)

+ − cl , m + 1 =  2 clm

Equations (5.24) and (5.25) yield:

+ − clm cl , m+ 1 = l(l + 1) − m(m + 1) (5.26)

Now take: l, m L− l, m + 1 = l, m Lx l, m + 1 − i l, m Ly l, m + 1

=  l, m + 1 L†x l, m + i l, m + 1 L†y l, m 

†

=  l, m + 1 Lx l, m + i l, m + 1 Ly l, m 

†

(5.27)

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Here, we used the fact that Lx and Ly are Hermitian. Eqn. (5.27) further simplifies to:



l, m L− l, m + 1 =  l, m + 1 L+ l, m 

( )

⇒ cl−, m+ 1 = cl+, m

*

†

(5.28)

Equations (5.26) and (5.28) can be combined to give:

2

+ clm = l(l + 1) − m(m + 1) (5.29a)

which yields:

+ clm = l(l + 1) − m(m + 1) (5.29b)

It is to be noticed that cl+, m is undetermined to an arbitrary phase factor; we can replace cl+, m , given above, by cl+, m e iφ where φ is real, and we still satisfy Eqn. (5.28). An arbitrary but convenient choice that cl+, m is real and positive has been made here. This is equivalent to choosing relative phases of the wave function kets l, m . Similar calculation yields:

(

− clm = cl+, m− 1

)

*

= l(l + 1) − m(m − 1) (5.30)

The inequality m2 ≤ λ lm implies that m has a maximum and a minimum possible value. The maximum value of m is denoted by l . To find minimum value, let us start to lower the value of m below its minimum value mmin . Since, there is no state below the state having mmin , we must have:

L− l, mmin = cl−, mmin l, mmin = 0 (5.31)

which implies that cl−, mmin = 0 . Therefore, l(l + 1) − mmin (mmin − 1) = 0 or mmin = −l. We therefore conclude that m takes a ladder of discrete values, each rung differing from its immediate neighbors by 1, with the top rung at l and the bottom rung at −l . The two possible choices for   l are as follows: (i) it is an integer allowing m to take values −l, …. − 2, −1, 0, 1, 2,….l , or (ii) it is a half-integer which allows m to take the values - l , …  −3 / 2,  −1/ 2, 0, 1/ 2, 3 / 2, …l. We will prove in the next section that l  can be assigned only take integer values. With the use of fundamental commutation relations along with the fact that Lx , Ly and Lz are Hermitian operators, we obtained that eigenvalues of L2 are given by l(l + 1) 2 , where l is an integer, or a half-integer. Also, we have shown that the eigenvalues of Lz can only take the values m where m lies in the range of −l, −l + 1,..., −1, 0, 1, ... l − 1, l . A normalized simultaneous wave function ket of L2 and Lz , belonging to the eigenvalues l(l + 1) 2 and m , respectively is l, m . With the use of Eqns. (5.18) and (5.29b), we write:

L+ l, m =  l(l + 1) − m(m + 1) l, m + 1 (5.32a)

Use of Eqns. (5.19) and (5.30) gives:

L− l, m =  l(l + 1) − m(m − 1) l, m − 1 (5.32b)

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5.3  Eigenfunctions of Orbital Angular Momentum Three components of orbital angular momentum, expressed in Cartesian coordinates ( x , y , z) , are:

 ∂ ∂  − z  (5.33a) Lx = − i  y ∂y   ∂z



∂  ∂ Ly = − i  z − x  (5.33b)  ∂x ∂z 

and,

 ∂ ∂ − y  (5.33c) Lz = − i  x ∂x   ∂y

Use of standard spherical polar coordinates: x = r cos φ sin θ, y = r sin φ sin θ, (5.34) z = r cos θ



gives:

 ∂ ∂ Lx = i  sin φ + cot θ cos φ  (5.35a) ∂θ ∂φ  



 ∂ ∂ Ly = − i  cos φ − cot θ sin φ  (5.35b) ∂θ ∂φ  



Lz = − i

∂ . (5.35c) ∂φ

The shift operators, L± and L2 are then expressed as follows:

 ∂ ∂ L± = ± e ± iφ  ± i cot θ  (5.36a) ∂φ   ∂θ

and

 1 ∂  1 ∂2  ∂ L2 = −  2   (5.36b)  sin θ  + sin 2 θ ∂φ2  ∂θ  sin θ ∂θ

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The eigenvalue equation for L2 is given by Eqn. (5.23). On combining Eqns. (5.23) and (5.36b), we get:

 1 ∂   ∂  1 ∂2  + l(l + 1)  l, m = 0 (5.37) sin θ  +  2 2 ∂θ  sin θ ∂φ   sin θ ∂θ  

Equation (5.37) suggests that l, m depends on l, m, θ and φ, which is represented by Ylm (θ, φ) . We then have:

  ∂  ∂  ∂2  2  sin θ sin θ  + 2  + l(l + 1)sin θ  Ylm (θ, φ) = 0 (5.38) ∂θ  ∂θ  ∂φ   

It is to be noted that a square integrable solution of Eqn. (5.38), which has to be single valued in real space, can exist if and only if l  is an integer. Also, by writing Ylm (θ, φ) = Θ(θ)Φ(φ) , Eqn. (5.38) becomes:

sin θ ∂  ∂Θ  1 ∂2 Φ 2 (5.39)  + l(l + 1)sin θ = −  sin θ Θ ∂θ ∂θ Φ ∂φ2

In Eqn. (5.39), the left-hand side is a function of θ while the right-hand side is a function of φ. This can be valid if and only if each side is equal to a constant value. Hence, we write: −

∂2 Φ ⇒ 2 + m2 Φ = 0 ∂φ

whose normalized solution is: Φ =



1 ∂2 Φ = m2 Φ ∂φ2

sin θ

(5.40)

1 imφ e . We thus have: 2π

∂  ∂Θ  2 2  sin θ  + l(l + 1)sin θ − m Θ = 0 (5.41) ∂θ  ∂θ 

{

}

By taking ξ = cos θ , Eqn. (5.41) converts to:

 d  m2  2 dP  (1 ) ( 1) + + − l l P = 0 (5.42) − ξ 1 − ξ 2  dξ  dξ  

whose solutions are the associated Legendre functions Plm (cos θ) . The solution of Eqn. (5.38) therefore is given by Ylm (θ, φ) = Clm Plm (cos θ)e imφ . Since L2 and Lz commute, Ylm (θ, φ) should also be the eigenfunction of Lz . We thus have:

LzYlm = − i

∂ Clm Plm e imφ = mYlm (5.43) ∂φ

(

)

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The normalization condition: π 2π

∫∫Y

* lm



(θ, φ)Yl′ m′ (θ, φ)sin θdθdφ = δ ll′ δ mm′ (5.44)

0 0

along with the requirement Yl , − m = (−1)m Ylm* , yields: Ylm (θ, φ) =



(2l + 1)(l − m)! (−1)m Plm (cos θ)e imφ (5.45) 4π(l + m)!

m is an integer that lies in the range −l ≤ m ≤ l . Thus, the wave function ψ ( r , θ, φ ) = R(r )Ylm ( θ, φ ) , where R(r ) is a radial function, has all the expected features of the wave function of a simultaneous eigenstate of L2 and Lz belonging to the quantum numbers l and m . Ylm (θ, φ) are known as spherical harmonics. Note that a spherical harmonic wave function Ylm (θ, φ) is symmetric about the z -axis (independent of φ) whenever m = 0 , and is spherically symmetric whenever l = 0 (since Y00 = 1/ 4π ). Therefore, by solving directly for the eigenfunctions of L2 and Lz , in Schrödinger’s representation, one can reproduce all the results of Section 5.2. However, the results of Section 5.2 are more generalized in nature than those obtained in this section, because they still apply when the quantum number l takes on half-integer values. Some of the spherical harmonics are as follows:

Y00 =

1 4π

 3  Y10 =    4π 

1/2

 5  Y20 =   16π 

1/2

( 3 cos

2

1/2

 3  sin θe iφ ;  Y1, −1 =    8π 

1/2

sin θe − iφ

)

θ−1

(5.46)

1/2

15 ( sin θ cos θ ) e iφ ;  Y2, −1 =   8π

1/2

 15  sin 2 θe 2 iφ ;  Y2, −2 =   32 π 

 15  Y21 = −    8π   15  Y22 =   32 π 

 3  cos θ ;  Y11 = −    8π 

1/2

1/2

( sin θ cos θ ) e − iφ

sin 2 θe −2 iφ

5.4  General Angular Momentum The theory of orbital angular momentum, as discussed in Section 5.2, suggests that quantum number m varies between −l   to   l , each value of m differing from immediate neighbors by unity, and there can be two possibilities of values of l . If angular momentum is taken as L = r × P , then the single-valued real space solution of L2Ylm (θ, φ) =  2 l(l + 1)Ylm (θ, φ) demands that l should take integer values. But, the general treatment presented in

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Angular Momentum

Section 5.2 suggests that l can be assigned the half-integer values as well, with m to take the values - l , …  −3 / 2,  −1/ 2, 0, 1/ 2, 3 / 2, …l. This would be true if particles have, in addition to orbital angular momentum ( r × P ), some intrinsic angular momentum that does not depend on spatial coordinates. There exists a wealth of experimental evidence which suggests that the quantum mechanics cannot be completely specified by giving the wave function ψ as a function of the spatial coordinates only. There exists intrinsic angular momentum called spin angular momentum, and the vector sum of orbital and spin angular momentums is known as the total angular momentum of system. Therefore, the general angular momentum is defined by J = L + S , where S represents the spin angular momentum. Such an angular momentum operator commutes with the Hamiltonian, and consequently is constant of motion. For a particle with spin, total angular momentum in the rest frame is non-vanishing. The commutation relations given by Eqns. (5.2) and (5.3) are redefined for J and its components as follows: [ J x , J y ] = iJ z , [ J y , J z ] = iJ x , and [ J z , J x ] = iJ y (5.47a)

and:

J × J = iJ (5.47b)



Equations (5.8) and (5.10) can also be generalized to:

[ J2 , J x ] = [ J2 , J y ] = [ J2 , J z ] = 0 (5.47c)



[ J2 , J + ] = [ J2 , J − ] = 0 (5.47d)

where J + = J x + iJ y and J − = J x − iJ y . The eigenvalue equations for J2 and J z are then given by: J2 j, m = j( j + 1) 2 j, m (5.48a)

and:

J z j, m = m j, m (5.48b)



where j can take both integer and half integer values. The half integer values are 1 3 5 1 1 3 0, , , ,…… and m varies between − j, − j + 1,  − j + 2, ……− , 0, ,  ,…. j − 1,  j. 2 2 2 2 2 2 The generalization of Eqns. (5.32) give:

J + j, m =  j( j + 1) − m(m + 1) j, m + 1 (5.49a)

and:

J − j, m =  j( j + 1) − m(m − 1) j, m − 1 (5.49b)

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The matrix elements for J2 , J z , J + , J − are given by:

j′ , m′ J2 j, m = j( j + 1) 2δ j′jδ m′m (5.50a)



j′ , m′ J z j, m = mδ j′jδ m′m (5.50b)



j ′ , m′ J + j, m =  j( j + 1) − m(m + 1)δ jj′ δ m′ , m+ 1 (5.50c)



j ′ , m′ J − j, m =  j( j + 1) − m(m − 1)δ jj′ δ m′ , m− 1 (5.50d)

The matrix element of J − J + is obtained from Eqn. (5.24) as: j, m J − J + j, m =  2 { j( j + 1) − m(m + 1)} (5.51)



(J =

+

+ J−

) and J

(J+ − J− ) . Hence, the matrix elements of J x and 2i 2 J y can be obtained from that of J + and J − . The matrix form of J2 , J z , J + , J − can be deduced with the use of Eqns. (5.50). As is obvious from these equations, matrices representing J2 and J z are diagonal matrices, while matrices that represent, J + , J − , J x  and  J y are not diagonal. Example matrices for J2 , J z , J + , J − for j = 1 , are given by: It is to be noted that J x

 22  J ≡ 0  0  2



  J ≡    −

0 2 2 0

0

0

2

0

0

2

0 0 2 2

y

      , Jz ≡  0   0 

=

0 0 0

0 0 −

 0   , J+ ≡    0     0

2 0 0

  2  and  0  0

0   0 . 0 

5.5  Spin Angular Momentum The Stern-Gerlach experiment pointed to another source of magnetic moment, which is proportional to the spin angular momentum M s ∝ S . From this and other experiments, it was concluded that each elementary particle has intrinsic angular momentum, known as spin angular momentum, and it is denoted by S , which is different from the orbital angular momentum. Spin is a new degree of freedom in addition to the spatial coordinates (x ,  y ,  z ). Unlike the spatial coordinates, spin only takes a discrete set of values. The proton also has spin of equal magnitude, but the magnetic momentum due to the proton spin is much smaller and it is neglected in the experiments. Since S is an angular momentum, it too satisfies the commutation relations:

Sx , Sy  = iSz , Sy , Sz  = iSx , and [ Sz , Sx ] = iSy (5.52a)

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Angular Momentum

and: S × S = iS (5.52b)



Spin angular momentum has many properties in common with orbital angular momentum. However, there is one important difference. Unlike orbital angular momentum, spin angular momentum operators cannot be expressed in terms of position (r ) and linear momentum ( p) operators, since this identification depends on an analogy with classical mechanics. The concept of spin has no analogy in classical physics; it is a purely quantum mechanical concept. Consequently, the restriction that the quantum number l must take integer values is lifted for spin angular momentum, and the quantum number s can take halfinteger values. All relations for angular momentum discussed in Sections 5.1 and 5.2 are applicable to S . Equations (5.47) for S are:

S2 s, sz = s( s + 1) 2 s, sz (5.53a)

and Sz s, sz = sz  s, sz (5.53b)



where quantum number s can, in principle, take integer or half-integer values, and the quantum number sz can only take the values − s, − s + 1....., −1, 0, 1,........s − 1, s. There is a fixed magnitude of spin vector for each elementary particle, which is given by the quantum number s . However, the projection of the spin onto one axis, generally chosen to be the z − axis, is needed in addition to the coordinates (or momenta) to fully specify the state of the particle. To have a complete description of spin, one requires relativistic quantum mechanics. The magnetic moment of a charged particle is contributed by spin angular momentum. It is predicted from relativistic quantum mechanics that a particle of charge q with spin must possess a magnetic moment µ s = gqS / 2 m , where g = −2.0023 for the electron. 5.5.1  Pauli Theory of Spin One-half Systems Particles such as electrons, protons, and neutrons have spin s =

1 . In such a case, a projection 2

1 1 and sz = − . One therefore introduces a 2 2 new variable ε , which takes two values +(↑) and −(↓) . Thus, a full set of coordinates to 1 describe a spin particle is represented by ( x , y , z, ε) , where spin operators S2 , Sz , S+ , S− 2 act only on the spin variable and therefore they commute with an operator that acts on the (x , y , z) -space. A complete set of commuting operators in spin variable space is formed by S2 and Sz . The eigenvectors of S2 and Sz are specified by:

onto the z-axis can take only two values, sz =





+ =

− =

1 1  1 1 ,  s = , sz =  (5.54a)  2 2 2 2

1 1  1 1 , −  s = , sz = −  (5.54b) 2 2  2 2

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The eigenvectors are orthonormal + − = 0 and + + = − − = 1. A generalized spin state is: χ = c + + + c − − (5.55)

Equations (5.53) are reduced to:

3 2  ± (5.56a) 4



S2 ± =



Sz ± = ±

 ± (5.56b) 2

By choosing + and − as basis vectors for spin space, we write:  c+  χ =  −  (5.57a)  c 



particularly when:  1   0  and − =  + =   (5.57b).  0   1 



Then, all spin operators S2 , Sz , S+ , S− are represented by 2 × 2 matrices. To find the matrix for Sz , we write:

Sz + =

 a  + ⇒ 2  c

b   1   1  = (5.58a) d   0  2  0 

Sz − = −

 a  − ⇒ 2  c

 0  b  0  =−  (5.58b)    d  1  2  1 

and:

which gives: a =  / 2 , d = −  / 2 and c = b = 0 . Hence:

 1 where σ z =   0

Sz =

 1 2  0

0   = σ z (5.58c) −1  2

0  . Equations (5.48) for spin angular momentum become: −1  S+ s, sz =  s( s + 1) − sz ( sz + 1) s, sz + 1 (5.59a)

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Angular Momentum

and S− s, sz =  s( s + 1) − sz ( sz − 1) s, sz − 1 (5.59b)

Therefore:

 a S+ + = 0 ⇒   c

b   1   0  = (5.60a) d   0   0 

 a S+ − =  − ⇒   c

 0  b   0  =   (5.60b) d   1   1 

and

Solution of Eqns. (5.60) gives:

 0 S+ =    0

1  (5.61a) 0 

 0 S− =    1

0  (5.61b) 0 

Similarly, we obtain:

Since S+ = Sx + iSy and S− = Sx − iSy , we have:

Sx =

 0 2  1

1   = σ x (5.62a) 0  2

Sy =

 0 2  i

−i   = σ y (5.62b) 0  2

and

 0 The matrices σ x =   1

 0 1  , σy =   0   i

 1 −i  , and σ z =   0   0

0  are known as Pauli −1 

matrices, because Pauli was the first person to recognize the need of two component state vectors to explain some of the features of experimentally observed atomic spectrum. It is easy to verify that:

σ 2x = σ 2y = σ 2z = 1, (5.63a)



σ x σ y = −σ y σ x = iσ z , σ y σ z = −σ z σ y = iσ x , and σ z σ x = −σ x σ z = iσ y (5.63b)

We thus see that Pauli matrices anticommute and the square of each matrix is equal to the unit matrix.

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5.6  Addition of Angular Momentum Addition of angular momentum operators is an important problem in quantum mechanics. It could be the addition of orbital angular momentum and spin angular momentum or the addition of general angular momentum of two particles (systems). We consider the case of addition of two sets of angular momentum operators J1 and J2 , which are Hermitian and obey the fundamental commutation relations:

J1 × J1 = iJ1 and J2 × J 2 = iJ 2 (5.64)



 J ni , J nj  = iε ijk J nk (5.65)

where n takes values 1 or 2 . The (i, j,  k ) stand for (x , y , z ) and the symbol εijk is equal to +1 if ijk take values in cyclic order (123, 231, 312), and it is equal to -1 when ijk take values not in cyclic order. It is equal to zero if any two of ijk are assigned the same value, for example 112. Also, we have:

[ J2n , J nx ] = [ J n2 , J ny ] = [ J n2 , J nz ] = 0 (5.66)

We further say that the two groups of operators correspond to different degrees of freedom of the system, and therefore:

 J1i , J 2 j  = 0 (5.67)

The J1 and J2 can also be an orbital angular momentum operator and a spin angular momentum operator, or the orbital angular momentum operators of two different particles in a multi-particle system. From Eqns. (5.47), we write:

J12 j1 , m1 = j1 ( j1 + 1) 2 j1 , m1 (5.68a)



J1z j1 , m1 = m1  j1 , m1 (5.68b)



J22 j2 , m2 = j2 ( j2 + 1) 2 j2 , m2 (5.69a)

and

J 2 z j2 , m2 = m2  j2 , m1 (5.69b)

1 3 5 where ji ( i = 1, 2 ) can take the values 0, , , ,…… and mi  ( i = 1, 2 )  varies between 2 2 2 1 1 3 − ji , − ji + 1,  − ji + 2, …… − , 0, ,  ,…. ji − 1,  ji . 2 2 2 Let us define the total angular momentum operator as:

J = J1 + J2 (5.70a)

with:

J × J = iJ (5.70b)

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Angular Momentum

J1 and J2 are Hermitian operators and therefore J too is a Hermitian operator. Thus, J possesses all the expected properties of an angular momentum operator, like: J2 j, m = j( j + 1) 2 j, m (5.71a)

and

J z j, m = m j, m (5.71b)



At this stage, we do not know how j and m can be expressed in terms of j1 , j2 , m1 and m2 . We 2 2 2 write J = J1 + J2 + 2 J1 .J2 . And then with the use of Eqns. (5.64) and (5.65), we have:  J2 , J12  =  J2 , J22  = 0 (5.72)



This implies that the quantum numbers j1 , j2 , and j can all be measured simultaneously, which means that one can know the magnitude of the total angular momentum together with the magnitudes of the component angular momenta. However:  J2 , J1z  ≠ 0, and  J2 , J 2 z  ≠ 0 (5.73)



which states that it is not possible to measure the quantum numbers   m1 and   m2 simultaneously with the quantum number j . Thus, we cannot determine the projections of the individual angular momenta along the z − axis and the magnitude of the total angular momentum simultaneously. From the above discussions, it became clear that we form two separate groups of mutually commuting operators: (i) one group is J12 , J22 , J1z , J 2 z and (ii) another group is J12 , J22 , J2 , J z , which are incompatible with one another. We can define simultaneous wave function kets for each operator group. Let us denote the simultaneous wave function kets for J12 , J22 , J1z , J 2 z by j1 , j2 , m1 , m2 . We then have:

(

(

)

(

)

)



J12 j1 , j2 , m1 , m2 = j1 ( j1 + 1) 2 j1 , j2 , m1 , m2 (5.74a)



J22 j1 , j2 , m1 , m2 = j2 ( j2 + 1) 2 j1 , j2 , m1 , m2 (5.74b)



J1z j1 , j2 , m1 , m2 = m1  j1 , j2 , m1 , m2 (5.74c)



J 2 z j1 , j2 , m1 , m2 = m2  j1 , j2 , m1 , m2 (5.74d)

(

)

Similarly, simultaneous wave function kets for J12 , J22 , J2 , J z are taken as j1 , j2 , j, m to write:

J12 j1 , j2 , j, m = j1 ( j1 + 1) 2 j1 , j2 , j, m (5.75a)



J22 j1 , j2 , j, m = j2 ( j2 + 1) 2 j1 , j2 , j, m (5.75b)



J2 j1 , j2 , j, m = j( j + 1) 2 j1 , j2 , j, m (5.75c)



J z j1 , j2 , j, m = m j1 , j2 , j, m (5.75d)

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Each wave function ket is complete and normalized. Also, j1 , j2 , m1 , m2 and j1 , j2 , j, m are mutually orthogonal. Since, the operators J12 and J22 are common to both operator groups, we assume that the quantum numbers j1 and j2 are known and hence we can always determine the magnitudes of the individual angular momenta. Additionally, one can either know the quantum numbers m1   and m2 , or the quantum numbers j and m , but one cannot know both pairs of quantum numbers at the same time. We write a conventional completeness relation for both sets of wave function kets:

∑

j1 , j2 , m1 , m2 j1 , j2 , m1 , m2 = 1 (5.76a)

m1 , m2

∑ j , j , j, m



1

2

j1 , j2 , j, m = 1 (5.76b)

j,m

The right-hand sides of Eqns. (5.76) denote the identity operator in the ket-space, corresponding to the states of given j1 and j2 . The summation is taken over all allowed values of m1 , m2 , j, and m . The incompatibility between two groups of operators means that if the system is in a simultaneous eigenstate of the former group, then, in general, it is not in an eigenstate of the latter, which also means that if the quantum numbers j1 , j2 , j, and   m are known with certainty, then a measurement of the quantum numbers m1 and m2 will give a range of possible values. The completeness relation (Eqn. 5.76a) allows us to write:

j1 , j2 , j, m =

∑

j1 , j2 , m1 , m2 j1 , j2 , j, m j1 , j2 , m1 , m2 (5.77)

m1 , m2

We thus find that wave function kets of the first group of operators are the weighted sum of wave function kets of the second group. 5.6.1  Clebsch-Gordon Coefficients and their Properties The weights j1 , j2 , m1 , m2 j1 , j2 , j, m are known as Clebsch-Gordon coefficients, which means that if the measurements on J12 , J22 , J2 and J z in a state of the system are bound to give results, j1 ( j1 + 1) 2 , j2 ( j2 + 1) 2 , j( j + 1) 2 , and m , respectively, then the measurements over J 1z 2 and J 2z will yield results m1 h and m2 h , with the probability of j1 , j2 , m1 , m2 j1 , j2 , m1 , m2 . The Clebsch-Gordon coefficients have many important properties. 1. The coefficients are zero unless m = m1 + m2 . To prove it note that:

( J z − J1z − J 2 z ) j1 , j2 , j, m = 0 ⇒ (m − m1 − m2 ) j1 , j2 , j, m = 0

(5.78)

On operating with j1 , j2 , m1 , m2 we get:

(m − m1 − m2 ) j1 , j2 , m1 , m2 j1 , j2 , j, m = 0 (5.79) which proves the assertion, suggesting that z -components of different angular momenta are added algebraically.

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2. Coefficients vanish unless j1 − j2 ≤ j ≤ j1 + j2 . We assume that j1 ≥ j2 . Since the largest values that can be assigned to m1 and m2 are j1 and j2 , the largest possible value m can take is j1 + j2 . Further, m varies between j and − j , and therefore j too can take the largest value equal to j1 + j2 . Also, the number of values m1 and m2 can take are (2 j1 + 1) and (2 j2 + 1) , respectively. Therefore, allowable independent wave function kets values of j1 , j2 , m1 , m2 to span the entire ket-space are (2 j1 + 1)(2 j2 + 1) . Since j1 , j2 , j, m span the same ket-space, (2 j1 + 1)(2 j2 + 1) must also be the number of independent j1 , j2 , j, m kets. As is seen: j1 + j2

∑(2 j + 1) = (2 j + 1)(2 j



1

2

+ 1) (5.80)

j1 − j2

which suggests that if j1 + j2 is the maximum value of j then j1 − j2 is its minimum 1/2 value. It should be noted from Eqn. (5.70a) that j = j12 + j22 + 2 j1 j2 cos θ , which takes a maximum value j1 + j2 when vectors are parallel and a minimum value j1 − j2 when vectors are anti-parallel. 3. The sum of the modulus squared of all of the Clebsch-Gordon coefficients is unity, which means:

(

∑



j1 , j2 , m1 , m2 j1 , j2 , m1 , m2

2

)

= 1 (5.81)

m1 , m2

From Eqn. (5.77), we write: j1 , j2 , j, m j1 , j2 , j, m =

∑

j1 , j2 , m1 , m2 j1 , j2 , j, m j1 , j2 , j, m j1 , j2 , m1 , m2

m1 , m2



⇒1=

∑

(5.82)

2

j1 , j2 , m1 , m2 j1 , j2 , j, m

m1 , m2

where we have made use of the fact that j1 , j2 , j, m are normalized kets.

5.6.2  Recursion Relations for Clebsch-Gordon Coefficients Two recursion relations are followed by Clebsch-Gordon coefficients. Consider the lowering and rising operators J − = J1− + J 2− = J1x − iJ1y + J 2 x − iJ 2 y and J + = J1+ + J 2+ = J1x + iJ1y + J 2 x + iJ 2 y . We write:

(



(

)

J − j1 , j2 , j, m =

∑ (J

m1′ , m2′

− 1

) (

)

(

)

+ J 2− ) j1 , j2 , m1′ , m2′ j1 , j2 , m1′ , m2′ j1 , j2 , j, m (5.83)

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or:



( j( j + 1) − m(m − 1)) j1 , j2 , j, m − 1 1/2 = ∑ ( j1 ( j1 + 1) − m1′ (m1′ − 1))  j1 , j2 , m1′ − 1, m2′

j1 , j2 , m1′ , m2′ j1 , j2 , j, m

m1′ , m2′

+

∑ ( j (j 2

2

+ 1) − m2′ (m2′ − 1))

1/2

(5.84)

 j1 , j2 , m1′ , m2′ − 1 j1 , j2 , m1′ , m2′ j1 , j2 , j, m

m1′ , m2′

Operating both sides from the left with j1 , j2 , m1 , m2 , we get:



( j( j + 1) − m(m − 1)) j1 , j2 , m1 , m2 j1 , j2 , j, m − 1 1/2 j1 , j2 , m1 + 1, m2 j1 , j2 , j, m = ( j1 ( j1 + 1) − m1 (m1 + 1)) 1/2 j1 , j2 , m1 , m2 + 1 j1 , j2 , j , m + ( j2 ( j2 + 1) − m2 (m2 + 1))

(5.85)

On replacing J − by J + in Eqn. (5.83) and then following a similar procedure we obtain:



( j( j + 1) − m(m + 1)) j1 , j2 , j, m1 , m2 j1 , j2 , j, m + 1 1/2 j1 , j2 , m1 − 1, m2 j1 , j2 , j, m = ( j1 ( j1 + 1) − m1 (m1 − 1)) 1/2 j1 , j2 , m1 , m2 − 1 j1 , j2 , j, m + ( j2 ( j2 + 1) − m2 (m2 − 1))

(5.86)

Equations (5.85) and (5.86) are two important recursion relations to compute the ClebschGordon coefficients. 5.6.3  Computation of Clebsch-Gordon Coefficients Let us write matrix elements j1 , j2 , m1 , m2 j1 , j2 , j, m as m1 m2 jm in short. The m1 m2 jm can have (2 j1 + 1)(2 j2 + 1) values because m1 and m2 have (2 j1 + 1) and (2 j2 + 1) values, respectively. Therefore, the number of elements in the matrix of Clebsch-Gordon coefficients will be (2 j1 + 1)(2 j2 + 1) . Depending upon the value of m , this matrix further breaks into submatrices of smaller size. For example, when j = j1 + j2 and m = j1 + j2 they are going to be a submatrix of 1 × 1 . There will a submatrix of 2 × 2 if m = j1 + j2 − 1 and j = j1 + j2 or j = j1 + j2 − 1 . The rank of these submatrices increases to reach a maximum value and then it decreases to 1. Generally, the first 1 × 1 submatrix is chosen to be +1: m1 m2 jm = j1 , j2 j1 + j2 , j1 + j2 = 1 (5.87)



To compute the next 2 × 2 submatrix, take m1 = j1 and m2 = j2 − 1 , m = j1 + j2 , and j = j1 + j2 . Then from Eqn. (5.85) we get:

( j1 + j2 )1/2

j1 , j2 − 1 ( j1 + j2 ),( j1 + j2 − 1) = ( j2 )

 j2  ⇒ j1 , j2 − 1 ( j1 + j2 ),( j1 + j2 − 1) =   j1 + j2 

1/2

1/2

j1 , j2 ( j1 + j2 ),( j1 + j2 ) (5.88)

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Angular Momentum

Similarly, when m1 = j1 − 1 and m2 = j2 , m = j1 + j2 , and j = j1 + j2 , we get:  j1  j1 − 1, j2 ( j1 + j2 ),( j1 + j2 − 1) =   j1 + j2 



1/2

(5.89)

The Eqns. (5.88) and (5.89) describe two elements for the case of j = j1 + j2 and m = j1 + j2 − 1. The other two elements are j1 , j2 − 1 ( j1 + j2 − 1), ( j1 + j2 − 1) and j1 − 1, j2 ( j1 + j2 − 1), ( j1 + j2 − 1) , which are evaluated with the use of the unitary nature of the transformation matrix, Eqn. (5.81). We obtain: j1 , j2 − 1 ( j1 + j2 − 1),( j1 + j2 − 1) ( j1 + j2 ),( j1 + j2 − 1) j1 , j2 − 1 +



( j1 + j2 ),( j1 + j2 − 1) j1 − 1, j2 j1 − 1, j2 ( j1 + j2 − 1),( j1 + j2 − 1) = 0

Since m1 m2 jm = jm m1 m2



*

(5.90)

matrix elements are real and positive, Eqn. (5.90) reduces to:

 j2  j1 , j2 − 1 ( j1 + j2 − 1),( j1 + j2 − 1)   j1 + j2 

1/2

 j1  j − 11 , j2 ( j1 + j2 − 1),( j1 + j2 − 1)   j1 + j2 

1/2

+ (5.91) =0

Conventionally, matrix elements are real and positive, and hence we can write:

 j2  j1 − 1, j2 ( j1 + j2 − 1),( j1 + j2 − 1) = −   j1 + j2 



 j1  j1 , j2 − 1 ( j1 + j2 − 1),( j1 + j2 − 1) =   j1 + j2 

1/2

(5.92)

1/2

(5.93)

The results of Eqns. (5.88), (5.89), (5.92), and (5.93) can be summarized as follows: jm m1 j1

j1 − 1

m2

j2 − 1

j2

j1 + j2 , j1 + j2 − 1

 j2   j + j  1 2

1/2

 j1   j + j  1 2

1/2

j1 + j2 − 1, j1 + j2 − 1

 j1   j + j  1 2

−

1/2

 j2   j + j  1 2

1/2

In a similar manner one can get the submatrices of 3 × 3 and 4 × 4 for the values of j and m .

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5.7  Solved Examples 1. Show that matrices:



Lx =

 0   1 2  0

1 0 1

0 1 0

  0 , L =   i  y 2   0

−i 0 i

0 −i 0

  1  , and L =   z   0   0

0 0 0

0 0 −1

   

satisfy the relation  Lx , Ly  = iLz . SOLUTION

Lx Ly − Ly Lx =

 0 2  1 2   0

1 0 1

0 1 0

=

 i 2  0 2   i

0 0 0

−i 0 −i

=

 2i 2  0 2   0

0 0 0

 0   i   0

0 0 −2i

−i 0 i

0 −i 0

  −i 2 −    2  0   i

0 0 0

  1  = i ×     0   0

  0 2 −    2  i   0 −i 0 i 0 0 0

−i 0 i

0 −i 0

 0   1   0

1 0 1

0 1 0

   

    0 0 −1

   

= iLz It can therefore be said that the above matrices give 3 × 3 matrix representation of components of orbital angular momentum.  0   0   1  2. Show that the vectors −1 =  0  , 0 =  1  , 1 =  0  are the wave func       1   0   0  tion kets for the matrix representing Lz in the above example 1. SOLUTION

The eigenvalue equation for Lz is Lz Y = m Y . We therefore have:





 1 Lz −1 =   0  0  1 Lz 0 =   0  0

0 0 0

0 0 0 0 0 −1

0 0 −1

 0   0    = −    0   0  = −  −1 (5.94a)   1   1 

 0   0    = 0×    1   1  = 0 ×  0 = 0 (5.94b)   0   0 

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Angular Momentum

and  1 Lz 1 =   0  0



0 0 0

0 0 −1

 1   1    =   (5.94c) 0    0 = 1   0   0 

3. For a system, the wave function ket φ satisfies L2 φ = l(l + 1) 2 φ and Lz φ = m φ . Calculate ∆Ly ==

{L

2 y

− Ly

}

2 1/2

for the state of φ .

SOLUTION

Since Lz is Hermitian, we have Lz φ = m φ and φ Lz = m φ . Therefore: Lz = φ Lz φ = m φ φ = m (5.95a)



[ Lz , Lx ] = iLy

implies: 1 φ Lz Lx − Lx Lz φ i 1 =  φ Lz Lx φ − φ Lx Lz φ  (5.95b) i

Ly =

= − im  φ Lx φ − φ Lx φ  =0 Further, symmetry in x − y plane permits us to write L2x = L2y and then: 1 2 Lx + L2y 2 1 2 = L − L2z 2 1 1 = φ L2 φ − φ L2z φ 2 2

L2y =

(

)

(

)

(5.96a)

Hence: L2y =



1 l(l + 1) − m2  2 (5.96b) 2

(

)

Therefore: ∆Ly =

{L

2 y

− Ly

}

2 1/2

 l(l + 1) − m2  =   2  

1/2

(5.97)

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1 e iφ sin θ + 2 cos θ f (r ) , where 2 2π 2 f (r ) r 2 dr = 1. (a) What are the possible results of a measurement on Lz in this

(

4. The state of an electron is described by ψ = ∞

∫

)

0

state? (b) Calculate the expectation value of Lz . (c) Is ψ normalized? SOLUTION

3 3 cos θ, and Y11 (θ, φ) = − sin θe iφ , 4π 8π

a. From Eqn. (5.46), we know that Y10 (θ, φ) = 1 3

which gives ψ =

(

)

2Y10 − Y11 f (r ) . Therefore, the given state has l = 1 and

m = 1 and 0 . The possible results of measurements on Lz are + and 0 . b. The expectation value of Lz is: Lz = ∫ ψ * Lz ψd 3 r ∞ π 2π

1 = 3 =

∫∫∫ f (r)

2

r 2 dr

0 0 0 π 2π

∫∫ (

 3

2Y10 − Y11

0 0 π 2π

 = 3

∫∫ Y

11

0 0

2

(

) ( *

2Y10 − Y11 Lz

) ( −Y *

11

)

2Y10 − Y11 sin θdθdφ

) sin θdθdφ

2 sin θdθdφ − 3

(5.98)

π 2π

∫∫Y Y

* 10 11

sin θdθdφ

0 0

2  Y11 Y11 − Y10 Y11 3 3  = 3 In deriving the above results, we have used the fact that spherical harmonics are normalized and orthonormal. =

c. The condition of normalization is ψ ψ = ∫ ψ * ψd 3 r = 1 ψ ψ = ∫ ψ * ψd 3 r = =

∞ π 2π

1 3

∫∫∫ f (r)

1 3

∫∫ (

1 = 3

2

r 2 dr

0 0 0 π 2π

2Y10 − Y11

0 0 π 2π

∫∫ 2 Y

10

0 0

2

(

2Y10 − Y11

)( *

)(

)

*

2Y10 − Y11 sin θdθdφ

)

2Y10 − Y11 sin θdθdφ

1 sin θdθdφ + 3

π 2π

∫∫ Y

11

0 0

2

2 sin θdθdφ − 3

2 1 2 = Y10 Y10 + Y11 Y11 − ( Y10 Y11 + Y11 Y10 3 3 3 =1 We thus find that ψ is normalized.

π 2π

∫∫ (Y Y

* 10 11

)

+ Y11*Y10 sin θdθdφ

0 0

) (5.99)

155

Angular Momentum

1 , consider an operator A = 3Sx + 4Sz , where Sx and Sz are 2 spin angular momentum operators. Express A in matrix form and then find the energy eigenvalues and eigenvectors of the particle.

