Problems and Solutions on Quantum Mechanics (Major American Universities Ph.D. Qualifying Questions and Solutions - Physics) [2 ed.] 9811256063, 9789811256066

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Major American Universities Ph.D. Qualifying Questions and Solutions - Physics

Second Edition

Problems Solutions QUANTUM MECHANICS and

on

Editors

Swee Cheng Lim Choy Heng Lai Leong Chuan Kwek

World Scientific

Second Edition

Problems Solutions and

on

QUANTUM MECHANICS

Major American Universities Ph.D. Qualifying Questions and Solutions - Physics ISSN: 1793-1487 Published Problems and Solutions on Quantum Mechanics Second Edition edited by Swee Cheng Lim, Choy Heng Lai and Leong Chuan Kwek (NUS, Singapore) Problems and Solutions on Thermodynamics and Statistical Mechanics Second Edition edited by Swee Cheng Lim, Choy Heng Lai and Leong Chuan Kwek (NUS, Singapore) Problems and Solutions on Mechanics (Second Edition) edited by Swee Cheng Lim, Choy Heng Lai and Leong Chuan Kwek (NUS, Singapore) Problems and Solutions on Thermodynamics and Statistical Mechanics edited by Yung-Kuo Lim (NUS, Singapore) Problems and Solutions on Optics edited by Yung-Kuo Lim (NUS, Singapore) Problems and Solutions on Electromagnetism edited by Yung-Kuo Lim (NUS, Singapore) Problems and Solutions on Mechanics edited by Yung-Kuo Lim (NUS, Singapore) Problems and Solutions on Solid State Physics, Relativity and Miscellaneous Topics edited by Yung-Kuo Lim (NUS, Singapore) Problems and Solutions on Quantum Mechanics edited by Yung-Kuo Lim (NUS, Singapore) Problems and Solutions on Atomic, Nuclear and Particle Physics edited by Yung-Kuo Lim (NUS, Singapore) More information on this series can also be found at https://www.worldscientific.com/series/mauqqsp

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Major American Universities Ph.D. Qualifying Questions and Solutions - Physics

Second Edition

Problems Solutions and

on

QUANTUM MECHANICS Editors

Swee Cheng Lim Choy Heng Lai National University of Singapore, Singapore

Leong Chuan Kwek CQT, National University of Singapore, and NIE, Quantum Science and Engineering Center, NTU, Singapore

World Scientific NEW JERSEY

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LONDON

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SINGAPORE

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BEIJING

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SHANGHAI

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HONG KONG

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TAIPEI

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CHENNAI

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TOKYO

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Control Number: 2022937746 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. Some new problems created by B. Rajesh, American College, Madurai. Major American Universities Ph.D. Qualifying Questions and Solutions - Physics PROBLEMS AND SOLUTIONS ON QUANTUM MECHANICS Second Edition Copyright © 2022 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN ISBN ISBN ISBN

978-981-125-606-6 (hardcover) 978-981-125-734-6 (paperback) 978-981-125-607-3 (ebook for institutions) 978-981-125-608-0 (ebook for individuals)

For any available supplementary material, please visit https://www.worldscientific.com/worldscibooks/10.1142/12829#t=suppl Desk Editor: Joseph Ang Typeset by Diacritech Technologies Pvt. Ltd. Chennai - 600106, India Printed in Singapore

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PREFACE This is the second edition of a former popular series on Problems and Solutions in various topics of physics ranging from Mechanics, Electromagnetism, Optics, Atomic, Nuclear and Particle Physics, Thermodynamics and Statistical Mechanics, Quantum Mechanics, to Solid State Physics, Relativity and Miscellaneous topics. We have greatly expanded the volumes. Each volume is divided into several subtopics, and there are new interesting problems and solutions. Altogether, we have compiled several thousand problems and solutions, spanning an entire undergraduate physics course. We have also included different genres of problems and questions: qualitative, quantitative, fill-in-the-blanks, and even multiple-choice. These questions are carefully chosen at the level of the PhD Qualifying Examinations at American universities. Moreover, these questions can also serve as an excellent resource for Physics Competitions, like the International Physics Olympiad or the regional Physics Olympiads. We believe that a good grounding in problem-solving is essential for the study of physics, and we believe that this compendium serves as an invaluable resource for the preparation of graduate study in physics, or competitions. We suggest that the student using the books should first attempt the problems before consulting the solutions. We have tried to elucidate solutions with figures and diagrams that illustrate how to set up the problems. Lim Swee Cheng, Lai Choy Heng and Kwek Leong Chuan Editors

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CONTENTS Preface

v

Part I Basic Principles and One-Dimensional Motions (1001–1072)

1

Part II Central Potentials (2001–2023)

111

Part III Spin and Angular Momentum (3001–3048)

155

Part IV Motion in Electromagnetic Fields (4001–4016)

237

Part V Perturbation Theory (5001–5083)

273

Part VI Scattering Theory & Quantum Transitions (6001–6061)

429

Part VII Many-Particle Systems (7001–7037)

553

Part VIII Miscellaneous Topics (8001–8040)

615

Index to Problems

689

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Part I

Basic Principles and OneDimensional Motions

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1 1001 Quantum phenomena are often negligible in the “macroscopic” world. Show this numerically for the following cases: a. The amplitude of the zero-point oscillation for a pendulum of length l = 1 m and mass m = 1 kg. b. The tunneling probability for a marble of mass m = 5 g moving at a speed of 10 cm / sec against a rigid obstacle of height H = 5 cm and width w = 1 cm. c. The diffraction of a tennis ball of mass m = 0.1 kg moving at a speed

υ = 0.5 m / sec by a window of size 1× 1.5 m2 . (Wisconsin) Sol: a. The theory of the harmonic oscillator gives the average kinetic energy as V = 12 E , i.e., 12 mω 2 A2 = 14 ω , where ω = g / l and A is the root-mean-square amplitude of the zero-point oscillation. Hence A=

 ≈ 0.41 × 10−17 m. 2mω

Thus the zero-point oscillation of a macroscopic pendulum is negligible. b. If we regard the width and height of the rigid obstacle as the width and height of a gravity potential barrier, the tunneling probability is  2w   1 T ≈ exp− 2m mgH − mυ 2        2  2mw  = exp− 2 gH − υ 2  ,   

3

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where

2mw 2 gH − υ 2 ≈ 0.9 × 1030. 

Hence 30

T ≈ e −0.9×10 ≈ 0 That is, the tunneling probability for the marble is essentially zero. c. The de Broglie wavelength of the tennis ball is

λ = h/ p = h / mυ = 1.3 × 10−30 cm, and the diffraction angles in the horizontal and the vertical directions are respectively

θ1 ≈ λ/D = 1.3 × 10−32 rad , θ 2 ≈ λ/L = 9 × 10−33 rad. Thus there is no diffraction in any direction.

1002 Express each of the following quantities in terms of  , e , c , m = electron mass, M = proton mass. Also give a rough estimate of numerical size for each. a. Bohr radius (cm). b. Binding energy of hydrogen (eV). c. Bohr magneton (choosing your own unit). d. Compton wavelength of an electron (cm). e. Classical electron radius (cm). f. Electron rest energy (MeV). g. Proton rest energy (MeV). h. Fine structure constant. i. Typical hydrogen fine-structure splitting (eV). (Berkeley) Sol: a. a =  2 /me 2 = 5.29 × 10−9 cm. b. E = me 4 /2 2 = 13.6 eV. c. µ B = e/2mc = 9.27 × 10−21 erg ⋅ Gs−1 . d. λ = 2π /mc = 2.43 × 10−10 cm.

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e. re = e 2 /mc 2 = 2.82 × 10−13 cm. f. Ee = mc 2 = 0.511 MeV. g. E p = Mc 2 = 938 MeV. h. α = e 2 /c = 7.30 × 10−3 ≈ 1/137. i. ∆E = e 8mc 2 /8 2c 4 = 18 α 4mc 2 = 1.8 × 10−4 eV.

1003 Derive, estimate, guess or remember numerical values for the following, to within one order of magnitude: a. The electron Compton wavelength. b. The electron Thomson cross section. c. The Bohr radius of hydrogen. d. The ionization potential for atomic hydrogen. e. The hyperfine splitting of the ground-state energy level in atomic hydrogen. f. The magnetic dipole moment of 3 Li7 ( Z = 3) nucleus. g. The proton-neutron mass difference. h. The lifetime of free neutron. i. The binding energy of a helium-4 nucleus. j. The radius of the largest stable nucleus. k. The lifetime of a π 0 meason. l. The lifetime of a µ − meason. (Berkeley) Sol: a. λe = h /me c = 2.43 × 10−2 Å. b. σ = 83π re2 = 6.56 × 10−31 m 2. c. a =  2 = 0.53 Å. me 2

e

d. I = 2e a = 13.6 eV. 2

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e. The splitting of the ground-state energy level is 2

 1  ∆E f = 13.6 ×   ≈ 10−4 eV.  137  The hyperfine splitting of the ground-state energy level is ∆Ehf ≈ ∆E f /103 ≈ 10−7 eV. −26 −1 f. µ = 1.67 × 10 J ⋅ T .

g. ∆m = m p − mn = −2.3 × 10−30 kg. 2 h. τ n ≈ 15 min = 9 × 10 s.

i. E = 4 × 7 MeV = 28 MeV. j. The radius r corresponds to a region of space in which nuclear force is effective. Thus 1

1

r ≈ 1.4 A 3 = 1.4 × (100)3 = 6.5 fm. k. τ = 8.28 × 10−17 s. l. The decay of µ − is by weak interaction, and so τ = 2.2 × 10−6 s.

1004 Explain what was learned about quantization of radiation or mechanical system from two of the following experiments: a. Photoelectric effect. b. Black body radiation spectrum. c. Franck–Hertz experiment. d. Davisson–Germer experiment. e. Compton scattering. Describe the experiments selected in detail, indicate which of the measured effects were non-classical and why, and explain how they can be understood as quantum phenomena. Give equations if appropriate. (Wisconsin) Sol: a. Photoelectric Effect This refers to the emission of electrons observed when one irradiates a metal under vacuum with ultraviolet light. It was found that the magnitude of the

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electric current thus produced is proportional to the intensity of the striking radiation provided that the frequency of the light is greater than a minimum value characteristic of the metal, while the speed of the electrons does not depend on the light intensity, but on its frequency. Einstein in 1905 explained these results by assuming light, in its interaction with matter, consisted of corpuscles of energy hv, called photons. When a photon encounters an electron of the metal it is entirely absorbed, and the electon, after receiving the energy hv, spends an amount of work W equal to its binding energy in the metal, and leaves with a kinetic energy 1 mυ 2 = hν − W . 2 This quantitative theory of photoelectricity has been completely verified by experiment. b. Black Body Radiation A black body is one which absorbs all the radiation falling on it. The spectral distribution of the radiation emitted by a black body can be derived from the general laws of interaction between matter and radiation. The expressions deduced from the classical theory are known as Wien’s law and Rayleigh’s law. The former is in good agreement with experiment in the short wavelength end of the spectrum only, while the latter is in good agreement with the long wavelength results but leads to divergency in total energy. Planck in 1900 succeeded in removing the difficulties encountered by classical physics in black body radiation by postulating that energy exchanges between matter and radiation do not take place in a continuous manner but by discrete and indivisible quantities, or quanta, of energy. He showed that by assuming that the quantum of energy was proportional to the frequency, ε = hv , he was able to obtain an expression for the spectrum which is in complete agreement with experiment: Ev =

8π hν 3 1 , hν c 3 e kT − 1

where h is a universal constant, now known as Planck’s constant. Planck’s hypothesis has been confirmed by a whole array of elementary processes and it directly reveals the existence of discontinuities of physical processes on the microscopic scale, namely quantum phenomena. c. Franck-Hertz Experiment The experiment of Franck and Hertz consisted of bombarding atoms with monoenergetic electrons and measuring the kinetic energy of the scattered

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electrons, from which one deduced by subtraction the quantity of energy absorbed in the collisions by the atoms. Suppose E0 , E1 , E2 , … are the sequence of quantized energy levels of the atoms and T is the kinetic energy of the incident electrons. As long as T is below ∆ = E1 − E0 , the atoms cannot absorb the energy and all collisions are elastic. As soon as T > E1 − E0 , inelastic collisions occur and some atoms go into their first excited states. Similarly, atoms can be excited into the second excited state as soon as T > E2 − E0 , etc. This was exactly what was found experimentally. Thus the Franck-Hertz experiment established the quantization of atomic energy levels. d. Davisson-Germer Experiment L. de Broglie, seeking to establish the basis of a unified theory of matter and radiation, postulated that matter, as well as light, exhibited both wave and corpuscular aspects. The first diffraction experiments with matter waves were performed with electrons by Davisson and Germer (1927). The incident beam was obtained by accelerating electrons through an electrical potential. Knowing the parameters of the crystal lattice it was possible to deduce an experimental value for the electron wavelength and the results were in perfect accord with the de Broglie relation λ = h/ p , where h is Planck’s constant and p is the momentum of the electrons. Similar experiments were later performed by others with beams of helium atoms and hydrogen molecules, showing that the wavelike structure was not peculiar to electrons. e. Compton Scattering Compton observed the scattering of X-rays by free (or weakly bound) electrons and found the wavelength of the scattered radiation exceeded that of the incident radiation. The difference ∆λ varied as a function of the angle θ between the incident and scattered directions:

∆λ = 2

h θ sin 2 , mc 2

where h is Planck’s constant and m is the rest mass of the electron. Furthermore, ∆λ is independent of the incident wavelength. The Compton effect cannot be explained by any classical wave theory of light and is therefore a confirmation of the photon theory of light.

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1005 In the days before Quantum Mechanics, a big theoretical problem was to “stop” an atom from emitting light. Explain. After Quantum Mechanics, a big theoretical problem was to make atoms in excited states emit light. Explain. What does make excited atoms emit light? (Wisconsin) Sol:

In the days before Quantum Mechanics, according to the Rutherford atomic model electrons move around the nucleus in elliptical orbits. Classical electrodynamics requires radiation to be emitted when a charged particle accelerates. Thus the atom must emit light. This means that the electrons would lose energy continuously and ultimately be captured by the nucleus. Whereas, in actual fact the electrons do not fall towards the nucleus and atoms in ground state are stable and do not emit light. The problem then was to invent a mechanism which could prevent the atom from emitting light. All such attempts ended in failure. A basic principle of Quantum Mechanics is that, without external interaction, the Hamiltonian of an atom is time-independent. This means that an atom in an excited state (still a stationary state) would stay on and not emit light spontaneously. In reality, however, spontaneous transition of an excited atoms does occur and light is emitted. According to Quantum Electrodynamics, the interaction of the radiation field and the electrons in an atom, which form two quantum systems, contains a term of the single-photon creation operator a+, which does not vanish even if there is no photon initially. It is this term that makes atoms in excited states emit light, causing spontaneous transition.

1006 Consider an experiment in which a beam of electrons is directed at a plate containing two slits, labelled A and B. Beyond the plate is a screen equipped with an array of detectors which enables one to determine where the electrons hit the screen. For each of the following cases draw a rough graph of the relative number of incident electrons as a function of position along the screen and give a brief explanation. a. Slit A open, slit B closed. b. Slit B open, slit A closed. c. Both slits open.

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d. “Stern–Gerlach” apparatus attached to the slits in such a manner that only electrons with s z = /2 can pass through A and only electrons with s z = −/2 can pass through B. e. Only electrons with s z = /2 can pass through A and only electrons with s z = /2 can pass through B. What is the effect of making the beam intensity so low that only one electron is passing through the apparatus at any time? (Columbia) Sol: a. The probability detected at the screen is that of the electrons passing through slit A:

Fig. 1.1 I1 = I A ( x ). b. The probability detected at the screen is that of the electrons passing through slit B: I 2 = I B ( x ).

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c. I c = I12 ( x ) = I1 + I 2 + interference term ≠ I1 + I 2 . d. The eigenstate of the electrons passing through slit A is different from that of the electrons passing through slit B, and so there is no interference term. The intensity on the screen is just the sum of the intensities of the single-slit cases: I d = I1 + I 2. e. Similar to (c), but the intensity is half that in (c): I e = I c /2. Because of the self-interference of the wave functions of the electrons, the answers above remain valid even when the incident electron beam intensity is so low that only one electron passes through at a time.

1007 A particle of mass m is subjected to a force F(r ) = −∇V (r ) such that the wave function ϕ (p, t ) satisfies the momentum-space Schrödinger equation (p2 /2m − a∇ 2p )ϕ (p, t ) = i ∂ϕ (p, t )/∂t , where  =1, a is some real constant and ∇ 2p = ∂2 / ∂ px2 +∂2 / ∂ py2 +∂2 / ∂ pz2. Find the force F(r). (Wisconsin) Sol:

The coordinate and momentum representations of a wave function are related by 3

 1 2 ψ (r , t ) =   ∫ ϕ (k , t )e ik ⋅r dk ,  2π  3

 1 2 ϕ (k , t ) =   ∫ψ (r , t )e − ik ⋅r dr ,  2π  where k = p . Thus (with  = 1) P 2ϕ (P, t ) → −∇ 2ψ (r , t ), ∇ 2pϕ (p , t ) → −r 2ψ (r , t ), and the Schrödinger equation becomes, in coordinate space,  −∇ 2 ∂ϕ (r , t ) 2  2m + ar  ϕ (r , t ) = i ∂t .

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Hence the potential is V (r ) = ar 2, and the force is F (r ) = −∇V (r ) = −

r d V (r ) = −2ar. r dr

1008 Consider the one-dimensional time-independent Schrödinger equation for some arbitrary potential V ( x ). Prove that if a solution ψ ( x ) has the property that ψ ( x ) → 0 as x →±∞ , then the solution must be nondegenerate and therefore real, apart from a possible overall phase factor. Hint: Show that the contrary assumption leads to a contradiction. (Berkeley) Sol:

Suppose that there exists another function φ ( x ) which satisfies the same Schrödinger equation with the same energy E as ψ and is such that lim x→∞ φ ( x ) = 0. Then

ψ ′′ / ψ = −2m( E − V ) /  2 , φ ″ / φ = −2m( E − V ) /  2 ,

and hence

ψ ″φ − φ ″ψ = 0, or

ψ ′φ − φ ′ψ = constant . The boundary conditions at x → ∞ then give

ψ ′φ − φ ′ψ = 0, or

ψ′ φ ′ = . ψ φ

Integrating we have ln ψ = ln φ + constant , or ψ = constant × φ . Therefore, ψ and

φ represent the same state according to a statistical interpretation of wave function. That is, the solution is nondegenerate. When V ( x ) is a real function, ψ * and ψ satisfy the same equation with the same energy and the same boundary condition lim x→∞ ψ * = 0. Hence ψ * = cψ , or

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2

ψ = c * ψ * , from which we have c = 1 , or c = exp(iδ ), where δ is a real number. If we choose δ = 0 , then c = 1 and ψ is a real function.

1009 Consider a one-dimensional bound particle. a. Show that

d dt

∫

∞ −∞

ψ *( x , t ) ψ ( x , t ) dx = 0.

(ψ need not be a stationary state). b. Show that, if the particle is in a stationary state at a given time, then it will always remain in a stationary state. c. If at t = 0 the wave function is constant in the region −a < x < a and zero elsewhere, express the complete wave function at a subsequent time in terms of the eigenstates of the system. (Wisconsin) Sol: a. Consider the Schrödinger equation and its complex conjugate

i ∂ψ / ∂t = −



−i ∂ψ * / ∂t = −

2 2 ∇ ψ + Vψ , 2m

(1)

2 2 ∇ ψ * +V ψ *. (2) 2m

Taking ψ * × (1) − ψ × (2) we obtain i

∂ 2 (ψ * ψ ) = − ∇ ⋅ (ψ * ∇ψ − ψ∇ψ *). ∂t 2m

For the one dimension case we have, integrating over all space, d ∞ i ∞ ∂  ∂ψ ∂ψ *  ψ * ( x , t )ψ ( x , t ) dx = −ψ ψ *  dx ∫ ∫ −∞ −∞ 2m dt ∂x  ∂x ∂x  i = [ψ * ∂ψ / ∂x − ψ ∂ψ * / ∂x ]∞−∞ . 2m If ψ is a bound state, then ψ ( x → ±∞) = 0 and hence1 d dt 1 

∫

∞ −∞

ψ * ( x , t )ψ ( x , t ) dx = 0.

The integral must be unity due to normalization of the wave function. Clearly the derivation is zero.

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b. Supposing the particle is in a stationary state with energy E at t = t0 , we have Hˆ ψ ( x , t0 ) = Eψ ( x , t0 ), where Hˆ does not depend on t explicitly. At any later time t, the Schrödinger equation i ∂ψ ( x , t )/∂t = Hˆ ψ ( x , t ) applies. As Hˆ does not depend on t explicitly, the Schrödinger equation has the formal solution

ψ ( x , t ) = exp[−iHˆ (t − t0 ) / ] ψ ( x , t0 ). Multiplying both sides by Hˆ from the left and noting the commutability between Hˆ and exp[−i(t − t0 )Hˆ / ] , we find  −iHˆ (t − t )  0 Hˆ ψ ( x , t ) = exp   Hˆ ψ ( x , t0 )     −iHˆ (t − t )  0 = E exp   Hˆ ψ ( x , t0 )    = E ψ ( x , t ). Hence ψ ( x , t ) represents a stationary state at any later time t. c. The wave function given for t = 0 can be written as C , ψ ( x , 0) =  0, where C is a constant. Normalization ∫

x < a, otherwise, a −a

ψ * ψ dx = 1 requires that C = ( 21a ) 2 . 1

Suppose the eigenfunction of the bound state is x n and Hˆ n = En n . Then

1=∑ n n , n

and | ψ ( x , 0) = ∑ | n n | ψ ( x , 0) , n

 E  | ψ ( x , t ) = ∑ | n n | ψ ( x , 0) exp−i n t  .    n Hence  E  ψ ( x , t ) = ∑anψn ( x )exp−i n t  ,    n

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with an = n | ψ ( x , 0) = =

1 2a

∫

a −a

∫

∞ −∞

ψn*( x )ψ ( x , 0)dx

ψn*( x )dx .

1010

ψ ( x , t ) is a solution of the Schrödinger equation for a free particle of mass m in one dimension, and

ψ ( x , 0) = A exp(− x 2 /a 2 ). a. At time t = 0 find the probability amplitude in momentum space. b. Find ψ ( x , t ). (Berkeley) Sol: a. At time t = 0 the probability amplitude in momentum space is ∞ −ipx/  1 ψ ( p, 0) = e ψ ( x , 0) dx ∫ −∞ 2π  ∞ A = exp(−x 2 /a 2 − ipx/ ) dx ∫ −∞ 2π  Aa exp(−a 2 p 2 /4  2 ). = 2 b. The Schrödinger equation in momentum space for a free particle, i ∂ψ ( p, t ) / ∂t = Hˆ ψ ( p, t ) = gives

p2 ψ ( p, t ), 2m

 −ip 2t  ψ ( p, t ) = B exp    2m 

At time t = 0 , we have B = ψ ( p , 0) . Hence

ψ( p, t ) =

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 a 2 p 2 ip 2t  Aa exp − 2 − , 2m  2  4

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ψ (x, t ) = =

1 (2π  )1/2

∫

∞

−∞

 ipx  ψ ( p, t ) dp exp    

Aa x2   exp  −  2  i t 2it  .   a 2 + a2 +  m  m  

We can also expand the wave function as a linear superposition of plane waves and get ∞ 1 ψ ( p , 0)e i ( kx −ω t ) dp ψ (x , t ) = 1/2 ∫−∞ (2π  ) =

1 (2π  )1/2

∫

∞

−∞

Aa  a2 p2  exp  − 2  2  4 

p p 2t   × exp i  x −  dp 2m     = =

Aa 2 π

 a 2 p 2 ip 2t ipx  exp  − 2 − dp + −∞ 2m    4

∫

∞

x2   Aa exp  − 2it   , 2it   a 2 + a2 +  m    m

which agrees with the previous result.

1011 Consider a particle in a state ψ = | ψ 1 > + | ψ 2 > at time t = 0, where ψ1 and ψ2 2 are the wave functions of the ground state and first excited state of a particle in an infinite potential well of length L. At what time will the final wave function be orthogonal to the initial state? Sol:

At time t, the wave function is given by ψf =

|ψ 1 > e − iE1t / + |ψ 2 > e − iE2t / 2

− iE1t / + < ψ 2 |ψ 2 > e − iE2t / For the orthogonality, = 0, therefore, < ψ 1 |ψ 1 > e 2 = 0 and all other cross terms are zero, and using the property of Kronecker delta,  cos −1 ( −1) and E2 = 4E1, so e − iE1t / = −e − iE2t /, which therefore leads to t = E2 − E1 2mL2 t= . 3π

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1012 A particle of mass m moves in a one-dimensional box of length l with the potential V = ∞, V = 0, V = ∞,

x < 0, 0 < x < l, x > l.

At a certain instant, say t = 0 , the wave function of this particle is known to have the form

ψ = 30/l 5 x ( l − x ),

0 < x < l,

ψ = 0,

otherwise.

Write down an expression for ψ ( x , t > 0) as a series, and expressions for the coefficients in the series. (Wisconsin) Sol:

The eigenfunctions and the corresponding energy eigenvalues are

ψn ( x ) =

2 πx  sin  n  ,  l  l

Thus

En =

2

2  π   n  , n = 1, 2, 3, … . 2m  l 

ψ =∑ n n ψ , n

where n ψ (t = 0) =

∫

l 0

2 πx  sin  n  ⋅ l  l 

30 x (l − x ) dx l5

3

 1  = 4 15   (1 − cos nπ )  nπ  = 4 15[1 − (−1)n ] (1/ nπ )3 , and hence ∞  E  ψ ( x , t ) = ∑ n ψ (t = 0) ψn ( x ) exp −i n t     n=1 ∞

= ∑8 n=0

2   2n+1  π t  l

 2n + 1  −i 2m 30 1 sin  πx e ⋅ 3 3  l  l (2n + 1) π

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1013 A rigid body with moment of inertia of I z rotates freely in the x − y plane. Let

φ be the angle between the x-axis and the rotator axis. a. Find the energy eigenvalues and the corresponding eigenfunctions. b. At time t = 0 the rotator is described by a wave packet ψ (0) = A sin2 φ . Find ψ (t ) for t > 0. (Wisconsin) Sol: a. The Hamiltonian of a plane rotator is H = −( 2 / 2 I z )d 2 /dφ 2 and so the Schrödinger equation is −( 2 / 2 I z )d 2ψ / dφ 2 = Eψ . Setting α 2 = 2 I z E /  2 , we write the solution as

ψ = A e iαφ + B e −iαφ , where A, B are arbitrary constants. For the wave function to be single-­ valued, i.e. ψ (φ ) = ψ (φ + 2π ), we require

α ≡ m = 0,

±1, ± 2, ….

The eigenvalues of energy are then Em = m 2  2 /2 I z ,

m = 0, ± 1, ….

and the corresponding eigenfunctions are 1 imφ ψm (φ ) = e , m = 0, ± 1, … , 2π after normalization

∫

2π

0

ψm*ψm dφ = 1 .

b. At t = 0

ψ (0) = A sin 2 φ = = A/ 2−

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A (1 − cos 2φ ) 2

A i 2φ −i 2φ (e + e ), 4

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19

which corresponds to m = 0 and m = ±2 . The angular speed is given by Em = 12 I zφ 2, or φ = mI . Hence we have for time t z

ψ (t ) =

A A i 2(φ −t /I z ) −i 2(φ+t /I z ) − [e +e ]. 2 4

1014 Consider a particle of mass m moving under the influence of potential V(x). 1/4

r Suppose its energy eigenstate is given by ψ =   exp(−r x 2 / 2) and its π  2 r , find the potential energy of the particle. energy is E = 2m Sol:

The time-independent Schrödinger equation is given by



ψ " ( x ) = r [−ψ ( x ) − xψ '( x )] = −r [ψ ( x ) + rx ( − xψ ( x ))]



So, the equation is given by

−2 ψ " ( x ) + V ( x )ψ ( x ) = Eψ ( x ) 2m 1/4 r ψ ' ( x ) =   exp(−r x 2 / 2)(− x ) = − xψ ( x )r π



V (x ) =

−2 ψ " ( x ) + V ( x )ψ ( x ) = Eψ ( x ) 2m

−  2r  2r (−rψ ( x ) + x 2r 2ψ ( x )) + V ( x )ψ ( x ) = ψ (x ) 2m 2m

2  2 x 2r 2 . (r + r 2 x 2 − r ) = 2m 2m

1015 Give the energy levels E

(a) n

well as the energy levels E

of the one-dimensional potential in Fig. 1.2(a) as

(b) n

of the potential in Fig. 1.2(b) (Wisconsin)

Fig. 1.2

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Problems and Solutions in Quantum Mechanics

Sol: a. Use coordinate system as shown in Fig. 1.3. The Schrödinger equation is

Fig. 1.3 −

 2 d 2ψ + V ψ = Eψ . 2m dx 2

where V =0

for x > a (region I) ,

V = −V0

for − a < x < a (region II) ,

V =0

for x < −a (region III) .

For bound states we require −V0 < E < 0 . Let k2 =

2m( E + V0 ) 2mE , k′ 2 = − 2 . 2  

The Schrödinger equation becomes d 2ψ 2 + k ψ = 0 for region II, dx 2 and d 2ψ − k′2ψ = 0 for region I and III, dx 2 which have solutions ψ = A sin kx + B cos kx

ψ = Ce

− k′x

+ De

k′x

for

−a < x < a,

for

x < −a and x > a.

The requirement that ψ → 0 as x → ±∞ demands that

ψ = Ce − k′x

for

x > a (region I) ,

ψ = De k′x

for

x < a (region III) .

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The boundary conditions that ψ and ψ ′ be continuous at x = ±a then give A sin ka + B cos ka = Ce − k ′a , − A sin ka + B cos ka = De − k ′a , Ak cos ka − Bk sin ka = −Ck′ e − k ′a , Ak cos ka + Bk sin ka = Dk′ e − k ′a ; or 2 A sin ka = (C − D )e − k ′a , 2 B cos ka = (C + D )e − k ′a , 2 Ak cos ka = −(C − D )k′e − k ′a , 2 Bk sin ka = (C + D )k′e − k ′a . For solutions for which not all A, B, C, D vanish, we must have either A = 0, C = D giving k tan ka = k′ , or B = 0, C = −D giving k cot ka = −k′. Thus two classes of solutions are possible, giving rise to bound states. Let ξ = ka, η = k′a . Class 1: ξ tanξ = η ,  2 2 2 ξ + η = γ , where γ 2 = k 2a 2 + k′2a 2 =

2mV0 a2 2

.

Since ξ and η are restricted to positive values, the energy levels are found from the intersections in the first quadrant of the circle of radius γ with the curve of ξ tan ξ plotted against ξ , as shown in Fig. 1.4. The number of discrete levels depends on V0 and a, which determine γ . For small γ only one solution is possible.

Fig. 1.4

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Class 2: ξ cot ξ = −η ,  2 2 2 ξ + η = γ . 2 A similar construction is shown in Fig. 1.5. Here the smallest value of V0a

gives no solution while the larger two give one solution each.

Fig. 1.5 Note that ξ = 0, η = 0 is a solution of ξ tan ξ = η and so no matter how small

γ is, there is always a class 1 solution, whereas γ has to be above a minimum for a class 2 solution to exist, given by ξ cot ξ = 0 which has a minimum solution ξ = π2 , i.e. γ = π2 or V0a 2 = π8m . 2 2

b. Use coordinates as shown in Fig. 1.6.

Fig. 1.6 The Schrödinger equation has solutions ψ = A sin kx + B cos kx

ψ = Ce ψ =0

− k′x

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for

0 < x < a,

for

x > a,

for

x < 0,

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23

satisfying the requirement ψ → 0 as x → ∞ . The boundary conditions at x = 0 and x = a then give B = 0, A sin ka = Ce − k ′a , Ak cos ka = −Ck′e − k ′a ; and finally

ξ cot ξ = −η , ξ 2 +η 2 = γ 2 , as for the class 2 solutions above.

1016 Consider the one-dimensional problem of a particle of mass m in a potential (Fig. 1.7) V = ∞, V = 0, V = V0 ,

x < 0, 0 ≤ x ≤ a, x > a.

a. Show that the bound state energies ( E < V0 ) are given by the equation tan

2mEa E =− . V0 − E 

b. Without solving any further, sketch the ground state wave function. (Buffalo) Sol: a. The Schrödinger equations for the two regions are

ψ ″+ 2mEψ/ 2 = 0,

0 ≤ x ≤ a,

ψ ″− 2m(V0 − E )ψ/ = 0, 2

x > a,

with respective boundary conditions ψ = 0 for x = 0 and ψ → 0 for x → +∞ . The solutions for E < V0 are then

ψ = sin( 2mEx/ ), ψ = Ae −

2m (Vo−E )x / 

0 ≤ x ≤ a;

, x > a,

dψ where A is a constant. The requirement that ψ and are continuous at dx x = a gives

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tan

(

)

2mEa /  = −[E / (V0 − E )]1/2 .

b. The ground-state wave function is as shown in Fig. 1.7.

Fig. 1.7 1017 Consider a particle confined to a region, −a < x < a with its wave function π x  given by ψ ( x , t ) = sin   exp(−iωt ). Find the potential in the region.  a  ∂ψ ( x , t ) ∂t

Sol:

Using the formula Eψ ( x , t ) = i



 πx Eψ ( x , t ) = i sin  (−iω )exp(−iω t ) so, from this converting this in terms of  a  wave function, Eψ ( x , t ) = ωψ ( x , t )



From this we get E = ω 



The time-independent Schrödinger equation is given by

(1) −2 ψ " ( x ) + V ( x )ψ ( x ) = Eψ ( x ) 2m 2

−2  πx π π  πx ψ " ( x ) + V ( x )ψ ( x ) = Eψ ( x ) where ψ ' ( x ) = cos   e − iω t and ψ " ( x ) = −   exp(−iω t )sin        a  a a a 2m 2 π   and converting this in terms of wave function we get, ψ " ( x ) = −   ψ ( x )  a

 2π 2 ψ ( x ) + V ( x )ψ ( x ) = Eψ ( x ) 2ma 2 2 π2 Substituting for E from Equation 1, V ( x ) = ω − . 2ma 2

So plugging this into the earlier equation, we get

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25

1018 Consider a free particle confined to a one-dimensional infinite potential well  0; 0 < x < L of length L, with potential V ( x ) =  . What is the probability of  ∞; elsewhere finding an electron in the nth state for the region x = 0 to x = L/n? Sol:

The wave function for the nth state is given by ψ n ( x ) =



The probability is given by P =



2 P= L



L /n

L /n

2 sin(nπ x / L ). L

∫ ψ ψ dx *

0

∫ sin (nπ x / L ) dx and using the formula for sin (x ) = 2

2

0

We get, P =

1  dx − L  ∫0 L /n

1 + cos2 x 2



L /n

∫ cos(2nπ x / L )dx  0

1  L    − 0 − (sin(2π ) − 0) L  n  



After applying the limit, it gives P =



So the probability is given by P = 1/n.

1019 Consider the following one-dimensional potential wells:

Fig. 1.8

Fig. 1.9

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Problems and Solutions in Quantum Mechanics

a. Can each well support a bound state for an arbitrarily small depth Vi (i = 1, 2) ? Explain qualitatively. b. For V1 = V2 , what is the relationship between the energies of the bound states of the two wells? c. For continuum states of a given energy, how many independent solutions can each well have? d. Explain qualitatively how it is possible to have bound states for which the particle is more likely to be outside the well than inside. (Wisconsin) Sol: a. For bound states, we must have −V < E < 0. Let 2m( E + V ) 2 2mE k2 = , k′ = − 2 , 2  

where V = V1 , V2 for the two cases, and set ξ = ka, η = k′a, γ = 2mV a .  The discussion in Problem 1015 shows that for the potential in Fig. 1.8, the solutions are given by

ξ cot ξ = −η , ξ 2 + η 2 = γ 2 . The energy levels are given by the intersection of the curve ξ cot ξ = −η with a circle of radius γ with center at the origin (Fig. 1.5) in the first quadrant. As the figure shows, γ must be greater than the value of ξ for which ξ cot ξ = 0 , i.e.

ξ≥

a 2mV1 π π . Hence for a bound state to exist, we require ≥ , or 2  2

V1 ≥

π 2 2 . 8ma 2

For the potential shown in Fig. 1.9, two classes of solutions are possible. One class are the same as those for the case of Fig. 1.5 and are not possible for arbitrarily small V2. The other class of solutions are given by ξ tanξ = η ,  2 2 2 ξ + η = γ . As the curve of ξ tan ξ = η starts from the origin, γ may be arbitrarily small and yet an intersection with the curve exists. However small V2 is, there is always a bound state.

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b. For V1 = V2 , the bound states of the potential of Fig. 1.8 are also bound states of the potential of Fig. 1.9. c. For continuum states of a given energy, there is only one independent solution for well 1, which is a stationary-wave solution with ψ = 0 at x = 0 ; there are two independent solutions corresponding to traveling waves in +x and −x directions for well 2. d. Let p1 , p2 denote respectively the probabilities that the particle is inside and outside the well. Consider, for example, the odd-parity solution ψ = A sin kx for 0 < x < a,

ψ = Ce − k′x where k =

2(mVi + E )  p1 = p2

0

2

∞

−2mE , for which 

(i = 1, 2) , k′ =

∫ A sin kx dx = A ∫ C e dx C a

a < x,

for

2

2

2

2 −2 k ′x

a

k′a  sin2ka  . 1 − e −2 k ′a  2ka 

The continuity of ψ at x = a gives A e − k ′a . = C sin ka Setting, as before, η = k′a, ξ = ka , we have p1 η  sin 2ξ  2η  sin 2ξ  . = 1− = 2 1 −  p2 sin ξ  2ξ  1 − cos 2ξ  2ξ  The odd-parity solutions are given by  ξ cot ξ = −η ,  2 2 2  ξ + η = γ ,

where γ 2 =

2mVi a 2 (i = 1, 2). 2

An analytic solution is possible if γ → (n + 12 ) π , or Vi a

2

(n + 12 ) →

2

π 2 2 , 2m

(n = 0, 1, 2, …)

for which the solution is ξ → (n + 12 ) π , η → 0 , and p1 →0 . p2 The particle is then more likely outside the well than inside.

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1020 Obtain the binding energy of a particle of mass m in one dimension due to the following short-range potential: V ( x ) = −V0δ ( x ). (Wisconsin) Sol:

The Schrödinger equation d 2ψ / dx 2 +

2m [E − V ( x )] ψ = 0, ( E < 0), 2

on setting k = 2m | E |  , U 0 = 2mV0  2 , can be written as

ψ ′′( x ) − k 2ψ ( x ) + U 0δ ( x ) ψ ( x ) = 0 . Integrating both sides of the above equation over x from −ε to ε , where ε is an arbitrarily small positive number, we get

ψ ′ (ε ) − ψ ′ (−ε ) − k 2 ∫ −ε ψ dx + U 0ψ (0) = 0 , ε

which becomes, by letting ε → 0,

ψ ′(0+ ) − ψ ′(0− ) + U 0ψ (0) = 0 . (1)

At x ≠ 0 (δ ( x ) = 0) the Schrödinger equation has solutions

ψ ( x ) ∼ exp (−kx )

for

x > 0,

ψ ( x ) ∼ exp ( kx )

for

x < 0.

It follows from Eq. (1) that

ψ ′ (0+ ) − ψ ′ (0− ) = −2kψ (0) . A comparison of the two results gives k = U 0 2 . Hence the binding energy is −E =  2 k 2 2m = mV02 2 2 .

1021 Consider a particle of mass m in the one-dimensional δ function potential V ( x ) = V0 δ ( x ) . Show that if V0 is negative there exists a bound state, and that the binding energy is mV02 2 2 .

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29

(Columbia) Sol:

In the Schrödinger equation d 2ψ dx 2 + 2m [ E − V ( x )] ψ  2 = 0, we set E < 0 for a bound state as well as k 2 = 2m | E | /  2 ,

U 0 = 2mV0 /  2 ,

and obtain d 2ψ dx 2 − k 2ψ − U 0δ ( x ) ψ = 0 . Integrating both sides over x from −ε to +ε , where ε is an arbitrarily small positive number, we obtain

ψ ′(ε ) − ψ ′(−ε ) − k 2 ∫ −ε ψ dx − U 0ψ (0) = 0 . ε

With ε → 0+ , this becomes ψ ′(0+ ) − ψ ′(0− ) = U 0ψ (0). For x ≠ 0 the Schrödinger equation has the formal solution ψ ( x ) ∼ exp (−k x ) with k positive, which gives

ψ ′( x ) ∼ −k and hence

− kx x − k x −ke , x > 0, e =  kx x ke , x < 0,

ψ ′ (0+ ) − ψ ′ (0− ) = −2kψ (0) = U 0ψ (0) .

Thus k = −U 0 2 , which requires V0 to be negative. The energy of the bound state 2 2

is then E = − 2mk = −mV02 2 2 and the binding energy is Eb = 0 − E = mV02 2 2. The wave function of the bound state is  mV  ψ ( x ) = A exp  2 0 | x | = −mV0  2 exp(mV0 | x |  2 ),    where the arbitrary constant A has been obtained by the normalization

∫

0 −∞

ψ 2 dx +

∫

∞ 0

ψ 2 dx = 1 .

1022 A particle of mass m moves non-relativistically in one dimension in a potential given by V ( x ) = −aδ ( x ), where δ ( x ) is the usual Dirac delta function. The particle is bound. Find the value of x 0 such that the probability of finding the particle with x < x 0 is exactly equal to 1/2. (Columbia)

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Problems and Solutions in Quantum Mechanics

For bound states, E < 0 . The Schrödinger equation  2 d 2  − aδ ( x ) ψ ( x ) = Eψ ( x ) − 2  2m dx  has for x ≠ 0 the solutions finite at x = ±∞ as follows,  A e kx ψ ( x ) =  − kx  A e

x < 0,

for

x > 0, for −2mE and A is an arbitrary constant. Applying limε→0+ where k = 

∫

ε −ε

dx to the

Schrödinger equation gives

ψ ′(0+ ) − ψ ′(0− ) = −



2ma ψ (0) (1) 2

since

∫

ε −ε

ε

ψ ( x )δ ( x ) dx = ψ (0), lim ∫ −ε ψ ( x ) dx = 0 ε →0

for finite ψ (0). Substitution of ψ ( x ) in (1) gives k=

ma . 2

Hence  ma | x |  ψ ( x ) = A exp − .  2  On account of symmetry, the probabilities are | A |2 (1 − e −2kx0 ) , 0 k ∞ | A |2 P(−∞ < x < ∞) = 2| A |2 ∫ e −2 kx dx = . 0 k P (| x | < x 0 ) = 2| A |2

∫

x0 −2 kx

e

dx =

As it is given 1 1 − e −2 kx0 = , 2 we have x0 =

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1023 A particle of mass m moving in one dimension is confined to the region 0 < x < L by an infinite square well potential. In addition, the particle experiences a delta function potential of strength λ located at the center of the well (Fig. 1.10). The Schrödinger equation which describes this system is, within the well, −

2 2 ∂ ψ ( x ) + λδ ( x − L 2) ψ ( x ) = Eψ ( x ) , 0 < x < L. 2m ∂ x 2

Fig. 1.10 Find a transcendental equation for the energy eigenvalues E in terms of the mass m, the potential strength λ , and the size L of the system. (Columbia) Sol:

Applying limε →0

∫



ψ ′ ( L 2 + ε ) − ψ ′ ( L 2 − ε ) = ( 2mλ  2 ) ψ ( L 2) ,

L /2+ε L /2−ε

dx to both sides of the Schrödinger equation, we get (1)

since

∫

L +ε 2 L −ε 2

L +ε  L L ψ ( x )δ  x −  dx = ψ   , lim ∫ L2−ε ψ ( x ) dx = 0. 0 ε → 2  2 2

Subject to the boundary conditions ψ (0) = ψ ( L ) = 0, the Schrödinger equation has solutions for x ≠ L2 :  A1 sin(kx ), ψ(x ) =   A2 sin[k( x − L )], where k =

0 ≤ x ≤ L 2 −ε L 2 +ε ≤ x ≤ L,

2mE and ε is an arbitrarily small positive number. The continuity 

of the wave function at L 2 requires A1 sin(kL 2) = − A2 sin(kL 2) , or A1 = − A2 . Substituting the wave function in (1), we get

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Problems and Solutions in Quantum Mechanics

A2 k cos(kL 2) − A1k cos(kL 2) = (2mλ A1  2 ) sin(kL 2), whence tan

2E  kL k 2 2mEL , which is the transcendental , or tan =− =− m λ 2 mλ 2

equation for the energy eigenvalue E.

1024 An infinitely deep one-dimensional square well potential confines a particle to the region 0 ≤ x ≤ L . Sketch the wave function for its lowest energy eigenstate. If a repulsive delta function potential, H ′ = λδ ( x − L 2)( λ > 0) , is added at the center of the well, sketch the new wave function and state whether the energy increases or decreases. If it was originally E0 , what does it became when λ →∞? (Wisconsin) Sol:

For the square well potential the eigenfunction corresponding to the lowest energy state and its energy value are respectively

φ0 ( x ) = 2 L sin(π x L ), E0 =

π 2 2 . 2mL2

A sketch of this wave function is shown in Fig. 1.11 With the addition of the delta potential H ′ = λδ ( x − L / 2) , the Schrödinger equation becomes

ψ ′′+ k 2 − αδ ( x − L / 2)ψ = 0, where k 2 = 2mE  2 , α = 2mλ  2 . The boundary conditions are

Fig. 1.11

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33

ψ (0) = ψ ( L ) = 0, (1)  L    L −  ψ ′   − ψ ′   = αψ ( L / 2),  2    2   +

(2)

 L +   L −  ψ    = ψ    . (3)  2    2  

Note that (2) arises from taking limε →0 ∫ L2 dx over both sides of the Schrödinger −ε L +ε

2

equation and (3) arises from the continuity of ψ ( x ) at x = L2 . The solutions for x ≠ L2 satisfying (1) are  A1 sin(kx ), ψ =  A2 sin[k( x − L )],

0 ≤ x ≤ L / 2, L / 2 ≤ x ≤ L.

Let k = k0 for the ground state. Condition (3) requires that A1 = − A2 = A , say, and the wave function for the ground state becomes  A sin(k0 x ), 0 ≤ x ≤ L / 2, ψ0 ( x ) =  L / 2 ≤ x ≤ L. − A sin[k0 ( x − L )], Condition (2) then shows that k0 is the smallest root of the transcendental equation cot(kL / 2) = −

mλ . k 2

As cot ( kL2 ) is negative, π / 2 ≤ k0 L / 2 ≤ π , or π / L ≤ k0 ≤ 2π / L. The new groundstate wave function is shown Fig. 1.12. The corresponding energy is π 2 2 π E =  2 k02 /2m ≥ E0 = , since k0 ≥ . Thus the energy of the new ground state 2 2mL 2 increases. Furthermore, if λ → +∞, k0 → 2π / L and the new ground-state energy E → 4 E0 .

Fig. 1.12

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1025 Consider a particle of mass m confined to a one-dimensional infinite square well. Assume the particle is in the ground state initially. If one of the wall at x = L is suddenly shifted to x = 2L, find the probability of finding the particle in the ground state of the new potential.

Sol:

The initial wave function is given by ψ i =

2 sin(π x / L ) and the final wave L

1 sin(π x / 2 L ), so the probability amplitude is L = ∫ ψ i *ψ f dx therefore the probability amplitude is given by

function is given by ψ f = given by Pamp L

2 sin(π x / L )sin(π x / 2 L ) dx L ∫0

where the limits are common to both systems. And using the formula

∫ sin(mx )sin(nx ) = So Pamp =

sin[(m − n)x ] sin[(m + n)x ] − where m = π/L and n = π/2L 2(m − n) 2(m + n) L

2  sin[(π / 2 L )x ] sin[(3π / 2 L )x ] 1 1  −  = Pamp = 2  +  = 4 2 / 3π  π /L π 3π L  3π / L 0

And the probability is given by P = | Pamp |2 =

32 . 9π 2

1026 An approximate model for the problem of an atom near a wall is to consider a particle moving under the influence of the one-dimensional potential given by V ( x ) = −V0δ ( x ), V ( x ) = ∞,

x > −d , x < −d ,

where δ ( x ) is the so-called “delta function”. a. Find the modification of the bound-state energy caused by the wall when it is far away. Explain also how far is “far away”.

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b. What is the exact condition on V0 and d for the existence of at least one bound state? (Buffalo)

Fig. 1.13 Sol: a. The potential is as shown in Fig. 1.13. In the Schrödinger equation

ψ ″+ (2m/ 2 ) [E + V0δ ( x )] ψ = 0, x > −d , let k = −2mE / , where E < 0 . This has the formal solutions ae kx + be − kx ψ ( x ) =  − kx e

for

−d < x < 0,

for

x > 0,

as ψ ( x ) is finite for x → ∞ . The continuity of the wave function and the discontinuity of its derivative at x = 0 (Eq. (1) of Problem 1020), as well as the requirement ψ ( x = −d ) = 0 , give a + b = 1, −k − (a − b)k = −2mV0 / 2 , ae − kd + be kd = 0. Solving these we find e 2 kd 1 , b= , 1 − e 2 kd 1 − e 2 kd mV k = 2 0 (1 − e −2 kd ).  a=−

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The wall is “far away” from the particle if kd  1, for which k ≈ mV0 / 2. A  better approximation is k ≈ (mV0 / 2 )1 − exp(−2mV0d / 2 ) , which gives the bound-state energy as 2  2mV0d   2k 2  2  mV0   ≈− E =−  2  1 − exp −     2m 2m   2  

≈−

 2mV0d  mV02  1 − 2 exp −  2   2   2 

≈−

 2mV0d  mV02 mV02 + 2 exp − . 2  2  2 

2

The second term in the last expression is the modification of energy caused by the wall. Thus for the modification of energy to be small we require d  1/k =  2 /mV0 . This is the meaning of being “far away”

Fig. 1.14 b. Figure 1.14 shows line 1 representing y = k and curve 2 representing y = ye [1 − exp(−2kd )] , where ye = mV0 / 2 . The condition for the equation k = mV0 [1 − exp(−2kd )] / 2 to have a solution is that the slope of curve 2 at the origin is greater than that of line 1: dy = 2mV0d/ > 1 . dk k=0 Hence if V0d > 2m , there is one bound state. 2

1027 The wave function of the ground state of a harmonic oscillator of force constant k and mass m is 2

ψ0 ( x ) = (α /π )1/4 e −α x /2, α = mω 0 / , ω 02 = k/m.

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Obtain an expression for the probability of finding the particle outside the classical region. (Wisconsin) Sol:

The particle is said to be outside the classical region if E < V ( x ) . For the ground state, E = ω0 /2 and the nonclassical region is 12 ω0 < 12 mω02 x 2 , i.e.,  1 , x >   1 α 2 x > = , or  ω0m α 1   x < − α . The probability of finding the particle in this nonclassical region is therefore P=

∫

=

∫

x>

1 a

ψ02 ( x ) dx α −α x 2 e dx + π α −α x 2 e dx π

− 1/α −∞

=2∫

∞ 1/α

= 2 ∫1

∞

∫

∞ 1/α

α −α x 2 e dx π

1 − t2 e dt  16%. π

1028 Consider a linear harmonic oscillator and let ψ0 and ψ1 be its real, normalized ground and first excited state energy eigenfunctions respectively. Let Aψ0 + Bψ1 with A and B real numbers be the wave function of the oscillator at some instant of time. Show that the average value of x is in general different from zero. What values of A and B maximize x and what values minimize it? (Wisconsin) Sol:

The orthonormal condition

∫ ( Aψ

0

+ Bψ1 )2 dx = 1

gives A2 + B 2 = 1. Generally A and B are not zero, so the average value of x, 〈x 〉 =

∫ x( Aψ

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0

+ Bψ1 )2 dx = 2 AB 〈ψ0 | x | ψ1 〉

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is not equal to zero. Rewriting the above as 〈 x 〉 = 1 − ( A2 + B 2 + 2 AB) 〈ψ0 | x | ψ1 〉 = 1 − ( A − B)2  〈ψ0 | x | ψ1 〉 1

and considering f = AB = A(1 − A 2 ) 2 , which has extremums at A = ±

1 2

,we see

that if A = B = 1/ 2, 〈 x 〉 is maximized; if A = −B = 1/ 2, 〈 x 〉 is minimized.

1029 Show that the minimum energy of a simple quantum harmonic oscillator is ω/2 if ∆x ∆p = /2, where ( ∆p )2 = 〈( p − 〈 p 〉)2 〉. (Wisconsin) Sol:

For a harmonic oscillator, 〈 x 〉 = 〈 p〉 = 0, and so (∆x )2 = 〈 x 2 〉, (∆p)2 = 〈 p 2 〉. Then the Hamiltonian of a harmonic oscillator, H = p 2 /2m + mω 2 x 2 /2, gives the average energy as H = p2

2m + mω 2 x 2

As for a, b real positive we have

(

2 = ∆p

a− b

)

2

2

2

2m + mω 2 ∆x

2.

≥ 0 , or a + b ≥ 2 ab ,

〈 H 〉 min = 〈∆p 〉 〈∆x 〉ω = ω / 2 .

1030 An electron is confined in the ground state of a one-dimensional harmonic oscillator such that

〈( x − 〈 x 〉)2 〉 = 10−10 m. Find the energy (in eV) required to

excite it to its first excited state. [Hint: The virial theorem can help.] (Wisconsin) Sol:

The virial theorem for a one-dimensional harmonic oscillator states that T = V . Thus E0 = 〈 H 〉 = 〈T 〉 + 〈V 〉 = 2〈V 〉 = meω 2 〈 x 2 〉 , or, for the ground state, ω = meω 2 〈 x 2 〉, 2 giving

ω=

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 . 2me 〈 x 2 〉

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As 〈 x 〉 = 0 for a harmonic oscillator, we have 〈( x − 〈 x 〉)2 〉 = 〈 x 2 〉 − 〈 x 〉 2 = 〈 x 2 〉 = 10−10 m . The energy required to excite the electron to its first excited state is therefore ∆E = ω = =

2  2c 2 = 2me 〈 x 2 〉 2me c 2 〈 x 2 〉 (6.58 × 10−16 )2 × (3 × 108 )2 = 3.8 eV. 2 × 0.51 × 10−20

1031 The wave function at time t = 0 for a particle in a harmonic oscillator potential V = 21 kx 2, is of the form   2 sin β ψ ( x , 0) = Ae −(α x ) 2 cos β H0 (α x ) + H2 (α x ) ,   2 2 where β and A are real constants, α 2 ≡ mk / , and the Hermite polynomials are normalized so that

∫

∞ −∞

e −α

2 2

x

[H x (α x )]2 dx =

π n 2 n! . α

a. Write an expression for ψ ( x , t ). b. What are the possible results of a measurement of the energy of the particle in this state, and what are the relative probabilities of getting these values? c. What is 〈 x 〉 at t = 0 ? How does 〈 x 〉 change with time? (Wisconsin) Sol: a. The Schrödinger equation for the system is i ∂t ψ ( x , t ) = Hˆ ψ ( x , t ), where ψ ( x , t ) takes the given value ψ ( x , 0) at t = 0. As Hˆ does not depend on t explicitly,

ψn ( x , t ) = ψn ( x )e −iEn t/ ,

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where ψn ( x ) is the energy eigenfunction satisfying Hˆ ψn ( x ) = Enψn ( x ). Expanding ψ ( x , 0) in terms of ψn ( x ) :

ψ ( x , 0) = ∑anψn ( x ), n

where an = ∫ ψn* ( x )ψ ( x , 0) dx . Thus

ψ ( x , t ) = ∑anψn ( x , t ) = ∑anψn ( x )e −iEn t/. n

n

For a harmonic oscillator,

ψn ( x ) = N ne −α

2 2

x /2

Hn (α x ),

so that an = ∫ N ne −α

2 2

x /2

H n (α x ) ⋅ Ae −α

2 2

x /2

sin β   H 2 (α x ) dx . × cos β H 0 (α x ) + 2 2   As the functions exp (− 12 x 2 ) Hn ( x ) are orthogonal, all an = 0 except a0 = AN 0 a2 = AN 2

π cos β , α2 π 2 2 sin β . α2

Hence

ψ (x , t ) = A

π  N 0 cos βψ 0 ( x )e − iE0t / α2 

+ 2 2 N 2 sin βψ 2 ( x )e − iE2t /  . 1

E t E t  π 4 −i 0 −i 2 = A  2  cos βψ 0 ( x )e  + sin βψ 2 ( x ) e   ,  α  

as N n are given by

∫ [ψ (x )] dx = 1 to be N = ( ) n

2

0

α2 π

1 4

, N2 = 1 2 2

( ) α2 π

1 4

.

b. The observable energy values for this state are E0 = ω / 2 and E2 = 5ω / 2, and the relative probability of getting these values is P0 / P2 = cos 2 β / sin 2 β = cot 2 β .

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c. As ψ ( x , 0) is a linear combination of only ψ0 ( x ) and ψ2 ( x ) which have even parity,

ψ (−x , 0) = ψ ( x , 0) . Hence for t = 0 , 〈 x 〉 = ∫ ψ ( x , 0) xψ ( x , 0) dx = 0 . It follows that the average value of x does not change with time.

1032 a. For a particle of mass m in a one-dimensional harmonic oscillator potential V = mω 2 x 2 /2 , write down the most general solution to the time-dependent Schrödinger equation, ψ ( x , t ), in terms of harmonic oscillator eigenstates φ n ( x ) . b. Using (a) show that the expectation value of x , 〈 x 〉 , as a function of time can be written as A cos ωt + B sinωt , where A and B are constants. c. Using (a) show explicitly that the time average of the potential energy satisfies 〈V 〉 = 21 〈E 〉 for a general ψ ( x , t ). Note the equality mω n +1 n xφ n = φ n+1 + φn−1 .  2 2 (Wisconsin) Sol: a. From the time-dependent Schrödinger equation i

∂  2 ∂2 ψ ( x , t ) ψ(x , t ) = − + mω 2 x 2 / 2 ∂t 2m ∂x 2

as Hˆ does not depend on time explicitly, we get

ψ ( x , t ) = e −iHt/ψ ( x , 0).

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We can expand ψ ( x , 0) in terms of φn ( x ) :

ψ ( x , 0) = ∑anφn ( x ), n

where an = 〈φn ( x )| ψ ( x , 0)〉, and φn ( x ) are the eigenfunctions of  1 Hφn ( x ) = Enφn ( x ) , with  En = n +  ω .  2 Hence

ψ ( x , t ) = ∑anφn ( x )e −iEn t/ . n

b. Using the given equality we have x = ∫ψ * ( x , t )xψ ( x , t )dx = ∑an*an ′ e − i ( En′ − En )t / ∫ φn* ( x )xφn ′ ( x ) dx n, n′

 n′ + 1  n′ = ∑an*an ′ e − i ( En′ − En )t /  δ n , n ′+1 + δ n , n ′−1 2 2   n, n′  n n + 1 − iω t  = ∑an*  an−1 e iω t + an+1 e  2 2   n

 mω

 mω

= A cos ω t + B sin ω t , where A=

  n n +1  + an+1 an* an−1 , ∑ mω n  2 2 

B=

  n n +1  − an+1 ian* an−1 , ∑ 2 2  mω n 

and we have used En+1 − En = ω . c. The time average of the potential energy can be considered as the time average of the ensemble average of the operator Vˆ on ψ ( x , t ) . It is sufficient to take time average over one period T = 2π /ω . Let 〈 A〉 and A denote the time average and ensemble average of an operator A respectively.

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As 1 mω 2 V |ψ 〉 = ω ⋅ x |ψ 〉 2  E t i n 1 mω mω = ω x ∑ an xφn ( x )e  2   1 mω ∞  n + 1 x ∑an n +1 = ω 2 2  n=0  +

 n n − 1  e − iω (n+1/2)t 2 

∞  (n + 1)(n + 2) 1 n+2 = ω ∑an  2 2 n=0 

 − i En t 1 n(n − 1)  | n − 2〉  e  , + n +  n +  2 2  we have V = 〈ψ | V |ψ 〉 ∞ ∞ 1 1 1  = ω ∑an*an  n +  + ω ∑an*+2  2 2 2 n=0 n=0 ∞ (n + 1)(n + 2) i 2ω t 1 e + ω ∑an 4 2 n=0

× an

(n + 1)(n + 2) − i 2ω t e 4 ∞ 1 1  = ω ∑an*an  n +   2 2 n=0 × an*+2

∞ 1 + ω ∑ | an*+2an | (n + 1)(n + 2) cos(2ω t + δ n ), 2 n=0

where δn is the phase of an*+1an . Averaging V over a period, as the second term becomes zero, we get 〈V 〉 =

1 T

∫

∞  1 1 V dt =  ω an*an n +  . ∑ 0  2 2 n=0 T

On the other hand, ∞  1 E = 〈ψ | H | ψ 〉 = ω ∑an*an n +  ,  2 n=0

and E = E . Therefore 〈V 〉 = 〈 E 〉 / 2.

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1033 Consider a particle of mass m in the one-dimensional potential V ( x ) = mω 2 x 2 /2, V ( x ) = V0 ,

| x | > b; | x | < b,

where V0   2 / mb 2  ω , i.e. a harmonic oscillator potential with a high, thin, nearly impenetrable barrier at x = 0 (see Fig 1.15).

Fig. 1.15 a. What is the low-lying energy spectrum under the approximation that the barrier is completely impenetrable? b. Describe qualitatively the effect on the spectrum of the finite penetrability of the barrier. (MIT) Sol: a. For the low-lying energy spectrum, as the barrier is completely impenetrable, the potential is equivalent to two separate halves of a harmonic oscillator potential and the low-lying eigenfunctions must satisfy the condition ψ ( x ) = 0 at x = 0 . The low-lying energy spectrum thus corresponds to that of a normal harmonic oscillator with odd quantum numbers 2n + 1, for which ψn ( x ) = 0 , at x = 0 and En = (2n + 3 / 2)ω , n = 0, 1, 2,… with a degeneracy of 2. Thus only the odd-parity wave functions are allowed for the low-lying levels. b. There will be a weak penetration of the barrier. Obviously the probability for the particle to be in | x | < b , where the barrier exists, becomes less than that for the case of no potential barrier, while the probability outside the barrier becomes relatively larger. A small portion of the even-parity solutions is mixed into the particle states, while near the origin the probability

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distribution of even-parity states is greater than that of odd-parity states. Correspondingly, a small portion of the energy En′ = (2n + 1/ 2)ω is mixed into the energy for the case (a). Since 〈ψ | barrier potential | ψ 〉 > 0, the energy levels will shift upwards. The level shifts for the even-parity states are greater than for odd-parity states. Furthermore, the energy shift is smaller for greater energies for states of the same parity.

1034 The Hamiltonian for a harmonic oscillator can be written in dimensionless units ( m =  = ω = 1) as Hˆ = aˆ + aˆ + 1/ 2 , where aˆ = ( xˆ + ipˆ ) / 2 , aˆ + = ( xˆ − ipˆ ) / 2 . One unnormalized energy eigenfunction is ψa = (2 x 3 − 3 x ) exp(− x 2 / 2) . Find two other (unnormalized) eigenfunctions which are closest in energy to ψa. (MIT) Sol:

+ In the Fock representation of harmonic oscillation, aˆ and aˆ are the annihilation

and creation operators such that ˆˆ +ψ n = (n + 1)ψ n , aˆψ n = nψ n−1 , aˆ +ψ n = n + 1 ψ n+1 , aa 1  En =  n +  ω , n = 0, 1, 2,… .  2 As x2 1 d  d  ˆˆ+ψa =  x +   x −  (2 x 3 − 3x )e − 2 aa 2  dx   dx  2 1 d  −x =  x +  (4 x 4 − 12 x 2 + 3)e 2 2  dx 

= 4(2 x 3 − 3x ) e

2

− x2

= (3 + 1)ψa ,

we have n = 3 . Hence the eigenfunctions closest in energy to ψ a have n = 2, 4 , the unnormalized wave functions being

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ψ2 =

1 1  d  3 − x 2 /2 aˆψa =  x +  (2 x − 3x ) e 3 6  dx  2

∼ (2 x 2 − 1) e − x /2 , 1 1  d  3 − x 2 /2 ψ4 = aˆ+ψa =  x −  (2 x − 3x ) e 2 2 2  dx  2

∼ (4 x 4 − 12 x 2 + 3) e − x /2 , where the unimportant constants have been omitted.

1035 A particle of mass “m” is in a one-dimensional infinite potential well with an 30 x ( a − x ) between 0 and a. Outside 0 and a, a5 the wave function would be zero. initial wave function ψ ( x ) =

Find the initial energy of the particle. Sol:

Normalized wave function, 30 x (a − x ) a5

ψ (x ) =

To find energy, we apply Schrödinger equation, −  2 d 2ψ + v ( x )ψ = ∈ψ 2m dx 2 − 2 30 × 5 × −2 = ∈ ψ 2m a a

∫ 0

2 m

a

30 30 × 5 x (a − x ) dx = ∫ ∈ ψ * ψ dx a 5 a 0   ψ(*x )

⇒

5 2 = ∈. ma 2

1036 Consider the one-dimensional motion of a particle of mass µ in the potential

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V ( x ) = V0 ( x /a )2 n , where n is a positive integer and V0 > 0. Discuss qualitatively the distribution of energy eigenvalues and the parities, if any, of the corresponding eigenfunctions. Use the uncertainty principle to get an order-of-magnitude estimate for the lowest energy eigenvalue. Specialize this estimate to the cases n = 1 and n →∞ . State what V ( x ) becomes in these cases and compare the estimates with your previous experience. (Buffalo) Sol:

Since the potential V ( x ) → ∞ as x → ∞ , there is an infinite number of bound states in the potential and the energy eigenvalues are discrete. Also, the mth excited state should have m nodes in the region of E > V ( x ) given by k∆x  (m + 1)π . ∆x Increases slowly as m increases. From the virial theorem 2T ∝ 2nV , we have k 2 ∝ (∆x )2n ∝ [(m + 1)π / k]2n , and so E ∝ k 2 ∝ (m + 1)2n/(n+1) . Generally, as n increases, the difference between adjacent energy levels increases too. Since V (− x ) = V ( x ), the eigenstates have definite parities. The ground state and the second, fourth, . . . excited states have even parity while the other states have odd parity. The energy of the particle can be estimated using the uncertainity principle px ∼ /2b, where b = (∆x )2 . Thus E∼

1 (/2b)2 + V0 (b/a)2n . 2µ

For the lowest energy let dE/db = 0 and obtain b = (h 2a 2n /8µnV0 )1/2(n+1) . Hence the lowest energy is E ∼ [(n + 1)V0 /a 2n ] ( 2a 2n /8µnV0 )n/(n+1) . For n = 1, V ( x ) is the potential of a harmonic oscillator, V ( x ) = V0 x 2 /a 2 = µω 2 x 2 /2 .