5. For a particle of spin

SOLUTION

The eigenvalue equation is A ψ = λ ψ . The matrices representing Sz and Sx are: Sz =



  0 0   and Sx = σ x =  −1  2 2 1

  1 σz =  2 2 0

1   (5.100) 0 

Therefore:

 0 A=3  2 1



 1 1  +4  0  2 0

  2  0 =   −1   3  2 

   (5.101)  −2    3 2

The eigenvalues are given by:

2 − λ

3 2

3 2

−2  − λ

=0

 9 2  = 0 (5.102) ⇒ − ( 2  − λ )( 2  + λ ) −   4 



25 2 4 5 ⇒λ=± 2 ⇒ λ2 =

Two eigenvalues are:

 x11  5 5 and λ 2 = − . The eigenvector that corresponds to λ 1 is X 1 =  , 2 2  x21  components of which are given by: λ1 =



3   x11  =0  −9   x21  (5.103) ⇒ − x11 + 3 x21 = 0 3 x11 − 9 x21 = 0

  −1  2 3

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2

2

which along with the condition of normalization x11 + x21 = 1 gives: x11 = ±

3 1   and  x21 = ± 10 10 1  3    10  1 

or   X 1 = ±

(5.104)

 x12  5 Similar calculations for the eigenvector X 2 =   that belongs to λ 2 = − 2  x22  yield: X2 = ±



1  1    (5.105) 10  3 

6. Components of arbitrary vectors A and B commute with those of σ . Show that:

(σ.A )(σ.B) = A.B + iσ.(A × B)

SOLUTION



Ax   0  + 0   iAy



0  Ax

( σ.A ) = 

0   Az  = − Az   A +

− iAy   Az  + 0   0

A− − Az

 , 

and  B z  B+

( σ.B ) = 

B− − Bz

  where A ± = Ax ± iAy ; B± = Bx ± iBy . We therefore have:   A B + A − B+ z z  A + Bz − Az B+

Az B− − A − Bz   (5.106) A + B− + Az Bz 

( σ.A )( σ.B ) = 



A.B = Ax Bx + Ay By + Az Bz  Ax Bx + Ay By + Az Bz = 0  



  Ax Bx + Ay By + Az Bz   0

and, iσ.(A × B) = i  σ x ( Ay Bz − Az By ) + σ y ( Az Bx − Ax Bz ) + σ z ( Ax By − Ay Bx )  0 i   0 1  i 0  =  ( Ay Bz − Az By ) +   ( Az Bx − Ax Bz ) +   ( Ax By − Ay Bx ).  i 0  −1 0   0 −i   i( Ax By − Ay Bx ) Az B− − Bz A− =  Az B+ − Bz A+ − i( Ax By − Ay Bx ) 

   

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Angular Momentum

Therefore:  A B + A − B+ z z A.B + iσ.(A × B) =   − B+ Az + Bz A +



− Bz A − + B− Az   (5.107) A + B− + Az Bz 

Hence (σ .A )(σ .B) = A.B + iσ .(A × B). 1 is described by the Hamiltonian 2 H = 2S1 .S2 , where S1 and S2 are two spin vectors. Calculate the energy eigenvalues of the Hamiltonian.

7. A system of two particles each having spin −

SOLUTION

Let us say that s, m is the eigenvector which satisfies S2 s, m = s( s + 1) 2 s, m , where s can take two values, 1 and 0 . s = 1 is a triplet and symmetric state, while s = 0 , is a singlet and anti-symmetric state. For triplet states, S2 s, m = 2  2 s, m and for a singlet state, S2 s, m = 0 . Now: S2 = (S1 + S2 )2 = S12 + S22 + 2S1 .S2

(



)

= S12x + S12y + S12z + S22x + S22y + S22z + 2S1 .S2

(5.108)

2

  = 6   + 2S1 .S2  2 Therefore: H s, m = 2S1 .S2 s, m 3   =  S2 −  2  s, m  2 



(5.109)

3   =  s( s + 1) 2 −  2  s, m  2  which gives the eigenvalue of H . It is equal to

2 for each of the triplet states and 2

3 2 for the singlet state. 2 There are four states, three of which belong to (s = 1, triplet states) and one 1 belongs to a singlet state ( s = 0). The energy of each of triplet state is  2 , which 2 corresponds to the eigenvectors 1, 1 ; 1, 0 and 1, −1 , respectively. For the singlet −

state, the energy is −

3 2 and the corresponding eigenvector is 0, 0 . 2

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8. Evaluate  L2x , Ly  and  Lx , L2y  . SOLUTION

 L2x , Ly  = L2x Ly − Ly L2x = L2x Ly − Lx Ly Lx + Lx Ly Lx − Ly L2x = Lx  Lx , Ly  +  Lx , Ly  Lx



= i ( Lx Lz + Lz Lx )

(5.110)

= i ( 2 Lx Lz + Lz Lx − Lx Lz ) = −  2 Ly + 2 iLx Lz and  Lx , L2y  = Lx L2y − L2y Lx = Lx L2y − Ly Lx Ly + Ly Lx Ly − L2y Lx =  Lx , Ly  Ly + Ly  Lx , Ly 



(

= i Lz Ly + Ly Lz

(5.111)

)

2

= −  Lx + 2 iLy Lz 9. Calculate Clebsch-Gordon coefficients for a system having J1 =

1 1 and J 2 = . 2 2

SOLUTION

We must add the angular momenta of two spin one-half systems, such as two 1 electrons at rest. We know that m1 = m2 = and 0 ≤ j ≤ 1 . Hence, j has two values, 2 1 and 0 . The combination of two spin one-half systems would give either a spinzero system or a spin-one system. The m has values 1, 0,  −1, for j = 1 , while m = 0 when j = 0 . Therefore, the Clebsch-Gordon coefficients form a 4 × 4 matrix. Also, the Clebsch-Gordon coefficients exist only when m = m1 + m2 . Non-zero ClebschGordon coefficients m1 , m2 j, m for j = 1 are: 1 1 1 1 1 1 1 1 , 1, 1 , , − 1, 0 , − , 1, 0 and − , − 1, −1 . When j = 0 and m = 0 , 2 2 2 2 2 2 2 2 1 1 1 1 two nonzero possible values are − , 0, 0 and , − 0, 0 . 2 2 2 2 Equation (5.87) yields

1 1 1 1 , 1, 1 = − , − 1, −1 = 1 . From Eqns. (5.88) and 2 2 2 2

1 1 1 1 1 1 (5.89), we have − , 1, 0 = and − , 1, 0 = . The Eqn. (5.93) gives 2 2 2 2 2 2

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Angular Momentum

1 1 1 1 1 1 − , 0, 0 = − and , − 0, 0 = . Other elements of the 4 × 4 matrix are 2 2 2 2 2 2 zero because for these we cannot fulfill the condition: m = m1 + m2 . Following Eqn. (5.77), the above results can also be represented by matrix equations as follow:

      



1, 1 1, 0 1, −1 0, 0

       =       

1 0 0 0

0 1 2 0 1 2

0 1 2 0 1 − 2

0   0   1   0  

           

1 1 , 2 2 1 1 ,− 2 2 1 1 − , 2 2 1 1 − ,− 2 2

       (5.112)     

10. Show that Sx Sy + Sy Sx = 0 . SOLUTION

We know that Sx =

  0 σx =  2 2 1

  0 1   and Sy = σ y =  0  2 2 i

Sx Sy + Sy Sx =



−i   . Therefore: 0 

2 σ x σ y + σ y σ x (5.113) 4

(

)

Let us evaluate:



 0 σxσy + σyσx =   1  i =  0

1  0  0  i

−i   0 + 0   i

0   −i + −i   0

−i   0  0  1

0  =0 i 

1   0 

(5.114)

Therefore Sx Sy + Sy Sx = 0 .

5.8 Exercises L+ L− L2z 1. For an axially symmetric rotator, the Hamiltonian is given by H = + . 2 I1 2 I 2 What are its eigenvalues? 2. The wave function of the particle moving under the influence of a spherically

(

)

1/2 −α x 2 + y 2 + z 2

symmetric potential is given by ψ (r ) = β ( x + y + 3 z ) e where α and β are constants. Calculate L2 ψ(r ) . Is ψ(r ) the eigenfunction of L2 ? If yes, find the value of l . [Hint: write ψ(r ) and L2 in terms of spherical polar coordinates.]

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A Textbook on Modern Quantum Mechanics

3. Use J z jm = m and J2 jm = j( j + 1) 2 and then show that J x = J y = 0 and 1 J x2 = J y2 =  j( j + 1) − m2   2 . 2  0 1  and show that tr(Sx ) = 0 and determi4. Calculate eigenvalues of Sx =  2  1 0  2 nant Sx = − . 4 1/2 −β( x 2 + y 2 + z 2 ) 5. The wave function ψ ( x , y , z) = Aze describes one quantum state of a system. Is it a definite angular momentum state? If yes, what are the values of l and m ? 6. Use Lx = ypz − zpy and evaluate [ Lx , x ] ,  Lx , y  , [ Lx , z ] ,  Lx , px  ,  Lx , py  and

(

)

 Lx , pz  . Show that  Lx , r 2  = 0 and  Lx , p 2  = 0 . 7. Use Eqns. (5.11), (5.23), (5.32a), and (5.32b) to find the matrix representations for L2 , L z , L x and L y for l = 1 . 2B 8. The state of an electron is described by the wave function ψ ( r , θ, φ ) = sin θ cos φf (r ), 4π ∞ with

∫ f (r ) 0

2

r 2 dr = 1. (a) Use the normalization condition to find the value of con-

stant, B. (b) Express the wave function in terms of spherical harmonics. (c) What are the possible results of measurement on Lz ? 9. Consider the electron described in Exercise 8. (a) Find the probability of obtaining each of the two possible results of measurement on Lz . (b) Calculate the expectation value of Lz .

6 Schrödinger Equation for Central Potentials and 3D System In Chapter 2, we solved the Schrödinger equation for one-dimensional problems. All physics problems cannot be understood using one-dimensional solutions. In this chapter, we would solve the Schrödinger equation and discuss it for three-dimensional systems such the hydrogen atom, the 3D potential well, and the cubic box. The hydrogen atom and three-dimensional potential wells are approximated by spherically symmetric potentials, which do not depend on the direction of the position vector.

6.1  Motion in a Central Field The Hamiltonian for a particle of mass m moving in a spherically symmetric potential is written as: H=



p2 + V (r ) (6.1) 2m

where V (r ) is the spherically symmetric potential. In Schrödinger’s representation, p = − i∇. For spherically symmetric systems, it is more convenient to work in spherical polar coordinates ( r , θ, φ ) . After expressing ∇ 2 in terms of spherical polar coordinates, we get:

H=−

∂ 2 1  ∂  2 ∂   1 ∂  1 ∂2   + + θ r sin   + V (r ) (6.2)       ∂θ  sin 2 θ ∂2 φ   2 m r 2  ∂r  ∂r   sin θ ∂θ 

With the use of Eqn. (5.36b), Eqn. (6.2) goes to:

H=

2  1 ∂  2 ∂  L2  − 2  + 2 2  + V (r ) (6.3)  r  ∂r  r  2 m  r ∂r

Three components of orbital angular momentum, L x , L y , and L z take the form of partial derivative operators in the angular coordinates, when written in terms of spherical polar coordinates; see Eqns. (5.35). As has been shown by Eqn. (5.8), L x , L y , and L z commute with L2 . Therefore, it follows from Eqn. (6.3) that L x , L y , and L z commute with the Hamiltonian, H , and hence:

[L, H ] = 0 (6.4) 161

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Also, L2 is a function of only ( θ, φ ) ; it does not depend on r . We conclude that:

L2 , H  = 0 (6.5)

This means that both orbital angular momentum L and its squared L2 are constants of the motion, which is expected for a spherically symmetric potential. Let us now take the energy eigenvalue problem:

H ψ = E ψ (6.6)

Since L2 and Lz commute with each other and with the Hamiltonian, it is always possible to represent the state of the system in terms of the simultaneous eigenvectors of L2 , Lz , and H . The most general form for the wave function, which would be a simultaneous eigenvector of L2 , Lz , and H is:

ψ (r , θ, φ) = Rl (r )Ylm (θ, φ) (6.7)

substitution of which in Eqn. (6.3) along with the use of L2Ylm (θ, φ) = l(l + 1) 2Ylm (θ, φ) gives:



  2  1 d  2 d  l(l + 1)   − 2 + V (r ) − E  Rl (r ) = 0 (6.8)  r  +    2 dr r   2 m  r dr 

Since there remains only one variable, the partial derivative with respect to r has been changed to a full derivative. Let us rewrite Eqn. (6.8) as:

1 d  2 dRl  2 m E − Veff (r ) Rl (r ) = 0 (6.9a) r + r 2 dr  dr   2

(

)

with:



 l(l + 1) 2  (6.9b) Veff (r ) =  V (r ) + 2 mr 2  

Veff (r ) is an effective potential, which is the sum of two terms. The term l(l + 1) 2 / 2 mr 2 represents the effect of centrifugal force in the equivalent one-dimensional problem. It is often called the centrifugal barrier potential and makes a dominatingly larger contribution to Veff (r ) for r → 0 . When V (r ) is the coulombic attractive potential, the centrifugal barrier potential is negligibly small at large values of r . Equation (6.9a), which is equivalent to the one-dimensional Schrödinger equation, is known as the Sturm-Liouville equation for the function Rl (r ) . From the general properties of this type of equation, it is well known that if Rl (r ) is required to be well behaved at r = 0 and as r → ∞ then solutions can only exist for a discrete set of values of E , which are the energy eigenvalues. In general, the energy eigenvalues can depend on the quantum number l but are independent of the quantum number m .

Schrödinger Equation for Central Potentials and 3D System

163

6.2  Energy Eigenvalues of the Hydrogen Atom We here consider the case of the hydrogen atom depicted in Fig. 6.1: V (r ) = −



e2 (6.10) 4πε 0 r

Thus for the hydrogen atom, we have:

 2  1 d  2 dRl  l(l + 1)Rl   e2  Rl = 0 (6.11) − 2 − E +  r  + 2 2µ  r dr 4πε 0 r  dr r  

Note that in writing radial Eqn. (6.9a) for the hydrogen atom, m is replaced by µ because of the following: In the case of the hydrogen atom, both the electron having mass me and the proton of mass mp rotate about a common center. This is equivalent to a particle of mass µ rotating about a fixed point, where µ = me mp /(me + mp ) is termed the reduced mass of the electron. Hydrogen also has two isotopes, known as deuterium and tritium, having reduced masses of the electron, µ = 2 me mp /(me + 2 mp ) and µ = 3me mp /(me + 3mp ) , respectively. Since me 0) , to avoid the unphysical behavior of f ( y ) for y → 0 . This is possible if and only if the expression under the bracket on the left-hand side of Eqn. (6.19) goes to zero at kmin so that no value of c k having k < kmin exists, which means that [ kmin ( kmin − 1) − l(l + 1)] = 0 , and hence the very first term in the series is c kmin y kmin . There are two possibilities: kmin = −l or kmin = l + 1. The choice kmin = −l predicts unphysical behavior of the wave function at y = 0 . We therefore conclude that kmin = l + 1 is the only physically acceptable solution. As is noted from Eqn. (6.17), there is a finite probability of finding the electron at the nucleus, for the state with l = 0 . Whereas for l > 0 state, there is zero probability of finding the elec2 tron at the nucleus ( ψ = 0 at r = 0 ). Also, note that it is only possible to obtain sensible behavior of the wave function for r → 0 , if l is an integer. c k +1 y The ratio of successive terms in the series (Eqn. 6.17) is . However, from Eqn. (6.19): ck



ck +1 ck

2µe 2 a 4πε 0  2 = (6.20) {k(k + 1) − l(l + 1)} 2k −

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Schrödinger Equation for Central Potentials and 3D System

∑

ck +1 y 2y 2 and → for k → ∞ . The series (2 y )k /k ! converges to e 2 y , and k ck k k 2y the ratio of two successive terms in the series of e 2 y is also equal to . We therefore conk e r /a 2y clude that f ( y ) → e as y → ∞ . It then follows from Eqns. (6.15) and (6.16) that R(r ) → r as r → ∞ . This does not permit us to have physically acceptable behavior of the wave func2 tion for r → ∞ , as ∫ ψ d 3 r will not be finite. The only way to avoid this unphysical behavior of the wave function is to terminate series (Eqn. 6.17) at some maximum value of k , which means c kmax + 1 = 0 . To make cn+1 zero, we find from the recursion relation (Eqn. 6.20) that: which tends to

µe 2 a = n (6.21) 4πε 0  2



And, the last term of the series is cn y n . Equation (6.21) with the use of a = −  2 /2µE gives the quantized energy eigenvalues: En = −



µe 4 (6.22) 2(4πε 0 )2  2 n2

and:

a=

4πε 0  2 n = 5.29n × 10−11 meters (6.23) µe 2

where n is a positive integer that must exceed the quantum number l ( n > l ); otherwise there would be no terms in the series (Eqn. 6.17), as is seen from nmin = l + 1 . Thus, the maximum value that can be assigned to l is n − 1 . Since l is a positive integer, the minimum value that can be assigned to n is 1. n takes values 1, 2, 3,…., in general. It is referred as a principal quantum number. A quantum state is completely specified by the set of the quantum numbers ( n, l, m ) , where quantum numbers are restricted to follow m ≤ l < n .  4πε 0  2  = 5.29 × 10−11 meters is known as the Bohr radius. The a0  = 2 µe  

6.3  Wave Functions of the Hydrogen Atom The Rl (r ) is a solution of radial Eqn. (6.11). Equation (6.16) with the use of Eqn. (6.21) becomes:

d2 f df l(l + 1) f 2 n + f = 0 (6.24) −2 − dy 2 dy y2 y

On substituting f ( y ) = y l + 1 w( y ) , we obtain:

 d2  d + 2(n − l − 1)  w( y ) = 0. (6.25)  y 2 + 2(l + 1 − y ) dy dy  

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Let us compare this equation with the equation that is satisfied by the associated Laguerre polynomial Lkp (ρ) of order k and degree ( p − k ) :  k  d2 d ρ dρ2 + ( k + 1 − ρ) dρ + ( p − k )  Lp (ρ) = 0. (6.26)  



where k and p are integers. Note that on taking 2y = ρ , 2l + 1 = k and n + l = p , and then replacing w(ρ) by  L2nl++l1 (ρ) Eqns. (6.25) become Eqn. (6.26). We therefore conclude that w(ρ) = L2nl++l1 (ρ) , which is the associated Laguerre polynomial of order (2l + 1) and of degree (n − l − 1) . Thus the radial function is: Rnl (ρ) = Ne −ρ/2 ρl L2nl++l1 (ρ) (6.27)



where N is the normalization constant. An associated Laguerre polynomial is related to a Laguerre polynomial as follows: p

 d Lpp − q ( x) = (−1)p   Lq ( x) (6.28)  dx 

where:

 d Lq ( x) = e x    dx 



q

( x e ) (6.29) q x

is the Laguerre polynomial. To determine N in Eqn. (6.27), we use the normalization condition:

∫ψ

* nlm

(r , θ, φ)ψ nlm (r , θ, φ)d 3 r = 1

∞



⇒

π 2π

∫

2

Rnl (r ) r 2 dr

0

∫∫

2

Ylm (θ, φ) sin θdθdφ = 1

(6.30)

0 0

π 2π

Since

∫∫ Y

lm

2

(θ, φ) sin θdθdφ = 1 , we get:

0 0

∞

∫R (r)r dr = 1 (6.31) 2 nl



2

0

ρa which on substituting Rnl (ρ) from Eqn. (6.27) with r = , where a is given by Eqn. (6.23), 2 goes to: 3



∞

 4πε 0 n 2  2 −ρ 2 l 2 l + 1 2 2  2µe 2  N e ρ Ln+ l ρ dρ = 1 (6.32a)

∫ 0

The associated Laguerre polynomials satisfy: ∞



∫

2

ρ2 l + 2 e −ρ L2nl++l1 (ρ) dρ = 2 n

0

{(n + l)!}3 (6.32b)

(n − l − 1)!

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Schrödinger Equation for Central Potentials and 3D System

Therefore:  4πε 0 n 2  2 2n  ( n + l )!  2µe 2  N (n − l − 1)! = 1 (6.33a) 3

3

which yields:

 2µe 2  3 ( n − l − 1)!   N = ±  3 2  4πε 0 n  2 n {( n + l )!}   

1/2

(6.33b)

and the normalized radial wave functions are given by:

 2µe 2  3 ( n − l − 1)!   Rnl (r ) =  3 2  4πε 0 n  2 n {( n + l )!}   

1/2

e −ρ/2 ρl L2nl++l1 (ρ) (6.34)

The normalized wave function for the hydrogen atom then is:



 2µe 2  3 ( n − l − 1)!   ψ nlm (r , θ, φ) =  3 2  4πε 0 n  2 n {( n + l )!}     2  3 ( n − l − 1)!   =  3   na0  2 n {( n + l )!}   

1/2

1/2

e −ρ/2 ρl L2nl++l1 (ρ)Ylm (θ, ϕ) (6.35) l

 2r  2l+1  2r  e − ( r/na0 )  Ln+ l  Ylm (θ, ϕ)  na0   na0 

The Ylm (θ, ϕ) are spherical harmonics given by Eqn. (5.45). The ground state of hydrogen belongs to n = 1 ,   l = 0, and m = 0 . Thus, the ground state of the hydrogen atom is a spherically symmetric, zero angular momentum state. The ground state energy eigenvalue is:

E1 = −

µe 4 = −13.6 electron volts. (6.36) 2(4πε 0 )2  2

and the corresponding eigenvector is ψ 100 or 1, 0, 0 . The ground state is thus a nondegenerate state. The next energy state corresponds to n = 2 . The other quantum numbers take the values l = 0; m = 0 and l = 1; m = −1, 0, 1 . Thus, there are 3-states with non-zero angular momentum for n = 2 . Note that the energy levels given by Eqn. (6.22) are independent of the quantum numbers l and m , though radial eigenfunctions given by Eqn. (6.34) depend on l . This is a special property of a 1/ r Coulomb potential. There are four eigenvectors, 2, 0, 0 ; 2, 1, −1 ; 2, 1, 0  and  2, 1, 1 , which correspond to the same energy eigenvalue µe 4 E2 = − . Thus, there is four-fold degeneracy for   n = 2 . For n  =  3 , there are 9 8(4πε 0 )2  2 states all together: l   =  0 gives one, l   =  1 gives 3 and l   =  2 yields 5 different m values. In fact there are n2 degenerate states for principal quantum number n. The n2 is the sum of the first n odd integers: n− 1



∑

(2l + 1) = 2

l= 0

n− 1

∑l + n = n(n − 1) + n = n (6.37) 2

l= 0

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The wave functions for some of the lower values of n, calculated with the use of Eqns. (6.35) and (5.45), are: 1/2



ψ 100 ( r ) =

1 1  a0  πa0 



ψ 200 ( r ) =

1  1  4 a0  2 πa0 

ψ 210 ( r , θ, φ ) =

1  1  4 a0  2 πa0 

1/2



1/2



1  1  ψ 21, ±1 (r , θ, ϕ) = 4 a0  πa03 



ψ 300 (r , θ, ϕ) =

e − r/a0 . 1/2

 r  − r/2 a0 .  2 − a  e 0  r  − r/2 a0 cos θ  a  e 0 re − r/2 a0 (sin θ)e ± iϕ

2  1  3 a0  12 πa0 

1/2

 3r 2 r 2  − r/3 a0 (6.38)  1 − 3na + 27 a 2  e 0 0

where a0 is the Bohr radius, defined above.

6.4  Radial Probability Density The probability of finding an electron in the volume element of d 3 r is given by:

2

2

2

ψ nlm (r , θ, φ) d 3 r = Rnl (r ) r 2 dr Ylm (θ, φ) sin θdθdφ (6.39)

The probability of finding the electron in a thin spherical shell having radii between r and r + dr is then given by: π 2π

2



P(r )dr = Rnl (r ) r 2 dr

∫∫ Y

lm

2

(θ, φ) sin θdθdφ

0 0

(6.40)

2

= Rnl (r ) r 2 dr where we used the property of normalization of spherical harmonics. Radial probability 2 P(r ) = Rnl (r ) r 2 is defined as the probability of finding an electron at distance r from nucleus. The maximum probability density for the ground state of a hydrogen atom exists at a dP distance where 10 = 0 . Hence: dr

(



)

(

)

d 4 d 2 −2 r/a0 2 R10 r 2 = 3 r e =0 dr a0 dr  2 r 2  −2 r/a0 =0 ⇒  2r − e a0  

(6.41)

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Schrödinger Equation for Central Potentials and 3D System

which gives r = a0 . Thus, the maximum probability of finding an electron in the ground state occurs at a distance equal to the Bohr radius from the nucleus. The plot of P ( r ) versus r has spherical distribution, which shows a maximum at r = a0 . Similarly, P ( r ) of other states in the hydrogen atom can be analyzed.

6.5  Free Particle Motion In Section 2.4.1 we discussed the motion of a free particle in one dimension. In the threedimensional case, the Schrödinger equation for the free particle is:



  2 2  − 2 m ∇ + V0  ψ = E ψ 2m ⇒ ∇ ψ (r ) + 2 ( E − V0 ) ψ (r ) = 0 

(6.42)

2

2m ( E − V0 ) ≥ 0 , 2 ik .r we find that ψ (r ) = Ce is one of the possible solutions of Eqn. (6.42). By restricting the domain of ψ(r ) to an arbitrarily large but finite cubic volume of side L, box normalization 1 2n π ik .r yields: ψ (r ) = . The values of k are then discretized: ki = i , where ni (i = x , y , z) 3/2 e L ( L) takes integer values 1, 2, 3,...... Equation (6.42) can be expressed in spherical polar coordinates (r , θ, φ) too. Following the procedure displayed in Section 6.2, we get the radial equation; see Eqn. (6.8): where V0 is a constant potential. We can choose V0 = 0 as well. By taking k 2 =



1 d  2 dRl   2 l(l + 1)  r  +k −  Rl (r ) = 0 (6.43) r 2 dr  dr   r2 

which on defining kr = ρ reduces to:



1 d  2 dRl   l(l + 1)  + 1− Rl (ρ) = 0 ρ 2   ρ2  dρ   ρ dρ  d 2 Rl (ρ) 2 dRl (ρ)  l(l + 1)  ⇒ + + 1− Rl (ρ) = 0 2 ρ dρ ρ2  dρ 

(6.44)

Taking Rl (ρ) = ξ(ρ)/ ρ , we obtain:

2 d 2 ξ(ρ) 1 dξ(ρ)   l + 1/ 2   + + − 1    ξ(ρ) = 0 (6.45) dρ2 ρ dρ ρ     

which is Bessel’s equation. Its solution is written as: ξ(ρ) = C1 J

1  l +  2



⇒ Rl (ρ) =

1 ρ

(ρ) + C2 J

 1 − l+   2

(ρ)

  (6.46) C1 J l + 1  (ρ) + C2 J −  l + 1  (ρ)   2    2  

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Here, C1 and C2 are constants and J ±  l + 1  (ρ) are Bessel functions. The spherical Bessel 

2

functions, jl (ρ) and spherical Neumann functions nl (ρ) are defined in terms of Bessel functions as follows:

 π jl (ρ) =    2ρ 

1/2

J

l+

1 2

(ρ) (6.47)

and:

nl (ρ) = ( −1)

l+1

 π  2ρ 

1/2

J

 1 − l+   2

(ρ) (6.48)

Therefore:

Rl (ρ) = Ajl (ρ) + Bnl (ρ) (6.49)

where A and B are new constants. The behavior of spherical Bessel and Neumann functions for ρ → 0 is approximated by:

jl (ρ) →

ρl (6.50) 1.3.5...(2l + 1)



nl (ρ) →

1.3.5...(2l − 1) (6.51) ρl + 1

Equation (6.51) suggests that B has to be zero; otherwise Rl (ρ) will go to ∞ , as ρ → 0 , which violates the requirement that wave function (probability) should have finite value every2 k 2 where. We therefore conclude that Rl (ρ) = Ajl (ρ) . Hence, free particle energy E = + V0 2m and the wave function for the angular momentum state of l is:

ψ l (r , θ, φ) = Al jl ( kr )Ylm (θ, φ) (6.52)

A more generalized solution for the free particle in the energy state E is given by:

ψ k (r , θ, φ) =

∑A j (kr)Y l l

lm

(θ, φ) (6.53)

l

6.6  Spherically Symmetric Potential Well This type of potential is frequently used to approximate several physics problems like that of nano-sized particles and of nuclear physics, where the exact nature of the potential is not known. It is also known as a 3D potential well, depicted in Fig. 6.2. The potential is defined by:

 −V0 V (r ) =   0

0 a (6.56) r 2 dr  dr    2 r2 

where µ is the reduced mass of the particle. The above two equations are to be solved separately, and then the two solutions and their first order derivative are to be matched at the boundary which occurs at r = a . It is to be noted that energy E can have a negative value ( E < 0 , a bound state) as well as a positive value, an unbound state. For both the 2µ cases, E + V0 is going to be positive, and hence we take α 2 = 2 (E + V0 ) . Then, on using  αr = ρ , Eqn. (6.55) reduces to Eqn. (6.44). We therefore can conclude that inside the sphere ( 0 < r < a ):

RlI (r ) = Ajl (αr ) (6.57)

We next take the exterior region of the sphere ( r > a ), where Eqn. (6.56) is applicable. There can be two scenarios: (i) E < 0 and (ii) E > 0 .

Case-I (E < 0): 2µ Let us take β 2 = 2 E , which is positive quantity. Then for E < 0 , we can rewrite Eqn. (6.56)  as follows:



1 d  2 dRl   2 l(l + 1)  r  − β +  Rl (r ) = 0 r 2 dr  dr   r2  1 d  2 2 dRl   l(l + 1)  ⇒ 2 2 β r −  1 + 2 2  Rl (r ) = 0   d(βr )   β r d(βr )  β r 

(6.58)

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On taking βr = − iρ , we get:

d 2 Rl (ρ) 2 dRl (ρ)  l(l + 1)  + + 1− Rl (ρ) = 0 (6.59) dρ2 ρ dρ ρ2  

which is exactly the same as is Eqn. (6.44). Therefore, its solution can be given by Eqn. (6.49) or any other linear combination of jl (ρ) and nl (ρ) . Let us write the solution as: RlII (ρ) = (B + C) jl (ρ) + i(B − C)nl (ρ)

= B { jl (ρ) + inl (ρ)} + C { jl (ρ) − inl (ρ)} (6.60) = Bhl(1) (ρ) + Chl(2) (ρ)

where B and C are constants. hl(1) (ρ) and hl(2) (ρ) are known as Hankel functions. Also, note that hl(1) (ρ) and hl(2) (ρ) are complex conjugates of each other. Writing solution in the form of Eqn. (6.60) is advantageous when asymptotic behavior of solution is discussed. The exterior region of the sphere includes r → ∞ , but it does not include r → 0 . From the properties of Hankel functions, it is well known that for ρ → ∞ :

hl(1) (ρ) ∝

e iρ e −βr ≈ C1 (6.61a) ρ r



hl(2) (ρ) ∝

e − iρ eβr ≈ C2 (6.61b) ρ r

where C1 and C2 are constants. It is obvious from Eqn. (6.61b) that the solution will tend to ∞ for r → ∞ , which is unacceptable. Hence, the solution in the exterior region of the sphere should not include hl(2) (ρ) . Therefore, the acceptable solution in the exterior region is:

RlII (r ) = Bhl(1) (iβr ) (6.62)

Thus, for a bound state case ( E < 0 ), Eqns. (6.57) and (6.62) are acceptable solutions in interior and exterior regions, respectively, of the spherically symmetric potential. Continuity from the interior to exterior or vice versa require that:

RlI ( a) = RlII ( a) ⇒ Ajl (αa) = Bhl(1) (iβa)

(6.63a)

and:



dRlI (r ) dR II (r ) = l dr r = a dr r = a (6.63b) ⇒ Ajl′(αa) = Bhl′(1) (iβa)

Division of Eqn. (6.63b) by (6.63a) gives:

hl(1) (iβa) jl′(αa) − jl (αa)hl′(1) (iβa) = 0 (6.64)

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Schrödinger Equation for Central Potentials and 3D System

The α and β are real and are dependent on n and l . Equation (6.64) explicitly determines the energy eigenvalues Enl for existing bound states that correspond to a given value l . The ∞

constants A and B can be determined from the normalization condition

∫R (r)r dr = 1 . 2 nl

2

0

Hankel functions of the imaginary argument are real apart from the factor il . Therefore, radial function:  Ajl (αr )  Rl (r ) =  (1)  Bhl (iβr )



0 8µ

π 2 2 9π 2  2 < V0 a 2 ≤ (6.68) 8µ 8µ

It is to be noted that unlike the case of the 1D potential well, where the bound state exists for any smallest value of V0 , for the spherically symmetric potential well, existence of a bound state is characterized by Eqn. (6.68), which is applicable to the s-state ( l = 0 ) only. For bound states with l > 0 , the width of the well should be even larger. To find bound states for the p-state ( l = 1 ) and for the d-state ( l = 2 ), we have to evaluate Eqn. (6.64) in terms of the spherical Bessel functions and Hankel functions for l = 1  and and l = 2 , and then solve the resulting equations graphically or numerically.

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Case-II (E > 0): For E > 0 (non-localized case), we can take ρ = βr and then from Eqns. (6.61) we find that both hl(1) (ρ) and hl(2) (ρ) can be included in the radial solution for the exterior region ( r > a ) of a spherical symmetric potential well. Therefore for r > a, we have: RlII (r ) = Bhl(1) (βr ) + Chl(2) (βr ) (6.69)



which means that at r = a, we should have: Ajl (αa) = Bhl(1) (βa) + Chl(2) (βa) (6.70)

and:

Ajl′(αa) = Bhl′(1) (βa) + Chl′(2) (βa) (6.71a)

where:



hl′(1) (βa) =

{

}

d (1) hl (βr ) dr

r=a

and  hl′(2) (βa) =

{

}

d (2) hl (βr ) dr

(6.71b) r=a

jl′(αa) is defined earlier. Equations (6.70) and (6.71a) can be solved to find B/A and C/A:



jl (αa)hl′(2) (βa) − jl′(αa)hl(2) (βa) B = (2) (6.72a) A hl (βa)hl′(1) (βa) − hl′(2) (βa)hl(1) (βa

and:



jl (αa)hl′(1) (βa) − jl′(αa)hl(1) (βa) C = (2) (6.72b) A hl (βa)hl′(1) (βa) − hl′(2) (βa)hl(1) (βa

Since hl(1) (βa) and hl(2) (βa) are complex conjugates of each other, B/A and C/A too are complex conjugates of each other, provided A is real. This suggests that Eqn. (6.69) gives a real solution for the exterior region of the spherical symmetric potential well. Equations (6.57) and (6.69) along with (6.72) uniquely define the wave functions for the spherically symmet∞

∫

ric potential for E > V0 . Further, evaluated values of Rl2 (r )r 2 dr are equal to some constant 0

but not unity. This clearly establishes that states are non-localized for E > 0 .

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Schrödinger Equation for Central Potentials and 3D System

6.7  Electron Confined to a 3D Box The potential energy can be taken to be zero inside the potential well. It is infinite outside the well. An electron in a conducting solid such as a metal is very well represented by a 3D infinite potential well. Inside the well, we have: −

2 2 ∇ ψ = Eψ 2m

 2  ∂2 ∂2 ∂2  ⇒− + + ψ ( x , y , z) = Eψ ( x , y , z) 2 m  ∂ x 2 ∂ y 2 ∂ z 2 

(6.73)

Separation of variables allows to write: ψ ( x , y , z) = ψ ( x)ψ ( y )ψ ( z) . Each of ψ( x) , ψ( y ) , and ψ( z) vanishes at boundaries of the well, which occur at ( x = 0, y = 0, z = 0) and at ( x = L, y = L, z = L) . Further, normalization of wave function demands: L L L

∫∫∫ψ (x, y, z)ψ(x, y, z)dxdydz = 1 *

0 0 0



L

⇒

L

∫ ψ(x) dx ∫ ψ(y) dy ∫ ψ(z) 0

(6.74)

L

2

2

0

2

dz = 1

0

This permits us to choose:



ψ ( x) =

2 sin( k x x); L

ψ(y) =

2 sin( k y y ); (6.75) L

ψ ( z) =

2 sin( k z z) L

Next, ψ (L, y , z) = 0 , ψ ( x , L, z) = 0 , and ψ ( x , y , L) = 0 force us to take k x = πnx /L , k y = πny /L , and k z = πnz /L , where nx , ny , and nz take nonzero integer values 1, 2, 3,.... . Substituting Eqn. (6.75) in Eqn. (6.73), we obtain:

E=

2 2 2 π 2 2 k x + k y2 + k z2 = nx + ny2 + nz2 (6.76) 2m 2 mL2

(

)

(

)

The lowest possible energy of the system is:

E1 =

3 2 π 2 (6.77) 2 mL2

And, the volume occupied by a single state is: 3



 π ∆k = ∆k x ∆k y ∆k z =   (6.78)  L

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(

)

Equation (6.76) gives the energy for a set of three quantum numbers nx , ny , nz in k -space. Hence, the number of states N (E) , the energy of which is less than or equal to E, will be equal to the volume of a sphere of radius of k divided by the volume occupied by a single k-state. Therefore:  4π 3  k    3 N (E) = ∆k =



4π  kL    3  π

3

(6.79) 3/2

4V  2 mE    3π 2   2  4V 3/2 = 2 3 ( 2 mE ) 3π  =

here we have used V = L3 . This is a useful relation to compute the number of electrons that occupy the quantum states up to a given energy level in the metal. The density of states, number of states per unit energy in the range of E to E + dE, is obtained as follows: dN dE d 4V = ( 2mE )3/2 dE 3π 2  3 2V 3/2 = 2 3 ( 2m) E π 

D(E) =

{



}

(6.80)

We thus find that the density of the states of a free electron is proportional to E .

6.8  Solved Examples 1. Show the Bohr radius is a0 = 0.0529 nm for both cases of an electron with actual 13.6 mass and its reduced mass in a hydrogen atom. Also, show that En = − 2 eV . n SOLUTION

Let us first calculate:

(

)

2

1.054 × 10−34 Js 4πε 0  2 = e2 8.988 × 109 Nm 2 / C2 1.602 × 10−19 C

(

)(

)

2

= 4.816 × 10−41 Js / Nm 2 .

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Schrödinger Equation for Central Potentials and 3D System

Hence, for an electron of mass me = 9.11 × 10−31 kg : a0 =

4πε 0  2 4.816 × 10−41 = 9.11 × 10−31 me e 2

= 0.5286 × 10−10  meter = 0.0529 nm.



The reduced mass of the electron in the hydrogen atom is:  mp  µ=  me  me + mp 

  1.673 × 10−24 me = −31 −24   9.11 × 10 + 1.673 × 10 

(6.81)

= 0.9995me The Bohr radius is: 4πε 0  2 4πε 0  2 0.5286 = = = 0.5289 × 10−10 m 2 2 0.9995me e 0.9995 µe = 0.0529 nm.

a0 =



The energy of nth level in the hydrogen atom is given by Eqn. (6.22): 2

En = −

 e2  µ µe 4 × 2 2 = − 2 2 2 2(4πε 0 ) n  2n   4πε 0 

(

= − 2.307 × 10−28



)

2

×

0.9995 × 9.11 × 10−31 J 2 n2 × (1.054 × 10−34 )2 (6.82)

−21.8115 × 10−19 J n2 −13.6 = eV n2 =

2. Consider an electron confined to the interior of a spherical cavity of radius a, with impenetrable walls (a 3D infinite potential well). The potential seen by the electron inside the cavity is −V0 . Calculate the energy and normalized wave function for the ground state of the system. SOLUTION

The walls of the cavity are impenetrable; hence the wave function of the electron vanishes at the wall ψ (r = a) = 0 . For inside the spherically symmetric cavity ( 0 ≤ r ≤ a ), V (r ) = −V0 . The radial Eqn. (6.8) for l = 0 , reduces to:

 d2 2 d 2m   dr 2 + r dr +  2 ( E + V0 )  R0 (r ) = 0 (6.83)  

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On taking P(r ) = rR0 (r ) , we have: d 2 P(r ) 2 m ( E + V0 ) P(r ) = 0 (6.84) + dr 2 2



when r > a, R0 (r ) = 0, as the electron cannot be found outside the cavity. The solution of Eqn. (6.84) can be given by: P(r ) = A sin(βr ) (6.85) A ⇒ R 0 (r ) = sin(βr )  r



1/2

 2 m(E + V0 )  with β =   , and A is constant. The condition ψ (r = a) = 0 yields  2 βa = nπ , where n is a positive integer, which gives:  2 m(E + V0 )  a   2



1/2

= nπ 2

 2  nπ  ⇒ E1 (n) =   − V0 2m  a 

(6.86)

Note that the energy levels for an electron in the ground state of a spherical cavity look very similar to those in a 1D potential infinite well. To fulfill the condition of ∞

∫

normalization: R02 (r )r 2 dr = 1. Let us evaluate: 0

a

∫

a

∫

R02 (r )r 2 dr = A 2 sin 2 (βr )dr

0

0 a



=A

2

∫ 0

= aA 2 which gives = 1 or  A = 2 is:

aA 2

1 − cos(2βr ) dr (6.87) 2

2

2 . The normalized wave function of the system then a

ψ 100 (r ) = R0 (r )Y00 =

{ }

1  1 nπr (6.88)   sin a 2 πa  r 

3. As an extension of problem 2, consider the case when a particle is constrained to move between two concentric impermeable spheres of radii r = a and r = b. Potential in the region between spheres is constant −V0 . Find the normalized wave function and the energy for ground state ( l = 0 ).

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Schrödinger Equation for Central Potentials and 3D System

SOLUTION

In this case, Eqn. (6.84) is applicable for a ≤ r ≤ b. The R0 (r ) = 0 at both r = a and r = b. Hence a normalization condition of the radial wave function is defined by: b

∫R (r)r dr = 1 (6.89) 2 0



2

a

and R0 (r ) does not exist when b <   r < a . To fulfill the requirement that wave function vanishes at r = a , we can choose: P(r ) = A sin {β(r − a)} (6.90)



and then we impose the requirement β(b − a) = nπ to make R0 (r ) = 0 at  r = b . We thus have: β(b − a) = nπ

⇒ E1 (n) =

2 (6.91)  2  nπ    − V0 2m b − a

We next evaluate: b

∫

1 = R02 (r )r 2 dr a

b

∫

= A 2 sin 2 {β(r − a)} dr

(6.92)

a

b

= A2

∫ a

=

1 − cos {2β(r − a)} dr 2

(b − a)A 2 2

2 . Therefore, the normalized wave functions and energies (b − a) for ground state (l = 0) are given by: which yields A =

ψ 100 (r ) = R0 (r )Y00 =



 nπ(r − a)  1  1  (6.93)   sin  2 π(b − a) r  (b − a) 

and:

E1 (n) =

n2  2 π 2 − V0 (6.94) 2 m(b − a)2

4. The wave function ket for the hydrogen atom is given as a linear combination of 1 2 ψ 100 + ψ 200 + 3 ψ 211 + ψ 21− 1 . Find s-orbital and p-orbitals such as φ = 15 the expectation value of the Hamiltonian of the hydrogen atom.

(

)

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SOLUTION

We know that energy eigenvalues and eigenvectors for the hydrogen atom satisfy: H ψ nlm = En ψ nlm (6.95a)

with:

En = −



µe 4 (6.95b) 2(4πε 0 )2 n2  2

where (nlm) are quantum numbers. The expectation value of the Hamiltonian is: E= φ H φ

1 2 ψ 100 + ψ 2100 + 3 ψ 211 + ψ 21− 1 2 H ψ 100 + H ψ 2100 + 3 H ψ 211 + ψ 21− 1 15 (6.96a) =

(

)(

)

With the use of Eqns. (6.95) and the orthogonality condition ψ nlm ψ n′l′m′ = δ nn′ δ ll′ δ mm′ , Eqn. (6.96a) simplifies to: E=



1 1 ( 4E1 + E2 + 9E2 + E2 ) = ( 4E1 + 11E2 ) (6.96b) 15 15

From Eqn. (6.82), we have E1 = −13.6eV and E2 = −3.4eV . Hence: E=−



1 ( 4 × 13.6 + 11 × 3.4 ) = −6.12 eV. 15

5. What is the probability of finding a 1s electron within a distance of a0 from the proton inside the hydrogen atom? SOLUTION

The probability of finding the electron in the volume element of d 3 r is given by:

2

Pd 3 r = Rnl (r )Ylm (θ, φ) r 2 dr sin θ dθ dφ (6.97) Hence the probability of finding a 1s electron at the distance of a0 from the nucleus is: π 2π

a0

P1s =

∫R

10

2

2

(r ) r dr

∫∫ Y

00

2

(θ, φ) sin θ dθ dφ

0 0

0

a0

=

∫R

10

2

(r ) r 2 dr

0



(6.98)

a0

4 = e −2 r/a0 r 2 dr ( a 0 )3

∫ 0

2 1 − x 2 − 2 x − 2 e − x  = 1 − 5e −2  0 2 = 0.323

=

(

)

(

)

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Schrödinger Equation for Central Potentials and 3D System

where we used R10 (r ) =

2 e − r / a0 . ( a0 )3/2 2

r2 − r

6. Calculate uncertainties ∆r =

p2 − p

and ∆p =

2

for the hydrogen

atom in the ground state and then evaluate ∆r∆p . SOLUTION

n = 1,  l = 0 and m = 0 for the ground state of the hydrogen atom. The, ψ 100 (r ) =  1  R10 (r )Y00 (θ, φ) =  3   πa0  observable A: A =

1/2

e − r/a0 = ψ * 100 (r ) . We define the expectation value of an

∫ ψ Aψd r *

3

∞ π 2π

=

∫∫∫ ( R (r)Y nl

lm

(θ, φ)) A ( Rnl (r )Ylm (θ, φ)) r 2 dr sin θdθdφ *

(6.99a)

0 0 0

Therefore r = x iˆ + y ˆj + z kˆ . With the use of x = r sin θ cos φ , we have: ∞



π

2π

0

0

1 x = 3 r 3 e −2 r/a0 dr sin θ sin θ dθ cos φ dφ (6.99b) πa0

∫

∫

0

∫

=0 Similarly, y = 0 and z = 0 giving r = 0 . Next take:

∫ ψ r ψd r (6.100a) 4 = r e dr a ∫

r2 =

* 2

3

∞



4 −2 r /a0

3 0

0

= 3 a02 p = px iˆ + py ˆj + pz kˆ  and,   px = −



=

i πa03

i πa04

∞ π 2π

∫∫∫e

− r /a0

0 0 0

∞ π 2π

∫∫∫e 0 0 0

∞

− r /a0

 d − r/a0  2  e  r dr sin θ dθ dφ dx

 x − r/a0  2  e  r dr sin θ dθ dφ r π

2π

0

0

i = 4 e −2 r/a0 r 2 dr sin 2 θ dθ cos φ dφ πa0

∫ 0

=0

∫

∫

.

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Similarly: ∞

py

π

2π

0

0

i = 4 e −2 r/a0 r 2 dr sin 2 θ dθ sin φ dφ = 0 πa0

∫

∫

0

∫

and

(6.100b) ∞

pz =

π

2π

0

0

i e −2 r/a0 r 2 dr cos θ sin θ dθ dφ = 0 πa04

∫

∫

0

∫

Hence, p = 0 . To evaluate p 2 , we take: p 2 = ∫ ψ * p 2 ψd 3 r

=−

∞ π 2π

2 πa03

∫∫∫e (∇ e ) r dr sin θ dθ dφ − r /a0

2 − r /a0

2

0 0 0

with: ∇2 =



1 ∂  2 ∂ 1 ∂  ∂ 1 ∂2 r sin θ + +     r 2 ∂r  ∂r  r 2 sin θ ∂θ  ∂θ  r 2 sin 2 θ ∂φ2

and obtain: p2 = − =

4 2 a03

2 a02

  2! 2 1  2 3 − 2  a0 ( 2 / a0 )   a0 ( 2 / a0 ) (6.101)

We then have: ∆r =



r2 − r

2

= 3 a0 , and ∆p =

p2 − p

2

=

 (6.102) a0

which gives ∆r∆p = 3 . 7. A particle of mass m oscillates under the spherically symmetric potential given 1 by: V (r ) = kr 2 , where k is some constant. Calculate the energy eigenvalue and 2 eigenfunction of the particle. SOLUTION

The time-independent Schrödinger equation for the particle is:

 2 2   − 2 m ∇ + V (r )  ψ (r ) = Eψ (r ) (6.103)  

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Schrödinger Equation for Central Potentials and 3D System

which is separable into three equations by writing: ψ (r ) = ψ 1 ( x)ψ 2 ( y )ψ 3 ( z) , and E = E1 + E2 + E3 . After taking ω = k / m , we get:

d 2 ψ 1 ( x) 2 m  1  + 2  E1 − mω 2 x 2  ψ 1 ( x) = 0 (6.104a) 2  dx   2



d 2 ψ 2 (y) 2m  1  + 2  E2 − mω 2 y 2  ψ 2 ( y ) = 0 (6.104b)  dy 2 2  



d 2 ψ 3 ( z) 2 m  1  + 2  E3 − mω 2 z 2  ψ 3 ( z) = 0 (6.104c) 2   dz  2 Each of Eqns. (6.104) is similar to the equation of the 1D harmonic oscillator whose Hamiltonian is defined by Eqn. (3.123). Hence, the method presented for the 1D harmonic oscillator in Section 3.19 applies to solve each the Eqns. (6.104). The particle under consideration is behaving like a 3D harmonic oscillator. We therefore get:



1  E1 =  n1 +  ω (6.105a)  2



1  E2 =  n2 +  ω (6.105b)  2



 E3 =  n3 + 

1  ω (6.105c) 2

n1 , n2 and n3 are integers, which takes values 0,1,2,3 ……. The wave functions that satisfy Eqns. (6.104) are then obtained from Eqn. (3.146), which are as follows: ψ 1 (ρ) =

 mω    n1 (2 n1 !)  π 

1/4



ψ 2 (σ ) =

 mω    n2 (2 n2 !)  π 

1/4



ψ 3 ( ε) =

 mω    n3 (2 n3 !)  π 

1/4



1

1

1

2

e −ρ

e −σ

e −ε

/2

2

2

H n1 (ρ) (6.106a)

/2

/2

H n2 (σ ) (6.106b)

H n3 (ε) (6.106c)

mω mω mω x, σ = y , and ε = z . We thus find that the energy and    wave function of the nth state of the particle are:

where ρ =



 En =  n + 

3  ω (6.107) 2

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A Textbook on Modern Quantum Mechanics

and:

ψ n (r ) =

1

{2 (n !)(n !)(n !)} n

1/2

1

2

3

 mω    π 

3/4

2

e − (ρ

+σ 2 +ε 2 )/2

H n1 (ρ)H n2 (σ )H n3 (ε) (6.108)

where n = n1 + n2 + n3 . 8. A NaCl crystal has certain negative ion vacancies behaving like a free electron, inside a volume having dimensions of the order of a lattice constant. Estimate the longest wavelength of electromagnetic radiation absorbed strongly by these electrons, when the crystal is at room temperature. SOLUTION

Energy levels for a free electron confined to a cubic box having each side of length L are given by Eqn. (6.76): E=



2 π 2 2 n1 + n12 + n32 (6.109) 2 mL2

(

)

where n1 , n2 , and n3 are positive integers. Taking L ∼ 1 Angstrom, the ground state energy is: 3 2 π 2 3(1.054 × 10−34 )2 (3.14)2 = = 1.8 × 10−17 Joules. 2 mL2 2 × 9.11 × 10−31 (10−10 )2



E111 =



E111 = 112.5eV. Electrons in a crystal at room temperature are almost as in the ground state. Hence, 6 2 π 2 3 2 π 2 the energy of the first excited states is: E211 = = . The longest wave 2 mL2 mL2 length for transition from the ground state to the first excited state is given by:

λ=

c ch = ν (E211 − E111 )

3 × 108 × 6.626 × 10−34 λ= = 1.104 × 10−8 m 1.8 × 10−17

(6.110)

or λ = 11.04 nm. 9. A nano-meter sized particle can be modeled as a spherical potential well of radius a, with impenetrable walls. Inside the nano-particle electrons move freely under a constant potential −V0 . What is the probability of finding an electron at distance a / 2 from the center of the nano-particle, for l = 0 . Calculate the first three energy gaps between the discrete states.

185

Schrödinger Equation for Central Potentials and 3D System

SOLUTION

As discussed in example 2, the wave function of an electron inside the spherical potential well of radius a, with impenetrable walls for l = 0 is given by:

{ }

2  1 nπr Y00 (6.111)   sin a  r a

ψ 100 (r , θ, φ) =



where n = 1, 2, 3,... is an integer. The probability of finding the electron at distance a / 2 from the center of the nano-particle is: a/2 π 2 π

P=

∫ ∫∫ ψ

2

100

(r , θ, φ d 3 r

0 0 0

1 = 2 πa

=

= =

2 a 1 a

a/2 π 2 π

1

∫ ∫∫ r

2

sin 2

0 0 0

∫ sin {

a/2

2

0

nπr 2 r dr sin θ dθ dφ a

}

nπr dr a

∫ 1 − cos {

a/2

{ }



0

(6.112)

}

2 nπr  dr a 

1 a 1 × = a 2 2

As discussed in example 2, the energy eigenvalues of the electron are discrete. We write:



En =

2

 2  nπ    − V0 (6.113) 2m  a 

The first three energy gaps between the discrete energy levels are:



∆E1 = E2 − E1 =

3 2 π 2 ; 2 ma 2

∆E2 = E3 − E2 =

5 2 π 2 ; (6.114) 2 ma 2

∆E3 = E4 − E3 =

72 π 2 2 ma 2

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A Textbook on Modern Quantum Mechanics

6.9 Exercises 1. The Hamiltonian for an electron in the ground state of hydrogen-like atoms is given by:  2  d2 2 d  Ze 2  + ψ (r ) = 0 (6.115) ψ (r ) +  E +   2 r dr  2µ  dr 4πε 0 r  



To choose ψ (r ) = Ae −αr as the eigenfunction, what should be the values of A, α and E ? Find the expectation values of potential energy V and of kinetic energy T . 2. Find the most probable radius of the electron in the 1s and 2p states of the hydrogen atom. 3. Calculate x , y , z , x 2 , y 2 and z 2 for the ground state of the hydrogen atom. What do you notice from these calculations? 4. Calculate px , py , pz , px2 , py2 and pz2 for the ground state of the hydrogen atom. Show that ∆x∆px > /2 . 5. A particle of mass m is confined to move within a two-dimensional rectangular system of lengths a and b along the x-axis and y-axis, respectively. The confining potential is:  V0 V(x, y) =   ∞



0 < x < a and 0 < y < b otherwise

.

Find out the wave function ψ( x , y ) and the energy E of the particle. 6. What is the probability of finding the particle described in exercise 5, in the region of 0 < x < a/2 and b/2 < y < b ? 7. A particle of mass m oscillates in the x-y plane under the potential defined by: 1 V ( x , y ) = k1 x 2 + k2 y 2 , where k1  and  k2 are constant. Calculate the energy eigen2 value and eigenfunction of the particle. 8. An electron in the 2p-state of the hydrogen atom is described by wave function ψ (r , θ, φ) = Are − r/2 a0 . Use the normalization condition to find the value of A and then calculate the probability of finding the 2p-electron within a distance of 2 a0 from the proton inside the hydrogen atom. 9. A GaAs-quantum wire can be modeled as a system in which electrons are confined along x-y directions and they move freely along the z-axis. The confining lengths along the x- and y-axes are equal to a, the potential seen by the electron inside the wire is zero, and the effective mass of the electron is m = 0.07 me . (a) Calculate the energy of an electron. (b) Find the frequency of the photon released when the electron makes transition from the third excited state to the ground state.

(

)

7 Approximation Methods In Chapters 2 and 6, we discussed some problems for which the Schrödinger equation can be solved exactly either in 1D or in 3D. However, the vast majority of physical and chemical problems cannot be represented by these prototype problems. There exist a large number of applications in physics and chemistry for which the Schrödinger equation cannot be solved exactly, and hence it is to be solved approximately. In this chapter, we will be discussing well known approximation methods developed to solve the Schrödinger equation approximately.

7.1  Perturbation Theory Perturbation theory is a mathematical technique commonly used in many areas of physics and chemistry, not just in quantum mechanics. In fact, it had been extensively and successfully used as a powerful tool by physicists long before the development of quantum mechanics. One of the procedures was a formulation of perturbation theory utilized by quantum chemists for estimating electron correlation. The basic concept of perturbation theory involves taking a physical system for which it is not possible to obtain exact solutions and separating the Hamiltonian of the system into two parts. The first part can be solved exactly, while the second part is the troublesome part that has no analytic solution. The Hamiltonian of a system is written as H = H 0 + H 1 , where H 0 is a simple Hamiltonian for which exact eigenvalues and eigenstates are known. An interesting physics is introduced by H 1 , but on adding H 1 to H 0 we are unable to find the exact energy eigenvalues and eigenvectors. However, if H 1 can be regarded as being small compared to H 0 , we can find the approximate eigenvalues and eigenvectors of the Hamiltonian H = H 0 + H 1 . Perturbation theory provides a mathematical means to study how the solutions evolve when an exact system undergoes the perturbation represented by H 1 . There exist many approaches; we here follow the Rayleigh-Schrödinger perturbation theory approach. Let H be the Hamiltonian of the system of interest. Energy eigenvalues and eigenfunctions for the state, specified by a composite indexn representing a set three quantum numbers (n, l, m) are the solutions to the Schrödinger equation:

H ψ n = En ψ n (7.1)

Further, there are two scenarios: (i) a non-degenerate case, where only one ψ n corresponds to one value of En , and (ii) a degenerate case, where more than one ψ n belong to the same En value. 187

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A Textbook on Modern Quantum Mechanics

7.1.1  Perturbation Theory for Nondegenerate States Let us further assume that for a system described by:

(7.2) H 0 ψ (0) = En(0) ψ (0) n n

solutions (eigenvalues and eigenfunctions) are already known or these can easily be obtained. We write the Hamiltonian as:

H = H 0 + λH 1 (7.3)

where the perturbation parameter is 0 ≤ λ ≤ 1. Thus, the Schrödinger equation for the system now is:

( H 0 + λH 1 ) ψ n = En ψ n (7.4)

The eigenfunctions and eigenvalues of H must depend on λ, because H depends on λ. Therefore, ψ n and En can be expanded in a power series with respect to λ, as follows:

En = En(0) + λEn(1) + λ 2 En(2) + ......... (7.5)



ψ n = ψ (0) + λ ψ (1) + λ 2 ψ (2) + ......... (7.6) n n n

where we used the following abbreviations in the expansion:



En( k ) =

ψ (nk ) =

1 ∂ k En k ! ∂λ k

(7.7) λ= 0

k 1 ∂ ψn k ! ∂λ k

(7.8) λ= 0

The En and ψ n of a system are the sum of the correction terms. It is obvious that the convergence of these power series is a key issue, because the series must be truncated in practice. Using Eqns. (7.5) and (7.6) in Eqn. (7.4), we get:



+ λ ψ (1) + λ 2 ψ (2) + .........  (H 0 + λH 1 )  ψ (0) n n n = En(0) + λEn(1) + λ 2 En(2) + .........   ψ (0) + λ ψ (1) + λ 2 ψ (2) + .........  n n n

(7.9)

the simplification of which gives terms with various powers of λ on both the sides. If equality is to hold for all 0 ≤ λ ≤ 1, then for a given power of λ, all terms on the left-hand side must equal to those on the right-hand side. This gives rise to:

H 0 ψ (0) = En(0) ψ (0) n n (7.10)

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Approximation Methods



H 0 ψ (1) + H 1 ψ (0) = En(0) ψ (1) + En(1) ψ (0) (7.11) n n n n



H 0 ψ (2) + H 1 ψ (1) = En(0) ψ (2) + En(1) ψ (1) + En(2) ψ (0) (7.12) n n n n n

Equations (7.11) and (7.12) can be generalized to: k



H0 ψ

(k) n

+ H1 ψ

( k − 1) n

=

∑E

(i) n

ψ (nk − i ) (7.13)

i= 0

Equation (7.10) provides zeroth order eigenvalues and eigenvectors. We have to find various corrections En( k ) to energy En , and ψ (nk ) to eigenvector ψ n . We assume that eigenvec(0) tors of H 0 are orthonormal, and hence ψ (0) = δ mn . The zeroth order energy is then m ψn obtained from: (0) En(0) = ψ (0) (7.14) n H0 ψ n



7.1.1.1  First Order Corrections to Energy and Wave Function Ket On operating with ψ (0) from the left on both the sides of Eqn. (7.11) we get: m



(1) (0) (1) (0) ψ (0) + ψ (0) = En(0) ψ (0) + En(1) ψ (0) m H0 ψ n m H1 ψ n m ψn m ψn (1) (0) (1) ⇒ Em0 ψ (0) + ψ (0) = En(0) ψ (0) + En(1)δ mn m ψn m H1 ψ n m ψn

(7.15)

(0) (1) (1) where we have used ψ (0) = δ mn and ψ (0) = Em(0) ψ (0) because H 0 is a m H0 ψ n m ψn m ψn Hermitian operator. We now consider the two possibilities: (i) m = n and (ii) m ≠ n. For the case of m = n Eqn. (7.15) simplifies to: (0) En(1) = ψ (0) (7.16) n H1 ψ n



and when m ≠ n, Eqn. (7.15) is rewritten as:



(1) (0) (1) Em(0) ψ (0) + ψ (0) = En(0) ψ (0) m ψn m H1 ψ n m ψn

(

)

(0) (1) ⇒ ψ (0) = En(0) − Em(0) ψ (0) m H1 ψ n m ψn

(7.17)

The Eqn. (7.11) can be rewritten as:

(H

0

)

(

)

− En(0) ψ (1) = En(1) − H 1 ψ (0) (7.18) n n

which suggests that ψ (1) can be constructed as a linear combination of exact solutions of n H 0 , which provide a convenient but not a unique choice for a complete orthonormal basis.

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A Textbook on Modern Quantum Mechanics

The various order corrections ψ (nk ) to the eigenstate can then also be constructed as a linear combination of exact solutions of H 0 . We therefore take: ψ (nk ) =



∑C

(k) nµ

ψ (0) (7.19) µ

µ

where expansion coefficients Cn( kµ) are to be determined. To find an expansion coefficient Cn(1)µ , we rewrite Eqn. (7.17) with the use of Eqn. (7.19) for k = 1 :

(E

(0) m

− En(0)

) ∑C

(1) nµ

(0) (0) ψ (0) = ψ (0) (7.20) m ψµ m H1 ψ n

µ

which yields:

(1) Cnm =

(0) ψ (0) m H1 ψ n

(E

(0) m

− En(0)

)

(7.21)

It is to be noted that Eqn. (7.20) is valid only for m ≠ n; it is not valid for m = n. We therefore obtain:

ψ (1) = n

(0) ψ (0) m H1 ψ n

∑ (E m≠ n

(0) m

− En(0)

)

ψ (0) m (7.22)

(

Further, we find that if ψ (1) satisfies Eqn. (7.18), then ψ (1) + A ψ (0) n n n

) also satisfies it, for

any constant A. This provides us freedom to subtract off the ψ (0) term, and therefore, n there is no need to include the m = n term in the sum for constructing ψ (1) . n

7.1.1.2  Second Order Corrections to Energy and Wave Function Ket On operating with ψ (0) from the left on both the sides of Eqn. (7.12) we have: m



(2) (1) ψ (0) + ψ (0) = m H0 ψ n m H1 ψ n (2) (1) (0) En(0) ψ (0) + En(1) ψ (0) + En(2) ψ (0) m ψn m ψn m ψn

(7.23)

When we take m = n, the first term on the left-hand side cancels out the first term on the (1) right-hand side, and the second term on the right-hand side vanishes, because ψ (0) =0 m ψn (1) due to non-inclusion of the m = n term in ψ n . We therefore get:

(1) En(2) = ψ (0) (7.24a) n H1 ψ n

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Approximation Methods

which with the use of Eqn. (7.22) gives:

En(2) =

(0) (0) ψ (0) ψ (0) m H1 ψ n n H1 ψ m

∑

(E

(0) m

m≠ n

=

− En(0)

(0) ψ (0) m H1 ψ n

∑ (E

(0) m

m≠ n

− En(0)

)

(7.24b)

2

)

On taking m ≠ n, Eqn. (7.19) gives: ψ (2) = n



∑C

(2) nµ

ψ (0) (7.25) µ

µ

Substituting Eqn. (7.25) into Eqn. (7.23), we get for m ≠ n:

(E

(0) m



) ∑C

(0) ψ (0) + m ψµ

(1) nµ

(0) µ

− En(0)

(2) nµ

µ

(1) n

=E

∑

C

ψ

(0) m

∑C

(1) nµ

(0) ψ (0) m H1 ψ µ

µ≠ n

ψ

(7.26)

µ≠ n

which gives: (1) − En(1)Cnm

(1) nµ

µ≠ n (0) m

(2) Cnm =



∑C (E

(0) ψ (0) m H1 ψ µ

− En(0)

(7.27)

)

(2) It is to be noted that Eqn. (7.27) cannot be used to calculate Cnn , because Eqn. (7.27) is (2) obtained for m ≠ n. The Cnn can be calculated by imposing the condition that the perturbed wave function be normalized, that is ψ n ψ n = 1. Retaining the terms up to λ 2 in Eqn. (7.6), we have:



(1) 2 (2)   (0) (1) 2 (2)   ψ (0)  n + λ ψ n + λ ψ n   ψ n + λ ψ n + λ ψ n  = 1 (7.28)

(0) With the use of the orthonormality condition ψ (0) = δ µν , Eqn. (7.28) simplifies to: µ ψν



 (1) (1) * (2) (2) * λ Cnn + (Cnn ) + λ 2  Cnn + (Cnn ) + 

(

)

∑ ν

2 Cn(1)ν  = 0 (7.29) 

(1) (1) * From discussions presented above in Section 7.1.1.1, we find that Cnn + (Cnn ) = 0 . Therefore:



(2) (2) * Cnn + (Cnn ) =−

∑C

(1) 2 nν

ν

(7.30)

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A Textbook on Modern Quantum Mechanics

The imaginary parts of the left-hand side of Eqn. (7.30) cancel out each other. Therefore, the real part of Cn(2), n is given by:

( )

(2) Re Cnn =−



1 2

∑C

(1) 2 nν

(7.31)

ν≠ n

We thus, have:

ψ (2) = n

(0) (0) (0)  ψ (0) H 1 ψ (0) ψ (0) ψ (0) ψ (0) n n H1 ψ n ν H1 ψ µ µ H1 ψ n  µ − 2  Eµ(0) − En(0) En(0) − Eν(0) Eµ(0 − ) − En(0) µ≠ n 

∑∑ ( ν≠ n

1 − 2

)(

(0) ψ (0) ν H1 ψ n

∑ (E ν≠ n

(0) ν

− En(0)

)

(

)

2

  ψ (0) µ  

(7.32)

ψ (0) n

)

2

7.1.1.3  kth Order Corrections to Energy and Wave Function Ket On operating with ψ (0) from the left in both sides, Eqn. (7.13) goes to: n k

(k) ( k −1) ψ (0) + ψ (0) = n H0 ψ n n H1 ψ n

∑E

( k −i) ψ (0) n ψn

(i) n

i= 0



(7.33)

k

=

∑E

(i) n 0, k − i

δ

i= 0

This gives: ( k − 1) (7.34) En( k ) = ψ (0) n H1 ψ n



which is a generalization of Eqns. (7.16) and (7.24a). We note from Eqn. (7.34) that in general the (k - 1)th order correction to the wave function ket is required to obtain the kth order energy correction. On operating with ψ (0) from the left, Eqn. (7.13) becomes for m ≠ n: m k



∑E

(i) n

( k −i) (k) ( k −1) ψ (0) = ψ (0) + ψ (0) m ψn m H0 ψ n m H1 ψ n

i= 0

(7.35)

(k) ( k −1) = Em0 ψ (0) + ψ (0) m H1 ψ n m ψn

or: k

∑∑C

( k − i) (i) nµ n



E

(0) ψ (0) = m ψµ

i = 0 µ≠ n (0) m

E

∑C

(k) nµ

µ≠ n

ψ

(0) m

ψ

(0) µ

+

∑C

( k − 1) nµ

µ≠ n

(7.36) ψ

(0) m

H1 ψ

(0) µ

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Approximation Methods

which simplifies to: k



∑∑C

( k − i) (i) nµ n mµ

= Em(0)

E δ

i = 0 µ≠ n

∑C

(k) nµ mµ

δ

µ≠ n

+

∑C

( k − 1) nµ

(0) ψ (0) (7.37a) m H1 ψ µ

µ≠ n

giving rise to: k

∑

(0) Cn( kµ− 1) ψ (0) − m H1 ψ µ



(k) nm

C

=

µ≠ n

(E

(0) n

− Em(0)

)

∑C

( k − i) (i) nm n

E

i=1

(7.37b)

Note that Eqn. (7.37b) is a generalization of Eqn. (7.27). 7.1.1.4  Anharmonic Oscillator We come across several realistic problems in physics where the potential energy of an oscillator has both a non-quadradic as well as a quadradic term. Such oscillators are termed anharmonic oscillators. The non-quadradic term can be treated as the perturbative term to the harmonic oscillator, if it is small as compared to the Hamiltonian of the harmonic oscillator. We here consider two such cases of (i) H 1 = Ax , and (ii) H 1 = Bx 4 , where A and B are constants, and calculate first order corrections to the energy and wave function of the nth state. The Hamiltonian 1/2 p2 1  mω  , of the unperturbed harmonic oscillator is H 0 = + mω 2 x 2 . On defining α =     2m 2 ip ip αx αx 1 1  a= + and a † = − we get H 0 =  a † a +  ω , and x = a† + a .   2 2α 2 α 2 2 α 2

(

)

For the nth eigenstate of the harmonic oscillator, we have H 0 n = En n , a n = n n − 1 , a † n = n + 1 n + 1 and N n = a † a n = n n . Also m n = δ mn . We then have: H1 =

(i)

A ( a † + a) (7.38) 2α

From Eqn. (7.16), we get: A n ( a † + a) n 2α A n a† n + n a n = 2α A n+1 n n+1 + n n n−1 = 2α A n + 1δ n, n+ 1 + nδ n, n− 1 = 2α =0

En(1) =

(



)

( (

)

)

(7.39a)

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because n cannot be equal to n + 1 or n − 1 . The first order correction to the wave function is:

ψ (1) = n = = =

(ii)

m ( a † + a) n m Em(0) − En(0)

A 2α

∑(

A 2α

∑ 

m≠ n

m≠ n

)

 (n + 1)δ m , n+ 1 + nδ m , n− 1   m  Em(0) − En(0)

(

)

(7.39b)

 A  (n + 1) n n+1 + 0 n−1   0 (0) (0) − − E E E E 2 α  n+ 1 n n− 1 n  A ω 2 α

(

(n + 1) n + 1 − n n − 1

)

B ( a † + a) 4 4α 4 2 B ( a † )2 + a † a + aa † + a 2  = 4  4α (7.40a) 2 B ( a † )2 + 1 + 2 N + a 2  = 4  4α B ( a † )4 + a 4 + ( a † )2 (1 + 2 N ) + ( a † )2 a 2 + (1 + 2 N )( a † )2 + (1 + 2 N )2 = 4α 4 

H1 =



+(1 + 2 N )a 2 + a 2 ( a †)2 + a 2 (1 + 2 N )  where we used a†a = N and  a, a †  = 1 . Hence:



En(1) =

B n ( a † )4 + a 4 + ( a † )2 (1 + 2 N ) + ( a † )2 a 2 + (1 + 2 N )( a † )2 + (1 + 2 N )2 4α 4 (7.40b) + (1 + 2 N )a 2 + a 2 ( a †)2 + a 2 (1 + 2 N )  n

With the argument that is used above in (i), we find:

B n ( a †)2 a 2 + a 2 ( a †)2 + (1 + 2 N )2  n 4α 4 B  n(n − 1) + (n + 1)(n + 2) + (1 + 2 n)2  = 4α 4  B = ( n + 1)2 . 4α 4

En(1) =

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Approximation Methods

Next we calculate: ψ (1) = n

∑ (E m≠ n



m H1 n m (0) (0) m − En

)

=

m ( a † )4 n m a4 n B   m m + 4 4α  m≠ n Em(0) − En(0) Em(0) − En(0) m≠ n 

+

∑ (E

∑(

m≠ n

+

+

∑(

)

(7.41a)

)

m ( a † )2 a 2 n m a 2 (1 + 2 N ) n m (1 + 2 N )a 2 n m m m + + Em(0) − En(0) Em(0) − En(0) Em(0) − En(0) m≠ n m≠ n

∑ (

)

∑ (

)

)

 m ( a † )2 (1 + 2 N ) n m (1 + 2 N )( a † )2 n  m m + Em(0) − En(0) Em(0) − En(0)  m≠ n

∑ ( m≠ n

)

m (1 + 2 N )2 n m a 2 ( a † )2 n m m + (0) (0) Em(0) − En(0) m − En m≠ n

∑( m≠ n

∑(

)

∑ (

)

)

which simplifies to:

ψ (1) = n

m ( a † )4 n m a4 n B   m m + 2 2 α 3  m≠ n Em(0) − En(0) Em(0) − En(0) m≠ n 

∑(

+



+

∑ (

)

(7.41b)

)

 m ( a † )2 (1 + 2 N ) n m (1 + 2 N )( a † )2 n  m m + Em(0) − En(0) Em(0) − En(0)  m≠ n

∑ ( m≠ n

)

m a 2 (1 + 2 N ) n m (1 + 2 N )a 2 n m + m (0) (0) (0) (0) E E − m − En m n m≠ n

∑ (E m≠ n

∑(

)

∑ (

)

)

Further simplification gives:

ψ (1) = n

{

B 1 ((n + 1)(n + 2)(n + 3)(n + 4))1/2 n + 4 4  4ωα  4 − ( n(n − 1)(n − 2)(n − 3)) − ( n(n − 1))

1/2

1/2

}

n − 4 + ((n + 1)(n + 2))

1/2

(3 + 2 n) n (7.42) +2

(2 n − 1) n − 2 

7.1.2  Perturbation Theory for Degenerate States As stated earlier, when there are more than one eigenvectors, the set of quantum numbers ( nlm) takes more than one value, and all belong to same energy eigenvalue, the state is called the degenerate state. As is seen from Eqns. (7.21), (7.27), and (7.37b), the perturba(0) tion series converges if ψ (0) a

Equation (8.109) goes to: a

µV0 (e 2 ikr ′ − 1)dr ′  1 (8.110) k 2

∫



0

which simplifies to: µV0  e 2 ika − 1  − a  1 k 2  2 ik 

µV0 ika e sin( ka) − ka  1 k 2 2 µV ⇒ 2 02 sin 2 ( ka) + ( ka)2 − 2 ka sin( ka)cos( ka) k  ⇒

(

{

)

(8.111)

}

1/2

1

For low energy scattering, ka  1, we write e ika sin( ka) − ka ≈ i( ka)2 to give:

µV0 a 2  1 (8.112) 2

which states that the potential should be weak enough not to bind the particle for scattering from the square well potential in the low energy regime.

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Quantum Theory of Scattering

8.7 Transformation from the Center of Mass Coordinate System to the Laboratory Coordinate System The theory of scattering is usually developed in the center of mass (CM) coordinate system, while experiments to measure the scattering cross-section are performed in the laboratory coordinate system. Therefore, a relation connecting the differential cross-section in CM and the laboratory frames is needed. The CM coordinate system is more convenient for making calculations, because there the two-body problem with 6-degrees of freedom reduces to two one-body problems with 3-degrees of freedom. In the laboratory system, the target or center of scattering is initially at rest, whereas in the CM system, the center of mass of two interacting particles (incident and target) is always at rest. To get the relation between two coordinate systems, we consider a particle of mass m1 moving with velocity v 1, let us say along x-axis, that collides with another particle of mass m2 at rest at the origin in the laboratory system. After collision, the incident particle is scattered with velocity v′1 in a direction of (α , β). The motion in the laboratory system is shown in Fig. 8.2. The velocity of the center of mass is then given by:

v cm =

m1 v 1 (8.113) m1 + m2

Now let us examine the same scattering with respect to an observer located at the center of mass. The observer sees that the particle of mass m2 is approaching him with velocity v cm from the left, while the particle of mass m1 is traveling from the right with velocity:

m1    m2  = v1  v c = v 1 − v cm = v 1  1 − (8.114)   m1 + m2  m1 + m2 

After collision, the two particles should scatter in opposite directions, to keep the center of mass at rest in the CM coordinate system. Let us say the direction of scattering in the CM coordinate system is (θ, φ). Also, elastic scattering demands that the magnitude of velocity of the two particles be equal. The collision process is illustrated in Fig. 8.3. Since two systems of coordinates move relative to each other with velocity v cm , v 1′ = v c + v cm, therefore:



v1′ cos α = vc cos θ + vcm v1′ sin α = vc sin θ (8.115) β=φ

FIGURE 8.2 Motion of a particle along the x-axis in the laboratory coordinate system; (a) before scattering and (b) after scattering.

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A Textbook on Modern Quantum Mechanics

FIGURE 8.3 Motion of a particle along the x-axis in the center of mass coordinate system; (a) before scattering and (b) after scattering.

We thus have:

tan α =

sin θ sin θ (8.116) = vcm cos θ + γ cos θ + vc

vcm m1 = . Hence, for m2  m1, θ  α and recoil is neglivc m2 gible. In such cases, the target behaves as a fixed system. The relationship between the differential scattering cross-section in the laboratory system σ L (α , β) and in the CM coordinate system σ C (θ, φ) is established by using the condition that the number of particles scattered into a given solid angle must be same in both the systems of coordinates:

From Eqns. (8.113) and (8.114) γ =



Nσ L (α , β)dΩL = Nσ C (θ, φ)dΩC ⇒ σ L (α , β)sin αdαdβ = σ C (θ, φ)sin dθdφ

(8.117a)

β = φ ⇒ dβ = dφ, and the square of both sides of Eqn. (8.116) gives:

sin 2 α sin 2 θ +1= +1 2 cos α (cos θ + γ )2

⇒ cos 2 α = ⇒ cos α =

(cos θ + γ )2 (cos θ + γ )2 = 2 2 (cos θ + γ ) + sin θ 1 + 2 γ cos θ + γ 2 (cos θ + γ )

(1 + 2 γ cos θ + γ )

2 1/2

(8.117b)

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Quantum Theory of Scattering

Differentiating both the sides, we get: − sin α dα =

− sin θdθ

(1 + 2 γ cos θ + γ )

2 1/2

⇒ sin α dα =

+

( γ cos θ + 1)

(1 + 2 γ cos θ + γ )

2 3/2

(cos θ + γ ) × γ sin θ dθ

(1 + 2 γ cos θ + γ )

2 3/2

(8.118)

sin θ dθ

Therefore:

(1 + 2 γ cos θ + γ ) (α , β) =

2 3/2

σ



L

( γ cos θ + 1)

σ C (θ, φ) (8.119)

As is seen for γ → 0, σ L (α , β) ≅ σ C (θ, φ) and for γ = 1, Eqn. (8.116) yields α = θ/2 and σ L (α , β) = 4 cos(2α)σ C (2α , β).