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In this case E equals ω/2 , consistent with the result of a precise calculation. For n = ∞ , V ( x ) is an infinite square-well potential, and E =  2 /8µ a 2, to be compared with the accurate result  2 π 2 /2µ a 2.

1037 Consider a particle in one dimension with Hamiltonian H = p 2 /2m + V ( x ) , where V ( x ) ≤ 0 for all x , V (±∞) = 0 , and V is not everywhere zero. Show that there is at least one bound state. (One method is to use the Rayleigh-Ritz variational principle with a trial wave function ψ ( x ) = (b / π )1/4 exp (−bx 2 / 2) .

However, you may use any method you wish.) (Columbia) Sol:

Method 1: Assume a potential V ( x ) = f ( x ) as shown in Fig. 1.16. We take a square-well potential V ′( x ) in the potential V ( x ) such that V ′( x ) = −V0 ,

| x |< a,

V ′( x ) = 0,

| x |> a,

V ′( x ) ≥ f ( x )

for all x .

Fig. 1.16 We know there is at least a bound state ϕ ( x ) in the well potential V ′( x ) for which

ϕ ( x )| H ′ | ϕ ( x ) = ϕ | p 2 / 2m + V ′( x )| ϕ = E0 < 0.

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We then have

ϕ | H | ϕ = ϕ p 2 / 2m + f ( x ) ϕ ≤ ϕ p 2 / 2m + V ′( x ) ϕ = E0 < 0. Let ψ n−1 ( x ) , ψ n ( x ), ... denote the eigenfunctions of H, and expand

ϕ ( x ) = ∑Cnψ n ( x ). n

As ∞

ϕ | H | ϕ = ∑ | Cn |2 ψn | H | ψn < 0 , n

there is at least an eigenfunction ψ i ( x ) satisfying the inequality 〈ψ i | H |ψ i 〉 < 0 . Hence there exists at least one bound state in V ( x ). Method 2: Let the wave function be

ψ ( x ) = (b / π )1/4 exp(−bx 2 / 2), where b is an undetermined parameter. We have H = b/π

  2 d 2  − bx 2 /2 − bx 2 /2 e e dx + V − ∫ −∞ 2  2m dx  ∞

=  2b / 4m + V , where V = (b / π )1/2 ∫ −∞ V ( x ) exp (−bx 2 ) dx , ∞

and thus  b 1/2 ∂〈 H 〉  2 1 = + 〈V 〉 −   π  ∂b 4m 2b =

∫

+∞ −∞

2

x 2V ( x )e − bx dx

1 2 + 〈V 〉 − 〈 x 2V 〉 = 0, 4m 2b

giving b=

〈V 〉

 2  2 〈 x 2V 〉 −  4m   Substitution in the expression for 〈 H 〉 yields

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 2 2  2 〈 x V 〉 −  〈V 〉 4m   . E = 〈H 〉 =  2 2  2 〈 x V 〉 −  4m   As V ( x ) ≤ 0 for all x, V(±∞ ) = 0 , and V is not everywhere zero, we have 〈V 〉 < 0, 〈 x 2V 〉 < 0 and hence E < 0, b > 0 . In fact, under the condition that the total energy has a certain negative value (which must be greater than 〈V 〉 to make 〈T 〉 positive), whatever the form of V a particle in it cannot move to infinity and must stay in a bound state.

1038 The wave function for a particle of mass M in a one-dimensional potential V ( x ) is given by the expression

ψ ( x , t ) = α x exp(−β x ) exp(iγ t / ), x > 0, = 0, x < 0, where α , β and γ are all positive constants. a. Is the particle bound? Explain. b. What is the probability density ρ ( E ) for a measurement of the total energy E of the particle? c. Find the lowest energy eigenvalue of V ( x ) in terms of the given quantities. (MIT ) Sol: a. The particle is in a bound state because the wave function ψ ( x , t ) satisfies lim ψ ( x , t ) = 0,

x →−∞

lim ψ ( x , t ) = lim α xe − β x e iγt/ = 0.

x →+∞

x →+∞

b, c. Substituting the wave function for x > 0 in the Schrödinger equation   2 ∂2  ∂ + V ( x ) ψ ( x , t ) i ψ ( x , t ) = − 2 ∂t  2 M ∂x 

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51

gives 2 ( β 2 x − 2 β ) + V ( x )x , 2M whence the potential for x > 0: −γ x = −

2 (β 2 − 2β /x ) . 2M As the stationary wave function of the particle in V ( x ) satisfies V ( x ) = −γ +

−

 2  d 2 2  2 − β + 2β / x  ψE ( x ) = ( E + γ ) ψE ( x ), ( x > 0) 2 M  dx 

or 2M d2 ψE ( x ) + 2 ( E′+ e 2 / x )ψE ( x ) = 0 dx 2  by setting E′ = E + γ − β 2  2 /2 M , e 2 = β  2 /M , →0 and ψ E ( x ) x → 0 , we see that the above equation is the same as that satis-

fied by the radial wave function of a hydrogen atom with l = 0. The corresponding Bohr radius is a =  2 /Me 2 = 1/β , while the energy levels are En′ = − Me 4 /2 2n 2 = −β 2  2 /2 Mn 2 , n = 1, 2, …. Hence En = −γ + (β 2  2 /2 M ) (1 − 1/n 2 ), n = 1, 2, … , and consequently the lowest energy eigenvalue is E1 = −γ with the wave function

ψ ( x , t ) = α x exp(−β x ) exp(iγ t/ ) ∝ ψE1 ( x ) exp(−iE1t/ ). The probability density ρ( E ) = ψ * ψ = ψE*1ψE1 is therefore 1 for E = −γ , ρ( E ) =  0 for E ≠ −γ .

1039 A particle of mass m is released at t = 0 in the one-dimensional double square well shown in Fig. 1.13 in such a way that its wave function at t = 0 is just one sinusoidal loop (“half a sine wave”) with nodes just at the edges of the left half of the potential as shown.

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Fig. 1.17 a. Find the average value of the energy at t = 0 (in terms of symbols defined above). b. Will the average value of the energy be constant for times subsequent to the release of the particle? Why? c. Is this a state of definite energy? (That is, will a measurement of the energy in this state always give the same value?) Why? d. Will the wave function change with time from its value at t = 0? If “yes”, explain how you would attempt to calculate the change in the wave function. If “no”, explain why not. e. Is it possible that the particle could escape from the potential well (from the whole potential well, from both halves)? Explain. (Wisconsin) Sol: a. The normalized wave function at t = 0 is ψ ( x , 0) = Hˆ

t =0

= −V0 − =

2 2 2m a

∫

2 a

sin πax. Thus

 π x  d2 πx  sin   2 sin   dx 0  a  dx  a  a

 2π 2 − V0 . 2ma

b. 〈 Hˆ 〉 is a constant for t > 0 since ∂ Hˆ

∂t = 0.

c. It is not a state of definite energy, because the wave function of the initial state is the eigenfunction of an infinitely deep square well potential with width a, and not of the given potential. It is a superposition state of the different energy eigenstates of the given potential. Therefore different measurements of the energy in this state will not give the same value, but a group of energies according to their probabilities.

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d. The shape of the wave function is time dependent since the solution satisfying the given conditions is a superposition state: 2 πx  sin   = ∑cnψn ( x ), a  a  n

ψ ( x , 0) =

ψ ( x , t ) = ∑cnψn ( x ) e −iEn t/ . n

The shape of ψ ( x , t ) will change with time because En changes with n. e. The particle can escape from the whole potential well if the following condition is satisfied:  2π 2 /2ma > V0 . That is to say, if the width of the potential well is small enough (i.e., the kinetic energy of the particle is large enough), the depth is not very large (i.e., the value of V0 is not very large), and the energy of the particle is positive, the particle can escape from the whole potential well.

1040 A free particle of mass m moves in one dimension. At time t = 0 the normalized wave function of the particle is

ψ ( x , 0, σ x2 ) = (2πσ x2 )−1/4 exp( − x 2 / 4σ x2 ), 2 2 where σ x = 〈 x 〉 . 2 2 a. Compute the momentum spread σ p = 〈 p 〉 − 〈 p 〉 associated with

this wave function. b. Show that at time t > 0 the probability density of the particle has the form 2

2

ψ ( x , t ) = ψ ( x , 0, σ x2 + σ p2t 2 / m2 ) . c. Interpret the results of parts (a) and (b) above in terms of the uncertainty principle. (Columbia) Sol: a. As  d  ψ * −i ψ dx  dx   x  − 2xσ22 ∞ 1 = −i ∫ − 2  e x dx = 0, 1/2  −∞ (2πσ x2 )  2σ x 

p =

∫

∞

−∞

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54

Problems and Solutions in Quantum Mechanics ∞  d2  p 2 = ∫ ψ *  −  2 2  ψ dx −∞  dx 

 1 x2  − +   1/2 −∞ ( 2πσ x2 )  2σ x2 4σ x4 

= −2 ∫ × e

σp =

−

p2

1

∞

x2 2 σ x2

dx =  2 / 4σ x2 ,  . = 2σ x

b. By Fourier transform, 1 e − ipx / ψ ( x , 0 ) dx ( 2π  )1/2 ∫ 1 1 − ipx /  = 1/2 ∫ e (2πσ x2 )1/4 ( 2π  )

ψ ( p, 0 ) =

× exp ( − x 2 / 4σ x2 ) dx

= ( 2πσ x2 ) 

1/4

Then

2π   exp  −σ x2 p 2  2  . 

ψ ( p, t ) = ψ ( p, 0) e − iEt/ ,

where E=

p2 2m

for a free particle. By inverse Fourier transformation

ψ ( x , t ) = ∫ e ipx /ψ ( p, t ) dp =

( 2πσ )

2 1/4 x

2 (π  )

 σ 2 p2  σ2 × exp  − x 2  dp =  x      2π 

1/4

∫e

ipx / 

 p2  t exp  −i  2m  1

t   2  σ x + i  2m 

1/2

x2   × exp  − .  t    4  σ x2 + i  2m    

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|ψ ( x , t ) |2 =

 x2 1 exp  − 2 2 σ 2t 2  σ pt   2 σ x2 + p 2π  σ x2 + 2   m2 m   1

(

55

   

)

= |ψ x , 0, σ x2 + σ p2t 2m −2 |2 . c. Discussion: i

The results indicate the width of the Gaussian wave packet at time t

(which was originally σ x at t = 0 ) is

σ x2 + σ p2t 2 m 2 , 2 2 2 where σ p =  4σ x .

ii As σ xσ p =  2 , the uncertainty principle is satisfied.

1041 A particle of mass m moves in one dimension under the influence of a potential V ( x ) . Suppose it is in an energy eigenstate ψ ( x ) = (γ 2 π ) exp ( −γ 2 x 2 2) 1/4

with energy E =  2γ 2 2m. a. Find the mean position of the particle. b. Find the mean momentum of the particle. c. Find V ( x ). d. Find the probability P ( p ) dp that the particle’s momentum is between p and p + dp. (Wisconsin) Sol: a. The mean position of the particle is x =

∫

∞ −∞

ψ * ( x ) xψ ( x ) dx =

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γ π

∫

∞ −∞

xe −γ

2 2

x

dx = 0 .

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b. The mean momentum is   d ψ * ( x )  ψ ( x )  dx   i dx γ  ∞ −γ 2 x 2 2 d −γ 2 x 2 2 = e dx = 0. ∫ e dx i π −∞

p =

∫

∞

−∞

(

)

c. The Schrödinger equation  2 d 2  + V ( x ) ψ ( x ) = Eψ ( x ) − 2  2m dx  can be written as −

2 d 2 ψ ( x ) = E − V ( x )ψ ( x ) . 2m dx 2

As −

2 2  2 d 2 −γ 2 x 2 /2 2 e =− (−γ 2 + γ 4 x 2 ) e −γ x /2 , 2 2m dx 2m

we have E − V (x ) = −

2 (−γ 2 + γ 4 x 2 ) , 2m

or V (x ) =

 2 4 2 2  2γ 2  2γ 4 x 2 (γ x − γ ) + 2m = 2m . 2m

d. The Schrödinger equation in momentum representation is  p 2  4γ 4 d 2  −   ψ ( p ) = Eψ ( p ). 2  2m 2m dp  Letting

ψ ( p) = Ne − ap

2

and substituting it into the above equation, we get 2 2 p 2 −ap2  4γ 4 e − −2a + 4a 2 p 2 ) e −ap = Ee −ap, ( 2m 2m

or 4a 2 p 2 − 2a =

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 1  p2 − 1 . 2 2  2 2  γ  γ 

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57

As the parameter a is independent of p, the above relation can be satisfied by a = 1/ 2 2γ 2 . Hence

ψ ( p) = N exp (− p 2 /2 2γ 2 ). This is the eigenfunction of the state with energy  2γ 2 /2m in the momentum

representation. Normalization gives N = (1/  2γ 2π ) . Thus the probability 1/4

that the particle momentum is between p and p + dp is 12

 1   p2  P( p ) dp = ψ ( p ) dp =  2 2  exp − 2 2  dp.  γ π   γ  2

Note that ψ ( p) can be obtained directly by the Fourier transform of ψ ( x ) :

ψ( p) = =

1/4

∫

γ 2  2 2 dx e −ip ⋅ x /   e −γ x /2 1/2 π π 2    ( )

∫

2 1/4  p dx  γ 2  iγ x    p 2   exp − exp −      1/2 2    2 2γ 2   2γ ( 2π  )  π  1/4

 1  2 =  2 2  e − p /2 2γ 2 .  γ π 

1042 In one dimension, a particle of mass m is in the ground state of a potential which confines the particle to a small region of space. At time t = 0, the potential suddenly disappears, so that the particle is free for time t > 0. Give a formula for the probability per unit time that the particle arrives at time t at an observer who is a distance L away. (Wisconsin) Sol:

Let ψ 0 ( x ) be the wave function at t = 0 . Then  −ip 2t  ψ ( x , t ) = x exp  ψ 0 (x )  2m  =∫

+∞

=∫

+∞

−∞

−∞

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x e − ip t /( 2m ) x ′ dx ′ x ′ ψ 0 2

x e − ip t /( 2m ) x ′ ψ 0 ( x ′ ) dx ′ , 2

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where  p 2t  x exp  −i x′  2m  = ∫∫

+∞

−∞

=∫

+∞

−∞

 p 2t  x p′ dp′ p′ exp  −i p dp p | x ′  2m 

1 px ′ p 2t   px exp i dp −i −i 2π  2m    

2 1 +∞ t ( x ′ − x ) m      2 m  exp −i  p = +   dp ⋅ exp i( x ′ − x ) ∫ −∞ 2π  2m 2t  2 t      

=

1 2m 2 m  +∞ − iq 2  exp i ( x ′ − x ) e dq 2π  2t  ∫−∞ t 

=

1− i m m   exp i( x ′ − x )2 . 2 π t 2t  

Thus

ψ (x , t ) =

1−i m 2 π t

∫

+∞ −∞

 2 m  expi ( x ′− x ) ⋅ ψ0 ( x ′)dx ′.  2t 

Represent the particle as a Gaussian wave packet of dimension a:

ψ 0 ( x ) = (π a 2 )

−1/4

exp ( − x 2 / 2a 2 ) .

The last integral then gives 1 (1 − i ) m ψ (x , t ) = 1 − i mt π 1/4 1 a1/2 ( ) π 2 ψ (x , t ) = 2  π t 2π 1/4a1/2    mx m   × expimx 2 1 −  m i m ht 2    ht + 2    2 ( ) 2 t × exp i 1 − 2 a   m i  2ht  2ht ( 2 t + 2a2 )   +∞ iξ 2 1 e dξ × ∫ +∞ −∞ iξ 2 m1 i × 2 t + 2a2 ∫ −∞ e dξ m i 2 t + 2 a 2  mx 2 1  1 m exp−mx 22 1 t , = 1/21 1/4 t m = a1/2 π1/4 m + iat 2 exp − 2a2 m + iat 2, a π m + i a2  2a m + i a2 

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59

whence the current density   2 xt p  1 j = Re ψ * x ψ  = 5 2   m  π a m 1 + 22t 24 m a

(

 2 x 1 × exp − 2  a 1 + t  ma2

( )

)

3/2

 . 2  

By putting x = L, we get the probability per unit time that the particle arrives at the observer a distance L away.

1043 A free particle of mass m moves in one dimension. The initial wave function of the particle is ψ ( x, 0) . a. Show that after a sufficiently long time t the wave function of the particle spreads to reach a unique limiting form given by

ψ ( x , t ) = m/t exp (−i π /4 ) exp (imx 2 /2t )ϕ ( mx /t ) ,

where ϕ is the Fourier transform of the initial wave function:

ϕ ( k ) = (2 π )

−1/2

∫ ψ ( x , 0) exp (−ikx ) dx .

b. Give a plausible physical interpretation of the limiting value of |ψ ( x , t ) |2 . Hint: Note that when α → ∞, exp ( − iα u2 ) → π /α exp ( − iπ /4 )δ (u ). (Columbia) Sol: a. The Schrödinger equation is i ∂/∂t + (h 2 /2m) d 2 /dx 2 ψ ( x , t ) = 0 .   By Fourier transform, we can write ∞ 1 ψ (k, t ) = dxe −ikxψ ( x , t ), ∫ −∞ 2π

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and the equation becomes  ∂ 2k2  i  − ψ (k , t ) = 0 .  ∂t 2m  Integration gives  k 2 t  ψ ( k , t ) = ψ ( k , 0) exp −i ,  2m  where 1 2π

ψ ( k , 0) =

∫

∞ −∞

dxe −ikxψ ( x , 0) ≡ ϕ ( k ) .

Hence  k 2 t  ψ ( k , t ) = ϕ ( k ) exp −i ,  2m  giving

ψ (x , t ) =

1 2π

∫

1 2π

∫

∞ −∞

dke ikxψ ( k , t )

 k 2ht  dkϕ ( k ) exp ikx − i  −∞ 2m    mx 2  ∞ 1 exp i =  ∫ dk 2π  2t  −∞  t  mx 2  × exp −i k −  ϕ (k ). t    2m 

=

∞

With ξ = k − mx/t , this becomes

ψ (x , t ) =

 mx 2  ∞ 1 exp i  ∫ dξ 2π  2t  −∞  t 2   mx  ξ ϕ ξ + × exp −i .  2m   t 

For α → ∞, 2

e −iαu →

 π π exp −i δ (u) ,  4 α

and so after a long time t (t → ∞),  π  t 2  2π m exp −i ξ → δ (ξ ) exp −i  ,  4  2m  t

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and  mx 2  ∞ 1 2π m exp i  ∫ −∞ dξ t 2π  2 t   mx   π × δ (ξ ) ϕ ξ +  exp −i    4 t  2  mx  2π m  π   mx  1 exp i exp −i ϕ  =    4   t  2π  2 t  t

ψ (x , t ) =

 mx 2   mx   π m exp −i  exp i . ϕ   4 t  2 t   t 

= b.

2

ψ (x , t ) =

2

m  mx  ϕ  . t  t 

Because ϕ (k ) is the Fourier transform of ψ ( x , 0) , we have

∫

+∞

| ϕ (k )|2 dk = −∞

+∞

∫ | ψ ( x , 0) |

2

dx .

−∞

On the other hand, we have

∫

+∞ −∞

ψ ( x , t ) dx =

∫

−∞

=

∫

−∞

2

+∞

+∞

2

m  mx  ϕ   dx t  t  | ϕ ( k ) |2 dk = ∫ | ψ ( x , 0)|2 dx , +∞

−∞

which shows the conservation of total probability. For the limiting case of t → ∞ , we have | ψ ( x , t ) |2 → 0 ⋅ | ϕ (0) |2 = 0, which indicates that the wave function of the particle will diffuse infinitely.

1044 The one-dimensional quantum mechanical potential energy of a particle of mass m is given by V ( x ) = V0δ ( x ), V ( x ) = ∞,

− a < x < ∞, x < −a ,

as shown in Fig. 1.18. At time t = 0 , the wave function of the particle is completely confined to the region − a < x < 0 . [Define the quantities k = 2mE / and α = 2mV0 / 2 ]

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a. Write down the normalized lowest-energy wave function of the particle at time t = 0 . b. Give the boundary conditions which the energy eigenfunctions

ψ k ( x ) = ψ kI ( x )   and  ψ k ( x ) = ψ kII ( x ) must satisfy, where the region I is − a < x < 0 and the region II x ≥ 0.

Fig. 1.18 c. Find the (real) solutions for the energy eigenfunctions in the two regions (up to an overall constant) which satisfy the boundary conditions. d. The t = 0 wave function can be expressed as an integral over energy eigenfunctions: ∞ ψ ( x ) = ∫ f ( k )ψk ( x )dk . −∞

Show how f ( k ) can be determined from the solutions ψ k ( x ) . e. Give an expression for the time development of the wave function in terms of f ( k ) . What values of k are expected to govern the time behavior at large times? (Wisconsin) Sol: a. The required wave function ψ ( x ) must satisfy the boundary conditions

ψ (−a) = ψ (0) = 0 . A complete orthonormal set of wave functions defined in −a < x < 0 and satisfying the Schrödinger equation consists of  2  sin ( nπa x ) , φn ( x ) =  a 0, 

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−a < x < 0, outside[−a, 0],

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63

where n = 1, 2,… , with 〈φn | H | φm 〉 = Enδmn , En = ( 2 /2m)(nπ /a)2 . The normalized lowest-energy wave function is given by n = 1 as  2  sin ( πax ) , ψn ( x ) =  a 0, 

−a < x < 0, outside[−a, 0].

b. The Schrödinger equation for x > −a is −

2 d 2 ψ + V0δ ( x )ψ = Eψ , 2m dx 2

or

ψ ″( x ) + k 2ψ ( x ) = αδ ( x )ψ ( x ) with 2mE 2mV , α= 2 0 . 2  The boundary conditions and the discontinuity condition to be satisfied are k2 =

ψ I (−a) = 0, ψ I (0) = ψ II (0) , ψ II (+∞) = finite, ψ II′ (0) − ψ I′ (0) = αψ I (0). The last equation is obtained by integrating the Schrödinger equation over a small interval [− ,  ] and letting  → 0 (see Problem 1020). c. In both the regions I and II, the wave equation is

ψ ″( x ) + k 2ψ ( x ) = 0 , whose real solutions are sinusoidal functions. The solutions that satisfy the boundary conditions are ψkI ( x ) = ck sin k( x + a), −a < x < 0,  II ψk ( x ) = ψk ( x ) = ck sin k( x + a) + Ak sin kx , x ≥ 0,  x < −a. 0, The discontinuity and normalization conditions then give cα Ak = k sin ka, k −1

 π  α sin 2ka  α sin ka  2   2 + ck =  1 +   .  k    k  2  d. Expand the wave function ψ ( x ) in terms of ψ k ( x ),

ψ (x ) =

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∫

∞ −∞

f (k )ψ k ( x ) dk ,

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Problems and Solutions in Quantum Mechanics

and obtain

∫

∞ −a

∫ ∫ f (k )ψ (x )ψ * (x )dkdx = ∫ f (k )δ (k − k′) dk = f (k′),

ψk*′ ( x )ψ ( x ) dx =

k

k′

∞

−∞

or

∫

−a

ψ ( x , 0) =

∫

f (k ) =

∞

ψk* ( x )ψ ( x ) dx .

e. As ∞ −∞

f (k )ψk ( x ) dk ,

we have

ψ(x , t ) =

∫

+∞ −∞

f (k )ψk ( x )e −iEk t/ dk.

At time t = 0, the particle is in the ground state of an infinitely deep square well potential of width a, it is a wave packet. When t > 0, since the δ ( x ) potential barrier is penetrable, the wave packet will spread over to the region x > 0. Quantitatively, we compute first

πx 2 dx sin a a  1 0   π  = ck cos  k −  x + ka ∫ −a  a 2a  

f (k ) =

∫

0

c sin k( x + a) ⋅

−a k

 π   − cos  k +  x + ka  dx  a   =

π 2 sin ka . 2 ck a a k 2 − ( πa )

and then

ψ(x , t ) =

π a

2 a

∫

−∞ −∞

ck

sin ka

k − ( πa ) 2

2

 sin k( x + a) × α  sin k( x + a) + k sin ka sin kx

  e −iEk t /dk , 

where Ek =  2 k 2 /2m . In the last expression the upper and lower rows are for regions I and II respectively. When t → ∞ , the oscillatory factor exp(−iEkt/ ) changes even more rapidly, while the other functions of the integrand behave quite normally (k = π /a is not a pole). Thus ψ ( x , t ) tend to zero for any given x. When t is very large,

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component waves of small wave number k play the principal role. At that time the particle has practically escaped from the region [−a, 0] .

1045 The radioactive isotope

83

Bi

212

decays to

81

Tl208 by emitting an alpha particle

with the energy E = 6.0 MeV . a. In an attempt to calculate the lifetime, first consider the finite potential barrier shown in Fig. 1.19. Calculate the transition probability T for a particle of mass m incident from the left with energy E in the limit T  1. b. Using the above result, obtain a rough numerical estimate for the lifetime of the nucleus Bi212. Choose sensible barrier parameters to approximate the true alpha particle potential. (CUS)

Fig. 1.19 Sol: a. If T  1, the incident wave is reflected at x = 0 as if the potential barrier were infinitely thick. We thus have

ψ ( x ) = e ikx + (t1 − 1)e −ikx , ψ ( x ) = t1e

− k ′x

x < 0, 0 < x < b,

,

where t1 is the amplitude transmission coefficient and k′ =

2m(V0 − E ) , 2

k=

2mE . 2

The continuity of ψ ′( x ) at x = 0 gives ik(2 − t1 ) = −k′t1 , or t1 =

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Consider the reflection at b. We have

ψ ( x ) = t1e − k ′b[e − k ′ ( x −b ) + (t 2 − 1)e k ′ ( x −b ) ],

0 < x < b,

ψ ( x ) = t1t 2e

x > b,

− k ′b ik ( x −b )

e

,

and so − k ′(2 − t 2 ) = ikt 2 , or t 2 = 2ik′/(k + ik′). Hence the transition probability is given by T = t1t 2e − k ′b to be | T |2 =

16k 2 k ′ 2 −2 k ′b 16 E(V0 − E ) −2 k ′b e = e . (k 2 + k ′ 2 )2 V02

b. To estimate the rate of α -decay of

83

Bi 212 , we treat, in first approximation,

the Coulomb potential experienced by the α -particle in the

81

Tl nucleus as

a rectangular potential barrier. As shown in Fig. 1.20, the width of the barrier r0 can be taken to be 2 Ze 2 2(83 − 2) e 2 c = E 6 c MeV 162 1 = × × 6.58 × 10−22 × 3 × 1010 6 137 = 3.9 × 10−12 cm.

r0 =

Fig. 1.20 The radius of the nucleus of Tl is 1

R = 1 × 10−13 × 208 3 = 6 × 10−13 cm,

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corresponding to a Coulomb potential height of V=

2 Ze 2 r0 ⋅ = 39 MeV . r0 R

An α -particle, moving with speed υ in the nucleus of Tl, makes v colli2R sions per second with the walls. Hence the lifetime τ of 83 Bi 212 is given by

τ | T |2 or

τ

υ ≈ 1, 2R

2R . υ | T |2

Taking for the rectangular potential barrier a height V0 ≈ 12 (39 − 6) + 6 = 22.5 MeV, b = r0 − V0 ≈ 12 (39 − 6) + 6 = 22.5 MeV, b = r0 − R = 33 × 10−13 cm (see Fig. 1.20), v =

2E = m

2×6 c ≈ 0.1 c , 940

we find 2k ′b =

2 2mc 2 (V0 − E ) 2 2 × 940 × 16.5 × 33 × 10−13 b= = 59, c 6.58 × 10−22 × 3 × 1010

τ=

2 × 6 × 10−13 22.52 × × e 59 3 × 109 16 × 6 × (22.5 − 6)

= 5.4 × 103 s.

1046 A particle of mass m is in an infinite potential well extending from 0 to a. If the initial wave function of the particle is ψ ( x ) = x ( a − x ) apart from normalization, what is the wave function after time t = Tsec? Sol:

∫ ψ 2 dx = 1



A2 ∫ x 2 (a − x ) = 1 2

a



A2 ∫x 2 ( a 2 − 2ax + x 2 ) dx = 1 0

A2

a5 =1 30

A=

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ψ ( x ,0) = ∫ 0

30 2 nπx sin x (a − x ) a5 a a a

=

30 2 nπ x x ( a − x ) sin dx a5 a ∫0 a

=

30 2 nπ x nπ x xa sin − ∫x 2 sin a5 a ∫0 a a 0

=

2 30 2  nπ x a nπ x  a   a  − x cos × + sin    5 a a  a nπ a nπ 

a

a

a

2 3  nπ x  a  nπ x  a   nπ x  a  2 sin 2cos x −  − x 2 cos + × + ×        a  nπ  a nπ a nπ 0 



=

0 0 3    a3 0  a  3 30 2  2a3 n a n h    1 0 1 1 0 2 − + − − + × + ( ) ( )   ( −1)   2 5 a a nπ (nπ )   nπ  nπ   

=

3 120 8 15 60   a   n n n 2 1 1 − − ( )    = 3 3 1 − ( −1)  = 3 3 1 − ( −1)  .  a6  nπ  n π nπ

ψ ( x ,0) = ∑ Cn ψ ( x ,t ) .

ψ ( x ,t ) =

8 15 1 nπ x − iET ∑ 3 sin e  . 3 a π n=odd n

1047 Consider a one-dimensional square-well potential (see Fig. 1.21) V ( x ) = 0, x < 0, V ( x ) = −V0 , 0 < x < a, V ( x ) = 0, x > a, where V0 is positive. If a particle with mass m is incident from the left with nonrelativistic kinetic energy E, what is its probability for transmission through the potential? For what values of E will this probability be unity? (Columbia)

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Fig. 1.21 Sol:

Let the wave function be

ψ ( x ) = e ikx + Re − ikx ,

x < 0,

ψ ( x ) = Se ,

x > a,

ikx

ψ ( x ) = Ae

ik ′x

+ Be

− ik ′x

,

0 < x < a,

where k=

2m( E + V0 ) 2mE . , k′ =  

The constants R , S , A, B are to be determined from the boundary conditions that ψ ( x ) and ψ ′( x ) are both continuous at x = 0 and x = a, which give       

1 + R = A + B, k(1 − R ) = k ′( A − B), Ae ik ′a + Be − ik ′a = Se ika , k ′( Ae ik ′a − Be − ik ′a ) = kSe ika .

Hence S=

4 kk ′e − ika (k + k ′ )2 e − ik ′a − (k − k ′ )2 e ik ′a

and the probability for transmission is j P = t = | S |2 ji =

4k 2k ′ 2 . 4(kk ′ cos k ′a) + (k 2 − k ′ 2 )2 sin 2 (k ′a)

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Resonance transmission occurs when k ′a = nπ , i.e., when the kinetic energy E of the incident particle is E = n 2 π 2  2 /2ma 2 − V0 . The probability for transmission, P, then becomes unity.

1048 Consider a one-dimensional square-well potential (see Fig. 1.22):

Fig. 1.22 V ( x ) = ∞, V ( x ) = −V0 , V ( x ) = 0,

x < 0, 0 < x < a, x > a.

a. For E < 0 , find the wave function of a particle bound in this potential. Write an equation which determines the allowed values of E. b. Suppose a particle with energy E > 0 is incident upon this potential. Find the phase relation between the incident and the outgoing wave. (Columbia) Sol:

The Schrödinger equations for the different regions are  2 d 2  − V0 − E  ψ ( x ) = 0, 0 < x < a, − 2  2m dx   2 d 2  − E  ψ ( x ) = 0, x > a. − 2  2m dx 

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a. E < 0. 〈i 〉 Consider first the case of V0 < − E , for which the wave function is 0,  ψ ( x ) =  A sinh(kx ),  − k ′x  Be , where k=

x < 0, 0 < x < a, x > a,

2m(−V0 − E ) 2m(− E ) . , k′ = 2  2

The continuity conditions of the wave function give A sinh(ka) = Be − k ′a , Ak cosh(ka) = −Bk′e − k ′a and hence k coth(ka) = −k′ . As coth x > 0 for x > 0 , there is no solution for this case. 〈ii 〉 For V0 > −E , ik → k , k = 2m(V0 + E )/ , and the equation determining the energy becomes k cot(ka) = − k ′ . The wave function is 0,  ψ ( x ) =  A sin(kx ),  − k ′x  Be ,

x < 0, 0 < x < a, x > a.

From the continuity and normalization of the wave function we get 1/2

  2 A = 1 2  ,  k ′ sin (ka) + a − 21k sin(2ka)  1/2

  k ′a 2 B = 1 2  e sin(ka) .  k ′ sin (ka) + a − 21k sin(2ka)  b. E > 0 . The wave function is 0,  ψ ( x ) =  A sin(kx ),   B sin(k ' x + ϕ ),

x < 0, 0 < x < a, x > a,

where k=

2m( E + V0 ) , k ′ = 2mE /  2 . 2

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As ∂ ln ψ/∂ ln x is continuous at x = a , (ka)cot(ka) = (k ′a)cot(k ′a + ϕ ), whence k  ϕ = arccot  cot(ka) − k ′a .  k′  For x > a,

ψ (x ) =

B − ik ′x −iϕ B ik ′x +iϕ e − e , 2i 2i

where

ϕ inc ( x ) ∝ e − ik ′x −iϕ , ϕ out ( x ) ∝ e ik ′x +iϕ . Hence the phase shift of the outgoing wave in relation to the incident wave is    k δ = 2ϕ = 2 arccot  cot(ka) − k′a.   k′  

1049 Consider a one-dimensional system with potential energy (see Fig. 1.23) V ( x ) = V0 , x > 0, V ( x ) = 0, x < 0, where V0 is a positive constant. If a beam of particles with energy E is incident from the left (i.e., from x = −∞ ), what fraction of the beam is transmitted and what fraction reflected? Consider all possible values of E. (Columbia)

Fig. 1.23 Sol:

For x < 0 , the Schrödinger equation is d2 2mE ψ + 2 ψ =0, 2 dx 

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whose solution has the form

ψ ( x ) = e ikx + re − ikx , where k=

2mE . 2

For x > 0, the equation is d2 2m( E − V0 ) ψ+ ψ = 0. dx 2 2 〈i 〉 If E < V0 , write the above as d2 2m(V0 − E ) ψ− ψ =0. dx 2 2 As ψ ( x ) must be finite for x → ∞ , the solution has the form

ψ ( x ) = te − k ′x , where k′ = The continuity conditions then give

2m(V0 − E ) . 2 1+ r = t,

ik − ikr = −tk ′ , whence r = (k ′ + ik ) / (ik − k ′ ) = (1 − ik ′ / k ) / (1 + ik ′ / k ). Therefore the fraction reflected is R = jref / jinc = | r |2 = 1, the fraction transmitted is T = 1 − R = 0. 〈ii 〉 E > V0 . For x > 0, we have  d 2 2m( E − V0 )   2+ ψ ( x ) = 0 . 2  dx  where

ψ ( x ) = te ik ′x, k′ =

2m( E − V0 ) . 2

Noting that there are only outgoing waves for x → ∞ , we have 1 + r = t , ik − ikr = ik ′t , and thus r = (k ′ − k ) / (k ′ + k ) . Hence the fraction reflected is R = [(k ′ − k ) / (k ′ + k )]2 , the fraction transmitted is T = 1 − R = 4 kk ′ / (k + k ′ )2 .

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1050 A particle of mass m and momentum p is incident from the left on the potential step shown in Fig. 1.24. Calculate the probability that the particle is scattered backward by the potential if a. p 2 /2m < V0 , b. p 2 /2m > V0 . (Columbia)

Fig. 1.24 Sol:

The Schrödinger equations are  d 2 2mE   2 + 2  ψ ( x ) = 0 for x < x 0 ,    dx  d 2 2m   2 + 2 ( E − V0 ) ψ ( x ) = 0 for x > x 0 .   dx  a. If E < V0 , we have e ik ( x − x0 ) + re − ik ( x − x0 ) , ψ ( x ) =  − k ′( x −x ) 0 , te

x < x0 , x > x0 ,

where k=

2mE , 2

k′ =

2m(V0 − E ) , 2

the condition that ψ ( x ) is finite for x → ∞ having been made use of. The continuity conditions give 1 + r = t , ik − ikr = − k ′t , whence r = (k ′ + ik ) / (ik − k ′ ). The probability of reflection is R = jr / ji =| r |2 = 1.

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b. If E > V0 . We have e ik ( x − x0 ) + re − ik ( x − x0 ) , ψ ( x ) =  ik ′ ( x − x ) 0 , te

x < x0 , x > x0 ,

where k=

2mE , 2

k′ =

2m( E − V0 ) , 2

noting that there is only outgoing wave for x > x 0 . The continuity conditions give 1 + r = t , ik − ikr = ik ′t , and hence r = (k − k ′ ) / (k + k ′ ) . The probability of reflection is then R =| r |2 = [(k − k ′ ) / (k + k ′ )]2 .

1051 Find the reflection and transmission coefficients for the one-dimensional potential step shown in Fig. 1.25 if the particles are incident from the right. (Wisconsin)

Fig. 1.25 Sol:

As the particles are incident from the right we must have E > V0 . And there are both incident and reflected waves in the region x > 0. The Schrödinger equation for x > 0,

ψ ″( x ) + k12ψ ( x ) = 0, where k1 = 2m( E − V0 ) /  , has solutions of the form

ψ = exp(−ik1 x ) + R exp(ik1x ) .

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There are only transmitted waves in the region x < 0, where the Schrödinger equation is

ψ ″( x ) + k22ψ ( x ) = 0 with k2 = 2mE / , and has the solution

ψ ( x ) = S exp(−ik2 x ). Using the continuity conditions of the wave function at x = 0 , we get 1 + R = S . From the continuity of the first derivative of the wave function, we get k1 (1 − R ) = k2 S . Hence R = (k1 − k2 ) / (k1 + k2 ) , giving the reflection coefficient 2

k −k V02 , |R| = 1 2 = k1 + k2 ( E + E − V0 )4 2

and the transmission coefficient | S |2 = 1− | R |2 = 1 −

V02 ( E + E − V0 )

4

.

1052 Consider, quantum mechanically, 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 and 3E / 4 for x > 0 . What fraction of the particles are reflected at x = 0? (Buffalo) Sol:

The Schrödinger equations are

ψ ″+ k 2ψ = 0

for

x ≤ 0,

ψ ″+ (k /2) ψ = 0

for

x > 0,

2

where k = 2mE / . As for x < 0 there will also be reflected waves, the solutions are of the form

ψ = exp(ikx ) + r exp(−ikx ), ψ = t exp(ikx / 2),

x ≤ 0, x > 0.

From the continuity conditions of the wave function at x = 0, we obtain 1 + r = t , k(1 − r ) = kt / 2, and hence r = 1/ 3. Thus one-ninth of the particles are reflected at x = 0.

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1053 Consider a particle beam approximated by a plane wave directed along the x-axis from the left and incident upon a potential V ( x ) = γ δ ( x ), γ > 0, δ ( x ) is the Dirac delta function. a. Give the form of the wave function for x < 0. b. Give the form of the wave function for x > 0. c. Give the conditions on the wave function at the boundary between the regions. d. Calculate the probability of transmission. (Berkeley) Sol: a. For x < 0, there are incident waves of the form exp(ikx) and reflected waves of the form R exp(−ikx ) . Thus

ψ ( x ) = exp(ikx ) + R exp(−ikx ),

x < 0.

b. For x > 0, there only exist transmitted waves of the form S exp(ikx ). Thus

ψ ( x ) = S exp(ikx ), x > 0. c. The Schrödinger equation is −

2 d 2 ψ ( x ) + γδ ( x )ψ ( x ) = Eψ ( x ) 2m dx 2

and its solutions satisfy (Problem 1020)

ψ ′(0+ ) − ψ ′(0− ) =

2mγ ψ (0) . 2

As the wave function is continuous at x = 0, ψ (0+ ) = ψ (0− ) . 2 d. From (a), (b) and (c) we have 1 + R = S , ikS − ik(1 − R ) = 2mγ S /  , giving

S = 1/ (1 + imγ /  2 k ). Hence the transmission coefficient is −1

−1

 m 2γ 2   mγ 2  , T = | S |2 = 1 + 4 2  = 1 + 2  k   2 E  where E =  2 k 2 /2m .

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1054 Consider a one-dimensional problem of a particle of mass m incident upon a potential of a shape shown in Fig. 1.26. Assume that the energy E at x → −∞ is greater than V0, where V0 is the asymptotic value of the potential as x → ∞. Show that the sum of reflected and transmitted intensities divided by the incident intensity is one. (Princeton)

Fig. 1.26 Sol:

As E > V0 we may assume the asymptotic forms

ψ → e ikx + re −ikx ψ → te

iβ x

for x → −∞, for x → +∞,

where r , t , k , β are constants. The incident intensity is defined as the number of particles incident per unit time: I = k/m. Similarly, the reflected and transmitted intensities are respectively R = | r |2 k / m, T = | t |2 β / m . Multiplying the Schrödinger equation by ψ *, −

2 * 2 ψ ∇ ψ + ψ *V ψ = Eψ *ψ , 2m

and the conjugate Schrödinger equation by ψ , −

2 ψ∇ 2ψ * + ψ *V ψ = Eψ *ψ , 2m

and taking the difference of the two equations, we have

ψ * ∇ 2ψ − ψ∇ 2ψ * = ∇ ⋅ (ψ * ∇ψ − ψ∇ψ *) = 0. This means that f ( x ) = ψ *dψ/dx − ψdψ * /dx is a constant. Then equating f (+∞ ) and f (−∞ ), we find k(1− | r |2 ) = β | t |2 .

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Multiplying both sides by  gives m I = R +T .

1055 A Schrödinger equation in one dimension reads (−∂2 / ∂ x 2 − 2sech2 x )ψ = εψ (  = 1, m = 1/ 2) . a. Show that exp(ikx )(tanh x + const) is a solution for a particular value of the constant. Calculate the S-matrix (transmission and reflection coefficients) for this problem. b. The wave function sech x happens to satisfy the Schrödinger equation. Calculate the energy of the corresponding bound state and give a simple argument that it must be the ground state of the potential. c. Outline how you might have proceeded to estimate the ground-state energy if you did not know the wave function. (Buffalo) Sol: a. Letting the constant in the given solution ψ be K and substituting ψ in the Schrödinger equation, we obtain k 2 (tanh x + K ) − 2(ik + K )sech 2 x = ε (tanh x + K ) . This equation is satisfied if we set K = −ik and ε = k 2 . Hence

ψ ( x ) = e ikx (tanh x − ik ) is a solution of the equation and the corresponding energy is k 2 . Then as tanh x → 1 for x → ∞ and tanh(− x ) = − tanh x we have

ψ = (1 − ik )e ikx

as

x → ∞,

ψ = −(1 + ik )e

as

x → −∞.

ikx

Since V ( x ) ≤ 0, ε > 0 , the transmission coefficient is T = 1 and the reflection coefficient is R = 0 as the particle travels through V ( x ). So the S-matrix is  1 −(1 − ik ) / (l + ik )   −(1 − ik ) / (l + ik ) 0 

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b. Letting ψ = sech x in the Schrödinger equation we have −ψ = εψ . Hence

ε = −1. Because sech x is a non-node bound state in the whole coordinate space, it must be the ground state. c. We might proceed by assuming a non-node bound even function with a parameter and obtain an approximate value of the ground state energy by the variational method.

1056 A monoenergetic parallel beam of nonrelativistic neutrons of energy E is incident onto the plane surface of a plate of matter of thickness t. In the matter, the neutrons move in a uniform attractive potential V. The incident beam makes an angle θ with respect to the normal to the plane surface as shown in Fig. 1.27. a. What fraction of the incident beam is reflected if t is infinite? b. What fraction of the incident beam is reflected if V is repulsive and V = E? Consider t finite. (CUS)

Fig. 1.27 Sol: a. Let k0 be the wave number of an incident neutron, given by k02 =

2mE . For 2

x < 0 , the wave function is

ψ1 ( x , y ) = e ik0 x cosθ +ik0 y sinθ + R e−ik0 x cosθ +ik0 y sinθ .

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With t infinite and the potential negative, for x > 0 the Schrödinger equation is −

2 2 ∇ ψ = ( E + V )ψ . 2m

Assuming a solution

ψ 2 ( x , y ) = Te

ikx x +ik y y

and substituting it the equation, we obtain kx2 + k y2 = 2m( E + V ) /  2 . The boundary conditions at x = 0 ψ 1 (0, y ) = ψ 2 (0, y ), ∂ψ 1 ∂x

= x =0

∂ψ 2 ∂x

, x =0

then give e ik0 y sinθ + Reik0 y sinθ = Te ik0 cosθ e

ik0 y sinθ

− Rik0 cosθ e

ik y y

ik0 y sinθ

,

= Tikx e

ik y y

,

As the potential does not vary with y , k y = k0 sinθ and the above become 1+ R = T, k0 (1 − R ) cosθ = kxT , which give R = ( k0cosθ − kx ) / ( k0cosθ + kx ). The probability of reflection is then P = | R |2 = (k0cosθ − kx )2 /(k0cosθ + kx )2, with kx2 = 2m ( E + V )/ 2 − k02 sin 2θ , k02 = 2mE /  2 kx2 = 2m ( E + V )/ 2 − k02 sin 2θ , k02 = 2mE /  2 . b. For x < 0 the wave function has the same form as that in (a). For 0 < x < t , E − V = 0, and the Schrödinger equation is − (  2 /2m) ∇ 2ψ = 0 .

As the potential is uniform in y we assume ψ = exp (ik ′y ) exp ( kx ), where k ′ = k0 sin θ . Substitution gives − k ′ 2 + k 2 = 0 , or k = ±k ′. Hence the wave function for 0 < x < t is

ψ 2 ( x , y ) = (ae k ′x + be − k ′x ) e ik ′y .

Writing ψ ( x , y ) = φ ( x ) e ik ′y, we have for the three regions x < 0, φ1 ( x ) = e ikx x + re − ikx x , 0 < x < t , φ 2 ( x ) = ae k ′x + be − k ′x , x > t , φ3 ( x ) = ce ikx x ,

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with kx =

2mE cos θ , k ′ = 2mE /  2 sin θ . 2

The boundary conditions φ1 (0) = φ 2 (0), φ 2 (t ) = φ3 (t ), dφ1 dx

= x =0

dφ 2 dx

, x =0

dφ 2 dx

= x =t

dφ3 dx

x =t

give 1 + r = a + b, ikx (1 − r ) = k ′ (a − b) ,

c exp(ikx t ) = a exp ( k ′t ) + b exp ( − k ′t ) ,

ikx c exp(ikx t ) = k ′a exp ( k ′t ) − k ′b exp ( − k ′t ) , whose solution is −1

 ik  a / b + 1    ikx  a / b + 1   r = x   − 1 ,  − 1    k′  a / b − 1    k′  a / b − 1   a b = ( k ′ + ikx ) ( k ′ − ikx ) ⋅ e −2 k ′t . Hence r=

e 2 k′t − 1 1 − β 2e 2 k′t

with

β=

k ′ − ikx , k ′ + ikx

and the fraction of neutrons reflected is R |2 = r |2 =

e

2 k ′t

e 2 k ′t + e −2 k ′t − 2 . + e −2 k ′t − 2 cos 4θ

Alternative Sol: The solution can also be obtained by superposition of infinite amplitudes, similar to the case of a Fabry–Perot interferometer in optics (see Fig. 1.28).

Fig. 1.28

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83

We need only consider the x-component of the waves. Let T12 , R12 denote the coefficients of amplitude transmission and reflection as a wave goes from medium 1 to medium 2, respectively. Let T21 , R21 denote coefficients of amplitude transmission and reflection from medium 2 to medium 1. Take the amplitude of incident wave as 1. Then the amplitude of the wave that is transmitted to medium 2 is the sum T = T12e −δ ⋅ T21 + T12e −δ ⋅ R21e −2δ R21 ⋅ T21 + T12e −δ ⋅ ( R21e −2δ R21 )2 ⋅ T21 + 

= T12e −δ T21[1 + R212e −2δ + ( R212e −2δ )2 +  = T12T21e −δ

1 . 1 − R212e −2δ

In the above exp ( −δ ) is the attenuation coefficient of a wave in medium 2, where

δ = k ′t with k ′ = 2mE (1 − cos 2θ )  = 2mE sin θ  . From (a) we have the coefficients of transmission and reflection R12 = ( k1x − k2 x ) ( k1x + k2 x ) ,

T12 = 1 + R12 = 2k1 ( k1x + k2 x ) . As k1x = 2mE cos θ  , k2 x = ik ′ = i 2mE sin θ  , we find 2 cos θ = 2 cos θ e − iθ , cosθ + i sin θ 2i sin θ T21 = 2k2 x ( k1x + k2 x ) = = 2i sin θ e − iθ , cos θ + i sin θ i sin θ − cos θ = e −2iθ , R21 = ( k2 x − k1x ) ( k1x + k2 x ) = i sin θ + cos θ T12 = 2k1x ( k1x + k2 x ) =

and hence T = T12T21e −δ

1 1 − R212e −2δ

4i cosθ sin θ e −2iθ ⋅ e − k ′t 1 − e −4iθ e −2 k 't 2i sin 2θ e −2iθ e − k ′t . = 1 − e −4iθ e −2 k ′t =

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The transmissivity is therefore 2

T =

4 sin 2 2θ e −2 k ′t (1 − e −2 k ′t cos 4 θ )2 + (e −2 k ′t sin 4θ )2

4 sin 2 2θ e −2 k ′t 1+ e − 2e −2 k ′ t cos 4θ 4 sin 2 2θ = 2 k ′t −2 k ′t , e +e − 2cos 4θ =

−4 k ′t

and the reflectivity is 2

4 sin 2 2θ e 2 k ′t + e −2 k ′t − 2 cos 4θ e 2 k ′t + e −2 k ′t − 2 , = 2 k ′t −2 k ′t e +e − 2 cos 4θ

1− T = 1−

where k ′ = 2mE sinθ  .

1057 Find the wave function for a particle moving in one dimension in a constant imaginary potential –iV where V  E. Calculate the probability current and show that an imaginary potential represents absorption of particles. Find an expression for the absorption coefficient in terms of V. (Wisconsin) Sol:

The Schrödinger equation is

i ∂ψ ∂t = ( pˆ 2 2m − iV ) ψ .

Supposing ψ = exp ( −iEt  ) exp (ikx ) , we have k 2 = ( 2mE /  2 ) (1 + iV / E ) . As V  E, k≈±

2mE  V  1 + i  , 2 2E   

and hence  2mE V    iEt  2mE  ψ± ( x , t ) = exp  ∓ x  ⋅ exp ±i x  exp −  , 2 2     2 E      where ψ + and ψ − refer to the exponentially attenuated right- and left-traveling waves respectively. The probability current is

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85

 k  2mE V   pˆ  k    j = Re ψ * ψ  = Re ψ * ψ  = Re  exp  ∓ x    m   2 E   m    m ≈±

  2mE 2mE V  exp  ∓ x . 2 m   2 E  

These are the exponentially attenuated currents in the respective directions. The absorption coefficient is then

µ= −

1 dj dlnj 2mE V =− = . j dx dx 2 E

The imaginary potential iV is responsible for the absorption of the particle, since the exponent in j would be imaginary. Hence there would be no absorption if V were real.

1058 Let the solution to the one-dimensional free-particle time-dependent Schrödinger equation of definite wavelength λ be ψ ( x , t ) as described by some observer O in a frame with coordinates ( x , t ) . Now consider the same particle as described by wave function ψ ′ ( x ′ , t ′ ) according to observer O ′ with coordinates ( x ′ , t ′ ) related to ( x , t ) by the Galilean transformation x ′ = x − υt, t′ =t . a. Do ψ ( x , t ) , ψ ′ ( x ′ , t ′ ) describe waves of the same wavelength? b. What is the relationship between ψ ( x , t ) and ψ ′ ( x ′ , t ′ ) if both satisfy the Schrödinger equation in their respective coordinates? (Berkeley) Sol: a. The one-dimensional time-dependent Schrödinger equation for a free particle i ∂t ψ ( x , t ) = ( −  2 / 2m) ∂2x ψ ( x , t )

has a solution corresponding to a definite wavelength λ

ψ λ ( x , t ) = exp i ( kx − ω t )

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with

λ = 2π / k = 2π  / p, ω = k 2 / 2m . As the particle momentum p is different in the two reference frames, the wavelength λ is also different. b. Applying the Galilean transformation and making use of the Schrödinger equation in the ( x ′ , t ′ ) frame we find i ∂t ψ ′ ( x ′ , t ′ ) = i ∂t ψ ′ ( x − υt , t )   = i ∂t ′ψ ′ ( x ′ , t ′ ) − υ ∂x ′ψ ′ ( x ′ , t ′ ) 2 2 ∂′x ψ ′ ( x ′ , t ′ ) − iυ ∂x ′ψ ′ ( x ′ , t ′ ) 2m 2 2 =− ∂x ψ ′ ( x − υt , t ) − iυ ∂x ψ ′ ( x − υt , t ) . (1) 2m =−

Considering

i ∂t e i(kx −ωt )ψ ′ ( x ′ , t ′ ) = ie i(kx −ωt ) (∂t ψ ′ − iωψ ′ ) and  2 2  i(kx−ωt ) ∂x e ψ ′ ( x ′, t ′) 2m   2 i(kx−ωt ) =− e (−k 2ψ ′+ 2ik ∂x ψ ′+∂2x ψ ′) 2m = ie i(kx−ωt ) (∂t ψ ′− iωψ ′) ,

−

making use of Eq. (1) and the definitions of k and ω , we see that  2 2  imυx / − imυ 2t /2  e ∂x e ψ ′ ( x − υt , t ) 2m  2 = i ∂t e imυx /e − imυ t /2 ψ ′ ( x − υt , t ).

−

This is just the Schrödinger equation that ψ ( x , t ) satisfies. Hence, accurate to a phase factor, we have the relation i  mυ 2   t . ψ ( x , t ) = ψ ′ ( x − υt , t ) exp  mυ x −    2  

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87

1059 A particle of mass m bound in a one-dimensional harmonic oscillator potential of frequency ω and in the ground state is subjected to an impulsive force pδ (t ). Find the probability it remains in its ground state. (Wisconsin) Sol:

The particle receives an instantaneous momentum p at t = 0 and its velocity changes to p/m instantaneously. The duration of the impulse is, however, too short for the wave function to change. Hence, in the view of a frame K ′ moving with the particle, the latter is still in the ground state of the harmonic oscillator ψ 0 ( x ′ ) . But in the view of a stationary frame K, it is in the state ψ 0 ( x ′ ) exp ( −ipx /  ) . We may reasonably treat the position of the particle as constant during the process, so that at the end of the impulse the coordinate of the particle is the same for both K and K ′ . Hence the initial wave function in K is

ψ0′ = ψ0 ( x ) exp (−ipx /  ) . Thus, the probability that the particle remains in its ground state after the impulse is  px  P0 = ψ 0 exp  −i  ψ 0   

α2 = π

+∞

∫e

− α 2 x 2 −i

px 

2

2

dx

−∞

 p2  α = exp  − 2 2   2α   π

∞

2  2 ip   exp − α x +    ∫   2α 2   dx −∞

2

 − p2  , = exp   2mω   where α = mω /  .

1060 An idealized ping pong ball of mass m is bouncing in its ground state on a recoilless table in a one-dimensional world with only a vertical direction. a. Prove that the energy depends on

m, g , h

according to:

2 α

ε = Kmg ( m g / h ) and determine α. 2

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b. By a variational calculation estimate the constant K and evaluate ε for m = 1 gram in ergs. (Princeton) Sol: a. By the method of dimensional analysis, if we have

[ε ] =

1+2α

[m]

1+α

[g]

,

[h]2α

or

[m][ L ] [T ]2

[m][ L ]

1−3α

2

=

,

[T ]2

then α = − 13 . Thus, provided α = − 13 , the expression gives the energy of the ball. b. Take the x coordinate in the vertical up direction with origin at the table. The Hamiltonian is H=

p2 2 d 2 + mgx = − + mgx , 2m 2m dx 2

taking the table surface as the reference point of gravitational potential. Try a ground state wave function of the form ψ = x exp ( − λ x 2 / 2) , where λ is to be determined. Consider H =

∫ ψ * Hψdx ∫ ψ *ψdx

∫ = =

∞ 0

xe − λ x

2

/2

(−

∫

)

2

+ mgx xe − λ x /2dx

2 d 2 2m dx 2 ∞ 2 −λx 2 0

xe

dx

3 2 2mg λ+ . 4m π λ 1/2 2

 4m 2 g  3 d H To minimize H , take = 0 and obtain λ =   . dλ  3 π 2  The ground state energy is then H = 3(3 / 4π )1/3 mg (m 2 g /  2 )−1/3 . giving K = 3(3 / 4π )1/3 .

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89

Numerically 1

 3 2  3 ε = 3mg  2   4π m g  1



 3 × 1.054 2 × 10−54  3 = 3 × 980 ×   4 π × 980   = 1.9 × 10−16 erg.

1061 The following theorem concerns the energy eigenvalues En ( E1 < E2 < E3 a,

V ( x ) = 0,

x ≤ a.

The number of bound states of U ( x ) is less than that of V ( x ), which for the latter is 2mωa 2 /π  + 1 . We can take the upper bound to the number of

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91

bound states of U ( x ) as 2mωa 2 /π  as for N  1 the term 1 can be neglected. Taken together, we get that the number of bound states is between mωa 2 /2 and 2mωa 2 /π  .

1062 For electronic states in a one-dimensional system, a simple model Hamiltonian is N

N

n=1

n=1

H = ∑E0 n n + ∑W { n n + 1 + n + 1 n } , where n are an orthonormal basis, n | n ′ = δ nn ′ ; E0 and W are parameters. Assume periodic boundary conditions so that N + j = j . Calculate the energy levels and wave functions. (Wisconsin) Sol:

From the fact that n form a complete set of orthonormal functions, where n = 1, 2, 3, … , N , and N

N

n=1

n=1

Hˆ = ∑E0 n n + ∑W { n n + 1 + n + 1 n } , or

Hˆ = E0 + W ( A + A+ ) ,

with N

N

n=1

n=1

A ≡ ∑ n n + 1 , A+ ≡ ∑ n + 1 n , and A n = n − 1 , A+ n = n + 1 , AA+ = A+ A = 1, or A+ = A−1 , we know that Hˆ , A and A+ have the same eigenvectors. Hence we only need to find the eigenvectors and eigenvalues of the operator A+ to solve the problem. As Ak′k = k′ A k = δk′ ,k−1 ,

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We have     A=     and so

    A−λ =    

0 0 0 . . 1

1 0 0 . . 0

0 . . 0   1 . . 0  0 1 . .  . . . .   . . 0 1  0 . . 0  N ×N

−λ 1 0 . . 0 0 −λ 1 . . 0 0 0 −λ 1 . . . . . . . . . . . . −λ 1 1 0 0 . . −λ

    ,      N ×N

i.e., det ( A − λ I ) = (− λ )N + (−1)N +1 = (−1)N ( λ N − 1) = 0, giving iθ

λj = e j , θ j =

2π j , j = 0, 1, 2, … , N − 1. N

If E are the same eigenvectors of the operators A and A+ , i.e., A E j = λ j E j , A+ E j = then

1 Ej , λj

  1  Hˆ E j = E0 + W  λ j +  E j λ j    = ( E0 + 2W cosθ j ) E j .

Hence the eigenvalues of Hˆ are 2π E j = E0 + 2W cosθ j , with θ j = j , ( j = 0, 1, 2, … , N − 1) . N The corresponding eigenfunctions can be obtained from the matrix equations

(A − λ ) E j

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j

=0.

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Thus

 l  iθ j  e 1  i 2θ j e Ei = N    i(N −l)θ j  e 

or Ej =

1 N

N

∑e n=1

i(n−1)θ j

93

    ,     n.

1063 Give a brief discussion of why there are energy bands in a crystalline solid. Use the ideas of quantum mechanics but do not attempt to carry out any complicated calculations. You should assume that anyone reading your discussion understands quantum mechanics but does not understand anything about the theory of solids. (Wisconsin) Sol:

A crystal may be regarded as an infinite, periodic array of potential wells, such as the lattice structure given in Problem 1065. Bloch’s theorem states that the solution to the Schrödinger equation then has the form u ( x ) exp (iKx ), where K is a constant and u ( x ) is periodic with the periodicity of the lattice. The continuity conditions of u ( x ) and du ( x ) / dx at the well boundaries limit the energy of the propagating particle to certain ranges of values, i.e., energy bands. An example is given in detail in Problem 1065.

1064 A particle of mass m moves in one dimension in a periodic potential of infinite extent. The potential is zero at most places, but in narrow regions of width b separated by spaces of length a (b  a ) the potential is V0 , where V0 is a large positive potential. [One may think of the potential as a sum of Dirac delta functions: V (x) =

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∞

∑ V bδ ( x − na).

n=−∞

0

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Alternatively one can arrive at the same answer in a somewhat more messy way by treating the intervals as finite and then going to the limit. ] a. What are the appropriate boundary conditions to apply to the wave function, and why? b. Let the lowest energy of a wave that can propagate through this potential be E0 =  2 k02 / 2m (this defines k0). Write down a transcendental equation (not a differential equation) that can be solved to give k0 and thus E0 . c. Write down the wave function at energy E0 valid in the region 0 ≤ x ≤ a (For uniformity, let us choose normalization and phase such that ψ ( x = 0) = 1) . What happens to the wave function between x = a and x = a+b ? d. Show that there are ranges of values of E, greater than E0 , for which there is no eigenfunction. Find (exactly) the energy at which the first such gap begins. (Berkeley) Sol: a. The Schrödinger equation is ∞  2 d 2  + ∑ V0bδ ( x − na)ψ ( x ) = Eψ ( x ) . − 2  2m dx n=−∞  Integrating it from x = a − ε to x = a + ε and letting ε → 0, we get

ψ ′ (a + ) − ψ ′ (a − ) = 2Ωψ (a) ,

where Ω = mV0b /  2 . This and the other boundary condition

ψ (a + ) − ψ (a − ) = 0 apply to the wave function at x = na , where n = −∞, … , − 1, 0, 1, … , +∞ . b. For x ≠ na , there are two fundamental solutions to the Schrödinger equation: u1 ( x ) = e ikx, u2 ( x ) = e −ikx, the corresponding energy being E =  2 k 2 / 2m .

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Let

ψ ( x ) = Ae ikx + Be − ikx , 0 ≤ x ≤ a . According to Bloch’s Theorem, in the region a ≤ x ≤ 2a ψ ( x ) = e iKa  Ae ik(x −a) + Be − ik(x −a)  , where K is the Bloch wave number. The boundary conditions give e iKa ( A + B ) = Ae ika + Be − ika ,

ike iKa ( A − B ) = ik ( Ae ika − Be − ika )

+ 2Ω ( Ae ika + Be − ika ) .

For nonzero solutions of A and B we require e iKa − e ika

e iKa − e − ika

ike iKa − (ik + 2Ω) e ika

−ike iKa + (ik − 2Ω) e − ika

= 0,

or cos ka +

Ω sin ka = cos Ka, k

which determines the Bloch wave number K. Consequently, the allowed values of k are limited to the range given by cos ka +

Ω sin ka ≤ 1, k

or 2

  Ω cos ka + sin ka  ≤ 1 .   k k0 is the minimum of k that satisfy this inequality, c. For E = E0 ,

ψ ( x ) = Ae ik0 x + Be −ik0 x , 0 ≤ x ≤ a, where k0 =

2mE0 

.

Normalization ψ ( x = 0) = 1 gives

ψ ( x ) = 2i A sin k0 x + e −ik0 x,

0 ≤ x ≤ a.

The boundary conditions at x = a give e iKa = 2i A sin k0a + e − ik0 a , or

2iA = (e iKa − e − ik0 a ) / sink0a .

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So

ψ ( x ) = (e iKa − e −ik0 a )

sin k0 x −ik0 x +e , sin k0a

0 ≤ x ≤ a.

For x ∈ a, a + b , the wave function has the form exp (±k1, x ) , where k1 = 2m (V0 − E0 ) /  . d. For ka = nπ + δ , where δ is a small positive number, we have Ω cos ka + sin ka k Ω = cos (nπ + δ ) + sin (nπ + δ ) k ≈ 1−

δ2 Ω + δ ≤ 1. 2 k

When δ is quite small, the left side ≈ 1 + Ωδ / k > 1 . Therefore in a certain region of k > nπ / a , there is no eigenfunction. On the other hand, ka = nπ corresponds to eigenvalues. So the energy at which the first energy gap begins satisfies the relation ka = π , or E = π 2  2 / 2ma 2 .

1065 We wish to study particle-wave propagation in a one-dimensional periodic potential constructed by iterating a “single-potential” V ( x ) at intervals of length l. V ( x ) vanishes for x ≥ l / 2 and is symmetric in x (i.e., V ( x ) = V ( − x ) ). The scattering properties of V ( x ) can be summarized as follows: If a wave is incident from the left, ψ + ( x ) = exp (ikx ) for x < − l / 2, it produces a transmitted wave ψ + ( x ) = exp (ikx ) for x > l / 2 and a reflected wave

ψ − ( x ) = exp ( −ikx ) for x < − l / 2 . Transmitted and reflected coefficients are

given by T (x) = R(x) =

ψ+ ( x )

1 2 ik x l = e e 2 iδe + e 2 iδ0  , x ≥ , 2 ψ+ (− x ) 2

ψ− ( − x )

1 2 ik x l ≈ e e 2 iδe − e 2 iδ0  , x ≥ , 2 ψ+ (− x ) 2

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97

and δ e and δ 0 are the phase shifts due to the potential V ( x ). Take these results as given. Do not derive them. Now consider an infinite periodic potential V∞ ( x ) constructed by iterating the potential V ( x ) with centers separated by a distance l (Fig. 1.30). Call the

points at which V∞ ( x ) = 0 “interpotential points”. We shall attempt to con-

struct waves propagating in the potential V∞ ( x ) as superpositions of left- and right-moving waves φ + and φ − .