8.8  Solved Examples 1. Calculate scattering amplitude and differential scattering cross-section, using the 2 λ Born approximation, for an attractive potential defined by V (r ) = − δ(r − a), where 2µ λ > 0 . Then, evaluate the total scattering cross-section for the low energy and small angle scattering case. SOLUTION

The scattering amplitude is given by Eqn. (8.100b): ∞ 2µ f ( k , θ) = − 2 r ′V (r ′)sin(qr ′)dr ′ , with q = 2 k sin(θ/2) . Substituting the value of q

∫ 0

V (r ′), we have:

a

λ f ( k , θ) = r ′δ(r ′ − a)sin(qr ′)dr ′ (8.120a) q

∫



0

which simplifies to: f ( k , θ) =



λa sin(qa) (8.120b) q

Therefore: 2



 λa  σ( k , θ) =   sin 2 (qa) (8.121)  q

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For low energy k  1/ a, and for small θ qa  1. We therefore get: σ( k , θ) = 2

 λa  2 2 2  q  sin (qa) ≈ (λa ) , which gives: σ tot = ∫ σ( k , θ)dΩ = 4πa 2 (λa)2 .



2. For the potential described in problem 1: (a) Find the minimum value of λ for which there exists a bound state. (b) Use partial wave analysis to calculate the phase shift and scattering cross-section when the particle incident on the potential has a low energy. (Similar Problem 10.2 in “Problems & Solutions in Quantum Mechanics,” Kyrikos Tamvakis.) SOLUTION

(a) The bound state of a single particle is described by ψ (r ) = Rl ( k , r )Ylm (θ, φ). Then, the radial equation is: d 2 Rl 2 dRl  2µE l(l + 1)  + + − + λδ(r − a)  Rl (r ) = 0 (8.122) dr 2 r dr   2 r2 



−2µ E . When r ≠ a, the equation is 2 an imaginary-variable spherical Bessel equation. The solution of the equation, for the bound state E < 0. We define α =

which is finite at r = 0, in the region of r < a is: Rl< (r ) = Ajl (iαr ) (8.123a)



As discussed in Section 6.6, the solution of Eqn. (8.122) for r > a can be given as: Rl> (r ) = Bhl(1) (iαr ) (8.123b)

where

hl(1) (iαr ) = jl (iαr ) + inl (iαr ). Continuity of wave function at r = a, demands: Ajl (iαa) = Bhl(1) (iαa)



⇒

A hl(1) (iαa) = B jl (iαa)

(8.124)

Integration of Eqn. (8.122) between a + ε → a − ε and then taking ε → 0 limit gives the discontinuity in the first-order derivative at the same point:

Rl′( a + ε → 0) − Rl′( a − ε → 0) = −λRl ( a) (8.125)

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Quantum Theory of Scattering

For low energy scattering l = 0 case is the most prominent. For l = 0 : R0< (r ) =



A sin(iαr ) (8.126a) iαr

R0> (r ) = − B



e −αr (8.126b) αr

and then: R0′ ( a + 0) =



R0′ ( a − 0) =



Be −αa αa

1   α +  (8.127a) a

A  α cosh(αa) sinh(αa)  −  (8.127b) a a2 α 

On substituting Eqns. (8.127) into Eqn. (8.125) and then using Eqn. (8.124), we get: aλ =



2αa (8.128) 1 − e −2 αa

When energy is low, 1 − e −2 αa ≈ 2αa − 2(αa)2 + ......., suggesting that the minimum value of aλ is unity. Hence, λ min = 1/a. 2µE (b)  To calculate phase shift, we take k = and obtain: 2 Rl< (r ) = Ajl ( kr )



Rl> (r ) = B  cos δ l jl ( kr ) − sin δ l nl ( kr ) 

(8.129)

Continuity of wave function at r = a then requires: Ajl ( ka) = B  cos δ l jl ( ka) − sin δ l nl (ka) 



⇒

A n ( ka) = cos δ − sin δ l l B jl ( ka)

(8.130)

Equation (8.125) gives: kB  cos δ l jl′( ka) − sin δ l nl′( ka)  − Akjl′( ka) = −λAjl ( ka)

⇒

A cos δ l jl′( ka) − sin δ l nl′( ka) = λ B jl′( ka) − jl ( ka) k

(8.131)

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A Textbook on Modern Quantum Mechanics

We thus have: 1 − tan δ l

nl ( ka) jl′( ka) − tan δ l nl′( ka) = λ jl ( ka) jl′( ka) − jl ( ka) k

(8.132)    nl ( ka)  jl′( ka) nl′( ka) ⇒ tan δ l  −  = 1− λ λ  jl ( ka) jl′( ka) − jl ( ka)  jl′( ka) − jl ( ka) k k   which yields:



λ 2 jl ( ka) k tan δ l = (8.133) λ  nl ( ka) jl′( ka) − jl ( ka)  nl ( ka) + nl′( ka)  k  −

which for l = 0 reduces to: ka ka − = −λa (8.134a) tan( ka + δ 0 ) tan( ka)

And for ka  1, we obtain:



ka − 1 = −λa tan δ 0 ⇒ sin δ 0 =

((ka)

ka 2

+ (1 − λa)2

(8.134b)

)

1/2

Hence, the total scattering cross-section is: σ tot =

  a2 4π 2 sin 4 δ = π 0 2 2 2   k  ( ka) + (1 − λa) 

4πa 2 ≈ (1 − λa)2

(8.135)

3. Calculate the energy dependent phase shift δ 0 for an attractive square well potential defined by:



 −V0 V (r ) =   0 and then show that at high energies δ 0 →

ra µaV0 . 2 k

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Quantum Theory of Scattering

SOLUTION

On taking χ(r ) = rR(r ), the radial equation for l = 0 is: χ ′′ + k ′ 2 χ = 0 for r < a (8.136a)

and:

χ ′′ + k 2 χ = 0 when r > a (8.136b)

where k 2 =

2µE 2µ and k ′ 2 = 2 (E + V0 ). The solution of Eqns. (8.136) can be written as: 2   A sin( k ′r ) χ(r ) =   B sin( kr + δ 0 )



ra

(8.137)

The continuity at r = a demands: A sin( k ′a) = B sin( ka + δ 0 ) (8.138a)

and:

k ′A cos( k ′a) = kB cos( ka + δ 0 ) (8.138b)

which yield:

k tan( k ′a) = k ′ tan( ka + δ 0 ) (8.139)



V   Since k ′ 2 = k 2  1 + 0  , at high energies we have:  E

(

)

kaV0   tan( k ′a) = tan ka (1 + V0 / E) ≈ tan  ka +  (8.140)  2E 



Also k ′ → k for E → ∞. Hence: kaV0   tan  ka +  ≈ tan( ka + δ 0 ) (8.141)  2E 

which gives:

δ0 =

kaV0 µaV0 = 2 (8.141) 2E  k

4. Calculate the total scattering cross-section using the Born approximation for attractive square well potential defined above in problem 3 and compare your results with that obtained using the partial wave method for very low energies.

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SOLUTION

From Eqn. (8.100a), we obtain: a

2µ f ( k , θ) = 2 r ′V0 sin(qr ′)dr ′  q

∫ 0



=

a 2µV0  r ′ 1  cos( ) − qr ′ + 2 sin(qr ′) 2 q  q q 0 

=

2µV0 2 q2

  (8.142) 0  a

1   sin(qa) − a cos(qa)  q  

which gives: 2



2

 2µV   1  σ( k , θ) =  2 20   sin(qa) − a cos(qa)  (8.143)   q  q  where q = 2 k sin(θ / 2) = 2

2µE sin(θ / 2). For low energies k → 0 ⇒ qa  1, we get: 2 2



 2µV   1     (qa)3 (qa)2 σ( k , θ) ≈  2 20    qa − + ....  + ... − a  1 − 3! 2!       q   q  2

2

 2µV   a(qa)2   2µV0 a 3  ⇒ σ( k , θ) ≈  2 20    ≈ 2   q   3   3 

2

2

(8.144)

Then the total scattering cross-section is: σ tot = ∫ σ( k , θ)dΩ

π 2π

⇒ σ tot =

∫∫ 0 0

2

 2µV0 a 3   2µV0 a 3   3 2  sin θ dθdφ = 4π  3 2 

2

(8.145)

From the partial wave analysis method, σ tot is given by: σ tot =



4π k2

∞

∑(2l + 1)sin δ , 2

l

l= 0

which for l = 0 , reduces to:

σ tot =

4π sin 2 δ 0 (8.146) k2

The δ 0 for the attractive square potential well is given by Eqn. (8.139) 2µV0 k tan( k ′a) = k ′ tan( ka + δ 0 ). For k → 0, k ′ → k0 = 2

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Quantum Theory of Scattering

where k0 =

2µV0 . We thus have: 2

 tan( k ′a) tan ( k0 a ) 1  ( k a )3 a( k0 a)2 + ...., if k0 a  1 (8.147a) ≈ ≈  k0 a + 0 + ..... ≈ a + 3 3 k′ k0 k0  

Also:

tan( ka + δ 0 ) δ ≈ a + 0 (8.147b) k k



Therefore δ 0 ≈

ka( k0 a)2 . Then, Eqn. (8.146) gives: 3 σ tot =

4π 4π sin 2 δ 0 ≈ 2 δ 20 2 k k

⇒ σ tot

 2µV0 a 3  a6 k04 ≈ 4π ≈ 4π  9  3 2 



2

(8.148)

which is the same as is given by Eqn. (8.145). 5. Calculate the phase shift using the Born approximation for l = 0 for scattering potential defined by V (r ) = V0 e − r/a . SOLUTION

The Born approximation for phase shift for a potential is given by Eqn. (8.80): ∞

δl (k ) = −



2 2µk V (r ) ( jl ( kr )) r 2 dr. 2

∫ 0

For l = 0 , we get: ∞

δ 0 (k ) = −

2µkV0 − r/a e ( j0 (kr))2 r 2 dr 2

∫ 0

∞

=−

2µV0 − r/a e sin 2 krdr k 2

∫ 0

∞

=−

µV0 − r/a e {1 − cos(2 kr )}dr k 2

∫ 0

µV = − 20 k

a  a − 2 2  4 k a + 1 

(8.149a)

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which simplifies to: δ0 = −



µV0 a  4 k 2 a 2  (8.149b) k 2  1 + 4 k 2 a 2 

The Born approximation is valid for high energies (ka  1) and weak potential. µV a Therefore: δ 0 ≈ − 02 . k 6. A particle of charge e, mass m, and wave vector k is incident perpendicular to the direction of a dipole made of two electric charges e and −e at a mutual distance of 2a. Calculate the scattering amplitude using the Born approximation and find the directions in which the scattering cross-section has maximum value. SOLUTION

On placing charge −e at −a and e at a, the potential created by the dipole is: V (r ) = −



e2 e2 + (8.150) 4πε 0 r + a 4πε 0 r − a

Substituting the above equation into Eqn. (8.98), we have: f ( k , θ) = −



m  1  2 π 2  4πε 0 

∫

 e2 e2  3 + e i( k − k ′ ).r ′  −  d r ′ (8.151)  r′ + a r′ − a 

Using q = k − k ′ and: 1 4π = r ′ ± a (2 π)3



∫

e − ik .(r ′± a ) 3 d K (8.152) K2

we get: ∫ e iq.r ′

1 4π d3r ′ = (2 π)3 r′ ± a =



4π (2 π)3

∫∫

e iq.r ′

e − ik .r ′  ik .a 3 3 e d r ′d K K2

e i(q − k ).r ′  ik .a 3 3 e d r ′d K K2

∫∫ δ(q − K) ∫ K e

4π = (2 π)3 4π = 2 e  iq.a q

2

(8.153)

 ik .a 3

d K

Therefore:

f ( k , θ) = −

2 me 2  e iq.a − e − iq.a  4ime 2 1 = − sin(q.a) (8.154a)   2 q 2  4πε 0  2 q 2 4πε 0

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ˆ we get: On taking k = kzˆ ; k′ = k sin θxˆ + k cos θzˆ and a = ax, q 2 = ( 2 k sin(θ / 2)) = 2 k 2 (1 − cos θ)  and q.a = − ka sin θ (8.154b) 2



which yields to: 2

2

 2 m2   1  sin 2 ( ka sin θ) σ ( k , θ ) = f (θ, φ) =  2 2   (8.155)   k   4πε 0  (1 − cos θ)2 2



π (2 n + 1), where n = 0, 1, 2, 3........ 2 7. Consider the scattering by the hard sphere potential, defined by Eqn. (8.59), and jl ( ka) show that tan δ l = , where a is the range of the potential. Calculate the scatnl ( ka) tering cross-section. Find out values of the phase shift and scattering cross-section for l = 0 . σ( k , θ) becomes maximum when ak sin θ =

SOLUTION

In the exterior region r > a, the solution of the radial equation is: Rl ( kr ) = Al jl ( kr ) + Bl nl ( kr )



= Cl  jl ( kr )cos δ l − sin δ l nl ( kr ) 

(8.156)

where Cl = Al2 + Bl2 , Al = Cl cos δ l , and Bl = −Cl sin δ l. The particle cannot penetrate into the region of r < a; hence Rl ( ka) = 0. This gives: jl ( ka)cos δ l = sin δ l nl ( ka)

⇒ tan δ l =

(8.157)

jl ( ka) nl ( ka)

To calculate the scattering cross-section, we need: sin 2 δ l =



jl2 ( ka) tan 2 δ l = (8.158) 1 + tan 2 δ l jl2 ( ka) + nl2 ( ka)

Thus: σ tot =



4π k2

∞

∑

(2l + 1)sin 2 δ l =

l= 0

4π k2

∞

jl2 ( ka)

∑(2l + 1) j (ka) + n (ka) (8.159) l= 0

2 l

2 l

For l = 0 : j0 ( ka) sin( ka) =− = − tan( ka) n0 ( ka) cos( ka) (8.160a) ⇒ δ 0 = − ka tan δ 0 =

and:

σ tot =

4π sin 2 ( ka) (8.160b) k2

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For ka  1, the above equation reduces to σ tot ≈ 4πa 2, while Eqn. (8.159) goes to: ∞

σ tot  4πa



2

∑ l= 0

4

1  2 l (l !)  ( ka)4l (8.160c) (2l + 1)  (2l)! 

8.9 Exercises 1. Use the Born approximation, Eqn. (8.100a), to calculate the scattering amplitude f ( k , θ) and differential scattering cross-section σ( k , θ) for the scattering of a parβ ticle of reduced mass µ from the potential described by V (r ) = − 2 , where r + a2 ∞ x sin(bx) π dx = e − ab. Analyze your result for a → 0. β  and  a are constants. Use x2 + a2 2

∫ 0

2. Consider the case of the scattering of a particle from an attractive potential β V (r ) = − 2 . (a) Calculate the phase shift for the s-state (δ 0) using the Born approxr imation, Eqn. (8.80). Does δ 0 depend on the energy of the incident particle? (b) Calculate the scattering cross-section and compare your results with that obtained above in exercise 1 for a → 0. (c) Compute δ 0 by taking µ equal to the mass of the ∞ sin 2 ( ax) πa dx = . electron and β = 3.82meV.(nm)2. Use x2 2 0 3. For a particle of mass m, which is scattered from a potential V (r ), the wave func2 mE tion can be written as ψ (r ) = e ikz + φ(r ). k 2 = 2 , where E is the energy of the par ticle. Derive the differential equation for φ(r ) within the first Born approximation. e ikr 4. Show that for r ≠ 0, φ(r ) = − satisfies the equation ∇ 2 + k 2 φ(r ) = 0 . r 5. Evaluate the validity condition for the Born approximation, Eqn. (8.109), for the 2 λ δ(r − a), where λ > 0 . What should attractive delta function potential V (r ) = − 2µ be the relation between λ and a to use ka  1? 6. Find the Green function for the 1D Schrodinger equation G0 ( k , x , x0 ) and then find the 1D form of the integral (Lippmann-Schwinger) equation. 7. Simplify the 1D Schrodinger integral equation obtained in the above exercise for x the asymptotic region ( x  x ′) and use k ′ = . (a) Deduce the expressions for scatx tering amplitude and for differential scattering cross-section. (b) Show that the

∫

(

m reflection coefficient can be expressed as R  2  k

2 ∞

∫e

)

2

2 ik ′x ′

V ( x ′)dx ′ .

−∞

8. For exercise 7, estimate the differential scattering cross-section and reflection coefficient for V ( x) = bδ( x), where b is a constant. Explain why your estimated value of m2 b 2 the scattering cross-section differs from its exact value σ( k ) = 2 2 . m b + 4 k 2

9 Quantum Theory of Many Particle Systems There are no restrictions in applying the general postulates of quantum mechanics to any kind of system. Though until now we have mainly discussed the quantum theory of single particle systems, the postulates are applicable to all quantum systems, including those containing many particles. Experimental information of considerable interest exists on systems of many particles. In this chapter, we will be discussing the quantum mechanical treatment of many particle systems.

9.1  System of Indistinguishable Particles The particles that can be interchanged without any change in the physical properties of a system are called indistinguishable or identical particles. For example, in a system of N-electrons one can assign N-position coordinates and N-momenta, but it is impossible to tell which electron occupies which position and momentum state. The Hamiltonian of the system remains unchanged on interchanging the position and momentum of any two electrons. Any two electrons in a many electron system are identical to one another and there exists no experiment that could possibly distinguish one from the other. A Hamiltonian for an N-particles system can be expressed as: N



H (1, 2, 3....., N ) =

pi2

∑ 2m + V(1, 2, 3,...., N ) (9.1) i=1

where the numbers 1, 2, 3,...., N represent composite coordinates (both position and spin) for particles 1, 2, 3,...., N, respectively. There exists no way (experimental or theoretical) to tell which particle out of N-particles is to be called particle 1 or 2 or so on. The above equation is invariant under the interchange of any two numbers. A quantum state representing the system at a given time t may in general be defined by ψ(1, 2, 3,....., N, t) . For brevity, let us consider a system of two particles. The Schrödinger equation is:

H (1, 2) ψ (1, 2 = E ψ (1, 2 (9.2a)

On interchanging 1 and 2 we have:

H (2, 1) ψ (2, 1 = E ψ (2, 1 (9.2b)

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Next, let us define the particle exchange operator by P12 , such that P12 f (1, 2) = f (2, 1) . We then have: P12 H (1, 2) ψ (1, 2) = P12 E ψ (1, 2) = E ψ (2, 1)



= H (2, 1) ψ (2, 1)

(9.3a)

= H (2, 1)P12 ψ (1, 2) which implies:

{P12 H (1, 2) − H (2, 1)P12 } ψ(1, 2)

= 0 (9.3b)

From Eqn. (9.1), we notice that H (2, 1) can be replaced by H (1, 2), as the interchanging of 1 and 2 does not change the Hamiltonian. Therefore:

[ P12 , H (1, 2)] ψ(1, 2)



= 0 (9.4)

Since ψ (1, 2) ≠ 0, we have [ P12 , H (1, 2)] = 0 saying that the exchange operator commutes with the Hamiltonian. This means that ψ(1, 2) is an eigenstate of P12 as well, and hence, we write: P12 ψ (1, 2) = λ ψ (1, 2) (9.5)



Operating both sides with P12 from the left, we obtain: P122 ψ (1, 2) = λP12 ψ (1, 2) = λ 2 ψ (1, 2)

⇒ ψ (1, 2) = λ 2 ψ (1, 2)

(

(9.6)

)

⇒ λ 2 − 1 ψ (1, 2) = 0 where we used P122 ψ (1, 2) = P12 ψ (2, 1) = ψ (1, 2) . Therefore, λ = ±1 and ψ (2, 1) = ±ψ (1, 2) . Hence, the eigenvalue of the particle exchange operator is either +1 or −1. Thus, there can exist only two types of many particle systems: (a) whose wave function is symmetric ψ (2, 1) = ψ (1, 2), and (b) which has an antisymmetric wave function ψ (2, 1) = −ψ (1, 2). The above conclusions can be generalized for a system of an N-particle, whose quantum state at a given time t is represented by the wave function ket ψ (1, 2, 3,.., α ,.., β,..., N , t) . An operator Pαβ , whose operation on ψ (1, 2, 3,.., α ,.., β,..., N , t) interchanges α and β , will commute with H (1, 2, 3,....., N ) and hence:

Pαβ ψ (1, 2, 3,.., α ,.., β,..., N , t) = λ ψ (1, 2, 3,.., α ,.., β ,..., N , t) ⇒ ψ (1, 2, 3,.., α ,.., β ,..., N , t) = λ 2 ψ (1, 2, 3,.., α ,.., β ,..., N , t)

(9.7)

which again yields λ = ±1. The α and β can be assigned any two values from 1 to N. Also, more than one exchange operator can be applied simultaneously to a wave function. Each operation will give rise to an eigenvalue λ , and the multiple of all values of λ will

Quantum Theory of Many Particle Systems

283

be either +1 or −1. Hence, ψ (1, 2, 3,.., α ,.., β,..., N , t) is either symmetric or antisymmetric. Further, in the time-dependent Schrödinger equation:

i

∂ ψ (1, 2,..., N , t) = H (1, 2,...., N ) ψ (1, 2,..., N , t) (9.8) ∂t

Since the Hamiltonian is always symmetric, the right-hand side is either symmetric or ∂ψ antisymmetric, depending on the wave function, which means that at a given time, ∂t ∂2 ψ is symmetric if the wave function is symmetric. This implies that will also be sym2 ∂t ∂ψ metric, because is symmetric. We thus find that a symmetric will remain symmetric ∂t at all the times. A similar argument applies to an antisymmetric wave function. Hence, the symmetry of wave function is independent of time. 9.1.1  Non-interacting System of Particles In the Schrödinger representation, the Hamiltonian is time-independent, which allows us to take: ψ (1, 2, 3,....., N , t) = ψ (1, 2, 3,....., N ) e − iEt/ . This gives:

H (1, 2,...., N ) ψ (1, 2,..., N ) = E ψ (1, 2,..., N ) (9.9)

As is obvious, one can generate N ! wave function kets by applying Pαβ to ψ(1, 2,..., N ) . And, all of these wave function kets will also be the solution of Eqn. (9.9), for the same energy eigenvalue E . This kind of degeneracy, which arises due to the interchanging of an indistinguishable particle, is known as exchange degeneracy. Any linear combination of all these solutions will also be the solution of Eqn. (9.9). For brevity, let us again consider a two particle system. As discussed above, both ψ(1, 2) and ψ(2, 1) are eigenfunctions of H (1, 2) with the eigenvalue E . Then, ψ (1, 2) ± ψ (2, 1) too will be an eigenfunction of H (1, 2) for energy eigenvalue E . The Hamiltonian H (1, 2) can be expressed as H (1, 2) = H (1) + H (2) + δH, where H (1) { H (2)} is the Hamiltonian for particle 1(2) in the absence of particle 2(1). The δH is interaction energy between the particles when both are present. Let us consider the case of δH = 0 , a non-interacting particle system. For the non-interacting case, we write H (1) φ1 (1) = E1 φ1 (1) and H (2) φ2 (2) = E2 φ2 (2) with H (1, 2) = H (1) + H (2) and E = E1 + E2 , where suffixes 1(2) to φ represent a set of quantum numbers ( n, l, m, ms ), specifying the quantum states of particle 1(2). The ψ(1, 2) and ψ(2, 1) are given by φ1 (1)φ2 (2) and φ1 (2)φ2 (1) , respectively. If φ1 (1) and φ2 (2) are normalized kets, φ1 (1) φ1 (1) = 1 = φ2 (2) φ2 (2) , then normalized linear combinations of ψ(1, 2) and ψ(2, 1) are:

ψ s (1, 2) =

and:

1 {φ1 (1)φ2 (2) + φ1 (2)φ2 (1)} (9.10a) 2

1 {φ1 (1)φ2 (2) − φ1 (2)φ2 (1)} 2 2 2 1 = ∈n1n2 φn1 (1)φn2 (2) (9.10b) 2 n =1 n =1 1 2 1 φ1 (1) φ1 (2) = 2 φ2 (1) φ2 (2)

ψ a (1, 2) =

∑∑

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where ∈n1n2 = − ∈n2 n1 and its value is one when n1 n2 take values in cyclic order such as 12 and 21, zero otherwise. Equations (9.10) give wave function kets of H (1, 2) for symmetric (λ = 1) and antisymmetric (λ = −1) cases. A generalized form of Eqn. (9.10b) for a system of N non-interacting particles is:



ψ a (1, 2,....N ) =

1 N!

φ1 (1)

φ1 (2)

φ1 (3)



φ1 ( N )

φ2 (1)

φ2 (2)

φ2 (3)



φ2 ( N )

φ3 (1)

φ3 (2)

φ3 (3)



φ3 ( N )

 φ N (1)

 φ N (2)

 φ N (3)

 

 φN (N )

(9.11a)

The determinant is known as the Slater determinant. The determinant expends as a linear combination of N ! terms, with each term having a multiplication of N -single particle wave functions, as follows:



ψ a (1, 2,....N ) =

1 N!

N

N

N

∑∑ ∑ ∈ ...

n1 = 1 n2 = 1

n1n2 .... nN

φn1 (1)φn2 (2)....φnN ( N ) (9.11b)

nN = 1

where ∈n1n2 ....nN takes one of the values 1, −1, 0 at a time, depending upon in which order n1 n2 ...nN are assigned the numbers. As is seen from Eqns. (9.10), ψ s (1, 2) is obtainable from ψ a (1, 2) by replacing the minus sign with a plus sign. In the same manner, ψ s (1, 2, 3,......N ) for a system of non-interacting particles is obtained by replacing all the minus signs with plus signs in a linear form of ψ a (1, 2, 3,......N ) . The total energy of the N-particles system is E = E1 + E2 + ........ + EN . The properties of a determinant remind us of the following: 1. If two rows of a determinant are identical, its value is zero, which implies that the 2 probability ψ a of finding more than one particle in a quantum state is zero. This is exactly the statement of Pauli’s exclusion principle. We thus see that a system represented by an antisymmetric wave function ket consists of Fermions, which 1 3 obey the Fermi-Dirac statistics and have half odd integer spins , ,.... . 2 2 2. The determinant also goes to zero if any two columns are identical. This means that no two Fermions can occupy the same position.

{

2

}

On the other hand, ψ s is nonzero even if more than one particle are allowed to occupy the same quantum state or the same position. The particles which are represented by a symmetric wave function ket are known as Bosons; they follow Bose-Einstein statistics and have integer spins ( 1, 2, 3,....) . The treatment of indistinguishable particles tells us all elementary particles found in nature fall into two categories, one that is described by a symmetric wave function ket and another which is represented by an antisymmetric wave function ket. Though the analysis presented in this section has been made for noninteracting particles, the conclusions drawn are of a general nature and these are applicable to a system of N interacting particles, too.

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9.1.2  Space and Spin Parts of Wave Function In the above section we said that numbers compositely represent both space and spin coordinates. The wave function of a single particle consists of both space and spin dependent parts. If the Hamiltonian H (1, 2, 3,....., N ) does not have a space-spin interaction term, then the single particle wave function ket is expressible as the product of the space-dependent and spin-dependent wave function φα = ϕ α (r ) χ ms . Here, composite suffix α represents a set of quantum numbers n, l, m. The spin part of the wave function ket χ ms describes the orientation of spin of the particle. Unlike position coordinates, a particle with spin s can only have (2 s + 1) spin orientations. As discussed in Section 5.5, χ ms are represented by column matrices having (2 s + 1) rows with zeros at all places except one place. Two spin  1 wave function kets of sz for a spin  ±  system are:  2

 1  χ↑ =   = ↑;  0 

ms =

1 , and 2

 0  χ↓ =   = ↓;  1 

ms = −

1 (9.12) 2

In case of two electron system, symmetric and antisymmetric wave function kets are written as follows: and:

 ϕ (r , r ) χ (1, 2) s 1 2 s  ψ s (1, 2) =   ϕ a (r1 , r2 ) χ a (1, 2)

(9.13a)

 ϕ (r , r ) χ (1, 2) a 1 2 s  ψ a (1, 2) =   ϕ s (r1 , r2 ) χ a (1, 2)

(9.13b)

where, spatial parts are given by:

ϕ s (r1 , r2 ) =

1 {φα (r1 ) φβ (r2 ) + φα (r2 ) φβ (r1 )} (9.14a) 2!

ϕ a (r1 , r2 ) =

1 {φα (r1 ) φβ (r2 ) − φα (r2 ) φβ (r1 )} (9.14b) 2!

and:

The spin parts of wave function kets are:



χa =    χs =    

{

}

1 ↑1 ↓2 − ↑2 ↓1 ms = 0; S = 0, singlet state (9.15a) 2 ↑1 ↑ 2

{

1 ↑1 ↓ 2 + ↑ 2 ↓1 2 ↓1 ↓ 2

   ms = 0  ; S = 1, triplet state (9.15b)  ms = −1  ms = 1

}

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The spin part of the wave function ket is an example of addition of angular momenta. Symmetric and antisymmetric wave function kets for a system of more than two electrons are constructed in a similar manner by taking the appropriate products of spatial and spin wave functions.

9.2  The Helium Atom The helium atom is the simplest many particle system. Calculation of the ground state energy of the helium atom with the use of the variation method is presented in Section 7.2.1, without taking into consideration the spin symmetry effect and exchange degeneracy. It has distinct sets of quantum numbers where two electrons can have an antiparallel spins state or a parallel spins state. However, Pauli’s exclusion principle does not permit parallel spin state in the ground state (1s2 ) of helium. In the ground state, n = 1, l = 0 and m = 0 for both the electrons; therefore two electrons must have antiparallel spins to obey the Pauli’s exclusion principle. This is depicted in Fig. 9.1. Let us rewrite the Hamiltonian of the helium atom given by Eqn. (7.65) as follows: H = H 0 + H ′;

H0 = − H′ =

2 2 2 2 2e 2 2e 2 ∇1 − − ∇2 − 2µ 4πε 0 r1 2µ 4πε 0 r2

(9.16)

e2 4πε 0 r1 − r2

where µ is the reduced mass of the electron. Since, helium has two identical electrons, which obey Puali’s exclusion principle, the wave function ket described by Eqn. (9.13b) has to be antisymmetric. As is seen, the Hamiltonian does not have any spin-dependent term, and hence the single particle wave function ket can be taken as a product of space-dependent and spin-dependent wave function kets. We aim to evaluate H ψ a (1, 2) = E ψ a (1, 2) , where ψ a (1, 2) is a twofold degenerate state, as has been described above by Eqn. (9.13b). We use the first order perturbation theory to estimate the E by treating H ′ as the perturbating part of the Hamiltonian. Following the procedure discussed in Section 7.1.2, we write:

ψ a (1, 2) = D1 ϕ a (r1 , r2 )χ s + D2 ϕ s (r1 , r2 )χ a (9.17a)

FIGURE 9.1 Schematic diagram of the helium atom.

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Quantum Theory of Many Particle Systems

and: H 0 ψ a (1, 2) = E 0 ψ a (1, 2) (9.17b)



It is to be noted here that if both the electrons occupy the same state (1s -state) , α = β then the antisymmetric wave function is simply given by ψ a (1, 2) = φα (1)φα (2)χ a (1, 2) . As is seen from Eqn. (9.16), H 0 is the sum of two Hamiltonians for hydrogen-like atoms. Therefore, we represent φα (r1 ) and φβ (r2 ) by hydrogen-like wave functions given by Eqn. (6.35). We have:



 2 2 2e 2  φi (r1 ) = Ei φi (r1 ) ; H (1) φi (r1 ) =  − ∇1 − 4πε 0 r1   2µ  2 2 2e 2  φi (r2 ) = Ei φi (r2 ) ∇2 − H (2) φi (r2 ) =  − 4πε 0 r2   2µ

, where i ≡ (α , β) (9.18)

Let us consider: H 0 {φα (r1 )φβ (r2 ) ± φα (r2 )φβ (r1 )}

= φβ (r2 )H (1)φα (r1 ) + φα (r1 )H (2)φβ (r2 ) ± φα (r2 )H (1)φβ (r1 ) ± φβ (r1 )H (2)φα (r2 ) = (Eα + Eβ )φα (r1 )φβ (r2 ) ± (Eβ + Eα )φα (r2 )φβ (r1 )

(9.19)

= (Eα + Eβ ) {φα (r1 )φβ (r2 ) ± φα (r2 )φβ (r1 )} which implies:

H 0 ψ a (1, 2) = D1 H 0 ϕ a (r1 , r2 )χ s + D2 H 0 ϕ s (r1 , r2 )χ a = ( Eα + Eβ ) ψ a (1, 2)

(9.20)

Therefore: E 0 = Eα + Eβ (9.21)



where each of Eα and Eβ is the energy of a state in the hydrogen atom with charge 2e at the nucleus. Therefore:

E0 = −

4µe 4 2(4πε 0 )2  2

 1 1  2 + 2  (9.22) n n β   α

where both nα and nβ take values 1, 2, 3,..... The first order correction to energy E ′ , which is obtained from Eqn. (7.49) for the degenerate system, is:

haa − E ′ hsa

has hss − E ′

= 0 (9.23)

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where:

haa = ϕ a (r1 , r2 ) H ′ ϕ a (r1 , r2 ) χ s χ s = ϕ a (r1 , r2 ) H ′ ϕ a (r1 , r2 ) (9.24a)



hss = ϕ s (r1 , r2 ) H ′ ϕ s (r1 , r2 ) χ a χ a = ϕ s (r1 , r2 ) H ′ ϕ s (r1 , r2 ) (9.24b)

and: has = hsa* = ϕ a (r1 , r2 ) H ′ ϕ s (r1 , r2 ) χ a χ s (9.24c)



Since χ a χ s = 0 = χ s χ a , has = 0 = hsa . Therefore, two values of E ′ are:

E−′ = haa = ϕ a (r1 , r2 )

e2 ϕ a (r1 , r2 ) (9.25a) 4πε 0 r1 − r2

E+′ = hss = ϕ s (r1 , r2 )

e2 ϕ s (r1 , r2 ) (9.25b) 4πε 0 r1 − r2

and:

After substituting Eqns. (9.14) into Eqn. (9.25), we rearrange as follows: e2 4πε 0

E±′ =



1  2

∫ ∫ φ (r ) α

1

2

φβ (r2 )

+

1 2

∫ ∫ φ (r )

±

1 2

∫∫

±

1 2

∫ ∫ φ (r )φ (r ) r − r

α

2

2

φβ (r1 )

φα* (r1 )φβ* (r2 ) * α

2

* β

2

2

1 d 3 r1 d 3 r2 r1 − r2

1 d 3 r1 d 3 r2 r1 − r2

1 φβ (r1 )φα (r2 )d 3 r1d 3 r2 r1 − r2 1

1

1

2

(9.26)

 φβ (r2 )φα (r1 )d 3 r1 d 3 r2  

As is seen, integration is done over both r1 and r2 ; hence, interchanging of r1 and r2 does not affect the value of any term. However, interchanging of r1 and r2 in the second and fourth terms makes these equal to first and third terms, respectively. We therefore obtain E±′ = Jαβ ± Kαβ , where:

Jαβ =

∫∫

2

φα (r1 ) φβ (r2 )

2

e2 d 3 r1 d 3 r2 (9.27a) 4πε 0 r1 − r2

and:

Kαβ =

∫∫

φ*α (r1 )φβ* (r2 )

e2 φβ (r1 )φα (r2 )d 3 r1 d 3 r2 (9.27b) 4πε 0 r1 − r2

The Jαβ is a direct electron-electron interaction contribution, whereas Kαβ is an exchange electron-electron contribution to the energy of the helium atom. Thus the two energies of helium atoms are: E+ = E 0 + Jαβ + Kαβ ; E− = E 0 + Jαβ − Kαβ , and their difference:

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Quantum Theory of Many Particle Systems

FIGURE 9.2 Schematic diagram for energy level splitting for the

100

to

nlm

transition in the helium atom.

∆E = E+ − E− = 2 Kαβ . Note in the case of α = β , there is no exchange electron-electron interaction and hence Kαβ = 0 . 2 2 In Eqn. (9.27a), − e φα (r1 ) d 3 r1 and − e φβ (r2 ) d 3 r2 represent the charge distribution in 3 3 infinitesimally small volumes d r1 and d r2 , respectively. Therefore, Jαβ is the quantum analogue of classical electrostatic interaction between two charges. However, there is no classical equivalence of Kαβ ; it is purely a quantum mechanical correction to the Coulomb interaction energy. Calculation of Jαβ and Kαβ with the use of hydrogen-like wave functions shows that both are positive. Hence, E+ > E− . Looking at Eqns. (9.13) to (9.15), we notice that (i) E+ corresponds to ϕ s and χ a , which represents the singlet state, and (ii)  E− belongs to ϕ a and χ s , which corresponds to the triplet state. Therefore, the singlet state has higher energy as compared to the triplet state, as is shown in Fig. 9.2. The physical interpretation is as follows: The singlet state is represented by the symmetric space-coordinate dependent part of the wave function, and therefore electrons have the tendency to come closer, making the electrostatic repulsion less serious, which results in higher energy, whereas, the triplet state involves the antisymmetric space-dependent part of the wave function, giving rise to stronger repulsive interaction and hence lower energy. Helium in the singlet state is known as parahelium, while in the triplet state it is termed orthohelium. In the ground state helium is parahelium only. It has been verified experimentally that the singlet state has more energy than the triplet state. 9.2.1  Ground State of Helium In the ground state both the electrons are in the 100 state (α = β); therefore, Kαβ = 0 and: J100 =

e2 4πε 0

∫∫ ψ

2

100

 e2  5 × =  4πε 0  4 a0

(r1 ) ψ 100 (r2 )

2

1 d 3 r1 d 3 r2 r1 − r2 (9.28)

5 = − E1 2 where, we have used the results reported in Section 7.2.1 (see Eqn. 7.74 with Z = 2) . The E1 ( = −13.6 eV ) is the ground state energy of hydrogen atom; Eqn. (6.22). The ground state energy of the helium atom then is:

Eg = 8E1 −

5 11 11 E1 = E1 = − × 13.6 = −74.8eV. (9.29) 2 2 2

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We thus see that the perturbation theory underestimates the ground state energy as compared to that evaluated with the use of the variational method ( Eg = −77.6 eV); see Eqn. (7.77). 9.2.2  Excited State of Helium In the excited state of helium two elections are in two different states. Let us consider one electron in α ≡ 100 (1s ) and the other electron in β ≡ nlm , which could be any of 2 s, 2 p, etc. In the excited state, both Jαβ and Kαβ will be nonzero, and therefore both para 2 and ortho forms exist. As is seen from Eqns. (9.14a) and (9.14b), computation of ϕ a (r1 , r2 ) 2 2 2 and ϕ s (r1 , r2 ) as a function of r1 − r2 , gives ϕ a (r1 , r2 ) → 0 and ϕ s (r1 , r2 ) equal to a maximum value at r1 = r2 . This too means that two electrons have the tendency to avoid each other when represented by ϕ a , whereas they like to reach closest to each other when represented by ϕ s . Hence, in the excited state, parahelium has more energy than ortho2 helium. The dip in ϕ a (r1 , r2 ) around r1 = r2 is called the Fermi hole, while an increase in 2 ϕ s (r1 , r2 ) is termed a Fermi heap. Each configuration, 1s − 2 s, 1s − 2 p, etc., will split into the para state and the ortho state, where the para state lies higher on the energy scale. It is important to note that though the Hamiltonian described by Eqn. (9.16) is independent of spin variables, yet there is a spin-dependent effect on energy, which arises from the FermiDirac statistics. This can be understood from the Heisenberg theory of ferromagnetism. The idea of the alignment of spins has been extended over microscopic distances in the helium atom to give a physical interpretation of spin-dependent energy. The Jαβ and Kαβ have been evaluated for 1s − 2 s and 1s − 2 p configurations. The values 6

68  2 E1 = 11.42 eV and K1s ,2 s = −   E1 = 1.20 eV; for  3 81 236 7 × 26 1s − 2 p configuration: J1s ,2 p = − E1 = 13.22 eV and K1s ,2 p = − E1 = 0.94 eV. The sin243 38 glet and triplet states are separated by 2.4 eV in 1s − 2 s configuration and by 1.88 eV in 1s − 2 p configuration.

for 1s − 2 s configuration are: J1s ,2 s = −

9.3  Systems of N-Electrons The Hamiltonian for an N-electron system such as a metal or a large atom is the generalization of Eqn. (9.16): N



H=

∑ i=1

 2 1 − ∇ 2i + Vext (ri ) + 2  2 m

∑ j≠ i

e2 4πε 0 ri − rj

where single electron potential at the position ri is Vext (ri ) = −

  (9.30) 

Ze 2 4πε 0

∑ r −1 R R

for a metal,

i

Ze 2 for 4πε 0 ri an atom. In the case of a metal, the Hamiltonian may also be written to have an additional

with R a position vector of a bare nuclei at a Bravais lattice point, and Vext (ri ) = −

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Quantum Theory of Many Particle Systems

constant term arising from nuclei-nuclei interactions. Since this does not affect the physics of electrons, it is dropped here. The N-electrons wave function ψ(r1 s1, r2 s2 , r3 s3 ,.......rN sN ), ri si ≡ i representing both position and spin coordinates, satisfies the Schrödinger equation: Hψ = Eψ (9.31)



The exact solution of Eqn. (9.31) is not possible, and hence one looks for an approximate solution that is most reasonable. An electron interacts with N-1 electrons, in addition to its interaction with nuclei. The methods based on the variational principle have been found most successful and reasonable approximations to solve the N-electrons Schrödinger equation. The variational principle assumes that the equation: δ δψ



{ψHψ

−E ψ ψ

} = 0 (9.32)

is equivalent to the Schrödinger equation. In other words, ψ that satisfies above equation is a solution of the Schrödinger equation or is an eigenfunction. More explicitly, if we take the functional derivative of:

H −E =

∫ ψ (r)Hψ(r)d r − E ∫ ψ (r)ψ(r)d r (9.33a) *

3

*

3

with respect to ψ * (r ) , we get:

δ H −E =

∫ {Hψ(r) − Eψ(r)} δψ (r)d r (9.33b) *

3

To make this stationary, the variation must be zero for all possible forms of δψ * (r ) , and hence we should have Hψ (r ) − Eψ (r ) = 0 , which is the Schrödinger equation. 9.3.1  Hartree Approximation We rewrite Eqn. (9.30) as follows: 

N

H=





∑  H + 21 ∑V  (9.34) i

i=1

ij

j≠ i

with: Hi = −



2 2 ∇ i + Vext (ri ) (9.35a) 2m

and: Vij =



e2 (9.35b) 4πε 0 ri − rj

We take the trial wave function:

ψ (1, 2, 3,.....N ) = φ1 (11 )φ2 (2)φ3 (3).........φ N ( N ) (9.36)

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with φi (ri si ) φ j (rj s j ) = δ ij . We then have: ψHψ N

=



∑ ∫ d r ∫ d r ...∫ d r ....∫ d r φ (r )φ (r )...φ (r )...φ 3

3

1

3

2

3

i

* N 1

1

* 2

2

* i

i

* N

(rN ) (9.37)

i=1

  × Hi + 

  Vij  φ1 (r1 )φ2 (r2 )...φi (ri )...φ N (rN )  j≠ i

∑

Here,

∫ d r ∫ d r ...∫ d r ....∫ d r φ (r )φ (r )...φ (r )...φ = d r φ (r )H φ (r ) ∫ 3

1

3

3

3

2

* i i

i

i i

3

i

* N 1

1

* 2

* i

2

i

* N

(rN )H i φ1 (r1 )φ2 (r2 )...φi (ri )...φ N (rN ) (9.38a)

i

Note that: because H i operates only on φi (ri ) and integration over each of the other coordinates is equal to one. The electron-electron interaction term simplifies to ∫ ∫ d 3 ri d 3 rj φ*i (ri )φ*j (rj )Vij φi (ri )φ j (rj ) . We then have: N

ψHψ =



∑∫ i=1

N   1   d 3 ri φ*i (ri )  H i φi (ri ) + d 3 rj φ*j (rj )Vij φ j (rj )φi (ri )  (9.38b) 2 j( ≠ i )= 1  

∑∫

To minimize ψ H ψ , we take the functional derivative with respect to φ*k (rk ) and obtain:

∫



N     d 3 rk δφ*k (rk )  H k φ k (rk ) + d 3 rj φ*j (rj )Vkj φ j (rj )φ k (rk )  = 0 (9.39a)   j( ≠ k )= 1

∑∫

1 before interaction term drops out because two equal 2 terms, one for i = k and the other for j = k , appear when we take the derivative with respect to φ*k (rk ) . Further notice that φ*k φ k = 1 , and therefore: It is to be noted that the factor of

∫

ε k d 3 rk δφ*k (rk )φ k (rk ) = 0 with k = 1, 2, 3,.....N (9.39b)



Here, ε k is a multiplying constant. Subtracting Eqn. (9.39b) from Eqn. (9.39a) we get:



∫

N     d 3 rk δφ*k (rk )  H k φ k (rk ) + d 3 rj φ*j (rj )Vkj φ j (rj )φ k (rk ) − ε k φ k (rk )  = 0 (9.40)   j( ≠ k )= 1

∑∫

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Quantum Theory of Many Particle Systems

To make it stationary, the variations are zero for all possible forms of δφ*k (rk ) ; hence we should have: N

H k φ k (rk ) +



∑ ∫ d r φ (r )V φ (r )φ (r ) = ε φ (r ) (9.41a) 3

j

* j

j

kj

j

j

k

k

k

k

k

j( ≠ k )= 1

More explicitly:

 2 e2  ∇ i2 + Vext (ri ) + − 4πε 0  2 m

N

∑∫

d 3 rj φ*j (rj )

j( ≠ i )= 1

 1  φ j (rj )  φi (ri ) = ε i φi (ri ) (9.41b) ri − rj 

which is an integro-differential equation, commonly known as the Hartree equation. It is an eigenvalue equation for an electron located at position ri and moving under the influence of an effective potential:

Veff (ri ) = Vext (ri ) + Vee (ri ) (9.42a)

with:

e2 Vee (ri ) = 4πε 0

∑∫ d r 3

j≠ i

j

φ j (rj ) ri − rj

2

(9.42b)

The simple interpretation of this is as follows: The ith electron interacts with both nuclei and the remaining N-1 electrons. The remaining electrons are treated as a smooth distribution of negative charge with charge density ρ(rj ) around the position rj ; then the potential energy due to the interaction between the ith electron and the charge in infinie ρ(rj ) 3 d rj . The contribution to charge density by a tesimally small volume d 3 rj is: − 4πε 0 ri − rj 2 2 single electron and by the N-1 electron would be − e φ j (rj ) and − e φ j (rj ) , respectively.

∑ j≠ i

Hence, Eqn. (9.42b) represents the potential energy due to the interaction of the ith electron with the remaining N-1 electrons. We thus find that the Hartree approximation converts the N-particle problem into a set of N-single particle equations that can be solved. It is to be noted that solving Eqn. (9.41b) for state φi (ri ) and eigenvalue ε i requires the prior knowledge of φ j (rj ) . Therefore, the equation is to be solved by iterative methods. One starts with a guessed single particle state to compute the Eqn. (9.42b). The computed Veff (ri ) is then used to compute a new state. If the computed state differs from that used to compute Vee (ri ), repeat the procedure until a self-consistency is not achieved. 9.3.2  Hartree-Fock Approximation Equation (9.41b) is a single electron equation that uses a potential obtained by averaging over the positions of the remaining N-1 electrons. It does not represent the way in which a particular electron is affected by the configuration of other N-1 electrons. In other words, Eqn. (9.41b) does not take care of the exchange effect, which demands that the

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wave function for a system of N-Fermions must be anti-symmetric. As discussed before, an anti-symmetric wave function is given by Eqn. (9.11b) not by Eqn. (9.36). However, the Hamiltonian of the system of N-electrons is still given by Eqn. (9.34). To derive the single electron equation within the Hartree-Fock approximation, we follow the procedure outlined for the Hartree approximation, with the use of the wave function given by Eqn. (9.11b), which is: ψ a (r1 , r2 ,....rN ) =



1 N!

N

N

N

∑∑...∑ ∈

n1n2 .... nN

n1 = 1 n2 = 1

φn1 (r1 )φn2 (r2 )....φnN (rN ) (9.43)

nN = 1

Let us first evaluate the matrix element of H i , defined in Eqn. (9.35a): ψ a Hi ψ a N  N N  1 * 3 3 3  ... = d r1 d r2 ......d rN ∈n1′ n2′ ....nN′ {φn1′ (r1 )φn2′ (r2 )....φnN′ (rN )}  × (9.44) N!  n1′ = 1 n2′ = 1 nN′ = 1 

∑∑ ∑

∫



N  N N  ... Hi  ∈n1n2 ....nN φn1 (r1 )φn2 (r2 )....φnN (rN )   n1 = 1 n2 = 1 nN = 1 

∑∑ ∑

There will be an integral over ( N !)2 terms, where each term involves a product of the N wave functions and an equal number of their complex conjugates. To understand the simplification of this equation, let us consider a case of two electrons situated at ri and rj . ψ a Hi ψ a

=

 2 2 1 d 3 ri d 3 rj  ∈n1′ n2′ φn1′ (ri )φn2′ (rj ) 2!  n1′ = 1 n2′ = 1

∑∑

∫

{

}

*

  2 2  (9.45a)  Hi  ∈n1n2 φn1 (ri )φn2 (rj )    n1 = 1 n2 = 1 

∑∑

Using ∈11 = ∈22 = 0 and ∈12 = − ∈21 = 1 , the equation reduces to: ψ a Hi ψ a

=

(9.45b) 1 d 3 ri d 3 rj φ*1 (ri )φ*2 (rj ) − φ*2 (ri )φ*1 (rj )  H i φ1 (ri )φ2 (rj ) − φ2 (ri )φ1 (rj )  2!

∫

which expands to: ψ a Hi ψ a =

2 1 3 d ri d 3 rj φ2 (rj ) φ*1 (ri )H i φ1 (ri ) + 2! 

(9.45c) ∫ ∫ ∫ d r ∫ d r φ (r ) φ (r )H φ (r ) 1  −  d r { d r φ (r )φ (r )} φ (r )H φ (r ) + d r { d r φ (r )φ (r )} φ (r )H φ (r )  ∫ ∫ ∫ ∫ 2!   3

i

3

* j 2

j

1

j

* 1

i

i 2

3

i

i

3

2

1

j

3

i

j

3

* 2

* j 1

i 2

i

j

2

i

j

* 2

i

i 1

i

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Quantum Theory of Many Particle Systems

The third and fourth terms are zero because of the orthonormality of φ1 and φ2 . The second term becomes identical to the first term on interchanging integrals over ri and rj . We therefore get:

∫ d r φ (r )H φ (r ) (9.46a) 3

ψ a Hi ψ a =



* i 1

i

i 1

i

The answer remains same, even when the two-electrons antisymmetric wave function is replaced by an N-electrons wave function to evaluate ψ a H i ψ a . However, in place of suffix 1, we can use a more generalized suffix ni to write: ψ a Hi ψ a =



∫ d rφ 3

* i ni

(ri )H i φni (ri ) (9.46b)

We next take the evaluation of ψ a Vij ψ a . Following the procedure of Eqns. (9.26) and (9.27), we find that for a two-electrons system:

ψ a Vij ψ a =

e2 4πε 0

∫∫ φ (r ) 1

e2 − 4πε 0

∫∫

i

2

φ2 (rj )

2

1 d 3 ri d 3 rj ri − rj

1 φ (ri )φ (rj ) φ2 (ri )φ1 (rj )d 3 ri d 3 rj ri − rj * 1

(9.47a)

* 2

A generalization for N-electrons can made by representing quantum states by ni and n j in place of 1 and 2: e2 4πε 0

ψ a Vij ψ a =

∫∫ φ

e2 − 4πε 0

∫∫

2

ni

(ri ) φn j (rj )

2

1 d 3 ri d 3 rj ri − rj

1 φ (ri )φ (rj ) φn j (ri )φni (rj )d 3 ri d 3 rj ri − rj * ni

(9.47b)

* nj

From Eqn. (9.34), we then have: N

ψa H ψa =

∑∫ i=1



2  φ* (r ) −  ∇ 2 + V (r )  φ (r )d 3 r ni i  i ext i  ni i i   2m 

+

−

1 2

∑ ∫∫ φ

1 2

∑ ∫∫ φ

2

ni

(ri ) φn j (rj )

* ni

(ri )φ*n j (rj )

i≠ j

i≠ j

2

e2 d 3 ri d 3 rj 4πε 0 ri − rj

 e2 φn j (ri )φni (rj )d 3 ri d 3 rj  4πε 0 ri − rj  

(9.48)

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On the right-hand side, the third term consists of φ*ni (ri )φni (rj ), while the second term 2 has φni (ri ) . Minimization with respect to φ*ni (ri ) following the procedure described for Eqns. (9.39) to (9.41) gives us:



 2 2  e2 ∇ i + Vext (ri )  φni (ri ) + − 4πε 0  2m  e2 − 4πε 0

   

∑∫ j≠ i

   

∑∫ φ j≠ i

nj

(rj )

2

 1  d 3 rj  φni (ri ) ri − rj 

 1  φ*n j (rj ) φni (rj )d 3 rj φn j (ri )  = ε ni φni (ri ) ri − rj 

(9.49)

This equation is known as the Hartree-Fock equation. On comparing Eqn. (9.49) with Eqn. (9.41b), we find that the third term on the left is originated from exchange interactions. Unlike the direct interactions term (second term), the exchange interactions term (third term) is nonlinear having the structure V (r )φ(r ) = ∫ U (r ′ , r )φ(r ′)d 3 r ′ , which is an integral operator 9.3.3  Thomas-Fermi Theory For a system of N-electrons in a stationary state, this theory tries to avoid the complicated many-electron wave function by using electron density n(r ) , which is physically observable, measured, calculated, and easily visualized. n(r ) is simply a probability density to find the particle near some position r and it is defined by n(r ) = φ* (r )φ(r ), if there is one particle only. For a system of N-particles, electron density to find a particle at or around r is given by:

∫ ∫

∫

n(r ) = N d 3 r2 d 3 r3 ....... d 3 rN ψ * (r , r2 , r3 ,......rN )ψ (r , r2 , r3 ,......rN ) (9.50)

Essentially, the Thomas-Fermi method concentrates on the particle density as a key variable by circumventing completely the discussion on the wave function. The Thomas-Fermi approach is semiclassical theory where certain ideas are borrowed from quantum mechanics. The condition to apply a semi-classical approach to a system is that the spatial variations of the de Broglie wavelength must be small; mathematically it means:

dλ( x)  1 (9.51) dx

Thomas-Fermi method borrows two ideas from quantum mechanics: (i) Fermi statistics–all the states up to some maximum energy and momentum, say pF , which may vary over the space are occupied, and (ii) The principle of uncertainty–every cell of phase space (of volume h3 ) can host up to two electrons with opposite spin directions. Thus in the ground state volume occupied by the electrons in the phase space would be 4πV 3 pF , where V is the volume of a system in real space. It is assumed that all electrons 3

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are accommodated up to the phase space sphere of radius pF . Therefore, the total number of electrons is:

 4πVpF3   h3  8πVpF3 / = N= (9.52) 3 h3  3   2 

and hence:

n(r ) =

pF3 N 8πpF3 = = 2 3 (9.53) 3 V 3h 3π 

which gives:

{

}

pF (r ) = k F (r ), with k F (r ) = 3π 2 n(r )

1/3

(9.54)

The pF and k F are known as Fermi momentum and Fermi wave vector, respectively. It is assumed here that pF and hence n(r ) vary with space coordinates over the region in which the condition, Eqn. (9.51), is fulfilled. Assuming that all electrons move as classical particles under the influence of common potential, the classical energy of fastest moving electrons is:

Emax =

pF2 (r ) + V (r ) (9.55) 2m

where both kinetic and potential energy parts may independently depend on r, and their sum in equilibrium remains constant. We next take total energy (T + U ) of the entire electrons distribution. With kinetic energy density t(r ), we write T = ∫ t(r )d 3 r . To evaluate t(r ), let us calculate the fraction of electrons that have momentum between p and p + dp:



   4πp 2 dp  p 2 dp F( p)dp =   = 2 3 when p < pF , zero otherwise. (9.56) 3  h  π   2 

The t(r ) at around r , when the classical expression for the kinetic energy of an electron is used, is: pF

t(r ) =

∫ 0



p2 F( p) dp 2m

1 = 2 mπ 2  3

pF

∫ p dp 4

0

5 F

p 1 × 2 mπ 2  3 5 2 2/3 4/3 3 (3) π = {n(r)}5/3 10m =

(9.57a)

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or: t(r ) = Ck n5/3 (r ), with Ck =



3 2 (3)2/3 π 4/3 (9.57b) 10m

The total kinetic energy therefore is:

∫

T = Ck n5/3 (r )d 3 r (9.58a)



The potential energy at r is due to the interaction with the external field and electrostatic interaction of the electron density with itself. Therefore: U=



∫

n(r )Vext (r )d 3 r +

1 e2 2 4πε 0

∫∫

n(r )n(r ′) 3 3 d rd r ′ (9.58b) r − r′

Ze 2 In the case of an atom with nucleus of charge Ze centered at r = 0 , Vext (r ) = − . 4πε 0 r The total energy of the system is:

∫

Etot = T + U = Ck n5/3 (r )d 3 r +

∫

n(r )Vext (r )d 3 r +

1 e2 2 4πε 0

∫∫

n(r )n(r ′) 3 3 d rd r ′ (9.59) r − r′

Next, we search for the electron density, which minimizes the total energy subject to the normalization condition ∫ n(r )d 3 r = N . On introducing a Lagrange multiplier µ, the method of the variational principle with respect to n(r ) gives:

δ(Etot − µN ) =

∫

5 e2 2/3  Ck n (r ) + Vext (r ) + 4πε 0 3

n(r ′)



∫ r − r ′ d r ′ − µ  δn(r)d r = 0 (9.60) 3

3

which yields:

µ=

5 e2 2/3 Ck {n(r )} + Vext (r ) + 3 4πε 0

n(r ′)

∫ r − r ′ d r ′ (9.61) 3

This equation is known as the Thomas-Fermi equation for determining the equilibrium ∂E distribution of the electron density. The µ = tot suggests that it is the chemical potential. ∂N Also, it is energy of the fastest moving electron, which is generally known as Fermi energy. 9.3.4  Thomas-Fermi Model of Atom Since the electrons cannot escape from the atom, µ = 0 for an atom in equilibrium. Hence:

  3 n(r ) =  − Veff (r )   5Ck 

3/2

(9.62a)

with:

Veff (r ) = Vext (r ) +

e2 4πε 0

n(r ′)

∫ r − r ′ d r ′ (9.62b) 3

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Because of spherical symmetry inside the atom, both electron density and Veff (r ) would −Veff (r ) be a function of r = r . The electrostatic potential and charge density ρ = − en(r ) e satisfy the Poisson’s equation: 1 2 en(r ) ∇ Veff (r ) = − e ε0



1 d  2 dVeff  e 2 n(r ) ⇒ 2 r =−   r dr  dr  ε0

(9.63a)

which with the use of Eqn. (9.62a) gives: e2 1 d  2 dVeff (r )  r = −   r 2 dr  dr  ε0



  3  − 5C Veff (r )  k  

3/2

(9.63b)

For r → 0 the leading term in Veff (r ) is the potential due to the nucleus and hence Ze 2 Ze 2 Veff (r ) → − . Therefore, it is convenient to introduce Veff (r ) = − χ(r ) . We then 4πε 0 r 4πε 0 r have: d 2 χ(r ) e 3 = dr 2 ε0



which on choosing x =

Z  3    4πε 0  5Ck 

5ε C  Z  r with b = 0 2 k   b 3e  4π 

χ 3/2 (9.64) r

1/3

, reduces to:

d 2 χ( x) {χ( x)} = dx 2 x



3/2

3/2

(9.65)

which is known as the Thomas-Fermi equation for an atom. Note that χ(0) → 1, while χ(∞) → 0 . 9.3.5  Density Functional Theory The Thomas-Fermi theory, which is the oldest theory to describe the electronic energy in terms of electron density distribution, was found quite useful to explain qualitative features of atoms. However, it is not of much use when one wants to study the electronic structures of large molecules and materials. A more generalized theory in terms of electron density distribution was derived by Hohenberg and Kohn, and later by Kohn and Sham, where it is found that the Thomas-Fermi theory is a special case of a generalized density functional theory (DFT). The two Hohenberg-Kohn theorems, which are the basis of DFT are as follows: Theorem-1: The ground state properties of a many-electron system depend only on the electronic density n(r ) and the external potential is determined by n(r ) . Theorem-2: The correct ground state density for a system is the one that minimizes the total energy through the functional E [ n(r )] and yields ∫ n(r )d 3 r = N .

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As compared to wave function dependent approaches, such the Hartree-Fock or post Hartree-Fock theories, DFT provides an alternative method to reformulate the many-body problem, where the density of electrons rather than a many-electrons wave function plays a central role. The wave function approach involves a large number of orbitals in terms of 3N coordinates of N-electrons, whereas DFT involves only one variable n(r ) that depends on 3-coordinates. The N-electrons Hamiltonian, Eqn. (9.30), is: N

H = T + Vee +



∑V

ext

(ri ) (9.66)

i=1

where T and Vee , kinetic energy and electron-electron interaction energy operators, are called universal operators, as they are the same for any of the N-electron systems. The third term, potential energy due to the interaction of the electron with an external potential (such as a positively charged nuclei), is system dependent. The N and Vext (r ) determine all the properties of the ground state. The functional, which is universal in the sense that it does not refer to any system or external potential Vext (r ) is defined as: F[n] = φ (T + Vee ) φ



min

(9.67)

The minimum is taken over all single particle wave functions φi , which constitute the Hartree wave function: ψ (r1 , r2 , r3 ,......rN ) = φ1 (r1 )φ2 (r2 )φ3 (r3 )........φ N (rN ) . We then write:

∫

E[n] = F[n] + Vext (r )n(r )d 3 r ≥ Eg (9.68)



for all N-representable n(r ) . Here, Eg is ground state energy. Many of the drawbacks of the Thomas-Fermi theory are due to the approximate treatment of kinetic energy. Use of a different separation introduced by Kohn and Sham greatly simplifies the task of finding good approximations to the energy functional:

E[n] = T0 [n] +

∫

 1 e2 n(r ) Vext (r ) + 2 4πε 0 

n(r ′)



∫ r − r ′ d r ′  d r + E 3

3

xc

(n) (9.69)

Here T0 [n] is the kinetic energy of a system with density n(r ) in the absence of electronelectron interactions. The Exc [n] defines the exchange-correlation energy. The exact treatment of T0 in this approach removes many of the deficiencies faced in the Thomas-Fermi approximation. Though T0 differs from the true kinetic energy T , magnitudes of two are comparable. All terms except Exc [n] in Eqn. (9.69) are evaluated exactly. Therefore, unavoidable approximations for Exc [n] play a central role in DFT calculations. The exact exchangecorrelation energy functionals either are not known or they are so complicated that it is not useful to compute these. Hence, various approximate exchange-correlation functionals (S-VWN, B3LYP, etc.) are used. Application of the variational principle to Eqn. (9.69) yields:

δE[n] δT0 [n] e2 = + Vext (r ) + δn(r ) δn(r ) 4πε 0

n(r ′)

∫ r − r′ d r′ + 3

δExc (n) = µ (9.70) δn(r )

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where µ is the Lagrange multiplier associated with the requirement of the constant particle number N. As is seen, mathematically the problem is identical to that described in the Thomas-Fermi theory, if effective potential is defined by: Veff (r ) = Vext (r ) +



e2 4πε 0

n(r ′)

∫ r − r′ d r′ + 3

δEex (n) (9.71) δn(r )

in place of that given by Eqn. (9.62b). A major step towards the solution of such an equation was introduced by Kohn and Sham by introducing an orbital method, which exactly evaluates: N

T0 [n] =



∑φ i=1

i

 2 2   − 2 m ∇ i  φi (9.72)

In order to evaluate the kinetic energy of N-noninteracting particles whose density distribution is known, they simply found the corresponding Veff (r ) and used the Schrödinger equation:   2 2  − 2 m ∇ + Veff (r ) φi (r ) = ε i φi (r ) (9.73)

such that:

N

n(r ) =



∑ φ (r) i

2

(9.74)

i=1

The φi (r ) here are ordered so that the energies ε i are non–decreasing, and the spin index is included in suffix i . Equations (9.71) and (9.73) are nonlinear and are to be solved selfconsistently subject to the condition (Eqn. 9.74), with the use of the iterative method. Much like the Hartree-Fock equation, exchange-correlation potential is used in Eqn. (9.73). Looking at Eqns. (9.61) and (9.71), it is said that the Thomas-Fermi theory is a special case of DFT.

9.4  Solved Examples 1. Consider a system of two non-interacting particles; both are independently attached to the origin by the force constants k1 and k2 , respectively. Each particle has mass m . What are the eigenfunction and eigenvalue for the system in the nth state. The single particle coordinates are x1 and x2 . SOLUTION

For a single particle case, the Schrödinger equation for a harmonic oscillator is:

−

2 d2 ψ 1 2 + kx ψ = Eψ (9.75) 2 m dx 2 2

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As discussed in Chapter 3, Section 3.19, the solution of Eqn. (9.75) for the nth state is given by: ψ ( y ) = N n H n ( y )e − y

2

/2

1/2

1 and En = (n + )ω , where ω = k / m , and y = αx 2

 mω  with α =  .    For the given system of two particles, we have:

  2 ∂2   2 ∂2 1 1 2 − + k x − + k2 x22  ψ ( x1 , x2 ) = Eψ ( x1 , x2 ) 1 1  2m ∂x 2 2 2 2 m ∂ x2 2 (9.76) 1  



⇒ [ H 1 ( x1 ) + H 2 ( x2 )] ψ ( x1 , x2 ) = Eψ ( x1 , x2 )

As is seen, the Hamiltonian is the sum of two independent Hamiltonians and hence the solution of Eqn. (9.76) can be taken as ψ ( x1 , x2 ) = φ1 ( x1 )φ2 ( x2 ) . We thus obtain: H 1 ( x1 )φ1 ( x1 )φ2 ( x2 ) + φ1 ( x1 )H 2 ( x2 )φ2 ( x2 ) = Eφ1 ( x1 )φ2 ( x2 ) (9.77a) 1 1 ⇒ H 1 ( x1 )φ1 ( x1 ) + H 2 ( x2 )φ2 ( x2 ) = E φ1 ( x1 ) φ 2 ( x2 )



Writing E = E1 + E2 , we obtain: H i φi ( xi ) = Ei φi ( xi ),(i = 1, 2) (9.77b)



The solution of Eqn. (9.77b) is: 1  − y 2 /2 Eni =  ni +  ω i ,  and  φi ( xi ) = N ni H ni ( y i )e i , with i = (1, 2) (9.78)  2



y1 = α 1 x1 and

where we used α 24 =

y 2 = α 2 x2 , with α 14 =

2

mk2  mω 2  = . We thus have:    2

mk1  mω 1  =    2

2

and

1 1   En1 , n2 =  n1 +  ω 1 +  n2 +  ω 2 (9.79a)   2 2

and:

ψ n1 , n2 ( y1 , y 2 ) = N n1 , n2 H n1 ( y1 )H n2 ( y 2 )e

(

− y12 /2 + y 22 /2

) (9.79b)

The total energy of the system is the sum of the energies of two non-interacting systems, which is in conformity to the statement made in Section 9.1.1. 2. Consider again the system of two oscillating particles, each of massm . Unlike problem 1, two particles have the same spring constant and they interact with each other. 1  2 ∂2  2 ∂2 The Hamiltonian of the system is H = − − + k( x12 + x22 + x1 x2 ) . 2 2 m ∂ x1 2 m ∂ x22 2 Find out the energy and wave function for the interacting system.

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SOLUTION

Let us introduce x1 = and x1 x2 =

1 1 ( z1 + z2 ) and x2 = ( z1 − z2 ) , which gives x12 + x22 = z12 + z22 2 2

1 2 z1 − z22 . This allows us to write: 2

(

)

1  ∂ ∂ ∂ ∂ z1 ∂ ∂ z2 ∂  = + = + (9.80)   ∂ x1 ∂ z1 ∂ x1 ∂ z2 ∂ x1 2 ∂ z1 ∂ z2 

and:

1  ∂ ∂ ∂ ∂ z1 ∂ ∂ z2 ∂  = + = − (9.81)  ∂ x2 ∂ z1 ∂ x2 ∂ z2 ∂ x2 2  ∂ z1 ∂ z2 

which yield:

1  ∂2 ∂2 ∂2 ∂ ∂  (9.82a) =  2 + 2 +2 2 ∂ x1 2  ∂ z1 ∂ z2 ∂ z1 ∂ z2 

and:

∂2 1  ∂2 ∂2 ∂ ∂  =  2 + 2 −2 (9.82b) 2 ∂ x2 2  ∂ z1 ∂ z2 ∂ z1 ∂ z2 

Therefore:

∂2 ∂2 ∂2 ∂2 + 2 = 2 + 2 (9.83a) 2 ∂ x1 ∂ x2 ∂ z1 ∂ z2



The Hamiltonian in terms of z1 and z2 is: H=−

 2 ∂2  2 ∂2 k k − + ( z12 + z22 ) + ( z12 − z22 ) 2 2 2 m ∂ z1 2 m ∂ z2 2 4

  2 ∂2 3k 2    2 ∂2 k  = − + z1  +  − + z22  2 2 2 m ∂ z 4 2 m z 4 ∂ 2 1    

(9.83b)

Taking ψ ( z1 , z2 ) = φ1 ( z1 )φ2 ( z2 ) , the eigenvalue equation Hψ = Eψ is:



H 1 ( z1 )φ1 ( z1 )φ2 ( z2 ) + φ1 ( z1 )H 2 ( z2 )φ2 ( z2 ) = Eφ1 ( z1 )φ2 ( z2 ) (9.84) 1 1 ⇒ H 1 ( z1 )φ1 ( z1 ) + H 2 ( z2 )φ2 ( z2 ) = E φ1 ( z1 ) φ 2 ( z2 ) Equations (9.84) and (9.77a) look similar. Thus, the Hamiltonian of two interacting oscillating particles is transformed to the Hamiltonian of two non-interacting oscillating particles.

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Defining α 14 =

3mk mk , and α 24 = 2 , a generalized solution of Eqn. (9.84b) is: 2 2 2 ψ n1 , n2 (ρ1 , ρ2 ) = N n1 , n2 H n1 (ρ1 )H n2 (ρ2 )e



(

− ρ12 /2 +ρ22 /2

) (9.85)

and: 1 1   En1 , n2 =  n1 +  ω 1 +  n2 +  ω 2 (9.86)   2 2



with: ρ1 = α 1 z1 =

1  3mk    2  22 

1/4

( x1 + x2 ); ρ2 = α 2 z2 =

1  mk    2  22 

k , with n1 , n2 = 0, 1, 2...... . On defining ω = 2m

and ω 2 =

En1 , n2 =



1/4

( x1 − x2 ); ω 1 =

3k 2m

k we get: m

1  3 + 1 3 n1 + n2 + ω (9.87) 2  2 

Note that Eqn. (9.87) endorses the statement of Section 9.1.1, “analysis and conclusions drawn for non-interacting particles are applicable to a system of N-interacting particles too.” 3. A central potential well has three possible single particle states, φ1 , φ2 , and φ3 . If there are two electrons, what would be possible wave functions? What would be the matrix element of two electrons interaction potential v(1, 2) when they interact? (Problem 7009 in “Problems & Solutions in Quantum Mechanics,” edited by Yung-Kuo Lim.) SOLUTION

Electrons obey the Pauli exclusion principle and therefore a many particle wave function is antisymmetric. All the possible antisymmetric wave functions from Eqn. (9.11) are: 1 [φ1 (1)φ2 (2) − φ1 (2)φ2 (1)] ; 2 1 ψ 23 (1, 2) = [φ2 (1)φ3 (2) − φ2 (2)φ3 (1)] ; (9.88) 2 1 ψ 31 (1, 2) = [φ3 (1)φ1 (2) − φ3 (2)φ1 (1)] 2 ψ 12 (1, 2) =



For a two electron interaction potential v(1, 2) , the matrix element can be given by: ψ ij v(1, 2) ψ kl

1 1 φi (1)φ j (2) v(1, 2) φ k (1)φl (2) + φi (2)φ j (1) v(1, 2) φ k (2)φl (1) 2 2 1 1 − φi (1)φ j (2) v(1, 2) φ k (2)φl (1) − φi (2)φ j (1) v(1, 2) φ k (1)φl (2) 2 2 =

(9.89)

Quantum Theory of Many Particle Systems

305

where each of the pairs ij and kl can be assigned 12, 23 and 31. Note that 1 and 2 represent both space and spin coordinates. 4. A system of two particles, each having mass m , is confined to volume V . Particles 2 interact with each other via the potential V (r1 , r2 ) = k r1 − r2 . Calculate the energy of the system. SOLUTION

The eigenvalue equation for a system of two interacting particles is: −



2 2 ∇12 + ∇ 22 ψ (r1 , r2 ) + k r1 − r2 ψ (r1 , r2 ) = Eψ (r1 , r2 ) (9.90) 2m

(

)

The Hamiltonian of the system is separable into three 1D Hamiltonians. We write: H = Hx + Hy + Hz

with:



Hx = −

 2  ∂2 ∂2  2 + k x1 − x2 , + 2 m  ∂ x12 ∂ x22 

Hy = −

2 ∂2   2  ∂2 + 2  + k y1 − y 2 , (9.91) 2  2 m  ∂ y1 ∂ y 2 

Hz = −

 2  ∂2 ∂2  2 + 2  + k z1 − z2 2  2 m  ∂ z1 ∂ z2 

which allows us to write ψ (r1 , r2 ) = φ( x1 , x2 )ϕ( y1 , y 2 )ξ( z1 , z2 ) . Substitution of these into Eqn. (9.90) yields the following three equations:

−



−



∂2   2  ∂2 2 + 2  φ( x1 , x2 ) + k ( x1 − x2 ) φ( x1 , x2 ) = E1φ( x1 , x2 ) (9.92a) 2  2 m  ∂ x1 ∂ x2 

2  2  ∂2 ∂2  + 2  ϕ( y1 , y 2 ) + k ( y1 − y 2 ) ϕ( y1 , y 2 ) = E2 ϕ( y1 , y 2 ) (9.92b) 2  2 m  ∂ y1 ∂ y 2 

−

 2  ∂2 ∂2  2 + 2  ξ( z1 , z2 ) + k ( z1 − z2 ) ξ( z1 , z2 ) = E1 ξ( z1 , z2 ) (9.92c) 2  2 m  ∂ z1 ∂ z2 

with E = E1 + E2 + E3 . The three equations are similar and hence we need to solve 1 only one of these. Let us solve Eqn. (9.92a). On defining s = ( x1 + x2 ) and 2 1 t= ( x1 − x2 ) , and then adopting the procedure of example 2, we get: 2

−

∂2  1  2  ∂2 + 2  φ( s, t) + kt 2 φ( s, t) = E1φ( s, t) (9.93a) 2  2m  ∂s 2 ∂t 

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Next, taking φ( s, t) = φ1 ( s)φ2 (t) we obtain: −



   2 ∂2 φ2 1 2  2 ∂ 2 φ1 φ2 (t) + − + kt φ2 (t)  φ1 ( s) = E1φ1 ( s)φ2 (t) (9.93b) 2 2 ∂ 2m ∂s 2 m t 2  

On writing: −



 2 ∂2 φ1 ( s) = E11φ1 ( s) (9.94a) 2m ∂s2

and: −



 2 ∂2 φ2 (t) 1 2 + kt φ2 (t) = E12 φ2 (t) (9.94b) 2 m ∂t 2 2

we get E1 = E11 + E12 . Solutions of Eqns. (9.94a) and (9.94b) are: φ1 ( s) = A sin ( k1 s ) and E11 =



 2 k12 (9.95a) 2m

and: φ2 (t) = N n1 H n1 (αt)e −α



2 2

t /2

1  and E12 =  n1 +  ω , n1 = 0, 1, 2,... (9.95b)  2

mω . Here, A and N n1 are constants and H n1 is the 2 Hermite polynomial. Since particles are confined to volume V, φ( x1 , x2 ) must vanish at the boundary, which means that if we take V = L3 then φ(L, x2 ) = φ( x1 , L) = 0 . This implies that: with ω = k / m and α 4 =

 x + x2   x1 − x2  φ( x1 , x2 ) = φ1  1  φ2    2   2 



 k ( x + x2 )   x1 − x2  = A sin  1 1  φ2    2 2 

(9.96)

should go to zero when any of x1 and x2 is equal to L. This would be fulfilled if k1 L = l1 π , where l1 = 1, 2, 3,... . We thus obtain: 2 E1 =



 2 π 2 l12  1 +  n1 +  ω (9.97a)  mL2 2

It is then straightforward to write:

E2 =

 2 π 2 l22  1 +  n2 +  ω (9.97b) 2  mL 2

E3 =

 2 π 2 l32  1 +  n3 +  ω (9.97c)  mL2 2

and:

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Therefore: El , n =



2 π 2l2  3 +  n +  ω (9.98) mV 2/3  2

with l 2 = l12 + l22 + l32 and n = n1 + n2 + n3 . The lowest value of energy is: E10 =



3 2 π 2 3 + ω (9.99) mV 2/3 2

5. Consider a system of N non-interacting particles whose Hamiltonian is written as: N

H=

∑H

i

where H i φi = ε i φi . Calculate the ground state energy of the system

i =1

1 for both the cases of Bosons (spin = 0) and Fermions (spin = ). Write explicitly the 2 ground state energy and wave function when N = 5 . SOLUTION

In the case of Bosons, all N-particles can occupy the one state. Thus if ε1 is the energy of lowest energy state, the ground state energy of the system would be E0b = Nε 1 . However, for Fermions, the Pauli exclusion principle allows a maximum of two particles of opposite spins to occupy the same state. Therefore f f E0 = 2(ε 1 + ε 2 + ε 3 + ....... ε N /2 ), if N is even, and E0 = 2(ε 1 + ε 2 + ε 3 + ....... ε( N − 1)/2 ) + ε N when N is odd. The wave function of five identical Bosons has to be symmetric with respect to interchanges of the particles. Hamiltonian is a sum of N-single particle Hamiltonians; the eigenfunction would be products of single-particle eigenfunctions. Therefore:

ψ (1, 2, 3, 4, 5) = φ1 (1) φ1 (2) φ1 (3) φ1 (4) φ1 (5) and E0b = 5ε 1 (9.100) In the case of five identical Fermions, wave function has to be antisymmetric with respect to the interchange of any two particles. In the ground state, two Fermions occupy the lowest energy ε1 state φ1 with opposite spins, the next two occupy the state φ2 of energy ε 2 , and the fifth Fermion occupies the state φ3 having energy eigenvalue ε 3 . Hence:

1 ψ (1, 2, 3, 4, 5) = 5!