Fig. 1.30 a. Write recursion relations which relate the amplitudes of the right- and left-moving waves at the nth interpotential point, φ±n , to the amplitudes

at the ( n − 1) th and ( n + 1) th interpotential points, φ±n−1 and φ±n+1.

b. Obtain a recursion relation for φ − or φ + alone by eliminating the other from part (a). c. Obtain an expression for the ratio of amplitudes of φ + to φ − at successive interpotential points. d. Find the condition on k, δ e and δ 0 such that traveling waves are allowed. e. Use this result to explain why it is “normal” for conduction by electrons in metals to be allowed only for bands of values of energy. (MIT) Sol:

For the wave incident from the left, the potential being V ( x ) , let 1 i 2δe i 2δ0 (e + e ) ; 2 1 ψr − = rφ − e −ikx , r = (e i 2δe − e i 2δ0 ) . 2

ψt+ = tφ + e ikx , t =

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For the wave incident from the right, let the transmission and reflection coefficients be t ′ and r ′ respectively. It can be shown that t ′ = t , r ′ = −r *t / t *. In the periodic potential, the transmission and reflection coefficients at adjacent interpotential points have relations tn = tn−1 and rn = rn−1exp (i 2kl ) . So the transmission coefficient can be denoted by a single notation t. a. The waves at adjacent interpotential points are as shown in Fig. 1.31. Obvin n−1 ously, only the reflection term of φ − and the transmission term of φ + n c­ ontribute to φ + :

φ+n = rn′−1φ−n + tφ+n−1.(1)

Similarly,

φ−n = rnφ+n + t ′φ−n+1.(2)

Thus we have

φ+n = e i 2 kl rn′φ−n + tφ+n−1 , (3)



φ−n = rnφ+n + tφ−n+1 φ+n−1 → ← φ−n−1

(4)

φ+n φ+n+1 → → ← ← φ−n φ−n+1

Fig. 1.31 b. With n replaced by n+1, Eq. (1) gives

φ +n+1 = rn′φ −n+1 + tφ +n . (5)



Equations (3), (4), (5) then give

t (φ +n−1 + e i 2 klφ +n+1 )

φ +n = Let

r0 = r .

Then

rn = r exp (i 2nkl ) .

rn′ = r ′ exp ( −i 2nkl ) . Hence

1 + t 2e i 2 kl − rnrn′e i 2 kl

φ +n =

1 + t 2e i 2 kl − rr ′e i 2 kl

φ −n =

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Assume

t (φ +n−1 + e i 2 klφ +n+1 )

Similarly,

. rn′ = −rn*t / t * .

Then

. (6)

t (φ −n+1 + e i 2 klφ −n−1 )

1 + t 2e i 2 kl − rr ′e i 2 kl

.

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c. As the period of the potential is l, if ψ ( x ) is the wave function in the region xn−1 , xn  , then ψ ( x − l ) exp (iδ ) is the wave function in the region xn , xn+1  . Thus φ n+1 = e i(δ −kl )φ n , + +  n+1 i δ +kl n ( ) φ − . (7) φ − = e



Let cn = φ +n / φ −n . From (4) and (5) we obtain respectively 1 = rncn + t



φ −n+1 , (8) φ −n

φ +n+1 φ −n+1 = r + tcn . (9) ′ n φ −n φ−n



Using (7), (9) can be written as cne − i 2 kl or, using (8), i.e.,

φ −n+1 φ −n+1 = r + tcn , ′ n φ −n φ−n

cne − i 2 kl (1 − rncn ) = rn′ (1 − rncn ) + t 2cn , rncn2 + (t 2e i 2 kl − rnrn′e i 2 kl − 1) cn + rn′e i 2 kl = 0 .

Solving for cn we have cn =

(1 + rr ′e

i 2 kl

− t 2e i 2 kl ) ± ∆ 2rn

,

where ∆ = (t 2e i 2 kl − rnrn′e i 2 kl − 1)2 − 4rnrn′e i 2 kl = (t 2e i 2 kl − rr ′e i 2 kl − 1)2 − 4rr ′e i 2 kl . d. The necessary condition for a stable wave to exist in the infinite periodic field is

φ +n+1 / φ +n = e iδ1 , where δ 1 is real and independent of n. If this were not so, when n → ∞ one n ( −n) of φ + and φ + would be infinite. From (7), we see that δ 1 = δ − kl . From (6),

we obtain  φ n−1 φ n+1  1 + t 2e i 2 kl − rr ′e i 2 kl = t  +n + e i 2 kl +n  φ+   φ+ = t e i(kl −δ ) + e i(kl +δ ) .

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Substituting r ′ = −r *t / t * in the above equation and using rr * + tt * = 1, we obtain te ikl + t *e −ikl = 2tt * cos δ , which means te ikl + t *e −ikl 2tt *

≤ 1,

or, using the definition of t, cos ( 2δe + kl ) + cos ( 2δ0 + kl ) 1 + cos 2 (δe − δ0 )

≤ 1,

i.e., cos (δ e + δ 0 + kl ) ≤1 . cos (δ e − δ 0 ) In general, only some of the values of k satisfy the above inequality, i.e., only energy values in certain regions are allowed while the others are forbidden. Thus we obtain the band structure of energy levels. e. In metals, the distribution of positive ions is regular and so the conduction electrons move in a periodic potential, (d) Shows that the electron waves can only have certain k values, corresponding to bands of electron energies.

1066 You are given a real operator  satisfying the quadratic equation Â2 – 3 + 2 = 0. This is the lowest-order equation that  obeys. a. What are the eigenvalues of Â? b. What are the eigenstates of Â? c. Prove that  is an observable. (Buffalo) Sol: a. As  satisfies a quadratic equation it can be represented by a 2 × 2 matrix. Its eigenvalues are the roots of the quadratic equation λ 2 − 3λ + 2 = 0, λ1 = 1, λ2 = 2 λ 2 − 3λ + 2 = 0, λ1 = 1, λ2 = 2 .

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b. Â is represented by the matrix 1 0  Aˆ =    . 0 2  The eigenvalue equation

a  1 0  a      = λ   0 2  b  b  then gives a = 1, b = 0 for λ = 1 and a = 0, b = 1 for λ = 2 . Hence the eigen-

1  0  states of  are   and   . 0  1  + c. Since Aˆ = Aˆ , Aˆ is Hermitian and hence an observable.

1067 If ψ q is any eigenstate of the electric charge operator Q corresponding to eigenvalue q, that is to say, Q ψq = q ψq , the “charge conjugation” operator C applied to |ψ q 〉 leads to an eigenstate |ψ − q 〉 of Q corresponding to eigenvalue –q:

C ψ q = ψ −q . a. Find the eigenvalues of the operator CQ + QC. b. Can a state simultaneously be an eigenstate of C and of Q? (Chicago) Sol: a. Let

ψ = ∑cq ψ q . q

Then

(CQ + QC ) ψ q = qC ψ q + Q ψ − q = q ψ − q − q ψ − q = 0 . Thus the eigenvalue of the operator CQ + QC is zero. b. As C is the charge conjugation transformation, CQC −1 = −Q , or CQ + QC = 0 , i.e., C and Q do not commute (they anticommute) they cannot have common eigenstates. (Unless q = 0 , in which case it is quite meaningless to introduce charge conjugation.)

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1068 A quantum-mechanical system is known to possess only two energy eigenstates denoted 1 and 2 . The system also includes three other observables (besides the energy), known as P, Q and R. The states 1 and 2 are normalized but they are not necessarily eigenstates of P, Q or R. Determine as many of the eigenvalues of P, Q and R as possible on the basis of the following sets of “experimental data”. [Warning: one data set is unphysical.] a.

1 P 1 = 1/ 2, 1 P 2 1 = 1/ 4 .

b.

1 Q 1 = 1/ 2, 1 Q 2 1 = 1/ 6 .

c.

1 R 1 = 1, 1 R 2 1 = 5 / 4, 1 R 3 1 = 7 / 4 . (MIT)

Sol:

We are given three observables P, Q, R, which satisfy the Hermiticity n P m = m P n * , and that the mechanical system has a complete set of energy eigenstates 1 and 2 . a. The completeness of the two states and the “experimental data”, give 1 P 1 = 1 +α 2 , 2 where α is a constant to be determined. The orthogonality of the eigenstates and the Hermiticity of P give 1 1 P 2 = 1 2 +α2 2 2 = α * . 2 So we have P 2 =α* 1 +β 2 , where β is to be determined. Then P 2 |1〉 = P ( P 1

As P 2 1 = 14

)

1 = P 1 + α P 2 = (1/ 4 + α * α ) 1 + (α / 2 + αβ ) 2 . 2 * according to experiment, α α = 0 and hence α = 0 .

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103

Therefore, P1 =

1 1 , 2

i.e., at least one of the eigenvalues of P is 1/2. b. Let Q1 =

1 1 +γ 2 , 2

where γ is to be determined. By a similar procedure, we get γ * γ = 1/ 6 − 1/ 4 < 0. So this data set is unphysical and the eigenvalue of Q could not be determined. c. As 1 R 1 = 1 , we can write R 1 = 1 +λ 2 , where λ is to be determined. Then 1 R 2 = 1 2 + λ* 2 2 = λ* , showing that R 2 = λ * 1 +η 2 . where η is to be determined. Consider R2 1 = R 1 + λ R 2 = 1 + λ 2 + λλ * 1 + λη 2 . Then as 1 R 2 1 = 1 + λλ * =

5 , 4

we have

λλ * =

1 , 4

and so 1 λ = exp (iδ ), 2 and 1 R 1 = 1 + e iδ 2 , 2 1 R 2 = e − iδ 1 + η 2 , 2 5 1 R 2 1 = 1 + (1 + η )e iδ 2 . 4 2

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It follows also 5 1 R 3 1 = R 1 + (1 + η ) e iδ R 2 4 2 5 5 iδ 1 = 1 + e 2 + (1 + η ) 1 4 8 4 1 + (1 + η )ηe iδ 2 . 2 5 1 7 3 7 Experimentally 1 R 1 = . Thus 4 + 4 (1 + η ) = 4 , giving η = 1. Hence on 4 the bases 1 and 2 the matrix of R is  1 −iδ  e  1 2  .  1 e iδ 1  2  To find the eigenvalues of R solve 1 −iδ e 2

1− λ 1 iδ e 2

1− λ

= 0,

i.e., (1 − λ )2 − 14 = (1 − λ − 12 )(1 − λ + 12 ) = 0 , and obtain the eigenvalues of R=1, 3 , 2 2

1069 For a charged particle in a magnetic field, find the commutation rules for the operators corresponding to the components of the velocity. (Berkeley) Sol:

Suppose the magnetic field arises from a vector potential A. Then the velocity components of the particle are υˆ = Pˆ /m − qA /mc . i

i

i

Hence υˆi , υˆj  = 1 Pˆi − q Ai , Pˆj − q A j  m 2  c c  =

q iq  ∂ A j ∂ Ai    [ pˆ j , Ai ] − [ pˆi , A j ]} = − 2 { mc mc 2  ∂xi ∂x j 

=

i q 3 ∑εijk Bk , m 2c k=1

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105

where εijk is the Levi-Civita density, use having been made of the correspondence  ∂ rule pˆi → i ∂x . i

1070 Using the coordinate-momentum commutation relation prove that

∑( E

2

n

− E0 )| n x 0 = constant ,

n

where En is the energy corresponding to the eigenstate n . Obtain the value of the constant. The Hamiltonian has the form H = p 2 /2M + V ( x ) . (Berkeley) Sol:

As H = p 2 /2 M + V ( x ), we have

[H , x ] =

1  2   p , x  = −ip/M , 2M

and so

[[H , x ], x ] = −

i [ p , x ] = −  2 /M . M

Hence m [ H , x ] , x  m = −

2 . M

On the other hand, m [ H , x ] , x  m = m Hx 2 − 2 xHx + x 2 H m = 2 Em m x 2 m − 2 m xHx m = 2 Em ∑ m x n

2

n

− 2∑En m x n

2

n

= 2∑( Em − En ) m x n

2

.

n

In the above we have used H m = Em m , m x2 m = ∑ m x n n x m n

=∑ m x n

2

n

m xHx m = ∑ m xH n n x m n

= ∑En m x n n x m n

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= ∑En m x n

2

.

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H m = Em m , m x2 m = ∑ m x n n x m n

106

∑

Problems and Solutions in Quantum = Mechanics mxn

2

n

m xHx m = ∑ m xH n n x m n

= ∑En m x n n x m n

= ∑En m x n

2

.

n

Equating the two results and setting m = 0 , we obtain

∑(E

n

− E0 ) n x 0

2

=  2 /2 M .

n

1071 a. Given a Hermitian operator A with eigenvalues an and eigenfunctions un ( x ) [ n = 1, 2, …, N ; 0 ≤ x ≤ L] , show that the operator exp (iA) is unitary. b. Conversely, given the matrix Umn of a unitary operator, construct the matrix of a Hermitian operator in terms of Umn. c. Given a second Hermitian operator B with eigenvalues bm and eigenfunctions υm ( x ) , construct a representation of the unitary operator V that transforms the eigenvectors of B into those of A. (Chicago) Sol: + a. As A = A, A being Hermitian, + −1 {exp (iA)} = exp ( −iA+ ) = exp ( −iA) = {exp (iA)} .

Hence exp (iA) is unitary. b. Let * Cmn = U mn + U nm = U mn + (U + )mn ,

i.e., C = U +U + . * As U ++ = U , C + = C . Therefore Cmn = U mn + U nm is the matrix representation

of a Hermitian operator.

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107

c. The eigenkets of a Hermitian operator form a complete and orthonormal set. Thus any um can be expanded in the complete set υn : um = ∑ υk υk | um ≡ ∑ υk Vkm , k

k

which defines Vkm ,

Vkm =

∫

L 0

υk* ( x ) um ( x ) dx .

Similarly,

υn = ∑ u j u j | υn j

= ∑ u j υn | u j

*

j

= ∑ u j Vnj* j

= ∑ u j V jn* j

= ∑ u j Vjn+ . j

Hence um = ∑∑ u j Vjk+Vkm = ∑ u j δ jm , j

k

j

or V +V = 1 , i.e., V + = V −1, showing that V is unitary. Thus V is a unitary operator transforming the eigenvectors of B into those of A

1072 Consider a one-dimensional oscillator with the Hamiltonian H = p 2 /2m + mω 2 x 2 /2 . a. Find the time dependence of the expectation values of the “initial position” and “initial momentum” operators x 0 = x cos ωt − ( p/mω ) sinωt , p0 = p cos ωt + mω x sinωt .

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b. Do these operators commute with the Hamiltonian? c. Do you find your results for (a) and (b) to be compatible? Discuss. d. What are the motion equations of the operators in the Heisenberg picture? e. Compute the commutator [ p0 , x 0 ] . What is its significance for measurement theory? (Princeton) Sol: a. Making use of the relation df 1  ∂f , =  f , H  + dt i ∂t we have d x0 1 = [ x cos ω t , H ] − ω x sin ω t dt i  p 1 p sin ω t , H  − cos ω t −  i  mω  m 1 = [ x , H ]cos ω t − ω x sin ω t i 1 1 1 p cos ω t = 0, − p, H ] sin ω t − [ mω i m d p0

1  p cos ω t , H  − p ω sin ω t dt i  1 + [mω x sin ω t , H ] + mω 2 x cos ω t i 1 = [ p , H ] cos ω t − ω p sin ω t i mω + [ x , H ] sin ω t + mω 2 x cos ω t = 0. ih =

Thus the expectation values of these operators are independent of time. b. Consider [x 0 , H ]= [x , H ]cos ωt −

[ p, H ] sin ωt mw

i p cos ωt + iω x sin ωt , m [ p0 , H ] = [ p, H ]cos ωt + mω [x , H ]sin ωt =

= −ihmω 2 x cos ωt + iω p sin ωt .

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[x 0 , H ]= [x , H ]cos ωt −

[ p, H ] sin ωt mw

ip Principles and One-Dimensional Motions Basic

=

cos ωt + iω x sin ωt ,

m [ p0 , H ]= [ p , H ]cos ωt + mω [x , H ]sin ωt

109

= −ihmω 2 x cos ωt + iω p sin ωt . Thus the operators x 0 , p0 do not commute with H. c. The results of (a) and (b) are still compatible. For while the expressions for x 0 and p0 contain t explicitly, their non-commutation with H does not exclude their being conserved. In fact dx 0 1 ∂x = [ x 0 , H ] + 0 = 0, dt i ∂t ∂ p0 dp0 1 = [ p0 , H ] + = 0, dt i ∂t showing that they are actually conserved. d. In the Heisenberg picture, the motion equation of an operator is 1 ∂A [ A, H ] + . i ∂t Thus the motion equations of x 0 and p0 are respectively dA / dt =

dx 0 / dt = 0, dp0 / dt = 0 . e. Using the expressions for x 0 and p0 , we have [ p0 , x 0 ] = [ p cos ωt + mω x sin ωt , x 0 ] = [ p, x 0 ]cos ωt +[x , x 0 ]mω sin ωt   p =  p , x cos ωt − sin ωt  cos ωt   mω   p sin ωt  mω sin ωt . + x , x cos ωt −   mω

= [ p , x ] cos 2 ωt − [ x , p ] sin 2 ωt = − [ x , p ] = −i , as [ x , x ] = [ p, p] = 0, [ x , p ] = i .

In general, if two observables A and B satisfy the equation

[ A, B] = i , then their root-mean-square deviations ∆A, ∆B , when they are measured simultaneously, must satisfy the uncertainty principle ∆A ⋅ ∆B ≥

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In the present case, the simultaneous measurements of position and momentum in the same direction must result in ∆x ⋅ ∆p ≥

 . 2

The relation shows ∆x 02 ∆p02 ≥  / 2 . It is a relation between possible upper limits to the precision of the two quantities when we measure them simultaneously.

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Part II

Central Potentials

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2 2001 An electron is confined in a three­dimensional infinite potential well. The sides parallel to the x­, y­, and z­axes are of length L each. a. Write the appropriate Schrödinger equation. b. Write the time­independent wave function corresponding to the state of the lowest possible energy. c. Give an expression for the number of states, N, having energy less than some given E. Assume N  1. (Wisconsin) Sol: a. The Schrödinger equation is i ∂ψ (r , t ) / ∂t = −( 2 /2m)∇ 2ψ (r , t ), 0 ≤ x , y , z ≤ L ,

ψ = 0,

otherwise.

b. By separation of variables, we can take that the wave function to be the product of three wave functions each of a one-dimensional infinite well potential. The wave function of the lowest energy level is

ψ111 ( x , y , z ) = ψ1 ( x )ψ1 ( y )ψ1 ( z ), where 2 π  sin  x  , etc. L L 

ψ1 ( x ) = Thus 3/2

2 πx  π y  πz  ψ111 ( x , y , z ) =   sin   sin   sin   . L  L   L   L  The corresponding energy is E111 = 3 2 π 2 /2mL2 .

113

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c. For a set of quantum numbers nx , n y , nz for the three dimensions, the energy is  2π 2 2 2 2 E= (nx + n y + nz ). 2mL2 Hence the number N of states whose energy is less than or equal to E is equal to the number of sets of three positive integers nx , ny , nz satisfying the inequality nx2 + n 2y + nz2 ≤

2mL2 E.  2π 2

Consider a Cartesian coordinate system of axes nx , ny , nz. The number N required is numerically equal to the volume in the first quadrant of a sphere of radius (2mL2 E / 2 π 2 )1/2, provided N ≥ 1. Thus 1 4 π  2mL2  N = .  2 2 E 8 3  π 

3/2

=

4 π  mL2  E  3  2 2 π 2 

3/2

.

2002 A ‘quark’ (mass = mp /3) is confined in a cubical box with sides of length 2 ­fermis = 2 × 10−15 m . Find the excitation energy from the ground state to the first excited state in MeV . (Wisconsin) Sol:

The energy levels in the cubical box are given by En1n2n3 =

 2π 2 2 2 2 (n1 + n2 + n3 ), ni = 1, 2, … 2ma 2

Thus the energy of the ground state is E111 = 3 2 π 2 /2ma 2 , that of the first excited state is E211 = 6 2 π 2 /2ma 2 = 3 2 π 2 /ma 2 . Hence the excitation energy from the ground state to the first excited state is 1.5π 2  2c 2 mc 2a 2 1.5π 2 (6.58 × 10−22 )2 × (3 × 108 )2 = = 461 MeV. ( 9383 ) × (2 × 10−15 )2

∆E = 3 2 π 2 /2ma 2 =

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2003 A NaCl crystal has some negative ion vacancies, each containing one electron. Treat these electrons as moving freely inside a volume whose dimensions are on the order of the lattice constant. The crystal is at room temperature. Give a numerical estimate for the longest wavelength of electromagnetic radiation absorbed strongly by these electrons. (MIT) Sol:

The energy levels of an electron in a cubical box of sides a are given by Enmk = (π 2  2 /2ma 2 )(n 2 + m 2 + k 2 ), where n, m and k are positive integers. Taking a ∼ 1Å , the ground state energy is E111 = 3 2 π 2 /2ma 2 ≈ 112 eV . For a crystal at room temperature, the electrons are almost all in the ground state. The longest wavelength corresponds to a transition from the ground state to the nearest excited state: ∆E = E211 − E111 =

3π 2  2 = 112 eV, ma 2

for which

λ=

c hc = = 110 Å. ν ∆E

2004 An electron is confined to the interior of a hollow spherical cavity of radius R with impenetrable walls. Find an expression for the pressure exerted on the walls of the cavity by the electron in its ground state. (MIT) Sol:

For the ground state, l = 0 , and if we set the radial wave function as R(r ) = χ (r )/r, then χ (r ) is given by d 2 χ 2µ E + 2 χ = 0 for r < R , dr 2  χ = 0 for r ≥ R , where µ is the electron rest mass. R(r) is finite at r = 0 , so that χ (0) = 0.

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The solutions satisfying this condition are 2 nπ r sin , (n = 1, 2, 3, …) R R

χn = for which

En =

π 2 2 2 n. 2µ R 2

The average force F acting radially on the walls by the electron is given by ∂E ∂ ˆ ∂Hˆ ∂Vˆ H =− . F= − =− =− ∂R ∂R ∂R ∂R As the electron is in the ground state, n = 1 and F = −∂E1 / ∂R = π 2  2 /µ R 3. The pressure exerted on the walls is p = F /4 π R 2 = π  2 /4µ R5.

2005 A particle of mass m is constrained to move between two concentric impermeable spheres of radii r = a and r = b. There is no other potential. Find the ground state energy and normalized wave function. (MIT) Sol:

Let the radial wave function of the particle be R(r ) = χ (r )/r. Then χ (r ) satisfies the equation

{

}

d 2 χ (r ) 2m l(l + 1) χ (r ) = 0. + 2 [E − V (r )] − 2 2 dr r  (a ≤ r ≤ b ) For the ground state, l = 0 , so that only the radial wave function is non-trivial. Since V (r ) = 0, letting K 2 = 2mE / 2 , we reduce the equation to

χ ″+ K 2 χ = 0, with

χ | r =a = χ | r =b = 0. χ (a) = 0 requires the solution to have the form χ (r ) = A sin[K (r − a)].

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Then from χ (b ) = 0, we get the possible values of K: K = nπ /(b − a),

(n = 1, 2, …)

For the particle in the ground state, i.e., n = 1 , we obtain the energy E =  2 K 2 /2m =  2 π 2 /2m(b − a)2 . From the normalization condition

∫

b a

R 2 (r )r 2 dr =

∫

b a

χ 2 (r ) dr = 1,

we get A = 2/(b − a) . Hence for the ground state, the normalized radial wave function is R(r ) =

π (r − a ) 2 1 sin , b−a r b−a

and the normalized wave function is

ψ (r ) =

1 4π

π (r − a ) 2 1 sin . b−a b−a r

2006 a. For a simple quantum harmonic oscillator with H = ( p 2 /m + kx 2 )/2 , show that the energy of the ground state has the lowest value compa­ tible with the uncertainty principle. b. The wave function of the state where the uncertainty principle mini­ mum is realized is a Gaussian function exp(−α x 2 ) . Making use of this fact, but without solving any differential equation, find the value of α . c. Making use of raising or lowering operators, but without solving any differential equation, write down the (non-normalized) wave function of the first excited state of the harmonic oscillator. d. For a three-dimensional oscillator, write down, in polar coordinates, the wave functions of the degenerate first excited state which is an eigenstate of lz. (Berkeley) Sol: a. The ground state of the harmonic oscillator has even parity, so that x = 0| x | 0 = 0,

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p = 0| p | 0 = 0;

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and so ∆p 2 = p 2 ,

∆x 2 = x 2 .

The uncertainty principle requires ∆p 2 ⋅ ∆x 2 ≥

2 . 4

It follows that p2 k 2 + x 2m 2 k ≥ p2 ⋅ x 2 m

E=

 k = ω /2 = E0 , 2 m = ω . Thus the energy of the ground state has the lowest value compat≥

as

k m

ible with the uncertainty principle. b. Using the given wave function we calculate x2 =

∫

∞

∫

2

e −2α x x 2 dx −∞

p 2 = −  2 ∫ e −α x ∞

−∞

2

∞

2

e −2α x dx = 1/4α , −∞

d 2 −α x 2 e dx dx 2

∫

∞ −∞

2

e −2α x dx = h 2α ,

and hence E= From

dE dα

2 k 1 α+ ⋅ . 2m 2 4α

= 0 we see that when α = km /2 = mω /2 the energy is minimum.

Therefore α = mω /2 . c. In the Fock representation of harmonic oscillation we define aˆ = i( pˆ − imω xˆ )/ 2mω , aˆ+ = −i( pˆ + imω xˆ )/ 2mω . Then [aˆ, aˆ+ ] = 1, H = (aˆ+aˆ + 1/2)ω . Denoting the ground state wave function by 0 . As H 0 = 12 ω 0 , the last

equation gives â + â| 0 = 0. It also gives

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1 H (â+ 0 ) = ωâ+ 0 + ωâ+ââ+ 0 2 1 = ωâ+ 0 + ωâ+ (â+â + 1) 0 2 3 = ω (â+ 0 ). 2 Hence 1 = â+ 0 = =

  mω 2  −i  ∂ x  −i + imω x  exp −    2  ∂x 2mω

 mω 2  2mω x exp − x   2  

in the coordinate representation. d. For a 3-dimensional oscillator, the wave function is

ψn1n2n3 (r ) = ψn1 ( x )ψn2 ( y )ψn3 ( z ). For the ground state, (n1 , n2 , n3 ) = (0,0,0) . For the first excited states, (n1 , n2 , n3 ) = (1, 0, 0); (0, 1, 0); (0, 0, 1).   1 ψ100 (r ) = N 02 N1 2α x exp − α 2r 2  ,  2    1 ψ010 (r ) = N 02 N1 2α y exp − α 2r 2  ,  2    1 ψ001 (r ) = N 02 N1 2α z exp − α 2r 2  .  2  Expanding x, y, z in spherical harmonics and recombining the wave functions, we get the eigenstates of lz  1  ψm (r ) = N mexp − α 2r 2  rY1m (θ , ϕ ) .  2  where N m−1 = 2/α 2 . Note that here α = mω /  , which is the usual definition, different from that given in (b).

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2007 The diagram (Fig. 2.1) shows the six lowest energy levels and the associated angular momenta for a spinless particle moving in a certain three-dimensional central potential. There are no “accidental” degeneracies in this energy spectrum. Give the number of nodes (changes in sign) in the radial wave function associated with each level. (MIT) Sol:

The radial wave function of a particle in a three-dimensional central potential can be written as R(r ) = χ (r ) / r . With a given angular quantum number l, the equation satisfied by χ (r ) has the form of a one-dimensional Schrödinger equation. Hence, if an energy spectrum has no “accidental” degeneracies, the role of the nodes in the radial wave function of the particle is the same as that in the one-dimensional wave function. For bound states, Sturm’s theorem remains applicable, i.e., χ (r ) obeys Sturm’s theorem: the radial wave function of the ground state has no node, while that of the nth excited state has n nodes. Thus, for a bound state of energy En, which has quantum number n = nr + l + 1, the radial wave has nr nodes. For angular quantum number l = 0, the numbers of nodes for the three energy levels (ordered from low to high energy) are 0, 1 and 2. Similarly, for l = 1, the numbers of nodes are 0 and 1; for l = 2, the number of nodes is 0. Thus, the numbers of nodes in the energy levels shown in Fig. 2.1 are 0, 1, 0, 0, 1, 2, from low to high energy.

Fig. 2.1

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2008 A particle of mass m and charge q is bound to the origin by a spherically symmetric linear restoring force. The energy levels are equally spaced at intervals ω 0 above the ground state energy E0 = 3ω 0 /2 . The states can be described alternatively in a Cartesian basis (three one-dimensional harmonic oscillators) or in a spherical basis (central field, separated into angular and radial motions). a. In the Cartesian basis, table the occupation numbers of the various states of the oscillators for the ground and first three excited levels. Determine the total degeneracy of each of these levels. b. In the spherical basis, write down (do not solve) the radial equation of motion. (Note that in spherical coordinates ∇ 2 = r12

∂ ∂r

(r ) − 2 ∂ ∂r

L2 r2

, where L2 is the

operator of total orbital angular momentum squared in units of  2.) Identify the effective potential and sketch it. For a given angular momentum, sketch the “ground state” radial wave function (for a given l value) and also the radial wave functions for the next two states of the same l. c. For the four levels of part (a), write down the angular momentum content and the parity of the states in each level. Compare the total degeneracies with the answers in (a). d. Does the second excited state ( E2 = 7ω 0 /2) have a linear Stark effect? Why or why not? Compare similarities and differences between this oscillator level and the second excited level ( n = 3) of the ­nonrelativistic hydrogen atom. (Berkeley)

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Sol: a. Table 2. 1 Energy level

Occupation Numbers

Degeneracy

E0

0,0,0

1

E1

1,0,0 , 0,1,0 , 0,0,1

3

E2

2,0,0 , 0, 2,0 , 0,0, 2

6

1,1,0 , 1,0,1 , 0,1,1 E3

3,0,0 , 0,3,0 , 0,0,3

10

2,1,0 , 0, 2,1 , 1,0, 2 1, 2,0 , 0,1, 2 , 2,0,1 1,1,1 b. Let

ψ (r ) = R(r )Ylm (θ , ϕ ) . The radial wave function R(r ) satisfies the equation  l(l + 1)  1 d  2 d   2m  m R  +   E − ω 2r 2  − r  R = 0,  2 r2  r 2 dr  dr    2  so that the effective potential is Veff = mω 2r 2 /2 +  2l(l + 1)/2mr 2 , which is sketched in Fig. 2.2, where r0 = [ 2l(l + 1) / m 2ω 2 ]1/4 . The shapes of the radial wave functions of the three lowest states for a given l are shown in Fig. 2.3.

Fig. 2.2

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Fig. 2.3 Note that the number of nodes of a wave function is equal to nr. c. Table 2.2 E

l

m

P

D

E0

0

0

+

1

E1

1

0, ± 1

−

3

E2

2

0, ± 1 ± 2

+

6

0 3

0 0, ± 1, ± 2, ± 3

−

10

1

0, ± 1

E3

Note: P = parity, D = degeneracy. d. The second excited state does not have a linear Stark effect because x is an operator of odd parity while all the degenerate states for E2 have even parity, with the result that the matrix elements of H ′ in the subspace of the energy level E2 are all zero. On the other hand, for the second excited level of the hydrogen atom, n = 3 , its degenerate states have both even and odd parities, so that linear Stark effect exists.

2009 a. A nonrelativistic particle of mass m moves in the potential V ( x , y , z ) = A( x 2 + y 2 + 2λ xy ) + B( z 2 + 2µ z ), where A > 0, B > 0,| λ |< 1, µ is arbitrary. Find the energy eigenvalues.

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b. Now consider the following modified problem with a new potential Vnew : for z > − µ and any x and y, Vnew = V , where V is the same as in part (a) above; for z < − µ and any x and y, Vnew = + ∞. Find the ground state energy. ( CUS) Sol: a. We choose two new variables µ , t defined by 1 ( x + y ), 2

t=

1 ( x − y ), 2

x = (µ + t ) / 2,

y=

1 (µ − t ), 2

µ= or

and write the potential as 1 1 V ( x , y , z ) = A (µ 2 + t 2 + 2µt ) + (µ 2 + t 2 − 2µt ) 2 2 1 2 2  (µ − t ) + B(z 2 + 2µ z ) 2 = A 1 + λ )µ 2 + (1 − λ )t 2  + B( z 2 + 2µ z ) +2λ ⋅

and the differentials as ∂ ∂ 1 ∂ 1 = + ⋅ , ∂x ∂µ 2 ∂t 2 ∂2 1  1 ∂2  ∂2 1 = 2 +  2 ∂x  ∂µ 2 ∂µ ∂t 2  2  ∂2 1 ∂2 1  1 + + 2 ;   ∂t ∂µ 2 ∂t 2  2 ∂ 1 ∂ ∂ 1 = − , ∂ y ∂µ 2 ∂t 2 ∂2  ∂2 1 ∂2 1  1 = −   ∂ y 2  ∂µ 2 2 ∂µ ∂t 2  2 ∂2 1  1  ∂2 1 − − 2  , 2  ∂t ∂µ 2 ∂t 2  ∇2 =

∂2 ∂2 ∂2 + 2 + 2. 2 ∂µ ∂t ∂z

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Then Schrödinger equation becomes  ∂2 ∂2 ∂2  2m  2 + 2 + 2  φ (µ , t , z ) + 2 [E − V (µ , t , z )]φ (µ , t , z ) = 0 . ∂t ∂z    ∂µ Let φ (µ , t , z ) = U (µ )T (t )Z ( z ). The above equation can be separated into ∂2 2m U (µ ) + 2 [E1 − A(1 + λ )µ 2 ]U (µ ) = 0, ∂µ 2  ∂2 2m T (t ) + 2 [E2 − A(1 − λ )t 2 ]T (t ) = 0, 2 ∂t  ∂2 Z ( z ) + (2m /  2 )[E3 − B( z 2 + 2µ z )] Z ( z ) = 0, ∂z 2 with E1 + E2 + E3 = E. By setting z ′ = z + µ , E3′ = E3 + Bµ 2 , all the above three equations can be reduced to that for a harmonic oscillator. Thus the energy eigenvalues are 1 2A  (1 + λ ); E1 =  n1 +  ω 1 , ω 1 =  2 m 1 2A  E2 =  n2 +  ω 2 , ω 2 = (1 − λ );   2 m 1 2B  E3 =  n3 +  ω 3 − Bµ 2 , ω 3 = .  m 2 (n1 , n2 , n3 = 0, 1, 2, 3, …) b. With a new potential Vnew such that for z < −µ , Vnew = ∞ and for z > −µ , Vnew is the same as that in (a), the wave function must vanish for z → −µ . The Z-equation has solution 2

Z ∼ H n3 (ζ )e −ζ /2 , where ζ = (2mB /  2 )1/4 ( z + µ ), Hn3 (ζ ) is the n3 th Hermite polynomial and has the parity of n3 . Hence n3 must be an odd integer. The ground state is the state for n1 = n2 = 0 and n3 = 1 , with the corresponding energy E = (ω1 + ω 2 )/2 + 3ω3 /2 − Bµ 2 .

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2010 Find the root mean squared (RMS) value of the momentum pˆ for the first three states for a particle trapped in a 1D quantum harmonic oscillator. Sol: (i) 

mω  + (a − a ) 2 2 −mω  + p2 = a − a) ( 2

0 p2 0

p=i

0 0 0 −mω  0 a+a+ − a+a − aa+ + aa 0 2 mω  mω  0 aa+ 0 = 1 0a1 = 2 2 mω  mω  = 1 1 00 = . 2 2

=

(ii)  1 p2 1 −mω  2 −mω  = 2 −mω  = 2 −mω  = 2

0

0

1 a + a + − a + a − aa + + aa 1

=

{− 1 a a 1 − 1 aa 1 } {− 1 1 a 0 − 2 1 a 2 } +

+

+

{−

}

1 1 11 − 2 2 11 =

3mω  . 2

(iii)     2 p2 2 −mω  2 −mω  = 2 −mω  = 2 −mω  = 2 5mω  . = 2

=

0

0

2 a + a + − a + a − aa + + aa 2

{− 2 a a 2 − 2 aa 2 } {− 2 2 a 1 − 3 2 a 3 } {−

+

+

+

2 2 2 −3 2 2

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2011 Assume that the eigenstates of a hydrogen atom isolated in space are all known and designated as usual by ψnlm ( r , θ , φ ) = Rnl ( r )Ylm (θ , φ ). Suppose the nucleus of a hydrogen atom is located at a distance d from an infinite potential wall which, of course, tends to distort the hydrogen atom. a. Find the explicit form of the ground state wave function of this hydrogen atom as d approaches zero. b. Find all other eigenstates of this hydrogen atom in half-space, i.e. d → 0, in terms of the Rnl  and  Ylm . (Buffalo) Sol: a. Choose a coordinate system with origin at the center of the nucleus and z-axis perpendicular to the wall surface as shown in Fig. 2.4. As d → 0, the solutions of the Schrödinger equation are still RnlYlm in the half-space z > 0 i.e., 0 < θ < π / 2, but must satisfy the condition ψ = 0 at θ = π / 2 where V = ∞ . That is, only solutions satisfying l + m = odd integer are acceptable. As | m | ≤ l , the first suitable spherical harmonic is Y10 = 3/4 π cosθ .

Fig. 2.4 Since n ≥ l + 1 , the ground state wave function is R21Y10.

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b. All the other eigenstates have wave functions Rnl Ylm where l + m = odd integer. For a given l, we have m = l − 1, 1 − 3, … , − 1 + 1 and hence a degeneracy l.

2012 A harmonic oscillator with constant “k” in ground state suddenly changes its constant to 2k. What is the probability that the oscillator remains in ground state after this sudden change? Sol: 1/2

2 2  α  ψ i =  i  e −α i x /2  π

 αf  ψf =  π 

1/2

e

αi =

− α f 2 x 2 /2

αf =

mω i  mω f 

ωi =

k m

ωf =

2k m

α f = 2 αi

Probability amplitude = ψ i ψ f α

 α  =∫ i   π 0 = =

1/2

( 2)

αi π

( 2)

αi

1/2

1/2

2 2

e −α i x

π

α

∫e

/2

 2α i  ×  π 

1/2 2 2

e −2α i x

/2

−3α 2 x 2 /2

0

×

π 3α 2

 2 =  3 

1/2

 2 Probability = Prob amp =   3  2

1/2 2

=

2 . 3

2013 The ground state energy and Bohr radius for the hydrogen atom are E0 = −e 2 /2a0 , a0 =  2 /me 2 , where m is the reduced mass of the system. [ me = 9.11 × 10−28 g, mp = 1.67 × 10−24 g, e = 4.80 × 10−10 e.s.u.,  = 1.05 × 10−27 erg sec.]

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a. Compute the ground state energy and Bohr radius of the positronium. b. What is the degeneracy of the positronium ground state due to electron spin? Write down the possible spin wave functions which have definite values of the total spin together with the corresponding eigenvalues. c. The ground state of positronium can decay by annihilation into photons. Calculate the energy and angular momentum released in this process and prove that there must be at least two photons in the final state. (Buffalo) Sol: a. The reduced mass m of the positronium is given by 1 = 1 + 1 , i.e., m = me /2. m me me Its use in the formulas gives 2

−e 2 mc 2  e 2  −0.51 × 106 =− E0 = = − 6.8 eV,   = 2a0 2  c  4 × 137 2 a0 = 1.05 × 10−8 cm. b. The degeneracy of the positronium ground state is 4. Denote the positron by 1 and the electron by 2, and let the spin eigenstates in the z direction of a single particle be α and β , corresponding to eigenvalues  /2 and −  /2 respectively. Then the ground eigenstates with definite total spin S = s1 + s2 and z-component of the total spin Sz = s1z + s2 z are

α (1) α (2), 1 [α (1)β (2) + β (1)α (2)], 2 β (1)β (2), 1 [α (1)β (2) − β (1)α (2)], 2

S = ,

Sz =  .

S = ,

Sz = 0.

S = ,

Sz = −  .

S = 0,

Sz = 0.

c. The energy released in the annihilation process mostly comes from the rest masses of the electron and positron, ∆E = 2me c 2 = 1.02 MeV . The released angular momentum depends on the state of the positronium before the annihilation. For the ground state S = 0 , no angular momentum is released. For the state S =  , the angular momentum released is ∆J = l(l + 1) = 2 . There must be at least two photons in the final state of an annihilation, for if there were only one photon produced in the annihilation of a positronium, the energy and momentum of the system could not both be conserved. This can be shown by a simple argument. If the single photon has energy E, it must

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have a momentum E /c at the same time. Thus the momentum of the photon cannot be zero in any reference frame. But in the positronium’s rest frame the momentum is zero throughout the annihilation. Hence we have to conclude that there must be at least two photons in the final state of an annihilation whose momenta cancel out in the rest frame of the positronium.

2014 Consider an electron moving in a spherically symmetric potential V = kr, where k > 0. a. Use the uncertainty principle to estimate the ground state energy. b. Use the Bohr-Sommerfeld quantization rule to calculate the ground state energy. c. Do the same using the variational principle and a trial wave function of your own choice. d. Solve for the energy eigenvalue and eigenfunction exactly for the ground state. (Hint: Use Fourier transforms.) e. Write down the effective potential for nonzero angular momentum states. (Berkeley) Sol: a. The uncertainty principle states that  ∆p∆r ≥ , 2 where ∆p = ( p − p )2 

1/2

= ( p 2 − 2 pp + p 2 )

1/2

= ( p 2 − p 2 )1/2 , ∆r = (r 2 − r 2 )1/2 . The potential is spherically symmetric, so we can take p = 0 , i.e. ∆p ≈ p 2 . For an estimate of the energy E=

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we shall also take ∆r ∼ r . Then E≈

(∆p )2 (∆p )2 k + k∆r ≥ + . 2m 2∆p 2m

For the ground state energy E, we have ∂E ∆p k − = 0, = ∂∆p m 2(∆p )2 giving  mk  ∆p =    2 

1/3

and 1/3

3  k 2 2  E=   . 2  4m  b. The Bohr-Sommerfeld quantization rule gives

∫ P dr = n h, r

∫ P dφ = n h . φ

r

φ

Choose polar coordinates such that the particle is moving in the plane θ = π / 2. The ground state is given by nr = 0, nφ = 1, and the orbit is circular with radius a. The second integral gives Pφ = Iω = ma 2ω =  . The central force is Fr = −

dV = −k = −mω 2a . dr

Combining we have a = ( 2 /mk )1/3, and hence 3 E0 = Pφ2 /2ma 2 + ka = (k 2  2 /m)1/3. 2 c. The notion in the ground state does not depend on θ and φ . Take a trial wave function ψ = exp(−λ r ) and evaluate H=

ψ | Hˆ | ψ ψ |ψ

,

where 2 2 Hˆ = − ∇ + kr . 2m

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As 2 ψ | Hˆ | ψ ∝ − 2m

∫

∞ − λr

e

0

1 d  2 d − λr  2 e  r dr r  r 2 dr  dr

+ k ∫ 0 r 3e −2 λr dr ∞

=

∞ 2  2 k3! λ ∫ 0  − λ  e −2 λr r 2 dr + r  (2λ )4 2m

=

2 3k + 4, 8mλ 8λ

ψ |ψ ∝

∫

∞ −2 λr 2 0

r dr =

e

2 1 = 3, 3 (2λ ) 4 λ

we have H=

 2 λ 2 3k + . 2m 2λ

For stable motion, H is a minimum. Then taking ∂H = 0, ∂λ we find 1/3

 3mk  λ = 2  ,  2  and hence 1/3

H=

9 k 3  9 k 2 2  =   . 4 λ 24 m 

d. The Schrödinger equation for the radial motion can be written as −

2 d 2 χ + (kr − E ) χ = 0, 2m dr 2

where χ = rR , R being the radial wave function. For the ground state, the angular wave function is constant. By the transformation 1/3

 2mk   E  y =  2  r −  ,     k the Schrödinger equation becomes the Airy equation d2 χ( y) − y χ ( y ) = 0, dy 2

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whose solutions are Ai(−x ) and Ai( x ), where x = − | y | , for y < 0 and y > 0 respectively. The boundary conditions that R(r ) and R′(r ) be continuous at r = Ek , i.e. y = 0 , are satisfied automatically as Ai( x ) = Ai(−x ) , Ai′( x ) = Ai′(−x ) for x → 0. The condition that R(r ) is finite at r → 0 requires that Ai(−x ) = rR(r ) → 0 as r → 0. The first zero of Ai(−x ) occurs at x = x 0 ≈ 2.35. Hence the ground state energy is 1

 2  3 E0 =   kx 0 ,  2mk  and the ground state eigenfunction is 1

 2mk  3  E  1 R(r ) = Ai(−x )  with  x =  2   0 − r  .     k r e. The effective potential for nonzero angular momentum is Veff = kr +  2l(l + 1) / 2mr 2.

2015 The interactions of heavy quarks are often approximated by a spinindependent nonrelativistic potential which is a linear function of the radial variable r, where r is the separation of the quarks: V ( r ) = A + Br . Thus the famous “charmonium” particles, the ψ and ψ ′, with rest energies 3.1 GeV and 3.7 GeV (1 GeV = 109 eV) , are believed to be the n = 0 and n = 1 bound states of zero orbital angular momentum of a “charm” quark of mass mc = 1.5 GeV/c 2 (i.e. E = 1.5 GeV) and an anti-quark of the same mass in the above linear potential. Similarly, the recently discovered upsilon particles, the ϒ and ϒ ′ , are believed to be the n = 0 and n = 1 zero orbital angular momentum bound states of a “bottom” quark and anti-quark pair in the same potential. The rest mass of bottom quark is mb = 4.5 GeV/c2 . The rest energy of ϒ is 9.5 GeV. a. Using dimensional analysis, derive a relation between the energy splitting of the ψ and ψ ′ and that of the ϒ and ϒ ′, and thereby evaluate the rest energy of the ϒ ′ . (Express all energies in units of GeV)

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b. Call the n = 2 , zero orbital angular momentum charmonium particle the

ψ ″ Use the WKB approximation to estimate the energy splitting of the ψ ′ and the ψ ″ in terms of the energy splitting of the ψ and the ψ ′, and thereby give a numerical estimate of the rest energy of the ψ ″ . (Princeton) Sol:

In the center-of-mass system of a quark and its antiquark, the equation of relative motion is  2 2  µ = mq / 2, − ∇ r + V (r )ψ (r ) = ERψ (r ),  2µ  where ER is the relative motion energy, mq is the mass of the quark. When the angular momentum is zero, the above equation in spherical coordinates can be simplified to  2 1 d  2 d    + V (r ) R(r ) = ER R(r ). r − 2  2µ r dr  dr   Let R(r ) = χ 0 (r ) / r. Then χ 0 (r ) satisfies d 2 χ 0 2µ + [ER − V (r )]χ 0 = 0, dr 2  2 i.e., d 2 χ 0 2µ + [ ER − A − Br ] χ 0 = 0. dr 2  2 a. Suppose the energy of a bound state depends on the principal quantum number n, which is a dimensionless quantity, the constant B in V(r), the quark reduced mass µ , and , namely E = f (n)B x µ y  z . As, [E] = [ M ][L]2[T ]−2 , −2 [ B] = [ M ][ L ][T ] , 2

[µ ] = [ M ]

−1

[  ] = [ M ][ L ] [T ] , we have 2 1 x =z= , y=− 3 3

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and hence E = f (n)( B )2/3 (µ )−1/3, where f (n) is a function of the principal quantum number n. Then ∆Eψ = Eψ ′ − Eψ = f (1) =

( B )2/3 ( B )2/3 − f (0) µc1/3 µc1/3

( B )2/3 [ f (1) − f (0)], µc1/3

and similarly ∆Eϒ =

( B )2/3 [ f (1) − f (0)]. µb1/3

Hence 1/3

1/3 1 ∆Eϒ  µc  =  =  . 3 ∆Eψ  µb 

As Eϒ ′ − Eϒ ≈ 0.42 GeV, Eϒ ′ = Eϒ + 0.42 = 9.5 + 0.42 ≈ 9.9 GeV. b. Applying the WKB approximation to the equation for χ 0 we obtain the Bohr-Sommerfeld quantization rule 2 ∫0

α

2µ ( ER − A − Br ) dr = (n + 3 / 4) with a =

ER − A , B

which gives, writing En for ER, 2/3

   3 3 n +  B / 4   4 En = A + . 1/3 (2µ ) Application to the energy splitting gives 23 23 9  ( B )2/3  21  Eψ ′ − Eψ = −  ,      16   (2µc )1/3  16  Eψ ″ − Eψ ′ =

23 23  21   ( B )2/3  33    −    , 1/3  16   (2µc )  16 

and hence Eψ ″ − Eψ ′ Eψ ′ − Eψ

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=

(33)2/3 − (21)2/3 ≈ 0.81. (21)2/3 − (9)2/3

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Thus Eψ ″ − Eψ ′ = 0.81 × ( Eψ ′ − Eψ ) = 0.81 × (3.7 − 3.1) ≈ 0.49 GeV, and Eψ ″ = 3.7 + 0.49 ≈ 4.2 GeV.

2016 Find the acceleration operator for a particle under the Hamiltonian H = AP2 + Bx2, where A and B are real constants. Sol:

Ehrenfest theorem Vx =

∂x dx 1 = [ x1 ,H ] + dt i ∂t ∂x 1 =  x1 , AP2 + Bx 2  + ∂t i 0 ∂x 1 =  x , AP2  +  x ,Bx 2  + ∂t i 1 = (A2P i + 0 + 0) = 2AP. i

{



}

dVx 1 ∂V = [Vx ,H ] + x dt i ∂t ∂V 1 =  2AP, AP2 + Bx 2  + x ∂t i ∂V 1 =  2AP, AP2  +  2AP, Bx 2  + x ∂t i 0 ∂V 1 = 2BA P x 2  + x ∂t i 1 = 2BA − 2x i = − 4ABx . i

ax =

{



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2017 Prove that in any single-particle bound energy eigenstate the following relation is satisfied in nonrelativistic quantum mechanics for a central fieldof-force potential V(r), | ψ (0)|2 =

m dV ( r ) 1 L2 − , 2π dr 2π r 3

where ψ (0) is the wave function at the origin, m the particle mass, and L2 the square of the orbital angular momentum operator (let  =1). Give a classical interpretation of this equation for the case of a state with angular momentum ≠ 0. ( Columbia) Sol: a. In the field of central force, the Schrödinger equation is 1  1 ∂  2 ∂  lˆ2   − r  − ψ + V (r )ψ = Eψ . 2m  r 2 ∂r  ∂r  r 2  Let 1 ψ (r , θ , ϕ ) = R(r ) Ylm (θ , ϕ ) ≡ u(r )Ylm (θ , ϕ ), r where u(r ) = rR(r ), and we have for the radial motion

{

u″+ 2m [ E − V (r )] −

}

l(l + 1) u = 0, (r > 0). r2

Multiplying the two sides of the above with u′(r ) and integrating from r = 0 to r = ∞ , we get

∫ u′(r )u″(r )dr + ∫

{2m [E − V (r )] − l(lr+1)} 12 u (r ) dr = 0. 2

′

2

For the eigenstates we may assume u′(∞) = 0, u(∞) = u(0) = 0. With u′(0) = [ R(r ) + rR′(r )]r =0 = R(0) , partial integration gives 1 1 dV (r ) 2 − R 2 (0) + 2m ∫ R 2 r dr − 2 2 dr

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∫

2l(l + 1) 2  R rdr  = 0.  r3

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Hence 1 2 R (0) 4π m dV (r ) = [ R(r )Ylm (θ , ϕ )] [ R(r )Ylm (θ , ϕ )] r 2drdΩ ∫ dr 2π Lˆ2 1 − [ R(r )Ylm (θ , ϕ )] 3 [ R(r )Ylm (θ , ϕ )] r 2drdΩ, ∫ 2π r

| ψ (0)|2 =

or | ψ (0)|2 =

m dV (r ) 1 Lˆ2 − . 2π dr 2π r 3

 2 = 0 , and so b. For l ≠ 0, | ψ (0)| dV (r ) 1 Lˆ2 = . dr m r3 Its corresponding classical expression is d 1 L2 V (r ) = . dr m r3

dV (r ) Here F = − is the centripetal force, and r dr

υ2 1 L2 | r × v |2 1 =m = m (v sin∠r , v )2 = m t , 3 3 mr r r r = mar , where υt is the tangential velocity along the spherical surface of r, is mass v2 multiplied by the centripetal acceleration ar = t . The equation thus r expresses Newton’s second law of motion.

2018 A spinless particle of mass m is subject (in 3 dimensions) to a spherically symmetric attractive square-well potential of radius r0. a. What is the minimum depth of the potential needed to achieve two bound states of zero angular momentum? b. With a potential of this depth, what are the eigenvalues of the Hamiltonian that belong to zero total angular momentum? (If necessary you may express part of your answer through the solution of a transcendental equation.)

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c. If the particle is in the ground state, sketch the wave function in the coordinate basis and the corresponding coordinate probability distribution. Explain carefully the physical significance of the latter. d. Predict the result of a (single) measurement of the particle kinetic energy in terms of this wave function. You may express your prediction through one-dimensional definite integrals. e. On the basis of the uncertainty principle, give a qualitative connection between parts (c) and (d) above. (Berkeley) Sol: a. The attractive potential may be represented by V = −V0 , where V0 is a positive constant. For bound states 0 > E > −V0 . Thus for l = 0 the radial wave function R(r ) = χ (r ) / r satisfies the equations −

2 d 2 χ − V0 χ = E χ , 2m dr 2

(0 < r < r0 )

−

2 d 2 χ = Eχ , 2m dr 2

(r0 < r < ∞)

with χ (0) = 0 and χ (∞) finite. To suit these conditions the wave function may be chosen as follows: χ (r ) = sin α r ,

χ (r ) = B exp(−β r ),

0 < r < r0 , r0 < r < ∞,

where α = 1 2m( E + V0 ), β = 1 −2mE . From the boundary condition that at r = r0 , χ and χ ′ should be continuous we get −α cot α r0 = β . Defining ξ = α r0 , η = β r0, we have

ξ 2 + η 2 = 2mV0r02 /  2 , −ξ cot ξ = η . Each set of the positive numbers ξ , η satisfying these equations gives a bound 2 2 2 state. In Fig. 2.5 curve 1 represents η = −ξ cot ξ and curve 2, ξ + η = 2 , for

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example. As shown in the figure, for a given value of V0 , to have two intersections in the quadrant we require 2

2mV0r02  3π  ≥  ,  2  2 or V0 ≥

9π 2  2 , 8mr02

which is the minimum potential depth needed to achieve two bound states of zero angular momentum.

Fig. 2.5 b. With a potential of depth given above, one intersection occurs at η = 0 , for which β = −2mE = 0 , i.e., E = 0 . The other intersection occurs at

( ) , i.e., ξ = 32π

ξ 2 + η 2 = 3π 2

2

2

 2β r0  1−  , and − ξ cot ξ = η , i.e.,  3π 

2 2   2β r0   2β r0   1  3π  3π 1− −   1−  cot   = β.  3π   3π   r0  2   2

Solving for β , we get the second eigenvalue of the Hamiltonian, E =−

β 2 2 . 2m

c. Setting the normalized ground state wave function as

χ (r ) = Asin α r ,

0 ≤ r ≤ r0 ,

χ (r ) = A sin ar0 exp β (r0 − r ) ,

r > r0 ,

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we have 4π

∫

∞ 0

R 2r 2 dr = 4 π

∫

∞ 0

u 2 dr

= 4 π A 2 ∫ 0 sin 2α r dr + 4 π A 2 sin 2α re 2 βr0 r0

× ∫ e −2 βr dr = 1, ∞

r0

or 1 1 1 = (α r0 − sin α r0 cos α r0 ) + sin 2α r0. 2 4 π A 2α 2β The wave function and probability distribution are shown in Figs 2.6(a) and 2.6(b) respectively. It can be seen that the probability of finding the particle is very large for r < r0 and it attenuates exponentially for r > r0 , and we can regard the particle as being bound in the square-well potential. d. The kinetic energy of the particle, ET = p 2 / 2m , is a function dependent solely on the momentum p; thus the probability of finding a certain value of the kinetic energy by a single measurement is the same as that of finding the corresponding value of the momentum p, | ψ (p )|2. Here ψ (p ) is the Fourier transform of the ground state coordinate wave function,

Fig. 2.6(a)

Fig. 2.6(b)

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1 (2π  )3/2

|ψ (p)|2 = |ψ (p )|2 = =

1 4π (2π  )3

1 = 2 3 2π 

∫

∞

0

1 χ (r ) − ip ⋅ r / e dr 4π r

∫

∞

π

2π

0

0

0

∫ ∫ ∫

2

χ (r ) − ipr cosθ / sinθ dθ dϕ r 2 dr e r

( ) ( ) pr

χ (r ) sin  2 r dr pr r 

2

2

.

The integration can be effected when the expression for χ (r ) in (c) is substituted in the integrand. The average kinetic energy ET is ET = ψ1

p2 ψ1 = E1 − ψ1 | V | ψ1 2m

1 ⋅ 4 π r 2 dr r2  sin2α r0  = E1 + 2π V0 A 2 r0 − .  2α  = E1 + V0 ∫ 0 A 2 sin 2α r ⋅ r0

e. From the above we see that the wave function in space coordinates in (c) gives the space probability distribution, whereas the wave function in p-space in (d) gives the momentum probability distribution. The product of uncertainities of one simultaneous measurement of the position and the momentum must satisfy the uncertainty principle ∆p∆r ≥  / 2. That is to say, the two complement each other.

2019 a. Given a one-dimensional potential (Fig. 2.7) V = −V0 , | x | < a, V = 0, | x | > a, show that there is always at least one bound state for attractive poten­ tials V0 > 0 . (You may solve the eigenvalue condition by graphical means.) b. Compare the Schrödinger equation for the above one-dimensional case with that for the radial part U(r)of the three-dimensional wave function when L = 0,

ψ (r ) = r −1U ( r )YLM (Ω),

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where ψ (r ) is the solution of the Schrödinger equation for the potential V = −V0 , V = 0,

r < a, r > a.

Why is there not always a bound state for V0 > 0 in the three-­ dimensional case? (MIT)

Fig. 2.7 Sol: a. For the bound state of a particle, E < 0. For | x | > a, V = 0 and the Schrödinger equation d 2ψ 2mE + 2 ψ =0 dx 2  has solutions  Ae − k ′x , ψ ( x ) =  k ′x  Be ,

x > a, x < −a,

where k′ = −

2mE . 2

For | x |< a , the Schrödinger equation d 2ψ 2m (V0 + E )ψ = 0 + dx 2  2 has solutions

ψ ( x ) ∼ cos kx , (even parity) ψ ( x ) ∼ sin kx, (odd parity)

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where k=

2m(V0 + E ) , 2

provided E > −V0 . Here we need only consider states of even parity which include the ground state. The continuity of the wave function and its derivative at x = ±a requires k tan ka = k′ . Set ka = ξ , k′a = η . Then the following equations determine the energy levels of the bound states: ξ tan ξ = η ,

ξ 2 + η 2 = 2mV0a 2 /  2 . These equations must in general be solved by graphical means. In Fig. 2.8(a), curve 1 is a plot of η = ξ tan ξ ,  and curve 2 plots ξ 2 + η 2 =1. The dashed line 3 is the asymtotic curve for the former with ξ = π / 2. Since curve 1 goes through the origin, there is at least one solution no matter how small is the value of V0a2. Thus there is always at least one bound state for a onedimensional symmetrical square-well potential.

Fig. 2.8(a)

Fig. 2.8(b)

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b. For r > a, the radial Schrödinger equation d 2U 2mE + 2 U = 0, dr 2  has solution U (r ) = A exp(−κ ′r ), where

κ′ = −

2mE . 2

For r < a, the equation is d 2U 2m (V0 + E ) U = 0 + dr 2  2 and the solution that satisfies the boundary condition U (r ) r → 0→ 0 is U (r ) = B sin κ r , where

κ=

2m(V0 + E ) . 2

The continuity of the wave function and its derivative at r = a , requires

κ cot κ a = −κ ′. Setting κ a = ξ , κ ′a = η we get ξ cot ξ = −η . ξ 2 + η 2 = 2mV0a 2 /  2 . These are again to be solved by graphical means. Fig. 2.8(b) shows curve 1 which is a plot of ξ cot ξ = −η , and the dashed line 2 which is its asymptotic if ξ = π . It can be seen that only when 2

π  ξ 2 + η 2 = 2mV0a 2 /  2 ≥   , 2 or V0a 2 ≥ π 2  2 /8m, can the equations have a solution. Hence, unlike the one-dimensional case, only when V0a 2 ≥ π 2  2 /8m can there be a bound state.

2020 a. Consider a particle of mass m moving in a three-dimensional square well potential V (| r |). Show that for a well of fixed radius R, a bound state exists only if the depth of the well has at least a certain minimum value. Calculate that minimum value.

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b. The analogous problem in one dimension leads to a different answer. What is that answer? c. Can you show that the general nature of the answers to (a) and (b) above remains the same for a well of arbitrary shape? For example, in the one-dimensional case (b) V ( x ) = λ f ( x ) < 0, V ( x ) = 0,

a ≤ x ≤ b, x < a or x > b ,

consider various values of λ while keeping f ( x ) unchanged. ( CUSPEA) Sol: a. Suppose that there is a bound state ψ (r ) and that it is the ground state (l = 0), so that ψ (r ) = ψ (r ). The eigenequation is −

 2 1 d  2 d ψ (r ) + V (r )ψ (r ) = Eψ (r ), r 2  2m r dr  dr

where E < 0, and V (r ) = 0,

r > R,

V (r ) = −V0 ,

0 < r < R,

with V0 > 0 , as shown in Fig. 2.9. The solution is

Fig. 2.9   Asin(kr ) / r , r < R , k = 2m( E + V0 ) ,  2 ψ (r ) =  −2mE  − k ′r , r > R , k′ =  Be / r , 2 

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where A and B are normalization constants. The continuity of ψ and ψ ′ at r = R , or equivalently

[ ln(rψ(r ))]′r=R = [ ln(rψ(r ))]′r=R , −

+

gives k cot (kR ) = −k′, while the definitions of κ , κ ′ require k 2 + k′ 2 =

2mV0 . 2

These equations can be solved graphically as in Problem 2018. In a similar way, we can show that for there to be at least a bound state we require 2

2mV0 R 2  π  ≥  ,  2 2 i.e., V0 ≥

π 2 2 . 8mR 2

b. If the potential is a one-dimensional rectangular well potential, no matter how deep the well is, there is always a bound state. The ground state is always symmetric about the origin which is the center of the well. The eigenequation is d2 2m ψ ( x ) + 2 [ E − V ( x )] ψ ( x ) = 0, dx 2  where, as shown in the Fig. 2.10, V ( x ) = −V0 ,

(V0 > 0),

V ( x ) = 0,

x < R / 2, x > R / 2.

Fig. 2.10

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For bound states, we require 0 > E > −V0 . As V ( x ) = V (−x ), the equation has solution  R  Acos(kx ), | x | < ,  2 ψ (x ) =  R  Be − k ′|x| , |x |> ,  2 where k, k’ have the same definitions as in (a). The continuity of ψ and ψ ′ at x = R gives 2 tan(kR / 2) = k′ / k . or sec 2 (kR / 2) =

V0 , E + V0

i.e., cos(kR / 2) = ±

E + V0 . V0

Since V0 > −E > 0 , there is always a bound-state solution for any V0. c. For a one-dimensional potential well of arbitrary shape, we can always define a rectangular potential well Vs(x) such that Vs ( x ) = − V0 ,

| x | < R / 2,

Vs ( x ) = 0,

| x | > R / 2,

and −V0 ≥ V ( x ) always (see Fig. 2.11). From (b) we see that there always exists a ψ0 ( x ) which is a bound eigenstate of Vs(x) for which

ψ0

p2 + Vs ( x ) ψ0 < 0 . 2m

Fig. 2.11

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Since

ψ0

p2 p2 + V ( x ) ψ0 < ψ0 + Vs ( x ) ψ0 , 2m 2m

we have

ψ0

p2 + V (x ) ψ 0 < 0 . 2m

This means that there is always a bound state for a one-dimensional well of any shape.

2021 Calculate Green’s function for a nonrelativistic electron in the potential V ( x , y , z ) = ∞, x ≤ 0, (any y , z ) V ( x , y , z ) = 0, x > 0, (any y , z ) and evaluate | G(r , r ′ , t )|2 . Describe the evolution in time of the pattern of probability and interpret physically the reason for this behavior. (Berkeley) Sol:

The potential in this problem can be replaced by the boundary condition G(r , r ′ , t ) = 0 and x = 0. The boundary problem can then be solved by the method of images. Suppose at r ′′ is the image of the electron at r′ about x = 0. Then

(i ∂t − H ) G(r , r ′, t ) = δ (t ) [δ (r − r ′) − δ (r − r ′′)] . (1) The Green’s function is zero for x ≤ 0 and for x ≥ 0 is equal to the x > 0 part of the solution of (1). Let





G(r , r ′, t ) =

1 (2π )4

∫d k∫ 3

∞ −∞

e ik⋅r−iωt G(k, r ′, ω )dω . (2)

2 2 We have i ∂t G = wG and H =  k , and the substitution of (2) in (1) gives 2m 1 −ik ⋅r ′ G( k , r ′, ω ) = − e −ik⋅r ′′ ), (3) 2 2 (e ω − 2mk Re-substituting (3) in (2) gives



G(r , r ′ , t ) =

− ik ⋅r ′ 1 − e − ik⋅r ′′ ) 3 ik ⋅r −iω t (e d k e dω . (4) 2 2 ∫Γ (2π )4 ∫ ω − 2mk

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We first integrate with respect to ω . The path Γ is chosen to satisfy the causality condition. Causality requires that when t < 0, G(r , r , t ) = 0. First let the polar point of ω shift a little, say by −iε , where ε is a small positive number. Finally letting

ε → 0, we get G(r , r ′, t ) =

−i (2π )3



k 2 

∫ exp ik ⋅ r − i 2m t (e

−ik ⋅r ′

− e −ik⋅r ′′ )d 3k

3/2  im(r − r ′′)2  1  m    im(r − r ′)2  =  exp   − exp   .    2π t    2t   2t 

(5)

Hence when both x and t are greater than zero, the Green’s function is given by (5); otherwise, it is zero. When x > 0 and t > 0, 3

1 m   2  2π t    im 2  [r′ − r ′′2 − 2r ⋅ (r′ − r′′)] . × 2 − 2 Re exp   2 t  

| G(r , r′, t )|2 =

If the potential V ( x , y , z ) were absent, the Green’s function for the free space, | G(r , r ′, t )|2, would be proportional to t–3. But because of the presence of the reflection wall the interference term occurs.

2022 An electron moves above an impenetrable conducting surface. It is attracted toward this surface by its own image charge so that classically it bounces along the surface as shown in Fig. 2.12. a. Write the Schrödinger equation for the energy eigenstates and energy eigenvalues of the electron. (Call y the distance above the surface.) Ignore inertial effects of the image.

Fig. 2.12

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b. What is the x and z dependence of the eigenstates? c. What are the remaining boundary conditions? d. Find the ground state and its energy. [Hint: they are closely related to those for the usual hydrogen atom). e. What is the complete set of discrete and/or continuous energy eigenvalues? ( Columbia) Sol: a. Figure 2.13 shows the electron and its image. Accordingly the electric energy for the system is V (r ) = 1 ∑ i qi Vi = 1 e ⋅ −e + (−e ) ⋅ e  = −e 2 /4 y . 2 2  2y 2y  The Schrödinger equation is then  2 2 e 2  ∇ − ψ ( x , y , z ) = Eψ ( x , y , z ). − 4y   2m b. Separating the variables by assuming solutions of the type ψ ( x , y , z ) ≡ ψn ( y )φ ( x , z ) ≡ ψn ( y )φ x ( x )φ z ( z ), we can write the above equation as

−

2 d 2 e2 ψn ( y ) − ψn ( y ) = E yψn ( y ), (1) 2 2m dy 4y

Fig. 2.13 h2 d 2 px2 x φ = φ x ( x ), ( ) x 2m dx 2 2m 2 d 2 pz2 − φ = φ z ( x ), ( z ) z 2m dz 2 2m

−

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with Ey +

px2 pz2 + = E. 2m 2m

2

Note that since V ( y ) = − 4e y depends on y only, px and pz are constants of the motion. Hence

φ ( x , z ) ≡ φ x ( x )φ z ( z ) ∼ e i ( px x+ pz z )/ , and

ψ ( x , y , z ) ≡ ψn ( y )e i ( px x+ pz z )/ . c. The remaining boundary condition is ψ ( x , y , z ) = 0 for y ≤ 0. d. Now consider a hydrogen-like atom of nuclear charge Z. The Schrödinger equation in the radial direction is −

 2 1 d  2 dR  Ze 2 l(l + 1) 2 R = ER. R r − +   2mr 2 2m r 2 dr  dr  r

On setting R = χ / r , the above becomes −

 2 d 2 χ Ze 2 l(l + 1) 2 − χ+ χ = Eχ . 2 2m dr r 2mr 2

In particular, when l = 0 we have

 2 d 2 χ Ze 2 − χ = E χ ,(2) 2m dr 2 r which is identical with (1) with the replacements r → y , Z → 1 . Hence the 4 solutions of (1) are simply y multiplied by the radial wave functions of the ground state of the atom. Thus −

Z  ψ1 ( y ) = yR10 ( y ) = 2 y   a 2 1  where a = 2 . With Z = , we have 4 me

3/2

e − Zy /a ,

3/2

 me 2   me 2 y  . ψ1 ( y ) = 2 y  2  exp − 2   4   4  Note that the boundary condition in (c) is satisfied by this wave function. The ground-state energy due to y motion is similarly obtained: Ey = −

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Z 2me 4 me 4 = − 2 2 32 2

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153

e. The complete energy eigenvalue for quantum state n is En , px , pz = −

me 4 1 + ( px2 + pz2 ), 2 2 32 n 2m

(n = 1, 2, 3, …)

with wave function i  ψn , px , pz (r) = AyRn 0 ( y )exp ( px x + pz z ),   where A is the normalization constant.

2023 A nonrelativistic electron moves in the region above a large flat grounded conductor. The electron is attracted by its image charge but cannot penetrate the conductor’s surface. a. Write down the appropriate Hamiltonian for the three-dimensional motion of this electron. What boundary conditions must the electron’s wave function satisfy? b. Find the energy levels of the electron. c. For the state of lowest energy, find the average distance of the electron above the conductor’s surface. ( Columbia) Sol: a. Take Cartesian coordinates with the origin on and the z-axis perpendicular to the conductor surface such that the conductor occupies the half-space z ≤ 0. 2 As in Problem 2022, the electron is subject to a potential V ( z ) = − e . Hence 4z the Hamiltonian is 1 e2 ( px2 + p y2 + p y2 ) − 2m 4z 2  2 2  ∂ ∂ ∂2  e 2 . =−  2 + 2 + 2− 2m  ∂x ∂y ∂z  4 z

H=

The wave function of the electron satisfies the boundary condition ψ ( x , y , z ) = 0 for z ≤ 0. b. As shown in Problem 2022 the energy eigenvalues are En =

me 4 1 1 . (n = 1, 2, 3 …) ( px2 + p y2 ) − 2m 32 2 n 2

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c. The ground state has energy E=

1 me 4 ( px2 + p y2 ) − , 2m 32 2

and wave function

ψ100 ( x , y , z ) =

Az − z /a′ e , a′3/2

where a′ =

4 2 , me 2

and A is the normalization constant. Hence

∫ ψ z ψ dx dy dz ∫ ψ ψ dx dy dz ∫ z e dz = ∫ z e dz * 100

z =

100

* 100

100

∞ 3 −2 z /a ′

0 ∞ 2 −2 z /a ′ 0

=

3!

( a2′ )

⋅ 4

( a2′ )

3

2!

6 2 3 = a′ = 2 . 2 me

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3 3001 Consider four Hermitian 2 × 2 matrices I, σ 1 , σ 2 , and σ 3 , where I is the unit matrix, and the others satisfy σ i σ j + σ j σ i = 2δij . You must prove the following without using a specific representation or form for the matrices. a. Prove that Tr(σ i ) = 0 . b. Show that the eigenvalues of σ i are ±1 and that det (σ i ) = −1. c. Show that the four matrices are linearly independent and therefore that any 2 × 2 matrix can be expanded in terms of them. d. From (c) we know that 3

M = m0 I + ∑mi σ i , i =1

where M is any 2 × 2 matrix. Derive an expression for mi (i = 0, 1, 2, 3) . (Buffalo) Sol: a. As

σ iσ j = −σ jσ i

(i ≠ j ),

σ jσ j = I ,

we have

σ i = σ iσ jσ j = −σ jσ iσ j , and thus Tr (σ i ) = −Tr (σ jσ iσ j ) = −Tr (σ iσ jσ j ) = −Tr (σ i ) . Hence Tr(σ i ) = 0 . b. Suppose σ i has eigenvector φ and eigenvalue λi , i.e.

σ iφ = λiφ .

157

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Then

σ iσ iφ = σ i λiφ = λiσ iφ = λi2φ . On the other hand,

σ iσ iφ = Iφ = φ . Hence

λi2 =1, or

λi = ±1 . As Tr(σ i ) = λ1 + λ2 = 0, the two eigenvalues of σ i are λ1 = +1, λ2 = −1 , and so Det(σ i ) = λ1λ2 = −1. c. If I , σ i , i = 1, 2, 3, were linearly dependent, then four constants m0 , mi could be found such that 3

m0 I + ∑miσ i = 0 . i=1

Multiplying by σ j from the right and from the left we have 3

m0σ j + ∑miσ iσ j = 0, i=1

and 3

m0σ j + ∑miσ jσ i = 0 . i=1

Adding the above two equations gives 2m0σ j + ∑ mi (σ iσ j + σ jσ i ) + 2m j I = 0, i≠ j

or m0σ j + m j I = 0 . Thus Tr (m0σ j + m j I ) = m0Tr (σ j ) + 2m j = 0 . As Tr(σ j ) = 0 , we would have m j = 0 , and so m0 = 0 . Therefore the four matrices I and σ i are linearly independent and any 2 × 2 matrix can be expanded in terms of them.

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159

d. Given M any 2 × 2 matrix, we can write 3

M = m0 I + ∑miσ i . i=1

To determine the coefficients m0 , mi , take the trace of the above: Tr ( M ) = 2m0 , or 1 m0 = Tr ( M ) . 2 Consider 3

σ j M = m0σ j + ∑miσ jσ i , i=1

and 3

Mσ j = m0σ j + ∑miσ iσ j . i=1

Adding the last two equations gives

σ j M + Mσ j = 2m0σ j + 2m j I . Thus Tr (σ j M + Mσ j ) = 2Tr (σ j M ) = 4m j , or

1 m j = Tr (σ j M ) . 2

3002 The three matrix operators for spin one satisfy s x s y − s y s x = i s z and cyclic permutations. Show that s z3 = s z , ( s x ± i s y )3 = 0. (Wisconsin) Sol:

2 The matrix forms of the spin angular momentum operator (s = 1) in the (s , sz ) 2 representation, in which s , sz are diagonal, are

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 0 1 0   0 −i 0    1  1  sx =  1 0 1  , sy =  i 0 −i  , 2 0 1 0  2 0 i 0       l 0 0    sz =  0 0 0  .  0 0 −l    Direct calculation gives 3

 l 0 0   1 0 0  3   sz = 0 0 0 =  0 0 0  = sz ,      0 0 − l   0 0 − l  3

  0 2 0   1   3 (sx + i s y ) =  0 0 2   = 0,   2   0 0 0    3

  0 0 0   1   (sx − i s y )3 =  2 0 0   = 0.  2    0 2 0   

3003 Three matrices Mx, My, Mz, each with 256 rows and columns, are known to obey the commutation rules [ Mx , My ] = iMz (with cyclic permutations of x, y and z). The eigenvalues of the matrix Mx are ±2, each once; ±3 2, each 8 times; ±1, each 28 times; ±1 2 , each 56 times; and 0, 70 times. State the 256 eigenvalues of the matrix M 2 = Mx2 + My2 + Mz2 . (Wisconsin) Sol:

M2 commutes with Mx. So we can select the common eigenstate M , M x . Then M 2 M , M x = m(m + 1) M , M x , M x M , M x = mx M , M x . For the same m, mx can have the values +m, m − 1, … , −m , while M2 has eigenvalue m(m + 1). Thus

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m

161

M 2 = m (m + 1)

mx

2

±2  ±1 each once  0 

3/ 2

±3 / 2  each 8 times ±1/ 2 

1

6

5 × 1 = 5 times

15 4

4 × 8 = 32 times

±1 each 27 times  0 

2

3 × 27 = 81 times

1/ 2

±1/ 2 each 48 times

3 4

2 × 48 = 96 times

0

0, each 42 times

0

1 × 42 = 42 times Total 256 eigenvalues

3004 A certain state ψ is an eigenstate of Lˆ2 and Lˆz : Lˆ2 ψ = l ( l + 1) 2 ψ , Lˆz ψ = m ψ . For this state calculate Lˆx and Lˆ2x . (MIT) Sol:

As Lˆ2 is a Hermitian operator, we have Lˆz ψ = m ψ → ψ Lˆz = m ψ . Then 1 1 Lˆx = ψ |  Lˆ y , Lˆz  ψ = ψ Lˆ y Lˆz − Lˆz Lˆ y ψ i i m = ψ Lˆ y ψ − ψ Lˆ y ψ = 0. i

(

)

Considering the symmetry with respect to x, y, we have 1 1 ˆ 2 ˆ2 Lˆ2x = Lˆ2y = Lˆ2x + Lˆ2y = L − Lz , 2 2

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and so 2 1  2 − L 2z ψ = 1 [l(l + 1) − m 2 ] 2 . L x = ψ L 2 2

It can also be calculated using the “raising” and “lowering” operators.

3005 The spin functions for a free electron in a basis where sˆz is diagonal can be  1

0

0

 1

written as   and   with eigenvalues of sˆz being +1 2 and −1 2 respectively. Using this basis find a normalized eigenfunction of sˆ y with eigenvalue −1 2 . (MIT) Sol:

In the diagonal representation of sˆ2 , sˆz , we can represent sˆy by 1 0 sˆy =  2 i

−i , 0

( = 1).

a  Let the required eigenfunction of sˆy be σ y =   . Then as b  1 0  2 i

− i  a  1 a    =−   , 0  b  2 b 

i  we have a = ib, and so σ y = b   . 1 Normalization i  σ +y σ y = b 2 (−i , 1)  = 2b 2 = 1, 1 gives b =

1 2

. Hence

σy =

1 i  .   2 1

3006 Consider a spinless particle represented by the wave function

ψ = K ( x + y + 2 z )e −αr,

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163

where r = x 2 + y 2 + z 2 , and K and α are real constants. a. What is the total angular momentum of the particle? b. What is the expectation value of the z-component of angular momentum? c. If the z -component of angular momentum, Lz, were measured, what is the probability that the result would be Lz = + ? d. What is the probability of finding the particle at θ , φ and in solid angle dΩ ? Here θ , φ are the usual angles of spherical coordinates. You may find the following expressions for the first few spherical harmonics useful: Y00 =

1 , 4π

Y10 =

3 15 cosθ , Y2±1 = ∓ sinθ cosθ e ±iφ . 4π 8π

Y1±1 = ∓

3 sinθ e±iφ , 8π

(CUS) Sol:

The wave function may be rewritten in spherical coordinates as

ψ = Kr (cosφ sinθ + sinφ sinθ + 2cosθ )e −αr , its angular part being

ψ (θ , φ ) = K ′(cos φ sinθ + sin φ sinθ + 2cosθ ), where K ′ is the normalization constant such that K ′2

∫

π θ

dθ

∫

2π 0

sinθ (cos φ sinθ + sin φ sinθ + 2cosθ )2 dφ = 1 .

Since 1 1 cosφ = (e iφ + e −iφ ), sinφ = (e iφ − e −iφ ), 2 2i we have 1  1 ψ (θ , φ ) = K ′ (e iφ + e −iφ )sinθ + (e iφ − e −iφ )sinθ + cos2θ , 2  2i  1 4π 0  8π −1 8π 1 1 Y1 . = K ′− (1 − i ) Y1 + (1 + i ) Y1 + 2 3 3 3 2   2

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Problems and Solutions in Quantum Mechanics m The normalization condition and the orthonormality of Yl then give

 1 8π 1 8π 4π  + ⋅ +4⋅ K ′2  ⋅ = 1, 2 3 2 3 3  or K′ =

1 , 8π

and thus  1 1 8π 1 ψ (θ , φ ) =  − (1 − i ) Y1 3  8π 2 4π 0  8π −1 1 + (1 + i ) Y1 + 2 Y1 . 3 3 2  a. The total angular momentum of the particle is L2 = l(l + 1) = 2 . as the wave function corresponds to l = 1. b. The z-component of the angular momentum is  1 8π 1 8π (− )(Y1−1 )2 ψ ∗ LZ ψ = K ′ 2  ⋅ ⋅ (Y11 )2 + ⋅ 2 3 2 3  4π (0)(Y10 )2  +4⋅  3 =

 1  1 8π 1 8π ⋅ (+ ) + ⋅ (− ) = 0.   8π  2 3 2 3

c. The probability of finding Lz = +  is P = Lz = +  | ψ (θ , φ ) =

2

1 1 8π 1 ⋅ ⋅ = . 8π 2 3 6

d. The probability of finding the particle in the solid angle dΩ at θ , ϕ is 1

∫ ψ (θ , ϕ )ψ (θ , ϕ )dΩ = 8π [sinθ (sinφ + cosφ ) + 2cosθ ] dΩ . *

2

3007 Consider a particle in spherically symmetric potential with orbital angular 1 momentum l =  and s = . 2

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165

Find the a. Eigen energy values associated with L-S coupling scheme b. Number of degenerate states in each energy level Sol: a. The Eigen functions of different angular momentum components are given by J 2ϕ =  2 j( j + 1)ϕ , L2ϕ =  2l(l + 1)ϕ , and S 2ϕ =  2 s(s + 1)ϕ ,(1) where j, l and s are quantum numbers and ϕ is the angular component of the wave function. In L-S coupling, the coupling energy is given by the equation 1 1 H = a( L.S ) = a(2 L.S + L2 + S 2 − L2 − S 2 ) = a( J 2 − L2 − S 2 )(2) 2 2 as J = L + S

the possible values of j are given by j = (l + s, …….l − s) 3 1 therefore j =  and (3) 2 2 from (1), (2) and (3) 2 Eso = a[ j( j + 1) − l(l + 1) − s(s + 1)] 2 2 3 For j = , Eso = a 2 2 1 j = , Eso = −  2a 2 b. Degeneracy of the energy states D = 2 j + 1  3  4, j =  2 D= 1  2, j =  2

3008 Consider the orbital angular momentum L = (Lx , Ly , Lz ) in Cartesian coordinates ( x , y , z ) and determine the commutators [[[Lx , Ly ], Lx ], Lx ], [[[Lx , Ly ], Lx ], Ly ], [[[Lx , Ly ], Lx ], Lz ]. Finally, determine the action of L2 = (Lx 2 + Ly 2 + Lz 2 ) on the combination x [Ly , z ] − y [Lx , z ] + z [Lx , y ] = i ( x 2 + y 2 + z 2 ) .

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Sol: To solve this problem, we need to consider the commutation rules of the angular momentum. In particular, we know that L × L = i L from which we get

[ Lx , L y ] = i  Lz [ Lz , Lx ] = i  L y [ L y , Lz ] = i  Lx

The above relations can be used to simplify the first commutator requested by the text. [[[Lx , L y ], Lx ], Lx ] = i [L z , Lx ], Lx ] = −  2[L y ,L x ] = i  3 Lz

When the last Lx is interchanged with L y , we get [[[Lx , L y ], Lx ], L y ] = i [L z , Lx ], L y ] = −  2[L y ,L y ] = 0 Also, when the last Lx is interchanged with Lz, we get [[[Lx , L y ], Lx ], Lz ] = i [L z , Lx ], Lz ] = −  2[L y ,L z ] = − i  3

As for the second point, from the relation [Lx , y ] = [ ypz − zp y , y] = i z

and its cyclic permutations, we find x[L y , z ] − y[Lx , z ] + z[Lx , y ] = i ( x 2 + y 2 + z 2 ).

Therefore, we see that the combination x[L y , z ] − y[Lx , z ] + z[Lx , y] = i ( x 2 + y 2 + z 2 ) is directly proportional to the square of the distance from the origin of co­­ordinates (r 2 ) and is independent of the angles, that is, when projected on the angular variables, it is proportional to the spherical harmonic Y0,0(θ, φ). Since the spherical harmonics Yl,m are eigenfunctions of L2 with eigenvalues  2l(l + 1), we find L2 (x[L y ,z] − y[Lx ,z] + z[Lx , y]) = 0

3009 Suppose an electron is in a state described by the wave function 1 ψ= ( e iφ sinθ + cosθ ) g( r ), 4π where

∫

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∞ 0

| g( r )|2 r 2 dr = 1,

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167

and φ , θ are the azimuth and polar angles respectively. a. What are the possible results of a measurement of the z-component Lz of the angular momentum of the electron in this state? b. What is the probability of obtaining each of the possible results in part (a)? c. What is the expectation value of Lz? (Wisconsin) Sol: a. As Y10 =

3 3 cosθ , Y1,±1 = ∓ sinθ e ±iφ , 4π 8π

the wave function can be written as

ψ=

1 (− 2Y11 + Y10 ) g (r ) . 3

Hence the possible values of Lz are +h, 0. b. Since

∫ |ψ |

2

π 1 ∞ | g (r )|2 r 2 dr ∫ dθ ∫ 0 0 4π 1 π = ∫ sinθ dθ = 1, 2 0

dr =

∫

2π 0

(1 + cos φ sin2θ )sinθ dφ

the given wave function is normalized. The probability density is then given by P = |ψ |2 . Thus the probability of Lz = + is of Lz = 0 is

( ) 1 3

2

2 3

2

or 2/3 and that

or 1/3.

c.

 1

* 2Y11 + Y10  

∫ ψ * L ψr sinθ dθ dφ dr = ∫  3 (− z

( )

2

)

  1 × Lˆz  − 2Y11 + Y10    3

(

)

× | g (r )|2 r 2dr sinθ dθ dφ 2π π 2 2 =  ∫ 0 dθ ∫ 0 Y112 dφ = . 3 3

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3010 If U ( β , yˆ ) refers to a rotation through an angle β about the y -axis, show that the matrix elements 〈 j , m | U ( β , yˆ )| j , m′〉, − j ≤ m, m′ ≤ j , are polynomials of degree 2 j with respect to the variables sin ( β 2) and cos ( β 2). Here | j , m〉 refers to an eigenstate of the square and z-component of the angular momentum: j 2 | j , m = j ( j + 1) 2 | j , m , j j , m = m j , m ′ 2 (Wisconsin) Sol:

We use the method of mathematical induction. If j = 0 , then m = m′ = 0 and the statement is obviously correct. If j = 1 2 , let

ˆj

   0 −i  = σy ≡   y 2 2  i 0 .

Consider Pauli’s matrices σ k, where k = x , y or z . Since 1 0  σ x2 = σ y2 = σ z2 =  , 0 1 the unit matrix, we have for α = constant iασ k (±iασ k )2 (±iασ k )3 + + + 1! 2! 3!  α2 α4  = 1 − + −  2! 4!  

exp (±iασ k ) = 1 ±

α α 3 α5  ±iσ k  − + −   1! 3! 5!  = cosα ± iσ k sinα . Thus  β  β β U (β , yˆ ) ≡ exp(−iβ ˆj y /  ) = exp−i σ y  = cos − iσ y sin  2  2 2 β 2ˆ β = cos − i j y sin , 2  2

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169

and 1 1 1 1 , m U , m′ = , m exp (−iβ ˆj y /  ) , m′ 2 2 2 2 1 β i2 1 β = δmm , cos − , m ˆj y , m′ sin . 2 2 2  2 ˆ As the matrix elements of j y in the second term are independent of β , 1 2 , m | U |1 2 , m′ is a linear homogeneous form of cos(β 2) and sin(β 2) and the statement is correct also. If the statement is correct for j, i.e., j , m |U | j , m′ = j , m |exp[−iβ ˆj y / ]| j , m 2j

 β = ∑ An cos   2 n=0

2 j −n

n

 β  sin  ,  2

where An depends on j , m, m′ , i.e., An = An ( j , m, m′),

  we shall prove that the statement is also correct for j + 1/ 2 . Let J = j + j1 , where the quantum numbers of j and j are j and 1/2 respectively. We can expand 1

J , m = j + 1/ 2, m in the coupling representation using terms of the uncoupling representation: 1 1 j + , m = C1 j , m + 2 2 + C2 j , m −

1 1 , − 2 2 1 2

1 1 , , 2 2

where C1 and C2 (independent of β ) are Clebsch-Gordan coefficients. Applying the expansion to 1 1 j + , m exp[−iβ ( j y + j1 y )/ ] j + , m′ , 2 2 we reduce the procedure to calculating the matrix elements 1 1 1 1 1 1 ,∓ j , m ± exp[−iβ ( j y + j1 y )/ ] j , m′ ± ∓ , 2 2 2 2 2 2 1 1 1 1 1 1 ,± . ,± j , m ∓ exp[−iβ ( j y + j1 y )/ ] j , m′ ∓ 2 2 2 2 2 2

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For example, 11 11 1 1 j , m + 1exp [−iβ ( j  + j  )/ ] j , m′ +′ 1 ,− 2 2 , −2 2 j , m +2 2 exp [−iβ (yj y + 1jy1 y )/ ] j , m +2 2 1 1 1 1 [−iβ j  / ] 1, − 1 == 1, −, − 1exp 2 2 2 2 exp [−iβ1jy1 y / ]2 2 , −2 2 1 11  ′ +′ + 1 ×× j ,jm exp[−[i−βiβj yj /y / ]] j ,j m ,m , m++2 exp 22 2 2j

2 j −n

11 11 ,− 2 2 , −2 2

n

 ββ ββ  2 j   ββ  2 j−n  ββ  n ==a1acos ++b1bsin  ∑ AA   sin  n  cos cos sin   1 2 2 1 2 2 n∑ n cos2 2    sin2 2  =0 n=0 2( i+ 12 ) 1 2( i+ 2 )

2( j+ 1 )−l

l

  11    ββ  2( 2j+ 12 )−l  ββ  l ′ ==∑ + B j , m , m   sin . n + B j , m , m   cos2   ∑ n  2 2 ′cos   sin2 2  .  l=0 2 l=0

Thus the statement is also valid for j + 1/ 2 . That is to say, the matrix elements j , m U (β , yˆ ) j , m′ = j , m exp (−iβ ˆj y /  ) j , m′ , are polynomials of degree 2 j with respect to the variables cos(β / 2) and sin (β /2) .

3011 An operator f describing the interaction of two spin-1/2 particles has the form f = a + bσ 1 ⋅ σ 2 , where a and b are constants, σ 1 and σ 2 are Pauli matrices. The total spin angu lar momentum is J = j1 + j2 = ( σ1 + σ 2 ) . 2 a. Show that f, J2 and Jz can be simultaneously measured. b. Derive the matrix representation for f in the J , M , j1 , j2 basis. (Label rows and columns of your matrix). c. Derive the matrix representation for f in the j1 , j2 , m1 , m2 basis. (Wisconsin) Sol: a. f, J2 and Jz can be measured simultaneously if each pair of them commute. We know that J2 and Jz commute, also that either commutes with a, a constant. From definition, J2 =

2 2 (σ 1 + σ 22 + 2σ 1 ⋅ σ 2 ), 4

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or

σ 1 ⋅σ 2 =

171

2J 2 1 2 − (σ 1 + σ 22 ). 2 2

Now for each particle

σ 2 = σ x2 + σ y2 + σ z2 = 3I , I being the unit matrix, so σ1 ⋅ σ 2 =

2J 2 −3. 2

Hence [J 2, f ] = [J 2, a] + b[J 2, σ 1 ⋅ σ 2 ]   2J 2 = b J 2, 2 − 3 = 0,     2J 2  [ J z , f ] = [ J z , a] + b  J z , 2 − 3 = 0.    Therefore, f, J 2 and J z can be measured simultaneously b. In the J , M , j1 , j2 basis, J , M , j1 , j2 f J ′, M ′, j1 , j2 = aδ JJ ′δMM ′ + b J , M , j1 , j2 σ 1 ⋅ σ 2 J ′, M ′, j1 , j2 2  = aδ JJ ′ δMM ′ + b 2 J ′( J ′+ 1) 2 − 3   × δ JJ ′ δMM ′ = [a + 2bJ ( J + 1) − 3b]δ JJ ′ δMM ′ , where J, M are row labels, J ′, M ′ are column labels. c. Denote the state of J = 0 and the state of J = 1 and J z = M as χ 0 and χ1 M , respectively. Since j1 = j2 = 1/ 2 , we can denote the state j1 , j2 , m1 , m2 simply as m1 , m2 . Then as  1 1 1 1 1 −− ,  ,−  χ0 = 2 2 22 2    1 1 1 1 1  χ10 =  ,− + − , 2 2 2 2 2     χ1, ±1 = ± 1 , ± 1 ,  2 2

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 ,   , 

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we have          

1 1 ± , ± = χ1, ±1 , 2 2 1 1 1 , − = ( χ 0 + χ10 ), 2 2 2 1 1 1 − , = (− χ 0 + χ10 ). 2 2 2

Using the above expressions and the result of (b) we can write the matrix elements m1 , m2 f m1′ , m′2 in the basis j1 , j2 , m1 , m2 as follows: 1 1 1 1 m1′ , m′2 1 1 1 1 , ,− − , − , − 2 2 2 2 2 2 2 2 m1, m2 1 , 2 1 , 2 1 − , 2 1 − , 2

1 2 1 − 2 1 2 1 − 2

 a+b 0 0 0  0 2 0 − a b b   0 2b a − b 0  0 0 a+b  0

     

3012 Consider the following two-particle wave function in position space: ψ (r1 , r2 ) = f ( r12 ) g( r22 ) [α ( a ⋅ r1 )(b ⋅ r2 ) + β (b ⋅ r1 )( a ⋅ r2 ) + γ ( a ⋅ b)(r1 ⋅r2 )], where a and b are arbitrary constant vectors, f and g are arbitrary functions, and α , β and γ are constants. a. What are the eigenvalues of the squared angular momentum for each particle ( L21 and L 22 )? b. With an appropriate choice of α , β and γ , ψ2 (r1 , r2 ) can also be an eigenfunction of the total angular momentum squared J 2 = (L1 + L 2 )2 . What are the possible values of the total angular momentum squared and what are the appropriate values of α , β and γ for each state? (MIT)

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Sol: a. We first note that r ∇f (r ) = f ′(r ) , r or

r × ∇f (r ) = f ′(r )

r×r = 0, r

and that r × ∇(a ⋅ r ) = r × a , (r × ∇ ) ⋅ (r × a ) = −2a ⋅ r. As L = r × p = −ir × ∇, we have

L21ψ (r1 , r2 ) = −  2 (r1 × ∇1 ) ⋅ (r1 × ∇1 ) { f (r12 ) g (r22 ) [α (a ⋅ r1 )( b ⋅ r2 ) + β ( b ⋅ r1 )(a ⋅ r2 ) + γ (a ⋅ b)(r1 ⋅ r2 )]}

= −  2 f (r12 ) g (r22 )(r1 × ∇I ) ⋅ (r1 × ∇1 ) ⋅[α (a ⋅ r1 )( b ⋅ r2 ) + β ( b ⋅ r1 )(a ⋅ r2 ) + γ (a ⋅ b)(r1 ⋅ r2 )] = −  2 f (r12 ) g (r22 )(r1 × ∇1 ) ⋅[α (r1 × a )( b ⋅ r2 ) + β (r1 × b)(a ⋅ r2 ) + γ (a ⋅ b)(r1 × r2 )] = 2 2 f (r12 ) g (r22 )[α (a ⋅ r1 )( b ⋅ r2 ) + β ( b ⋅ r1 )(a ⋅ r2 ) +γ (a ⋅ b)(r1 ⋅ r2 )] = 1(1 + 1) 2ψ (r1 , r2 ), and similarly L22ψ (r1 ⋅ r2 ) = − 2 (r2 × ∇ 2 ) ⋅ (r2 × ∇ 2 )ψ (r1 ⋅ r2 ) = 1(1 + 1) 2ψ (r1 , r2 ). 2 2 2 Hence the eigenvalues of L1 and L2 are each equal to 2 , and so each parti-

cle has the quantum number l = 1 b. We further note that (r × ∇ ) ⋅ (a ⋅ r )e = (r × a ) ⋅ c , (a × b) ⋅ (c × d ) = (a ⋅ c )( b ⋅ d ) − (a ⋅ d )( b ⋅ c ).

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Thus we require L1 ⋅ L 2ψ (r1 , r2 ) = − 2 (r1 × ∇1 ) ⋅ (r2 × ∇ 2 ){ f (r12 ) g (r22 )[α (a ⋅ r1 ) ×( b ⋅ r2 ) + β ( b ⋅ r1 )(a ⋅ r2 ) + γ (a ⋅ b )(r1 ⋅ r2 )]} = − 2 f (r12 ) g (r22 )(r1 × ∇1 ) ⋅ (r2 × ∇ 2 )[α (a ⋅ r1 )( b ⋅ r2 ) +β ( b ⋅ r1 )(a ⋅ r2 ) + γ (a ⋅ b)(r1 ⋅ r2 )] = − 2 f (r12 ) g (r22 )(r1 × ∇1 ) ⋅[α (a ⋅ r1 ) (r2 × b) +β ( b ⋅ r1 )(r2 × a ) + γ (a ⋅ b)(r2 × r1 )] = − 2 f (r12 ) g (r22 )[α (r1 × a ) ⋅ (r2 × b) +β (r1 × b) ⋅ (r2 × a ) + γ (a ⋅ b)(2r1 ⋅ r2 )] = − 2 f (r12 ) g (r22 )[−β (a ⋅ r1 )( b ⋅ r2 ) − α ( b ⋅ r1 )(a ⋅ r2 ) +(α + β + 2γ )(a ⋅ b)(r1 ⋅ r2 )] = − 2 λψ (r1 , r2 ) for ψ (r1 , r2 ) to be an eigenfunction of L1 ⋅ L 2 . This demands that −β = λα , −α = λβ , α + β + 2γ = λγ , which give three possible values of λ : 3 λ = −1, α = β = − γ ; 2 λ = +1, α = −β , γ = 0; λ = 2, α = β = 0. Therefore the possible values of the total angular momentum squared, J 2 = L21 + L22 + 2L1 ⋅ L 2 , and the corresponding values of α , β and γ are  2 2(2 + 1) ,  J 2 = 2 2 + 2 2 − 2 2 λ =  1(1 + 1) 2 ,  0.

3 (α = β = − γ ) 2 (α = − β , γ = 0) (α = β = 0)

3013 A quantum-mechanical state of a particle, with Cartesian coordinates x, y and z, is described by the normalized wave function

ψ( x , y , z ) =

α 5/2 z exp −α ( x 2 + y 2 + z 2 )1/2  . π

Show that the system is in a state of definite angular momentum and give the values of L2 and Lz associated with the state. (Wisconsin)

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Sol:

Transforming to spherical coordinates by x = r sinθ cos ϕ , y = r sin θ sin ϕ ,

175

z = r cosθ ,

we have

ψ (r , θ , ϕ ) =

α 5/2 r cosθ e −αr = f (r )Y10 . π

Hence the particle is in a state of definite angular momentum. For this state, l = 1, L2 = l(l + 1) 2 = 2 2 , Lz = 0 .

3014 A free atom of carbon has four paired electrons in s-states and two more electrons with p-wave orbital wave functions. a. How many states are permitted by the Pauli exclusion principle for the latter pair of electrons in this configuration? b. Under the assumption of L-S coupling what are the “good” quantum numbers? Give sets of values of these for the configuration of the two p-wave electrons. c. Add up the degeneracies of the terms found in (b), and show that it is the same as the number of terms found in (a). (Buffalo) Sol: a. Each electron can occupy one of the (2l + 1) (2s + 1) = 3 × 2 = 6 states, but it is not permitted that two electrons occupy the same state. So the number of permitted states is C26 = 15 . The “good” quantum numbers are L2, S 2, J 2 and J z. Under the assumption of L-S coupling, the total spin quantum numbers for two electrons, S = s1 + s 2, are S = 0, 1 and the total orbital quantum numbers, L = 11 + 12 , are L = 0, 1, 2. Considering the symmetry of exchange, for the singlet S = 0, L should be 1 1 even: L = 0, 2 , corresponding to S0 , D2 respectively; for the triplet S = 1, L

S = 1, L odd: L = 1, corresponding 3 P0,1,2 . b. The degeneracy equals to 2 J + 1. For J = 0, 2 and 0, 1, 2 in (b), the total number of degeneracies is 1 + 5 + 1 + 3 + 5 = 15.

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3015 a. Determine the energy levels of a particle bound by the isotropic potential V ( r ) = kr 2 /2, where k is a positive constant. b. Derive a formula for the degeneracy of the Nth excited state. c. Identify the angular momenta and parities of the Nth excited state. (Columbia) Sol: a. The Hamiltonian is Hˆ = −( 2 /2m)∇ 2 + kr 2 /2 . 2

Choose ( Hˆ , l , l z ) to have common eigenstates. The energy levels of bound states are given by E = (2nr + l + 3/2)ω0 = ( N + 3/2)ω0 ,

ω0 = k / m.

b. The degeneracy of the states is determined by nr and l. As N = 2nr + l , the odd-even feature of N is the same as that of l. For N even (i.e., l even), the degeneracy is N

∑

f=

(2l + 1) = 4

l =0( l even)

 l 1  N /2  1  ∑  + 4  = 4∑l′+ 4  l =0( l even) 2 l ′=0 N

1 = ( N + 1)( N + 2). 2 For N odd (i.e., l odd), f=

N

∑

N

(2l + 1) =

l =1( l odd )

∑

2 (l − 1) + 3 = 4

( N −1)/2

l =1( l odd )

∑ l ′=0

 3 l′+   4

1 = ( N + 1)( N + 2). 2 Hence the degeneracy of the Nth excited state is f = ( N + 1)( N + 2) / 2. c. In the common eigenstates of ( Hˆ , l 2 , lz ), the wave function of the system is

ψnr l (r , θ , ϕ ) = R(r )Ylm (θ , ϕ ), and the eigenenergy is Enr l = (2nr + l + 3 / 2)ω0 . As N = 2nr + l , the angular momentum l of the Nth excited state has N2 +1 values 0, 2, 4, 6 ... , N for N even, or 12 ( N + 1) values 1, 3, 5, ... , N for N odd. Furthermore, the parity is P = (−1)l = (−1)N .

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3016 The ground state of the realistic helium atom is of course nondegenerate. However, consider a hypothetical helium atom in which the two electrons are replaced by two identical, spin-one particles of negative charge. Neglect spin-dependent forces. For this hypothetical atom, what is the degeneracy of the ground state? Give your reasoning. (CUS) Sol:

The two new particles are Bosons; thus the wave function must be symmetrical. In the ground state, the two particles must stay in Is orbit. Then the space wave function is symmetrical, and consequently the spin wave function is symmetrical too. As s1 = 1 and s2 = 1, the total S has three possible values: S = 2, the spin wave function is symmetric and its degeneracy is 2S + 1 = 5. S = 1 , the spin wave function is antisymmetric and its degeneracy is 2S + 1 = 3. S = 0, the spin wave function is symmetric and its degeneracy is 2S + 1 = 1. If the spin-dependent forces are neglected, the degeneracy of the ground state is 5 + 3 + 1 = 9.

3017 Consider an electron in the quantum state given by 1 Ψ(r ,θ ,ϕ ) = (2i sinϕ sinθ + cosθ )g (r ) 4π ∞

Where ∫ g (r ) r 2 dr = 1 2

0

a. Find the possible values for the z-component of the angular momentum Lz. b. Find the probabilities for each of the possible values of Lz. c. Find the expectation value of Lz. Sol: a. The expression for ψ can be rewritten using the expressions for spherical harmonics Y1,0 =

3 3 3 sin θ e iϕ and Y1,−1 = sin θ e − iϕ as cosθ , Y1,1 = − 8π 8π 4π

Ψ (r , θ , ϕ ) =

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1 ((e iϕ − e − iϕ )sin θ + cosθ ) g (r ) 4π

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=

1 2 2 Y1,0 − Y1,1 − Y1,−1(1) 3 3 3

From (1), the possible values of Lz are +  ,0 and −  b. For a normalized wavefunction, the sum of the squares of coefficients should 3 be equal to 1, therefore normalization constant is and 5

ψ normalized =

 3 1 2 2 Y1,0 − Y1,1 − Y1,−1   5 3 3 3 

1 2 2 The probability of Lz = 0 is , Lz = +  is and Lz = −  is 5 5 5 c. Expectation value of Lz *

 3 1  3 1   2 2 2 2 Y1,0 − Y1,1 − Y1,−1   × Lz   Y1,0 − Y1,1 − Y1,−1     ∫ 3 3 3 3     Lz =  5  3  5  3 2

× g (r ) r 2dr sinθ dθ dϕ Lz = 0

3018 a. Consider a system of spin 1/2. What are the eigenvalues and normalized eigenvector of the operator Asˆ y + Bsˆz , where sˆy , sˆz are the angular momentum operators, and A and B are real constants. b. Assume that the system is in a state corresponding to the upper eigenvalue. What is the probability that a measurement of sˆy will yield the value /2 ? The Pauli matrices are  0 1 0 − i  1 0  σ x =  , σ y =  , σ z = . 1 0  i 0   0 − 1 (CUS) Sol: a. Using the definition of angular momentum operators let 1 1 Tˆ = Asˆy + Bsˆz = A σ y + B σ z . 2 2

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Then 1 1 (Tˆ )2 =  2 ( A 2 + B 2 + AB{σ y , σ z }) =  2 ( A 2 + B 2 ), 4 4 as 1 0  σ i2 =   = I , i = 1, 2, 3, 0 1 and {σ i , σ j } ≡ σ iσ j + σ jσ i = 2δij . Hence the two eigenvalues of T are 1 1 T1 =  A2 + B 2 , T2 = −  A2 + B 2 . 2 2 2 In the representation of sˆ and sˆz ,    B − iA  Tˆ = ( Aσ y + Bσ z ) =  . 2 2 iA − B  a  Let the eigenvector be   . Then the eigenequation is b  a    B − iA  a    =T  , 2 iA − B  b  b  where   ± A2 + B2 T=  2 0 

0 ∓ A2 + B2

 ,  

or  2 2  B∓ A +B  iA 

−iA −B ∓ A2 + B2

  a  = 0.  b  

Hence a : b = iA : B ∓ A 2 + B 2 , and so the normalized eigenvector is 12

  a  1   b  =  2 2 2  A + (B ∓ A + B ) 

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 +iA   B ∓ A2 + B 2

 . 

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b. In the representation of sˆ2 and sˆz , the eigenvector of sˆy is sy =

 1 −i  =  . 2 2 1 

Hence the probability of finding s y =  / 2 is P∓ =

2

2

a  1 ( B ∓ A 2 + B 2 − A)2 . (i1)  = (ia + b )2 = b 2 2 2 2  2 2 2[( B ∓ A + B ) + A ]

1

Note that P− is the probability corresponding to the system in the state of eigenvalue T =  A 2 + B 2 /2 , and P + is that corresponding to the state of T = − A 2 + B 2 /2.

3019 A system of three (non-identical) spin one-half particles, whose spin operators are s1, s2 and s3, is governed by the Hamiltonian H = As1 ⋅ s 2 /  2 + B(s1 + s 2 ) ⋅ s 3 /  2. Find the energy levels and their degeneracies. (Princeton) Sol:

2 Using a complete set of dynamical variables of the system ( H , s12 , s 2 , s3 ), where s12 = s1 + s 2 , s = s12 + s3 = s1 + s 2 + s3 , the eigenfunction is | s12 s3sms 〉, and the stationary state equation is Hˆ s s sm = E s s sm . 12 3

s

12 3

s

As 3 1 0  s12 = s 22 = s32 =  2  , 4 0 1 1 2 − s12 − s 22 ) , s1 ⋅ s 2 = ( s12 2 1 2 (s1 + s 2 ) ⋅ s3 = ( s 2 − s12 − s32 ) , 2 we have A B Hˆ = 2 s1 ⋅ s 2 + 2 (s1 + s 2 ) ⋅ s3   A  1 2 3 3 =  2 s12 − −  2  4 4 B 1 1 2 3 +  2 s 2 − 2 s12 − . 2  4 

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Now as the expectation value of s is s(s + 1), etc., we have  A 3 Hˆ | s12 s3 sms =  s12 (s12 + 1) −  2 2 3  B + s(s + 1) − s12 (s12 + 1) −  s12 s3 sms , 4  2 and hence E=

A 3 s12 (s12 + 1) −   2 2  3 B + s(s + 1) − s12 (s12 + 1) − . 2 4

It follows that for s12 = 0, s = 1/ 2 : E = −3 A / 4 , the degeneracy of the energy level, 2s + 1 , is 2; for s12 = 1, s = 1/ 2 : E = A / 4 − B , the degeneracy of the energy level, is 2; for s12 = 1, s = 3 / 2 : E = A / 4 + B / 2, the degeneracy of the energy level is 4.

3020 A particle of spin one is subject to the Hamiltonian H = As z + Bs x2 , where A and B are constants. Calculate the energy levels of this system. If at time zero the spin is in an eigenstate of s with s z = + , calculate the expectation value of the spin at time t. (Princeton) Sol:

We first find the stationary energy levels of the system. The stationary Schrödinger equation is Eψ = Hˆ ψ = ( As + Bs 2 )ψ , z

x

where u1    ψ = u2    u3  is a vector in the spin space. As 0 0 0    sx = 0 0 − i  ,   0 i 0 0 − i 0    sz =  i 0 0  ,   0 0 0  0 0 0    2 sx =  0 1 0   2 ,    0 0 1 xxxxxxxxxxxxx_Chapter_03.indd  Page 181

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182

0 0 0    sx = 0 0 − i  ,   0 i 0 0 − i 0    sz = Mechanics  i 0 0  , Problems and Solutions in Quantum   0 0 0  0 0 0    2 sx =  0 1 0   2 ,    0 0 1 we have  0 − iA′ 0 2 ˆ H = Asz + Bsx =  iA′ B′ 0  ,    0 0 B′  where A′ = A , B′ = B 2. The energy levels are given by the eigenvalues of the above matrix, which are roots of the equation 0   − E − iA′  Det iA′ B′ − E 0  = 0,    0 0 B′ − E  i.e., ( E − B′)( E 2 − B′E − A′) = 0 . Thus the energy levels are E0 = B′, E± = ( B′′± ω ) / 2, where ω = B′ + 4 A′ / , B″ = B′/ = B . The corresponding eigenfunctions are 2

2

0   E0 = B′ : ϕ S0 = 0  ,   1   1    B′ B′ + 4 A′ ω − B″  ω + B″  E+ = + : ϕ s+ = , i 2 2 2ω  ω − B″    0   2

2

1    ω + B″  ω − B″  B′ B′ + 4 A′ −i : ϕ S− = . E− = + 2ω  ω + B″  2 2   0   2

2

The general wave function of the system is therefore  iB′   E+  ϕ S (t ) = C1ϕ s0 exp − t  + C2ϕ S+exp −i t        E  +C3ϕ s−exp −i − t .   

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Initially, szϕ s (0) = ϕ s (0). Let α    ϕ s (0) =  β   .   γ  The above requires  0 −i 0  α  α        i 0 0  β  = β  ,  0 0 0  γ  γ       i.e.,

β = iα , γ = 0 . Thus we can take the initial wave function (normalized) as 1  1   ϕ s(0) = i . 2   0  Equating ϕ s (0) with C1ϕ S 0 + C2ϕ S+ + C3ϕ S− gives (ω + B″)1/2 + (ω − B″)1/2 , 2ω 1/2 (ω + B″)1/2 − (ω − B″)1/2 C3 = . 2ω 1/2 C1 = 0, C2 =

We can now find the expectation value of the spin: sx = ϕ S+ (t )sx ϕ S (t ) = 0, where we have used the orthogonality of ϕ S 0 , ϕ S+ and ϕ S−. Similarly, s y = 0,  2B2 2 2  sz = ϕ S+ (t )szϕ s(t ) = 1 − sin (ωt / 2) . 2 ω  

3021 A system of two particles each with spin 1/2 is described by an effective Hamiltonian H = A( s1z + s2 z ) + Bs1 ⋅ s 2 ,

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where s1 and s2 are the two spins, s1z and s2z are their z-components, and A and B are constants. Find all the energy levels of this Hamiltonian. (Wisconsin) Sol:

We choose χ SMS as the common eigenstate of S 2 = (s1 + s 2 )2 and Sz = s1z + s2 z . For S = 1, M S = 0, ± 1, it is a triplet and is symmetric when the two electrons are exchanged. For S = 0, M S = 0 , it is a singlet and is antisymmetric. For stationary states we use the time-independent Schrödinger equation H χ SMS = E χ MS. As S 2 χ1 MS = S(S + 1) 2 χ1 MS = 2 2 χ1 M , S 2 χ 00 = 0, S 2 = (s1 + s 2 )2 = s12 + s 22 + 2s1 ⋅ s 2 =

3 2 3 2 +  + 2s1 ⋅ s 2 , 4 4

we have  S2 3   2 3  s1 ⋅ s 2 χ1 M =  −  2  χ1 MS =   2 −  2  χ1 Ms = χ1 MS ,  4  4 2 4   3  3 s1 ⋅ s 2 χ 00 = 0 −  2  χ 00 = −  2 χ 00 ,  4  4 and Sz χ1 MS = (s1z + s2 z ) χ1 MS = M S χ1 MS , sz χ 00 = 0. Hence for the triplet state, the energy levels are E = M S A +

2 B, with  M S = 0, ± 1, 4

comprising three lines 2 2 h2 B, E2 = B, E3 = −A + B. 4 4 4 For the singlet state, the energy level consists of only one line 3 E0 = −  2 B. 4 E1 = A +

3022 Suppose an atom is initially in an excited 1 S0 state (Fig. 3.1) and subsequently decays into a lower, short-lived 1 P1 state with emission of a photon γ 1

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(Fig. 3.2). Soon after, it decays into the 1 S0 ground state by emitting a second photon γ 2 (Fig. 3.3). Let θ be the angle between the two emitted photons. a. What is the relative probability of θ in this process? b. What is the ratio of finding both photons with the same circular polarization to that of finding the photons with opposite circular polarizations?

Fig. 3.1

Fig. 3.2

Fig. 3.3

It may be of some help to know the rotation matrices dm′m, which relate one angular momentum representation in one coordinate system to another angular momentum representation in a rotated coordinate system, given below: m =1

m=0

m = −1

 1+ cos α 1 1− cos α  − sinα 2 2  2  1 1 − dm′m = m′ = 0  sinα cos α sinα 2  2  1− cos α 1 1+ cos α  sinα m′ = −1 2 2 2  m′ = 1

        

where α is the angle between the z-axis of one system and the z′ -axis of the other. (Columbia) Sol:

1 The atom is initially in the excited state S0 . Thus the projection of the atomic angular momentum on an arbitrary z direction is Lz = 0 . We can take the direc-

tion of the first photon emission as the z direction. After the emission of the 1 photon and the atom goes into the P1 state, if the angular momentum of that photon is Lz = ± , correspondingly the angular momentum of the atomic state

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P1 is Lz = ∓ , i.e., mz = ∓1 . If we let the direction of emission of the second pho-

ton be the z ′ -axis the projection of the eigenstate of the z-component of angular momentum on the z ′ direction is equivalent to multiplying the initial state with the dm′m matrix. Only atoms that are in states m′z = ±1 can emit photon (as 1 Lz′ = ± must be satisfied) in z ′ direction and make the atom decay into the S0

state (m′z = 0) . Then the transitions are from m = ±1 to m′ = ±1 , and we have C1 = 〈mz ′ = −1| dm ′m | mz = +1〉  1 + cosθ  2   1 = (0, 0, 1)  sin θ 2   1 − cosθ  2 

1 sin θ 2

−

cosθ 1 sin θ 2

1 − cosθ 2 1 − sin θ 2 1 + cosθ 2

        

 1  0    0

1 − cosθ , 2 C2 = mz ′ = +1| dm ′m | mz = +1 =

 1 + cosθ  2   1 sin θ = (1, 0, 0)   2  1 − cos θ  2  =

−

1 sin θ 2 cosθ 1 sin θ 2

1 − cosθ 2 1 sin θ − 2 1 + cos θ 2

        

 1  0    0

1 + cosθ , 2

1 − cosθ , 2 1 + cosθ . C4 = mz ′ = −1| dm ′m | mz = −1 = 2 a. The relative probability of θ is C3 = mz ′ = +1| dm ′m | mz = −1 =

P(θ ) ∝| C1 |2 + | C2 |2 + | C3 |2 + | C4 |2 = 1 + cos 2θ . b. The ratio of the probability of finding both photons with the same circular polarization to that of finding the photons with opposite circular polarizations is (| C2 |2 + | C4 |2 ) / (| C1 |2 + | C3 |2 ) = (1 + cosθ )2 / (1 − cosθ )2.

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3023 Consider an electron in a uniform magnetic field in the positive z direction. The result of a measurement has shown that the electron spin is along the positive x direction at t = 0 . Use Ehrenfest’s theorem to compute the probability for t > 0 that the electron is in the state (a) s x = 1/ 2 , (b) s x = −1/ 2 , (c) s y = 1/ 2 , (d) s y = −1/ 2 , (e) s z = 1/ 2 , (f ) s z = −1/ 2 . Ehrenfest’s theorem states that the expectation values of a quantum mechanical operator obey the classical equation of motion. [Hint: Recall the connection between expectation values and probability considerations]. (Wisconsin) Sol:

In the classical picture, an electron spinning with angular momentum s in a magnetic field B will, if the directions of s and B do not coincide, precess about the direction of B with an angular velocity w given by ds = s ×ω, dt where ω = mce B , m being the electron mass. Ehrenfest’s theorem then states that in quantum mechanics we have d e s = s × B. dt mc This can be derived directly as follows. An electron with spin angular momentum s has a magnetic moment µ = mce s and consequently a Hamiltonian e B sz , mc taking the z axis along the direction of B. Then d s 1 ˆ eB ˆ ˆ +s Y ˆ [s x X = [s, H] =− y + s z Z, s z ] dt i imc eB =− {[sx , sz ] xˆ +[s y , sz ] yˆ} imc e =− B − s y xˆ + sx yˆ mc e = s × B, mc H = −µ ⋅ B = −

(

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in agreement with the above. Note that use has been made of the commutation relations [sx , s y ] = isz , etc. Initially

sx = 1/2, s y = sz = 0 . and so we can write for t > 0, sx = (cosωt )/2, s y =

t > 0, sx = (cosωt )/2, s y = (sinωt )/2, sz = 0 . Let the probability for t > 0 of the electron being in the state sx = 1/ 2 be P and being in the state sx = −1/ 2 be 1–P since these are the only two states of sx . Then 1  1 1 P   + (1 − P )−  = cos ωt , 2  2 2 giving P = cos 2 (ωt/2), 1 − P = sin 2 (ωt/2). Similarly, let the probabilities for the electron being in the states s y = 1/ 2, s y = −1/ 2 be P and 1–P respectively. Then 1  1 1 π  1 P   + (1 − P ) −  = sinωt = cos  − ωt  , 2  2 2 2  2 or π   ωt π  1 P = 1 + cos  − ωt  = cos 2  −  , 2   2 4 2 2 and hence 1 − P = sin ( ω2t − π4 ) . Lastly for (e) and (f), we have

1 1 1 P − = 0 , giving  P = , 1 − P = . 2 2 2

3024 A particle with magnetic moment µ = µ 0 s and spin s, with magnitude1/ 2, is placed in a constant magnetic field pointing along the x-axis. At t = 0 , the particle is found to have s z = +1/ 2 . Find the probabilities at any later time of finding the particle with s y = ±1/ 2 . (Columbia) Sol:

The Hamiltonian (spin part) of the system is 1 Hˆ = −µ0 sˆ ⋅ B = − σˆ x Bµ0 , 2

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as s = 12 σ x , being in the x direction. In the σ x representation, the Schrödinger  a1  equation that the spin wave function   satisfies is a 2  i or

0 1 a1  d a1  1   + µ0 B     = 0, a dt  2  2 1 0  a2   d 1 i a1 + µ0 Ba2 = 0, 2 dt  i d a + 1 µ Ba = 0.  dt 2 2 0 1

Elimination of a1 or a2 gives d2 µ 2 B2 a + 0 2 a1, 2 = 0, 2 1, 2 dt 4 which have solutions a1, 2 = A1, 2e iωt + B1, 2e −iωt , where

ω=

µ0 B 2h

and A1,2 , B1,2 are constants. As 1 1 1 0  sˆz = σ z =   2 2 0 − 1 and 1  1 1  sˆz   =   , 0  2 0  1  the initial spin wave function is   , i.e. a1 (0) = 1, a2 (0) = 0 . The Schrödinger 0  equation then gives da1 (0) da2 (0) µ0 B = 0, =i = iω . 2 dt dt These four initial conditions give  A + B = 1, A2 + B2 = 0, 1 1   ω ( A1 − B1 ) = 0, ω ( A2 − B2 ) = ω ,

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with the solution A1 = A2 = B1 = −B2 = 1/ 2. Substitution in the expressions for a1, 2 results in al (t )  cos ωt   = . a2 (t ) i sin ωt  As the eigenstate of s y = +1/ 2 is s y (+) =

1 1   2 −i 

s y (+) =

1 1  , 2 −i 

and that of s y = −1/ 2 is

the probability of finding s y = +1/ 2 is P(+) = s y (+) ψ (t ) =

2

cosωt  1 (1 − i )  2 i sin ωt 

2

1 = (1 + sin2ωt ). 2 Similarly the probability of finding s y = −1/ 2 is 1 P(−) =| 〈s y (−)| ψ (t )〉 |2 = (1 − sin2ωt ) . 2

3025 The Hamiltonian for a spin −

1 2

particle with charge +e in an external magnetic

field is H =−

ge s ⋅B . 2mc

Calculate the operator ds/dt if B = Byˆ . What is s z (t ) in matrix form? (Wisconsin) Sol:

In the Heisenberg picture, ge ds 1 = [s , H ] = − [s , s ⋅ B] . i 2mc dt i

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As [s , s ⋅ B] = [sx , s ⋅ B]xˆ +[s y , s ⋅ B]yˆ +[sz , s ⋅ B]zˆ and [sx , s ⋅ B] = [sx , sz ]Bx +[sx , s y ]By +[sx , sz ]Bz = i(sz By − s y Bz ) = i(B × s)x , etc we have [s, s ⋅ B] = −is × B , and hence ds ge =+ s × B. dt 2mc If B = B yˆ , the above gives dsx (t ) geB =− sz (t ), dt 2mc dsz (t ) geB = sx (t ), dt 2mc and so 2

d 2 sz (t )  geB  +  s (t ) = 0,  2mc  z dt 2 with the solution sz (t ) = c1 cos( gωt ) + c2 sin( gωt ), where ω = eB/2mc . At t = 0 we have sz (0) = c1 ,

sz′ (0) = c2 gω = gω sx (0),

and hence  geB   geB  t  + sx (0)sin  t. sz (t ) = sz (0)cos   2mc   2mc 

3026 Two electrons are tightly bound to different neighboring sites in a certain solid. They are, therefore, distinguishable particles which can be described in terms of their respective Pauli spin matrices σ (1) and σ (2) . The Hamiltonian of these electrons takes the form (2) (1) (2) H = − J(σ (1) x σ x + σ y σ y ),

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where J is a constant. a. How many energy levels does the system have? What are their energies? What is the degeneracy of the different levels? b. Now add a magnetic field in the z direction. What are the new energy levels? Draw an energy level diagram as a function of Bz (Chicago) Sol: a. The Hamiltonian of the system is (2) H = − J[σ x(1)σ x(2) + σ (1) y σy ]

 (σ (1) + σ (2) )2 − σ (1)2 − σ (2)2 = −J  2  2 2 (σ z(1) + σ z(2) )2 − σ z(1) − σ z(2)  −  2 

 (σ (1) + σ (2) )2 − 3 − 3 = −J  2  −

(σ z(1) + σ z(2) )2 − 1 − 1   2 

 (σ (1) + σ (2) )2 − (σ z(1) + σ z(2) )2  = −J  − 2 2   2J = − 2 (s 2 − sz2 −  2 ),  where s = s(1) + s(2) =

h (1) σ + σ (2) ) ( 2

is the total spin of the system and  sz = sz(1) + sz(2) = (σ z(1) + σ z(2) ) 2 2 is its total z-component. s , sz and H are commutable. Using the above and the 2 coupling theory of angular momentum, we have (noting the eigen value of s 2 is s(s + 1) )

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s 1 0

sz

number of states

energy

1 2 1

0 -2 J 0

2

2J

 1   0  −1  0

193

a. As seen from the table, the system has three energy levels −2 J , 0, 2 J , each with a degeneracy of 2. Note that if the electrons are indistinguishable, the second and fourth rows of the table would be different from the above. b. In the presence of a magnetic field | B | = Bz , (2) H = − J[σ x(1)σ x(2) + σ (1) y σ y ] − µ ⋅B ,

where

µ =−

e s, mc

–e and m being the electron charge and mass respectively. Thus e sz Bz mc  s 2  eB = −2 J s(s + 1) − z2 − 1 + z sz .    mc

(2) H = − J[σ x(1)σ x(2) + σ (1) y σ y ]+

s 2, sz and H are still commutable, so the new energy levels are: 2 J , eBz /mc , − eBz / 2 J , eBz /mc , − eBz /mc , − 2 J . The energy level diagram is shown in Fig. 3.4 as a function of Bz (lines 1, 2, 3 and 4 for the above levels respectively).

Fig. 3.4

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3027 Consider a one electron atom in n = 2 state. a. Find out the possible spectroscopic states for this atom. 



b. What is the angle between l and s for the state 2P3/2. Sol: a. For n = 2 state, the orbital quantum number l can take values of 0 and 1. The value of spin quantum number S for electron is ½. So, the multiplicity is 2S + 1 = 2. The possible values of the total quantum number j is given by j=l±S For l = 0, j = and

1 2

3 1 , 2 2 Hence, the possible spectroscopic states are For l = 1, j =

2S

1/2 ,

2P 3/2

b. For the state 2P3/2 we have, l = 1, s =

and 2P1/2

3 1 and j = 2 2

  The angle between l and s is given by

 j( j + 1) − l(l + 1) − s(s + 1) cos( l .s ) = 2 l(l + 1) s(s + 1) =

1 6

  1  ∴ l .s = cos −1  = 65.9°  6    The angle between l and s for the state 2P3/2 is 65.9°.

3028 −

A negatively charged π meason (a pseudoscalar particle: zero spin, odd parity) is initially bound in the lowest-energy Coulomb wave function around

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a deuteron. It is captured by the deuteron (a proton and a neutron in a 3 S1 state), which is converted into a pair of neutrons:

π − + d → n + n. a. What is the orbital angular momentum of the neutron pair? b. What is their total spin angular momentum? c. What is the probability for finding both neutron spins directed opposite to the spin of the deuteron?

d. If the deuteron’s spin is initially 100% polarized in the R direction, what is the angular distribution of the neutron emission probability (per unit solid angle) for a neutron whose spin is opposite to that of the initial deuteron? You may find some of the first few (not normalized) spherical harmonics useful: Y00 = 1,

Y1±1 = ∓ sinθ e±iφ ,

Y10 = cosθ ,

Y2±1 = ∓ sin2θ e±iφ . (CUS)

Sol: a, b. Because of the conservation of parity in strong interactions, we have p(π − ) p(d )(−1)L1 = p(n) p(n)(−1)L2 , where L1, L2 are the orbital angular momenta of π − + d and n+n respectively. As the π −, being captured, is in the lowest energy state of the Coulomb ­potential before the reaction, L1 = 0 . Since p(π − ) = −1, p(d ) = 1, p(n) p(n ) = 1, we have (−1)L2 = −1, and so L2 = 2m + 1, m = 0, 1, 2, … − The deutron has J = 1 and π has zero spin, so that J = 1 before the reaction takes place. The conservation of angular momentum requires that after the reaction, L 2 + S = J . The identity of n and n demands that the total wave function be antisymmetric. Then since the spacial wave function is antisymmetric, the spin wave function must be symmetric, i.e. S = 1 and so L2 = 2, 1 or 0. As L2 is odd, we must have L2 = 1, S = 1.

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Fig. 3.5 The total orbital angular momentum and the total spin angular momentum are both 1(1 + 1) = 2 . c. Assume that the deuteron spin is in the direction J z = 1 before the reaction. If both neutrons had spins in the reversed direction, we would have Sz = −1 , Lz = 2 , which is impossible since L2 = 1 . Hence the probability is zero.  direction. Then the initial state is J , J = 1, 1 . In d. Take the z-axis along the R z

the noncoupling representation, the state is | L , Lz , S , Sz 〉, with L = 1, S = 1. Thus 1, 1 =

2 2 1, 0, 1, 1 − 1, 1, 1, 0 . 2 2

The state 1, 1, 1, 0 = Y11 (θ , φ )|1, 0 = − 3 8π sinθ e iφ 1, 0 has Sz = 0 and so there must be one neutron with sz = − / 2 . Hence the probability distribution required is 1 3 3 dP(θ , φ )/ dΩ = ⋅ sin 2θ = sin 2θ . 2 8π 16π

3029 −

An Ω hyperon (spin 3/2, mass 1672 MeV/c2, intrinsic parity +) can decay via the weak interaction into a Λ hyperon (spin 1/2, mass 1116 MeV/c2, intrinsic parity + ) and a K − meson (spin 0, mass 494 MeV/c2, intrinsic parity –), i.e., Ω− → Λ + K −.

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a. What is the most general form of the angular distribution of the K − mesons relative to the spin direction of the Ω− for the case when the Ω− has a maximum possible component of angular momentum along the 3/2 z-axis, i.e., the initial state | Ωmj 〉 = | Ω+3/2 〉. (Assume that the Ω− is at rest).

b. What restrictions, if any, would be imposed on the form of the angular distribution if parity were conserved in the decay process? (Berkeley) Sol: a. The initial state of the system is 3 / 2, 3 / 2 , where the values are the orbital and spin momenta of the Ω− . The spin part of the final state is 1/ 2, sz 0, 0 = 1/ 2, sz , and the orbital part is Ylm (θ , ϕ ) = | l , m〉. Thus the total final state of this system is l , m〉 |1/ 2, sz . By conservation of angular momentum l = 1, 2; m = 3/2 − sz . Thus the final state is a p wave if l = 1 , the state being 1, 1 1/ 2, 1/ 2 ; a d wave if l = 2 , the state being a combination of 2, 2 1/ 2, − 1/ 2 and 2, 1 1/ 2, 1/ 2 . Hence the wave functions are 1 ψ p = Y11 (θ , ϕ )   ,  0  ψd =    − =  − 

4 1 1 1 1 1  | 2, 1〉 , −  | 2, 2〉 , − − 5 2 2  5 2 2  1 Y21 (θ , ϕ )  5 ,  4 Y22 (θ , ϕ )  5

and

ψ = adψd + a pψ p   1 a p Y11 (θ , ϕ ) − ad Y21 (θ , ϕ ) 5 . =   4   adY22 (θ , ϕ )   5

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Therefore

ψ *ψ =

3 sin 2θ [| a p |2 + | ad |2 − 2 Rea*pad cosθ ], 8π

i.e., the intensity of the emitted particles is I ∝ sin 2θ (1 + α cosθ ), where

α = −2 Re a*pad / (| a p |2 + | ad |2 ). − This is the most general form of the angular distribution of the K mesons.

b. If parity were conserved in the decay process, the final state would have positive parity, i.e. (−1)l PK PΛ = +1 . Since PK PΛ = (−1)(+1) = −1, we get l = 1. It would follow that 1  ψ f = Y11 (θ , ϕ )   , 0  and

ψ *f ψ f =

3 3 sin 2θ = (1 − cos 2θ ). 8π 8π

Hence parity conservation would impose an angular distribution of the form I ∝ (1 − cos 2θ ).