φ1↑ (1)

φ1↑ (2)

φ1↑ (3)

φ1↑ (4)

φ1↑ (5)

φ1↓ (1)

φ1↓ (2)

φ1↓ (3)

φ1↓ (4)

φ1↓ (5)

φ2↑ (1)

φ2↑ (2)

φ2↑ (3)

φ2↑ (4)

φ2↑ (5)

φ2↓ (1)

φ2↓ (2)

φ2↓ (3)

φ2↓ (4)

φ2↓ (5)

; E0f = 2(ε 1 + ε 2 ) + ε 3

φ3↑ (1) φ3↑ (2) φ3↑ (3) φ3↑ (4) φ3↑ (5) (9.101)

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6. Two non-interacting electrons are confined to a one dimensional infinite potential well:  0 V ( x) =   ∞



0a

Find the energy eigenvalue and wave function for the ground state. SOLUTION

Inside the well, each of the electrons obeys the free particle Schrödinger equation. Therefore, the single particle eigenvalue equation for the nth eigenstate is: −



 2 d 2 φni ( xi ) = ε ni φni ( xi ) with i ≡ (1, 2)for 0 < x < a (9.102) 2 m dxi2

along with the condition φni (0) = φni ( a) = 0 . Therefore two solutions are: φn1 ( x1 ) =

2  πn x  sin  1 1  ;  a  a

 2 π 2 n12 ε n1 = 2 ma 2

(9.103)

and: φn2 ( x2 ) = ε n2

2  πn x  sin  2 2  ;  a  a

 2 π 2 n22 = 2 ma 2

(9.104)

where n1 and n2 are integers and take values 1, 2, 3,...... . Total wave function has to be antisymmetric for a system of electrons. In the ground state both electrons occupy the lowest energy state with opposite spin orientations. Therefore, from Eqns. (9.13) to (9.15), we find that the space-dependent part of wave function should be symmetric and the spin-dependent part is antisymmetric. We thus have: ψ 11 ( x1 , x2 ) = =

2×2 1  πx   πx  ↑1 ↓2 − ↓1 ↑2  sin  1  sin  2  ×     a a 2a 2 2 π  π  cos  ( x1 − x2 ) − cos  ( x1 + x2 )  ↑1 ↓2 − ↓1 ↑2  a  a  a 

(9.105a)

and:

Eg = 2 ε1 =

2 π 2 (9.105b) ma 2

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7. In some of the metallic systems it is found that Vext (ri ) , which is generated by distribution of positively charged ions, is equal and opposite to the potential created by the distribution of negatively charged electrons. For such systems, Eqn. (9.49) reduces to:

−

2 2 e2 ∇ i φi (ri ) − 2m 4πε 0

   

∑∫

φ*j (rj )

j≠ i

 1  φi (rj )d 3 rj φ j (ri )  = ε i φi (ri ) (9.106) ri − rj 

where φi (ri ) can be given by a plane wave. Such systems are called free electron systems. Calculate single particle energy for a free electron system. SOLUTION



For N-free electrons confined to volume V, the single electron wave function can be given by: 1 ik i .ri φi (ri ) = e (9.107) V We then have: −



2 2  2 ki2 ∇ i φi (ri ) = φi (ri ) (9.108a) 2m 2m

To evaluate the exchange interaction energy term, let us write the Coulomb potential in the Fourier transformed form: e2 e2 = 4πε 0 ri − rj ε 0V



∑e

iq .( ri − rj )

q2

q

e2 = ε 0 (2 π)3

∫d q 3

e

iq .( ri − rj )

(9.108b)

q2

which yields: e2 4πε 0 =



= = = =

∑ ∫ φ (r ) r − r * j

1

j

i

j≠ i

2

j iq .( ri − rj )

∑ q ( ) ∫∫ e 1 e d rd q ∑ ∫ ∫ q ( 2π V )

1 e ε0 2π V e2 ε0

φi (rj )d 3 rj φ j (ri ) d 3 rj d 3 q

3

2

e

i ( k i − k j ).rj ik j .ri

3

3

j

3

iq .ri

i ( k i − k j − q ).rj ik j .ri

2

kj

∑∫d q

e2 1 ε 0 V 3/2

∑k

3

e

i (q + k j ).ri

q2

kj

kj

∑k kj

e

kj

e2 1 ε 0 V 3/2

e2 1 ε0 V

e

e ik i .ri i

− kj

1 i

− kj

2

2

φi (ri )

δ(k i − k j − q)

e

(9.109)

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Substitution of Eqns. (9.108a) and (9.109) into Eqn. (9.106) gives:  2 2 2   ki − e 1  2m ε0 V 



∑ kj

  2 φ i (ri ) = ε i φ i (ri ) (9.110) k i − k j  1

On changing k i → k and k j → k ′ , we have: ε( k ) =



2 k 2 e2 1 − 2m ε0 V

∑ k′

1 2 k 2 e2 = − 2 2m ε0 k − k′

∫

d3 k ′ 1 (9.111) 3 (2 π) k − k ′ 2

Since one value of k can accommodate a maximum of two electrons, distribution of electrons occupies the states having k -values from 0 to k F . Choosing spherical polar coordinates and taking k along the polar axis, we obtain:

ε( k ) =

e2 2 k 2 − 2 m (2 π)3 ε 0

π

kF

∫

2π

∫

∫ (k

k ′ 2 dk ′ sin θ dθ dφ

0

0

0

1 2

2

+ k ′ − 2 kk ′ cos θ

)

(9.112)

Evaluation of integrals gives: π

kF

∫

2π

∫

∫ (k

I = k ′ 2 dk ′ sin θ dθ dφ 0

0

1 2

0

+ k ′ 2 − 2 kk ′ cos θ

π

kF

∫

∫

= 2 π k ′ 2 dk ′ sin θdθ 0

0

kF

(k

1 2

2

+ k ′ − 2 kk ′ cos θ

 1 = 2 π k ′ 2 dk ′  ln k 2 + k ′ 2 − 2 kk ′ cos θ 2 kk ′  0



∫

(

kF

  dk ′ 

 k + k ′ = 2 π k ′ ln   k − k ′ 0

∫

(

)

k k F2 − k 2 k + kF = 4π  F + ln 4k k − kF  2

)

)

)

π

  (9.113) 0 

  

Therefore:

ε( k ) =

2 k 2 e2 − 2 2m 2π ε0

(

)

 kF k F2 − k 2 k + kF   +  (9.114) ln k − kF  4k  2 

8. Two spinless identical particles are confined to a one dimensional infinite potential well of width a , described by:

 0 V ( x) =   ∞

0≤x≤a 0>x>a

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Quantum Theory of Many Particle Systems

If the particles interact with each other via a potential bδ( x1 − x2 − a / 2) , find the approximate ground state energy of the system with the use of the first order perturbation theory. b is constant. SOLUTION

The Hamiltonian of the system for 0 ≤ x ≤ a: H=−



∂2   2  ∂2 + 2  (9.115) 2  2 m  ∂ x1 ∂ x2 

The interaction potential can be used as the perturbative term. The unperturbed single particle wave functions for a state of an infinitely deep potential well are:

φn (1) =

2  nπx1  sin   and   φl (2) =  a  a

2  lπx2  sin  , where(n, l) = 1, 2, 3,... (9.116)  a  a

Therefore, the unperturbed wave function and energy are as follows: ψ (1, 2) =

2  lπx2   nπx1  sin  sin  ;  a   a  a

2 π2 2 3 = n +l 2 ma 2

(

(0) n,l

E

)

(9.117)

The first order correction to the energy is: En(1),l = b ψ (1, 2) δ(x1 − x2 − a/2) ψ (1, 2)

 2 = b   a  2 = b   a

2 a

a

 lπx2   nπx1  δ( x1 − x2 − a/2)sin 2  dx2 dx1 sin 2   a   a 

∫ ∫ 0

(9.118a)

0

2 a

 nπx2 nπ  2  lπx2  +  sin   2 a a 

∫dx sin  2

2

0

For the ground state, n = 1 and l = 1 . Taking y =

πx2 , we get: a

π

(1) E11 =

4b dy cos 2 ( y ) sin 2 ( y ) πa

∫ 0

=

b πa

π

∫ cos (2y)dy 2

0

π b  = dy + 2 πa  0 b (1) ⇒ E11 = 2a

∫

π

∫ 0

 cos(4 y )dy   

(9.118b)

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Hence, ground state energy is:

E11 =



2 π 2 b + (9.119) ma 2 2 a

9. Combining Eqns. (9.54) and (9.55), the particles density (number of particles per unit volume) within the Thomas-Fermi theory, is given by: 3/2 1 n(r ) = 2 3  2 m ( µ − Vext (r ))  for µ > Vext (r ) and zero otherwise. µ is called 3π  the chemical potential or Fermi energy. Calculate the total number of particles N = ∫ n(r )d 3 r when Vext (r ) = αr 2 . SOLUTION

Note that for r > r0 , n(r ) will become imaginary, which is unphysical. Therefore, µ the number density must go to zero at µ = Vext (r0 ) = αr02 or at r0 = . We thus have: α r0

N=

∫ 0

=

π

2π

0

0

3/2 1  2 m µ − αr 2  r 2 dr sin θ dθ dφ  3π 2  3 

(

4  2 mµ    3π   2 

)

3/2 r0

∫

∫ (1 − r /r )

2 3/2 0

2

∫

(9.120)

r 2 dr

0

Taking r/r0 = sin x , we get: r0

∫ {1 − ( r/r ) } 0

2 3/2

π/2 2

∫ cos (x)sin (x)dx

3 0

4

r dr = r

2

0

0

r3 = 0 16



=

π

∫ (1 + cos(y)) sin (y)dy 2

0

3 0

πr π  µ =   32 32  α 

3/2

which gives:



N=

1  2m    24  α 2 

3/2

µ 3 (9.121)

This shows that µ ∝ N 1/3 is determined by the total number of particles in a system.

Quantum Theory of Many Particle Systems

313

9.5 Exercises 1. Two Bosons occupy the n = 4 and n = 3 states of a 1D infinite potential well of width a . Write the properly normalized wave function for the system. 2. Consider the two electrons confined to move freely within a solid of volume V . They occupy the momentum states k and −k and their spin functions are represented by α and β . Use Eqn. (9.15) and write the wave function for the singlet state (S = 0) and triplet states (S = 1). 3. 1D infinite potential well described by V ( x) = 0 for 0 < x < a and V ( x) = ∞ for 0 > x > a , contains three electrons. If the width of the well is 0.5 nm, calculate the total energy of the system by neglecting the Coulombic interaction between the electrons. 4. Suppose that the Schrödinger equation for a singly ionized helium atom is solved by considering it as a hydrogen-like atom with charge 2e at the nucleus. (a) How does the Bohr radius change in the ψ N (r ) , where N (= nlm) is a composite quantum number, as compared to that in wave functions of the hydrogen atom? (b) Include the spin parts of wave functions α and β for up and down spins, respectively. Now treat the helium atom having two electrons without any interaction between them and then write down the typical two electron wave function. Do not choose ground state. (c) Show that your wave functions comply with Pauli’s exclusion principle. 5. Write down the Hamiltonian for a system of two particles having mass m1 and m2 . They interact via potential V ( x ) which depends only on relative position x = x1 − x2 .  2 ∂2  2 ∂2 Show that the Hamiltonian can be transformed to H = − − + V ( x) , 2 M ∂ X 2 2µ ∂ x 2 m x + mx2 1 1 1 where X = 1 1 , M = m1 + m2 , and = + . m1 + m2 µ m1 m2 6. Consider a pair of electrons which are constrained to move along the x-axis in a spin S = 1 state. Electrons interact through an attractive potential V ( x1 , x2 ) = −V0 for 0 < x1 − x2 < a and V ( x1 , x2 ) = ∞ for 0 ≥ x1 − x2 ≥ a. Find the lowest energy E0 and corresponding eigenstate ψ 0 ( x1 , x2 ) in the case where total momentum is zero. 7. Two non-interacting and indistinguishable particles, each of mass m , are oscillating with frequency ω in a 1D harmonic potential. One of the particles is in the ground state while the other particle is in the first excited state. (a) Construct the antisym2 metric spatial part of the wave function of the system and (b) calculate  ( x2 − x1 ) .

10 Time-dependent Perturbations and Semi-classical Treatment of Interaction of Field with Matter In the preceding chapters, we discussed solutions and the applications of the Schrödinger wave equation having time-independent potentials. Schrödinger quantum mechanics assume that operators and the Hamiltonian are independent of time. And the time evolution of a system is taken care of by wave function. The time independence of the Hamiltonian allows to us to factorize the wave function into space- and time-dependent parts. Therefore, in Schrödinger representation we write:

i

− iEt  ∂ψ (r , t)   2 2 = − ∇ + V (r )  ψ (r , t),  with   ψ (r , t) = ϕ(r )e  (10.1) ∂t  2m 

This allows us to obtain an eigenvalue equation:

 2 2   − 2 m ∇ + V (r )  ϕ(r ) = Eϕ(r ) (10.2)   2

where E is the energy eigenvalue. The ψ (r , t) , which is interpreted as probability density, is constant in time. In addition to this, expectation values of operators (observables) are also constant in time. Equation (10.2), which involves time-independent potential, is solved exactly or approximately, depending upon the nature of the potential V (r ) .

10.1  Time-dependent Potentials The Hamiltonians of a large number of systems depend on time through the time-dependent potentials. When, the Hamiltonian is time-dependent, factorization of the wave function into space- and time-dependent parts is not possible and hence the Schrödinger picture is inadequate to describe the system. In this chapter, we are going to deal with the Hamiltonian that has the form H (r , t) = H 0 (r ) + H 1 (r , t) , where H 0 (r ) is time-independent and it is assumed to be an unperturbed part, whose energy eigenvalues and eigenstates are known exactly. H 1 (r , t) is the time-dependent part of the Hamiltonian, and it is treated as a perturbation applied to the system. A perturbation theory can be used if H 1  H 0 . Because of the time dependence of H 1 (r , t) , calculation for stationary energy eigenstates is not expected to be of any importance. ψ (r , t) for a time-dependent Hamiltonian cannot be found in the Schrödinger representation. The interaction picture of quantum mechanics, which utilizes some elements of the Heisenberg picture and some elements of the Schrödinger picture have been used to find ψ (r , t) for the time-dependent Hamiltonian. 315

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However, we are not going to use the interaction picture here. The approach adopted by us is as follows. Let us say H 0 satisfies the eigenvalue equation: H 0 φn = En φn (10.3)



where En and φn are the energy eigenvalue and eigenstate of the nth state of the system in the absence of perturbation. Let us also assume that φn form a complete set. Unlike time-independent perturbation theory, we here used En in place of En(0) , because here we are not going to determine energy corrections, which are irrelevant in the case of time-dependent perturbations. The eigenstate φn can be taken as basis vectors of

∑c (0) φ

Hilbert space to write ϕ =

n

, where cn (0) is a constant in time. In the absence of

n

n

H 1 (r , t) , we are dealing with stationary state problems and hence we have: ψ (r , t) =



∑

cn (0)e

−

iEnt 

φn (r ) (10.4)

n

Now, let us consider the situation where initially, in the absence of H 1 (r , t) , only one of the eigenstates, say φi , is populated. But as time goes on and H 1 (r , t) is turned on, states other than φi get populated. We are then no longer dealing with stationary state problems and hence the time evolution of ψ (r , t) is not as simple as given by Eqn. (10.4). The time-dependence of H 1 (r , t) causes transition to states other than φi . The basic question we would like to address is: What change in ψ (r , t) is expected on introducing H 1 (r , t) ? Or, how does an arbitrary state ket change as time goes on, where the Hamiltonian is timedependent? We can expect that ψ (r , t) is still represented in the same form as in Eqn. (10.4), provided cn is made time-dependent: We therefore write: ψ (t) =



∑c (t)e

−

n

iEnt 

φn (10.5)

n

Note that in this procedure, time-dependence of cn (t) is solely arising due to H 1 (r , t) and it must go to cn (0) whenever H 1 (r , t) = 0 . Then the probability of finding a particle in the 2 2 state φn is given by cn (t) in place of cn (0) . The time-dependence of cn (t) is thus determined from the Schrödinger equation having the Hamiltonian H (r , t) = H 0 (r ) + H 1 (r , t) . We then have: ∂ ψ (t) i = ( H 0 + H 1 ) ψ (t) (10.6) ∂t which with the use of Eqn. (10.5) gives: ∂ ∂t

∑

−

⇒ i

∑

−

i



cn (t)e

iEnt 

n

cn (t)e

iEnt 

n

=

∑E c (t)e

−

n n

∑

cn (t)e

n

φn +

∑

cn (t)e

∑E c (t)e n n

n iE t − n 

−

iEnt 

φn

iEnt 

∑c (t)e

φn + H 1 (r , t)

−

n

iEnt 

φn

n

−

iEnt 

∑

φn = H 1 (r , t)

cn (t)e

n

−

iEnt 

φn

(10.7)

n

n

⇒ i

φn = ( H 0 + H 1 )

φn

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Time-dependent Perturbations and Interaction of Field

iEm t

On taking the inner product with φm e

i

∑

cn (t)e

i ( Em − En )t 

and then defining ω mn =



φm φn =

n

∑

cn (t)e

i ( Em − En )t 

( Em − En ) , we get:

φm H 1 (r , t) φn

n

1 ⇒ cm (t) = i

∑h

mn

(t)cn (t)e



(10.8a)

iω mnt

n

Here, we have used φm φn = δ m , n and defined hmn (t) = φm H 1 (r , t) φn . We thus obtained a matrix equation:



 c1  d  c2 dt  .   .

 h11 e iω11t    iω t  = 1  h21 e 21  i  .   .  

h12 e iω12 t

.

iω 22 t

. . .

h22 e . .

.   c1  .   c2  .  .  .   .

   (10.8b)   

which yields coupled equations that are solved to find c1 (t), c2 (t).....cn (t)..... . After knowing c1 (t), c2 (t).....cn (t)..... , one determines the probability of finding the system in any particular state at a later time. Note that we have not used any approximation until now and hence all treatments are exact. Exactly solvable problems with time-dependent potentials are rather rare. In most of the cases, perturbation expansion is used to solve coupled equations. However, there are problems of enormous practical importance such as nuclear magnetic resonance and MASERs, which are exactly solvable.

10.2  Exactly Solvable Time-dependent Two-state Systems The matrix Eqn. (10.8b) is solved exactly for two-state systems perturbed by a periodic external field. The real physical systems may have more than two states, but for some important cases two of the states are very weakly coupled to other states, and hence twostate analysis becomes relevant. One of the examples is the ammonia MASER. A two-state system problem with sinusoidal perturbing potential is described as follows:

ψ = c1 (t)e

− iE1t 

φ1 + c2 (t)e

− iE2 t 

φ2 (10.9)

H 0 = E1 φ1 φ1 + E2 φ2 φ2 , with E2 > E1 (10.10a)

and:

H 1 = γe iωt φ1 φ2 + γe − iωt φ2 φ1 (10.10b)

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where γ and ω are real positive. We thus have time-dependent perturbing potential that connects the two states and the transition between two states takes place. Note that Eqn. (10.10a) is consistent with Eqn. (10.3). The matrix elements h11 , h12 , h21 , and h22 are: h11 = φ1 H 1 φ1 = γe iωt φ1 φ1 φ2 φ1 + γe − iωt φ1 φ2 φ1 φ1



⇒ h11 = γe iωt δ 21 + γe − iωt δ 12 = 0

(10.11a)

Similarly, we find that h22 = 0 . We next take: h12 = φ1 H 1 φ2 = γe iωt φ1 φ1 φ2 φ2 + γe − iωt φ1 φ2 φ1 φ2



⇒ h12 = γe iωt

(10.11b)

and: h21 = φ2 H 1 φ1 = γe iωt φ2 φ1 φ2 φ1 + γe − iωt φ2 φ2 φ1 φ1



⇒ h21 = γe − iωt

(10.11c)

On substituting Eqns. (10.11) into Eqn. (10.8a), we get:

i

dc1 (t) = γe iωt e iω12 t c2 (t) = γe i(ω+ω12 )t c2 (t) (10.12a) dt

and:

i

where, ω 12 = −ω 21 =

dc2 (t) = γe − iωt e iω 21t c1 (t) = γe − i(ω−ω 21 )t c1 (t) (10.12b) dt

( E1 − E2 ) . The two first order differential equations are combined to

 give one second order differential equation:

i

dc (t) d 2 c2 (t) d = γe − i(ω−ω 21 )t c1 (t) = − i(ω − ω 21 )γe − i(ω−ω 21 )t c1 (t) + γe − i(ω−ω 21 )t 1 (10.13a) 2 dt dt dt

(

)

which with the use of Eqns. (10.12a) and (10.12b) goes to: 2



d 2 c2 (t) dc (t)  γ  + i(ω − ω 21 ) 2 +   c2 (t) = 0 (10.13b) 2   dt dt

This is a standard second order differential equation. We solve it with the use of a trial solution, c2 (t) = c2 (0)e iΩt , which satisfies Eqn. (10.13b) if:

2  2 γ  Ω + Ω ( ω − ω ) − 21     = 0 (10.14)   

whose solution gives: 1/2



 ω − ω 21  2  γ  2  1 Ω = − (ω − ω 21 ) ±   +   2 2      

(10.15)

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Time-dependent Perturbations and Interaction of Field

Therefore, a general solution of Eqn. (10.13b) is: c2 (t) = e



−i

( ω−ω 21 )t 2

 i  Ae  

2

2

2

 ω−ω 21   γ    +  t 2   

+ Be

2

 ω−ω 21   γ  t −i  +  2    

  (10.16)  

if initially at t = 0 only the lower state φ1 is populated so that c1 (0) = 1 and c2 (0) = 0 . This implies A = − B and from Eqn. (10.12b):

i

dc2 (t) = γ (10.17) dt t = 0

Differentiating Eqn. (10.16) with respect to t and then using Eqn. (10.17), we get: A=−



{ ( ω − ω 2

γ

)2 + 4 γ 2 }

1/2

21

(10.18)

The probability that both the states φ1 and φ2 are populated at later time t is then given by: 2

c2 (t) =



2

2  ω − ω 2   γ   21  2  + sin t     (10.19)     2   2         2  ω − ω 21  2       +γ  2    

γ2

2

and c1 (t) = 1 − c2 (t) . Equation (10.19) is the famous Rabi’s formula. I.I. Rabi is known as the father of molecular beam technologies. π 2 The plot of c2 (t) as a function of ω at t = is displayed in Fig. 10.1. 2γ

FIGURE 10.1 Plot of c 2 ( t ) versus ω at t = 2

π . 2γ

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A Textbook on Modern Quantum Mechanics

As is seen from Fig. 10.1, a maximum occurs at: ω = ω 21 =



E2 − E1 (10.20) 

which is known as the Resonance condition. The full width at half maximum of the curve is 4γ . Note that the weaker time-dependent potential (smaller γ ) makes the resonance peak  narrower. 2 2 The c2 (t) and c1 (t) are plotted as the function of t at ω = ω 21 (resonance condition) in Fig. 10.2. 2 2 As is seen from the figure, both c2 (t) and c1 (t) exhibit oscillatory behavior. The behav2 2 2 2 ior of c2 (t) is the opposite of that of c1 (t) . At t = 0 , c1 (t) = 1 and c2 (t) = 0 , which means the initially lower state φ1 is populated and the upper state φ2 is empty. As time grows, 2 the system absorbs the energy from the time-dependent perturbation H 1 (r , t) and c1 (t) π 2 2 2 decreases while c2 (t) increases. At t = , c2 (t) = 1 and c1 (t) = 0 the upper state is 2γ π π 2 populated and the lower state is empty. Then between t = and t = , c1 (t) increases 2 γ γ 2 and c2 (t) decreases, which means that the system gives up the excess energy back to the perturbing field during this time interval. This cycle of absorption and emission is repeated infinitely, as is seen from Fig. 10.2. We thus see that time-dependent perturbation is causing transitions from φ1 to φ2 (absorption) and then from φ2 to φ1 (emission) as time increases. 2 The absorption-emission cycle takes place even if ω ≠ ω 21 . However, in that case, c2 (t) 2 no longer reaches to 1 and c1 (t) does not reduce down to zero. Transition from φ2 to φ1 starts even if the state φ1 is not completely empty. There are many applications of the general time-dependent two state problem, such as nuclear magnetic resonance, spin-magnetic resonance, MASER (microwave amplification of stimulated emission of radiations), the atomic clock, and optical pumping. It is amazing

FIGURE 10.2 Plot c1 ( t ) and c2 ( t ) as a function t at ω = ω 21 (resonance condition). 2

2

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Time-dependent Perturbations and Interaction of Field

to know that four Nobel Prizes have been awarded to those who exploited the applications of a timedependent two state system in some form. We have seen that an oscillating field can drive a collection of molecules from the ground state to an excited state. In an ammonia MASER, a stream of ammonia molecules travels down with known velocity through a tube of definite length. The tube has an oscillating microwave field, so that all (or almost all) molecules emerging at the other end of the tube are in the first excited state. Then, the excited outgoing molecules will decay on application of a small amount of electromagnetic radiation of the same frequency, generating an intense and coherent radiation because of the shorter period for all to decay.

10.3  Time-dependent Perturbation Theory Except for a few problems such as two state time-dependent systems, the exact solution to differential equations that determine cn (t) is not possible. One then attempts a perturbative approach to solve Eqn. (10.7). Similar to the case of time-independent perturbation theory, we include the small dimensionless parameter λ to write the Hamiltonian H = H 0 + λH 1 and expand the cn (t) in terms of λ as follows:

cn (t) = cn(0) (t) + λcn(1) (t) + λ 2 cn(2) (t) + λ 3 cn(3) (t) + ...... (10.21)

Substituting Eqn. (10.21) into Eqn. (10.8a) and replacing H 1 by λH 1 we get:



cm(0) (t) + λcm(1) (t) + λ 2 cm(2) (t) + ........ 1 = hmn (t) λcn(0) (t) + λ 2 cn(1) (t) + λ 3 cn(2) (t) + +...... e iω mnt i n

∑

{

}

(10.22)

To satisfy this, the coefficient of each power of λ must vanish. We therefore have: dcm(0) (t) = 0 (10.23a) dt



dcm(1) (t) 1 = dt i (2) m

dc (t) 1 = dt i

∑h

mn

(t)cn(0) (t)e iω mnt (10.23b)

n

∑h

mn

(t)cn(1) (t)e iω mnt (10.23c)

n

and so on. Equation (10.23a) implies that cm(0) (t) is constant in time, which is because of the fact that the zeroth order Hamiltonian H 0 is independent of time. 10.3.1  First Order Perturbation The first order contribution to cm (t) is given by Eqn. (10.23b). We obtain:

cm(1) (t) =

1 i

∫ ∑h

mn

n

(t ′)cn(0) (t ′)e iω mnt′ dt ′ (10.24)

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The majority of problems of practical interest are definable by assuming that the system evolves according to H 0 until t = 0 , and then a time-dependent perturbation is turned on. If the system initially at t = 0 is in only one unperturbed state φn then cn(0) is non-zero (equal to 1) only for one value of n belonging to φn , and it is zero for all other values of n. When t > 0 but very small, such that we still have cn(0) (t)  1 , we then can drop all terms on right side of Eqn. (10.24) except one term. We get:

cm(0) (t) = δ mn (10.25a)



1 c (t) = hmn (t ′)e iω mnt′ dt ′ , m ≠ n (10.25b) i

t

∫

(1) m

0

The perturbation H 1 (r , t) has introduced transitions to other states. If we confine ourselves only to first order perturbation theory, then the probability that the transition from φn to φm has occurred after time t is given by: 2

2

cm (t) = cm(1) (t) . (10.26)



To see that the first order perturbation theory gives a reasonably correct answer, notice that if all elements in the column matrix on the right side of Eqn. (10.8b) are zero except one element cn (t) that is equal to unity, we have the equation:

dcm (t) 1 = hmn e iω mnt , with m ≠ n (10.27) dt i

This is equivalent to saying that initially at t = 0 only the φn state is populated, and other states are empty. The integration of Eqn. (10.27) between 0 → t yields: t



1 cm (t) = hmn (t ′)e iω mnt′ dt ′ (10.28) i

∫ 0

which has right hand side identical to that in Eqn. (10.25b), justifying the use of first order perturbation theory.

10.4  Harmonic Perturbation Let us consider a system initially in the state of φn perturbed by a periodic potential that is described by:

{

}

H 1 (r , t) = 2 H 1 (r )cos(ωt) = H 1 (r ) e iωt + e − iωt (10.29)

which is switched on at t = 0 . An example of this is an atom or a molecule exposed to electromagnetic radiation (harmonic perturbation). The probability amplitude for an atom in

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the initial state φn to be in state φm after time t is then obtained from Eqn. (10.25b) with the use of:

(

)

hmn (t ′) = H mn e iωt′ + e − iωt′ , where H mn = φm H 1 (r ) φn (10.30)

We thus have: t

∫ (e

H c (t) = mn i (1) m



i ( ω+ω mn )t ′

)

+ e − i(ω−ω mn )t′ dt ′

0

 e − i(ω−ω mn )t − 1 e i(ω+ω mn )t − 1  ⇒ c (t) = H mn  −   ( ω + ω mn )    ( ω − ω mn )

(10.31)

(1) m

where the first term on the right-hand side will reach to its maximum value when ( Em − En ) → ω , while the second term will tend to a maximum value for ( Em − En ) → − ω . On application of an harmonic perturbating field, the system either receives energy from the field (absorption) or it transfers the energy to the field (induced or stimulated emission). We thus find that the first term on the right-hand side represents absorption, while the second corresponds to stimulated emission. Since both absorption and stimulated emission cannot take place simultaneously, only one term is to be discussed at a time. We retain the first term for further discussions and obtain: cm(1) (t) = H mn

e − i(ω−ω mn )t − 1  ( ω − ω mn )

=

H mn e  ( ω − ω mn )

i ( ω mn −ω )t 2

=

2 iH mn e  ( ω − ω mn )

i ( ω mn −ω )t 2

− i ( ω mn −ω )t  i(ω mn2−ω )t  2 e − e   (10.32)  

sin

{

(ω mn − ω )t 2

}

10.4.1  Transition Probability If the system was in the φn state initially at t = 0 , then the probability to find the system in the state φm after time t is given by:

2

Pn→ m (t) = cm(1) (t) =

2

{

}

H mn 2 (ω mn − ω )t (10.33) 2 sin 2 2  ω mn − ω      2

sin αt (ω − ω ) → t , its maximum value, when α = mn → 0 or ( Em − En ) → ω . Thus, α 2 2 2 H mn t ( Em − En ) , Pn→ m (t) exhibits a peak of height equal to and width equal to 2 π/t at ω = 2  (ω mn − ω ) 1  when plotted as a function of ω . The Pn→ m (t) shows more peaks at t = n+  π  2 2 where n is an integer. A plot Pn→ m (t) versus ω mn − ω is displayed in Fig. 10.3. Note that

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FIGURE 10.3 Plot of Pn→ m (t) as a function of ( ω mn − ω ) at fixed t.

A central peak has the largest height, as is seen in Fig. 10.3. As t increases, the height of the central peak enhances as t 2 , while its width decreases as 1/t. For t → ∞ , one expects that the function: sin 2 αt → πtδ(α) (10.34) α2

10.4.2  Fermi’s Golden Rule

We next consider the transitions from one of the initial discrete states to a final state that is part of a continuum, with density of states ρ(Em ) . For example consider the case of ionization of an atom in which the initial state may be one of the discrete bound states and the final state includes a free electron momentum eigenstate that is part of a continuum of non-normalizable states. As has been shown above, the probability of transition between discrete states exhibits periodic dependence in time. But, if the final state is part of a continuum, an integral over all final states is required to get a resultant transition probability. A transition rate is associated with such a probability function. The final transition rate is given by Fermi’s Golden Rule. An approximation to the sum of transition probabilities over final states is given by the integral: which gives:

P(t) =

∑P

n→ m

m

P(t) =

∫

(t) =

∫P

n→ m

(t)ρ(Em ) dEm (10.35a)

2 sin 2 ( αt ) H mn ρ(Em )dEm (10.35b) 2 α2

As discussed above, when transitions are allowed for a longer time, the width of the central peak becomes very narrow, and therefore a very small energy range is expected 2 to contribute to the integral (Eqn. 10.35b). In this range of energy, H mn and ρ(Em ) almost remain the same, and they are treated independent of energy in the continuum of states. We thus have: sin 2 ( αt ) H mn ρ(Em ) dEm (10.36) 2  α2 2



P(t) =

∫

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If all final states do not necessarily have the same matrix element, then H mn 2 replaced by an average value of H mn . Note that: α=



( ω mn − ω ) = Em − En

2 ⇒ dEm = 2 dα

2

−

ω 2

2

must be

(10.37)

Since the integral is contributed by a very small energy range, our answer will not be changed even if the limit of integration over α is extended to ±∞ . We therefore have: 2



P(t) =

2 H mn ρ(Em ) 

∞

∫

−∞

sin 2 ( αt ) 2π 2 dα = t H mn ρ(Em ) (10.38) α2 

where we made use of: ∞

∫



−∞

2

 sin x    dx = π (10.39) x 

Note that the total transition probability is proportional to t for larger values of t . This linearity in t is a consequence of the fact that total transitional probability is proportional to the area under the main peak whose height varies as t 2 and width varies as 1/t. In obtaining Eqn. (10.38), we restricted the integral to the central peak only, which yields our answer 90% correct. Inclusion of more peaks from to the left and right of the central peak improves accuracy further. P(t) The probability of transition per unit time , also termed the transition rate is: t

W=

P(t) 2 π 2 = H mn ρ(Em ) (10.40) t 

This equation is known as Fermi’s Golden Rule. Equation (10.40) shows that total transition 2 d probability is independent of time, or cm(1) (t) is constant in time, provided that the dt m first order perturbation theory is valid. Note that for larger t , Eqn. (10.33), with the use of Eqn. (10.34), gives:

∑

2

wn→ m (t) =

Pn→ m (t) π H mn = δ(α) t 2 2

=

π H mn  ω −ω δ  mn   2 2

(10.41)

which is the rate of transition from state φn to φm . The total transition probability is then obtained from the integration ∫ wn→ m ρ ( Em ) dEm = 2 ∫ wn→ m ρ ( Em ) dα , which yields the same results as are given by Eqn. (10.40). Equation (10.41) also is called Fermi’s Golden Rule. It agrees well with experimental results when applied to atomic systems, and it is of great practical importance.

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10.5  Constant Perturbation Let us consider that a constant perturbation is turned on at t = 0 :

 0 H 1 (r , t) =   H ′(r )

t0

.

Use the first order perturbation theory to calculate the probability of finding the atom in the 2s state for t  τ . SOLUTION

The time-dependent perturbation is: H 1 = V0 x 2 + y 2 e − t/τ



for t > 0 (10.117)

For the transition from the ground state (at t ≤ 0 ) to the 2s state (at t > 0 ) and t  τ , we have: t

1 c (t) = h21 (t ′)e iω 21t′ dt ′ (10.118) i (1) 2



∫ 0

For the transition: ω 21 =



1 µe 4 3µe 4  1  (10.119) [E2 − E1 ] = − 2 3  2 − 1 = 2   8 ( 4πε 0 )  3 2 ( 4πε 0 )   2

And:

h21 (t) = V0 2, 0, 0

x 2 + y 2 1, 0, 0 e − t/τ , (10.120)

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We thus have: c2(1) (t) =

t

V0 2, 0, 0 i

x 2 + y 2 1, 0, 0

∫e

iω 21t ′ − t ′/τ

e

dt ′

0



  (10.121) iω 21t − t/τ   e e − 1 x 2 + y 2 1, 0, 0  1   iω 21 −  τ  

V = 0 2, 0, 0 i

For t  τ , e − t/τ is very small and we therefore have: c2(1) (t) =



V0 2, 0, 0 

x 2 + y 2 1, 0, 0

1   ω 21 +

i  τ

(10.122)

To evaluate the matrix element, we use from Section 6.3: 1 1  a0  πa0 



ψ 100 ( r ) =



1  1  ψ 200 ( r ) = 4 a0  2 πa0 

1/2

1/2

e − r/a0 (10.123)

 r  − r/2 a0 . (10.124)  2 − a  e 0

With the use of spherical polar coordinates, we obtain: ∞

2, 0, 0

π

2π

0

0

 1 r x + y 1, 0, 0 = r 3 e − r/a0  2 −  e − r/2 a0 dr sin 2 θ dθ dφ 3  a 4 2 πa0 0 2

∫

2

=

πa0 4 2

∞

∫

0

 r

∫  a  0

0

3

 r e − r/a0  2 −  e − r/2 a0 d(r / a0 )  a0 

∫

(10.125)

4 πa0  3! 4!  πa0  2  = 2 − = −     4 2  (3 / 2)4 ( 3 / 2 )5  2  3

Therefore:

c2(1) (t) = −

πa0V0  2     2  3

4

1 (10.127) ω ( 21 + i/τ )

The probability of finding the atom in the 2s state is:

2

P1s→ 2 s = c2(1) (t) =

2

8

1  πV0 a0   2  1 (10.128)     1 2     3  2  ω 21 + 2  τ

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5. An electron confined to a cubical box of length L is subjected to the time-dependent perturbation H 1 ( z, t) = V0 e i( kz z −ωt ) . The perturbation is turned on at t = 0 . Calculate the transition rate, transition probability per unit time, using first order perturbation theory. What is its largest value? SOLUTION

The transition probability from state ψ k at t = 0 to state ψ k ′ at time t is given 2 by: Pk → k ′ = c(1) k ′ (t) , where: t

c(1) k ′ (t) =



1 hk ′k (t ′)e iω k ′k t′ dt ′ (10.129) i

∫ 0

The wave function and energy of the electron confined to a cubic box are given by:

ψ k (r ) =

1 ik .r e (10.130) L



Ek =

2 k 2 (10.131) 2m

(

3/2

)

π ˆ ˆ + ˆjn + kn with k = in , where each of nx , ny , and nz takes values 1, 2, 3....... To x y z L evaluate:

hk ′k (t) = V0 ψ k ′ e ikz z ψ k e −ωt (10.132) we calculate: ψ k ′ e ikz z ψ k =



⇒ ψ k′ e

ik z z

1 L3

∞

∫

∞

∞

∫

e i( kx − kx′ ) x dx e

−∞

−∞

i ( k y − k y′ ) y

∫

dy e i(2 kz − kz′ ) z dz (10.133)

ψ k = δ kx , kx′ δ ky , ky′ δ 2 kz , kz′

−∞

Then: V c (t) = 0 i (1) k′

= =

V0 i V0 i

t

∫δ

k x , k x′

δ ky , ky′ δ 2 kz , kz′ e − iωt′ e iω k ′k t′ dt ′

0

t

∫

δ kx , kx′ δ ky , ky′ δ 2 kz , kz′ e − iωt′ e

0 t

∫e

− iωt ′

e

i

3 k z2 t ′ 2m

dt ′

0

⇒ c(1) k ′ (t) =

V0 e i(α−ω )t − 1  (ω − α )

i

(

)

 k x′ 2 − k x2 + k y′ 2 − k y2 + k z′ 2 − k z2 t ′ 2m

dt ′ (10.134)

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with α =

3k z2 . Therefore transition probability per unit time is: 2m wk → k ′ =

Pk → k ′ t

2  V  1 − cos {(ω − α)t} (10.135) = 2 0  2   t (ω − α )



2 2  V  sin {(ω − α)t/2} = 0    {(ω − α)t/2}2

2

V  The largest value of wk → k ′ is  0  that occurs for ω → α .   6. Calculate the Einstein coefficients for the transition 1, 0, 0 to 2, 1, 0 in the hydrogen atom. SOLUTION

Einstein coefficients A and B are given by Eqns. (10.90):

A2→1 =

2 2e 2 ω 321 2, 1, 0 r 1, 0, 0 (10.136a) 3ε 0 πc 3



B2→1 =

2 2 πe 2 2, 1, 0 r 1, 0, 0 (10.136b) 3ε 0  2

The ω 21 is given by Eqn. (10.119) and: 2, 1, 0 r 1, 0, 0 = iˆ 2, 1, 0 x 1, 0, 0 + ˆj 2, 1, 0 y 1, 0, 0 + kˆ 2, 1, 0 z 1, 0, 0 (10.137)



Evaluations of 2, 1, 0 x 1, 0, 0 , 2, 1, 0 y 1, 0, 0 and 2, 1, 0 z 1, 0, 0 , with the use of 100  and  210 defined in Section 6.3, Eqn. (6.38) yield: ∞

π

2π

0

0

1 2, 1, 0 x 1, 0, 0 = r 4 e − r/a0 e − r/2 a0 dr sin 2 θ cos θdθ cos φdφ = 0 (10.138a) 4 2 πa04



∫

∫

∞

π

2π

0

0

0

2, 1, 0 y 1, 0, 0 =



∫

1 r 4 e − r/a0 e − r/2 a0 dr sin 2 θ cos θdθ sin φdφ = 0 (10.138b) 4 2 πa04

∫

∫

0

∫

and, ∞



π

2π

0

0

1 2, 1, 0 z 1, 0, 0 = r 4 e − r/a0 e − r/2 a0 dr sin θ cos 2 θdθ dφ 4 2 πa04

∫

∫

0

5

 2 = 4 2 a0   = 0.745 a0  3

∫

(10.138c)

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Therefore: B2→1 =



3πe 2 3.33π 2 2 × (0.745 a ) = a0 = 1.91 × 1021 meter/kg (10.139a) 0 2ε0 2 me

and: A2→1 =



2e 2 1.48a0 3 2 ω 321 ( 0.745 a0 ) = ω 21 3 3ε 0 πc me c 3 3

=

 1.48a0  3µe 4  2 3  me  8 ( 4πε 0 )  c 

=

1.48 × 1.05 × 10−34 × .529 × 10−10  16.32  × × 107   3 × 1.05  9.11 × 10−31

(10.139b) 3

= 12.54 × 108 / sec

10.10 Exercises 1. A particle of mass m and charge q is oscillating under a 1D harmonic potential. 2 q2 2 Show that , where n and n′ represent the quantum ( En′ − En ) n′ qx n = 2m n′

∑

states. Calculate the absorption cross-section using Eqn. (10.77b). Comment on absorption frequency. − eE0 τz 2. A time-dependent perturbation H 1 = is applied to a hydrogen atom. πω t 2 + τ 2 The atom is in its ground state at t = −∞ . Use the first order time-dependent perturbation theory to calculate the probability of finding the atom in its first excited state at t = ∞ . 3. Consider a particle in a 1D infinite potential well of width- a . The particle is in its ground state at time t = 0 . It is subjected to a traveling pulse represented by the time-dependent perturbation H 1 = Aδ( x − ct) . Calculate the transition probability of finding the particle in the nth excited state. 4. Take the expression of the transition probability derived in exercise 3 and compute it for the transition from the first excited state to the ground state by taking the mass of the particle equal to the mass of the electron, the width of well 1 nm, and 3 A = 3.2 × 10−23  Joules ( meter ) . 5. The energy levels of an electron confined to a nano-sized particle of radius a are  2 π 2 n2 given by En = . The nano-particle is in its first excited state and is placed 2 me a 2

(

)

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in a cavity. At what temperature of the cavity is the ratio of the stimulated emis1 sion rate to the spontaneous emission rate equal to , for the transition from the 3 excited state to the ground state. Take a = 4 nm . 1 6. Consider a system consisting of a spin- particle, with no orbital angular momen2 tum, placed in a uniform field B kˆ directed along the z-axis. The Hamiltonian of it 0

geB0 is given by H 0 = − Sz . The system is then subjected to a small time-dependent 2m magnetic field rotating in the x-y plane at the angular frequency ω . The perturbageB1 − tive part of the Hamiltonian is H 1 = γ e iωt S− + e − iωt S+ , with γ = − . S and S+ 4m are lowering and rising operators. The B0 and B1 are constants with B1  B0 . The spin up and spin down states of the system are represented by + and − , respec tively. Use the equations presented in Section 5.5.1 and show that Sz ± = ± ± , 2 S + + = S − − = 0, S + − =  + and S− + =  − .

(

)

7. Consider again the system described in exercise 6 and show that (a) it is an exactly geB0 2 solvable two-state system with ω 21 = , and (b) c2 (t) is given by Eqn. (10.19). 2m

11 Relativistic Quantum Mechanics The Schrödinger equation describes the quantum mechanics of nonrelativistic particles. Some of the inherent limitations of the Schrödinger equation are the following: (i) The Hamiltonian does not include the spin motion of particle(s); (ii) The Hamiltonian involves the first order derivative with respect to time but the second order derivatives in space coordinates indicating that time and space coordinates are not treated at par; (iii) The Schrödinger formulation of quantum mechanics is not Lorentz–invariant, which requires that the laws of physics must be formulated in such a way that the description should not allow one to differentiate between inertial frames (frames of reference) that are moving relative to each other with a constant uniform velocity. The Schrödinger equation is applicable when the kinetic energy of the particle is given 2 p2 2 by instead of ( pc ) + m0 c 2 and its energy is not comparable with its rest mass 2 m0 energy m0 c 2, where c is the velocity of light and m0 is the rest mass of the particle. If space and time coordinates of an event are represented by (r , t) and (r′ , t ′) in two inertial frames, then the Lorentz–invariance requires the linear transformation c 2 (dt)2 − (dr )2 = c 2 (dt ′)2 − (dr ′)2, which connects space-time coordinates ( x , y , z, t ) of one reference frame to that of another ( x ′ , y ′ , z ′ , t ′ ). The concept of Minkowski space or 4-vector space is used in the special theory of relativity and relativistic mechanics. Some of the well-known 4-vectors in covariant representation are the position vector xµ ≡ ( x1 , x2 , x3 , x4 ) with x4 = ict; linear momentum pµ ≡ ( p1 , p2 , p3 , p4 ) with p4 = iE/c where E is energy; current density jµ = ( j1 , j2 , j3 , j4 ) with j4 = icρ where ρ is charge density; and wave vector kµ ≡ ( k1 , k2 , k3 , k 4 ) with k 4 = iω / c where ω is angular frequency.   ∂   ∂ The first order derivative in the 4-vector space is defined as  , ∇  or  , −∇  . The  c ∂t   c ∂t  continuity equation in 4-vector space is:

(



)

∂ρ  + ∇.J = 0 ∂t ∂(icρ) ∂ jx ∂ jy ∂ jz ∂ j1 ∂ j2 ∂ j3 ∂ j4 ⇒ + + + = + + + = 0 (11.1) ∂(ict) ∂ x ∂ y ∂ z ∂ x1 ∂ x2 ∂ x3 ∂ x4 4

⇒

∂ jν

∑ ∂x ν= 1

ν

=0

Here ( x , y , z) are represented by ( x1 , x2 , x3 ) and ict by x4 .

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11.1  The Klein-Gordon Equation Klein-Gordon proposed a relativistic wave equation which is the quantum version of the energy momentum relation. It is of second order in both space and time and manifestly Lorentz-invariant. It relates to the Schrödinger equation as well. p2 ∂ In the nonrelativistic case, we replace p → − i∇ and E → i in = E to get the free ∂t 2m 2  ∂ψ (r , t) particle Schrödinger equation − ∇ 2 ψ (r , t) = i . For the relativistic case, the ∂t 2m 2 2 energy and momentum relation is E 2 = ( pc ) + m0 c 2 . Replacing E and p by operators and then operating both sides on ψ (r , t), one gets:

(



−2

)

∂2 ψ (r , t) = −  2 c 2 ∇ 2 ψ (r , t) + m0 c 2 ∂t 2

(

)

2

ψ (r , t) (11.2)

which is known as the Klein-Gordon equation. One of the solutions of the equation is: ψ (r , t) =



1 i(p.r )/ − iEt/ e e (11.3a) V

where V is volume. It is a second-order equation with respect to both time and space coordinates. Therefore: ψ (r , t) =



1 i(p.r )/ iEt/ e e (11.3b) V

is also an equally acceptable solution. Since E = ± c p 2 + ( m0 c ) , one of the solutions corresponds to E > 0, while the other belongs to E < 0. Note that for each value of p, there are negative as well as positive values of E. There is no mechanism to have a transition from positive energy to negative energy. In the presence of some external potential, E and p will be altered to include the effect of field, and then the solutions ψ (r , t) can be expressed as superposition of free particle solutions, provided that the free particle solutions form a complete set. Formation of a complete set does not permit us to discard the negative energy solutions. 2

11.1.1  Probability Density and Probability Current Let us rewrite Eqn. (11.2): 2



 m0 c 2  ∂2 ψ (r , t) 2 2 = ∇ ψ ( , − r ) c t    ψ (r , t) (11.4a) ∂t 2

the complex conjugate of which is: 2



 m0 c 2  * ∂2 ψ * (r , t) 2 2 * ( , = ∇ ψ r ) − c t    ψ (r , t) (11.4b) ∂t 2

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On operating from the left on the first equation by ψ * (r , t) and then on the second equation by ψ (r , t) and then subtracting the second from the first, we get: ∂  * ∂ψ (r , t) ∂ψ * (r , t)  2 * * − ψ (r , t) ψ (r , t)  = c ∇. ψ (r , t)∇ψ (r , t) − ψ (r , t)∇ψ (r , t) ∂t  ∂t ∂t 

{

 

⇒

∂  i  ∂t  2 m0 c 2

}

 * ∂ψ (r , t) ∂ψ * (r , t)   r r ( , ) ( , ) ψ − ψ t t   ∂t ∂t   

(11.5)

 i  − ∇. ψ * (r , t)∇ψ (r , t) − ψ (r , t)∇ψ * (r , t)  = 0 2 m  0 

{

}

Defining probability density P and probability current (flux) S by:

P=

i 2 m0 c 2

 * ∂ψ (r , t) ∂ψ * (r , t)  − ψ (r , t) ψ (r , t)  (11.6a) ∂t ∂t  

and:

S=−

i ψ * (r , t)∇ψ (r , t) − ψ (r , t)∇ψ * (r , t) (11.6b) 2 m0

{

}

we get the continuity equation:

∂P + ∇.S = 0 (11.7) ∂t

In the case of the Schrödinger equation, P is associated with the single particle complex 2 valued wave function so that Pd 3 r = ψ d 3 r is the probability of finding the particle in 2 3 volume d r. Interpretation of P as probability density requires that ψ > 0 . Therefore, if relativistic quantum mechanics is to be constructed in analogy with nonrelativistic quantum mechanics, then the relativistic wave function must be able to construct bilinear forms that can be interpreted as probability density and probability current, which satisfy the continuity equation. And, the probability density must be positive. It demands that the P defined by Eqn. (11.6a) must be positive at all values of time t. In addition to this, P should 2 be the fourth component of a 4-vector density and it must transform like P → P/ 1 − ( v/c ) 3 to make Pd r invariant under Lorentz transformations. The S defined by Eqn. (11.6b) is exactly the same as that defined in the case of nonrelativistic quantum mechanics. However, there is difficulty in interpreting P as probability density. The Klein-Gordon equation consists of a second order derivative with respect time. As stated above, both Eqns. (11.3a) and (11.3b) provide equally acceptable solutions for a given situation. Use of Eqn. (11.3a) in Eqn. (11.6a) gives P = E/Vm0 c 2, while with the use of Eqn. (11.3b) we get P = −E/Vm0 c 2. This implies that P can be positive or negative. Also, the solutions for E < 0 cannot be omitted because this violates the condition of formation of a complete set of solutions. It then appears that either the interpretation of P defined in Eqn. (11.6a) as probability density is to be abandoned or the Klein-Gordon equation is to be discarded.

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To overcome this problem, Pauli and Weisskopf suggested that Eqn. (11.5) can be multiplied throughout with electric charge q and then ρ = qP and J = qS can be treated as charge density and current density, respectively. q can take both positive and negative values. The ρ and J satisfy the continuity equation and ρ can be assigned both positive as well as negative values. Then, the problem of interpretation would be dissolved.

11.1.2  The Klein-Gordon Equation in a Coulombic Field In the presence of an electromagnetic field, the energy and momentum of a particle are modified according to E → E − qφ and p → p + qA, where φ and A are scaler and vector potentials and q is a charge on the particle. We then have:

(E − qφ)2 = c 2 (p + qA)2 + ( m0 c 2 ) (11.8) 2



When φ is the coulombic potential qφ = − 2

Ze 2 and a magnetic field is absent A = 0, we get: 4πε 0 r

 Ze 2  2 2 2  E + 4πε r  ψ (r , t) = c p ψ (r , t) + m0 c 0



(

)

2

2

ψ (r , t)

 ∂ Ze 2  ψ (r , t) = −  2 c 2 ∇ 2 ψ (r , t) + m0 c 2 ⇒  i +  ∂t 4πε 0 r 

(

)

(11.9) 2

ψ (r , t)

Taking ψ (r , t) = ϕ(r )e − iEt/ , we obtain: 2

 Ze 2  2 2 2 2  E + 4πε r  ϕ(r ) = −  c ∇ ϕ(r ) + m0 c 0



(

)

2

ϕ(r ) (11.10)

Note that E is an operator in Eqn. (11.9), while it is energy eigenvalue in Eqn. (11.10). Writing: ∇2 =



1 ∂  2 ∂ 1  1 ∂  ∂ 1 ∂2  (11.11a)  r  + 2   sin θ  + 2 ∂r ∂θ sin 2 φ ∂φ2  r ∂r r  sin θ ∂θ

and then writing ( θ, φ ) dependent part in terms of L2 , with the use of Eqn. (5.36b), we get: ∇2 =



1 ∂  2 ∂ L2 r − (11.11b)   r 2 ∂r  ∂r  r 2  2

Thus: 2

 Ze 2  L2  2 2 2 1  d  2 d  ϕ = − − E + ( r )  c r   ϕ(r ) + m0 c    4πε 0 r  r 2  dr  dr   2 



(

)

2

ϕ(r ) (11.12)

Writing ϕ(r ) = Rl (r )Ylm (θ, φ) and then using L2Ylm (θ, φ) = l(l + 1) 2Ylm (θ, φ), we have:

1 r2

2

2 d  2 d  Ze 2  1   m0 c  Rl (r ) (11.13)   r  Rl (r ) − 2 2  E +  − l(l + 1)  Rl (r ) =  dr   c  4πε 0 r   dr 

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Taking γ =

Ze 2 , we rewrite the equation as: 4πε 0 c  m02 c 4 − E 2  1 d  2 dRl   2Eγ l(l + 1) − γ 2  − r − R + l      2 c 2  Rl = 0 (11.14) r 2 dr  dr   cr r2

Equation (11.14) resembles Eqn. (6.11) and on writing E = E ′ + m0 c 2 and then taking  Ze 2  2  E ′ + 4πε r   m0 c , it reduces to a nonrelativistic equation for the hydrogen atom. 0 Therefore, the well behaved solution of the equation can be found following the procedure described in Section 6.2. After defining:

λ=

Eγ 2 4 0

m c − E2

 and  s( s + 1) = l(l + 1) − γ 2 (11.15)

we find that a well behaved solution exists if: λ = υ + s, with υ = 1, 2, 3,….. (11.16)



where s should be positive to make Rl (r ) finite at r = 0. Solving for s yields: s=−



2  1  1  ±  l +  − γ 2  2  2  

1/2

(11.17)

Since γ is very small for all practical purposes, positive values of s are possible for l > 0 with the plus sign. Thus, we have: 2  1  1  = υ − +  l +  − γ 2  2 4 2 2  2  m0 c − E 

Eγ



= υ−

2 1  1   1  +  l +  1 − γ 2 /  l +    2  2  2 

≈υ+l−

To obtain E , we take υ + l = n and n −

1/2

1/2

(11.18)

γ2  1 2l +   2

γ2  2l + 

1  2

= C:

E2 γ 2 = m02 c 4 − E 2 C

 γ2  ⇒ E 2  1 +  = m02 c 4 C   γ2  ⇒ E = m0 c 2  1 +  C 

−1/2

(11.19)  γ 2 3γ 4  ≈ m0 c 2  1 − + 2C 8C 2  

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Expanding with the use of the binomial series yields:



 1  = n − C  

  γ2   1 2l +    2  

−2

1 ≈ 2 n

  1 +   

 3 γ 4  + (11.20a)  1  4n2  1  2  n l +  + l    2 2   γ2

and:



   1  γ2 = n −  C2   1  2l +   2   

−4

    2 4 1 2γ 5 γ  ≈ 4 1 + + 2 2 (11.20b) n   1  2n  1   n l +   l +     2 2  

and then rearranging the terms, we obtain:   γ2 γ4 E = m0 c 1 − 2 − 4 2n 2n   2



  3   n − (11.21)  1 4   l +    2

which finally gives: 4



 Ze 2  Z 2 e 4 m0 m0 E ′ = E − m0 c = − 2 2 2 −  2  4 n4 c 2 πε 4  πε n 4  ( 0) 0 2

   n 3 −  (11.22)    l + 1  4    2 

The first term on the right-hand side is simply nonrelativistic energy and the second is the relativistic corrections to energy.

11.2  The Dirac Equation The Klein-Gordon equation is quite satisfactory when it is interpreted properly and therefore it had been used to explain several physical phenomena. However, there are valid reasons to discard it for the description of Fermions. Like the Schrödinger equation, it does not accommodate the spin behavior of a Fermion in a natural manner. Also, if we look closely at the problem of interpreting the probability density, we find that it is originated from the fact that the Klein-Gordon equation is second order in time and hence both the solutions given by Eqns. (11.3a) and (11.3b) are to be accepted. This suggests that the problem of interpreting the probability density can be avoided if the relativistic equation involves only the first order time derivative. And then to maintain the parity between time and space coordinates, the equation should have first order derivatives in space coordinates as well. It can then be inferred that a relativistic quantum mechanical equation should incorporate the spin of a particle, and it should have first order derivatives in both time and space coordinates. In 1920, P.A.M. Dirac successfully derived a wave equation

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Relativistic Quantum Mechanics

∂ . With the use of the equation, he ∂t had been able to show that derived probability density is always positive. The Dirac equation is considered the only correct relativistic wave equation.

starting with the condition that it should be linear in

11.2.1  Derivation of the Dirac Equation In the case of nonrelativistic quantum mechanics, the operator representing kinetic energy of an electron with rest mass m0 is: T=



p2 2 2 =− ∇ (11.23) 2 m0 2 m0

0  are −1  (σ.p)(σ.p) . Pauli matrices, which are discussed in Section 5.5.1, see Eqn. (5.63), we get T = 2 m0 In the solved example 6 of Chapter 5, it was shown that for any two operators A and B, ( σ.A )( σ.B ) = A.B + iσ.(A × B), which gives (σ.p)(σ.p) = p 2 . This suggests that replacing p by σ.p leaves kinetic energy unchanged. Next let us consider the effect of magnetic field on the kinetic energy of an electron. In the presence of a magnetic field, the effective linear momentum of the electron would be p′ = p − eA , where −e is the charge on the electron and A is the vector potential. Again on replacing p′ by σ.p′ the kinetic energy operator in the presence of the magnetic field is:  0 On replacing p by σ. p , where σ 1 =   1



T′ =

 0 1  ; σ2 =   0   i

 1 −i  and σ 3 =   0   0

(σ.p′)(σ.p′) {σ.(p − eA)}{σ.(p − eA)} (11.24) = 2 m0 2 m0

which with the use of the identity ( σ.A )( σ.B ) = A.B + iσ.(A × B) goes to: T′ = =

(p − eA)2 i + σ.(p − eA) × (p − eA) 2 m0 2 m0 (p − eA)2 ie − σ. ( p × A + A × p ) 2 m0 2 m0

(11.25)

The operation of p × A + A × p on a function gives: (p × A + A × p) f (r ) = − i {∇ × (Af ) + A × (∇f )}

= − i {(∇ × A) f − A × (∇f ) + A × (∇f )} (11.26)

= − iBf (r ) where B is a magnetic field. We thus have:

T′ =

(p − eA)2 e − σ.B (11.27) 2 m0 2 m0

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The second term on the right-hand side is nothing but the energy of the electron due to the interaction of the spin of the electron with a magnetic field, while the first term is the square of the effective linear momentum p′ divided by 2 m0. An important point emerging here is that if linear momentum is replaced by the scalar product of σ with linear momentum, the spin behavior of the electron is automatically accommodated in the kinetic energy operator. This suggests that if we follow a similar procedure, electron spin is incorporated in the relativistic expression for the kinetic energy operator. Starting from the relativistic energy momentum relation for a free electron: 2

2  E 2   = p + ( m0 c ) c



2 E  E  ⇒  − p .  + p = ( m0 c ) c  c 

(11.28)

we then replace p by σ.p to get: 2 E E   − σ.p  + σ.p = ( m0 c ) (11.29) c c



∂ where E = i and p = − i∇ are operators. A second order differential equation of a par∂t ticle then is:    ∂ ∂ 2  i ∂ x + iσ.∇   i ∂ x − iσ.∇  φ = ( m0 c ) φ (11.30) 0 0



∂ ∂ = . φ is a two component wave function. Our aim is to obtain c ∂t ∂ x0 ∂ an equation that is linear in . The relativistic covariance demands that the wave equa∂t ∂ tion linear in should also be linear in ∇. Let us define two component wave functions ∂t φ = φ and where we have used

L



φR =

 i  ∂ − σ.∇  φ  m0 c  ∂ x0

(11.31)

The total number of components in the wave has now gone to 4 from 2. We then have:



 i  ∂ + σ.∇  φR = φL    m0 c ∂ x0  ∂  ⇒ i  + σ.∇  φR = m0 cφL  ∂ x0 

(11.32a)

and:

 ∂  i  − σ.∇  φL = m0 cφR (11.32b)  ∂ x0 

In this manner, a second order Eqn. (11.30) is equivalently written in two first order Eqns. (11.32a) and (11.32b). The subscripts R and L are chosen to represent the fact that

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Relativistic Quantum Mechanics

for m0 → 0 , φR and φL describe a right-handed (spin parallel to momentum direction) and 1 left-handed (spin anti-parallel to momentum direction) state of the particle having spin − . 2 On taking the sum and difference of these two equations we obtain:

i



i

∂ (φR + φL ) + i ( σ.∇ ) (φR − φL ) = m0 c (φR + φL ) (11.33a) ∂ x0

∂ (φR − φL ) + i ( σ.∇ ) (φR + φL ) = − m0 c (φR − φL ) (11.33b) ∂ x0

Defining ψ A = φR + φL and ψ B = φR − φL , we get: ∂ψ A − σ.pψ B = m0 cψ A ; ∂ x0 (11.34) ∂ψ B − σ.pψ A = − m0 cψ B i ∂ x0 i

Note that:  p3 σ.p = σ 1 p1 + σ 2 p2 + σ 3 p3 =   p1 + ip2



p1 − ip2   (11.35)  − p3

 ψA suggests that each ψ A and ψ B must have two components. Hence ψ =   ψ B components. Equation (11.34) can be written as a matrix equation:  ∂  I    i  ∂ x0  0 

0   0 − I   σ.p

σ.p    ψ A  0    ψ B 

  I  = m0 c    0

0  ψA  −I   ψB 

  has four 

  (11.36) 

where σ.p = σ 1 p1 + σ 2 p2 + σ 3 p3 is the sum of three 2 × 2 matrices. Therefore:  0    σ.p

σ.p   = α 1 p1 + α 2 p2 + α 3 p3 , 0 

 0 with α i =   σi

σi   0 

and i = 1, 2, 3 (11.37)

Since σ 1, σ 2 , and σ 3 are 2 × 2 matrices, each of α 1, α 2 , and α 3 will be a 4 × 4 matrix. This also suggests that I in Eqn. (11.36) is a 2 × 2 unit matrix. Equation (11.36) thus takes the form:

i

∂ψ = cα.p + βm0 c 2 ψ (11.38) ∂t

(

)

with:



  β=  

1 0 0 0

0 1 0 0

0 0 −1 0

0 0 0 −1

   (11.39)  

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and:    ψ=   



ψ1   ψ2  (11.40) ψ3   ψ4  

It is to be noted here that though ψ has four components, it not a 4-vector because it has nothing to do with the 4-dimensional nature of space-time. It is commonly known as a bispinor or Dirac spinor. The matrices α 1 , α 2 , α 3 and β anti-commute with each other:

 

 0 α 1α 2 + α 2 α 1 =   σ 1

σ1   0  0   σ 2

 σ 1σ 2 + σ 2 σ 1 = 0 

σ2   0 + 0   σ 2

σ2   0  0   σ 1

 0  σ 1 σ 2 + σ 2 σ 

σ1   0 

(11.41a)

Since σ 1 σ 2 = iσ 3 and σ 2 σ 1 = − iσ 3 , we get α 1α 2 + α 2 α 1 = 0. Similar algebra shows that α 2 α 3 + α 3 α 2 = 0; α 3 α 1 + α 1α 3 = 0, and α 1β + βα 1 = 0. Also, we see that:

 0 α 2i =  2  σi 

 σ 2i   = 1 and β 2 =  1 0   0

0   = 1 (11.41b) 1 

The properties of matrices can be summarized by α i α j + α j α i = 2δ ij and α iβ + βα i = 0. The α matrices and the β matrix are Hermitian and traceless. 11.2.2  Covariant Form of the Dirac Equation Equation (11.38) can be written in covariant form. On dividing each term with c and multiplying by β and then using β 2 = 1, we get:

 ∂ m0 c  β ∂(ict) − iβα.∇ +   ψ (r , t) = 0 (11.42)  

 0 Writing ict = x4 ; β = γ 4 and γ k = − iβα k =   iσ k form of the Dirac equation:   



4

∑γ µ= 1

µ

− iσ k   ; k = 1, 2, 3 , we obtain the covariant 0 

∂ m c + 0  ψ = 0 (11.43) ∂ xµ  

which is also expressed as: 4



 4  ∂  m c  γ µ )ij +  0  δ ij  ψ j = 0 (11.44) ( ∂ xµ      µ= 1  

∑∑ j=1

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11.2.3  Probability Density and Probability Current Multiplying Eqn. (11.38) from the left with the conjugate adjoint of Eqn. (11.40): ψ † =  ψ *1 



ψ *2

ψ *3

ψ *4  (11.45) 

we get: iψ †



∂ψ = − icψ † α.∇ψ + ψ †βm0 c 2 ψ (11.46a) ∂t

We next multiply the conjugate adjoint of Eqn. (11.38) with ψ from the right, to obtain: − i



∂ψ † ψ = ic ∇ψ † .α † ψ + ψ †β † m0 c 2 ψ (11.46b) ∂t

(

)

α i and β are Hermitian and m0 c 2 scalar, subtraction of Eqn. (11.46b) from (11.46a) yields:



or

{

) }

∂ ψ † ψ = − ic ψ † α. ( ∇ψ ) + ∇ψ † .αψ ∂t ∂ ⇒ ψ † ψ + ∇. cψ † αψ = 0 ∂t i

(

)

(

)

(

(

)

(11.47)

∂P + ∇.S = 0, with probability density (P) and probability current (S): ∂t P = ψ † ψ (11.48a)

and:

(

)

Sj = c ψ † α j ψ (11.48b)

2

2

2

2

Thus P = ψ 1 + ψ 2 + ψ 3 + ψ 4 , which is contributed by four probable states of an electron, is positive definite. The probability current has three components: S1 = cψ † α 1 ψ

= c  ψ *1 

ψ *2

ψ *3

ψ *4

      

0 0 0 1

ψ *4

       

ψ4   ψ3  ψ2   ψ1  

= c  ψ *1 

(

ψ *2

ψ *3

= c ψ *1 ψ 4 + ψ *2 ψ 3 + ψ *3 ψ 2 + ψ *4 ψ 1

)

0 0 1 0

0 1 0 0

1 0 0 0

      

ψ1   ψ2  ψ3   ψ4  

(11.49a)

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S2 = cψ † α 2 ψ

= c  ψ *1 

ψ *2

ψ *3

ψ *4

      

ψ *4

 − iψ 4    iψ 3   − iψ 2   iψ 1 



= c  ψ *1 

ψ *2

ψ *3

(

0 0 0 i

= − ic ψ *1 ψ 4 − ψ *2 ψ 3 + ψ *3 ψ 2 − ψ *4 ψ 1

0 0 −i 0

)

0 i 0 0

−i 0 0 0

      

ψ1   ψ2  ψ3   ψ 4  (11.49b) 

0 0 −1 0

0 0 0 −1

      

ψ1   ψ2  ψ3   ψ4  

      

and S3 = cψ † α 3 ψ

= c  ψ *1 

ψ *2

ψ *3

ψ *4

      

ψ *4

 ψ1    ψ2   −ψ 3   −ψ 4 

= c  ψ *1 

(

2

ψ *2

2

ψ *3

2

= c ψ1 + ψ2 − ψ3 − ψ 4

2

1 0 0 0

0 1 0 0

)

      

(11.49c)

The continuity equation in terms of 4-vector probability current is: 4



∂Sµ

∑ ∂x µ= 1

µ

(

2

2

2

= 0, where S4 = icP = ic ψ 1 + ψ 2 + ψ 3 + ψ 4

2

) (11.50)

Equation (11.38) is the Dirac equation for a free particle. In the presence of potential energy V (r ) , the Dirac Hamiltonian is:

H = cα.p + βm0 c 2 + V (r ) (11.51)

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Relativistic Quantum Mechanics

11.3  Free Particle Solutions of the Dirac Equation For a freely moving particle, we can write:

 ψA ψ=  ψ B

 1  χ A (p)  =  V  χ B (p)

 i( p.r − Et )/ (11.52) e 

where V is volume and χ A (p) and χ A (p) are spinors. Substitution of Eqn. (11.52) into Eqn. (11.38) yields:  0 c  σ.p

σ.p   χ A (p)  0   χ B (p)

  I 2  + m0 c    0

0   χ A (p)  − I   χ B (p)

  I  = E   0

0   χ A (p)  I   χ B (p)

  

(11.53) which gives two equations:

cσ.pχ B + m0 c 2 χ A = Eχ A ⇒ χ A =

(

cσ.p χ B (11.54a) E − m0 c 2

)

and:

cσ.pχ A − m0 c 2 χ B = Eχ B ⇒ χ B =

(

cσ.p χ A (11.54b) E + m0 c 2

)

From Eqn. (11.35) we notice that σ.p is a 2 × 2 matrix and hence each of χ A and χ B should have two components. E takes both positive and negative values. Note that Eqn. (11.54a) cannot be used for E > 0, because it leads to an unacceptable solution when E → m0 c 2 . Similarly, Eqn. (11.54b) is applicable only for E < 0. For E = c p 2 + m0 c 2 > 0, we take, apart  1   0  from the normalization constant,   and  1  for χ A (p). The corresponding two 0     components of χ B (p), call these χ B1 (p) and χ B 2 (p), are:

χ B1 (p) =



=

(

cσ.p  1    E + m0 c 2  0 

(

c E + m0 c 2

)

( (

)

 p3   p1 + ip2

 cp / E + m c 2 3 0 =  cp + / E + m c 2 0 

) )

   

p1 − ip2   1     0  (11.55a) − p3

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and:

χ B 2 (p) =

=



(

cσ.p  0    E + m0 c 2  1 

(

c E + m0 c 2

) )

 p3   p1 + ip2

 cp − / E + m0 c 2  = 2  − cp3 / E + m0 c

( (

) )

p1 − ip2   0    − p3   1 

(11.55b)

   

where p + = p1 + ip2 and p − = p1 − ip2 . Thus, two solutions for E > 0 are:



   χ(1) (p) = N  cp3 /   +  cp / 

1 0

(E + m c ) (E + m c ) 0

0

Here N is the normalization  1 procedure and choosing   0

2

2

  0   1    , and χ(2) (p) = N  cp − / E + m c 2 0     2   − cp3 / E + m0 c  

( (

    (11.56a)    

) )

constant that is to be determined later. Following a similar   0   and  1  for χ B (p), we obtain another two solutions for   

the case of E = − c p 2 + m0 c 2 < 0:



  − cp3 /  + χ(3) (p) = N  − cp /    

(E +m c ) (E +m c ) 0

0

1 0

2

2

(

  − 2   − cp / E + m0 c   2  , and χ(4) (p) = N  cp3 / E + m0 c   0     1  

(

)

)

    (11.56b)    

Four values of χ(l ) (p) with l = 1, 2, 3, 4 represent four possible states: (i) E > 0 and spin up, (ii) E > 0 and spin down, (iii) E < 0 with spin up, and (iv) E < 0 with spin down. Since i

ψ (r , t) = χ(l ) (p)e 

(p.r − Et )

satisfies the free particle Dirac Eqn. (11.38), the spinors χ(l ) (p) satisfy

( cα.p + βm c ) χ 0

2

(l )

(p) = Eχ(l ) (p) (11.57)

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Relativistic Quantum Mechanics

(

)

†

The spinors are orthogonal: χ(l′ ) (p) χ(l ) (p) = 0 . To see it, let us take:

(χ

(1)

)

†

(p) χ(2) (p)

= N 2  1 

(

cp3 / E + m0 c 2

0

)

(

cp − / E + m0 c 2

 0  1   − 2   cp / E + m0 c  2  − cp3 / E + m0 c 

)

( (

) )

     (11.58)   

 c 2 p3 p − − c 2 p3 p −  = N 2 0 + 0 + =0 2 2   E m c + 0  

(

)

Similarly, it can be proved for other values of l ′ and l. The normalization of wave function requires ∫ ψ † ψd 3 r = 1 . Two approaches have been adopted to normalize the relativistic wave function:

(

)

†

a. χ(l ) (p) χ(l ) (p) = 1 , which for l = 1 reduces to:

(χ

(1)

)

†

(p) χ(1) (p)

= N 2  1 

(

cp3 / E + m0 c 2

0

( (

 c 2 p32 + p12 + p22  = N 1+ 0+ 2  E + m0 c 2  2

)

)

(

cp − / E + m0 c 2

)  = N

2

 

 c2 p2 1 +  E + m0 c 2 

(

 2 p 2 c 2 + 2 m2 c 4 + 2Em c 2  2 0 0  = 2N E = N2  2   E + m0 c 2 E + m0 c 2  

(

)

(

      cp3 /  +  cp / 

)

)

2

1 0

(E + m c ) (E + m c ) 0

0

2

2

       (11.59a) 

   

)

This yields:



N=

( E + m c ) (11.59b) 0

2E

2

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Another method of normalization used in some of the textbooks on quantum mechanics is as follows:

(

)

†

b. χ(l ) (p) χ(l ) (p) = E / m0 c 2 , which gives:

(



2N 2 E E = 2 m0 c 2 E + m0 c

⇒N=

)

(E +m c ) 0

2

(11.59c)

2 m0 c 2

The second normalization method says that χ † χ transforms like the zeroth component of a 4-vector, which appears somewhat artificial. However, it has been used in the literature. In the nonrelativistic limit p  m0 c, E ≈ m0 c 2 and N → 1. Therefore, for the nonrelativistic case, we have from Eqns. (11.56) to (11.59):



χ(1)

  =  

1 0 0 0

  0   0   0    1   0   0   ; χ(2) =   ; χ(3) =   ; χ(4) =   (11.60)   0   1   0    0   0   1 

11.3.1  Positive and Negative Energy Eigenvalues The two energies E + = p 2 c 2 + m02 c 4 and E − = − p 2 c 2 + m02 c 4 with 0 ≤ p ≤ ∞, suggests that m0 c 2 ≤ E + ≤ ∞ and −∞ ≤ E − ≤ − m0 c 2 . Therefore, there are two regions of energy; one starts from m0 c 2 and extends up to ∞, while the other begins at −m0 c 2 and extends down to −∞. Two regions of energy are separated by a forbidden energy gap Eg = 2 m0 c 2 and no energy states exist in the gap region. Fig. 11.1 illustrates the energy regions and the gap.

FIGURE 11.1 Energy levels for the free particle Dirac equation.

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Relativistic Quantum Mechanics

There is a problem in imaging negative energy states because an electron in a positive energy (bound or free) state should be able to emit a photon and make a transition to a negative energy state. The process could continue giving off an infinite amount of energy, which does not happen in reality. Dirac postulated a solution to this problem, which says that all of the negative energy states are completely filled and hence Pauli’s exclusion principle does not permit the positive energy electrons to make any transitions to negative energy states. The negative energy states cannot have any physically observable effect. It was further postulated that electrons in completely filled negative energy states can make transitions to positive energy states if they pick up energy which is greater than or equal to the forbidden energy gap. Completely filled negative energy states are also referred to as vacuum states. When an electron makes the transition to positive energy states, it leaves behind a vacancy in the negative energy spectrum. The vacancy is termed a hole, whose behavior is opposite that of an electron or the particles in negative energy states. A hole has a positive charge and its momentum and spin are in the opposite direction to that of vacuum (negative) states. The transition of an electron from a negative to a positive energy state is the pair creation, which produces a positron, a pair of an electron and a hole. The discovery of the positron provided a big support to the theory of the hole. The interpretation of negative energy states is not the hypothetical but has many elements of truth. When the Dirac field is quantized, one no longer needs the idea of an infinite negative energy sea; the electrons and positrons behave as if they were there.

11.4  The Dirac Equation and the Constants of Motion For an observable A, the Heisenberg equation is: dA i = [ A, H ] (11.61) dt 



where H is the Hamiltonian. If [ A, H ] = 0, we say that A is the constant of motion described by the Hamiltonian H . In the case of nonrelativistic quantum mechanics, where H is the Schrödinger Hamiltonian, it was shown that linear momentum p, orbital angular momentum L, and L2 are the constants of motion. What are the constants of motion when motion is described by the relativistic Dirac equation? Let us answer this. a. Linear and Angular Momenta We first take the case of linear momentum. The free particle Dirac Hamiltonian H = cα.p + βm0 c 2 gives:

[p, H ] = pH − Hp = p(cα.p + βm0 c 2 ) − (cα.p + βm0 c 2 )p (11.62) Since β does not depend on space coordinates pβm0 c 2 = βm0 c 2 p. This yields:



(

ˆ [p, H ] = c ipˆ 1 + ˆjp2 + kp 3

3

3

) ∑α p − c∑α p (ipˆ + ˆjp j

j=1

j

j

j=1

j

1

2

)

ˆ = 0 (11.63) + kp 3

stating that the linear momentum is a constant of motion even when the motion is described by the relativistic Dirac equation.