3030 Given two angular momenta J1 and J2 (for example, L and S) and the corresponding wave functions, where j1 = 1 and j2 = 1/ 2. Compute the Clebsch–Gordan coefficients for the states with J = J1 + J 2 , m = m1 + m2 , where: a. j = 3 / 2, m = 3 / 2, b. j = 3 / 2, m = 1/ 2 . Consider the reactions

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K − p → Σ −π + , → Σ +π − , → Σ 0π 0 , K − n → Σ −π 0 , → Σ 0π − . Assume they proceed through a resonance and hence a pure I-spin state. Find the relative rates based on I-spin conservation: c. for an I = 1 resonance state, d. for an I = 0 resonance state. Use the Clebsch–Gordan cofficients supplied. The I-spins for K , n, Σ, and π are 1/2, 1/2, 1, and 1 respectively. (Berkeley) Sol: a. As j1 = 1, j2 = 12 , we have | 3 / 2, 3 / 2 = 1, 1 1/ 2, 1/ 2 . b. Defining the operator J − = J1− + J 2− , we have J − 3 / 2, 3 / 2 = ( J1− + J 2− ) 1, 1 1/ 2, 1/ 2 , or, using the properties of J − (Problem 3008),  3 3 / 2, 1/ 2 =  2 1, 0 1/ 2, 1/ 2 +  1, 1 1/ 2, − 1/ 2 , and hence 3 / 2, 1/ 2 = 2 / 3 1, 0 1/ 2, 1/ 2 + 1/ 3 1, 1 1/ 2, − 1/ 2 . To calculate the relative reaction cross sections, we use the coupling representation to describe the initial and final I-spin states: K − p = 1/ 2, − 1/ 2 1/ 2, 1/ 2 = 1/ 2 1, 0 − 1/ 2 0, 0 , Σ− π + = 1, − 1〉 |1, 1 = 1/ 6 2, 0 − 1/ 2 1, 0 + 1/ 3 0, 0 , Σ+ π − = 1, 1 1, − 1 = 1/ 6 2, 0 + 1/ 2 1, 0 + 1/ 3 0, 0 , Σ0 π 0 = 1, 0 1, 0 = 2 / 3 2, 0 − 1/ 3 0, 0 , K −n = 1/ 2, − 1/ 2 1/ 2, − 1/ 2 = 1, − 1 , Σ− π 0 = 1, − 1 1, 0 = 1/ 2 2, − 1 − 1/ 2 1, − 1 , Σ0 π − = 1, 0 1, − 1 = 1/ 2 2, − 1 + 1/ 2 1, − 1 .

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K p = 1/ 2, − 1/ 2 1/ 2, 1/ 2 = 1/ 2 1, 0 − 1/ 2 0, 0 , Σ− π + = 1, − 1〉 |1, 1 = 1/ 6 2, 0 − 1/ 2 1, 0 + 1/ 3 0, 0 , Σ+ π − = 1, 1 1, − 1 = 1/ 6 2, 0 + 1/ 2 1, 0 + 1/ 3 0, 0 , Σ0 π 0 = 1, 0 1, 0 = 2 / 3 2, 0 − 1/ 3 0, 0 ,

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Problems and− Solutions in Quantum Mechanics

K n = 1/ 2, − 1/ 2 1/ 2, − 1/ 2 = 1, − 1 ,

Σ− π 0 = 1, − 1 1, 0 = 1/ 2 2, − 1 − 1/ 2 1, − 1 , Σ0 π − = 1, 0 1, − 1 = 1/ 2 2, − 1 + 1/ 2 1, − 1 . − To K p reactions going through the resonance state I = 1 , the final states

Σ− π + contributes

1 2

1, 0 , Σ+ π − contributes

1 2

1, 0 , while Σ0 π 0 does

not contribute. Hence

σ ( Σ− π 0 ) : σ ( Σ+ π − ) : σ ( Σ0 π 0 ) = 1:1: 0 . − Similarly for K n reactions we have

σ ( Σ− π 0 ) : σ ( Σ0 π − ) = 1:1 . −

Only the K p reactions go through the I = 0 resonance state. A similar consideration gives the following reaction cross section ratios:

σ ( Σ− π + ) : σ ( Σ+ π − ) : σ ( Σ0 π 0 ) = 1:1: 1.

3031 a. Compute the Clebsch-Gordan coefficients for the states with J = J1 + J2 , M = m1 + m2 , where j1 = 1 and j2 = 1/ 2 , and j = 3 / 2, M = 1/ 2 for the various possible m1 and m2 values. b. Consider the reactions:

π + p → π + p (i)



π − p → π − p (ii)



π − p → π 0 n (iii)

These reactions, which conserve isospin, can occur in the isospin I = 3 / 2 state ( ∆ resonance) or the I = 1/ 2 state ( N * resonance). Calculate the ratios of these cross-sections, σ i: σ ii : σ iii , for an energy corresponding to a ∆ resonance and an N * resonance respectively. At a resonance energy you can neglect the effect due to the other isospin states. Note that the pion is an isospin I = 1 state and the nucleon an isospin I = 1/ 2 state. (Berkeley) Sol: a. As M = m1 + m2 = 1/ 2, (m1 , m2 ) can only be (1, − 1/ 2) or (0, 1/ 2). Consider 3 / 2, 3 / 2 = 1, 1 1/ 2, 1/ 2 .

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As M − 3 / 2, 3 / 2 = 3 3 / 2, 1/ 2 , and M − 3 / 2, 3 / 2 = ( M1− + M 2− ) 1, 1 1/ 2, 1/ 2 = 2 1, 0 1/ 2, 1/ 2 + 1, 1 1/ 2, − 1/ 2 , we have 1, 1, 1/ 2, − 1/ 2| 3 / 2, 1/ 2 = 1/ 3 , 1, 0, 1/ 2, 1/ 2| 3 / 2, 1/ 2 = 2 / 3 . b. As

π + = 1, 1, , π 0 = 1, 0 , π − = 1, − 1 , p = 1/ 2, 1/ 2 , n = 1/ 2, − 1/ 2 , we have

π + p = 1, 1 1/ 2, 1/ 2 = 3 / 2, 3 / 2 , π − p = 1, − 1 1/ 2, 1/ 2 = a 3 / 2, − 1/ 2 + b 1/ 2, − 1/ 2 , π 0n = 1, 0 1/ 2, − 1/ 2 = c 3 / 2, − 1/ 2 + d 1/ 2, − 1/ 2 .

From a table of Clebsch-Gordan coefficients, we find a = 1/ 3, b = − 2 / 3, c = 2 / 3, d = 1 a = 1/ 3, b = − 2 / 3, c = 2 / 3, d = 1/ 3 . For the ∆ resonance state, I = 3 / 2 and the ratios of the cross sections are

σ i : σ ii : σ iii = 1 : | a |4 : | ac |2 = 1 :

1 2 : . 9 9

For the N * resonance state, I = 1/ 2 , and the ratios are 4 2 σ i : σ ii : σ iii = 0 : | b |4 : | bd |2 = 0 : : . 9 9

3032 Consider an electron in a uniform magnetic field along the z direction. Let the result of a measurement be that the electron spin is along the positive y direction at t = 0 . Find the Schrödinger state vector for the spin, and the average polarization (expectation value of s x ) along the x direction for t > 0 . (Wisconsin)

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Sol: As we are only interested in the spin state and the magnetic field is uniform in space, we can leave out the space part of the wave function. Then the Hamiltonian can be taken to be H = −µ ⋅ B = µe σ ⋅ B = ωσ z , where µ = −µ eσ , ω = µe B /  = eB / 2mc, µe being the Bohr magneton e . As 2mc the electron is initially along the y direction, the initial spin wave function is 1 1 ψ (t = 0) =  . 2 i  α  Let the spin wave function at a later time t be   . The Schrödinger equation β  i dψ/dt = Hψ then gives α  1 i    = ω   0 β 

0  α   , − 1  β 

or

α = −iωα , β = iωβ , with the solution α (t ) e −iωtα 0  1 e −iωt   =  . ψ (t ) =   =  2 ie −iωt   β (t ) e iωt β0  Hence Sx = 〈ψ (t )| sˆx | ψ (t )〉  = 〈ψ | σ x | ψ 〉 2

1 0  e −iωt   1  = ⋅ (e iωt − ie −iωt )   2 2 0 − 1 ie −iωt  i = (e 2iωt − e −2iωt ) 4  = − sin(2ωt ). 2

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3033 Consider an electron in a uniform magnetic field pointing along the z direction. The electron spin is measured (at time t 0 ) to be pointing along the positive y-axis. What is the polarization along the x and z directions (i.e. the expectation values of 2 s x and 2 s z ) for t > t 0 ? (Wisconsin) Sol:

The Schrödinger equation for the spin state vector is a(t ) a(t ) ∂ a(t ) i   = − µ ⋅ B  = µeσ ⋅ B  , ∂t b(t ) b(t ) b(t ) where µe = | e |  / 2me c is the magnitude of the magnetic moment of an electron. As B is along the z direction, the above becomes a(t ) 1 0  a(t ) i ∂t   = µe B  ,  0 − 1 b(t ) b(t ) or i ∂t a(t ) = µe Ba(t ),  i ∂t b(t ) = − µe Bb(t ). The solutions are a(t ) = a(t0 )e

− i µe B ( t −t0 )

b(t ) = b(t0 )e

i µ B ( t −t 0  e

)

,

.

At time t0 , the electron spin is in the positive y direction. Thus a(t0 )  0 − i  a(t0 )  a(t0 ) sy  =  =  ,  b(t0 ) 2  i 0  b(t0 ) 2 b(t0 ) or −ib(t0 ) = a(t0 ),   ia(t0 ) = b(t0 ). The normalization condition | a(t0 )|2 + | b(t0 )|2 = 1, then gives 1 | a(t0 )|2 = | b(t0 )|2 = . 2

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As

b(t ) 0 = i , we can take a(t ) 0 a(t0 ) = 1/ 2, b(t0 ) = i / 2 .

Hence for time t > t0, the polarizations along x and z directions are respectively 0 1 a  〈2sx 〉 = (a*, b*)    1 0  b   2µ  = (a * b + b * a) = −  sin  e B(t − t0 ) ;    0 1 a  〈2sz 〉 = (a*, b*)    1 0  b  = (a * a − b * b) = 0.

3034 Two spin − 1 particles form a composite system. Spin A is in the eigenstate 2 Sz = +1/ 2 and spin B in the eigenstate Sx = +1/ 2 . What is the probability that a measurement of the total spin will give the value zero? (CUS) Sol: In the uncoupling representation, the state in which the total spin is zero can be written as 1  1 1 1 1  0 = SBz = − − SAz = − SBz =  SAz = , 2 2 2 2  2 where SAz and SBz denote the z-components of the spins of A and B respectively. As these two spin − 12 particles are now in the state | Q 〉 = SAz = +1/ 2 SBx = +1/ 2 , the probability of finding the total spin to be zero is P = | 0| Q |2 . 2 In the representation of S and S z , the spin angular momentum operator S x is defined as   0 1 Sˆx = σ x =  . 2 2 1 0 

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Solving the eigenequation of S x, we find that its eigenfunction Sx = +1/ 2 can 2 be expressed in the representation of S and Sz as | Sx = +1/ 2 =

1 (| Sz = 1/ 2 + | Sz = −1/ 2 ) . 2

Thus 1 1 1  1 1 = Sx = +  Sz = − Sz = 2 2 2 2 2 1 1  1 , + z = − Sz = −  = 2 2  2 Sz = −

and hence 0| Q =

1  1 1  SAz = SAz = + 2 2 2 − SAz = −

=

1 1 SAz = + 2 2

SBz = − SBz =

1 1 SBx = + 2 2

1 1  SBx = +  2 2 

1 1 1 1 SBz = − SBx = + = . 2 2 2 2

Therefore P = 0| Q

2

1 = = 25% . 4

3035 a. An electron has been observed to have its spin in the direction of the z-axis of a rectangular coordinate system. What is the probability that a second observation will show the spin to be directed in x − z plane at an angle θ with respect to the z-axis? b. The total spin of the neutron and proton in a deuteron is a triplet state. The resultant spin has been observed to be parallel to the z-axis of a rectangular coordinate system. What is the probability that a second observation will show the proton spin to be parallel to the z-axis? (Berkeley)

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Sol: a. The initial spin state of the electron is 1  | ψ0 =   . 0  The state whose spin is directed in x − z plane at an angle θ with respect to the z-axis is  θ cos  2 . |ψ =   sin θ   2 Thus the probability that a second observation will show the spin to be directed in x − z plane at an angle θ with respect to the z-axis is P(θ ) = ψ | ψ0

2

 θ θ  1  = cos , sin     2 2  0 

2

θ  = cos 2   . 2 b. The initial spin state of the neutron-proton system is | ψ0 = |1, 1 = |1/ 2, 1/ 2 n |1/ 2, 1/ 2 p . Suppose a second observation shows the proton spin to be parallel to the z-axis. Since the neutron spin is parallel to the proton spin in the deuteron, the final state remains, as before, |ψ

f

= |1/ 2, 1/ 2 n |1/ 2, 1/ 2 p .

Hence the probability that a second observation will show the proton spin to be parallel to the z-axis is 1.

3036 The deuteron is a bound state of a proton and a neutron of total angular momentum J = 1. It is known to be principally an S( L = 0) state with a small admixture of a D( L = 2) state. a. Explain why a P state cannot contribute. b. Explain why a G state cannot contribute. c. Calculate the magnetic moment of the pure D state n − p system with J = 1. Assume that the n and p spins are to be coupled to make the total

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spin S which is then coupled to the orbital angular momentum L to give the total angular momentum J. Express your result in nuclear magnetons. The proton and neutron magnetic moments are 2.79 and –1.91 nuclear magnetons respectively. (CUS) Sol: a. The parities of the S and D states are positive, while the parity of the P state is negative. Because of the conservation of parity in strong interaction, a quantum state that is initially an S state cannot have a P state component at any later moment. b. The possible spin values for a system composed of a proton and a neutron are 1 and 0. We are given J = L + S and J = 1. If S = 0, L = 1, the system would be in a P state, which must be excluded as we have seen in (a). The allowed values are then S = 1, L = 2, 1, 0. Therefore a G state ( L = 4) cannot contribute. c. The total spin is S = s p + sn . For a pure D state with J = 1 , the orbital angular momentum (relative to the center of mass of the n and p) is L = 2 and the total spin must be S = 1 . The total magnetic moment arises from the coupling of the magnetic moment of the total spin, µ , with that of the orbital angular momentum, µ L , where µ = µ p + µn , µ p , µn being the spin magnetic moments of p and n respectively. The average value of the component of µ in the direction of the total spin S is µs =

(g µ p

s + g n µ N sn ) ⋅ S

N p

S

2

1 S = ( g p + g n )µ N S, 2

where

µN = 1 as s p = sn = S . 2

e , g p = 5.58, g n = −3.82, 2m pc

The motion of the proton relative to the center of mass gives rise to a magnetic moment, while the motion of the neutron does not as it is uncharged. Thus µL = µN L p , where L p is the angular momentum of the proton relative to the center of mass. As L p + Ln = L and we may assume L p = Ln , we have L p = L / 2 (the

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center of mass is at the mid-point of the connecting line, taking m p ≈ mn ). Consequently, µ L = µ N L / 2 . The total coupled magnetic moment along the direction of J is then 1  1  2 µ N L ⋅ J + 2 ( g p + g n )µ N S ⋅ J  J . µT = J ( J + 1) Since J = L + S, S ⋅ L = 12 ( J 2 − L2 − S 2 ) . With J = 1, L = 2, S = 1 and so J 2 = 2, L2 = 6, S 2 = 2 J 2 = 2, L2 = 6, S 2 = 2 , we have S ⋅ L = −3 and thus L ⋅ J = 3, S ⋅ J = −1. Hence 1  1 µT =  µ N ⋅ 3 + ( g p + g n )µ N (−1) J / 2 2  2  1 1 = 1.5 − ( g p + g n ) µ N J = 0.31µ N J .  2 2 Taking the direction of J as the z-axis and letting J z take the maximum value J z = 1, we have µT = 0.31µ N .

3037 A preparatory Stern-Gerlach experiment has established that the z-component of the spin of an electron is − / 2. A uniform magnetic field in the x-direction of magnitude B (use cgs units) is then switched on at time t = 0 . a. Predict the result of a single measurement of the z component of the spin after elapse of time T. b. If, instead of measuring the z-component of the spin, the x-component is measured, predict the result of such a single measurement after elapse of time T. (Berkeley) Sol: Method 1 a  The spin wave function   satisfies b   a  e B  a  a  0 1 a  i ∂t   = σ x   = ω     = ω   , 1 0  b  b  2mc b  b 

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where ω = eB / 2mc, or ia = ωb,  ib = ωa. Thus a = −iωb = −ω 2a, the solution being a = Ae iωt + Ce −iωt , i b = a = − ( Ae iω t − Ce −iωt ), ω where A and C are arbitrary constants. From the initial condition a(0) = 0 and b(0) = 1 as the initial spin is in the –z direction, we get A = −1/ 2, C = 1/ 2 . Hence  1 −iωt iωt a = 2 (e − e ) = − i sin ωt ,  b = 1 (e iwt + e −iωt ) = cos ωt ,  2 and the wave function at time t is −i sin ωt  ψ (t ) =    cos ωt  a. At t = T , −i sin ωT  1  0 ψ (T ) =   = − i sin ωT   + cos ωT   .  cos ωt  0 1  0  0  As 1  and 1  are the eigenvectors for σ z with eigenvalues +1 and −1     respectively the probability that the measured z-component of the spin is 2 2 positive is sin ωT ; the probability that it is negative is cos ωT . b. In the diagonal representation of σ z , the eigenvectors of σ x are

ψ (σ x = 1) =

1 1 1 −1 .   , ψ (σ x = −1) =   2 1 21 

As we can write −i sin ωT  1 −iωT 1 1 1 iωT 1 −1 ψ (T ) =  e ⋅ e ⋅ =  +   2 2 1 2 2 1  cos ωT  1 −iωT 1 iωT e ψ (σ x = 1) + e ψ (σ x = −1), = 2 2

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the probabilities that the measured x-component of the spin is positive and is negative are equal, being 2

2

e −iωT e iωT 1 = = . 2 2 2 Method 2 The Hamiltonian for spin energy is H = −µ ⋅ B = eBσ x /2mc . The eigenstates of σ x are 1 1  , 2 1

1 1  . 2 −1

We can write the initial wave function as 0  1  1 1 1  1  ψ (t = 0) =   =   −   . 2  2 1 2 −1 1  The Hamiltonian then gives 1 1 eB 1 1  ψ (t ) =   e −iωt −   e iωt , ω = . 2 1 2 −1 2mc a. As −i sin ωt  ψ (t ) =  ,  cos ωt  the probabilities at t = T are Pz↑ = sin 2ωT , Pz↓ = cos 2ωT . b. As in method 1 above, 2  1 −iωt 1 e = ,  Px↑ = 2 2  1   Px↓ = 2 .

3038 An alkali atom in its ground state passes through a Stern–Gerlach apparatus adjusted so as to transmit atoms that have their spins in the + z direction. The atom then spends time τ in a magnetic field H in the x direction. At the end of this time what is the probability that the atom would pass through a

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Stern–Gerlach selector for spins in the − z direction? Can this probability be made equal to unity? if so, how? (Berkeley) Sol:

The Hamiltonian | e | H |e| H Hˆ = −µ ⋅ H = σ x ≡ ωσ x , ω = , 2mc 2mc gives the equation of motion  ψ   0 1   ψ1   d  ψ1    = ω  2  , = ω  −   ψ1  i dt  ψ2   1 0   ψ2    or iψ1 = ωψ2 ,  iψ 2 = ωψ1 . and hence ψ  The solution is ψ =  1  , with 2 ψ 

ψ1 + ω 2ψ1 = 0.

ψ1 (t ) = ae iω t + be −iωt ,   i iωt −iωt ψ2 (t ) = ψ1 = −ae + be .  ω The initial condition 1  ψ (t = 0) =   0  then gives a = b = 1/2. Thus ψ1  ψ = = ψ 2 

iωt −ivt 1  e + e   cos ωt   iωt −iωt  =   2 −e + e  −i sin ωt 

1  0  = cos ωt   − i sin ωt   . 0  1  0  In the above   is the eigenvector σ z for eigenvalue −1 . Hence the probability 1  that the spins are in the –z direction at time τ after the atom passes through the Stern–Gerlach selector is

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 cos ωτ  (0 1)   −i sin ωτ 

2

= sin 2ωτ =

1 − cos 2ωτ . 2

The probability equals 1, if 1 − cos 2ωτ = 2, or cos 2ωτ = −1, i.e. at time

τ=

(2n + 1)π mcπ = (2n + 1) . 2ω |e| H

Hence the probability will become unity at times τ = (2n + 1)mcπ / | e | H .

3039 A beam of particles of spin 1/2 is sent through a Stern–Gerlach apparatus, which divides the incident beam into two spatially separated components depending on the quantum number m of the particles. One of the resulting beams is removed and the other beam is sent through another similar apparatus, the magnetic field of which has an inclination α with respect to that of the first apparatus (see Fig. 3.6). What are the relative numbers of particles that appear in the two beams leaving the second apparatus? Derive the result using the Pauli spin formalism. (Berkeley) Sol: For an arbitrary direction in space n = (sinθ cos ϕ , sinθ sin ϕ , cosθ ) the spin operator is

Fig. 3.6

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σ ⋅ n = σ x sin θ cos ϕ + σ y sin θ sin ϕ + σ z cosθ cosθ sinθ e −iϕ   , =  iϕ  sinθ e − cosθ  where 0 1 0 − i  1 0  σx = , σy = , σz = , 1 0  i 0  0 − 1 a  are Pauli’s spin matrices. Let its eigenfunction and eigenvalue be   and λ b  respectively. Then a  a  σ ⋅ n  = λ   , b  b  or a(cosθ − λ ) + be −iϕ sin θ = 0, ae iϕ sin θ − b(λ + cosθ ) = 0. For a, b not to vanish identically, cosθ − λ e −iϕ sin θ = λ 2 − cos 2 θ − sin 2 θ = 0, e iϕ sin θ − ( λ + cosθ ) or λ 2 =1, i.e., λ = ±1, corresponding to spin angular momenta ± 12  . Also, normalization requires a  (a*b* )  = | a |2 + | b |2 = 1 . b  For λ = +1, we have b 1 − cosθ sin θ2 iϕ = = e , a e −iϕ sinθ cos θ2 and, with normalization, the eigenvector  θ  cos  2  ↑n〉 =  . e iϕ sin θ   2 For λ = −1, we have cos θ b cosθ + 1 = − −iϕ = − −iϕ 2 θ a e sinθ e sin 2

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and, with normalization, the eigenvector  −iϕ θ −e sin  2. |↓ n =   cos θ   2  For the first Stern–Gerlach apparatus take the direction of the magnetic field as the z direction. Take n along the magnetic field in the second Stern–Gerlach apparatus. Then ϕ = 0, θ = α in the above. If the particles which are sent into the second S − G apparatus have spin up, we have ↑ z = c | ↑ n〉 + d | ↓ n ,  1 c = ↑ n | ↑ z = (cos(α / 2), sin(α / 2))   = cos(α / 2),  0  1 d = ↓ n | ↑ z = (− sin(α / 2), cos(α / 2))   = − sin(α / 2).  0 Therefore, after they leave the second S − G apparatus, the ratio of the numbers of the two beams of particles is | c |2 cos 2 (α / 2) α = = cot 2 . 2 2 |d | sin (α / 2) 2 If the particles which are sent into the second S − G apparatus have spin down, we have ↓ z = c ↑n + d ↓n ,  0 c = ↑ n ↓ z = (cos(α / 2), sin(α / 2))   = sin(α / 2),  1  0 d = ↓ n ↓ z = (− sin(α / 2), cos(α / 2))   = cos(α / 2),  1 and the ratio of the numbers of the two beams of the particles is

α sin 2 (α / 2) = tan 2 . 2 cos (α / 2) 2

3040 The magnetic moment of a silver atom is essentially equal to the magnetic moment of its unpaired valence electron which is µ = − γ S , where γ = e / mc and s is the electron’s spin.

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Suppose that a beam of silver atoms having velocity V is passed through a Stern–Gerlach apparatus having its field gradient in the z direction, and that only the beam with ms =  / 2 is to be considered in what follows. This beam then enters a region of length L having a constant magnetic field B0 directly along the axis of the beam (y-axis). It next enters another Stern–Gerlach apparatus identical to the first as shown in Fig. 3.7. Describe clearly what is seen when the beam exits from the second Stern–Gerlach apparatus. Express the intensities of the resulting beams in terms of V , L, B0 and the constants of the problem. Use quantum mechanical equations of motion to derive your result. (Berkeley)

Fig. 3.7 Sol: If we took a picture at the exit of the second S − G apparatus, we would see two black lines arising from the deposition of the two kinds of silver atoms with ms =  / 2 and ms = − / 2. Denote the state of the system in the region L by | t . If we consider only atoms of ms =  / 2 in the beam that enters the region L at t = 0, then 1  t =0 =  . 0  The Hamiltonian of the system in the region of length L is γ B0 H = −µ ⋅ B = σ y, 2 and so

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 i  t = exp − Ht  | t = 0     iγ B t  1  = exp − 0 σ y     2  0   γB t γ B t  1  = cos 0 − iσy sin 0     2 2  0  = cos

γ B0t 1  γ B t 0    + sin 0   . 2 0  2 1 

Hence at the exit of the region, the intensities of beams with ms = 2 and ms = − 2 are respectively γ B t  γ B t  I + = I 0cos 2  0  , I − = I 0 sin 2  0  ,  2   2  where I 0 is the intensity of the beam that enters the region. When the beam leaves the region L , t = L / V . So the ratio of intensities is cot 2 (γ B0 L / 2V ). The splitting of the beam is seen when it exists from the second Stern–Gerlach apparatus.

3041 Two oppositely charged spin − 21 particles (spins s1 and s 2 ) are coupled in a system with a spin-spin interaction energy ∆E . The system is placed in a uniform magnetic field H = Hzˆ . The Hamiltonian for the spin interaction is Hˆ = ( ∆E / 4)(σ 1 ⋅ σ 2 ) − (µ1 + µ2 ) ⋅ H where σ i = gi µ 0 si is the magnetic moment of the i th particle. The spin wave functions for the 4 states of the system, in terms of the eigenstates of the z-component of the operators σ i = 2s i , are

ψ1 = α1α 2 , ψ2 = sβ1α 2 + cα1β2 , ψ3 = c β1α 2 − sα1β2 , ψ4 = β1β2 , where (σ z )i α i = α i , (σ z )i βi = −βi , s = (1/ 2 ) ⋅ (1− x / 1+ x 2 )1/2 ,

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c = (1/ 2 )(1+ x / 1+ x 2 )1/2 , x = µ 0 H ( g2 − g1 ) / ∆E . a. Find the energy eigenvalues associated with each state ψi . Discuss the limiting cases µ 0 H / ∆E  1 and µ 0 H / ∆E  1. b. Assume that an initial state ψ (0) is prepared in which particle 1 is polarized along the field direction z, but particle 2 is unpolarized. Find the time dependence of the polarization of particle 1: P1z (t ) = ψ (t )| σ 1z | ψ (t ) . Again discuss the limiting cases µ 0 H / ∆E  1 and µ 0 H / ∆E  1. (Columbia) Sol: a. σ , α , β may be represented by matrices 0 1 0 − i  1 0  σx = , σy = , σz =  1 0  i 0   0 − 1 1  0  α = , β = , 0  1  for which the following relations hold: σ ixαi = βi , σ ix βi = αi , σ iyαi = iβi , σ iy βi = −iαi ,

σ izαi = αi ,

σ iz βi = −βi .

Then as Hˆ = (∆E / 4)(σ 1 ⋅ σ 2 ) − (µ1 + µ 2 ) ⋅ H = (∆E / 4)(σ 1xσ 2 x + σ 1 yσ 2 y + σ 1zσ 2 z ) 1 − µ0 H ( g 1σ 1z + g 2σ 2 z ), 2 where we have used µ = g µ0 s = 12 g µ0 s, we have 1 Hˆ ψ1 = Hˆ α1α 2 = (∆E / 4)(β1β 2 − β1β 2 + α1α 2 ) − µ0 H ( g 1 + g 2 )α1α 2 2     g1 + g 2 g1 + g 2 =  ∆E / 4 − µ0 H  α1α 2 =  ∆E / 4 − ⋅ µ0 H  ψ1 ,     2 2

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and hence 1 E1 = ∆E / 4 − ( g 1 + g 2 )µ0 H . 2 Similarly, ∆E Hˆ ψ2 = [s(α1β 2 + α1β 2 − β1α 2 ) + c(β1α 2 + β1α 2 − α1β 2 )] 4 1 + ( g 1 − g 2 )µ0 H (sβ1α 2 − cα1β 2 ) 2 = [(∆E/4)(2c − s ) − (∆E/2)xs]β1α 2 +[(∆E/4)(2s − c ) + (∆E/2)xc]α1β 2 = (∆E/4)(2c / s − 2 x − 1)sβ1α 2 + (∆E/4)(2s / c + 2 x − 1)cα1β 2 . Then as 2 c s − 2x = 2

(1 + x / 1 + x 2 )1/2 (1 − x / 1 + x 2 )1/2

− 2x = 2 1 + x 2

= 2( 1 + x 2 − x ) + 2 x = 2s / c + 2 x , we have Hˆ ψ2 = (∆E 4)(2 1 + x 2 − 1)ψ2 , or E2 = (∆E 4)(2 1 + x 2 − 1) . By the same procedure we obtain E3 = (−∆E 4)(2 1 + x 2 + 1), g +g E4 = (∆E 4) + 1 2 µ0 H . 2 b. As particle 2 is unpolarized and can be considered as in a mixed state, its state ξ 2 can be expanded in terms of α 2 and β 2 :

ξ 2 = aα 2 + bβ 2 , 2

2

where a = b = 1 2 . Then the initial total wave function is

ψ (0) = α1ξ 2 = α1 (aα 2 + bβ 2 ) s c = aψ1 + b 2 2 ψ2 − b 2 2 ψ3 c +s c +s = aψ1 + bcψ2 − bsψ3 , as 1 x  c 2 + s 2 = 1 + + 2 1+ x 2 

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1 x  1 −  = 1. 2 1+ x 2 

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Hence  E   E  ψ (t ) = aψ1 exp −i 1 t  + bcψ2 exp −i 2 t         E  − bsψ3 exp −i 3 t  .    Then using the relations σ 1zψ1 = σ 1zα1α 2 = α1α 2 ,

σ 1zψ2 = σ 1z (sβ1α 2 + cα1β 2 ) = −sβ1α 2 + cα1β 2 , σ 1zψ3 = σ 1z (cβ1α 2 − sα1β 2 ) = −(cβ1α 2 + sα1β 2 ), we obtain P1z (t ) = ψ (t ) σ 1z ψ (t ) 2

2

= a + b exp(−iE2 t  ) c(sβ1α 2 + cα1β 2 ) − exp(−iE3 t  ) s(cβ1α 2 − sα1β 2 ) exp(−iE2 t  ) × c(−sβ1α 2 + cα1β 2 ) + exp(−iE3 /  ) s(cβ1α 2 + sα1β 2 ) 1 1 = + [(s 2 − c 2 )2 + 4 s 2c 2 cos( E2 − E3 )t/] 2 2 1 1 [x 2 + cos( 1 + x 2 ∆Et/ )] = + 2 2(1 + x 2 ) 1 sin 2 ( 1 + x 2 ∆Et/2 ). = 1− 1+ x 2 c. In the limit µ0 H/∆E  1 , i.e., x  1, we have ∆E g 1 + g 2 1 − µ0 H ≈ − ( g 1 + g 2 )µ0 H , 4 2 2 1 ∆E (2 1 + x 2 − 1) ≈ (∆E/4) × 2 x = ( g 2 − g 1 )µ0 H , E2 = 2 4 1 ∆E 2 E3 = − (1 + 2 1 + x ) ≈ − ( g 2 − g 1 )µ0 H , 2 4 1 ∆E g 1 + g 2 E4 = µ0 H ≈ ( g 1 + g 2 )µ0 H , + 4 2 2 P1z (t ) ≈ 1. E1 =

When µ0 H /∆E  1 , i.e., x  1 , we have E1 ≈ E4 ≈ ∆E/4, E2 =

∆E ∆E (2 1 + x 2 − 1) ≈ , 4 4

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E3 ≈ −

3∆E , 4

P1z (t ) ≈ 1 − sin 2 (∆Et / 2 ) .

3042 A hydrogen atom is in a P1/2 state with total angular momentum up along the z-axis. In all parts of this problem show your computations and reasoning 2

carefully. a. With what probability will the electron be found with spin down? b. Compute the probability per unit solid angle P(θ , ϕ ) that the electron will be found at spherical angles θ , ϕ (independent of radial distance and spin). c. An experimenter applies a weak magnetic field along the positive z-axis. What is the effective magnetic moment of the atom in this field? d. Starting from the original state, the experimenter slowly raises the magnetic field until it overpowers the fine structure. What are the orbital and spin quantum numbers of the final state? [Assume the Hamiltonian is linear in the magnetic field.] e. What is the effective magnetic moment of this final state? (Berkeley) Sol: a. For the state 2 P1/2 , l = 1, s = 1/ 2, J = 1/ 2, J z = 1/ 2 . Transforming the coupling representation into the uncoupling representation, we have J , J z = 1/ 2, 1/ 2 = 2 / 3 1, 1 1/ 2, − 1/ 2 − 1 / 3 1, 0 1/ 2, 1/ 2 .

Therefore P↓ = 2 / 3. b. As | J , Jz =

1  2Y11   , 3  Y10 

we have 1 P(θ , ϕ )dΩ = (2Y11* Y11 + Y10* Y10 ) dΩ. 3

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Hence the probability per unit solid angle is  1 3 3 1 sin 2 θ + cos 2 θ  = . P(θ , ϕ ) =  2 ×   3 8π 4π 4π c. In the weak magnetic field, J and Jz are “good” quantum numbers and the state remains unchanged. The effective magnetic moment is e 1 1 1 1 e 1 1 1 1 , Jz , = g ⋅ µ = , µz , = g , 4mc 2 2 2 2 2mc 2 2 2 2 Where m is the electron mass and J ( J + 1) − l(l + 1) + s(s + 1) 2 g = 1+ = . 2 J ( J + 1) 3 Hence µ = e 6mc . d. In a strong magnetic field, the interaction of the magnetic moment with the field is much stronger than the coupling interaction of spin and orbit, so that the latter can be neglected. Here l and s are good quantum numbers. The Hamiltonian related to the magnetic field is W = −µ ⋅ B − µ ⋅ B = eBlz 2mc + eBsˆ mc . l

s

z

When the magnetic field is increased slowly from zero, the state remains at the lowest energy. From the expression of W, we see that when the magnetic field becomes strong, only if lz = − , sz = − / 2 can the state remain at the lowest energy. Thus the quantum numbers of the final state are l = 1, lz = −1, s = 1/ 2, sz = −1/ 2 l = 1, lz = −1, s = 1/ 2, sz = −1/ 2 . e. the effective magnetic moment of the final state is

µ = µlz + µ sz = − e / 2mc − e / 2mc = − e / mc.

3043 Consider a neutral particle with intrinsic angular momentum

s( s + 1), where

s =  / 2 , i.e., a spin-1/2 particle. Assume the particle has a magnetic moment M = γ s, where γ is a constant. The quantum-mechanical state of the particle can be described in a spin space spanned by the eigenvectors |+

and | − representing alignments

parallel and antiparallel to the z-axis:   sˆz + = + , sˆz − = − − . 2 2

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At time t = 0 the state of the system is ψ (t = 0) = + . The particle moves along the y-axis through a uniform magnetic field B = B yˆ oriented along the y-axis. a. As expressed in the + , − basis, what is ψ (t ) ? b. What would be the expectation values for measurements of the observables sx, sy, sz as functions of time? (CUS) Sol: a. The Hamiltonian of the particle is Hˆ = −M ⋅ B = −γ sˆy B . In the representation of sˆ2, sˆz ,   0 −i  sˆy =  , 0 2 i and so the two eigenstates of sˆy are s y = /2 =

1 −i  1 i    , s y = − /2 =  . 21 2 1

As 1 1 1 Hˆ s y =  = −γ B  s y =  , 2 2 2 1 1 1 Hˆ s y = −  = γ B  s y = −  , 2 2 2 any state of the particle can be expressed as  1  1 ψ (t ) = c1 s y =  exp +i γ Bt   2  2  1  1 +c2 s y = −  exp −i γ Bt  .  2  2 Then the initial condition | ψ (t = 0) = sz =  / 2

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gives 1 1 1 sz =  = c1 s y =  + c2 s y = −  , 2 2 2 and so +

1 1 1 −i  1   = s y sz =  =     2 2 2  1  0  1  1 1 (i 1)  = i, = 2 2 0

c1 =

+

1 1 1  i  1  c2 = −  = s y s z =  =     2 2 2 1 0  1  1 1 (−i 1)  = − i. = 2 2 0 Therefore

ψ (t ) =

 1  1 1 −i  i   exp i γ Bt   2  2 21  1  1 1 i  i −   exp −i γ Bt   2  2 2 1

1  1  1 1 = cos  γ Bt  sz =  − sin  γ Bt  sz = −  . 2  2  2 2 b.  1  1  1 1 sz = cos  γ Bt  − sin  γ Bt     2  2  0  2

0  − 1

 1   cos  γ Bt      1  1  1  2 × =  (cos  γ Bt  sin  γ Bt   2  2  1  2  − sin  γ Bt    2    1   cos  γ Bt      1 2 × =  cos(γ Bt ).  1  2 sin γ − Bt    2   s y = 0 , because s y = 0 at t = 0 and sy is conserved.

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 1  1  1 0 1 sx = cos  γ Bt  − sin  γ Bt      2  2 1 0   2  1   cos  γ Bt      1  1  2 × = − sin  γ Bt  cos  γ Bt   2  2  1   − sin  γ Bt    2    1   cos  γ Bt   2  1 1 ×  = −  sin(γ Bt ).   1  2 2 − sin  γ Bt  2  

3044 A particle of spin 1/2 and magnetic moment µ is placed in a magnetic field B = B0 zˆ + B1 cos ωt xˆ − B1 sinωt yˆ , which is often employed in magnetic resonance experiments. Assume that the particle has spin up along the + z -axis at t = 0 ( ms = +1/ 2) . Derive the probability to find the particle with spin down ( ms = −1/ 2) at time  t > 0 . (Berkeley) Sol: The Hamiltonian of the system is H = −µσ ⋅ B . Letting

ω0 = µ B0 /  , ω1 = µ B1 /  , we have H = −µ ( B0σ z + B1σ x cos ωt − B1σ y sin ωt ) 0 e iωt   , = −ω0σ z − ω1  −iωt 0 e where 0 σx = 1

0 1 , σ y = 0 i

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1 −i , σz = 0 0

0 . − 1

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Let the wave function of the system be a(t ) t = . b(t ) The Schrödinger equation i ∂t t = H t , or 0 0  a     − ω1  iωt − 1 b  e

a  1 i   = −ω0  0 b 

e iωt  a     . 0  b 

gives a = iω0a + iω1e iωt b,  b = −iω0b + iω1e −iωt a. Try a solution of the type a = α exp(iω0t ),  b = β exp(−iω0t ). Substitution in the above equations gives α = iω1 exp[i(−2ω0 + w )t ]β ,  β = iω1 exp[−i(−2ω0 + ω )t ]α . Assume that α and β have the forms α = A1 exp[i(−2ω0 + ω + Ω)t ],  iΩt β = A2e , Where A1, A2, and Ω are constants. Substitution gives (−2ω0 + ω + Ω)A1 − ω1 A2 = 0,  −ω1 A1 + ΩA2 = 0. For this set of equations to have nontrivial solutions the determinant of the coefficients of A1, A2 must be zero, i.e., (−2ω0 + ω + Ω)Ω − ω12 = 0, giving  ω Ω± = − −ω0 +  ± (−ω0 + ω / 2)2 + ω12 .  2 Therefore the general form of β is

β = A2+ exp(iΩ+t ) + A2− exp(iΩ−t ),

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and that of α is

β exp[i(−2ω0 + ω )t ] iω1 1 = exp[i(−2ω0 + ω )t ][Ω+ A2+exp(iΩ+t ) ω1 + Ω− A2−exp(iΩ−t )].

α=

Initially the spin is up along the z-axis, so 1  a(0) α (0) t =0 = = = , 0  b(0)  β (0) giving 1 (Ω+ A2+ + Ω− A2− ) = 1, ω1 A2+ + A2− = 0. The solution is A2+ = − A2− = ω1 / (Ω+ − Ω− ) ω1 = . 2 (ω0 − ω / 2)2 + ω12 Hence b(t ) = exp(−iω0t )β (t ) = exp(−iω0t )A2+ ×[exp(iΩ+t ) − exp(iΩ−t )] = exp(−iωt/2)2iA2+ × sin( (ω0 − ω/2)2 + ω12 t ) =

iω1 exp[−i(ω/2)t | (ω0 − ω/2)2 + ω12 × sin( (ω0 − ω/2)2 + ω12 t ).

The probability that the particle has spin down along the z-axis at time t is P = z ↓| t 2

= b(t ) =

2

ω12 sin 2 ( ω0 − ω / 2)2 + ω12 t ) . (ω0 − ω / 2)2 + ω12

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3045 A spin- 21 system with magnetic moment µ = µ 0σ is located in a uniform time-independent magnetic field B0 in the positive z direction. For the time interval 0 < t < T an additional uniform time-independent field B1 is applied in the positive x direction. During this interval, the system is again in a uniform constant magnetic field, but of different magnitude and direction z′ from the initial one. At and before t = 0 , the system is in the m = 1/ 2 state with respect to the z-axis. a. At t = 0 + , what are the amplitudes for finding the system with spin projections m′ = ±1/ 2 with respect to the z′ direction? b. What is the time development of the energy eigenstates with respect to the z′ direction, during the time interval 0 < t < T ? c. What is the probability amplitude at t = T of observing the system in the spin state m = −1/ 2 along the original z-axis? [Express your answers in terms of the angle θ between the z and z′ axes and the frequency ω 0 = µ 0 B0 / .] (Berkeley) Sol: a. In the representation of sz, the eigenvectors of sz ′ are  θ cos  2,   sin θ   2

 θ − sin  2,   cos θ   2 

corresponding to the eigenvalues sz ′ = 1/ 2 and −1/ 2 respectively. Then the probability amplitudes for m′ = ±1/ 2 are respectively  θ θ  1  θ C+ = cos sin    = cos ,  2 2  0  2  θ θ  1  θ C− = − sin cos    = − sin .   2 2 0  2

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b. The Hamiltonian in the interval 0 < t < T is H = − µ ⋅ B = − µ0 ( B0σ z + B1σ x )  B0 = − µ0   B1

B1  . − B0 

The initial eigenfunctions are  θ  θ cos  − sin  2 2,  , χ − (0) =  χ + (0) =  θ θ  sin  cos   2  2  where B  θ = tan−1  1  .  B0  Substitution in the Schrödinger equation H χ ± (0) = ±E χ ± (0) gives E = −µ0 B0 / cosθ = −µ0 B, where B = B02 + B12 . At a later time t in 0 < t < T , the eigenstates are

χ ± (t ) = exp( ∓iEt/ ) χ ± (0) = exp(±iµ0 Bt/ ) χ ± (0) . c. The probability amplitude at t = T is 1  C− (T ) = (0 1) exp(−iHT / )  0  = (iB1 / B02 + B12 )sin(µ0T B02 + B12 /  ) = i sin θ sin(µ0 BT /  ). An alternative way is to make use of 1  θ θ ψ (0) =   = cos χ + (0) − sin χ − (0), 2 2 0  and so

θ ψ (t ) = χ + (0)cos exp(iµ0 Bt/ ) 2 θ − χ − (0)sin exp(−iµ0 Bt/ ), 2

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to get

θ θ C− (T ) = β +ψ (T ) = cos sin 2 2 × {exp(iµ0 BT / ) − exp(−iµ0 BT / )}, = i sin θ sin

0  µ0 BT , where β =   .  1 

3046 A spin- system of magnetic moment µ is placed in a dc magnetic field H0 e z 1 2

in which the energy of the spin state | +1/ 2 is ω 0 , that of | −1/ 2 being taken as 0. The system is in the state | −1/ 2 when at t = 0 , a magnetic field H (e x cos ω 0t + e y sinω 0t ) is suddenly turned on. Ignoring relaxation find the energy of the spin system as a function of ω 0 , H, c and t, where c = +1/ 2 | µ x + iµ y | −1/ 2 . Why is the energy of the spin system not conserved? (Columbia) Sol:

The Hamiltonian is Hˆ = −µ ( H + H 0 ) = −µσ ⋅ ( H + H 0 ) = −µ ( Hσ x cos ω0t + Hσ y sin ω0t + H 0σ z )  H0 H exp(−iω0t ) = −µ   H exp(iω0t ) −H 0 

 .  

In the Schrödinger equation ψ , ih ∂ψ / ∂t = H setting a(t ) ψ = , b(t ) we get  da µi  = [H 0a + H exp(−iω0t )b], dt   db  = µi [H exp(iω t )a − H b]. 0 0  dt 

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Try a solution of the type   1   a = A exp−i Ω + ω0  t  ,   2     1   b = B exp−i Ω − ω0  t  ,   2   where A, B and Ω are constants. Substitution gives   1 Ω + ω0 + ω  A + ω ′B = 0,   2  Ω − 1 ω − ω  B + ω ′A = 0,  0   2 where

µ H0 ,  µH ω′ = .  ω=

For nontrivial solutions we require 1  Ω +  ω0 + ω  2 

ω′

ω′

1  Ω −  ω0 + ω  2 

= 0,

giving 2

1  Ω = ± ω ′2 +  ω0 + ω  = ±Q , 2  2

1  where Q = | ω ′2 +  ω0 + ω  | . Hence 2   ω  ψ (t ) = ( A1e −iQt + A2e iQt )exp −i 0 t  α  2   ω  + ( B1e −iQt + B2e iQt )exp i 0 t  β ,  2  where  1  Ω1, 2 + ω + ω0   2  , B1, 2 = − A1, 2 ω′

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the subscripts 1, 2 corresponding to the values of Ω with + , − signs respectively, and 1  0  α = , β =  . 0  1  0  At t = 0 the system is in the − 12 state and ψ =   . Thus B1 + B2 = 1, A1 + A2 = 0 . 1  Then as   1 Q + ω0 + ω  A1 + ω ′B1 = 0,   2   1 −Q + ω0 + ω  A2 + ω ′B2 = 0,   2 we have Q( A1 − A2 ) + ω ′ = 0, 1   ω0 + ω  ( A1 − A2 ) + ω ′( B1 − B2 ) = 0, 2  giving

ω′ ω′ , A2 = , 2Q 2Q  1  Q + ω + ω0   2 , B1 = 2Q  1  Q − ω + ω0   2 . B2 = 2Q A1 = −

Therefore the wave function of the system is  ω  ω′ ψ (t ) = i sin(Qt )exp −i 0 t  α  2  Q    1  ω + ω0    ω     2 sinQt  exp i 0 t  β , + cosQt − i Q   2     and the energy of the system is E = ψ | H | ψ = −µ −H 0cos 2Qt 2    ω  1  ω ′2 H 0 − ω + ω0  H 0 − 2ω ′H ω + 0     2  2  sin 2 Qt . +  Q2  

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Note that as 1 1 c = + µ x + iµ y − 2 2 1 1 = µ + σ x + iσ y − 2 2 0 2  0  = µ (10)    = 2µ ,  0 0  1  c µ= . 2 As the energy of the system changes with time t, it is not conserved. This is because with regard to spin it is not an isolated system.

3047 A beam of neutrons of velocity υ passes from a region (I) (where the magnetic field is B = B1e z ) to a region (II) (where the field is B = B2 e x ). In region (I) the beam is completely polarized in the + z direction. a. Assuming that a given particle passes from (I) to (II) at time t = 0, what is the spin wave function of that particle for t > 0 ? b. As a function of time, what fraction of the particles would be observed to have spins in the + x direction; the + y direction; the + z direction? c. As a practical matter, how abrupt must the transition between (I) and (II) be to have the above description valid? (Wisconsin) Sol: a. Considering only the spin wave function, the Schrödinger equation is i ∂| χ / ∂t = Hˆ | χ , where Hˆ = −µ ⋅ B = −µn B2σ x , with µn = −1.9103µ N being the anomalous magnetic moment of the neutron,

µ N = e/2m p c the nuclear magneton, mp the mass of a proton. Thus d i χ = µn B2σ x χ ≡ −iω 2σ x χ , dt 

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where ω 2 =

|µn |B2 

233

a  . Let | χ =   . The above gives b  a = −iω 2b ,  b = −iω 2a.

The initial condition is a(0) = 1, b(0) = 0 as the beam is initially polarized in the +z direction. Solving for a and b and using the initial condition we obtain for t > 0 cos ω 2t  | χ = , −i sin ω 2t  b. The mean value of the spin in the state | χ , i.e., the polarized vector for neutron, is p = χ | σ | χ = χ | σ xex + σ y e y + σ z e z | χ = (0, − sin 2ω 2t , cos2ω 2t ) Thus in the region (II), the neutron spin is in the yz plane and precesses about the x direction with angular velocity 2ω 2. c. For the descriptions in (a) and (b) to be valid, the time of the transition between (I) and (II) must satisfy 2π h . t = w 2 | µn | B2 3 For example if B2 ∼ 10 Gs, then t  0.7µ s.

If the kinetic energy of the incident neutrons is given, we can calculate the upper limit of the width of the transition region.

3048 + −

The Hamiltonian for a (µ e ) atom in the n = 1, l = 0 state in an external magnetic field B is H = asµ ⋅ se +

|e| |e| se ⋅ B − sµ ⋅ B . me c mµ c

a. What is the physical significance of each term? Which term dominates in the interaction with the external field? b. Choosing the z-axis along B and using the notation (F, M), where F = s µ + s e , show that (1, + 1) is an eigenstate of H and give its eigenvalue.

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c. An rf field can be applied to cause transitions to the state (0, 0). Describe qualitatively how an observation of the decay µ + → e +ν eν µ could be used to detect the occurrence of this transition. (Wisconsin) Sol: a. The first term in Hˆ is due to the magnetic interaction between µ and e, the second and third terms respectively account for the magnetic interactions of µ and e with the external field B. Of the latter, the term | e | se ⋅ B / me c is dominant as me ≈ mµ / 200 . b. As F = sµ + se 1 eB eB Hˆ = a[F 2 − s µ2 − s e2 ]+ sez − sµ z . 2 me c mµ c Consider the state 1  1  (1, + 1) =     . 0 e 0 µ 2 2 2 2 As the eigenvalues of F , sµ , se , sµ z , sez are 1 (1 + 1) , 1 2

1 2

( 12 + 1)  2 , 12 ( 12 + 1)  2 , 12 , 12 

( 12 + 1)  2 , 12 ( 12 + 1)  2 , 12 , 12  respectively, we have  1  Hˆ (1, + 1) =  a 2 2 − 2 ⋅   2

3 e  e B− B (1, + 1) +  4  2me c 2mµ c 

1 e e  =  a 2 + B− B (1, + 1). 2me c 2mµ   4 Thus (1, +1) is an eigenstate of Hˆ , with the eigenvalue a 2 /4 + eB/2me c − eB/2mµ c . + + c. The decay µ → e ν eν µ can be detected through the observation of the + − annihilation of the positronium e e → 2γ . For the state (1, +1), the total + − + − angular momentum of the e e system is 1, and so e e cannot decay into

2γ whose total angular momentum is 0. For the state (0, 0) , the total angular

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+ − momentum of the e e system is 0 and so it can decay into 2γ . Hence, + − + + detection of e e → 2γ implies the decay µ → e ν eν µ of the Coulomb

potential before the reaction, (µ +e − ) system in the state (0, 0) , as well as the transition (1, + 1) → (0, 0) .

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Part IV

Motion in Electromagnetic Field

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4 4001 We may generalize the semi-classical Bohr-Sommerfeld relation

!∫ P ⋅dr = (n+1/ 2)h, (where the integral is along a closed orbit) to apply to the case where electromagnetic field is present by replacing P with p − eA/c, where e is the charge of the particle. Use this and the equation of motion for the linear momentum p to derive a quantization condition on the magnetic flux of a semi-classical electron which is in a magnetic field B in an arbitrary orbit. For electrons in solids this condition can be restated in terms of the size S of the orbit in k-space. Obtain the quantization condition on S in terms of B (Ignore spin effects). (Chicago) Sol:

In the presence of an electromagnetic field, the mechanical momentum P is P = p − eA/c , where p is the canonical momentum, e is the charge of particle. The generalized Bohr-Sommerfeld relation becomes e 



∫ P ⋅ dr = ∫  p − c A  ⋅ d r = (n + 1/2)h, or e

∫ p ⋅ d r − c φ = (n + 1/2)h, where

φ = ∫B ⋅ ds = ∫ (∇ × A ) ⋅ ds = ∫ A ⋅ d r , S

S

239

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using Stokes theorem. The classical equation of the motion of an electron in a constant magnetic field B, dp e dr =− × B, dt c dt gives p = − er × B/c and e

e

∫ p ⋅ dr = − ∫ c (r × B) ⋅ d r = ∫ c ∇ × (B × r ) ⋅ ds S

e = ∫ 2B ⋅ d s = 2eφ/c . c S Hence

φ = (n + 1/2)φ0 , where φ0 = hc/e . Defining k by p = k = −er × B/c, we have, assuming r is perpendicular to B, ∆k = − Be∆r/c , or ∆r = − c∆k/Be . Therefore, if the orbit occupies an area Sn in k-space and an area An in r-space, we have the relation An = (c/Be )2 Sn . As

 c  φ = ∫ BdAn =    Be 

2

2

 c  BSn = (n + 1/2)hc/e , n =  Be 

∫ BdS

we have Sn = 2π Be(n + 1/2)/c .

4002 A particle of charge q and mass m is subject to a uniform electrostatic field E. a. Write down the time-dependent Schrödinger equation for this system. b. Show that the expectation value of the position operator obeys ­ ewton’s second law of motion when the particle is in an arbitrary state N ψ (r , t ) .

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c. It can be shown that this result is valid if there is also a uniform magnetostatic field present. Do these results have any practical application to the design of instruments such as mass spectrometers, particle accelerators, etc.? Explain. (Buffalo)

Sol: a. The Schrödinger equation for the particle is i

∂ψ 2 2 =− ∇ ψ − qE ⋅ rψ . ∂t 2m

b. The Hamiltonian of the particle is H=

p2 − qE ⋅ r. 2m

Then dx 1 p 1  p2  p 1 = [x , H ] = x , x  = x [x , px ] = x , dt i i  2m  m i m 1 dpx 1 qE = [ px , H ] = [ px , − Ex x ] = − x [ px , x ] = qEx , dt i i i and hence d 〈r 〉 〈p 〉 = , dt m d 〈p 〉 = qE. dt Thus 1d p d2 , 〈r 〉 = dt 2 m dt or m

d2 r = qE, dt 2

which is just Newton’s second law of motion. c. These results show that we could use classical mechanics directly when we are computing the trajectory of a charged particle in instruments such as mass spectrometers, particle accelerators, etc.

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4003 The Hamiltonian for a spinless charged particle in a magnetic field B = ∇ × A is 2

 e 1  H=  p − A(r ) , c 2m   where e is the charge of particle, p = ( px py , pz ) is the momentum conjugate to the particle’s position r. Let A = − B0 ye x , corresponding to a constant magnetic field B = B0 e z . a. Prove that px and pz are constants of motion. b. Find the (quantum) energy levels of this system. (MIT) Sol:

The Hamiltonian for the particle can be written as H=

2

1  1 2 1 2 eB0  y + py + pz .  px +  2m c 2m 2m

a. As H does not depend on x and z explicitly, the basic commutation relations in quantum mechanics [xi , p j ] = iδij , [ pi , p j ] = 0, require [ px , H ] = 0, [ pz , H ] = 0, which show that px, pz are constants of the motion. b. In view of (a) we can choose {H , px , pz } as a complete set of mechanical variables. The corresponding eigenfunction is

ψ ( x , y , z ) = e i ( xpx + zpz )/φ ( y ), where px , pz are no longer operators but are now constants. The Schrödinger equation Hψ ( x , y , z ) = Eψ ( x , y , z ) then gives 1  eB  px + 0 2m  c

2  d2  y  −  2 2 + pz2 φ ( y ) = Eφ ( y ),  dy 

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or 2

−

2   2 d 2φ m  eB0   cpx  p2  + φ = E − z φ.    y + 2   2m dy 2 mc eB0  2m  

Setting

ω=

| e | B0 cp p2 , y ′ = y + x , E′ = E − z , mc eB0 2m

we can write the equation as −

 2 d 2φ m 2 2 + ω y ′ φ = E′φ , 2m dy ′2 2

which is the energy eigenequation for a one-dimensional harmonic oscillator. The energy eigenvalues are therefore E′ = E − pz2 /2m = (n + 1/2)ω , n = 0, 1, 2, .... Hence the energy levels for the system are En = pz2 /2m + (n + 1/ 2)ω , n = 0, 1, 2, ....

4004 An electron of mass m and charge –e moves in a region where a uniform magnetic field B = ∇ × A exists in z direction. a. Set up the Schrödinger equation in rectangular coordinates. b. Solve the equation for all energy levels. c. Discuss the motion of the electron. (Buffalo) Sol: a. The Hamiltonian is

2

1 ˆ e  Hˆ = P + A  . 2m  c 

As ∂ Az ∂ Ay − = 0, ∂y ∂z ∂ Ax ∂ Az − = 0, ∂z ∂x ∂ Ay ∂ Ax − = B, ∂x ∂y

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we can take Ax = Az = 0, Ay = Bx , i.e. A = Bxyˆ , and write the Schrödinger equation as 2  1  ˆ 2  ˆ eBx  Hˆ ψ = + Pˆz2 ψ = Eψ . Px +  Py +   2m  c   b. As [Pˆy , Hˆ ] = [Pˆz , Hˆ ] = 0, Py and Pz are conserved. Choose H, Py , Pz as a com-

plete set of mechanical variables and write the Schrödinger equation as  1 ˆ 2 1  ˆ eBx  2   Pz2  ψ. Px +  Py +  ψ =  E −   2m c 2m   2m  Let ξ = x + cPy /eB, Pˆξ = Pˆx . Then [ξ , Pˆξ ] = i and 2

1 ˆ 2 m  eB  2 1 ˆ 2 Hˆ = Pξ +   ξ + Pz . 2m 2  mc  2m ˆ2 The above shows that Hˆ − Pz is the Hamiltonian of a one-dimensional har2m

eB monic oscillator of angular frequency ω = mc . Hence the energy levels of the

system are E = (n + 1/ 2)ω + Pz2 /2m, n = 0, 1, 2, …. Because the expression of E does not contain Py explicitly, the degeneracies of the energy levels are infinite. c. In the coordinate frame chosen, the energy eigenstates correspond to free motion in the z direction and circular motion in the x − y plane, i.e. a helical motion. In the z direction, the mechanical momentum mv z = Pz is conserved, describing a uniform linear motion. In the x direction there is a simple harmonic oscillation round the equilibrium point x = −cPy / eB. In the y direction, the mechanical momentum is mv y = Py + eBx/c = eBξ /c = mωξ and so there is a simple harmonic oscillation with the same amplitude and frequency.

4005 Write down the Hamiltonian for a spinless charged particle in a magnetic field. Show that the gauge transformation A(r ) → A(r ) + ∇f (r ) is equivalent to multiplying the wave function by the factor exp [ief (r ) / c ] . What is the significance of this result? Consider the case of a uniform field B directed along the z-axis. Show that the energy levels can be written as

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E = ( n + 1/2)

245

|e|  2 k z2 . B+ mc 2m

Discuss the qualitative features of the wave functions. Hint: use the gauge where Ax = − By , Ay = Az = 0. (Wisconsin) Sol:

The Hamiltonian for the particle is 2

1  e  Hˆ =  pˆ − A  , 2m c where A is related to the magnetic field by B = ∇ × A. The Schrödinger equation is then 2

1  e   pˆ − A  ψ (r ) = Eψ (r ). 2m c Suppose we make the transformation A(r ) → A′(r ) = A(r ) + ∇f (r ),

ψ (r ) → ψ ′(r ) = ψ (r )exp

{ iec f (r )} ,

and consider  e  e   ie  e  pˆ − A ′ψ ′(r ) = pˆ ψ ′(r ) −  A + ∇f (r ) exp  f (r )ψ (r ) c c   c   c   ie  e  = exp f (r )  pˆ − A ψ (r ), c    c  2

2

 e   ie  e   pˆ − A ′ ψ ′(r ) = exp f (r )  pˆ − A  ψ (r ),  c  c   c  where we have used    ie  e     ie pˆ ψ ′(r ) = ∇ exp f (r )ψ (r ) = exp f (r )  ∇f (r ) + pˆ ψ (r ). i   c   c c   Substitution in Schrödinger’s equation gives 2

1  e   pˆ − A ′ ψ ′(r ) = Eψ ′(r ). 2m  c 

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This shows that under the gauge transformation A′ = A + ∇f , the Schrödinger equation remains the same and that there is only a phase difference between the original and the new wave functions. Thus the system has gauge invariance. Now consider the case of a uniform field B = ∇ × A = Bez , for which we have Ax = − By ,

Ay = Az = 0.

The Hamiltonian can be written as 2  1  ˆ eB  Hˆ =  px + y  + pˆ y2 + pˆ z2  . 2m  c   Since [ pˆ x , Hˆ ] = [ pˆ z , Hˆ ] = 0 as H does not depend on x, z explicitly, we may choose the complete set of mechanical variables ( pˆ , pˆ , Hˆ ) . The corresponding eigenx

z

state is

ψ ( x , y , z ) = e i ( px x + pz z )/ χ ( y ). Substituting it into the Schrödinger equation, we have 2 2  eB  1  2 ∂ + pz2  χ ( y ) = E χ ( y ).  px + y  −  2   c ∂y 2m   Let cpx /eB = − y0 . Then the above equation becomes −

2

2 m  eB  χ ″+   ( y − y0 )2 χ = ( E − pz2 /2m) χ , 2m 2  mc 

which is the equation of motion of a harmonic oscillator. Hence the energy levels are 2 2  1  |e | B kz +  n +   E= , n = 0, 1, 2, …,  2m 2  mc where kz = pz /h , and the wave functions are

ψ px pzn ( x , y , z ) = e i ( px x + pz z )/ χ n ( y − y0 ), where    |e | B  |e | B χ n ( y − y0 ) ∼ exp− ( y − y0 )2  Hn  ( y − y0 ) ,   c  2c  Hn being Hermite polynomials. As the expressions for energy does not depend on px and pz explicitly, there are infinite degeneracies with respect to px and pz.

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4006 A point particle of mass m and charge q moves in spatially constant crossed magnetic and electric fields B = B0 zˆ , E = E0 xˆ . a. Solve for the complete energy spectrum. b. Evaluate the expectation value of the velocity v in a state of zero momentum. (Princeton) Sol: a. Choose a gauge A = B0 x yˆ , ϕ = − E0 x so that ∇ × A = B0zˆ , − ∇⋅ ϕ = E0. Then H=

2 2  q  q 1  1  2   2 p − A + q ϕ = p + p − B x     y x 0  + pz − qE0 x .  2m c 2m  c 

As H does not depend on y and z explicitly, py and pz each commutes with H, so that py and pz are conserved. Thus they can be replaced by their eigenvalues directly. Hence cp y c 2mE0  1 2 q 2 B02  H= px + − − x 2m 2mc 2  qB0 qB02 

2

1 2 mc 2 E02 cp y E0 1 2 pξ pz − − = 2 2m 2 B0 2m B0 1 2 mc 2 E02 cp y E0 m , − pz − + ω 2ξ 2 + 2 2m 2 B02 B0

+

where pξ = px , ξ = x −

cp y qB0

−

mc 2 E0 qB02

are a new pair of conjugate variables. Let ω = | q | B0 / mc . By comparing the expression of H with that for a one-dimensional harmonic oscillator, we get the eigenvalues of H: En = (n + 1/2)ω + pz2 /2m − mc 2 E02 /2 B02 − cp y E0 /B0 , n = 0, 1, 2, .... The fact that only py and pz, but not y and z, appear in the expression for energy indicates an infinite degeneracy exists with respect to py and to pz. b. A state of zero momentum signifies one in which the eigenvalues of py and pz as well as the expectation value of px are all zero. As velocity is defined as

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v=

q  1 1 p mec =  p − A  , m m c 

its expectation value is v =

q q qB 1 p− A = − A = − 0 x yˆ . m c mc mc

Then as x = ξ +

cp y qB0

+

mc 2 E0 mc 2 E0 = , qB02 qB02

since ξ = 0 for a harmonic oscillator and p y = 0 , we have v =−

cE0 yˆ . B0

4007 Determine the energy levels, their degeneracy and the corresponding eigenfunctions of an electron contained in a cube of essentially infinite volume L3. The electron is in an electromagnetic field characterized by the vector potential A = H 0 x eˆ y (| eˆ y | = 1) .

(Chicago) Sol:

As A = H 0 x eˆ y, we have the Schrödinger equation 1 ˆ2 ˆ2 ˆ Hˆ ψ = [ px + pz + ( p y − H 0 xe / c )2 ]ψ = Eψ , 2m where e is the electron charge (e < 0). As [Hˆ , pˆ y ] = [Hˆ , pˆ z ] = 0,[ pˆ y , pˆ z ] = 0 , we can choose Hˆ , pˆ y , pˆ z as a complete set of mechanical variables, the corresponding eigenfunction being i ( p y + p z )/

ψ = e y z ψ0 ( x ), where py, pz are arbitrary real numbers. Substitution of ψ in the Schrödinger equation gives 1 ˆ2 [ px + (eH 0 /c )2 ( x − cp y /eH 0 )2 ]ψ0 = E0ψ0 , 2m

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2 where E0 = E − pz /2m, or

m 2 2 d ψ0 /dx 2 + ( H 0e /cm)2 ( x − x 0 )2 ψ0 = E0ψ0 , 2m 2 where x 0 = cp y /eH 0 . −

The last equation is the energy eigenequation of a one-dimensional oscillator of natural frequency ω0 = − H 0e/mc and equilibrium position x = x 0 , the energy eigenvalues being E0 = (n + 1/2)ω0 , n = 0, 1, 2, …, or E = pz2 /2m − (n + 1/2)H 0e / mc , n = 0, 1, 2, …. The corresponding eigenfunctions are  eH   eH  ψ0n ∼ exp  0 ( x − x 0 )2  Hn  − 0 ( x − x 0 ) ,   2c   c where Hn are Hermite polynomials. As no py terms occur in the expression for energy levels and py can be any arbitrary real number, the degeneracies of energy levels are infinite. The eigenfunctions for the original system are therefore   i( p y y + pz z ) eH 0 ψ ( x ) = Cn exp + ( x − x 0 )2  2 c     eH  × Hn  − 0 ( x − x 0 ) ,  c  where Cn is the normalization constant.

4008 Consider a hydrogen atom placed in a uniform constant magnetic field B. Find the energy spectrum of this atom. Sol:

The Hamiltonian of a particle in a magnetic field is given by H=

2 1   e  (1) A p +  c  2µ 

The Schrodinger wave equation is Hψ = Eψ

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2 1   e  A ψ = Eψ p + 2 µ  c 



1   e    e   p + A  pψ + Aψ  = Eψ  2 µ  c  c



e   e   1  −i∇ + A  −i∇ψ + Aψ  = Eψ  2 µ  c  c



 −2 2 ie  ie e2 A 2ψ = Eψ ∇ψ − A.∇ψ − (∇. A)ψ + 2µ µc µc 2mc 2



 −2 2 ie  e2 A 2ψ = Eψ (∵ ∇. A = 0) ∇ψ − A.∇ψ + 2 2µ µc 2mc

(2)



For a constant magnetic field B, we can choose the direction along z axis and  1  A= r ×B 2



To evaluate equation (2), we use  i     r × B.∇ψ = − B.r × ∇ψ = − B.Lψ    2 2 2   2 (r × B ) = r B − (r . B )



and write





−2 2 e   e2   ∇ψ− B.Lψ + [r 2 B 2 − (r .B)2 ]ψ = Eψ 2 2µ 2µc 8mc

(3)

For small values of r, the third term in equation (3) is very small and can be neglected. Therefore, the Schrodinger wave equation for a particle in a uniform constant magnetic field becomes e   −2 2 ∇ψ− B.Lψ = Eψ 2µ 2 µc e   e In the above equation, the additional term Bz .Lz is called the B.L = 2µc 2µc Zeeman term. The Hamiltonian of the Hydrogen atom in a magnetic field is the sum of the Hamiltonian of a free Hydrogen atom and Zeeman term, ( H Hydrogen + H Zeeman )ψ nlm = ( En + mµ B B)ψ nlm (4)

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where µB is the Bohr magneton and m is the magnetic Quantum number. The energy spectrum of the system is discrete and given by equation (4).

4009 a. Assuming that nonrelativistic quantum mechanics is invariant under time reversal, derive the time reversed form of the Schrödinger wave function. b. What is the quantum mechanical Hamiltonian for a free electron with magnetic moment µ in the external constant magnetic field Hz in the z-direction, in the reference frame of the electron? c. Suppose that an extra constant magnetic field Hy is imposed in the y-direction. Determine the form of the quantum mechanical operator for the time rate of change of µ in this case. (Buffalo) Sol: a. Consider the Schrödinger equation ∂ i ψ (t ) = Hˆ ψ (t ). ∂t Making the time reversal transformation t → −t , we obtain ∂ −i ψ (−t ) = Hˆ (−t )ψ (−t ), ∂t or i

∂ ψ *(−t ) = Hˆ *(−t )ψ *(−t ). ∂t

If Hˆ * (−t ) = Hˆ (t ), then the Schrödinger equation is covariant under time reversal and the time reversed form of the wave function is ψ *(−t ) . b. Let –e be the charge of the electron. Then µ = − e σ and in the reference 2mc frame of the electron, e Hˆ = −µ ⋅ H = −µ z H z = σ z H z. 2mc c. The magnetic field is now H y yˆ + H z zˆ , and so e (σ z H z + σ y H y ), Hˆ = 2mc

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so 2

dµ 1 1  e  = [µ , Hˆ ] =   [−σ x xˆ − σ y yˆ − σ z zˆ , dt ih i  2mc  2

2  e  σ z Hz +σ y H y ] =   [(σ y H z − σ z H y )xˆ − σ x H z yˆ   2mc  2

2  eh  +iσ x H y zˆ] =   σ ×H   2mc  e = H× µ, mc where use has been made of the relations σ xσ y = iσ z , σ yσ z = iσ x , σ zσ x = iσ y .

4010 An electron is subject to a static uniform magnetic field B = B 0 z and occupies the spin eigenstate |↑>. At a given moment t = 0, an additional timedependent, spatially uniform magnetic field B1(t ) = B1(cos ωt x + sinωt y ) is turned on. Calculate the probability of finding the electron with its spin along the negative z-axis at time t > 0. Ignore spatial degrees of freedom. Sol: The Schrodinger equation for the system is i

d e ψ (t ) = − ( B0 Sz + B1 cos ω tSx + B1 sin ω tS y ) ψ (t ) dt me

Setting we obtain

ψ (t ) = (a(t ), b(t ))

T

da e =i ( B0a + B1e − iω t b) dt 2me db e =i (− B0b + B1e − iω t a) dt 2me

Introducing

ω0 =

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we obtain the above equations in the form da = − iω 0a − iω 1e − iω t b dt db = iω 0b − iω 1e iω t a dt At this point, let us substitute the trial solutions a(t ) = exp(−iω t /2 + iΩtA), b(t ) = exp(iω t /2 + iΩtB) We immediately obtain Ω=±

1 1 (ω − 2ω 0 )2 + 4ω 12 = ± γ 2 2

and B=−

±γ − (ω − 2ω 0 ) A 2ω 1

Finally, we obtain a(t ) = e − iω t /2 ( A+ e iγ t /2 + A− e − iγ t /2 ) b(t ) = e iω t /2 (−

γ − (ω − 2ω 0 ) γ + (ω − 2ω 0 ) A+ e iγ t /2 + A− e − iγ t /2 ) 2ω 1 2ω 1

Applying the initial condition a(0) = 1, b(0) = 0, we arrive at A± =

γ ± (ω − 2ω 0 ) 2γ

and

γ t ω − 2ω 0 γt ω γt  sin ), −2ie − iω t /2 1 sin  ψ (t ) =  e − iω t /2 (cos + i  γ 2 2 γ 2

T

The probability of finding the spin of the electron pointing along −z is P↓ =

4ω 12 γt sin 2 ω − 2ω 0 + 4ω 12 2

Note that for ω = 2ω 0, this probability exhibits the resonance phenomenon. For this choice, the probability of spin flip is P↓ω = 2ω 0 = sin 2 ω 1t and becomes unity at t = (2n + 1)π /2ω 1 .

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4011 In a recent classic table-top experiment, a monochromatic neutron beam o

( λ = 1.445 A) was split by Bragg reflection at point A of an interferometer into two beams which were recombined (after another reflection) at point D (see Fig. 4.1). One beam passes through a region of transverse magnetic field of strength B for a distance l. Assume that the two paths from A to D are identical except for the region of the field.

Fig. 4.1 Find the explicit expressions for the dependence of the intensity at point D on B, l and the neutron wavelength, with the neutron polarized either parallel or anti-parallel to the magnetic field. (Chicago) Sol:

This is a problem on spinor interference. Consider a neutron in the beam. There is a magnetic field B in the region where the Schrödinger equation for the (uncharged) neutron is  2 2  ∇ − µ σ ⋅ B ψ = Eψ . −  2m  Supposing B to be constant and uniform, we have ψ (t1 ) = exp[−i Hˆ (t1 − t0 )/]ψ (t0 ), where t0, t1 are respectively the instants when the neutron enters and leaves the magnetic field. Write ψ (t ) = ψ (r , t )ψ (s , t ) , where ψ (r , t ) and ψ (s , t ) are respectively the space and spin parts of ψ . Then

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 i  2 2   ψ (r , t1 ) = exp − − ∇ (t1 − t0 ) ψ (r , t0 ),    2m   which is the same as the wave function of a free particle, and i  ψ (s , t1 ) = exp  µ σ ⋅ B(t1 − t0 ) ψ (s , t0 ).   The interference arises from the action of B on the spin wave function. As ψ (r , t ) is the wave function of a free particle, we have t1 − t0 = l/v = ml/k and

ψ (s , t1 ) = exp[i 2πµmlλ σ ⋅ B/h 2 ]ψ (s , t0 ),

where k = 2π = mυ is the wave number of the neutron. The intensity of the inter λ ference of the two beams at D is then proportional to | ψD(1) (r , t ) ψD(1) (s , t ) + ψD(2) (r , t )ψD(2) (s , t )|2

∝| ψD(1) (s , t ) + ψD(2) (s , t )|2 = | ψ (2) (s , t0 ) + ψ (2) (s , t1 )|2 .

As 2πµmlλ B B  2πµmlλ  exp  i σ ⋅ B = cos + iσ ⋅ 2 2   h h B × sin

2πµmlλ B , h2

and σ ⋅ B = ±σ B depending on whether σ is parallel or anti-parallel to B, we have 2   2πµmlλ ψ (2) (s , t0 ) + ψ (2) (s , t1 ) = 1 + exp i σ ⋅ B 2   h

= 1 + cos

2

ψ (s , t 0 )

2πµmlλ B 2πµmlλ B ± iσ sin 2 h h2

2

2

2

 2πµmlλ B  = 1 + cos    h2 2πµmlλ B πµmlλ B + sin 2 = 4cos 2 . 2 h h2 Therefore, the interference intensity at D ∝ cos 2 (πµmlλ B/h 2 ) , where µ is the intrinsic magnetic moment of the neutron (µ < 0).

4012 A neutron interferometer beam splitter plus mirrors as shown in Fig. 4.2 has been built out of a single crystal.

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Fig. 4.2 a. By varying the thickness of a thin plastic sheet placed in the beam in one arm of the interferometer one can vary the relative phase and hence shift the fringes. Give a brief qualitative explanation of the origin of the phase shift. b. By inserting in one arm a magnetic field which is normal to the beam, time independent and very nearly uniform so that the force on the neutrons can be neglected, and by choosing the field so each neutron spin vector precesses through just one rotation, one finds the relative phase of the two beams is shifted by π radians, or one-half cycle. Explain, with appropriate equations, why this is so. (Princeton) Sol: a. When a neutron passes through the thin plastic sheet, it is under the action of an additional potential, and so its momentum changes together with its de Broglie wavelength. The phase change of the neutron when it passes through the plastic sheet is different from that when it passes through a vacuum of the same thickness. If the thickness of the plastic sheet is varied, the relative phase of the two beams (originating from the same beam) also changes, causing a shift in the fringes. b. The neutron possesses an anomalous magnetic moment µ n= −µnσ and its Schrödinger equation is ( p2 /2mn + µnσ ⋅ B)ψ = Eψ .

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We may neglect the reflection that occurs when a neutron wave is incident on the “surface” of the sheet-like magnetic field as the action of the field on the neutron is rather weak. Under such an approximation we may show (by solving the above two-spin-component Schrödinger equation for a one-­dimensional square well): The wave function ψinc for a neutron incident normally on the sheet-like magnetic field is related to the transmitted wave function ψout out of the field by a unitary transformation

ψout = exp(−iσ ⋅ ρ/2)ψinc , where ρ = ω Lτ e B, with ω L = 2µn B/ being the Larmor frequency, τ = Lmn /k the time taken for the neutron to pass through the magnetic field of thickness L, e B the unit vector in the direction of B, k the wave number of the incident neutron. If a neutron is polarized in the (θ , ϕ ) direction before entering the field, i.e., its polarized vector is

ψinc | σ | ψinc = {sinθ cos ϕ , sinθ sinϕ , cosθ }, then we can take   e −iϕ /2 ψinc =   iϕ /2  e 

θ 2 θ sin 2

cos

   e i k ⋅x ,   

where θ is the angle the polarized vector makes with the direction of the magnetic field. Taking the latter as the z direction, we have ρ ⋅ σ = ρσ z . Then as  ρ  ρ ρ ρ 1 0  exp −i σ z  = cos − iσ z sin = cos    2  2 2 2 0 1 0  e −iρ /2 0  ρ 1 , − i sin  = 2 0 − 1  0 e iρ /2  we have  −i (ϕ +ρ )/2 θ  cos  e e −iρ /2 0  2  e i k ⋅x .   ψout =  ψ = inc iρ /2  θ e 0  ( )/2 i ϕ + ρ   sin  e  2 By adjusting B (or L) so that ρ = 2π , we make the polarized vector of a neutron precess through one rotation as it traverses the region of the magnetic field. Then

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  e −iϕ /2 ψout =   iϕ /2  e 

θ 2 θ sin 2

cos

   e i k⋅x+π ,   

i.e., the phase of the transmitted wave increases by π . Hence, compared with the wave traversing the other arm (without magnetic field), the relative phase of the beam changes by a half-cycle.

4013 a. A hydrogen atom is in its 2P state, in a state of Lx = + . At time t = 0 a strong magnetic field of strength | B | pointing in the z direction is switched on. Assuming that the effects of electron spin can be neglected, calculate the time dependence of the expectation value of Lx. b. How strong must the magnetic field in part (a) be so that the effects of electron spin can actually be neglected? The answer should be expressed in standard macroscopic units. c. Suppose that, instead, the magnetic field is very weak. Suppose, further, that at t = 0 the atom has Lx = + and s x = 21 h , and the magnetic field is still oriented in the z direction. Sketch how you would calculate the time dependence of the expectation value of Lx in this case. You need not do the full calculation, but explain clearly what the main steps would be. Note: All effects of nuclear spin are to be ignored in this problem. (Princeton) Sol: a. The initial wave function of the atom is

ψ (r , t = 0) = R21 (r )Θ(θ , ϕ ), where 1 Θ(θ , ϕ ) = (Y11 + Y1−1 + 2Y10 ), 2

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is the eigenstate of Lx =  . At t = 0 a strong magnetic field Be z is switched on. Then for t ≥ 0 the Hamiltonian of the system is p 2 eBlz e 2 B 2 ( x 2 + y 2 ) e 2 Hˆ = + + − . 8me c 2 2me 2me c r For a not too strong magnetic field B ∼ 105 Gs, we can neglect the B 2 term and take as the Hamiltonian p 2 eBlz e 2 Hˆ = + − . 2me 2me c r The Schrödinger equation i ∂ψ/∂t = Hˆ ψ then gives the eigenstate solutions

ψn (r , t ) = Rnl (r )Ylm (θ , ϕ )e −iEnlmt/ , where Enlm = Enl +

eB m. 2me c

Thus the general solution is  E  ψ (r , t ) = ∑anψnlm (r )exp −i nlm t  .    n ,l ,m For t = 0 , we then have

∑a ψ n

n, l , m

or

nlm

1  1 1 (r ) = R21 (r ) Y11 + Y1−1 + Y10  , 2 2 2 

1 a2ψ211 (r) = R21 (r )Y11, etc. 2

Hence 1  E  ψ (r , t ) = R21 (r ) Y11exp −i 211 t   2    E  1 + Y1−1exp −i 21−1 t   2    E  1 Y10exp −i 210 t  . +    2

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ψ (r , t )| Lx | ψ (r , t ) . As

The expectation value of Lx is given by Lx = ( L+ + L− )/2, L+Ylm =  (l + m + 1)(l − m)Yl , m+1 , L− Ylm =  (l − m + 1)(l + m)Yl , m−1 , we have

  Y10 , LxY1−1 = Y10 , 2 2  LxY10 = (Y11 + Y1−1 ), 2 LxY11 =

and hence Lx (t ) = ψ (r , t )| Lx | ψ (r , t ) = cos

eBt . 2me c

b. The effects of electron spin can be neglected if the additional energy due to the strong magnetic field is much greater than the coupling energy for spin-orbit interaction, i.e., e B −3  ∆Espin-orbit ≈ 10 eV , 2me c or B ≥ 106 Gs. Thus when the magnetic field B is greater than 106 Gs, the effects of electron spin can be neglected. c. If the magnetic field is very weak, the effects of electron spin must be taken into consideration. To calculate the time dependence of the expectation value of Lx , follow the steps outlined below. i. The Hamiltonian is now p2 e 2 eB e2 eB ˆ sˆz , Hˆ = − + 2 2 3 (sˆ ⋅ Lˆ ) + jz + 2me r 2me c r 2me c 2me c

which is the Hamiltonian for anomalous Zeeman effect and we can use the coupling representation. When calculating the additional energy due to the sˆx term, we can regard sˆx as approximately diagonal in this representation. ii. Write down the time-dependent wave function which satisfies the initial condition Lx = + and sx = 12  . At time t = 0 the wave function is

ψ0 (r , sx ) = R21 (r )Θ(θ , ϕ )φ s ,

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where Θ and φ s are the eigenfunctions of Lx =  and sx = /2 in the representations (l 2, lz ), (s 2, sz ) respectively. Explicitly, 1 1 ψ 0 (r , sx ) = R21 (r ) [Y11 + Y1−1 + 2Y10 ] (α + β ) 2 2 R12 (r ) (Y11α + Y11β + Y1−1α + Y1−1β = 2 2 + 2Y10α + 2Y10 β ). As 3 2 3 1 φ j = , m j = = Y11α , φ 3 1 = Y11β + Y10α , 22 2 2 3 3

φ 3−1 = 2 2

2 1 Y1−1α + Y10 β , φ 3 − 3 = Y1−1β , 2 2 3 3

ψ0 can be written in the coupling representation as ψ0 ( r , s x ) =

1 R21 (r ) φ 3 3 + 3φ 3 1 22 22 2 2  + 3φ 3 1 + φ 3 3  , − − 2 2 2 2

(

where φ jm j is the eigenfunction of ( j 2, jz ) for the energy level Enljm j. Therefore, the time-dependent wave function for the system is

ψ(r , s , t ) =

  E21 3 3  R21 (r ) ⋅ φ 3 3 exp −i 2 2 t  2 2   2 2   1

 E21 3 1  + .3φ 3 1 exp −i 2 2 t  2 2     E21 3 − 1  + .3φ 3 − 1 exp −i 2 2 t  2 2     E21 3 − 3   + φ 3 − 3 exp −i 2 2 t   . 2 2     iii. Calculate the expectation value of Lx in the usual manner:

ψ (r , s , t )| Lx | ψ (r , s , t ) .

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4014 Consider the one-dimensional motion of an uncharged particle of spin 1/2 and magnetic moment µ = −2µ 0 s/h. The particle is confined in an infinite square well extending from x = −L to x = L. In region I ( x < 0) there is a uniform magnetic field in the z direction B = B0 e z ; in region II ( x > 0) there is a uniform field of the same magnitude but pointing in the x direction B = B0 e x . Here ex and ez are unit vectors in the x and z directions. a. Use perturbation theory to find the ground state energy and ground state wave function (both space and spin parts) in the weak field limit B0  ( /L)2 /2mµ 0. b. Now consider fields with B0 of arbitrary strength. Find the general form of the energy eigenfunction ψI (both space and spin parts) in region I which satisfies the left-hand boundary condition. Find also the form ψII that the eigenfunction has in region II which satisfies the right-hand boundary condition (Fig. 4.3). c. Obtain an explicit determinantal equation whose solutions would give the energy eigenvalues E. (MIT)

Fig. 4.3 Sol: a. In the absence of magnetic field, H = H 0 and the energy eigenfunctions (space part) and eigenvalues are respectively

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ψn = 1/L sin En =

263

nπ ( x + L ) , 2L

π 2  2n 2 , n = 1, 2, 3, …. 8mL2

As for the spin part, we know that each energy level has a degeneracy of 2. When a magnetic field is present, H = H 0 + H ′ , where  µ B σ , −L ≤ x ≤ 0,  0 0 z 2µ0s ⋅ B H ′ = −µ ⋅ B = = µ0 σ ⋅ B =  µ0 B0σ x , 0 ≤ x ≤ L ,   elsewhere.  0 1  0  If the field is weak, let u1 = ψ1 ( x )  , u2 = ψ1 ( x )  be the base vectors. 0  1  Then 1 H11′ = u1 | H ′ | u1 = µ0 B 0*(1 0) 0 × ∫ − L ψ1* ( x )ψ1 ( x )dx = 0

′ = H12′ = u1 | H ′ | u2 H 21 0 = µ0 B0 (1 0)  1

0  1    − 1 0 

µ0 B0 , 2

1  0    0  1 

∫

L 0

ψ1* ( x )ψ1 ( x ) dx =

µ0 B0 , 2

′ = µ2 | H ′ | µ2 H 22 1 = −µ0 B0 (0 1) 0

0  0 0 µB    ∫ − L ψ1* ( x )ψ1 ( x ) dx = − 0 0 , − 1 1  2

and from det( H ′ − E (1) I ) = 0 we get  µ0 B0  µ B  µ 2 B2 − E (1)  − 0 0 − E (1)  − 0 0 = 0,   2  2  4 or E (1) = ±

1 µ0 B0. 2

The ground state energy level is therefore E0 =

π 2 2 1 µ0 B0. − 8mL2 2

From a  0  ( H ′ − E (1) I )  =   , b   0 

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we get the ground state wave function 1 − ϕ 0 = au1 + bu2 = ψ1 ( x )  1

2  . (unnormalized) 

b. The space part of the wave function in region I is  Asink1 ( x + L ) + B cosk1 ( x + L ), −L ≤ x ≤ 0, ψIk1 =  x < −L.  0, The continuity condition of the wave function gives B = 0 . In region I, the 0  1  spin is aligned to the z direction, the eigenvectors being   for z ↓ and   1  0  for z ↑. Hence  0  ψI k1 z↓ = sink1 ( x + L )  ,  1   1   ψI k1 z↑ = sink1 ( x + L )0  , 

E=

 2 k12 − µ0 B0 ; 2m

E=

 2 k12 + µ0 B0 . 2m

In a similar way we obtain the eigenfunctions for region II (0 ≤ x ≤ L )  1 ψII k2 x↓ = sink2 ( x − L )  ,  −1  1  ψII k2 x↑ = sink2 ( x − L )1 , 

E=

 2 k22 − µ0 B0 ; 2m

E=

 2 k22 + µ0 B0 . 2m

c. Considering the whole space the energy eigenfunction is  Aψ + Bψ , −L ≤ x ≤ 0, I k1 z↓ I k1′ z↑  ψE =  CψII k2 x↓ + DψII k2′ x↑ , 0 ≤ x ≤ L ,  elsewhere.  0, Thus HψE =           

  2 k12    2 k′ 2  − µ0 B0  AψIk1 z↓ +  1 + µ0 B0  BψIk1′z↑ ,   2m   2m 

−L ≤ x ≤ 0,

  2 k22    2 k′ 2  − µ0 B0 CψIIk2 x↓ +  2 + µ0 B0  DψIIk2′ x↑ , 0 ≤ x ≤ L ,   2m   2m  0.

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elsewhere.

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265

From HψE = EψE for each region we have  2 k12  2 k1′2  2 k22 − µ0 B0 = + µ0 B0 = − µ0 B0 2m 2m 2m  2 k2′2 = + µ0 B0 , 2m

E=

and so k1 = k2 = k , k1′ = k2′ = k′. Then the continuity of the wave function at x = 0 gives B sin k′L = −C sin kL − D sin k′L , A sin kL = C sin kL − D sin k′L , and the continuity of the derivative of the wave function at x = 0 gives Bk′ cos k′L = Ck cos kL + Dk′ cos k′L , Ak cos kL = −Ck cos kL + Dk′ cos k′L. To solve for A, B, C, D, for nonzero solutions we require 0 sin k′L sin kL sin k′L − sin kL sin kL 0 sin k′L = 0, 0 k′ cos k′L −k cos kL −k′ cos k′L k cos kL 0 k cos kL −k′ cos k′L i.e., k sin kL cos k′L − k′ sin k′L cos kL = 0 . This and E=

 2k 2  2 k′ 2 − µ0 B0 = + µ0 B0 2m 2m

determine the eigenvalues E.

4015 Consider an infinitely long solenoid which carries a current I so that there is a constant magnetic field inside the solenoid. Suppose in the region outside the solenoid the motion of a particle with charge e and mass m is described by the Schrödinger equation. Assume that for I = 0, the solution of the equation is given by

ψ0 ( x , t ) = e iE0t ψ0 ( x ).

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(  = 1)

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a. Write down and solve the Schrödinger equation in the region outside the solenoid for the case I ≠ 0. b. Consider a two-slit diffraction experiment for the particles described above (see Fig. 4.4). Assume that the distance d between the two splits is large compared to the diameter of the solenoid. Compute the shift ∆S of the diffraction pattern on the screen due to the presence of the solenoid with I ≠ 0 . Assume l  ∆S.

Fig. 4.4 Hint: Let

ψ ( x , t ) = ψ0 ( x , t )ψ A ( x ), where   e  ∇ − i A(x) ψ A (x) = 0. c  

(  = 1).

(Chicago) Sol: a. In the presence of a vector potential A , p → p − eA /c . In the absence of electromagnetic field the Schrödinger equation is i

 1  ∂ ψ0 ( x , t ) =  p 2 + V ( x )ψ0 ( x , t ),  2m  ∂t

where, as below, we shall use units such that  = 1 . The Schrödinger equation in the presence of an electromagnetic field can thus be obtained (using the minimum electromagnetic coupling theory) as 2  1   ∂ e  i ψ ( x , t ) =  −i∇ − A  + V ( x )ψ ( x , t ), ∂t c   2m  

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267

where A is given by ∇ × A = B . Let  xe  ψ ( x , t ) = ψ1 ( x , t )exp i ∫ A ⋅ d x  .   c Then the above becomes i

 1  ∂ ψ1 ( x , t ) =  p 2 + V ( x )ψ1 ( x , t ),  2m  ∂t

which is the Schrödinger equation for zero magnetic field. Hence

ψ1 ( x , t ) = ψ0 ( x , t ) = e iE0t ψ0 ( x ), and so  x e  ψ ( x , t ) = e iE0t ψ0 ( x )exp i ∫ A ⋅ dx  .  c  b. This is a problem on the Aharonov-Bohm effect. When I = 0 , for any point on the screen the probability amplitude f is f = f+ + f− , where f + and f− represent the contributions of the upper and lower slits respectively. When the current is on, i.e., I ≠ 0 , we have the probability amplitude f ′ = f+′+ f−′ with   xe f+′ = exp i ∫ c+ A ⋅ dx  f+ ,   c  xe  f−′ = exp i ∫ c− A ⋅ dx  f− ,   c where c+ and c− denote integral paths above and below the solenoid respectively. Thus

(

)

x  xe  f ′ = f+′+ f−′ = exp i ∫ c+ A ⋅ d x  f+ + exp i ∫ c− A ⋅ d x f−   c  e  ∼ exp i ∫ A ⋅ d x  f− + f+ ,  c  x

on dividing the two contributions by a common phase factor exp(i ∫ ec A ⋅ dx ), c+

which does not affect the interference pattern. The closed line integral, to be taken counterclockwise along an arbitrary closed path around the solenoid, gives e e e eφ ∫ c A ⋅ d x = c ∫ ∇ × A ⋅ ds = c ∫ B ⋅ ds = c , where φ is the magnetic flux through the solenoid.

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Thus the introduction of the solenoid gives a phase factor eφ/c to the probability amplitude at points on the screen contributed by the lower slit. Using a method analogous to the treatment of Young’s interference in optics, we see that the interference pattern is shifted by ∆S . Assuming l  d and l  ∆S, we have d e ∆S ⋅ ⋅ k = φ , l c k being the wave number of the particles, and so elφ elφ . ∆S = = cd k cd 2mE0 Note the treatment is only valid nonrelativistically.

4016 a. What are the energies and energy eigenfunctions for a nonrelativistic particle of mass m moving on a ring of radius R as shown in the Fig. 4.5.?

Fig. 4.5 b. What are the energies and energy eigenfunctions if the ring is doubled (each loop still has radius R) as shown in Fig. 4.6?

Fig. 4.6 c. If the particle has charge q, what are the energies and energy eigen-­ functions if a very long solenoid containing a magnetic flux passes the rings in (a) as shown in the Fig. 4.7.? and in (b)? Assume the system does not radiate electromagnetically. (Columbia)

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269

Fig. 4.7 Sol: a. As 2 P 2 d 2 ˆ H= = − , 2mR 2 2mR 2 dθ 2 we have the Schrödinger equation −

2 d 2 Ψ (θ ) = E Ψ (θ ), 2 I dθ 2

where I = mR 2 , or d 2 Ψ (θ ) 2 + n Ψ (θ ) = 0, dθ 2 with n2 =

2 IE 2

Thus the solutions are Ψ n (θ ) = Ae inθ. For single-valuedness we require Ψ (θ + 2π ) = Ψ (θ ), i.e., n = 0, ± 1, ± 2, …. Normalization requires A * A = 1, or A =

1 . 2π

Hence the eigenfunctions are 1 inθ Ψ n (θ ) = e , n = 0, ± 1, ± 2, … , 2π

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and the energy eigenvalues are En (θ ) =

n2 2 . 2I

b. The same Hamiltonian applies, and so we still have the same Schrödinger equation −

2 d 2 Ψ (θ ) = E Ψ (θ ) . 2 I dθ 2

However, the single valuedness of the solutions now requires Ψ (θ + 4 π ) = Ψ (θ ) . Hence the normalized eigenfunctions and the energy eigenvalues are now 1 i n2θ e , 4π

Ψ n (θ ) =

n = 0, ± 1, ± 2, … ,

and En (θ ) =

n2 2 . 8I

c. The Hamiltonian in the presence of a magnetic field is 2

 1 iq 2  Hˆ = (P − qA( x ))2 = −  ∇ − A ( x ) .  2m 2m   In the region where the particle moves, B = ∇ × A = 0 and we can choose A = ∇ϕ . From the symmetry, we have A = Aθ eθ , Aθ = constant . Then

∫ A ⋅ d l = ∫

2π 0

Thus A=

Aθ R dθ = 2π RAθ = φ , say .

φ eθ = ∇(φθ / 2π ), 2π R

and we can take ϕ = φθ /2π , neglecting possibly a constant phase factor in the wave functions. The Schrödinger equation is 2

 2  iqφ Hˆ Ψ = − ∇θ  Ψ ∇ −  2m  2π  2  qφ  2   iqφ    θ  ∇ exp − θ  Ψ  = EΨ . exp i =−  2π h    2π    2m On writing

 qφ  ψ ′(θ ) = exp −i θ  Ψ (θ ),  2π h 

it becomes −

2 d 2 ψ ′(θ ) = Eψ ′(θ ), 2 I dθ 2

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with solutions

271

 2 IE  ψ ′(θ ) = exp ±i θ . 2  

Hence

ψ (θ ) = c exp (iαθ )ψ ′(θ ) = c exp [i(α ± β )θ ],

where

α=

qφ , 2π 

β=

2I E , 2

c = a constant.

For the ring of (a), the single-valuedness condition Ψ (θ + 2π ) = Ψ (θ ), requires

α ± β = n, i.e.,

n = 0, ± 1, ± 2, …

qφ ± 2 I E/ 2 = n. 2π 

Hence En =

2

2  qφ  n − , 2 I  2π  

and

ψn (θ ) =

1 inθ e . 2π

where n = 0, ± 1, ± 2, …. Similarly for the ring of (b), we have En =

2

2

 2  n qφ   2  qφ   −  = n − , 2 I  2 2π   8 I  π  

and

ψn (θ ) =

 n  1 exp i θ  ,  2  4π

where n = 0, ± 1, ± 2, ….

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Part V

Perturbation Theory

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5 5001 a. Show that in the usual stationary state perturbation theory, if the Hamiltonian can be written H = H0 + H ′ with H0φ0 = E0φ0 , then the correction ∆E0 is ∆E0 ≈ φ0 | H ′ | φ0 . b. For a spherical nucleus, the nucleons may be assumed to be in a  0, r < R , spherical potential well of radius R given by Vsp =   ∞ , r > R . For a slightly deformed nucleus, it may be correspondingly assumed that the nucleons are in an elliptical well, again with infinite wall height, that is:  x2 + y2 z2  0, inside the ellipsoid + 2 = 1, Vel =  b2 a  ∞ , otherwise,  where a ≅ R(1 + 2β / 3), b ≅ R(1 − β / 3), and β  1. Calculate the approximate change in the ground state energy E0 due to the ellipticity of the non-spherical nucleus by finding an appropriate H ′ and using the result obtained in (a). HINT: Try to find a transformation of variables that will make the well look spherical. (Buffalo) Sol: a. Assuming that H ′ is very small compared with H 0 so that the wave function Ψ can be expanded as Ψ = φ0 + λ1 φ1 +  + λn φn +  ,

275

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where λ1 λn  are small parameters. The Schrödinger equation is then

(

)

( H ′ + H 0 ) φ0 + λ1 φ1 +  + λn φn + 

(

)

= ( E0 + ∆E0 ) φ0 + λ1 φ1 +  + λn φn +  . Considering only the first order correction, we have

(

)

H ′ φ0 + H 0 λ1 φ1 +  + λn φn + 

(

)

= ∆E0 φ0 + E0 λ1 φ1 +  + λn φn +  . Multiplying both sides of the equation by φ0 and noting the orthonormality of the eigenfunctions we get ∆E0 = φ0 H ′ φ0 . b. For the stationary state, 2  = −  ∇2 + V , H 2m

where  x2 + y2 z2  0 inside the ellipsoid + 2 = 1, V= b2 a  ∞ otherwise.  Replacing the variables x, y, z by

b R

ξ , Rb η , Ra ζ respectively, we can write the

equation of the ellipsoid as ξ 2 + η 2 + ζ 2 = R 2 and H≡−

∂2 ∂2   2  ∂2 + 2 + 2 2  2m  ∂ x ∂ y ∂z 

=−

 2  R 2 ∂2 R 2 ∂2 R 2 ∂2  + + 2m  b 2 ∂ξ 2 b 2 ∂η 2 a 2 ∂ζ 2 

≈−

∂2 ∂2   2  ∂2 + 2 + 2 2  2m  ∂ξ ∂η ∂ζ 

− =−

∂2 ∂2   2 β  ∂2 + 2 −2 2  2  ∂ζ  3m  ∂ξ ∂η ∂2 ∂2   2 2  2 β  ∂2 ∇′ − + 2 −2 2 . 2  ∂ζ  2m 3m  ∂ξ ∂η

The second term in H can be considered a perturbation as β  1. Thus ∆E0 = φ0 H ′ φ0 = φ0 −

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∂2 ∂2   2 β  ∂2 + 2 − 2 2  φ0 , 2  ∂η ∂ζ  3m  ∂ξ

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277

where φ0 is the ground state wave function for the spherical potential well,

φ0 =

2 sin πRr , γ 2 = ξ 2 + η2 + ζ 2. R r

As φ0 is spherically symmetric,

φ0

∂2 ∂2 ∂2 φ φ φ φ φ0 , = = 0 0 0 0 ∂ξ 2 ∂η 2 ∂ζ 2

and so ∆E0 = 0.

5002 Calculate the first-order energy correction to the ground state and first excited state of an infinite square well of width “a” whose portion has been sliced off as shown in the figure, where V0 x/a is the perturbed potential.

Fig. 5.1 Sol: The wave function of a particle inside an infinite potential well of width a is given by 2 ψ= sin(nπ x / a)(1) a The correction to the energy in first order is given by ∫ dx ψ × Hψ a

2 / a ∫ dx sin 2 (nπ x / a)V0 x / a 0

a

For ground state n = 1, so the integral evaluates to 2V0 / a 2 ∫ x dx sin 2 (π x / a ) 0

Using the identity sin2 x = [1 − cos(2πx)]/2 So the integral becomes a

V0 / a 2 ∫ x dx[1 − cos ( 2π x / a ] = V0 / a 2 0

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{∫ x dx − ∫ cos(2π x / a)dx} a

a

0

0

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where the second term in the integral vanishes and the correction to the energy is given by E11 = V0/2 In a similar manner, the correction to the first-order energy is given by E21 = V0/2

5003 A particle of mass m moves one-dimensionally in the oscillator potential V ( x ) = 21 mω 2 x 2 . In the nonrelativistic limit, where the kinetic energy T and momentum p are related by T = p 2 / 2m, the ground state energy is well known to be 21 ω . Allow for relativistic corrections in the relation between T and p and compute the ground state level shift ∆E to order

1 c2

(c = speed of light). (Buffalo)

Sol:

In relativistic motion, the kinetic energy T is T ≡ E − mc 2 = m 2c 4 + p 2c 2 − mc 2 1

 p2  2 = mc  1 + 2 2  − mc 2  mc  2

 p2 p4  ≈ mc 2  1 + − 4 4  − mc 2 2 2  2m c 8m c  = to order

1 c2

. The

− p4 8 m3 c 2

p2 p4 − 3 2 2m 8m c

term may be considered as a perturbation. Then the energy

shift of the ground state is ∆E = −

 − p 4  ∞ p4 * = φ ∫−∞ 0  8m3c 2  φ0 dx 8m3c 2 1/4

∞  mω   mω x 2  =∫   exp  − −∞  π    2  1

  4 ∂ 4   mω  4 × − 3 2 4    exp  8m c ∂ x   π   =−

 − mω x 2  dx  2 

15 (ω )2 . 32 mc 2

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279

5004 An electron moves in a Coulomb field centered at the origin of coordinates. With neglect of spin and relativistic corrections the first excited level ( n = 2) is well known to be 4-fold degenerate: l = 0, ml = 0; l = 1, ml = 1, 0, − 1. Consider what happens to this level in the presence of an additional non-central potential Vpert : Vpert = f ( r ) xy , where f ( r ) is some central function, wellbehaved but not otherwise specified (it falls off rapidly enough as r → ∞). This perturbation is to be treated to first order. To this order the originally degenerate n = 2 level splits into several levels of different energies, each characterized by an energy shift ∆E and by a degeneracy (perhaps singly degenerate, i.e., nondegenerate; perhaps multiply degenerate). a. How many distinct energy levels are there? b. What is the degeneracy of each? c. Given the energy shift, call it A ( A > 0), for one of the levels, what are the values of the shifts for all the others? (Princeton) Sol: With V = f (r )xy = f (r )r 2 sin 2 θ sinϕ cosϕ treated as perturbation, the unperturbed wave functions for energy level n = 2 are l = 0, ml = 0,

R20 (r )Y00 ,

l = 1,

ml = 1,

R21 (r )Y11 ,

l = 1,

ml = 0,

R21 (r )Y10 ,

l = 1,

ml = − 1, R21 (r )Y1,−1 .

As they all correspond to the same energy, i.e., degeneracy occurs, we have first to calculate H l′′m′lm = 〈l ′m′ | V | lm〉 = ∫ R2l ′ (r )R2l (r )r 2 f (r )Yt*′m′ sin 2 θ sinϕ cosϕYlm dV . The required spherical harmonics are 1

 1 2 Y00 =   ,  4π  1

 3 2 Y10 =   cosθ ,  4π 

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1

 3 2 Y11 =   sinθ e iϕ ,  8π  1

 3 2 Y1,−1 =   sinθ e − iϕ .  8π 

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Considering the factor involving ϕ in the matrix elements H l ′m′lm we note that all such elements have one of the following factors:

∫

2π

0

∫

sinϕ cosϕ dϕ = 0,

2π

0

e ± i 2ϕ sinϕ cosϕ dϕ = 0 ,

except H1,′ −1,1,1 and H1,1,1, ′ −1, which have nonzero values H1,′ −1,1,1 =

3 [R21 (r )]2 r 4 f (r )dr ∫ 8π ×

∫

2π

0

∫

π

0

sin5 θ dθ

sinϕ cos ϕ e −2iϕ dϕ = iA,

H1,1,1, ′ −1 = −iA, with A =

1 5

∫ [R(r )] r

2 4

f (r )dr .

We then calculate the secular equation     

0 0 0 0 0 0 0 −iA

0 0 0 0

0 iA 0 0

 ∆E 0  0 ∆E  − ∆EI = 0 0   0 −iA

0 0 0 iA = 0, ∆E 0 0 ∆E

whose solutions are ∆E = 0, ∆E = 0, ∆E = A, ∆E = − A. Thus with the perturbation there are three distinct energy levels with n = 2 . The energy shifts and degeneracies are as follow.  A, one-fold degeneracy,  ∆E =  − A, one-fold degeneracy,  0, two-fold degeneracy.  Thus there are three distinct energy levels with n = 2 .

5005 A particle moves in a one-dimensional box with a small potential dip (Fig. 5.2):

Fig. 5.2

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281

V = ∞ for x < 0 and x > l V = − b for 0 < x < (1/ 2)l , V = 0 for (1/ 2)l < x < l . Treat the potential dip as a perturbation to a “regular” rigid box ( V = ∞ for x < 0 and x > I , V = 0 for 0 < x < l ). Find the first order energy of the ground state. (Wisconsin) Sol:

For the regular rigid box, the energy and wave function of the ground state are respectively E (0) =

2 πx π 22 , ψ (0) ( x ) = sin . 2 2ml l l

The perturbation is H (1) = − b, 0 ≤ x ≤ 2l . Hence, the energy correction of first order perturbation is 1 2

E (1) = ∫ φ (0) ( x )(−b)ϕ (0) ( x )dx ∗

0

2 2 πx sin  (−b)dx  l  l b 12  2π x  b = − ∫  1 − cos  dx = − . l 0 l  2

=∫

1 2

0

Thus the energy of the ground state with first order perturbation correction is E = E (0) + E (1) =

 2π 2 b − . 2ml 2 2

5006 An infinitely deep one-dimensional square well has walls at x = 0 and x = L . Two small perturbing potentials of width a and height V are located at x = L / 4, x = (3 / 4)L , where a is small ( a  L /100, say) as shown in Fig. 5.3. Using perturbation methods, estimate the difference in the energy shifts between the n = 2 and n = 4 energy levels due to this perturbation. (Wisconsin)

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Fig. 5.3 Sol:

The energy levels and wave functions for a one-dimensional infinite potential well are respectively En(0) =

π 22 2 n , 2 µ L2

ψ n (x ) =

2 nπ sin x , L L

n = 1,2, … .

(1) ′ , according to first order perturbation is The shift of the energy level n, En = H nn given by

2  πn  V ⋅ sin 2  x dx  L  L 3 L /4+a/2 2  πn  x dx . +∫ V ⋅ sin 2  3 L /4−a/2  L  L

Hnn ′ =∫

L /4+a/2

L /4−a/2

As a  L/100, we can apply the mean value theorem to the integrals and obtain Hnn ′ = =

2Va  2  π n L   π n 3L   + sin 2  ⋅ sin  ⋅  4 4   L 4   L  3π n  2Va  2 π n + sin 2  sin . L 4 4 

Therefore, the change of energy difference between energy levels n = 2 and n = 4 is 2Va  2 π  2 3π − sin 2 π − sin 2 3π   sin + sin  L  2 2 4Va = . L

E2(1) − E4(1) =

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283

5007 A particle of mass m moves in a one-dimensional potential box    V(x) =    

∞

for | x |> 3| a |,

0 0 V0

for a < x < 3a , for − 3a < x < − a , for − a < x < a ,

as shown in Fig. 5.4. Consider the V0 part as a perturbation on a flat box ( V = 0 for −3a < x < 3a , V = ∞ −3a < x < 3a , V = ∞ for | x | > 3| a |) of length 6a. Use the first order perturbation method to calculate the energy of the ground state. (Wisconsin)

Fig. 5.4 Sol: The energies and wave functions of a particle in a flat box of length 6a are respectively E (0) =

π 2  2n 2 , n = 1, 2, …, 72ma 2

ψ (0) ( x ) =

1 nπ x cos , n = odd integer, 3a 6a

ψ (0) ( x ) =

1 nπ x sin , 3a 6a

n = even integer.

Particularly for the ground state, we have 1 πx cos 3a 6a 2 2 π  . E1(0) = 72ma 2

ψ 1(0) ( x ) =

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The energy correction of first order perturbation is given by E (1) = (ψ (0) ( x ), Vˆψ (0) ( x )), 1

1

where Vˆ = V0 for −a ≤ x ≤ a . Thus E (1) = ∫

a

−0

1 V0 3  πx cos 2  dx = V0  + .  6a  3a  3 2π 

Hence, the energy of ground state given by first order perturbation is E = E (0) + E (1) =

1 π 22 3 + V0  +  . 72ma 2 3 2 π 

5008 Consider a free particle confined to a one-dimensional infinite potential box of length L. A uniform electric field is applied along the x direction. Find the first-order correction to the energy to the first three states. Sol: The X component of electric field is given by Eȋ. The corrections to the first three states are given by E11, E21, and E31. We know that the force on a charged particle is given by F = qEȋ(1) The work done is stored in terms of potential energy, which acts as a perturbation. Work is defined as W = F.X which gives W = qEȋ . xȋ = qEx Now, the first-order correction is given as = where q and E are constants, so = qE The is given by ∫ ψ × X ψ dx with the system limits 0 → L and the wave function is given by ψ = 2 / L sin (nπ x / L ) where n takes values 1, 2, and 3. L

For the ground state n = 1, < X > = 2 / L ∫ dx sin 2 (π x / L )x 0

Using the identity sin 2 x = [1 − cos(2π x )]/ 2

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< X > = 1/ L

285

{∫ dx  x + ∫ dx cos(2π x / L)} L

L

0

0

The second integral is zero after integrating and applying the limits, only the first part survives and = L/2. So the energy correction is given by = qEL/2 In a similar manner, the energy corrections for the second and third states are given by qEL/2. The energies of the first three levels after corrections are given by E1 + qEL/2 for the ground state. 4E1 + qEL/2 for the first excited state. 9E1 + qEL/2 for the third excited state, where E1 = π 2ħ2/2mL2.

5009 A perfectly elastic ball is bouncing between two parallel walls. a. Using classical mechanics, calculate the change in energy per unit time of the ball as the walls are slowly and uniformly moved closer together. b. Show that this change in energy is the same as the quantum mechanical result if the ball’s quantum number does not change. c. If the ball is in the quantum state with n = 1, under what conditions of wall motion will it remain in that state? (Chicago)

Fig. 5.5 Sol: a. In classical mechanics, the energy of the ball is E=

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dE p dp . = dt m dt

so

At a certain instant t, the walls are separated by L and the ball moves to the right with speed υ 2′ . Because the collision is perfectly elastic, the speed of the ball relative to the right wall before and after bouncing from it remains the same: υ 2′ + υ1 = υ 2 + (−υ1 ), where υ 2 is the speed of the ball after bouncing. Thus,

υ 2′ − υ 2 = − 2υ1 , ∆p = m(υ 2′ − υ 2 ) = − 2mυ1 , dE / dt =

p dp p ∆p p υ pυ υ ≈ = − 2mυ1 ⋅ 2 = − 1 2 , m dt m ∆t m 2L L

2L is the time interval between two successive collisions. As the υ2 right wall moves very slowly, pυ p p dL dE /dt = − 2 υ1 = − ⋅ ⋅ L L m dt 2 p 2 dL 2 E dL =− . =− , L 2m dt L dt

where ∆t =

which is the rate of change of the energy of the ball according to classical mechanics. b. As the motion of the right wall is very slow, the problem can be treated as one of perturbation. If the wall motion can be neglected, we have En =

nπ 2  2 . 2mL2

If n does not change, dEn /dt =

2E 1 dL n 2π 2  2 (−2) 3 = − n dL /dt , 2m L dt L

same as the classical result. c. If the energy change during one collision is much smaller than E2 − E1, the ball can remain in the state n = 1 (analogous to adiabatic processes in thermodynamics). More precisely, as E=

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287

we have ∆E =

p 2E ∆p = − ⋅ 2mυ1 = − 2 2mEυ1. m m

The condition E2 − E1  ∆E then gives

π 22 2 2 π 22 (2 − 1 ) 2 2 m  |υ1 | , 2mL2 2mL2 or 3π  . 4mL 3π  . This means that the speed of the right wall should be much smaller than 4mL

υ1 

5010 Consider a free particle confined to a one-dimensional box of length a (from x = 0 to x = a), if it is subjected to a perturbation V ( x ) = V0 cos(π x / a ) in the range 0 < x < a/2. Find the first-order correction to the energy of the ground state. Sol:

2 sin(π x / a) and the first-order correction to the energy is given by a < ψ | Hˆ |ψ >

ψ gs ( x ) =

E11 = 2V0 / a ∫

a/2

0

cos(π x / a)sin 2 (π x / a) dx

The limits are taken only up to the perturbed part. This integral can be done by substitution. Let sin(π x / a) = t ,cos(π x / a) = (a / π )dt where the limits change from At x = 0 → t = 0 At x = a / 2 → t = 1 1

So the integral changes to (a / π ) ∫ dt t 2 , which gives a/3π . 0

Now, E11 = 2V0/a × (a/3π ) = 2V0/3π .

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5011 A charged particle is bound in a harmonic oscillator potential V = 21 kx 2 . The system is placed in an external electric field E that is constant in space and time. Calculate the shift of the energy of the ground state to order E 2. (Columbia) Sol: Take the direction of electric field as the x-direction. The Hamiltonian of the system is 2 d 2 1 2 + kx − qEx = Hˆ 0 + Hˆ ′ , Hˆ = − 2m dx 2 2 where Hˆ ′ = − qEx is to be treated as a perturbation. The wave function of the ground state of a harmonic oscillator is

ψ (x ) ≡ x 0 =

a  1  exp  − α 2 x 2  . 1/2  2  π

where

α=

mω k , ω= . m 

As ψ 0 is an even function, the first order correction 0 H ′ 0 = 0 and we have to go to the second order. For the harmonic oscillator we have n′ x n =

 1  n′ n′ + 1 δ n ,n′−1 + δ n ,n′+1  ,  2 α 2 

and hence H 0,′ n = − qE 0 x n = − (qE / 2α )δ n ,1 . Thus the energy correction for the ground state to order E 2 is q2 E 2

(2) 0

∆E

2 ′ |H′ | ′ 2 δ n ,1 = ∑ (0) 0,n (0) = ∑ 2α E0 − En −nω n n

=− where the partial sum

∑

l n

q2 E 2 q2 E 2 , = − 2ωα 2 2mω 2

excludes n = 0 .

5012 For a one-dimensional harmonic oscillator, introduction of the dimensionless 1/2 coordinate and energy variables y = x ( mω 0 /  ) and ε n = 2En / ω 0 gives a

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Schrödinger equation with kinetic energy operator T = −

289

d2 and potential dy 2

energy V = y 2. a. Using the fact that the only non-vanishing dipole matrix element is n + 1 (and its Hermitean conjugate), find values for all n +1 y n = 2 the non-vanishing matrix elements of y 3 that connect to the ground state 0 . b. The oscillator is perturbed by an harmonic potential V ′ = α y 3 . Find the correction to the ground state energy in the lowest non-vanishing order. (If you did not get complete answers in part(a), leave your result in terms of clearly defined matrix elements, etc.) (Berkeley) Sol: a. As myn =

m n δ n ,m−1 + δ m ,n−1 , 2 2

the non-vanishing matrix elements connected to 0 , m y3 0 = ∑ m y k

kyl

l y0 ,

k ,l

are those with m = 3, k = 2, l = 1, and with m = 1, and k = 0, l = 1, or k = 2, l = 1, namely 3 3 δ m,1. δ m,3 and 2 2 2 b. Because ψ 0 is an even function,

0 y 3 0 = 0 , or the first order energy

correction 〈0| α y 3 |0〉 is zero, and we have to calculate the second order energy correction:  | 0 y 3 1 |2 | 0 y 3 3 | 0 α y 3 n |2 = | α |2  + 1 − εn 1− ε3  1 − ε1 n≠ 0

∆E2 = ∑

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 . 

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As

εn =

2 En = 2n + 1, ω 0

 | 〈0| y 3 1 |2 | 〈0| y 3 3 |2  11 2 ∆E = | α |2  +  = − 16 | α | .  −2 −6

5013 Consider a one-dimensional harmonic oscillator of frequency ω 0. Denote the energy eigenstates by n, starting with n = 0 for the lowest. To the original harmonic oscillator potential a time-independent perturbation  = V ( x ) is added. Instead of giving the form of the perturbation V ( x ), we shall give explicitly its matrix elements, calculated in the representation of the unperturbed eigenstates. The matrix elements  are zero unless m and n are even. A portion of the matrix is given below, where ε is a small dimensionless constant. [Note that the indices on this matrix run from n = 0 to 4.]  1  0   ε ω 0 − 1/ 2  0   3/ 8 

0 − 1/ 2 0 0 0 0

1/ 2 0

0 − 3 /16

   0 − 3 /16   0 0  0 3/ 8   0 0

3/ 8 0

a. Find the new energies for the first five energy levels to first order in perturbation theory. b. Find the new energies for n = 0 and 1 to second order in perturbation theory. Sol: a. The energy levels to first order in perturbatiosswn theory are En′ = En + Hnn ′ . where En = (n + 12 ) ω 0 , Hnn ′ = n  n . Thus the energy for the first five energy levels are

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291

1 1  E0′ = ω 0 + ε ω 0 =  + ε  ω 0 , 2  2 3 E1′ = ω 0 , 2 5 1  E2′ =  + ε  ω 0 , 2 2  7 E3′ = ω 0 , 2 9 3  E4′ =  + ε  ω 0 . 2 8  b. The energies for n = 0 and 1 to the second order in perturbation theory are | H k 0 |2 k ≠0 E0 − Ek

E0′′= E0 + H 00′ + ∑

1 1 | H k′0 |2 = ω 0 + ε ω 0 + ∑ 2 k ≠ 0 − k ω 0 1  1 3 = ω 0  + ε − ε 2  + +   ,   4 32 2 | H k′1 |2 k ≠0 E1 − Ek 3 1 | H k′1 |2 = ω 0 + 0 + ∑ 2 k ≠1 (1 − k )ω 0

E1′′= E1 + H11′ + ∑

3  3 = ω 0  + 0 + 0 = ω 0 . 2  2

5014 A mass m is attached by a massless rod of length l to a pivot P and swings in a vertical plane under the influence of gravity (see Fig. 5.6).

Fig. 5.6

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a. In the small angle approximation find the energy levels of the system. b. Find the lowest order correction to the ground state energy resulting from inaccuracy of the small angle approximation. (Columbia) Sol: a. Take the equilibrium position of the point mass as the zero point of potential energy. For small angle approximation, the potential energy of the system is 1 V = mgl(1 − cosθ ) ≈ mglθ 2 , 2 and the Hamiltonian is 1 1 H = ml 2θ 2 + mglθ 2 . 2 2 By comparing it with the one-dimensional harmonic oscillator, we obtain the energy levels of the system 1  En =  n +  ω ,  2 where ω = g /l . b. The perturbation Hamiltonian is 1 H ′ = mgl (1 − cosθ ) − mglθ 2 2 1 1 mg 4 4 x , ≈ − mglθ = − 24 24 l 3 with x = lθ . The ground state wave function for a harmonic oscillator is

ψ0 =

α  1  exp  − α 2 x 2  1/2  2  π

with α = mω . The lowest order correction to the ground state energy  resulting from inaccuracy of the small angle approximation is 1 mg E′ = 0 H ′ 0 = − 0 x4 0 . 24 l 3 As 0 x4 0 =

α π

E′ = −

∫

∞

−∞

x 4 exp(−α 2 x 2 )dx =

3 , 4α 4

2 . 32ml 2

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293

5015 A quantum mechanical rigid rotor constrained to rotate in one plane has moment of inertia I about its axis of rotation and electric dipole moment µ (in the plane). This rotor is placed in a weak uniform electric field ε , which is in the plane of rotation. Treating the electric field as a perturbation, find the first nonvanishing corrections to the energy levels of the rotor. (Wisconsin)

Fig. 5.7 Sol:

Take the plane of rotation of the rotor as the xy plane with the x-axis parallel to ε as shown in Fig. 5.7. In the absence of external electric field the Hamiltonian of rotor is H0 = −

 2 ∂2 , 2 I ∂θ 2

and the eigenequation is −

 2 ∂ 2ψ = Eψ , 2 I ∂θ 2

which has solutions

ψ=

1 imθ e , m = 0, ± 1, ± 2, … , 2π

corresponding to energy levels Em(0) =

 2m 2 . 2I

When the external electric field acts on the system and may be treated as perturbation, the Hamiltonian of perturbation is H ′ = − µ ⋅ ε = − µε cosθ .

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The first order energy correction is E (1) = m H ′ m = −

µε 2π

∫

2π

0

cosθ dθ = 0

The second order energy correction is 2

E

(2)

m′ H ′ m = ∑ ′ (0) . Em − Em(0)′ m ′≠m

As

µε 2π i (m−m ′ )θ e cosθ dθ 2π ∫0 µε 2π i (m−m ′+1)θ i (m−m ′−1)θ e  dθ =− +e 4π ∫0  µε = − (δ m ′ ,m+1 + δ m ′ ,m−1 ) , 2

m′ H ′ m = −

we have E (2) = =

µ 2ε 2 2 I  1 1  ⋅ 2 2 + 2 2 2  4   m − (m − 1) m − (m + 1)  µ 2ε 2 I 1 . ⋅  2 4m 2 − 1

5016 The polarization of a diatomic molecule in weak electric fields may be treated by considering a rigid rotator with moment of inertia I and electric dipole moment d in a weak electric field E. a. Ignoring the motion of the center of mass write down the Hamiltonian H of the rigid rotator in the form of H0 + H ′ . b. Solve the unperturbed problem exactly. How are the levels degenerate? c. Calculate the lowest order correction to all the energy levels by nondegenerate perturbation method. d. Explain why is nondegenerate perturbation method applicable here and how are the levels degenerate. The following relation may be useful: cosθ Ylm =

( l + 1− m)( l + 1+ m) (2l + 1)(2l + 1)

Yl +1,m +

( l + m)( l − m) (2l + 1)(2l − 1)

Yl −1,m . (Buffalo)

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295

Sol: a. Take the z-axis in the direction of the field. The Hamiltonian of the rotator is H=

J2 J2 − d ⋅ E = − dE cos θ . 2I 2I

Considering −dE cos θ as a perturbation, we have H0 =

J2 , H ′ = −dEcosθ . 2I

b. The eigenfunctions of the unperturbed system are

ψ jm = Yjm (θ, ϕ) ,

where m = - j, (- j + 1),º , ( j - 1), j,

and the energy eigenvalues are

j( j + 1) 2 E(0) = . The levels are (2 j + 1) -fold degenerate since E (jm0) does not jm 2I depend on m at all. c. The first order energy correction is jm −dE cosθ jm = −dE jm cosθ jm = −dE ∫

π

0

∫

2π

0

cosθ sinθ dθ dϕ = 0.

For the second order correction we calculate 2

(2) n

E

n H′ i = ∑′ (0) , En − Ei(0) i

Where n, i denote pairs of j, m and the prime signifies exclusion of i = n in the summation. As the non-vanishing matrix elements are only j + 1, m −dE cosθ jm = −dE

( j + 1 − m)( j + 1 + m) , (2 j + 1)(2 j + 3)

j − 1, m −dE cosθ jm = −dE

( j + m)( j − m) , (2 j + 1)(2 j − 1)

the lowest order energy correction is 2 Id 2 E 2 2 ( j + 1 − m)( j + 1 + m)  ×  (2 j + 1)(2 j + 3)[ j( j + 1) − ( j + 1)( j + 2)]

E ( 2) =

+ =

 ( j + m)( j − m )  (2 j + 1)(2 j − 1)[ j( j + 1) − j( j − 1)]  Id 2 E 2  j( j + 1) − 3m 2 

 2 j( j + 1)(2 j − 1)(2 j + 3)

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d. Because H ′ is already diagonal in the j subspace, the nondegenerate perturbation theory is still applicable. However even with the perturbation, degeneracy does not completely disappear. In fact states with same j but opposite m are still degenerate.

5017 A rigid rotator with electric dipole moment P is confined to rotate in a plane. The rotator has moment of inertia I about the (fixed) rotation axis. A weak uniform electric field E lies in the rotation plane. What are the energies of the three lowest quantum states to order E2? (MIT) Sol: The Hamiltonian for the free rotator is

Fig. 5.8 H0 =

1 2 2 d 2 . Lφ = − 2I 2 I dφ2

A Schrödinger equation then gives the eigenvalues and eigenfunctions Em(0) =

 2 m2 1 imφ , ψm(0) (φ) = e . 2I 2π

(m = 0,

± 1, ± 2, …)

When a weak uniform electric field E is applied, the perturbation Hamiltonian is H ′ = −E ⋅ P = λ cosφ , where λ = −EP. The matrix elements of the perturbation Hamiltonian is λ 2π i (m−n )φ λ n H′ m = e cosφ dφ = (δ n ,m+1 + δ n ,m−1 ) . 2π ∫0 2 Define Em(0) and E by H 0 m = Em(0) | m , (H 0 + H ′) | = E | ,

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297

and expand | in m ≡ ψ m(0) (φ ) : ∞

∑C

| =

m = −∞

m .

m

Then (H 0 + H ′)∑Cm m = ∑Cm E m . m

or

m

∑C ( E − E ) m − ∑C (0) m

m

m

m

m

H ′ m = 0.

Multiplying both sides by n we have, on account of the orthonormality of m and the property of n H ′ m given above,

∑C ( E − E ) δ m

(0) m

m

mn

−

λ ∑C ( δ + δ ) = 0, 2 m m m , n −1 m , n + 1

or

( E − E )C (0) n

n

λ λ − Cn −1 − Cn +1 = 0. 2 2

Expanding E and Cn as power series in λ: ∞

∞

ρ =0

ρ =0

E = ∑E ( ρ ) λ ρ , Cn = ∑Cn( ρ ) λ ρ , and substituting in the above, we obtain perturbation equations of different orders:

λ0 : λ1 : λ2 :

(E (E (E

(0)

− En(0) )Cn(0) = 0,

1 1 (0) − En(0) )Cn(1) + E (1)Cn(0) − Cn(0) −1 − Cn+1 = 0, 2 2 (0) (0) (2) (1) (1) (2) (0) − En )Cn + E Cn + E Cn

(0)

1 1 − Cn(1)−1 − Cn(1)+1 = 0, 2 2 … 0 To find the energy level Ek , we first see that the zeroth order equation ( Ek(0) − En(0) )Cn(0) = 0 requires Cn(0) = 0 if k ≠ n . Hence, we write Cn(0) = akδ n ,k + a− kδ n ,− k . (1) Substitution in the first order equation gives

(E

(0) k

− En(0) )Cn(1) + E (1) ( akδ n ,k + a− kδ n ,− k )

1 − (akδ n−1,k + a− kδ n−1,− k + akδ n+1,k + a− kδ n+1,− k ) = 0. 2

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When n = ± k , we have E (1) = 0 . When n ≠ ± k , we have 1 1 aδ + a− k δn −1, − k + ak δn +1, k + a− k δn +1, − k (0) 2 En − Ek(0) k n −1, k 1 1 C (0) + Cn(0+)1 . =− (0) 2 En − Ek(0) n −1

(

Cn(1) = −

(

)

)

Substituting Cn(1) in the second order equation gives for n = k

(E

(0) k

1 1 − En(0) )Cn(2) + E (2)Cn(0) − Cn(1)−1 − Cn(1)+1 = 0 , 2 2

i.e.,

1 (1) E (2) C (0) = 1(Ck(1) −1 + Ck +1 ) (2) k(0) E Ck = 2 (Ck(1)−1 + Ck(1)+1 ) 2 1 1 1 (0) = 1− 1 (0) 1 (0) (Ck(0) − 2 + Ck(0)) (0) − E = 2 −2 Ek(0) C + C ( k−2 k ) −1 k 2  2 Ek−1 − Ek(0)  1 1 − 1 (0) 1 (0) (Ck(0) + Ck(0)  +2 )  (0) (0) − −2 Ek(0) + E C C ( k k+2 ) +1 k 2 Ek+1 − Ek(0)  1 1 (0) (0) = − 1 (0) 1 (0) (Ck(0) +C = −4 Ek(0) (Ck−−22 + Ckk(0))) −1 − Ek 4  Ek−1 − Ek(0)  1 + (0) 1 (0) (Ck(0) + Ck(0) . +2 )  (0) (0) (2) +Ek(0) . +1 − Ek(0) (Ck + Ck + 2 ) Ek+1 − Ek  For the ground state k = 0, Eq. (1) gives C0(0) ≠ 0, C−(0)2 = C2(0) = 0 , and so Eq. (2)

becomes  (0) 1 1 1 E (2)C0(0) = −  (0) +  C0 . 4  E−1 − E0(0) E1(0) − E0(0)  Hence, E0(2) = −

1 1 I =− 2 2 E1(0) − E0(0) 

For the first exited state k = ± 1, Eq. (1) gives C±(0)1 ≠ 0 , and Eq. (2) becomes 1 1 E (2)C1(0) = −  (0) (C−(0)1 + C1(0) ) 4  E0 − E1(0) +

 1 (C1(0) + C3(0) ) E2(0) − E1(0) 

1  1 1  (0) 1  = −  (0) − C1 − (0) C−(0)1  4  E2 − E1(0) E1(0)  E1 

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1 1 E (2)C1(0) = −  (0) (C−(0)1 + C1(0) ) 4  E0 − E1(0) +

E

(0) 2

 1 C (0) + C3(0) )  (0) ( 1 − E1 

Perturbation Theory

299

1  1 1  (0) 1  = −  (0) − C1 − (0) C−(0)1  4  E2 − E1(0) E1(0)  E1  and   1  1 1 1  E (2)C−(0)1 = − − (0) C1(0) +  (0) − (0)  C−(0)1  (0) 4  E−1  E−2 − E−1 E−1    1  1 1 1   = − − (0) C1(0) +  (0) − (0)  C−(0)1  , (0) − 4  E1 E E E  2 1 1   as E0(0) = 0, C±(0)3 = 0, or  1  1 1  1 1 (0) − (0)  − E (2)  C1(0) + C−1 = 0 ,  −  (0) (0) − 4 E E E 4 E1(0)   2 1 1   and  1 1 (0)  1  1 1  − (0)  − E (2)  C−(0)1 = 0. C1 +  −  (0) (0) − 4 E1(0) 4 E E E 1 1    2  These are homogeneous equations in C10 and C−01. Solving the secular equation, we obtain for the first excited state two energies E1(2) + =

1 1 1 1 1  5I −  (0) − (0)  = 2 , (0) (0) E1  6 4 E1 4  E2 − E1

E1(2) − =−

I 1 1 1 1 1  − − =− 2. 4 E1(0) 4  E2(0) − E1(0) E1(0)  6

For the second exited state k = ± 2 , Eq. (1) gives C±(0)2 ≠ 0 and Eq. (2) becomes  (0) 1 1 1 E (2)C±(2)2 = −  (0) + (0) C±2 . (0) (0)  4  E1 − E2 E3 − E2  Thus  1 1 1 I E2(2) = −  (0) + = . 4  E1 − E2(0) E3(0) − E2(0)  15 2 Therefore, the energy correction to the second order perturbation for the ground state is E0 = λ 2 E0(2) = −

I ( Ep)2 , 2

for the first exited state is E1+ =

 2 5 I ( Ep)2 + , 2I 6  2

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E1− =

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and for the second exited state is E2 =

2 2 1 I ( Ep)2 + . I 15  2

5018 Consider a particle of mass m in a potential V(X) = (mω 2X2/2) + gcos(kX). Calculate the change in the ground-state energy compared to the harmonic oscillator V(X) = mω 2X2/2 to the first order in g. Sol: The perturbed Hamiltonian Hˆ ′ = gcos(kX). The first order correction to the ground state is given by , where |n> is the basis. ∞

In integral form, < Hˆ > = ∫ dx ψ 0* Hˆ ′ ψ 0, where ψ 0 is given by (α 2 / π )1 4 exp(α 2 X 2 /2). −∞

So the correction term is given by ∞ < Hˆ ′> = (α 2 / π )1/2 g ∫ dx exp (α 2 X 2 ) cos ( kX ) . −∞

∞

Now here is the standard integral ∫ dx exp(−aX )2 cos bX = (π / a)1/2 exp(−b 2 / 4a). −∞

Using this standard integral, the above integral evaluates to ( π /α 2)1/2 exp(−k2/4α 2). So the correction to the first order energy is = gexp(−k2/4α 2).