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We next take [ L, H ] = iˆ [ L1 , H ] + ˆj [ L2 , H ] + kˆ [ L3 , H ], where:

[ L1 , H ] = L1 H − HL1

= c ( L1α.p − α.pL1 ) + ( L1β − βL1 ) m0 c 2 = c ( x2 p3 − x3 p2 )



3

3

∑

α j pj − c

j=1

 = c ( x 2 p 3 − x 3 p 2 )  

∑α p ( x p j

j

2 3

− x3 p2 )

j=1

 α j p j + i   j=1 3

∑

3

∑

αj

j=1

(11.64)

 ∂   ( x2 p3 − x3 p2 )   ∂ x j  

Since β does not depend on space coordinates, we have used L1β − βL1 = 0. Then:   − i 

3

∑

αj

j=1

∂   ( x2 p3 − x3 p2 ) ∂ x j 

∂ ∂ ∂   = − i α 1 x2 p3 − x3 p2 ) + α 2 x2 p3 − x3 p2 ) + α 3 x2 p3 − x3 p2 )  ( ( ( ∂ x2 ∂ x3  ∂ x1  ∂ ∂ ∂   = − i α 1 − α 2 x3 p2 x2 p3 − x3 p2 ) + α 2 p3 + α 2 x2 p3 (  ∂ x2 ∂ x2   ∂ x1

(11.65a)

 ∂ ∂  − i α 3 x2 p3 − α 3 p2 − α 3 x3 p2  ∂ x3 ∂ x3   = α 1 p1 ( x2 p3 − x3 p2 ) − iα 2 p3 + α 2 x2 p3 p2 − α 2 x3 p22 + α 3 x2 p32 + iα 3 p2 − α 3 x3 p2 p3 = α 1 p1 ( x2 p3 − x3 p2 ) + ( x2 p3 − x3 p2 ) α 2 p2 + ( x2 p3 − x3 p2 ) α 3 p3 − i(α 2 p3 − α 3 p2 ) = ( x2 p3 − x3 p2 )

3

∑ α p − i(α p j

j

2 3

− α 3 p2 )

j=1

Therefore:

[ L1 , H ] = ic(α 2 p3 − α 3 p2 ) (11.66) Adopting a similar procedure we get:



[ L2 , H ] = ic(α 3 p1 − α 1 p3 ),

and [ L3 , H ] = ic(α 1 p2 − α 2 p1 ) (11.67)

Note that we obtain [ L2 , H ] from [ L1 , H ] and [ L3 , H ] from [ L2 , H ] by changing suffixes in cyclic order. We finally get:

[L, H ] = ic {iˆ(α 2 p3 − α 3 p2 ) + ˆj(α 3 p1 − α 1 p3 ) + kˆ(α 1 p2 − α 2 p1 )} ⇒ [ L, H ] = ic(α × p)

(11.68a)

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Relativistic Quantum Mechanics

which yields: dL = c(α × p) ≠ 0 (11.68b) dt



suggesting that L is not the constant of motion when we use the relativistic Dirac equation to describe the motion. Now, let us consider an operator:   σ Σ=  2 2 0

S=



0  (11.69) σ 

We then calculate [ S, H ] = iˆ [ S1 , H ] + ˆj [ S2 , H ] + kˆ [ S3 , H ]. Take:  2

[S1 , H ] = S1 H − HS1 = ( Σ 1 H − HΣ 1 ) (11.70a)

where:  Σ 1 H − HΣ 1 = cΣ 1  

3

∑ j=1

  α j p j + βm0 c  − c   



3

∑ α p + βm c Σ j

j

0

1

j=1

= c ( Σ 1α 1 − α 1 Σ 1 ) p1 + ( Σ 1α 2 − α 2 Σ 1 ) p2 + ( Σ 1α 3 − α 3 Σ 1 ) p3 + ( Σ 1β − α 1β ) m0 c  (11.70b) Now: ( i)

 σ1 Σ 1α 1 − α 1 Σ 1 =   0

0  0 σ 1   σ 1

 0 =  1− 1  σ1 (ii) Σ 1α 2 − α 2 Σ 1 =   0

σ1   σ1 0   0

0  σ 1 

σ2   0 − 0   σ 2

σ2   σ1 0   0

0  σ 1 

1− 1  = 0; 0  0  0 σ 1   σ 2

 0 = σ σ − σ2 σ1 1 2   σ1 (iii) Σ 1α 3 − α 3 Σ 1 =   0

σ1   0 − 0   σ 1

 0 σ 1σ 2 − σ 2 σ 1  = 2  0 σ 1σ 2  

0  0 σ 1   σ 3

σ3   0 − 0   σ 3

σ 1σ 2   = 2 iα 3 ; (11.70c) 0 

σ3   σ1 0   0

0  σ 1 

= −2 iα 2 ; (iv)

 σ1 Σ 1β − βΣ 1 =   0

0  I σ 1   0

 σ1 − σ1 = 0 

0   I − − I   0

 0 =0 −σ 1 + σ 1 

0   σ1 − I   0

0  σ 1 

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Therefore:

[ S1 , H ] = − ic(α 2 p3 − α 3 p2 ) (11.71a)

Similarly, we get:

[ S2 , H ] = − ic(α 3 p1 − α 1 p3 ), and [S3 , H ] = − ic(α 1 p2 − α 2 p1 ) (11.71b)



which yields:

[S, H ] = − ic (α × p) , and



dS = − c ( α × p ) (11.72) dt

Combining Eqns. (11.69) and (11.72), we obtain: dL dS dJ + = = c ( α × p ) − c ( α × p ) = 0 (11.73) dt dt dt



stating that J = L + S is the constant of motion. The S is called spin angular momentum and J is known as total angular momentum. [ J, H ] = 0 implies that  J2 , H  = 0 = [ J 3 , H ]. b. Operator K

 β has been found very useful in solving the 2 Dirac equation for the central potential problem. We will show that K too is a constant of motion. Let us take: An operator defined by K = βΣ.J −

[βΣ.J, H ] = βΣ.JH − HβΣ.J = βΣ.JH − βHΣ.J − HβΣ.J + βHΣ.J = −β [ H , Σ ] .J − [ H , β ] Σ.J



(11.74)

From Eqn. (11.72), we have [ Σ , H ] = −2 ic ( α × p ) ⇒ [ H , Σ ] = 2 ic ( α × p ). And:

[ H , β ] = cα.pβ − cβα.p + βm0 c 2β − βm0 c 2β  = c 





3

∑(α β − βα ) p  j

j

j

(11.75)

j=1

= −2 cβα.p Hence:

[βΣ.J, H ] = −2icβ (α × p) .J + 2cβ (α.p) ( Σ.J ) (11.76) The identity ( σ.A )( σ.B ) = A.B + iσ. ( A × B ) can also be written as ( Σ.A )( Σ.B ) = A.B + iΣ. ( A × B ) . Also, matrices Σ j can be transformed to α j though α j = −γ 5 Σ j , γ 5 is a 4 × 4 matrix, whose explicit expression is not needed here. To simplify the second term in Eqn. (11.76), we take:



(α.p) ( Σ.J ) = −γ 5 ( Σ.p) ( Σ.J ) = −γ 5 p.J + iΣ. (p × J ) (11.77)

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Relativistic Quantum Mechanics

which yields:



[βΣ.J, H ] = −2icβ (α × p) .J − 2cβγ 5 p.J + iΣ. (p × J ) = −2 icβ ( α × p ) .J − 2 cβγ 5 ( p.J ) + 2 icβα. ( p × J ) = −2 cβγ 5 ( p.J )    = −2 cβγ 5 p.  L + Σ   2 

(11.78)

= −2 cβγ 5 p.L + cβ ( α.p ) = cβ ( α.p )

 where we have used p.L = p.r × p = −p × p.r = 0. Then, using cβ ( α.p ) = − [ H , β ], 2 from Eqn. (11.75), we obtain: 

[βΣ.J, H ] + 2 [ H , β ] = 0

  Hβ − βH = 0 2 2 (11.79)       ⇒ β  Σ.J −  H − Hβ  Σ.J −  = 0   2 2

⇒ βΣ.JH − HβΣ.J +

⇒ KH − HK = 0  β  = β ( Σ.L +  ). We thus find that K is the constant of where K = β  Σ.J −  = βΣ.J −  2 2 motion. [ J, H ] = 0 and [ K, H ] = 0 imply that H , K, J2 , and J 3 commute with each other.

11.5 Spin Magnetic Moment (the Dirac Electron in an Electromagnetic Field) We would show that the Dirac equation includes the intrinsic spin magnetic moment of an electron. In the presence of an electromagnetic field, the energy and momentum of a free particle change: E → E − eφ and p → p − eA , where φ and A are scalar and vector potentials. The free particle Dirac equation in the presence of an electromagnetic field is then given by:

E − eφ − cα.(p − eA) − βm0 c 2  ψ = 0 (11.80)

Here, E and p are energy and momentum operators. To get the equation that is quadratic in E, we operate both sides from the left with E − eφ + cα.(p − eA) + βm0 c 2  and obtain:

E − eφ + cα.(p − eA) + βm0 c 2  E − eφ − cα.(p − eA) − βm0 c 2  ψ = 0 (11.81)

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which expands to:



 (E − eφ)2 − c(E − eφ)α.(p − eA) − (E − eφ)βm c 2 + cα.(p − eA)(E − eφ) 0   − c 2α.(p − eA)α.(p − eA) − cα.(p − eA)βm0c 2 + βm0c 2 (E − eφ)   −βm0c 2 cα.(p − eA) − m02 c 4 

  ψ = 0 (11.82)   

which with the use of α jβ + βα j = 0 and (E − eφ)β = β(E − eφ) simplifies to:



 (E − eφ)2 − c(E − eφ)α.(p − eA) + cα.(p − eA)(E − eφ) − m2 c 4 0   − c 2α.(p − eA)α.(p − eA) 

 ψ = 0 (11.83)  

The use of the identity (α.A)(α.B) = A.B + i ∑ .(A × B) gives:

{ {(p − eA) {(p − eA)

}

c 2α.(p − eA)α.(p − eA)ψ = c 2 (p − eA)2 + i ∑ .[(p − eA) × (p − eA)] ψ = c2



= c2

}

2

− ie ∑ .[ A × p + p × A ] ψ

2

− e ∑ .B ψ

}

(11.84)

where we used p × Aψ = ( p × A ) ψ − ( A × pψ ) and ∇ × A = B. Here B is the magnetic field. Since operator E depends on t and p depends on r they are interchangeable, and hence  ∂A  ψ and ( pφ − φp ) ψ = − i ( ∇φ ) ψ . On Epψ = pEψ . Further, note that ( EA − AE ) ψ = i   ∂t  substituting these into Eqn. (11.83), we obtain:    ∂A  2 2 4 2 2 (E − eφ) + iceα.  ∂t  + iceα. ( ∇φ ) − m0 c − c (p − eA) − eΣ.B  ψ = 0  

{



∂A   ⇒ (E − eφ) − c (p − eA) − m c  ψ + ec Σ.Bψ + iceα. ∇φ + ψ = 0 ∂t   2

2

In the Lorentz gauge E = −∇φ −



}

2

2 4 0

(11.85)

2

∂A , where E is the electric field. Therefore: ∂t

(E − eφ)2 − c 2 (p − eA)2 − m02 c 4  ψ + ec 2 Σ.Bψ − iceα.Eψ = 0 (11.86)

The (E − eφ)2 − c 2 (p − eA)2 − m02 c 4  ψ = 0 is the Klein-Gordon equation in the presence of an electromagnetic field. An additional two terms represent the interaction of electron spin with magnetic and with electric fields, respectively. To understand the physical significance of additional terms, let us take the nonrelativistic limit of Eqn. (11.86). The non-relativistic limit yields the well-known Schrödinger–Pauli equation. In the nonrelativistic limit,

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E ′ 0. For E > 0, we take two possible values  1   0   0  and  1  for χ a . Then from Eqn. (11.138b), we get:    



  (1) χ ( p1 ) = N    cp1 /

1 0 0

(E + m c ) 0

2

     and χ(2) ( p ) = N  1   cp1 /    

0 1

(E + m c ) 2

0

0

   (11.139a)   

and for E < 0, we have from Eqn. (138a):



  − cp1 / χ(3) ( p1 ) = N   

0

(E +m c ) 0

2

1 0

    − cp1 /  and χ(4) ( p ) = N  1      

(E +m c ) 0

0 0 1

2

   (11.139b)   

where N is the constant of normalization. The two energies of the particle are: E ± = ± p12 c 2 + m02 c 4 . ∂P + ∇.S = 0, where 4. Show that the conservation law for the Dirac probability current ∂t P = ψ † ψ and S = cψ † αψ , also holds in the presence of an electromagnetic field.

(

)

SOLUTION

In the presence of an electromagnetic field, momentum and the Hamiltonian are modified as p → p − eA and H → H + eφ , where φ and A are scalar and vector potentials, respectively. We therefore have:

i

∂ψ =  cα. ( − i∇ − eA ) + βm0 c 2 + eφ  ψ (11.140a) ∂t 

the conjugate adjoint of which is:

− i

∂ψ † = ic∇ψ † .α † + ψ †  − ceα.A + βm0 c 2 + eφ  (11.140b) ∂t

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Operate on Eqn. (11.140a) from the left with ψ † and on Eqn. (11.140b) with ψ from the right and then subtract Eqn. (11.140b) from Eqn. (11.140a):

(

) )

 ∂ψ ∂ψ †  = − ic ψ † ( α.∇ψ ) + α.∇ψ † ψ i  ψ † +ψ ∂t ∂t  

(

) ( ∂ ⇒ ( ψ ψ ) + c∇. ( ψ αψ ) = 0 ∂t

(

)(

− ce ψ † α.Aψ − ψα.Aψ † + βm0 c 2 + eφ ψ † ψ − ψψ †



†

)

(11.141)

†

which yields: ∂P + ∇.S = 0 (11.142) ∂t

2

2  E 5. Prove that ( α.p + βm0 c ) =   .  c

SOLUTION



(α.p + βm0 c )2 = (α 1 p1 + α 2 p2 + α 3 p3 + βm0 c )2 2 2 = ( α 1 p1 + α 2 p2 + α 3 p3 ) + (βm0 c ) + ( α 1 p1 + α 2 p2 + α 3 p ) β + β ( α 1 p1 + α 2 p2 + α 3 p ) = ( α 1 p1 + α 2 p2 + α 3 p3 ) + β 2 m02 c 2 + 2

(11.143)

3

∑(α β + βα ) p j

j

j

j=1

Since β 2 = 1 and α jβ + βα j = 0 , we get:

(α.p + βm0 c )2 = (α 1 p1 + α 2 p2 + α 3 p3 )2 + m02 c 2 = p 2 + (α 1α 2 + α 2 α 1 )p1 p2 + (α 2 α 3 + α 3 α 2 )p2 p3 + (α 3 α 1 + α 1α 3 )p3 p1 + m02 c 2



(11.144)

= p 2 + m02 c 2  E =   c

2

Here we used α 12 = α 22 = α 23 = 1 and α i α j + α j α i = 0 when i ≠ j. 6. Show that an alternative choice for matrices α j and β in the Dirac equation can be:  −σ j αj =   0

0 σj

  0  and β =    I

unit matrix, respectively.

I  , where σ j and I are Pauli matrices and the 0 

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Relativistic Quantum Mechanics

SOLUTION

The necessary conditions on α j and β matrices is that they should satisfy α 12 = α 22 = α 23 = 1 = β 2 ; α i α j + α j α i = 0 when i ≠ j, and α jβ + βα j = 0 . We see that:  −σ j α 2j =   0

  −σ j    0

0 σj

 0 value of j. Also, β 2 =   I

0 σj

  σ 2j  =   0 

0 σ 2j

  = 1 because σ 2j = 1 for any  

I  0 0   I

I   1 = 0   0

0  = 1. 1 

 0    I

I   0 + 0   I

I   −σ j  0   0

We next take:  −σ j α jβ + βα j =   0



0 σj

−σ j   0  + 0   −σ j

 0 =  σ j

0 σj

  

(11.145)

σj   =0 0 

Further:



 −σ j α jα i + α iα j =   0  σ jσi = 0 

  −σ i    0

0 σj 0

σ jσi

0   −σ i  + σi   0

  σiσ j  + 0  

0 σiσ j

0   −σ j  σ i   0

  σ jσi + σiσ j  = 0  

0 σj

   0

σ jσi + σiσ j

  

(11.146) Since σ j σ i + σ i σ j = 0 for i ≠ j, therefore α j α i + α i α j = 0. We thus find that the given values of α j and β are the correct alternative choice for the Dirac equation. 7. Show that Σ.p commutes with the free particle Dirac Hamiltonian H = cα.p + βm0 c 2 . SOLUTION

The commutator

[ Σ.p, H ] = Σ.pH − HΣ.p = [ Σ.p, cα.p ] + m0 c 2 [ Σ.p, β ]

Take:  σ.p

[ Σ.p, cα.p] = c 



0

 0  σ.p   σ.p 0

 0 σ.p   − c  σ.p 0 

(11.147)

σ.p   σ.p  0   0

  σ.p  0

 0  0 p2  p2   − c  =0 = c  p 2  p 2 0  0  (11.148a)

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and:  σ.p

[ Σ.p, β ] =  



 I  σ.p   0 0

0

 σ.p =  0

0   σ.p  − I   0

0   I − − I   0

  σ.p  − −σ.p   0

  =0 −σ.p 

0

0

  σ.p  0

(11.148b)

Therefore:

[ Σ.p, H ] = 0. 8. Apply the radial Eqns. (11.115) to the spherically symmetric infinite potential well defined by:  0 r ω(q) and it is negative (attractive) when ω < ω(q) . The positive value of v(q) causes resistivity and metals exist a normal conducting state. However, a phase transition from the normal conducting state to the superconductivity state occurs in a metal when v(q) becomes negative. Further details on this topic can be found in specialized textbooks on condensed matter physics.

12.5  Solved Examples  iε  1. Define C(r , t) =  0  E(r , t) and then evaluate commutator  A(r , t), C † (r′ , t)  .  2  SOLUTION

We have:

C(r , t) =

1 2 V

∑∑ k

α

ε0ω akα (t)εˆ α (k)e ik .r − ak†α (t)εˆ α (k)e − ik .r (12.88) 2

{

}

Then:  A ( r , t ) , C† ( r ′ , t ) 

= A (r , t ) C† (r ′, t ) − C† (r ′, t ) A (r , t ) =

1 4V

∑ ∑ ∑ ∑ εˆ {a α

k

{

α

k′

kα

α′

} {

(t)e ik.r + ak† α (t)e − ik.r εˆ α ′ ak† ′α ′ (t)e − ik ′.r ′ − ak ′α ′ (t)e ik ′.r ′

} {

}

− εˆ α ′ ak† ′α ′ (t)e − ik ′.r ′ − ak ′α ′ (t)e ik ′.r ′ εˆ α akα (t)e ik .r + ak†α (t)e − ik .r  1 = εˆ α εˆ α ′  akα , ak† ′α ′  e ik .r e − ik ′.r ′ − [ akα , ak ′α ′ ] e ik .r e ik ′.r ′ 4V k α k ′ α ′

∑∑∑∑

{

+  ak†α , ak† ′α ′  e − ik .r e − ik ′.r ′ −  ak†α , ak ′α ′  e − ik .r e ik ′.r ′

}

} (12.89)

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Quantization of Fields and Second Quantization

With the use of  ak†α , ak† ′α ′  = 0 = [ akα , ak ′α ′ ] ,  akα , ak† ′α ′  = δ k , k ′ δ α ,α ′ = −  ak† ′α ′ , akα  , we get:

∑∑∑∑εˆ εˆ

1  A ( r , t ) , C † ( r ′ , t )  = 4V

δ k , k ′ δ αα ′ e ik .r e − ik ′.r ′ + δ k ′ , k δ αα ′ e − ik .r e ik ′.r ′ 

α α′

α

k

∑ e

1 = 2V

k′

α′

ik .( r − r ′ )

k

(12.90)

+ e − ik .(r − r ′ ) 

= δ(r − r ′)

where we used

{b

k

,b

}

∑e

ik . x

k

= δ(x ) , δ(x ) = δ(− x ) and ( εˆ α ) = ( εˆ 1 ) + ( εˆ 2 ) = 2 . 2

2

2

d 3 k † − ik .r bk e , where (2 π)3 † † † = (2 π)3 δ(k − k ′) show that ψ (r ), ψ (r ′) = δ(r − r ′) and ψ (r )ψ (r ′) = 0 .

2. Consider † k′

1 V

the

Fermionic

field

operator

{

}

ψ † (r ) =

∫

SOLUTION

We have ψ (r ) =

∫

{ψ(r), ψ

†

d3 k bk e ik .r . Thus: (2 π)3

}

(r ′) = ψ (r )ψ † (r ′) + ψ † (r ′)ψ (r ) 1 (2 π)6 1 = (2 π)6 1 = (2 π)3 =



∫∫ d kd k ′ (b b e e + b b e ∫∫ d kd k ′(2π) δ(k − k′)e e ∫ e d k = δ(r − r ′) 3

3

3

3

† ik .r − ik ′ .r ′ k k′

† − ik ′ .r ′ ik .r k′ k

e

) (12.91)

ik .r − ik ′ .r ′

3

ik .( r − r ′ ) 3

and:

{ψ

†

}

(r ), ψ † (r ′) = ψ (r )† ψ † (r ′) + ψ † (r ′)ψ † (r ) 1 (2 π)6 1 = (2 π)6 =



(

)

∫∫ d kd k ′ (b b ∫∫ d kd k ′ (b b 3

3

† † − ik .r − ik ′ .r ′ k k′

3

3

† † k k′

e

e

+ bk† ′ bk† e − ik ′.r ′ e − ik .r

)

(12.92)

)

+ bk† ′ bk† e − ik .r e − ik ′.r ′

Since bk† bk† ′ + bk† ′ bk† = 0 , we get ψ (r )† ψ † (r ′) + ψ † (r ′)ψ † (r ) = 0 , which implies that ψ † (r )ψ † (r ′) = 0.

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A Textbook on Modern Quantum Mechanics

3. A quantized scalar field is given by: φ(r , t) =



∑c

  ak (t)e ik .r + ak† (t)e − ik .r  (12.93) 2ωV 

k

with: 2

ω  m c = k 2 +  0  and ak (t) = ak (0)e − iωt (12.94)    c



ak and ak† are Bosonic operators. Show that: 1 ∂φ ∂t

[φ(r , t), π(r , t)] = iδ(r − r ′) , where π = c 2



(12.95)

SOLUTION

Let us calculate: π( r , t ) =



1 ∂φ(r , t) i =− c 2 ∂t c

∑ k

ω  ak (t)e ik .r − ak† (t)e − ik .r  (12.96) 2V 

Then:

[φ(r , t), π(r ′, t)] = φ(r , t)π(r ′, t) − π(r ′, t)φ(r , t) =

∑ ∑  −2Vi  { a (t)e k

k



k′

ik .r

+ ak† (t)e − ik .r   ak ′ (t)e ik ′.r ′ − ak†′ (t)e − ik ′.r ′ 

−  ak ′ (t)e ik ′.r ′ − ak†′ (t)e − ik ′.r ′   ak (t)e ik .r + ak† (t)e − ik .r   − i  =  2V 

∑ ∑{[ a , a k

k′

k

+  a , ak ′  e † k

k′

− ik .r ik ′ .r ′

e

}

(12.97)

] e ik.r e ik′.r ′ −  ak , ak†′  e ik.r e − ik′.r ′

−  ak† , ak†′  e − ik .r e − ik ′.r ′

}

Since [ ak , ak ′ ] = 0 =  ak† , ak†′  and  ak , ak†′  = δ kk ′ = −  ak† , ak ′  , we get:

[φ(r , t), π(r ′, t)] =  2V  ∑∑ {δ k , k ′ e ik.r e − ik′.r ′ + δ k , k ′ e − ik.r e ik′.r ′ } i



k

 i  =  2V  Here we have used

1 V

∑e k

ik .( r − r ′ )

k′

∑(e

ik .( r − r ′ )

+e

− ik .( r − r ′ )

) = iδ(r − r ′)

k

= δ(r − r ′) and δ(r − r ′) = δ(r ′ − r ) .

(12.98)

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Quantization of Fields and Second Quantization

4. Consider the Bosonic field operator ψ (r ) =

∫

d3 p a e ip.r , where (2 π)3 p

 ap , ap† ′  = ap ap† ′ − ap† ′ ap = (2 π)3 δ(p − p′) (12.99)



Show that  ψ (r ), ψ † (r ′)  = δ(r − r ′) and  ψ † (r ), ψ † (r ′)  = 0 . SOLUTION

The conjugate adjoint field operator is: ψ † (r ) =



∫

d 3 p′ † − ip′.r ap e (12.100) (2 π)3 ′

Then:  ψ (r ), ψ † (r ′)  = ψ (r )ψ † (r ′) − ψ (r )ψ † (r ′) 1 d 3 pd 3 p ′ ap ap† ′ − ap† ′ ap e i(p.r − p′.r ′ ) = 6 2 π ( ) p p′

∑∑ ∫∫



=

1 ( 2π )6

1 = ( 2 π )3

(

)

∑∑ ∫∫ d pd p′ ( 2π ) δ 3

p

3

3

p , p′

e i ( p. r − p ′ . r ′ )

(12.101)

p′

∑d pe 3

ip.( r − r ′ )

p

= δ(r − r ′)

Here we used  ap , ap′  = (2 π)3 δ(p − p′) . Next:



 ψ † (r ), ψ † (r ′)  = ψ (r )† ψ † (r ′) − ψ † (r )ψ † (r ′) (12.102a) 1 d 3 pd 3 p ′ ap† ap† ′ − ap† ′ ap† e i(p.r − p′.r ′ ) = 6 ( 2 π ) p p′

∑∑ ∫∫

(

(

)

)

Since  ap† , ap† ′  = ap† ap† ′ − ap† ′ ap† = 0 , we have:

 ψ † (r ), ψ † (r ′)  = 0 (12.102b) ∂ 5. Consider the case of a 1D harmonic oscillator and prove that  a, f ( a † )  = † f ( a † ) ∂a † † † † ∂ and  a † , f ( a)  = − f ( a) . Show that  a, e αa  = αe αa and e −αa αe −αa = α + a.   ∂a

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SOLUTION

For a 1D harmonic oscillator, a and a † are defined in terms of the position operator x and the momentum operator p as follows: a=



mωx + ip mωx − ip and a † = (12.103) 2 mω 2 mω

Therefore:  a, f ( a † )  ψ = af ( a † )ψ − f ( a † )aψ



∂   † † †  mωxf ( a )ψ +  ∂ x f ( a )ψ  − f ( a )aψ

=

1 2 mω

=

  ∂ f (a† ) 1 † † ∂ †  mωxf ( a )ψ +  ∂ x ψ + f ( a ) ∂ x ψ  − f ( a )aψ 2 mω  

=

f (a† ) ∂ f (a† )  mωx + ip ) ψ + ψ − f ( a † )aψ ( 2 mω 2 mω ∂ x

=

∂ f (a † )  ψ= 2 mω ∂ x

=

{

}

(12.104a)

∂ f (a † ) ∂a †  ψ 2 mω ∂ a † ∂ x

∂ f (a † ) ψ ∂a†

which implies: ∂  a, f ( a † )  = f ( a † ) (12.104b) ∂a

Similarly, we get:

∂  a † , f ( a)  = − f ( a) (12.104c) ∂a

Hence:

( )

 a, e αa†  = ∂ e αa† = αe αa†   ∂a†



†

†

†

⇒ ae αa − e αa a = αe αa . †

On operating with e −αa , we obtain: †



†

†

†

†

e −αa ae αa − e −αa e αa a = e −αa αe αa †

†

⇒ e −αa ae αa = a + α

†

(12.105)

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Quantization of Fields and Second Quantization

6. The number operator is defined as N = ∫ d 3 rψ † (r )ψ (r ) , where the Bosonic field operators satisfy  ψ (r ), ψ † (r ′)  = δ(r − r ′) , [ ψ (r ), ψ (r ′)] = 0 and  ψ † (r ), ψ † (r ′)  = 0 . Show that Nψ † = ψ † ( N + 1) and Nψ = ψ ( N − 1) . SOLUTION

Let us take:

[ N , ψ(r)] = Nψ(r) − ψ(r)N

∫ ψ (r ′)ψ(r ′)d r ′ψ(r) − ψ(r)∫ ψ (r ′)ψ(r ′)d r ′ = ψ (r ′)ψ (r ′)ψ (r )d r ′ − ψ (r )ψ (r ′)ψ (r ′)d r ′ ∫ ∫ =



†

3

†

3

†

3

†

3

(12.106a)

[ ψ(r), ψ(r ′)] = 0 implies ψ(r ′)ψ(r) = ψ(r)ψ(r ′) and hence: [ N , ψ(r)] = ∫ ψ † (r ′)ψ(r)ψ(r ′)d3 r ′ − ∫ ψ(r)ψ † (r ′)ψ(r ′)d3 r ′

∫ ψ (r ′)ψ(r) − ψ(r)ψ (r ′) ψ(r ′)d r ′ = − δ(r ′ − r )ψ (r ′)d r ′ ∫ =



†

†

3

(12.106b)

3

= −ψ (r )

which means Nψ (r ) = ψ (r )( N − 1) . We next take:  N , ψ † (r )  = Nψ † (r ) − ψ † (r )N

∫ ψ (r ′)ψ(r ′)d r ′ψ (r) − ψ (r)∫ ψ (r ′)ψ(r ′)d r ′ = ψ (r ′)ψ (r ′)ψ (r )d r ′ − ψ (r )ψ (r ′)ψ (r ′)d r ′ ∫ ∫ = ψ (r ′) {ψ (r )ψ (r ′) + δ(r ′ − r )} d r ′ − ψ (r )ψ (r ′)ψ (r ′)d r ′ ∫ ∫ = ψ (r ′)δ(r ′ − r )d r ′ ∫ =

†

3

†



†

†

†

†

†

3

†

†

†

3

†

3

3

†

†

3

(12.107)

3

= ψ † (r ) which implies that Nψ † (r ) = ψ † (r )( N + 1) . This means that N decreases (increases) by one on interchanging its order with ψ(r ) ψ † (r ) .

(

)

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A Textbook on Modern Quantum Mechanics

12.6 Exercises

( )

1. Show that for a system of Bosons (a) ai ai† i ≠ j . N i and N j are number operators.

2

(

)

= ai† 2 + ai† ai and (b)  N i , N j  = 0 for

∑

2. For a system of Bosons, the interaction Hamiltonian is H 1 = dijkl ai† a †j ak al . Find the commutator [ H 1 , ai ]. i , j , k ,l 3. For a system of independent harmonic oscillators, the Hamiltonian is given by H=

∑ε α α . Determine the equation of motion for the annihilation and crei

† i

i

i

ation operators for both the cases of Bosons and Fermions, with the use of the Heisenberg equation. 4. Show that the Bosonic number operator N = Hamiltonian H =

∑a t

† j jk k

j,k

a +

∑a a v † † j k

∑a a

† i i

commutes with the

i

a a.

jklm m l

j , k ,l , m

5. Consider a system of non-interacting two particles which can be Bosons or Fermions. The state and the Hamiltonian are defined by α , β = aα† aβ† 0 and H=

∑ε a a

† α α α

for both the cases. Find the energy Eαβ of the system.

α

6. For a system of Fermions, the matrix representation of the annihilation operator bl , the creation operator bl† and the state 1l are given as follows:

 0 bl = (−1)vα   0

 0 1  † , bl = (−1)vα   0   1

 0  0  and 1l =   . 0   1 

Calculate N l = bl† bl and show that bl bl† + bl† bl = 1 and N l2 1l = N l 1l .

∑

7. For a system of Bosons, N n1 , n2 , n3 ...... = N i n1 , n2 , n3 ...... = n n1 , n2 , n3 ...... . Show that: i a. Na † n1 , n2 , n3 ...... = (n + 1)a † n1 , n2 , n3 ...... , and b. Na n1 , n2 , n3 ...... = (n − 1)a n1 , n2 , n3 ...... . 8. A particle of mass m with charge q moves under the influence of a uniform 2 1 magnetic field B = B0 kˆ . The Hamiltonian of the particle is H = p − qA ) , ( 2m B where A = 0 − yiˆ + xj . Show that the Hamiltonian can be expressed as: 2 pz2  † 1 H= +  a a +  ω , with aa † − a † a = 1 . What are the values of a and ω ? 2m  2

(

)

Annexure A: Useful Formulae

I. Table of Integrals ∞

∫

1. x n e − ax dx = 0

n! , n ≥ 0 and a > 0. a n+ 1

∞

∫

2

2. x 2 n+ 1 e − ax dx = 0

∞

∫

2

3. x 2 n e − ax dx = 0

n! , n ≥ 0 and a > 0. 2 a n+ 1

(2 n − 1)!! π , where (2l + 1)!! = 1.3.5.....(2l + 1), 0!! = 1, and (−1)!! = 1, 2 n+ 1 a n a

a > 0. ∞

4.

∫ 0

∞

5.

∫

−∞

e − ax dx = x

π . a

e − iqx π − aq dx = e , a ≠ 0 , q is real. a + x2 a 2

6. Integration by parts: b

b

∫FdG = [ FG] − ∫GdF, where both F and G are function of x and integration is done b a

a

a

from x = a to x = b.

II. Series and Expansions 1. Taylor Series: ∞

f ( x) =



∑ (x −nx! ) 0

n= 0

n

d n f ( x) . dx n x = x0

2. Exponential Series: ∞



ex =

∑ k=0

xk . k! 409

410

A Textbook on Modern Quantum Mechanics

3. Geometric Series: ∞

(1 − x)−1 =



∑x

k

k=0

4. Trigonometric series: ∞

sin x =



∑

(−1)k

k=0

x2 k +1 , (2 k + 1)!

∞

cos x =



∑

(−1)k

k=0

tan x = x +



x2 k  , ( 2 k )!

1 3 2 5 17 7 x + x + x + ....... 3 15 315

5. Binomial Expansion: n



( a + x )n =

∑ k !(nn−! k)! x a

k n− k

,  n > 0

k=0

q



q

(1 + x) =

∑ k !(q − k)! x ,  q > 0. k=0

III. Basic Functional Relations 1. e ix = cos x + i sin x. 2. e x = cosh x + sinh x. 3. sin( x ± y ) = sin x cos y ± sin y cos x. 4. cos( x ± y ) = cos x cos y  sin x sin y. tan x ± tan y 5. tan( x ± y ) = . 1  tan x tan y

q!

k

411

Annexure A: Useful Formulae

IV. Coordinate Systems 1. V = Vx xˆ + Vy yˆ + Vz zˆ in Cartesian coordinate system. 2. V = Vr rˆ + Vθ θˆ + Vφ φˆ in spherical polar coordinate system. ( xˆ , yˆ , zˆ ) and rˆ , θˆ , φˆ are sets of orthonormal vectors in Cartesian and spherical polar

(

)

coordinate systems, respectively.

3. x = r sin θ cos φ y = r sin θ sin φ z = r cos θ   4.   

xˆ   sin θ cos φ   yˆ  =  sin θ sin φ cos θ zˆ  

cos θ cos φ cos θ sin φ − sin θ

− sin φ cos φ 0

     

rˆ   θˆ  .  φˆ  

5. Volume element, dxdydz = r 2 dr sin θ dθ dφ , with −∞ ≤ ( x , y , z) ≤ ∞ and 0 ≤ r ≤ ∞ , 0 ≤ θ ≤ π , 0 ≤ φ ≤ 2 π . 6. The gradient operator: ∇ = xˆ = rˆ

∂ ˆ ∂ ˆ ∂ +y +z ∂x ∂y ∂z ∂ θˆ ∂ φˆ ∂ + + . ∂r r ∂r r sin θ ∂φ

7. The Laplacian operator: ∇2 = =

∂2 ∂2 ∂2 + 2 + 2 2 ∂x ∂y ∂z 1 ∂  2 ∂ 1 1 ∂2 ∂  ∂ sin + . + θ r     ∂θ  r 2 sin 2 θ ∂φ2 r 2 ∂r  ∂r  r 2 sin θ ∂θ 

Annexure B: Dirac Delta Function The Dirac delta function does not fit into the usual definition of a function that has value at each and every point within its domain. It is a function on the real line, which has infinite value at origin and zero value everywhere else. It is defined by:  ∞ δ( x) =   0



x=0 (B.1) x≠0

when origin is taken at x = 0 and by:  ∞ δ( x − a) =   0



x=a (B.2) x≠a

for change of origin to x = a. Dirac delta function is constrained to satisfy: ∞

∫δ(x) dx = 1



∞

∫δ(x − a) dx = 1 (B.3)

and

−∞

−∞

Another defining characteristic of delta function is:

∞

∞

−∞

−∞

∫ f (x)δ(x)dx = f (0),  and ∫ f (x)δ(x − a)dx = f (a) (B.4)

where f ( x) is a well-defined function both at x = 0 and x = a. The range of integration in Eqn. (B4) need not be from –∞ to ∞, it can be over any domain surrounding the critical point at which the delta function is not zero. Generalization of definition of Dirac delta function to three-dimensional space is made as follows:

δ ( r − a ) = δ( x − ax ) δ( y − ay ) δ( z − az ) (B.5)

with the constrain:

∫ f (r)δ(r − a) dr = f (a) (B.6) (

)

where ranges of integration include ax , ay , az .

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414

A Textbook on Modern Quantum Mechanics

I. Properties of Delta Function 1.   xδ( x) = 0 2.   xδ( x − a) = aδ( x − a) 3.   δ( x) = δ(− x) 4.   f ( x)δ( x − a) = f ( a)δ( x − a) 5.   δ( xa) =

1 δ( x), a > 0 a

6.   δ( x 2 − a 2 ) =

(B.7)

1 [δ(x − a) + δ(x + a)] 2a n

7.   δ  f ( x)  =

∑ δ(fx′(−xx) ) , with f ′(x ) = dx i

i=1

df

i

i

and f ( xi ) = 0 x = xi

∫

8.  δ( x − a)δ( x − b) dx = δ(b − a) 

II. Representation of Delta Function Delta function is viewed in various forms. Some of forms are: 1. Limiting value of Rectangle function:  1  Rσ ( x) =  2 σ  0 



−σ ≤ ( x − a) ≤ σ

(B.8a)

−σ > ( x − a) > σ

δ( x − a) = lim Rσ ( x) (B.8b)



σ→ 0

2. Limiting value of Gaussian function: 2 2 1 e − ( x − a ) /2 σ , σ > 0 (B.9) σ→ 0 σ 2 π

δ( x − a) = lim



3. Limiting value of Lorentzian function: δ( x) = lim



ε→ 0

1 ε (B.10) π x2 + ε2

(

)

4. Fourier series: ∞



δ( x) =

∑

1 e inx  on the interval of [ −π , π ] (B.11) 2 π n=−∞

415

Annexure B: Dirac Delta Function

5. Fourier transformation: ∞

δ( x) =



1 e ikx dk (B.12a) 2π

∫

−∞

A three dimensional generalization of which is: δ(r ) =



1 (2 π)3

∫e

ik .r

d 3 k (B.12b)

6. Differentiation of Heaviside Step function:  0 θ( x − a) =   1



x