5019 a. State all the energy levels of a symmetric top with principal moments of inertia I1 = I2 = I ≠ I3 . b. A slightly asymmetric top has no two I’s exactly equal, but I1 − I2 = ∆ ≠ 0, I1 + I2 = 2I , ( ∆ / 2I )  1. Compute the J = 0 and J = 1 ener-

gies through 0 ( ∆ ) .

(Berkeley) Sol: a. Let ( x , y , z ) denote the rotating coordinates fixed in the top. The Hamiltonian of the system is

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301

2 1  J2 Jy J2  H=  x + + z 2  I1 I 2 I 3 

=

1 2 2 1 2 Jx + J y + Jz 2I 2I3

=

1 2 1  1 1 2 J +  −  Jz . 2I 2  I3 I 

(

)

Hence, a state with quantum numbers J, m has energy E=

2 2  1 1  J ( J + 1) +  −  m 2 , 2I 2  I3 I 

which gives the energy levels of the symmetric top. b. For the slightly asymmetric top the Hamiltonian is H=

1 2 1  1 1 2 ∆ 2 2 J +  −  Jz + 2 ( J y − Jx ) 2I 2  I3 I  4I

= H0 + H ′, ∆ 2 2 where H ′ = 2 ( J y − J x ), to be considered as a perturbation. 4I Defining J ± = J x ± iJ y we have 1 J x2 − J y2 = ( J +2 + J −2 ) . 2 Noting that J + | jm = ( j + m + 1)( j − m)| j , m + 1 , J − | jm = ( j − m + 1)( j + m)| j , m − 1 ,

J ±2 | jm = J ± ( J ± | jm ) , we have

J +2 00 = J −2 |00 = 0, J +2 10 = J −2 |10 = 0, J +2 1,1 = 0, J −2 1, − 1 = 0,

J +2 |1, − 1 = 2 2 |11 , J −2 |11 = 2 2 |1, −1 .

Hence, for the perturbed states: i.

J = 0, m = 0, (nondegenerate): E0′ = E0(0) + 00 H ′ 00 = E0(0) = 0 .

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ii. J = 1, m = 0, (nondegenerate): E1′ = E1(0) + 10 H ′ 10 = E1(0) = iii. J = 1, m = ± 1, (two-fold degenerate):

2 , I

As degeneracy occurs, we first calculate 1, − 1 H ′ 1, − 1 = 1,1 H ′ 11 = 0, 1, −1 H ′ 11 = 11 H 1, −1 = We then form the secular equation   0 λ1 −   ∆ 2  2  4I

∆ 2 . 4I 2

∆ 2   4I 2  = 0,  0  

i.e.,

λ

−

∆ 2 − 2 4I

∆ 2 4I 2

= 0.

λ

This equation has two solutions

λ± = ±

∆ 2 , 4I 2

which means that the energy of the states J = 1, m = ± 1, E1,±1 , splits into two levels: E1,±1 =

 2  2 ∆ 2 . + ± 2I 2I3 4 I 2

5020 a. Using a simple hydrogenic wave function for each electron, calculate by perturbation theory the energy in the ground state of the He atom associated with the electron-electron Coulomb interaction (neglect exc­ hange effects). Use this result to estimate the ionization energy of helium. b. Calculate the ionization energy by using the variational method, with the effective charge Z in the hydrogenic wave function as the variational

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303

parameter. Compare the results of (a) and (b) with the experimental ionization energy of 1.807E0 , where E0 = α 2 mc 2 / 2 . Note: Z3 2 exp( − Zr / a ), a = , 0 0 π a03 me 2

ψ 1s ( r ) =

∫ ∫d rd r e 3

3

1

2

− α (r1+r2 )

/ r1 − r2 = 20π 2 / α 5. (Columbia)

Sol: a. The unperturbed Hamiltonian of the system is H0 = −

2 1 Ze 2 Ze 2 ∇ 2 + ∇ 22 ) − − . ( 2m r1 r2

The wave functions have the form

Φ = φ ( r1 , r2 ) χ 0 ( s1z , s2 z ).

For the ground state,

φ (r1 ,r2 ) = ψ 100 (r1 )ψ 100 (r2 ), where Z3 exp{− Zr / a0 } π a03

ψ 100 (r ) =

e2 as a | r1 − r2 | perturbation, the energy correction to first order in perturbation theory is

with a0 =  2 / me 2 . Treating the electron-electron interaction

∆E = H ′ = e 2

∫∫

d 3 r1d 3 r2 2 2 ψ 100 (r1 ) ψ 100 (r2 ) r1 − r2

 Z3  =e  3  πa 

2

  2Z d 3 r1d 3 r2 exp  − (r1 + r2 ) r1 − r2   a0

∫∫

2

0

2

 Z 3  20π 2 5 Ze 2 = e2  3  ⋅ . 5 = 8a0  π a0  2aZ

( ) 0

The energy levels of a hydrogen-like atom are given by En = −

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and so the energy of the ground state of the system excluding the electronelectron Coulomb interaction is E0 = − 2

e2Z 2 e2Z 2 =− . 2a0 a0

Hence the corrected energy of the ground state of helium is e 2 Z 2 5 Ze 2 11e 2 , E=− + =− a0 8a0 4α 0 with Z = 2 for helium nucleus. The ionization energy is the energy required to remove the two electrons of helium atom to infinity. Thus for the ground state I=−

e 2 Z 2  e 2 Z 2 5 Ze 2  3e 2 − − + = 2a0  a0 8a0  4a0

with Z = 2 for helium nucleus, i.e., I = 1.5 E0 , with 2

E0 =

e2 1  e2  1 =   mc 2 = α 2mc 2, 2a0 2  c  2

α being the fine structure constant. b. The Hamiltonian of He with electron-electron interaction is H=−

2 2 (∇1 + ∇ 22 ) − Ze 2 / r1 − Ze 2 / r2 + e 2 / r12 2m

with Z = 2, r12 = r1 − r2 . For the ground state use a trial wave function

φ (r1 , r2 , λ ) =

λ 3 − λ ( r1 +r2 ) e . π

Let u(r ) = e − λr . Then

λ2  1 2 λ  − ∇ −  u(r ) = − u(r ), 2 r 2 i. e.,

 2 2 λ2  λ 22  − 2m ∇ − mr  u(r ) = − 2m u(r ).

Setting Ze 2 −

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λ2 =σ , m

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305

we have, using the above results,   2 2  2 2 Ze 2 Ze 2 e 2  H = ∫∫ d 3r1d 3r2 Φ*  − ∇1 − ∇2 − − + Φ 2m r1 r2 r12   2m = ∫∫ d 3r1d 3r2 Φ* (− =−

λ 22 σ σ e2 − − + )Φ m r1 r2 r12

λ 2  2 2σλ 3 e −2 λr1 3 e2λ 6 d r1 + 2 − ∫ π π m r1

∫∫ d r d r 3

1

3

2

e −2 λ ( r1 +r2 ) . r12

As

∫

−2 λr1 ∞e e −2 λr1 3 4π r12 dr1 = π /λ 2 , d r1 = ∫ 0 r1 r1

λ 22 λ 6 20π 2 − 2σλ + 2 π (2λ )5 m λ 22  5 = −  2Z −  e 2 λ . 8 m 

H=−

Letting

∂H = 0 , we get ∂λ

λ=

5 me 2  2Z −  . 2  2  8

Therefore, the energy of the ground state is 2

2

2 5  me 4   27  e E = −Z −  = − ,     16  a0 16   2

2 as Z = 2 a0 =  , and the ground state ionization energy is me 2 I=−

2 2  e2 Z 2e 2  5  e 2   27  +Z−  =    − 2  = 1.695 E0 2a0  16  a0   16   a0

Thus the result from the variational method is in better agreement with experiment.

5021 A particle of mass m moves in one dimension in the periodic potential  2π x  V ( x ) = V0 cos   α  . We know that the energy eigenstates can be divided into classes characterized by an angle θ with wave functions φ ( x ) that obey φ ( x + a ) = e iθφ ( x ) for all x. For the class θ = π , this becomes φ ( x + a ) = − φ ( x ) (antiperiodic over length a).

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a. Even when V0 = 0 , we can still classify eigenstates by θ . For which values of k does the plane wave φ ( x ) = e ikx satisfy the antiperiodic condition over length a? What is the energy spectrum of the class θ = π for V0 = 0 ? b. When V0 is small (i.e. V0   2 / ma 2 ) , calculate the lowest two energy eigenvalues by first order perturbation theory. (MIT) Sol: a. For the plane wave ψ ( x ) = e ikx, we have

ψ ( x + a) = e ik ( x +a ) = e ikaψ ( x ) . If k satisfies ka = (2n + 1)π , (n = 0, ± 1, ± 2, …) the plane wave satisfies the antiperiodic condition ψ ( x + a) = − ψ ( x ). The corresponding energy spectrum is En = b. If V0 

 2π 2 (2n + 1)2 . (n = 0, ± 1, ± 2, …) 2ma 2

2 , one can treat ma 2

 2π x  H ′ = V0 cos   a  as a perturbation imposed on the free motion of a particle. For the ground state, the eigenvalue and eigenfunction of the free particle are respectively (n = 0, − 1; i.e., ka = π , − π ) E (0)0 =

(−1)

 2π 2 , 2ma 2

1 iπ x /a e , a 1 − iπ x /a ψ −(0)1 ( x ) = e . a

ψ 0(0) ( x ) =

Let 2π = β =and consider m H ′ n . We have a

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307

V0 a  2π x  dx = 0, cos   a  a ∫0 V a −1 H ′ 0 = 0 H ′ − 1 = 0 ∫ e ± iβ x (e iβ x + e − iβ x )dx 2a 0 V = 0. 2

−1 H ′ − 1 = 0 H ′ 0 =

Hence, for ground state,   0 H′ =   V0  2

V0  2  ,  0  

and the secular equation for first order perturbation is E (1) V0 2

V0 2 E

(1)

= 0,

giving E (1) = ±

V0 . 2

Thus the ground state energy level splits into two levels E1 =

 2π 2 V0  2π 2 V0 − , E2 = + . 2 2ma 2 2ma 2 2

These are the lowest energy eigenvalues of the system.

5022 a. A rigid rotor in a plane is acted on by a perturbation Ĥ’= (V0/2) (3cos2φ − 1). Calculate the first-order correction to the energy of the ground state, where ψ (ϕ ) = eimϕ / 2π is the normalized wave function and m takes integer values. Consider a free particle of mass m confined to move in a one-dimensional infinite box of length a. If the system is subjected to perturbation V(x) = V0δ(x − a/2), find the first order correction to the energy of the 2 nπ x sin is the normalized wave function a a where n takes integer values. ground state if ψ n( x ) =

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Sol: a. For the ground state, m = 0,ψ 0 = 1/ 2π . 2π So the correction is given by E01 = = (V0 / 4π ) ∫ dϕ (3cos 2ϕ − 1) 0

2π 2π E01 = V0 / 4π  ∫ dϕ 3cos 2ϕ − ∫ dϕ  = V0 / 4 0  0 

where we have used the formula cos2 x = (1 + cos2 x)/2. b. The perturbed Hamiltonian is given by Hˆ ′= V0δ ( x − a / 2). The correction to the energy is given by a < Hˆ ′> = ( 2V0 / a ) ∫ dx sin 2 (π x / a) δ ( x − a / 2). 0

Using the property of Dirac delta integral,

∫

b

a

f ( x )δ ( x − c ) dx = f (c ), where [a < c < b]

< Hˆ ′> = (2V0 / a)sin 2 (π x / a)|x =a/2 = 2V0 / a .

5023 Consider the one-dimensional motion of an electron confined to a potential  well V ( x ) = 1 kx 2 and subjected also to a perturbing electric field F = F x. 2 a. Determine the shift in the energy levels of this system due to the electric field. b. The dipole moment of this system in state n is defined as Pn = − e x n , where x

n

is the expectation value of x in the state n. Find the dipole

moment of the system in the presence of the electric field. (Wisconsin) Sol: a. The Hamiltonian of the system is 2 2 1 2 ∇ + kx − qFx 2m 2 2 2 2 1  qF  q 2 F 2 =− ∇ + k x −  − 2  2k 2m k 

H=−

=−

2 2 1 2 q2 F 2 , ∇′ + kx ′ − 2 2k 2m

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Perturbation Theory

where x ′ = x −

309

qF . k

Hence the energy shift due to the perturbing electric field F xˆ is E′ =

q2 F 2 e 2 F 2 = . 2k 2k

b. The expectation value of x in state n is x

n

= x′ +

qF qF qF = x′ + = . k k k

Therefore the dipole moment of the system is Pn = − e

qF F = e 2 , (q = − e ) . k k

5024 If a very small uniform-density sphere of charge is in an electrostatic potential V(r), its potential energy is U (r ) = V (r ) + 1 r02∇ 2V (r ) +  , where r is the position 6 of the center of the charge and r0 is its very small radius. The “Lamb shift” can be thought of as the small correction to the energy levels of the hydrogen 2 atom because the physical electron does have this property. If the r0 term of U is treated as a very small perturbation compared to the Coulomb interaction V (r) = − e 2 / r ., what are the Lamb shifts for the 1s and 2p levels of the hydrogen atom? Express your result in terms of r0 and fundamental constants. The unperturbed wave functions are 1 −5/2 − r /2 aB m ψ 1s (r ) = 2aB−3/2 ⋅ e − r / aB Y00 ; ψ 2 pm (r ) = aB re Y1 , 24 2 2 where aB =  / me e .

(CUS) Sol: The state 1s is nondegenerate, so the energy correction is ∆E = 1s H ′ 1s . As r02 2 r2 1 ∇ V (r ) = 0 (−e 2 )∇ 2 r 6 6 r02 π 2 = (−e 2 )(−4π )δ (r ) = r02e 2δ (r ), 6 3

H′ =

1

 1 2 Y =  ,  4π  0 0

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we have 2π 2 2 r0 e δ (r)|ψ 1s (x)|2 dx 3 2π 2 2 2 r 2e 2 = r0 e |ψ 1s (0)|2 = 0 3 . 3 3 aB

∆E = ∫

The perturbation H ′ , being a δ -function, has an effect only if ψ (0) ≠ 0. As

ψ 2 pm (0) = 0, H ′ has no effect on the energy, i.e., ∆E2 pm = 0 .

5025 Positronium is a hydrogen atom but with a positron as “nucleus”, instead of a proton. In the nonrelativistic limit, the energy levels and wave functions are the same as for hydrogen, except for scale. a. From your knowledge of the hydrogen atom, write down the normalized wave function for the 1s ground state of positronium. Use spherical coordinates and the hydrogenic Bohr radius a0 as a scale parameter. b. Evaluate the root-mean-square radius for the 1s state in units of a0. Is this an estimate of the physical diameter or the radius of positronium? c. In the s states of positronium there is a contact hyperfine interaction 8π Hint = − µe ⋅ µ pδ (r ), 3 where µe and µ p are the electron and positron magnetic moments e   s .  µ = g 2mc  For electrons and positrons, | g | = 2. Using first order perturbation theory compute the energy difference between the singlet and triplet ground states. Determine which state lies lowest. Express the energy splitting in GHz (i.e., energy divided by Planck’s constant). Get a number! (Berkeley) Sol: a. By analogy with the hydrogen atom the normalized wave function for the 1s ground state of positronium is

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1 1  ψ 100 (r ) =  π  2a0 

311

3/2

e − r /2a0

with a0 =  2 , m being the electron rest mass. Note that the factor 2 in front me of a0 is to account for the fact that the reduced mass is µ = 1 m . 2 b. The mean-square radius for the 1s state is ∞ 1 r2 = e − r /a0 r 2 ⋅ r 2 dr 8π a03 ∫0 =

a02 8π

∫

∞

0

e − x x 4 dx =

3a02 , π

and the root-mean-square radius is r2 =

3 a. π 0

This can be considered a physical estimate of the radius of positronium. c. Taking the spin into account a state of the system is to be described by n, l , m, S , Sz , where S and Sz are respectively the total spin and the z-component of the spin. Thus 100S ′Sz′ |Hint |100 S Sz  8π  * = ∫ d 3rψ 100 (r)  −  δ (r)ψ 100 (r)χ s+ , (Sz′ )µe ⋅ µ p χ s (Sz )  3  =

2

8π  e  2 +   ψ 100 (0) χ s ′ (Sz′ )se ⋅ s p χ s (Sz ) 3  mc  2

1  e2  e2 1 3 =   ⋅ ⋅  S(S + 1) − δ SS′δ Sz Sz′ , 3  c  a0  2 4 where we have used S = se + s p and so 1 se ⋅ s p = S 2 − (se2 + s 2p ) 2 1  1  1   =  S(S + 1) − 2    + 1  .  2  2   2 For the singlet state, S = 0, Sz = 0, 2

1  e2  e2 ∆E0 = −   < 0. 4  c  a0

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For the triplet state, S = 1, Sz = 0, ± 1, 2

∆E1 =

1  e2  e2 > 0. 12  c  a0

Thus the singlet ground state has the lowest energy and the energy splitting of the ground state is 2

4

2 2 2  1 1  e  e 1 e  ∆E1 − ∆E0 =  +    =   mc 2  12 4   c  a0 3  c  4

1 1  6 −4 =   ⋅ 0.51 × 10 = 4.83 × 10 eV, 3  137  corresponding to v = ∆E / h = 1.17 × 1011 Hz = 117 GHz .

5026 Consider the proton to be a spherical shell of charge of radius R. Using first order perturbation theory calculate the change in the binding energy of hydrogen due to the non-point-like nature of the proton. Does the sign of your answer make sense physically? Explain. Note: You may use the approximation R  a0 throughout this problem, where a0 is the Bohr radius. (MIT) Sol: If we consider the proton to be a spherical shell of radius R and charge e, the potential energy of the electron, of charge –e, is  e2  − ,  R V (r ) =  2  −e ,  r 

0 ≤ r ≤ R, R ≤ r ≤ ∞.

2 Take the difference between the above V (r ) and the potential energy − e due to r a point-charge proton as perturbation:

 e2 e2  − , 0 ≤ r ≤ R, H′ =  r R  0, R ≤ r ≤ ∞. 

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313

The energy correction given by first order perturbation is R

R  e 22 ee 22  22 22 ∆E = ∫  e − R10dr dr ∆ E= −  rr R 10  r R R  00  r 4 RR  e 22 ee 22  22 −−22rr //aa0 = 43 ∫  e − −  rr ee 0 dr dr = aa03 00  rr R R  0

44 RR  ee 22 ee 22  22 22ee 22 R R 22 ≈ ≈ a33 ∫0  r − − R  rr dr = 3a33 .. (( R R  aa00 )) dr = a00 0  r R  3a00 As ∆E > 0, the ground state energy level of the hydrogen atom would increase due to the non-point-like nature of the proton, i.e., the binding energy of the hydrogen atom would decrease. Physically, comparing the point-like and shellshape models of the proton nucleus, we see that in the latter model there is an additional repulsive action. As the hydrogen atom is held together by attractive force, any non-point-like nature of the proton would weaken the attractive interaction in the system, thus reducing the binding energy.

5027 Assume that the proton has a nonzero radius rp ≈ 10 −13 cm and that its charge is distributed uniformly over this size. Find the shift in the energy of the 1s and 2p states of hydrogen due to the difference between a point charge distribution and this extended charge. (Columbia) Sol: The Coulomb force an electron inside the sphere of the proton experiences is 3

r 1 e2 F = −e   2 er = − 3 rer . rp  rp  r 2

The electrical potential energy of the electron is V1 =

e2 2 r +C 2rp3

V2 = −

e2 r

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for

r ≤ rp ,

for

r ≥ rp ,

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Continuity at rp requires that V1 (rP ) = V2 (rp ) , giving C = −

3 e2 . Thus 2 rp

 e2  − , r > rp ,  r 2 V (r ) =  2      e  r − 3 , r ≤ r . p    2rp  rp      The Hamiltonian of the system is Hˆ = Hˆ 0 + Hˆ ′ , where  0, r > rp ,  2   Hˆ ′ =  e 2  r  2rp  2r  r  + r − 3  , r ≤ rp   p  p    2 2 e Pˆ Hˆ 0 = − . 2m r Hence Enl′ = nlm | H ′ | nlm = nl | H ′ | nl ∞

= ∫ Rnl* Rnl H ′(r )r 2 dr 0

rp

= ∫ Rnl* (r )Rnl (r ) 0

2  e 2  r  2rp  × − 3  r 2dr ,   + r 2rp  rp    

where R10 =

2 − r /a e , a3/2

R21 =

1 r × e − r /2a 3/2 a 2 6a

2

with a = me 2 . As rp  a we can take e − r /a ≈ 1 . Integrating the above gives the energy shifts of 1s and 2p states: E10′ = 10 H ′ 10 ≈ E21 ′ = 21 H ′ 21 ≈

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2e 2rp2 5a3 e 2rp4

,

1120a5

.

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5028 Consider a free particle of mass m confined in a three-dimensional (3D) cubic box of length “a,” where the potential V(X) = 0 if 0 < X, Y, Z < a and infinite elsewhere. If the system is subjected to a perturbation that is given by Ĥ’ = V0δ(x − a/2)δ(y − a/4), evaluate the first order correction to the groundstate energy. Sol: The ground-state wave function for a particle in 3D box is given by

ψ gs = (2 / a)3/2 sin(π x / a) sin(π y / a) sin(π z / a) a

a

a

0

0

0

< Hˆ ′> = V0 (2 / a)3 ∫ dx sin 2 (π x / a)δ ( x − a / 2) ∫ dy sin 2 (π y / a)δ ( x − a / 4) ∫ dz sin 2 (π z / a) a a a < Hˆ ′> = V0 (2 / a)3 ∫ dx sin 2 (π x / a)δ ( x − a / 2) ∫ dy sin 2 (π y / a)δ ( x − a / 4) ∫ dz sin 2 (π z / a) . 0

0

0

These are three separate integrals, and using the property of Dirac delta integrals, i.e.,

∫

b

a

f ( x )δ ( x − c ) dx = f (c ), where [a ≤ c ≤ b] the first two integrals can be

evaluated. The third integral can be evaluated by Using the identity sin2 x = [1−cos(2π x)]/2 so,

∫

a

∫

a

∫

a

0

0

0

dx sin 2 (π x / a)δ ( x − a / 2) = sin 2 (π x / a)|x =a/2 = 1 dy sin 2 (π y / a)δ ( x − a / 4) = sin 2 (π y / a)|x =a/4 = 1/ 2 a

dz sin 2 (π z / a) = (1/ 2) ∫ dz (1 − cos2π z / a) = a / 2 0

So the correction to the energy of the ground state is given by 8/a031 × (1/2) × (a/2) = 2V0/a2.

5029 A hydrogen atom subjected to a perturbation Vpert = aL2, where L is the angular momentum. Evaluate the shift in the energy level of the 2P state when the effects of spin is neglected up to second order in a. Sol: The given perturbation is Hˆ ′ = aL2. The first order correction to the energy is given by = a.

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We know that L2|l,m > = ħ2l(l + 1)|l,m> which is an eigenvalue equation, where the eigenvalue turns out to be ħ2l(l + 1). Now, for a P state, l = 1 so, L2|l,m> = 2ħ2|l,m> and sandwiching 0  and the energy levels shift up on account of the perturbation. The shifts of energy levels of s states are larger than those of p and d states becuase a muon in s state has a greater probability of staying in the  r ∼ 0  region than a muon in p and d states. Besides, the larger the quantum number l, the greater is the corresponding orbital angular momentum and the farther is the spread of µ cloud from the center, leading to less energy correction. In Fig. 5.9, the solid lines represent unperturbed energy levels, while the dotted lines represent perturbed energy levels. It is seen that the unperturbed energy level of d state almost overlaps the perturbed energy level. b. The energy shift of 1s state to first order perturbation is given by ∆E1s = 1s H ′ 1s .

Fig. 5.9 As R  aµ , we can take e ∆E1s ≈

− r /aµ

4 N 02 Ze 2 4π

≈ 1 . Thus

∫

R

0

1 1  3 1 r2   4π r 2 dr  −  − 2   r R 2 2 R  

2

2  R  Ze 2 =   . 5  aµ  aµ

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c. By the same procedure, 2

1 Ze 2  R  ∆E2 s ≈ , 20 aµ  aµ  ∆E2 p ≈ 0, and so 2

1 Ze 2  R  , ∆E2 s − ∆E2 p ≈ ∆E2 s = 20 aµ  aµ  2

=

3

1 Ze 2  R   mµ  , 20 a0  a0   me 

where a0 is the Bohr radius. Thus by measuring the energy shift, we can deduce the value of R. Or, if we assume R = 10−13 cm, Z = 5 , we get ∆E2 s − ∆E2 p ≈ 2 × 10−2 eV . d. In the calculation in (b) the approximation R  aµ is used. If R is not much smaller than aµ , the calculation is not correct. In such a case, the actual energy shifts of p and d states are larger than what we obtain in (b) while the actual energy shifts of s states are smaller than those given in (b). In fact the calculation in (b) overestimates the probability that the muon is located

(

2

)

inside the nucleus  probability density ∝ ψ 1s (0) .

5032 a. Using an energy-level diagram give the complete set of electronic quantum numbers (total angular momentum, spin, parity) for the ground state and the first two excited states of a helium atom. b. Explain qualitatively the role of the Pauli principle in determining the level order of these states. c. Assuming Coulomb forces only and a knowledge of Z = 2 hydrogenic wave functions, denoted by 1s , 2 s , 2 p , etc., together with associated Z = 2 hydrogenic energy eigenvalues  E1s , E2 s , E2 p ,… ,  give perturbation formulas for the energies of these helium states. Do not evaluate integrals, but carefully explain the notation in which you express your result. (Berkeley)

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321

Sol: a. Figure 5.10 shows the ground and first two excited states of a helium atom in para (left) and ortho states with the quantum numbers (J, S, P). b. Pauli’s exclusion principle requires that a system of electrons must be described by an antisymmetric total wave function. For the two electrons of a helium atom, as the triplet states have symmetric spin wave functions the space wave functions must be antisymmetry. Likewise, the singlet states must have symmetric space wave functions. In the latter case, the overlap of the electron clouds is large and as the repulsive energy between the electrons is greater (because  r1 − r2  is smaller). So, the corresponding energy levels are higher.

Fig. 5.10 c. The Hamiltonian of a helium atom is  2 2  2 2 2e 2 2e 2 e2 Hˆ = − ∇1 − ∇2 − − + . r1 r2 2m 2m r1 − r2 Treating the last term as perturbation, the energy correction of 1s1s state is ∆E1s = 1s1s

e2 1s1s . r1 − r2

The perturbation energy correction of spin triplet states is  1 e2 ∆Enl(3) = ( 1snl − nl1s ) 1snl − nl1s )  ( 2 r1 − r2  =

1 e2 1 e2 1snl 1snl 1snl − nl1s 2 r1 − r2 2 r1 − r2 −

1 1 e2 e2 1snl nl1s + nlls nl1s 2 2 r1 − r2 r1 − r2

= 1snt

e2 e2 1snl − 1snl nl1s . r1 − r2 r1 − r2

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The perturbation energy correction of spin singlet states is ∆Enl(1) = 1snl

e2 1snl r1 − r2

+ 1snl

e2 nl1s . r1 − r2

The first term of the above result is called direct integral and the second term, exchange integral.

5033 A particle of mass m is confined to a circle of radius a, but is otherwise free. A perturbing potential H = Asinθ cosθ is applied, where θ is the angular position on the circle. Find the correct zero-order wave functions for the two lowest states of this system and calculate their perturbed energies to second order. (Berkeley) Sol: The unperturbed wave functions and energy levels of the system are respectively 1 inθ e , 2π n22 En = , n = ±1, ± 2, … 2ma 2

ψ n (θ ) =

The two lowest states are given by n = ±1, which correspond to the same energy. To first order perturbation, we calculate for the two degenerate states n = ±1 A 2π sinθ cosθ dθ = 0, ±1| H | ±1 = 2π ∫0 A 2π −2iθ +1| H | −1 = e sinθ cosθ dθ 2π ∫0 A 2π = ( cos2θ − i sin2θ )sin2θ dθ 4π ∫0 −iA = , 4 iA −1| H | +1 = . 4

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323

The perturbation matrix is thus  iA   0 −4   .  iA  0   4  Diagonalizing, we obtain ∆E (1) = ± A . Hence the two non-vanishing wave 4 functions and the corresponding energy corrections are A ψ 1′ = ( −1 + i 1 ) / 2, ∆E1(1) = . 4 A (1) ∆E2 = − . ψ 2′ = ( 1 + i −1 ) / 2, 4 To second order perturbation, the energy correction is given by

ψ k′ H ′ n ∆E (2) = ∑′ Ek − En n≠ k

2

.

As A sin2θ n 2 A 1 = (ei 2θ − e −i 2θ )einθ 4i 2π A 1 e i (n+2)θ − e i (n−2)θ  = 4i 2π  A A = n+2 − n−2 , 4i 4i

H′ n =

we have 2

2 ψ 1′ | n + 2 − ψ 1′ | n − 2  A ∆E = ∑ ′ ×  2  4  A n≠±1 (1 − n 2 ) + 2 2ma 4 1 2  A 2 ×  = 2  A  4 1 − (−3)2  + 2  2ma 4 1 2  A 2 + 2 ×     (1 − 32 ) + A4 4 2ma 2 ma 2 A2 ≈− , 64  2 (2) 1

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and similarly  ∆E2(2) ≈ −

ma 2 A2 . 64  2

Therefore A ma 2 A 2 2 + − , 2 2ma 4 64  2 2 A ma 2 A2 E2 = − − . 2 2ma 4 64  2 E1 =

5034 An electron at a distance x from a liquid helium surface feels a potential K V(x) = − , x > 0, K = constant, x V ( x ) = ∞, x ≤ 0. a. Find the ground state energy level. Neglect spin. b. Compute the Stark shift in the ground state using first order perturbation theory. (Berkeley) Sol: a. At  x ≤ 0 , the wave function is ψ ( x ) = 0. At  x > 0 , the Schrödinger equation is  2 d 2 K   − 2m dx 2 − x  ψ ( x ) = Eψ ( x ). In the case of the hydrogen atom, the radial wave function R(r) satisfies the equation   2 1 d  2 d  l(l + 1) 2  .  − 2m r 2 dr  r dr  + 2mr 2 + V (r ) R = ER   Let  R(r ) = χ (r )/ r . For  l = 0, the equation becomes  2 d 2 e 2   − 2m dr 2 − r  χ (r ) = E χ (r ). This is mathematically identical with the Schrödinger equation above and both satisfy the same boundary condition, so the solutions must also be the same (with r ↔ x , e 2 ↔ K ).

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325

As the wave function and energy of the ground state of the hydrogen atom are respectively me 4 , 2 2 2r χ10 (r ) = 3/2 e − r /a0 with a0 =  2 /me 2 , a0 E10 = −

the required wave function and energy are mK 2 , 2 2 2 ψ 1 ( x ) = 3/2 xe − x /a , a =  2 /mK . a E1 = −

b. Suppose an electric field ε e is applied in the x direction. Then the perturbation potential is V ′ = eε e x and the energy correction to the ground state to first order in perturbation theory is ∆E1 = ψ 1 V ′ ψ 1 ∞

2 2 xe − x /a ⋅ eε e x ⋅ 3/2 xe − x /a dx 3/2 a a 0

=∫

3 2eε e 3 . = eε e a = 2 2mK

5035 Discuss and compute the Stark effect for the ground state of the hydrogen atom. (Berkeley) Sol:

Suppose the external electric field is along the z-axis, and consider its potential as a perturbation. The perturbation Hamiltonian of the system is H ′ = eε ⋅ r = eε z . As the ground state of the hydrogen atom is nondegenerate, we can employ the stationary perturbation theory. To first order perturbation, the energy correction is E (1) = n = 1, l = 0,m = 0| eε z | n = 1, l = 0,m = 0 . l For the hydrogen atom the parity is (−1) , so the ground state (l = 0)  has even parity. Then as z is an odd parity operator,  E (1) = 0 .

To second order perturbation, the energy correction is given by E

(2)

=e ε 2

2

∑ n≠1 l ,m

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1, 0, 0 z n, l , m E1 − En

2

.

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As  En = E1 /n 2 , where  E1 = −

e2 2 , a = 2 , we have  E1 − En < 0, (n ≠ 1). Thus the 2a me

energy correction E (2)  is negative and has a magnitude proportional to ε 2 . Hence, increasing the electric field strength would lower the energy level of the ground state. We can easily perform the above summation, noting that only matrix elements with l = ± 1 m = 0  are non-vanishing.

5036 Describe and calculate the Zeeman effect of the hydrogen 2p state. (Berkeley) Sol: The change in the energy levels of an atom caused by an external uniform magnetic field is called the Zeeman effect. We shall consider such change for a hydrogen atom to first order in the field strength H. We shall first neglect any interaction between the magnetic moment associated with the electron spin and the magnetic field. The effect of electron spin will be discussed later. A charge e in an external magnetic field H has Harmiltonian 2

1  e  Hˆ =  P − A  + eφ , 2m c which gives the Schrödinger equation i

 ∂ψ  1 2 ie ie e2 2 =− ∇ + A ⋅∇ + ∇⋅ A + A + eφ  ψ .  ∂t  2m mc 2mc 2mc

As H is uniform it can be represented by the vector potential 1 A = H×r 2

1 since  H = ∇ × A. Then ∇⋅ A = (r ⋅∇ × H − H ⋅∇ × r ) = 0  and so the only terms 2 involving A that appear in the Hamiltonian for an electron of charge −e and reduced mass  µ  are −

ie e2 e e2 2 A ⋅∇ + A H r = × ⋅ + ( ) P (H × r ) ⋅ (H × r ) µc 2 µc 2 2 µc 8 µc 2 =

e e2 H 2r 2 sin 2θ , H⋅L + 2 2 µc 8 µc

where  L = r × P and  θ  is the angle between r and H.

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To first order in H, we can take the perturbation Hamiltonian as e Hˆ ′ = H ⋅ L. 2 µc Taking the direction of the magnetic field as the z direction we can choose for the energy eigenfunctions of the unperturbed hydrogen atom the eigenstates of Lz with eigenvalues  m , where m is the magnetic quantum number. Then the energy correction from first order perturbation is  ′ | m = e Hm . Wl ′= m | H 2 µc Thus the degeneracy of the 2l + 1 states of given n and l is removed in the first order. In particular, for the 2p state, where l = 1, the three-fold degeneracy is removed. We shall now consider the effect of electron spin. The electron has an intrinsic magnetic moment in the direction of its spin, giving rise to a magnetic moment operator −(e / mc )S. For a weak field, we shall consider only the first order effects of H. The Hamiltonian is 2 2 Hˆ = − ∇ + V (r ) + ξ (r )L ⋅ S + ε (lz + 2sz ) 2m

with

ε=

eH , 2mc

where the field is taken to be along the z-axis. 2 We choose the following eigenfunctions of  J  and  J z as the wave functions:      2 P3  2     

m= 1 2

3 (+ )Y1,1 , 2 − 3 2  2 2 (+ )Y1,0 + (− )Y1,1  , 1

1 2 3 − 2

−

1

− 3 2  2 2 (− )Y1,0 + (+ )Y1,−1  , 1

1

(− )Y1,−1 ,

 1 1 − 12  3  −(+ )Y1,0 + 2 2 (− )Y1,1  ,  m=  2 2 P1  2 1 1 −1  − 3 2 (− )Y1,0 + 2 2 (+ )Y1,−1  ,  2  1 (+ )Y0,0 ,  m=  2 2 S1  2  −1 (− )Y0,0 ,  2

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where Y0,0 ,Y1,0 ,Y1,1  and  Y1,−1  are spherical harmonic functions, + ( )  and  (− )  are spin wave functions. It can be shown that the magnetic energy  ε (lz + 2sz ) = ε ( J z + sz ) has non-vanishing matrix elements between states of different j, but not between states of the same j and different m. We can neglect the former because of the relatively large energy separation between states of different j. Thus the magnetic energy is diagonal with respect to m for each j and shifts the energy of each state above by its expectation value for the state. In each case,  J z  is diagonal, and so its expectation value is m . The expectation value of sz  for the  P3/2  state with m = 1/ 2 , for example, is 1 1 1 −1 −1 † † ∫∫ 3 2  2 2 (+ ) Y1,0* + (− ) Y1,1*  2 σ z 3 2  2 2 (+ )Y1,0 + (− )Y1,1  sinθ dθ dφ 1 1  = ∫ ∫  2 2 (+ )† Y1,0* + (− )† Y1,1*   2 2 (+ )Y1,0 − (− )Y1,1  sin θ dθ dφ   6   = (2 − 1) = . 6 6

 1 1 2 Hence, the magnetic energy of this state is ε   +  = ε .  2 6 3 This and similar results for the other states can be expressed in terms of the Landé g− factor as ε mg, with g=

4 for 3

2

P3/2 , g =

2 for 3

2

P1/2 , g = 2 for

2

S1/2.

5037 Explain why excited states of atomic hydrogen can show a linear Stark effect in an electric field, but the excited states of atomic sodium show only a quadratic one. (MIT) Sol: The potential energy of the electron of the atom in an external electric field E is H ′ = eE ⋅ r . If we make the replacement  r → − r  in  l ′ | H ′ | l , as the value of the integral does not change we have l ′ | H ′ | l (r ) = l ′ | H ′ | l (− r ) = (−1)l ′+l+1 l ′ | H ′ | l (r ). This means that if the  l ′  and l states have the same parity (i.e. l and  l ′  are both even or both odd) we must have  l ′ H ′ l = 0.

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329

If the electric field is not too small, we need not consider the fine structure of the energy spectrum caused by electron spin. In such cases, an exited state of the hydrogen atom is a superposition of different parity states, i.e. there is degeneracy with respect to l and the perturbation theory for degenerate states is to be used. Because of the existence of non-vanishing perturbation Hamiltonian matrix elements, exited states of the hydrogen atom can show a linear Stark effect. For exited states of atomic sodium, each energy level corresponds to a definite parity, i.e., there is no degeneracy in l. When we treat it by nondegenerate perturbation theory, the first order energy correction  l ′ | H ′ | l  vanishes. We then have to go to second order energy correction. Thus the exited states of atomic sodium show only quadratic Stark effect.

5038 The Stark effect. The energy levels of the n = 2  states of atomic hydrogen are illustrated in Fig. 5.11.

Fig. 5.11 The S1/2  and  P1/2  levels are degenerate at an energy ε 0  and the  P3/2  level is degenerate at an energy  ε 0 + ∆ . A uniform static electric field E applied to the atom shifts the states to energies ε1 , ε 2  and  ε 3. Assuming that all states other than these three are far enough away to be neglected, determine the energies ε1 , ε 2 and ε 3 to second order in the electric field E. (Princeton) Sol: Suppose the matrix elements of the perturbation Hamiltonian  H ′ = −eE ⋅ r  are P3/2 P1/2 S1/2 P3/2

0

0

b

P1/2

0

0

a

S1/2

b∗

a∗

0

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since  l ′ | H ′ | l = 0  for  l ′ , l  states of the same parity (Problem 5037). Then for energy level  P3/2 , we have P3/2 | H ′ | S1/2

EP 3/2 = EP(0)3/2 +

2

EP(0)3/2 − ES(0) 1/2

| b |2 . ∆ For energy levels  P1/2  and  S1/2 , we diagonalize the Hamiltonian in the correspond­ = ε0 + ∆ +

ing subspace, i.e, solve −λ a = 0. a* − λ The roots are  λ = ± | a |, which give the new wave functions 1 =

 a P + | a | S1/2 1  a P1/2 + S1/2  = 1/2 ,  2  |a| 2 |a| 

| 2〉 =

 −a P1/ 2 + | a || S1 2 1  a , − P + S 12 12  = 2 |a| 2  | a | 

with energies E1 = E1(0) + | a | +

1| H ′ | P3/2 E1(0) − EP(0)3/2

= ε0 + | a | +

| b |2 2(−∆ )

= ε0 + | a | −

| b |2 , 2∆

E2 = E2(0) − | a | + = ε0 − |a | −

2

2| H ′ | P3/2

2

E2(0) − EP(0)3/2

| b |2 . 2∆

5039 The Stark effect in atoms (shift of energy levels by a uniform electric field) is usually observed to be quadratic in the field strength. Explain why. But for some states of the hydrogen atom the Start effect is observed to be linear in

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the field strength. Explain why. Illustrate by making a perturbation calculation of the Stark effect to lowest non-vanishing order for the ground and first excited states of the hydrogen atom. To within an uninteresting overall constant, the wave functions are

ψ 100 = 4 2a0 e − r / a0 , ψ 200 = (2a0 − r )e − r /2 a0 , ψ 21±1 = ± re − r /2 a0 sinθ e ± iφ / 2 , ψ 210 = re − r /2 a0 cosθ . (Wisconsin) Sol: The electric dipole moment of an atomic system with several electrons is d = − ∑ei ri j

In general, the energy levels do not have degeneracy other than with respect to lz. The energy depends on the quantum numbers n and l. As the perturbation Hamiltonian  H ′ = −d ⋅ E  is an odd parity operator, only matrix elements between opposite parity states do not vanish. Thus its expectation value for any one given state is zero, i.e., nlm | − d ⋅ E | nlm′ = 0. This means that first order energy corrections are always zero and one needs to consider energy corrections of second order perturbation. Hence, the energy corrections are proportional to E2. As regards the hydrogen atom, degeneracy occurs for the same n but different l. And so not all the matrix elements  nl ′ | H ′ | nl  between such states are zero. So, shifts of energy levels under first order (degenerate) perturbation may be nonzero, and the Stark effect is a linear function of the electric field E. Write the perturbation Hamiltonian in spherical coordinates, taking the z-axis in the direction of E. As H ′ = eEz = eEr cosθ , the ground state, which is not degenerate, has wave function

ψ 100 = 4 2a0e − r /a 0, and so (Problem 5037) V (1) = 100| H ′ |100 = 0 .

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The second order energy correction is | H ′ |2 V (2) = ∑′ (0) n 0 (0) n≠1 E1 − En 2

=e E

2

nl 0| z |100

∞

∑

2

1  e2   1 − 2  n 2a

n= 2

9 9 = 2aE 2 ⋅ a 2 = a3 E 2 . 8 4 Note that for  H n0 ′ ≠ 0  we require  ∆l = ±1. The first exited state  n = 2  is four-fold degenerate, the wave functions being ψ 200 , ψ 210 , ψ 21,±1. As (l − m + 1)(l + m + 1) 3a0n 2 2 ⋅ n −l (2l + 1)(2l + 3) 2 = −3eEa0

eE n, l + 1, m z n, l , m = −eE

are the only non-vanishing elements of H ′  for  n = 2 , we have the secular equation

H ′ − E (1) I =

− E (1)

−3eEa0

0

−3eEa0

− E (1)

0

0 0

0 0

−E 0

0 0 (1)

0 − E (1)

= 0,

which gives the energy corrections E (1) = ± 3eEa , 0,0 . Therefore, the energy level for  n = 2  splits into −

e2 1 ⋅ + 3eEa0 , 2a0 22 −

e2 − 3eEa0 , 8a0

2 . The splitting of the first excited state n = 2  is me 2 shown in Fig. 5.12. Note that the two other states are still degenerate. where a0  is the Bohr radius 

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Fig. 5.12 5040 Consider an ionized atom ( Z , A)  with only a single electron remaining. Calculate the Zeeman splitting in the  n = 2  state in a “weak” magnetic field a. for an electron b. for a hypothetical  spin = 0 particle with electron mass. c. Calculate the first-order Stark effect (energy levels and wave functions) for an electron in the  n = 2  state. (After you define the radial integrals you can express the term by a parameter; you need not evaluate them. The same holds for nonzero angular integrals.) (Berkeley) Sol: a. Take the direction of the external magnetic field as the z direction. For an electron and a weak external magnetic field, in comparison with its effect the spin-orbit coupling cannot be neglected, which gives rise to anomalous Zeeman effect. The Hamiltonian of the system P2 Ze 2 eB  Hˆ = − + (l z + 2sˆz ) + ξ (r )sˆ ⋅ l ß 2me r 2me c can be written as

eB  eB Hˆ = H 0 + j + sˆz , 2me c z 2me c

with H0 ≡

p2 Ze 2 − + ξ (r )ˆs ⋅ l, j z = l z + sˆz . r 2me

Before applying the magnetic field, we have 1  H 0ψ nl jm j = Enl jψ nl jmj ⋅  j = l ±   2

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If we neglect the term 

(

)

eB sˆz , Lˆ2 , ˆJ 2 ,ˆjz  are still conserved quantities. Then 2me c

jm j | j z | jm j = m j   and the energy of the system is Enl j + m j ω L , where

ωL =

eB . 2me c

When the weak magnetic field is applied, the contribution of the term eB sˆz  is (Problem 5057) 2me c ω L jm j | σˆ z | jm j 2  mj 1 ω L , j=l+ ,  j 2 2  =  − m j ω , j = l − 1 .  2j+2 L 2 

ω L sz =

Hence,

Enl jm j

   = Enl j +    

 1  1 + 2 j  m j ω L ,

1 j=l+ , 2

 1   1 − 2 j + 2  m j ω L ,

1 j=l− . 2

For n = 2 , we have  1  E20 12m j = E20 12 + 2m j ω L , m j = ± , 2  4 3 1   E2132m j = E2132 + m j ω L , m j = ± , ± , 3 2 2  2 1   E2112mj = E2112 + 3 m j ω L , m j = ± 2 .  b. When spin = 0, there is no spin-related effect so that Pˆ 2 eB  Hˆ = + V (r ) + lz . 2me 2me c The eigenfunction is ψ nlm (r , θ , ϕ ) = Rnlm (r )Ylm (θ , ϕ ),

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and the energy eigenvalue is Enlm = Enl +

eB m . 2me c

For n = 2, E200 = E20 , E210 = E21 , E21,±1 = E21 ±

eB . 2me c

c. See the solution of Problem 5042.

5041 Stark showed experimentally that, by applying an external weak uniform electric field, the 4-fold degeneracy in the n = 2  level of atomic hydrogen could be removed. Investigate this effect by applying perturbation theory, neglecting spin and relativistic effects. Specifically: a. What are the expressions for the first order corrections to the energy level? (Do not attempt to evaluate the radial integrals). b. Are there any remaining degeneracies? c. Draw an energy level diagram for n = 2  which shows the levels before and after application of the electric field. Describe the spectral lines that originate from these levels which can be observed. (Chicago) Sol:

Write the Hamiltonian of the system as  H = H 0 + H ′ , where e2 L2 + , r 2mr 2 H ′ = eEz ,

H0 = −

taking the direction of the electric field E as the z direction. For a weak field, H ′  H 0  and we can treat  H ′  as perturbation. Let (0, 0), (1, 0), (1,1)  and  (1, − 1)  represent the four degenerate eigenfunctions (l ,m)  of the state  n = 2  of the hydrogen atom.

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The matrix representation of  H ′  in the subspace is    H′ =    where

0

l, 0| H ′ |1, 0

l, 0| H ′ |0, 0

0

0 0

0 0

0 0   0 0 ,  0 0  0 0 

1, 0| H ′ |0, 0 = 0, 0| H ′ |1, 0 * * (r )rcosθ u200 (r )d 3r = eE ∫ u210 = −3eEa0 , a0 =

2 me 2

being the Bohr radius. Note that  l ′ | H ′ | l = 0  unless the  l ′ , l  states have opposite parties. Solving the secular equation −w1

0,0 H ′ 1,0

0

0

1,0 H ′ 0,0

−w1

0

0

0

0

− w1

0

0

0

0

−w1

= 0,

we get four roots w1(1) = 3eEa0 , w1(2) = w1(3) = 0, w1(4) = 3eEa0 . a. The first order energy corrections are thus  3eEa0 ,   0, ∆E = w1 =   0,  −3eEa0 .  (2) (3) b. As  ω 1 = ω 1 = 0 , there is still a two-fold degeneracy. c. Figure 5.13 shows the n = 2  energy levels. The selection rules for electric dipole transitions are ∆l = ±1, ∆m = 0, ± 1, which give rise to two spectral lines: hν1 = 3ea0 E , hν 2 = 2 × 3ea0 E ,

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ν1 = 3ea0 E /h; ν 2 = 6ea0 E /h.

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Fig. 5.13 5042 Consider the n = 2 levels of a hydrogen-like atom. Suppose the spins of the orbiting particle and nucleus to be zero. Neglect all relativistic effects. a. Calculate to lowest order the energy splittings in the presence of a uniform magnetic field. b. Do the same for the case of a uniform electric field. c. Do the same for both fields present simultaneously and at right angles to each other. (Any integral over radial wave functions need not be evaluated; it can be replaced by a parameter for the rest of the calculation. The same may be done for any integral over angular wave function, once you have ascertained that it does not vanish.) (Berkeley) Sol: a. Take the direction of the magnetic field as the z direction. Then the Hamiltonian of the system is eB ˆ 1 2 p + V (r ) + lz , 2me 2me c eB  e2 l z  as perturbation, the eigenfuncwhere V (r ) = − . Considering  H ′ = 2me c r tions for the unperturbed states are H=

ψ nlm (r , θ , ϕ ) = Rnl (r )Ylm (θ , ϕ ), with n = 1, 2, 3 … ,

l = 0,1, 2,…, n − 1. m = −l , − l + 1, ... l − 1, l .

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As ( H , l 2 , lz )   are still conserved quantities,  nlm | l z | nlm = m  and the energy splittings to first order for  n = 2  are given by eB m E2lm = E2l + 2mc c  E20 ,   eB  ,   2me c   =   E21 +  0,  −eB  ,    2me c  

l = 0;  m = 1,  l = 1  m = 0,  m = −1. 

b. The energy level for n = 2  without considering spin is four-fold degenerate. The corresponding energy and states are respectively E2 = −

Z 2e 2 1 , ψ 200 ,ψ 210 ,ψ 211 ,ψ 21−1. 2a0 22

Suppose a uniform electric field is applied along the z-axis. Take as perturbation 2 H ′ = eε z = E0V ′, where  E0 = eε a0 , V ′ = z / a0 = r cosθ / a0 , a0 =  2 . Since me e cosθYlm =

(l + 1)2 − m 2 Yl+1,m (2l + 1)(2l + 3) l 2 − m2 + Yl−1,m (2l + 1)(2l − 1)

,

Hnl′ ′m′ ,nlm ≠ 0  for only  ∆l = ±1, ∆m = 0. Hence, the non-vanishing elements of the perturbation matrix are * H ′ψ d 3 x = ψ H ′ψ d 3 x, ( H ′ )200,210 = ∫ψ 200 210 ∫ 200 210 * H ′ψ d 3 x = ψ H ′ψ d 3 x. ( H ′ )210,200 = ∫ψ 210 200 ∫ 210 200 Let  ( H ′ )200,210 = ( H ′ )210,200 = E ′ , i.e., H 01 = H10 = E ′ , and solve the secular equation det H µυ − E (1)δ µυ = 0 . The roots are E (1) = ± E ′, 0, 0 . Hence, the energy state n = 2  splits into three levels: E2 ± E ′ , E2

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( two-fold degeneracy for E2 ) .

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c. Assuming that the magnetic field is along the z-axis and the electric field is along the x-axis, the perturbation Hamiltonian of the system is eB  H′ = l z + eε x = β lz /  + 2γ x / 3a, 2me c where

β = eB /2me c , γ = 3eε a/ 2, a = a0 / Z . The non-vanishing matrix elements of x are ( x )ll−, 1,mm−1 = (x)ll−,m1,−m1 3 (n 2 − l 2 )(l − m + 1)(l − m) = n a, 4 (2l + 1)(2l − 1) ( x )ll−,m1,m−1 = ( x )ll−,m1,m−1 3 (n 2 − l 2 )(l + m − 1)(l + m) =− n a. 4 (2l + 1)(2l − 1) Thus, for  n = 2, 11 x 00 =−

x100−1 =

3 a = x1100 , 2

3 1−1 a = x 00 , 2

and the perturbation matrix is l =1   m = 1  l =1 m=0

  

 l =1  m = −1 

 β   0  0   −γ

0

0

0 0 0 −β 0

γ

−γ   0  γ   0 

l=0   m = 0  The secular equation

β − E (1) det

0 −γ

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−γ

0

β−E γ

(1)

γ −E

=0 (1)

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has roots (1) E1(1) = 0, E2,3 = ± β 2 + 2γ 2 .

Hence, the energy state  n = 2  splits into three levels, of energies E2 , E2 ± β 2 + 2γ 2 .

5043 A nonrelativistic hydrogen atom, with a spinless electron, is placed in an electric field   in the z direction and a magnetic field   in the x direction. The effect of the two fields on the energy levels are comparable. a. If the atom is in a state with n, the principal quantum number, equal to two, state which matrix elements in the first-order perturbation calculation of the energy shifts are zero. b. Now obtain an equation for the energy shifts; once you have the determinantal equation you need not go through the algebra of evaluating the determinant. Do not insert the precise forms of the radial wave functions; express your results in terms of matrix elements of r n  (where n is an appropriate power) between radial wave functions. (  x ± i  y ) , m = {(  ∓ m)(  ± m + 1)} , m ± 1 . (Berkeley) Sol: a. The perturbation Hamiltonian is H′ =

eB  l x + eε zˆ. 2mc

Let the state vectors for  n = 2  be  200 , 210 , 211 , 21, − 1 . As ( x ± i y )  ,m = {( ∓ m)( ± m + 1)}  ,m ± 1 , we have l x 0,0 = 0, l x 1,1 = l x 1, −1 = 2  1,0 , 2 2 l x 1,0 = { 1,1 + 1, −1 }. 2

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As  z = r cosθ , we have 1 r 3 ∞ with  r = ∫ r 3 R20 R21 dr , other matrix elements of z being zero. Hence, the 210| rcosθ | 200 = 200| rcosθ | 210 =

0

perturbation matrix is      H ′ =      

1 eε r 3

0

0

2eB 4mc

0

2eB 4mc

0

0

2eB 4mc

0

0 1 eε r , 3

   2eB   4mc  .  0    0  0

b. The secular equation  H − λI = 0, i.e., −λ α

α −λ

0

β β

0

0 0 β β = 0, −λ 0 0 −λ

2eB , has roots λ = 0,0, ± 2β 2 + α 2 , which are where α = 1 eε r , β = 3 4mc the energy shifts. Note that a two-fold degeneracy still remains.

5044 Two non-identical particles, each of mass m, are confined in one dimension to an impenetrable box of length L. What are the wave functions and energies of the three lowest-energy states of the system (i.e., in which at most one particle is excited out of its ground state)? If an interaction potential of the form V12 = λδ ( x1 − x 2 ) is added, calculate to first order in λ  the energies of these three lowest states and their wave functions to zeroth order in λ . (Wisconsin) Sol: Both particles can stay in the ground state because they are not identical. The energy and wave function are respectively

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E11 =

2 πx  2π 2 πx , ψ 11 = sin 1 sin 2 . 2 mL L L L

If one particle is in the ground state, the other in the first exited state, the energies and corresponding wave functions are 5  2π 2 , 2mL2 5  2π 2 E21 = , 2mL2

2 πx 2π x 2 ψ 12 = sin 1 sin , L L L 2 2π x1 πx sin 2 . ψ 21 = sin L L L

E12 =

When both particles are in the single-particle ground state, i.e., the system is in the ground state, we have the energy correction E (1) = (ψ 11 ,V12ψ 11 ) =

L 4 3λ  πx  λ ∫ sin 4  1  dx1 = , 2 0   L L 2L

and the wave function to zeroth order in λ φ11 = ψ 11 . When one particles is in the ground state and the other in the first excited state, the energy level is two-fold degenerate and we have to use the perturbation theory for degenerate states. We first calculate the elements of the perturbation Hamiltonian matrix:

∫ ∫ψ * V ψ 12 12

12

* V12ψ 21dx1dx 2 dx1dx 2 = ∫ ∫ψ 21

L πx λ 4 2π x1 λ ∫ sin 2 1 sin 2 dx1 = , 2 0 L L L L λ ∫ ∫ψ 12* V12ψ 21dx1dx 2 = ∫ ∫ψ 21* V12ψ 12dx1dx 2 = L .

=

We then solve secular equation

λ − E (1) L det λ L

λ L

λ − E (1) L

= 0,

and obtain the roots E+(1) =

2λ , L

E−(1) = 0,

which are the energy corrections. The corresponding zeroth order wave functions are 1 φ12 = (ψ 12 ± ψ 21 ). 2

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5045 Consider a three-level system described by the Hermitian Hamiltonian H = H0 + λH1, where λ  is a real number. The eigenstates of H0  are  |1〉,| 2〉  and  | 3〉 , and H0 1 = 0, H0 2 = ∆ 2 , H0 3 = ∆ 3 . a. Write down the most general 3 × 3  matrix representation of H1  in the

{ 1 , 2 , 3 }  basis.

b. When the spectrum of H is computed using perturbation theory, it is found that the eigenstates of H to lowest order in λ are 1 1,± ≡ ( 2 ± 3 )  and that the corresponding eigenvalues are 2

λ2 + O( λ 3 ), ∆ λ2 E+ = ∆ + λ + + O( λ 3 ), ∆ E− = ∆ − λ + O( λ 3 ). E1 = −

Determine as many of the matrix elements of H1  from part (a) as you can. (Buffalo) Sol: a. Since λ  is a real number, the Hermition perturbation Hamiltonian matrix has the form   a d e   H1 =  d * b f ,  e* f * c    where a,b,c are real numbers.

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b. To first order approximation, energy eigenvalue is the expectation value of the Hamiltonian with respect to the selected state vectors. Thus E + = + | H 0 + λ H1 | + = + | H 0 | + + λ + | H1 | + .

λ2 + O( λ 3 ) ∆ Comparing the coefficients of λ  gives = ∆+λ+

+ | H1 | + = 1. Similarly − | H1 | − = −1. As the energy levels corresponding to  2  and  3  are degenerate, we ­transform to the following state vectors in whose representation the degeneracy disappears, 1 ± = ( 2 ± 3 ). 2 Thus H1  is transformed to a representation in basic vectors  1 , + , − :    1   0    0 

0 1 2 1 2

 a    d ∗ + e∗ = 2   d ∗ − e∗  2 

  0  1  2  −1  2  d+e 2 1 0

  a d e   f   d* b  e* f * c   

   1   0    0 

0 1 2 1 2

  0  1  2  −1  2 

d−e   2   0 .   −1  

In the above we have used 1 + | H1 | + = (b + f + f * +c ) = 1, 2 1 − | H1 | − = (b − f − f * +c ) = −1 2 and chosen the solution b = c = 0, f = f * = 1.

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Perturbation theory for nondegenerate states gives Em = Em(0) + λ Hmm ′ + λ 2 ∑′ n≠m

2

Hmn ′ + O(λ 3 ) . Em − En

Thus 2

2

λ2 d + e λ2 d − e + + O( λ 3 ) 2(0 − ∆ ) 2(0 − ∆ )

E1 = 0 + λ a +

 d+e 2 d−e 2  = λa − λ 2  + + O(λ 3 ), 2 ∆   2∆ 2

λ2 d + e + O(λ 3 ), 2∆ 2 λ2 d − e E3 = ∆ − λ + + O(λ 3 ). 2∆

E2 = ∆ + λ +

Identifying  E1 , E2 , E3  with the given energies E1 , E + , E−  and comparing the coefficients of λ  and  λ 2  give  a = 0  and | d + e |2 + | d − e |2 = 2, | d + e |2 = 2, | d − e |2 = 0, or  d + e = 2e iδ , d − e = 0, where δ  is an arbitrary constant. iδ Hence  a = 0, d = e = e  and 2     H1 =     

0 1 − iδ e 2 1 − iδ e 2

1 iδ e 2 0 1

1 iδ  e  2   1    0  

is the representation in state vectors  1 , 2 , 3 .

5046 1 Two identical spin −  fermions are bound in a three-dimensional isotropic 2 harmonic oscillator potential with classical frequency ω . There is, in addition, a weak short-range spin-independent interaction between the fermions. a. Give the spectroscopic notation for the energy eigenstates up through energy 5ω  (measured from the bottom of the well).

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b. Write the appropriate approximate (i.e., to lowest order in the interaction) wave functions of the system, expressed in terms of singleparticle harmonic oscillator wave functions, for all states up through energy  4 ω . c. For a specific interparticle interaction V12 = − λδ 3(r1 − r2 ) , find the energies of the states of (b) correct to first order in λ . You may leave your result in the form of integrals. (Wisconsin) Sol: a. For a three-dimensional harmonic oscillator, 3  En =  n +  ω ,  2

n = 2nr + l ,

where nr  and l are integers not smaller than zero. For the system of two identical fermions in harmonic oscillator potential, we have, from the Hamiltonian, 3 3   EN =  n1 +  ω +  n2 +  ω = ( N + 3)ω , N = n1 + n2 .    2 2 Consequently, for E0 = 3ω , (l1 , l2 ) = (0, 0), 1

there is only one state  S0 ; for  1 1 E1 = 4 ω , (l1 , l2 ) = (0, 0) or (0,1), (s1 , s2 ) =  ,  ,  2 2 there are two states  1 P1  and  3 P210 ; for E2 = 5ω , and 1. (n1 , n2 ) = (2, 0)  or (0, 2),  (l , l ) = (0,0), there are two states 1S0 , 3 S1 ;  1 2  1 3  (l1 , l2 ) = (2,0) or (0,2), there are two states D2 , D321 ; 2. (n1 , n2 ) = (1,1), (l1 , l2 ) = (1,1), there are three states  1 S0 , 1D2 , 3 P210 .

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b. Let ψ 0  be the ground state and ψ 1m  the first exited state of a single-particle system, where m = 0, ± 1 , and χ 0  and  χ1m  be spin singlet and triplet states. With the states labeled as  NLLz SSz , the wave functions required are 00000 = χ 0ψ 0 (1)ψ 0 (2), state 1S0 ; 1 11m00 = χ 0 (1 + P 12 )ψ 0 (1)ψ 1m (2), state 1 P1 ; 2 1 11m1M = χ1 M (1 − P 12 )ψ 0 (1)ψ 1m (2), state 3 P210 ; 2

where  m, M = 0, ± 1 ( Lz = m, Sz = M ) .

c. For the ground state  1 S0 , the energy correction to first order in λ  is 1

〈1 S0 V12 S0 〉 ≈ − λ ∫ d r1 d r2δ (r1 − r2 )[ψ 0 (r1 )ψ 0 (r2 )]2  α  = − λ ∫ d rψ 04 (r ) = − λ   2π 

3

with α = mω / . Hence the ground state energy is E

(

1

 mω  S0 = 3ω − λ   2π  

)

3/2

.

The first exited state consists of 12 degenerate states (but as there is no spin in V12 , 〈1 P1 | V12 |3 P1 〉 = 0) . As the spatial wave function is antisymmetric when S = 1 , the expec­tation

value of − λδ 3 (r1 − r2 )  equals to zero, i.e., 〈11m′1M ′ | V12 |11m1M 〉 = δ M ′M 〈1m′ | V12 |1m〉 = 0

11m′1M ′ | V12 |11m1M 〉 = δ M ′M 〈1m′ | V12 |1m〉 = 0 . As E

(

3

)

P210 = 4 ω ,

11m′00| V12 |11m00 = − λ ∫

4 drψ 02 (r)ψ 1*m ′ (r)ψ 1m (r ) 2

= −2λ ∫ dr |ψ 0 (r )ψ 1m (r )|2 δ m ′m 3

 α  δ , = −λ   2π  m ′m we have E

(

1

 mω  P1m = 4 ω − λ   2π  

)

3/2

,

Where m is the eigenvalue of Lz .

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5047 The Hamiltonian for an isotropic harmonic oscillator in two dimensions is H = ω ( n1 + n2 + 1), where  ni = ai+ ai , with [ ai , a +j ] = δ ij  and  [ ai , a j ] = 0 . a. Work out the commutation relations of the set of operators  {H , J1 , J2 , J3 } where 1 i J1 = ( a2+ a1 + a1+ a2 ), J2 = ( a2+ a1 − a1+ a2 ), 2 2 1 J3 = ( a1+ a1 − a2+ a2 ). 2 b. Show that  J2 ≡ J12 + J22 + J32  and  J3  form a complete commuting set and write down their orthonormalized eigenvectors and eigenvalues. c. Discuss the degeneracy of the spectrum and its splitting due to a small perturbation V . J where V is a constant three-component vector. (Buffalo) Sol: a. The system can be considered a system of bosons, which has two singleparticle states. The operators ai+  and  ai   are respectively creation and destruction operators. As among their commulators only [ai , ai+ ]  is not zero, we can use the relation [ab, cd] = a[b, c]d + ac[b, d] + [a, c]bd + c[a, d]b to obtain [a1+ a1 , a1+ a1 ] = [a2+ a2 , a2+ a2 ] = [a1+ a1 , a2+ a2 ] = 0, [a1+ a1 , a2+ a1 ] = −[a2+ a2 , a2+ a1 ] = − a2+ a1 , [a1+ a1 , a1+ a2 ] = −[a2+ a2 , a1+ a2 ] = a1+ a2 , [a2+ a1 , a1+ a2 ] = a2+ a2 − a1+ a1 . Hence, [H , J1 ] = [H , J 2 ] = [H , J 3 ] = 0, [ J1 , J 2 ] = iJ 3 , [ J 2 , J 3 ] = iJ1 , [ J 3 , J1 ] = iJ 2 . b. The above commutation relations show that  J1 , J 2 , J 3  have the same properties 2 as the components of the angular momentum L. Hence J  and  J 3  commute

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and form a complete set of dynamical variables of the two-dimensional system. The commutation relations of  a, a + , [ai , a +j ] = δ ij , [ai , a j ] = 0, can be satisfied if we define a1 | n1 , n2 〉 = n 1 | n1 − 1, n2 〉,

a2 | n1 , n2 〉 = n 2 | n1 , n2 − 1〉

+ 1

a2+ | n1 , n2 〉 = n2 + 1 | n1 , n2 + 1〉

a | n1 , n2 〉 = n1 + 1 | n1 + 1, n2 〉, and thus −

1

| n1 , n2 〉 = (n1 !n2 !) 2 (a1+ )n1 (a2+ )n 2 |0, 0〉 . These can be taken as the common normalized eigenvectors of the complete 2 set of dynamical variables  J  and  J . As 3

1 J 2 = {(a2+ a1 + a1+ a2 )2 − (a2+ a1 − a1+ a2 )2 + (a1+ a1 − a2+ a2 )2 } 4 1 + + = {2a2 a1a1 a2 + 2a1+ a2a2+ a1 + a1+ a1a1+ a1 4 + a2+ a2a2+ a2 − a1+ a1a2+ a2 − a2+ a2a1+ a1 } 1 = {a2+ a1a1+ a2 + a1+ a2a2+ a1 + a2+ a2[a1 , a1+ ] 4 + a1+ a1[a2 , a2+ ] + a1+ a1a1+ a1 + a2+ a2a2+ a2 } 1 = {a1+ a2a1+ a2 + a2+ a1a2+ a1 + a2+ a2 + a1+ a1 + a1+ a1a1+ a1 + a2+ a2a2+ a2 }, 4 where use has been made of a2+ a1a1+ a2 = a2+ a1a2a1+ = a2+ a2a1a1+ , etc., and a1+ a2 | n1 , n2 〉 = (n1 + 1)n2 | n1 ,n2 〉, a1+ a1 | n1 , n2 〉 = n1 a1+ | n1 − 1, n2 〉 = n1 | n1 , n2 〉 , etc, we find 2 1 J | n1 , n2 〉 = [(n1 + 1)n2 + (n2 + 1)n1 + n12 + n22 + n1 + n2 ]| n1 , n2 〉 4 1 1 = (n1 + n2 )  (n1 + n2 ) + 1 | n1 , n2 〉 2 2 

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and J | n , n 〉 = 1 (a + a − a + a )| n , n 〉 z 1 2 1 1 2 2 1 2 2 1 = (n1 − n2 )| n1 , n2 〉. 2 2  Thus the eigenvalues of J , J z are respectively 1 1 J 2 = (n1 + n2 )  (n1 + n2 ) + 1 2 2  1 J z = (n1 − n2 ). 2 Furthermore with n1 = j + m, n2 = j − m, the above give

2 J | j , m〉 = j( j + 1)| j , m〉,

J z | j , m〉 = m | j , m〉. c. Energy levels with the same value of J are degenerate. The situation is exactly analogous to that of the general angular momemtum. Adding the perturbation V . J will remove the degeneracy because the different energy levels have different value of JV in the direction of the vector V.

5048 Consider a system in the unperturbed state described by a Hamiltonian  1 0  H ( 0) =  . The system is subjected to a perturbation of the form  0 1   δ H′ =   δ

δ  . Find the energy eigenvalues of the system using first order δ 

perturbation theory. Sol:  1 0  , the eigenvalues of H(0) give the unperturbed energies. H (0) =   0 1   As this is a diagonal matrix the eigenvalue is the diagonal elements of this matrix. The unperturbed energies are given by (1,1)

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351

 δ δ  δ  , the eigenvalue H 1 gives =  the first  order correction to the energies,  δ   δ δ   δ −λ δ  and to find the eigenvalues, H ′ − λ I = 0, i.e.,  = 0, δ − λ   δ  δ H1 =   δ

which gives the values 0, 2δ Therefore, the final energies after the correction is just the sum of unperturbed and perturbed eigenvalues, which is given by 1, 1+2δ.

5049 A particle of mass m moves (nonrelativistically) in the three-dimensional potential 1 V = k ( x 2 + y 2 + z 2 + λ xy ) . 2 a. Consider λ  as a small parameter and calculate the ground state energy through second order perturbation theory. b. Consider λ  as a small parameter and calculate the first excited energy levels to first order in perturbation theory. Formulas from the standard solution of the one-dimensional harmonic oscillator: 1 1 ω = ( k /m) 2 , En = ( n + )ω , n = 0,1,2, …, 2 1

  2 ( a + a + ), aψ n = nψ n−1 , x =  2mω  [ a , a + ] = 1, a +ψ n = n + 1ψ n+1. (Berkeley) Sol: a. The ground state has wave function ψ 000 ( x , y , z ) = ψ 0 ( x )ψ 0 ( y )ψ 0 (z ) 3 and energy  w . Consider 1 λ xy  as perturbation, the first order energy 2 2 correction is E (1) = 000

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as the integral is an odd function with respect to x or y. The second order energy correction is E (2) = ∑ 〈000| n1 ,n2

=−

2

kλ xy | n1n2n3 〉 /(−n1 − n2 )ω 2

λ2 ω , 32

as 〈0,0,0|

kλ λ xy | n1n2n3 〉 = ωδ 1n1 δ 1n2 δ 0n3 . 2 4

Therefore, the ground state energy corrected to second order is  3 λ2  E0(1) = ω  −  .  2 32  b. The first excited energy level E1 = 5 ω  is three-fold degenerate, the three 2 states being |1,0,0〉, |0,1,0〉, |0,0,1〉 . λk The matrix of the perturbation  xy  is 2  0 1 0  λ ω  1 0 0 ,  4   0 0 0  and the secular equation E1(1)

λ ω 4 0

λ ω 4

0

E1(1)

0

0

E1(1)

=0

has roots E1(1) = 0,

λ ω λ ω , − . 4 4

Thus the first excited energy level splits into three levels 5 5 λ  +  ω , ω , 2 4 2

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353

5050 Consider the following model for the Van der Waals force between two atoms. Each atom consists of one electron bound to a very massive nucleus by a potential V ( ri ) = 21 mω 2 ri 2 . Assume that the two nuclei are d    apart mω along the x-axis, as shown in Fig. 5.14, and that there is an interaction  x x e2 V12 = β 1 23 . Ignore the fact that the particles are indistinguishable. d

Fig. 5.14 a. Consider the ground state of the entire system when β = 0. Give its energy and wave function in terms of r1  and  r2. b. Calculate the lowest nonzero correction to the energy,  ∆E , and to the wave function due to V12 . c. Calculate the r.m.s. separation along the x direction of the two electrons to lowest order in  β .  1  ψ 0 ( x ) = 〈 x | 0〉 =   π 

1/2

e

−

x 2 mω 2

 1 2mω  ψ 1( x ) = 〈 x |1〉 =   π  

;

1/2

xe

−

x 2 mω 2

,

〈n | x | m〉 = 0, for n − m ≠ 1, 〈n − 1| x | n〉 = ( n /2mω )1/2 , 〈n + 1| x | n〉 = (( n + 1) /2mω ) , 1/2

(Wisconsin) Sol: a. The Schrödinger equation of the system is  2 2 1 x x e2  (∇1 + ∇ 22 ) + mω 2 (r12 + r22 ) + β 1 32 ψ = Eψ . − 2 d   2m

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When  β = 0  the system is equal to two independent three-dimensional harmonic oscillators and the energy and wave function of the ground state are 3 3 E0(0) = ω + ω = 3ω , 2 2 r r Ψ (0) = ψ x ψ ( , ) ( ) 0 1 2 0 1 0 ( y1 )ψ 0 ( z1 )ψ 0 ( x 2 )ψ 0 ( y 2 )ψ 0 ( z 2 ) 3

mω  1  − 2  ( r12 +r22 ) e = ,   π

where  r12 = x12 + y12 + z12 , etc. b. Treating H′ =

βe2 x1 x 2 d3

as perturbation we have the first order energy correction

βe2 00| x1 x 2 |00 d3 βe2 = 3 ψ 0 ( x1 )| x1 |ψ 0 ( x1 ) ψ 0 ( x 2 )| x 2 |ψ 0 ( x 2 ) = 0 d

∆E0(1) =

as  n | x | k = 0  for  k ≠ n ± 1. For the second order energy correction, we have

βe2 0| x1 | n1 0| x 2 | n2 d3 βe2  = 3 δ n ,1δ n ,1 , d 2mω 1 2

00| H ′ | n1n2 =

and hence ∆E0(2) =

| 〈00| H ′ | n1n2 〉 |2 E0 − En1n2 n1 ,n2 ≠0

∑′

2

| 〈00| H ′ |11〉 |2 1  βe2   =− = , E0 − E11 2ω  d 3 2mω  3 3   as  En1n2 =  n1 +  ω +  n2 +  ω . Thus the energy corrected to lowest   2 2 order is 2

1  e2   E0 = 3ω −  3  2 3 β 2, 8 d  m ω

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355

and the corrected wave function is Ψ 0 = Ψ (0) 0 + = Ψ (0) 0 −

00| H ′ |11 (0) Ψ1 E0 − E11

βe2 1 Ψ1(0) , 4d 3 mω 2

where  Ψ1(0) = ψ 1 ( x1 )ψ 0 ( y1 )ψ 0 (z1 )ψ 1 ( x 2 )ψ 0 ( y 2 )ψ 0 (z 2 ) . c. Let  S12 = x 2 − x1. Then  S12 = x 2 − x1 = 0 as Ψ 0 remains the same when 1 and 2 are interchanged showing that  x1 = x 2 . Consider S122 = x12 + x 22 − 2 x1 x 2 = 2 x12 − 2 x1 x 2 . We have (0) (0) x1 x 2 = Ψ (0) x1 x 2 Ψ (0) 0 − λΨ1 0 − λΨ1

{

(0) = − λ Ψ (0) + complex conjugate 0 x1 x 2 Ψ 1

= −2λ (〈0| x |1〉)2 = − λ

}

 , mω

where

λ=

βe2 1 , 4d 3 mω 2

and (0) (0) x12 = Ψ (0) x12 Ψ (0) 0 − λΨ1 0 − λΨ1

= 0| x 2 |0 + λ 2 1| x 2 |1 = 0| x 2 |0 + O(λ 2 ). Also according to the virial theorem 1 1 mω 2 0| x 2 |0 = ω , 2 4 or 2 x12 =

 + O( λ 2 ) . mω

Hence S122 = 2 x12 − 2 x1 x 2 =

2   λ + O( λ 2 ) ≈ + (1 + 2λ ). mω mω mω

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Thus the root-mean-square distance between the two electrons in x direction is (d + S12 )2 = d 2 + 2d S12 + S122 ≈ d2 +

   βe2  (1 + 2λ ) = d 1 + 1 + . mω mω d 2  2mω 2d 3 

5051 The first excited state of three-dimensional isotropic harmonic oscillator (of natural angular frequency ω 0  and mass m) is three-dold degenerate, Use the perturbation method to calculate the splitting (to the first order) of this three-fold degenerate state due to a small perturbation of the form  H ′ = bxy , where b is a constant. Give the first-order wave functions of the three split levels in terms of the wave functions of the unperturbed three-dimensional harmonic oscillator, given that, for a one-dimensional harmonic oscillator, n | x | n +1 =

( n + 1) . 2mω 0 (Wisconsin)

Sol: Write the unperturbed energy eigenstate as | nx n y nz 〉 = nx n y nz , where  | n〉  is the nth eigenstate of a one-dimensional harmonic oscillator. The first excited state of the 3-dimensional isotropic harmonic oscillator is degenerate in the states |ψ 1 〉 = 100 , ψ 2 = 010 , ψ 3 = 001 . Calculating the matrix elements Hij′ = b ψ i xy ψ j , we find H11′ = b 〈100| xy |100〉 = b〈1| x |1〉 〈0| y | 0〉 = 0 = H 22 ′ = H 33 ′, H12′ = b 〈100| xy | 010〉 = b 〈1| x | 0〉 〈0| y |1〉 = b 〈0| x |1〉 * 〈0| y |1〉 =

b = H 21 ′, 2mω o

H13′ = b 〈100| xy | 001〉 = b〈1| x | 0〉 〈0| y | 0〉 = 0 = H 31 ′, H 23 ′ = b〈010| xy | 001〉 = b〈0| x | 0〉 〈1| y | 0〉 = 0 = H 32 ′.

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357

Thus  0 l 0  b  H′ = 1 0 0 .  2mω 0   0 0 0  The secular equation det H ′ − E (1) = 0 has roots E (1) = 0, E (1) = ± b . The unperturbed oscillator has energy  2mω 0 (0) 3 En = n + ω . The first excited state,  n = 1, now splits into three levels with the 2

( )

wave functions indicated: 5 5 E0(1) = ω + 0 = ω , ψ 0′ =|ψ 3 〉 =|001〉. 2 2  b 5 1 E±(1) = ω ± , ψ ±(1) = (|100〉 ± |010〉 ). 2 2mω 0 2

5052 A quantum mechanical system is described by the Hamiltonian  H = H0 + H ′ , where  H ′ = i λ [ A, H0 ]  is a perturbation on the unperturbed Hamiltonian H0 ,  A is a Hermitian operator and λ  is a real number. Let B be a second Hermitian operator and let  C = i[ B , A] . a. You are given the expectation values of the operators A, B and C in the unperturbed (and nondegenerate) ground states; call these  A 0 , B

0

and  C 0 . With perturbation switched on, evaluate the expectation value of B in the perturbed ground state to first order in λ . b. Test this result on the following three-dimensional problem. 3  p2 1  H0 = ∑  i + mω 2 x i2  , H ′ = λ x 3 , 2  i =1  2 m

by computing the ground state expectation value  x i , (i = 1, 2, 3)  to lowest order in λ . Compare this result with an exact evaluation of  x i . (Princeton)

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Sol: a. Assume that the eigenstates and the corresponding energies of an unperturbed system are respectively | k 〉(0) , Ek(0) , then Hˆ 0 | k 〉(0) = Ek(0) | k 〉(0) . The matrix elements of the perturbation Hamiltonian are (0)

 ′ |0 Hn′0 = (0) n | H

=(0) n | iλ AH 0 − iλ H 0 A |0

= iλ ( E0(0) − En(0) )

(0)

n | A |0

(0)

(0)

.

Then, the ground state wave function with first order perturbation correction is 0 =0

(0)

=0

(0)

Hn′0 n − En(0) n≠ 0 E ∞

+∑

(0)

(0) 0

∞

+ ∑iλ (0) n | A |0

(0)

n

(0)

n≠ 0

∞

= (1 − iλ A 0 )|0〉(0) + ∑iλ (0) n | A |0

(0)

n

n=0

(0)

.

Hence 〈0| B |0〉 =  (0) 〈0|(1 + iλ 〈 A〉0 ) + (−iλ ) ∞ (0)  × ∑ ( (0) 〈n | A |0〉 )* 〈n | B (1 − iλ 〈 A〉0 )|0〉(0) n= 0  (0)  + iλ ∑ ∞ 〈m | A |0〉(0) | m〉(0)  m= 0  (0) ≈ 〈 B〉0 − λ 〈0| iAB − iBA |0〉(0)

= 〈 B〉0 + λ (0) 〈0| C |0〉(0) = 〈 B〉0 + λ 〈C 〉0 , (to first order in λ ). (0) Note that the completness of | k 〉 ,

∑ | k〉

(0)

〈k |(0) = 1,

k

has been assumed in the above calculation, b. The given Hamiltonians 3  p2 1  H 0 = ∑  i + mω 2 xi2  , H ′ = λ x 3  2 i =1  2m

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satisfy  H ′ = iλ[ A, H 0 ] = λ x 3  if we set  A = have the following: For  B = x1 , as C1 = i[B, A] =

359

p3 . Using the results of (a) we mω 2 

i [x1 , p3 ] = 0, mω 2 

we have x1 = B ≈ B 0 + λ C1

0

= x1 0 + λ C1

0

= 0.

Note that  x1 = 0  as the integral is an odd function of xi . For B = x 2 , a similar calculation gives  x 2 = 0 . For  B = x 3 , as p3  1 C3 = i [ B, A ] = i  x 3 , =− , mω 2   mω 2  and so C3

0

=−

1 , mω 2

we have x 3 = B ≈ x 3 0 + λ C3

0

λ . =− mω 2 For an accurate solution for  Hˆ = Hˆ 0 + H ′ , write 3  p2 1  Hˆ = ∑  i + mω 2 xi2  + λ x 3   m 2 2 i =1

λ  λ2  = Hˆ 01 ( x1 ) + Hˆ 02 ( x 2 ) + Hˆ 03  x 3 + − , 2  mω  2mω 2 2 d 2 1 where  Hˆ 0i ( xi ) = − + mω 2 xi2  is the Hamiltonian of a onedimen2m dxi2 2 sional harmonic oscillator. As the constant term − λ 2 / 2mω 2  does not affect the dynamics of the system, the accurate wave function of the ground state is just that for an isotropic harmonic oscillator:  mω  |0〉 =   π  

3/4

 mω 2   mω 2  exp  − x exp  − x  2 1   2 2 

2  mω  λ   × exp  − x +  3  . mω 2    2 

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It follows that x1 = 0, x 2 = 0, x 3 = −

λ . mω 2

These results are exactly the same as those obtained in a)

5053 A particle of mass m is moving in the three-dimensional harmonic oscillator potential  V ( x , y , z ) = 21 mω 2 ( x 2 + y 2 + z 2 ) . A weak perturbation is applied in the form of the function ∆V = kxyz + kω x 2 y 2 z 2 , where k is a small constant. 2

Note the same constant k appears in both terms. a. Calculate the shift in the ground state energy to second order in k. b. Using an argument that does not depend on perturbation theory, what is the expectation value of x in the ground state of this system? Note: You may wish to know the first few wave functions of the onedimensional harmonic oscillator: ground state  mω  ψ 0(x) =   π  

1/4

 mω 2  exp  − x ,  2 

first excited state  mω  ψ 1( x ) =   π  

1/4

2mω  mω 2  x exp  − x ,  2  

second excited state  mω  ψ 2(x) =   π  

1/4

1  2mω 2   mω 2  x − 1 exp  − x .    2   2 (Princeton)

Sol:

The ground state of a particle in the potential well of three-dimensional harmonic oscillator is Φ0 ( x , y , z ) = ψ 0 ( x )ψ 0 ( y )ψ 0 (z )  mω  =  π  

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3/4

 mω 2 2  exp  − ( x + y 2 + z 2 ) . 2   

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The first order energy correction is  k2 2 2 2  ∆E 1 = ∫ Φ0 ( x , y , z ) kxyz + x y z  Φ0 ( x , y , z )d 3 x   ω  mω  =  π  

3/2

k 2  +∞ 2  mω 2   x exp  − x  dx      ω  ∫−∞

3

3

2    k = .  2mω  ω

While the perturbation  ∆V ′ = kxyz  does not give rise to first order correction, it is to be considered for second order perturbation in order to calculate the energy shift accurate to  k 2 . Its perturbation Hamiltonian has matrix elements n | ∆V ′ |0 = ∫ Φn ( x , y , z ) kxyz Φ0 ( x , y , z )d 3 x +∞

= k ∫ ψ n1 ( x )xψ 0 ( x )dx −∞

∫

+∞

−∞

ψ n2 ( y ) yψ 0 ( y )dy

+∞

× ∫ ψ n 3 (z )zψ 0 (z )dz −∞

3/2

   = k δ (n1 − 1)δ (n2 − 1)δ (n3 − 1),  2mω  where  n = n1 + n2 + n3 , and so the second order energy correction is ∆E 2 = ∑ n≠0

n | ∆V ′ |0

2

E0 − En 3

   k2  3 2  2mω     k = = − .   2mω  3ω E0 − E3 Therefore the energy shift of the ground state accurate to k 2  is ∆E = ∆E 1 + ∆E 2 =

3

2 k2      . 3 ω  2mω 

b. V + ∆V  is not changed by the inversion  x → − x , y → − y , i.e., H ( x , y , z ) = H (− x , − y , z ) . Furthermore the wave function of the ground state is not degenerate, so ψ (− x , − y , z ) = ψ ( x , y , z ) and, consequently,

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x = (ψ , xψ ) +∞

+∞

−∞

−∞

= ∫ dz ′ ∫ +∞

= − ∫ dz −∞

∫

+∞

−∞

ψ * ( x ′ , y ′ , z ′ )x ′ψ ( x ′ , y ′ , z ′ ) dx ′ dy ′

+∞

+∞

−∞

−∞

∫ ∫

ψ * ( x , y , z ) ⋅ x ⋅ψ ( x , y , z ) dx dy = − x ,

where we have applied the transformation  x ′ = − x , y ′ = − y , z ′ = z . Hence x = 0. In the same way we find  y = 0, z = 0. Thus  (x) = 0.

5054 1 A spin −  particle of mass m moves in spherical harmonic oscillator potential 2 1 V = mω 2 r 2  and is subject to an interaction  λ σ ⋅ r  (spin orbit forces are to be 2 ignored). The net Hamiltonian is therefore H = H0 + H ′ , where H0 =

P2 1 + mω 2 r 2 , 2m 2

H ′ = λ σ ⋅r .

Compute the shift of the ground state energy through second order in the perturbation H ′ . (Princeton) Sol:

The unperturbed ground state is two-fold degenerate with spin up and spin down along the z-axis. Using the perturbation method for degenerate states, if the degeneracy does not disappear after diagonalizing the perturbation Hamiltonian, one has to diagonalize the following matrix to find the energy positions: n | V | n′ + ∑ m

n | V | m m | V | n′ ≡ n |W | n′ . En(0) − Em(0)

Let  nx n y nz ↑  and  nx n y nz ↓   be the unperturbation quantum states, where nx ,n y  and  nz  are the oscillation quantum numbers in the x, y and z direction, ↑ (↓)  represents the spin up (down) state. As 3  En(0) =  nx + n y + nz +  ω , x n y nz  2

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the matrix has elements

λ 2 〈000 ↑| σ ⋅ r |001 ↑〉〈001 ↑| σ ⋅ r |000 ↑〉 3 5 ω − ω 2 2 2 λ 〈000 ↑| σ ⋅ r |100 ↓〉 〈100 ↓| σ ⋅ r |000 ↑〉 + 3 5 ω − ω 2 2 2 λ 〈000 ↑| σ ⋅ r |010 ↓〉 〈010 ↓| σ ⋅ r |000 ↑〉 + 3 5 ω − ω 2 2  | 〈000 ↑| σ z z |001 ↑〉 |2 | 〈000 ↑| σ x x |100 ↓〉 |2 = λ2  + − ω − ω 

〈000 ↑|W |000 ↑〉 =

| 〈000 ↑| σ y y |010 ↓〉 |2   − ω  2 3λ =− , 2mω 2 3λ 2 〈000 ↓|W |000 ↓〉 = − , 2mω 2 〈000 ↑|W |000 ↓〉 = 0. +

In the above calculation we have used the fact that (ni + 1) , 2mω

〈ni | xi | ni + 1〉 =

xi = x , y , z ,

all other elements being zero. It is seen that a two fold degeneracy still exists for −3λ 2 the eigenvalue  . This means that the degeneracy will not disappear until at 2mω 2 least the second order approximation is used. The ground state, which is still 2 degenerate, has energy  3 ω − 3λ . 2 2mω 2

5055 Consider a spinless particle of mass m and charge e confined in a spherical cavity of radius R: that is, the potential energy is zero for  x ≤ R  and infinite for x > R . a. What is the ground state energy of this system? b. Suppose that a weak uniform magnetic field of strength  | B |  is switched on. Calculate the shift in the ground state energy.

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c. Suppose that, instead, a weak uniform electric field of strength | E |  is switched on. Will the ground state energy increase or decrease? Write down, but do not attempt to evaluate, a formula for the shift in the ground state energy due to the electric field. d. If, instead, a very strong magnetic field of strength | B |  is turned on, approximately what would be the ground state energy? (Princeton) Sol:

The radial part of the Schrödinger equation for the particle in the potential well is 2 l(l + 1)   R ′′ + R ′ +  k 2 − 2  R = 0, (r < R ), r r   where  k = 2mE /  2 , the boundary condition being  R(r )|r = R0 = 0 . Introducing a dimensionless variable  ρ = kr, we can rewrite the equation as d 2 R 2 dR  l(l + 1)  + + 1− R = 0. d ρ 2 ρ d ρ  ρ 2  This equation has solutions  jl ( ρ )  that are finite for  ρ → 0, jl ( ρ )  being related to Bessel’s function by 1

 π 2 jl ( ρ ) =   J 1 ( ρ ) .  2 ρ  l+ 2 Thus the radial wave function is Rkl (r ) = Ckl jl (kr ), where Ckl  is the normalization constant. The boundary condition requires jl (kR0 ) = 0 which has solutions kR0 = α nl ,

n = 1, 2, 3 …

Hence the bound state of the particle has energy levels Enl =

2 α nl2 , 2mR02

For the ground state,  ρ = 0  and  j0 ( ρ ) =

n = 1, 2, 3 … sinρ , so that α 10 = π  and the energy ρ

of the ground state is E10 =  2π 2 /2mR02.

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b. Take the direction of the magnetic field as the z direction. Then the vector potential A has components Ax = −

B B y , Ay = x , Az = 0, 2 2

and the Hamiltonian of the system is 2 2 1  eB   eB  2 Hˆ =  px + y  +  p y − x  + pz  + V (r ) 2m  2c 2c 

=

1  2 eB e 2 B2 2  ( ) ( x + y 2 )  + V (r ) p − xp − yp + y x  2 2m  4c c 

1  2 eB e 2 B2  p − lz + 2 ( x 2 + y 2 ) + V (r ).  2m  4c c  eB e 2 B2 As the magnetic field is weak we can treat − lz + 2 ( x 2 + y 2 )  as a c 4c perturbation. When the system is in the ground state l = 0, lz = 0  we only e 2 B2 2 need to consider the effect of the term  ( x + y 2 ). The wave function of 8mc 2 the ground state is =

2k 2 j0 (kr )Y(00) (θ , ϕ ) R0

ψ (r ,θ ,ϕ ) =

k2 sin(kr ) , j0 (kr ) = 2π R0 2π R0 r

=

and the first order energy correction is E ′ = ψ (r ,θ ,ϕ ) =

e 2 B2 2 ( x + y 2 ) ψ (r ,θ ,ϕ ) 8mc 4

e 2 B2 1 ⋅ 2 8mc 2π R0

∫

R0

0

r 2 sin 2 (kr )dr

∫

θ

0

2π sin3θ dθ

1  e 2 B 2 R02 1 . = − 2  3 2π  12mc 2 Note that in the above calculation we have used x 2 + y 2 = r 2 sin 2θ , sin(kR0 ) = 0 or kR0 = π . c. Suppose a weak uniform electric field E is applied in the z direction, instead of the magnetic field. The corresponding potential energy of the particle is V ′ = −eEz , which is to be taken as the perturbation. The shift of the ground state energy is then Ee′ = 〈ψ (r , θ , ϕ )| −eEr cosθ |ψ (r , θ , ϕ )〉. As  Ee′  is negative, the energy of the ground state decreases as a result.

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d. If a strong magnetic field, instead of the weak one, is applied then P2 e 2 B2 2 Hˆ = + (x + y 2 ) 2m 8mc 2 and the B 2  term can no longer be considered as a perturbation. The particle is now to be treated as a two-dimensional harmonic oscillator with e 2 B2 1 eB mω 2 = , or ω = . 2 8mc 2 2mc Hence, the ground state energy is approximately eB E0 = ω = . 2mc

5056 A particle of mass m and electric charge Q moves in a three-dimensional isotropic harmonic-oscillator potential  V = 21 kr 2 . a. What are the energy levels and their degeneracies? b. If a uniform electric field is applied, what are the new energy levels and their degeneracies? c. If, instead, a uniform magnetic field is applied, what are the energies of the four lowest states? (Columbia) Sol: a. The Hamiltonian of the system is H=−

2 2 1 2 ∇ + kr = H x + H y + H z , 2m 2

where Hi = −

 2 ∂2 1 2 + kxi (i = x , y , z ). 2m ∂ xi2 2

The energy levels are given by EN = ( N + 3 / 2)ω 0 , where  ω 0 = k / m , N = nx + n y + nz .

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The degeneracy of state N is N 1 f = ∑ ( N − nx + 1) = ( N + 1)( N + 2) . 2 nx =0

b. Take the direction of the uniform electric field as the z direction. Then the Hamiltonian is ˆ 2 /2m + 1 kr 2 − QEz Hˆ = P 2  pˆ x2 1 2   2 1 = + kx  +  pˆ y /2m + ky 2   2m 2   2  1 +  pˆz2 /2m + k(z − QE /k )2  − Q 2 E 2 /2k. 2   Comparing this with the Hamiltonian in (a), we get EN = ( N + 3/2)ω 0 − Q 2 E 2 /2k , 1 f = ( N + 1)( N + 2). 2 c. Consider the case where a magnetic field, instead of the electric field, is applied. In cylindrical coordinates, the vector potential has components 1 Aϕ = Bρ , Aρ = Az = 0 . 2 Thus the Hamiltonian is 2

1  Q  Hˆ =  pˆ − A  + V 2m c =

1 2 Q Q2 1 1 pˆ − pˆ ⋅ A + A 2 + k ρ 2 + kz 2 , 2m mc 2mc 2 2 2

where we have used r 2 = ρ 2 + z 2 , and ∇⋅ A = 0  which means  pˆ ⋅ A = A ⋅ pˆ . Write  Q ˆ  2 2 1   pˆ 2 1 Hˆ =  − ∇t + mω ′ 2 ρ 2  +  z + kz 2  − ω Lz 2 m 2 2 m 2     |Q | = Hˆ + Hˆ ∓ ω Lˆ , z

t

where

z

∇t2 = ∇ 2x + ∇ 2y ,ω = | Q | B/2mc ,ω ′ 2 = ω 2 + ω 02 , Lˆz = pˆz p,

and the symbols  ∓  correspond to positive and negative values of Q. Of the partial Hamiltonians, Hˆ  corresponds to a two-dimensional harmonic oscillator t

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normal to the z-axis,  Hˆ z  corresponds to a one-dimensional harmonic oscillator along the z-axis. Therefore, the energy levels of the system are given by 1  Enρnz m = (2nρ + 1+ | m |)w ′ +  nz +  ω 0 ∓ mω  2 1   =  ω ′ + ω 0  + 2nρ ω ′+ | m | ω ′ ∓ mω + nz ω 0 ,   2 where n p = 0,1, 2, …, nz = 0,1, 2, …, m = 0, ± 1, ± 2, …. The four lowest energy levels are thus 1 E000 = ω ′ + ω 0 , 2 1 1 E001 = ω ′ + ω 0 + (ω ′ − ω ) = 2ω ′ − ω + ω 0 , 2 2 1 3 E010 = ω ′ + ω 0 + ω 0 = ω ′ + ω 0 , 2 2 1 1 E002 = ω ′ + ω 0 + 2(ω ′ − ω ) = 3ω ′ − 2ω + ω 0 . 2 2

5057 a. Describe the splitting of atomic energy levels by a weak magnetic field. Include in your discussion a calculation of the Landé g -factor (assume LS couplings). b. Describe the splitting in a strong magnetic field (Paschen-Back effect). (Columbia) Sol: In LS coupling the magnetic moment of an atomic system is the sum of contributions of the orbital and spin angular momenta:

µˆ = µ0 ( g l l + g s sˆ) = µ0[ g l j + ( g s − g l )sˆ], where  µ0  is the Bohr magneton. Taking the direction of the magnetic field as the z direction, the change of the Hamiltonian caused by the magnetic field is Hˆ = − µ ⋅ B = − g µ Bj − ( g − g )µ Bs . 1

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l

0

z

s

l

0

z

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a. The Hamiltonian of system is ˆ ˆ = Hˆ + Hˆ = pˆ 2 /2m + V (r ) + ξ (r )s ⋅ I + Hˆ . HH 0 1 1 Considering  −( g s − g l )µ0 Bsz  as perturbation and operating ( H 0 − g l µ0 Bjz ) 2 2 ˆ on the common state of  I , j  and  jz  we have ( Hˆ 0 − g l µ0 Bjˆz )ψ nljmj = ( Enlj − gιµ0 Bm j )ψ nljm j . If B very weak, then  mj 2j   sz = 〈 jm j | σˆ z | jm j 〉 =  2  −m j  2( j + 1)

1 for j = l + , 2 1 for j = l − , 2

where we have used the relations l ± m j + 12

| jm j 〉 =

2l + 1

1 1 mj − , ∓ 2 2

l ∓ m j + 12 2l + 1

1 1 mj + ,− 2 2

for  j = l ± 12 , and

σ z mj ∓

1 1 1 1 , ± = ± mj ∓ , ± . 2 2 2 2

Hence, the energy level of the system becomes Enljm j ≈ Enlj − gιµ0 Bm j mj  2j ,  −( g s − g l )µ0 B   −m j ,  (2 j + 2)

1 j=l+ , 2 1 j=l− . 2

As  g l = −1, g s = −2  we can write Enljm j ≈ Enlj − g µ0 Bm j , where  j( j + 1) + s(s + 1) − l(l + 1)  g = − 1 +  2 j( j + 1)   is the Landé g-factor. Thus an energy level of the atom splits into (2 j + 1) levels.

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b. If the magnetic field is very strong, we can neglect the term ξ (r )s ⋅1  and the Hamiltonian of the system is Hˆ = pˆ 2 / 2m + V (r ) + Hˆ 1 = Hˆ 0 + Hˆ 1 . 2 Operating on the common eigenstate of  ( Hˆ 0 ,I ,l z , sˆ 2 , sz ), we get

Hˆψ nlmt ms = Enlmpmsψ nlmlms , where Enlmιms = Enl − g ι µ0 Bml − g s µ0 BmS = Enl + µ0 B(ml + 2mS ), 1 as  g l = −1, g s = −2 . For an electron,  mS = ± . Then, due to the selection rule 2 1 ∆mS = 0, transitions can only occur within energy levels of mS = +  and 2 1 within energy levels of mS = − . The split levels for a given l are shown in 2 Fig. 5.15. For the two sets of energy levels with  ml + 2mS = −l + 1  to  l − 1  (one set with mS = 1 , the other with mS = − 1 , i.e.,  2l − 1  levels for each set), there 2 2 is still a two-fold degeneracy. So the total number of energy levels is  2(2l + 1) − (2l − 1) = 2l + 3  as shown.

Fig. 5.15 5058 Consider an atom with a single valence electron. Its fine-structure Hamiltonian is given by ξ  L. S. a. Determine the difference in the energies of the levels characterized by J = L + 1/ 2  and  J = L − 1/ 2  (fine structure interval) in terms of  ξ .

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371

b. This atom is placed in a weak external magnetic field of magnitude H. Use perturbation theory to determine the energy splitting between adjacent magnetic (Zeeman) sublevels of the atom. c. Describe qualitatively how things change in a very strong magnetic field. (Columbia) Sol: a. In the representation of (J , Lˆ , jz ), 2

2

2 2 1 2 S ⋅ L = (J − L − S ) 2 2 = [ j( j + 1) − l(l + 1) − s(s + 1)] . 2 3 2  =  j( j + 1) − l(l + 1) −  . 2 4 As  l + 1 l + 3 − l − 1 l + 1 = 2l + 1, the energy difference is 2 2 2 2

( )( ) ( )( ) ∆E = E

1 j =l + 2

−E

1 j =l − 2

= ξ (r ) ⋅

2 (2l + 1). 2

b. If the atom is placed in a weak magnetic field of magnitude H whose direction is taken to be the z direction, the Hamiltonian of the system is eH  pˆ 2 Hˆ = + V (r ) + ( L z + 2S z ) + ξ (r )S ⋅ L 2m 2mc pˆ 2 eH  eH ˆ = + V (r ) + j z + ξ (r )S ⋅ L ⋅+ Sz 2m 2mc 2mc eH  Sz, ≡ Hˆ 0 + 2mc  z + S z . where  J z = L 2 2 Let ψ nljm j  be a common eigenstate of (J , L , jz ) for the unperturbed Hamilto-

nian Hˆ o . Then eH    Hˆ 0ψ nljm j =  Enlj + m j  ψ nljm j .  2mc  eH sˆz  use spherical coordiTo consider the effect of the perturbation term  2mc nates and write ψ nljm j = Rn (r )ϕ ljm j (θ , ϕ ), where the angular wave functions are

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j + mj

ϕ ljrn j =

2j +

ϕ ljm j = −

αY

j − mj 2j

1 1 j − ,mj − 2 2

βY

1 1 j − ,rn j + 2 2

1 for j = l + , 2

j − mj + 1 αY 1 1 j + ,m j − 2j+2 2 2 j + mj +1

1 for j = l − , 2 h  α , β  being eigenstates of  S z  of eigenvalues  and −  respectively. Thus for 2 2 1 j=l+ , 2 j + mj  S zϕ ljm j ≡ S z jm j = αY 1 1 j − ,mj − 2j 2 2 2 +

2j+2

βY

−

1 1 j + ,mj + 2 2

j − mj  βY 1 1 2 j 2 j − 2 ,m j + 2 ,

and j + mj  j − mj   mj 〈 jm j | S z | jm j 〉 = − = , 2j 2 2j 2 2 j 1 and for j = l − 2  mj jm j S z jm j = − . 2 ( j + 1) Hence,  m /j , eH eH   j 〈Sz 〉 =  2mc 4mc  −m /( j + 1),  j

1 j=l+ , 2 1 j=l− , 2

and so

Enljm j

 1  1 +  m j , eH   2 j  = Enlj +  2mc  1   1 − 2 j + 2  m j , 

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373

Therefore,   1  µB H  1 +  ,  2j  ∆E =  µ H  1 − 1  ,  B  2 j + 2   e where  µ B =  is the Bohr magneton. 2mc

1 j=l+ , 2 1 j=l− , 2

c. If the magnetic field is very strong, ξ (r )S ⋅ L  µ B B  and the Hamiltonian of the system is eH  p2 Hˆ = + V (r ) + (l z + 2S z ). 2m 2mc

2 2 Since  ( Hˆ 0 , Lˆ , Lˆz , Sˆ , Sz )  form a complete set of dynamical variables,

eH (m + 2ms ) 2mc = Enl + µ B H (m ± 1),

Enlmms = Enl + as  ms = ± 12 . Hence

∆E = µ B H.

5059 Positronium is a hydrogen-like system consisting of a positron and an electron. Consider positronium in its ground state ( l = 0). The Hamiltonian H can be written:  H = H0 + HS + HB , where H0  is the usual spin-independent part due to the Coulomb forces,  HS = As p ⋅ s e  is the part due to the interaction of the spins of the positron and the electron, and  HB = −( µ p + µe ) ⋅ B  is the part due to the interaction with an externally applied magnetic field B. a. In the absence of an externally applied field what choice of spin and angular momentum eigenfunctions is most convenient? Calculate the energy shifts for each of these states due to HS . b. A very weak magnetic field is applied  (HB  HS ) . What are the allowed energies for this system in this case? c. Now suppose the applied magnetic field is increased such that  HB  HS . What kind of eigenfunctions are now most appropriate? What are the energy shifts for each of these states due to HB ?

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d. Indicate how you would solve this problem for the energies and the corresponding eigenfunctions in the general case; however, do not try to carry out the algebra unless you have nothing better to do. No long essays, please. (Berkeley) Sol: a. In the absense of the external field B take H s  as perturbation. The total spin of the system is  S = s p + se  and so 1 1 s p ⋅ se = (S 2 − s 2p − se2 ) = [S(S + 1) − s p (s p + 1) − se (se + 1)] . 2 2 It is most convenient to choose the eigenfunctions in the form | lmSSz 〉 . The ground state l = 0  consists of four states  S = 1, Sz = 0, ±1; S = 0, Sz = 0 . For l = 0, S = 1, as As p ⋅ se lm1Sz =

A 3 1(1 + 1) −   2 lm1Sz , 2  2

we have Es = 〈lm1Sz | H s | lm1Sz 〉 =

A 2 , (Sz = 0, ± 1) , 4

which is the energy shift for the triplet state. For  l = 0, S = 0, as H S lm00 = the energy shifts is  − 3 A 2 . 4

A 3 2  0 −   | lm00〉 , 2 2

b. The external field B is switched on, but as H B  H S  we can consider the effect of the external field as a perturbation  ( H = ( H 0 + H s ) + H B ). Taking  lmSSz as the state vector and the direction of B as the z direction, i.e., we have B = Be z , H B = −( µ p + µe ) ⋅ B = and (cf. Problem 5066 (b)) eB |0010〉, H B 0000 = mc eB |0000〉, H B 0010 = mc

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eB  (s ez − s pz ), mc H B 0011 = 0, H B 001, − 1 = 0.

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Hence, lmSSz | H B | lmSSz = 0 and the energy levels do not change further for first order perturbation, in addition to the splitting into singlet and triplet states described in (a). We have 2

(0) n

En = E

n | HB |i + ∑′ (0) . En − Ei(0) i ≠n

As only the following matrix elements are nonzero: eB 0010 H B 0000 = mc eB 0000 H B 0010 = , mc the energy levels from second order perturbation are, for  0000 : 3  eB  E1 = − A 2 +   mc  4

2

 3 2 1 2  − A − A  4 4 2

3 1  eB  = − A 2 −   , 4 A  mc  for  0011  and  001, −1 : 1 E2 = E4 = A 2 , 4 for 0010 : 1  eB  E3 = A 2 +   mc  4

2

 1 2 3 2  A + A  4 4 2

1 1  eB  = A 2 +   . 4 A  mc  c. As now H B  H S , we can neglect the H s   term and consider only a eB (sˆez − sˆpz ). It is then convenient to choose  perturbation consisting of  H B = mc 1 1 lmsez s pz  as the eigenfunctions. Then for states  lm, ± , ± , we have 2 2 1 1 1 1 1 1 (sˆez − sˆpz ) ± , ± = sˆez ± , ± − sˆpz ± , ± 2 2 2 2 2 2      1 1 = ± −  ±   ± , ± = 0  2  2  2 2

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1 1 , we and the corresponding energy shifts are zero. For states lm, ± , ∓ 2 2 have 1 eB 1 1 1 HB ± , ∓ =± ± ,∓ , 2 2 mc 2 2 and hence the energy shifts  ±

eB . mc

d. In the general case, take  lmSSz  as the state vector and  H ′ = H S + H B  as the perturbation Hamiltonian. Then treat the problem using the perturbation method for degenerate states, i.e., solve the secular equation det H mn ′ − Eδ mn = 0  to find the energy corrections and the corresponding wave functions.

5060 An atom is in a state of total electron spin S, total orbital angular momentum L, and total angular momentum J. (The nuclear spin can be ignored for this problem). The z component of the atomic total angular momentum is Jz. By how much does the energy of this atomic state change if a weak magnetic field of strength B is applied in the z direction? Assume that the interaction with the field is small compared with the fine structure interaction. The answer should be given as an explicit expression in terms of the quantium numbers J, L, S, Jz and natural constants. (Princeton) Sol: The Hamiltonian and eigenfunctions before the introduction of magnetic field are as follows: Hˆ = Hˆ , 0

ψ nLJM J = RnLJ (r1 ,…, rn )φSLJM J , where the subscripts of r, 1, 2, …, n, represent the different electrons in the atom, and φSLJM J is the common eigenstate of (L2 , S 2 , J 2 , J z ), i.e,

φSLJM J = ∑ LM L S , M J − M L | JM J YLM L Θ S ,M L − M L , ML

LM L S , M J − M L | JM J

being Clebsch-Gordan coefficients. The corresponding

unperturbed energy is EnSLJ .

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After switching on the weak magnetic field, the Hamiltonian becomes eB ˆ eB ˆ Hˆ = Hˆ 0 + jz + Sz . 2me c 2me c As B is very small, we can still consider (L2 , S 2 , J 2 , J z ) as conserved quantities and take the wave function of the system as approximately ψ nLJMJ . The energy change eB J z is ∆E1 = M J  2eB caused by the term τ nc as J z has eigen value M J  . The 2mc eB Sz is diagonal in the subspace of the 2 J + 1 state vectors for the matrix of 2mc energy EnLJ , and hence the energy change caused by it is ∆E2 =

eB  eB Sz = JM J | S z | JM J , 2mc 2mc

where 2 JM J | S z | JM J = ∑( M J − M L )  LM L S , M J − M L | JM J  . ML

The total energy change is then 2

∆E = M J ω l − w l + ∑( M J − M L ) ⋅  LM L S , M J − M L | JM J  , ML

where wl =

eB . 2mc

5061 The deuteron is a bound state of a proton and a neutron. The Hamiltonian in the center of mass system has the form H=

p2 x x 1   + V1( r ) + σ p ⋅σ nV2 ( r ) +  σ p ⋅   σ n ⋅  − (σ p ⋅σ n ) V3 ( r ). 2µ r  r 3  

Here x = x n − x p , r = x , σ p and σ n are the Pauli matrices for the spins of the proton and neutron, µ is the reduced mass, and p is conjugate to x a. The total angular momentum J 2 = J( J + 1) and parity are good quantum numbers. Show that if V3 = 0 , the total orbital angular momentum L2 = L( L + 1), total spin S2 = S( S + 1) and S = 21 σ p + 21 σ n are good quantum numbers. Show that If V3 ≠ 0, S is still a good quantum number. (It may help to consider interchange of proton and neutron spins.)

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b. The deuteron has J = 1 and positive parity. What are the possible values of L? What is the value of S? c. Assume that V3 can be treated as a small perturbation. Show that in zeroth order (i.e. V3 = 0) the wave function of the state with J z = +1 is of the form ψ 0 ( r ) α , α where α , α is the spin state with s pz = snz = +1/ 2 . What is the differential equation satisfied by ψ 0 ( r )? d. What is the first order shift in energy due to the term in V3? Suppose that to first order the wave function is ψ 0 ( r ) α , α + ψ 1(x) α , α + ψ 2 (x)( α , β + β , α ) +ψ 3 (x) β , β , where β is a state with Sz = −1/ 2 and ψ 0 is as defined in part (c). By selecting out the part of the Schrödinger equation that is first order in V3 and proportional to α , α find the differential equation statisfied by ψ 1( x ). Separate out the angular dependence of ψ 1 ( x ) and write down a differential equation for its radial dependence. (MIT) Sol: a. Use units such that  = 1. First consider the case where V3 = 0 . As P2 = −

1 ∂2 1 r − 2 ∇θ2 ,ϕ , L2 = −∇θ2 ,ϕ , 2 r ∂r r

we have [L2, P 2 ] = 0, [L2 , V1 + (σ p ⋅σ n )V2 ] = [L2 ,V1 ] + (σ p ⋅σ n )[L2, V2 ] = 0 + (σ p ⋅σ n ) ⋅ 0 = 0, and so [L2 , HV3 =0 ] = 0 . Thus L2 is a good quantum number. Now consider the total spin S 2. As 1 1 1 3 1 3 1  3 s p ⋅ sn = (S 2 − s 2p − sn2 ) =  S 2 − ⋅ − ⋅  =  S 2 −  , 2 2 2 2 2 2 2  2 1 1 sp = σ p, sn = σ n , 2 2

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we have 1 [S 2 ,σ p ⋅σ n ] = [σ p ⋅σ n ,σ p ⋅σ n ] = 0, 2 Furthermore, [S 2 , P 2 ] = 0,

[S 2 ,V1 (r )] = 0.

Hence, [S 2 , HV3 =0 ] = 0 and S 2 is also a good quantum number. If V3 ≠ 0 , we have to consider also 2 2 2 x x  1  x  x  x    σ p ⋅   σ n ⋅  =  (σ p + σ n ) ⋅  −  σ p ⋅  −  σ n ⋅   . r r 2   r r r 

As σ p + σ n = 2(s p + s n ) = 2S and 2

2

x  x x x x x    σ p ⋅  =  σ p ⋅   σ p ⋅  = ⋅ = 1 =  σ n ⋅  , r r r r r r using the formula (σ ⋅ A )(σ ⋅ B) = A ⋅ B + iσ ⋅ (A × B), the above becomes 2

x x   x  σ p ⋅   σ n ⋅  = 2  S ⋅  − 1. r r r Then as  2  x  xi  2  S ,  S ⋅ r   = ∑[S , Si ] r  = 0,   i we have x S   2   S ,  σ p ⋅ r   σ n ⋅ r   = 0,   and so [S 2 , H ] = 0,

[S , H ] = 0 .

Hence, S is still a good quantum number if V3 ≠ 0 . b. The parity of the deuteron nucleus is P = P( p ) ⋅ P(n) PL = (+1) ⋅ (+1) ⋅ (−1)L = (−1)L. Since the parity is positive, L = even . Then as S can only be 0 or 1 and J = 1, we must have S equal to 1 and L equal to 0 or 2. c. In zeroth order perturbation V3 = 0, L , S are good quantum numbers.

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For the lowest state, L = 0 and so Lz = 0 . Then for the state J z = Lz + Sz = 1, Sz = 1 and hence S = 1, s pz = + 12 , snz = + 12 . Because L = 0, the spatial wave function is spherically symmetric i.e., ψ 0 = ψ 0 (r ). Thus the wave function of state J z = 1 is ψ 0 (r ) , and Hψ 0 (r ) α ,α = Eψ 0 (r ) α ,α . As rp ⋅ rn = 4s p ⋅ s n = 2S 2 − 2s 2p − 2s n2 = 2S 2 − 3, and so rp ⋅ rn α ,α = (2S 2 − 3) α , α = (2 × 2 − 3) α , α = α , α , we have  2 1 ∂  2 ∂   Hψ 0 (r ) α ,α =  −  r  + V1 (r ) 2 ∂r  2 µ r ∂r  × ψ 0 (r ) α , α + V2 (r )ψ 0 (r ) α , α . Hence the differential equation satisfied by ψ 0 is

{

}

1 d  2 dψ 0  2 µ r  + 2 [ E − V1 (r ) − V2 (r )] ψ 0 = 0 . r 2 dr  dr   For L ≠ 0 , the wave functions of states with J z = 1 do not have the above form. d. In first order approximation, write the Hamiltonian of the system as H = H0 + H ′, where x x 1   H ′ =  σ p ⋅   σ n ⋅  − σ p ⋅σ n V3 (r ) ,     r r 3  

H 0 = HV3 =0 ,

and the wave function as

ψ = ψ 0 α ,α . The energy correction is given by ∆E = ψ | H ′ |ψ . As  cosθ x σ ⋅ = σ x sinθ cosϕ + σ y sinθ sinϕ + σ z cosθ =  iϕ r  sinθ e  cosθ x α σ ⋅ α = ( 1 0 ) iϕ r  sinθ e

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sinθ e − iϕ  , −cosθ 

sinθ e − iϕ   1    = cosθ , cosθ   0 

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we have x x 1 1  α ,α  σ p ⋅   σ n ⋅  − σ p ⋅σ n α ,α = cos 2θ − ,  r  r 3 3 and hence ∆E = 〈ψ | H ′ |ψ 〉 1  = ∫ |ψ 0 |2 V3 (r ) cos 2 θ −  dx  3 ∞

π

0

0

= ∫ V3 (r )|ψ 0 |2 r 2 dr ∫ = 0.

∫

2π

0

1  2  cos θ −  sinθ dθ dϕ 3

Note that S as is conserved and L is not, the wave function is a superposition of the spin-triplet states (S = 1):

ψ ( x ) = ψ 0 (r ) α ,α + ψ 1 ( x ) α ,α + ψ 2 ( x )( α , β + β ,α ) + ψ 3 ( x ) β , β , and Hψ = ( H 0 + H ′ )ψ = ( E + E (1) + )ψ . Therefore, in first order approximation, making use of E (1) = ∆E = 0 and H 0ψ 0 α ,α = Eψ 0 α ,α , we obtain H 0 ψ 1 α ,α + ψ 2 ( α , β + β ,α ) + ψ 3 β , β ] + H ′ψ 0 α ,α = E[ψ 1 α ,α + ψ 2 ( α ,α + α ,α ) + ψ 3 β , β  . To calculate the perturbation term H ′ψ 0 α ,α : x x 1   H ′ψ 0 α ,α = V3 (r )ψ 0  σ p ⋅   σ n ⋅  − (σ p ⋅σ n ) α ,α r  r 3    cosθ   cosθ  1 = V3 (r )ψ 0     − αα  sinθ e iϕ  p  sinθ e iϕ  n 3 

 .  

As  cosθ   1   0  = cosθ  + sinθ e iϕ   iϕ    0   1   sinθ e  = α cosθ + β sinθ e iϕ ,

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1 H ′ψ 0 α ,α = V3 (r )ψ 0 cos 2θ −  α ,α + V3 (r )ψ 0 sin 2θ e i 2ϕ β , β . 3  By considering the terms in the above equation that are proportional to α ,α , we can obtain the equation for the wave function ψ 1 (x): −

2  2  1 ∂  2 ∂  L  −  2  r  ψ 1 ( x ) + V1 (r )ψ 1  2 µ  r ∂r ∂r   2 r 2 

1  +V2 (r )ψ 1 + V3 (r )ψ 0 (r ) cos 2θ −  = Eψ 1 .  3 Writing ψ 1 ( x ) = R1 (r ) Φ1 (θ , ϕ ) , we can obtain from the above 1 1 16π Φ1 (θ ,ϕ ) = cos 2θ − = Y20 (θ , ϕ ), 3 3 5  2 Φ = 2(2 + 1) 2 Φ . The equation for R is and thus L 1 1 1 1 d  2 dR1  2 µ  6 E − V1 (r ) − V2 (r ) − 2  R1 r + r 2 dr  dr   2  r  =

2µ V3 (r )ψ 0 (r ) . 2

Here, it should be noted, even though the normalization factor of Φ1 will affect the normalization factor of R1, their product will remain the same. It is noted also that ψ 1 ( x ) corresponds to L = 2, Lz = 0 . By considering the term in H ′ψ 0 αα that is proportional to β , β , we see that ψ 3 ( x ) corresponds to L = 2, Lz = 2 . Then from the given J z = 1, we can see that ψ 2 ( x ) corresponds to L = 2, Lz = 1 (note that J z is also conserved if V3 ≠ 0) . In other words, the existence of V3 requires the ground state of deuteron to be a combination of the L = 0 and L = 2 states, so that J = 1, S = 1, J z = 1 and parity = positive .

5062 Consider the bound states of a system of two non-identical, nonrelativistic, spin one-half particles interacting via a spin-independent central potential. Focus in particular on the 3 P2 and 1 P1 levels ( 3 P2 : spin-triplet, L = 1, J = 2; 1P1 : spin-singlet, L = 1, J = 1). A tensor force term H ′ = λ [3σ (1) ⋅ rσ (2) ⋅ r − σ (1) ⋅σ (2)] is added to the Hamiltonian as a perturbation, where λ is a constant, r is a unit vector along the line joining the two particles, σ (1) and σ (2) are the Pauli spin operators for particles 1 and 2.

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a. Using the fact that H ′ commutes with all components of the total angular momentum, show that the perturbed energies are independent of m, the eigenvalues of J z . b. The energy is most easily evaluated for the triplet state when the eigenvalue of J z takes on its maximum value m = j = 2 . Find the perturbation energy ∆E ( 3 P2 ). c. Find ∆E (1 P1 ). (Princeton) Sol: a. Use units for which  = 1. As S = 12 [σ (1) + σ (2)], the perturbation Hamiltonian H ′ can be written as  3 H ′ = λ   (2S ⋅ r )2 − (σ (1) ⋅ r )2 − (σ (2) ⋅ r )2    2 1 −  4 S 2 − σ (1)2 − σ (2)2  2 = λ  6(S ⋅ r )2 − 2S 2  ,

}

where we have also used the relation (σ ⋅ r )2 = (σ x + σ y + σ z )(σ x + σ y + σ z ) = σ x2 + σ y2 + σ z2 = σ 2 , on account of

σ iσ j + σ jσ i = 2δ ij . To prove that H ′ commutes with all components of the total angular momentum J, we show for example [ J z , H ′] = 0 . As [Sz , S 2 ] = 0, [Lz , S 2 ] = 0 , we have [ J z , S 2 ] = 0. Also, as [Sx , S y ] = iSz , etc, Lz = −i ∂∂ϕ , we have [ J z , S ⋅ r ] = [Sz , S ⋅ r ] + [Lz , S ⋅ r ] = [Sz ,sinθ cosϕ Sx + sinθ sinϕ S y + cosθ Sz ] + [Lz ,sinθ cosϕ Sx + sinθ sinϕ S y + cosθ Sz ] = i sinθ cosϕ S y − i sinθ sinϕ Sx − i

∂ sinθ cosϕ Sx + sinθ sinϕ S y + cosθ Sz ∂ϕ

(

)

= 0.

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and hence [ J z ,(S ⋅ r )2 ] = (S ⋅ r )[ J z , S ⋅ r ] + [ J z , S ⋅ r ](S ⋅ r ) = 0 Combining the above results, we have [ J z , H ′] = 0 . Similarly we can show [ J x , H ′] = [ J y , H ′] = 0 . It follows that J + = J x + iJ y also commutes with H ′. J + has the property J+ j, m = a j, m +1 . Where a is a constant. Suppose there are two unperturbed states j , m1 and j , m2 , where m2 = m1 + 1, which are degenerate and whose energies to first order perturbation are E1 and E2 respectively. Then 〈 j , m2 |[ J + , H 0 + H ′]| j , m1 〉 = ( j , m2 | J + ( H 0 + H ′ )| j , m1 〉 − 〈 j ,m2 |( H 0 + H ′ ) J + | j , m1 〉 = ( E1 − E2 )〈 j , m2 | J + | J , m1 〉 = a( E1 − E2 ) = 0 as ( H 0 + H ′ ) j , m1 = E1 j , m1 , ( H 0 + H ′ ) J + j , m1 = ( H 0 + H ′ )a j , m2 = E2a | j , m2 〉 = E2 J + j , m1 . Since the matrix element a ≠ 0, E1 = E2 , i.e., the perturbation energies are independent of m. b. The perturbation energy is ∆E(3 P2 ) = j = 2, m = 2 H ′ j = 2, m = 2 . Since j = 2, m = 2 = l = 1, ml = 1 S = 1, mS = 1 , ∆E( P2 ) = ∫ dΩY11* (θ , ϕ ) S = 1, ms = 1 H ′ S = 1, mS = 1 Y11 (θ , ϕ ) 3

3 λ sin 2 θ (3cos 2 θ − 1)2π sinθ dθ 8π ∫ 3 π = λ ∫ sin3 (3cos 2 θ − 1)dθ 4 0 2 = − λ. 5 =

c. For the state 1 P1, as S = 0, ms = 0 and so H ′ = 0, we have ∆E(1 P1 ) = 0 .

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5063 A hydrogen atom is initially in its absolute ground state, the F = 0 state of the hyperfine structure. (F is the sum of the proton spin I, the electron spin s and the orbital angular momentum L.) The F = 0 state is split from the F = 1 state by a small energy difference ∆E. A weak, time-dependent magnetic field is applied. It is oriented in the z direction and has the form Bz = 0, t < 0, Bz = B0 exp( −γ t ), t > 0. Here B0 and γ are constants. a. Calculate the probability that in the far future when the field dies away, the atom will be left in the F = 1 state of the hyperfine structure. b. Explain why, in solving this problem, you may neglect the interaction of the proton with the magnetic field. (Princeton) Sol: a. When considering the hyperfine structure of the hydrogen atom, we write the Hamiltonian of the system as H = H 0 + f (r )σ p ⋅σ e , where H 0 is the Hamiltonian used for considering the fine structure of the hydrogen atom, f (r )σ p ⋅σ e is the energy correction due to the hyperfine structure, σ p and σ e being the Pauli spin matrices for the proton and electron respectively.

1 When the atom is in its absolute ground state with L = 0, j = s = , the hyper2 fine structure states with F = 0 and F = 1 respectively correspond to spin parallel and spin antiparallel states of the proton and electron. The initial wave function of the system is ψ (r , F ) = R10Y00 (θ ,ϕ )Θ00 .  1   0 Letting α , β represent the spinors   ,   , we have the spin wave  0  1  function Θ00 =

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1 (−α p β e + α e β p ), 2

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which makes ψ antisymmetric. When t > 0, a weak magnetic field Bz = B0e −γ t acts on the system and the Hamiltonian becomes eBz Hˆ = H 0 + f (r )σ p ⋅σ e + σˆ ez , 2 µc neglecting any interaction between the magnetic field and the proton. Suppose the wave function of the system at t > 0 is ψ (r , F , t ) = R10 (r )Y00[C1 (t )Θ00 + C2 (t )Θ11 + C3 (t )Θ10 + C4 (t )Θ1,−1 ], where Θ11 = α pα e , Θ1,−1 = β p β e , Θ1,0 =

1 (α p β e + α e β p ). 2

Then the probability that the system is at hyperfine structure state F = 1 at time t is 2

A(t ) = 1 − C1 (t ) , and the initial conditions are C1 (0) = 1, C2 (0) = C3 (0) = C4 (0) = 0 . As

σ xα = β ,

σ xβ = α ,

σ y α = iβ ,

σ y β = −iα ,

σ zα = α ,

σ z β = −β ,

we have rp ⋅ re Θ00 = (σ pxσ ex + σ pyσ ey + σ pz σ ez ) = −3Θ00 ,

1 (−α p β e + α e β p ) 2

and similarly rp ⋅σ e Θ1m = Θ1m . Finally, from the Schrödinger equation i ∂∂t ψ = Hˆψ we obtain dC dC dC  dC  iR10 (r )Y00  1 Θ00 + 2 Θ11 + 3 Θ10 + 4 Θ1,−1   dt  dt dt dt = R10 (r )Y00 {[ E10 − 3 f (a0 )]C1 (t )Θ00 + [ E10 + f (a0 )][C2 (t )Θ11 + C3 (t )Θ10 + C4 (t )Θ1,−1 ]} + R10 (r )Y00 µ0 Bz [C3 (t )Θ00 + C2 (t )Θ11 + C1 (t )Θ10 − C4 (t )Θ1,−1 ].

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Comparing the coefficient of Θ00 and of Θ10 on the two sides of the equation, we get  d −γ t i dt C1 (t ) = [ E10 − 3 f (a0 )]C1 (t ) + C3 (t )µ0 B0e ,  i d C3 (t ) = [ E10 + f (a0 )]C3 (t ) + C1 (t )µ0 B0e −γ t .  dt As the energy of the hyperfine structure, f (a0 ) , is much smaller than E10 , energy of the fine structure, it can be neglected when calculating C1 (t ). Then the above equations give

{

}

1 − i E10 t iµ0B0 (1−e − γ t ) − iµγ0B0 (1−e − γ t ) . C1 (t ) = e  e γ  +e 2 As t → ∞ , µ B  2 C1 (t ) → cos 2  0 0  .  γ  Hence µ B  µ B  A(t )|t→∞ = 1 − cos 2  0 0  = sin 2  0 0  ,  γ   γ  which is the probability that the hydrogen atom stays in state F = 1. b. The interaction between the magnetic moment of the proton and the magnetic field can be neglected because the magnetic moment of the proton is 1 only of the magnetic moment of the electron. 1840

5064 A spinless nonrelativistic particle in a central field is prepared in an s-state, which is degenerate in energy with a p-level ( m = 0, ± 1). At time t = 0 an electric field E = E0 sinω t zˆ is turned on. Ignoring the possibility of transitions to other than the above-mentioned states but making no further approximations, calculate the occupation probability for each of the four states at time t, in terms of the nonzero matrix elements of zˆ . (Wisconsin) Sol: Choose the four given states as the state vectors and the level of energy such that the degenerate energy E = 0. The Hamiltonian of the system is Hˆ = Hˆ 0 + Hˆ ′ , Hˆ ′ = − gE ⋅ r = − gE0 z sinω t .

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To find the perturbation energy matrix one notes that only elements l + 1,ml z l ,ml are non-vanishing. Thus we have 00 00

   Hˆ ′ = 1, − 1   11  10

1, − 1

10 0

1

0

l 0 0

0 0 0

0 0 0

11 0   0  (− gE 00 z 10 sinω t ). 0 0  0 

Suppose the wave function is ψ = ( x1 , x 2 , x 3 , x 4 ), and initially ψ (t = 0) = (1, 0, 0, 0) ψ (t = 0) = (1, 0, 0, 0), where x1 , x 2 , x 3 , x 4 are the four state vectors. The Schrödinger e­ quation i ∂ ψ = Hˆψ can then be written as ∂t

i





 0 1   x1  d  x1    = − λ sinω t    , (1) dt  x 2   1 0   x 2  i

d  x3    = 0, (2) dt  x 4 

where λ = gE0 00 z 10 is a real number. Equation (2) and the initial condition give  x3   x3   0  ,   =  =  x 4   x 4   0  t =0 which means that the probability that the system occupies the states ml = ±1 of the p-level is zero. To solve Eq. (1), we first diagonalize the matrix by solving the secular equation

λ 1 = 0, 1 λ which gives λ = ±1. Hence, the first two components of ψ are to be transformed to

ψ=

1  x1 + x 2  2  x1 − x 2

  a  ≡  .   b 

Then Eq. (1) becomes i

 1 0  a  d a  = − λ sinω t  ,   dt  b   0 −1   b 

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389

subject to the initial condition  a 1  1  b  = 2  1 . t =0 Solving the equation we find   iλ   exp  (1 − cosω t )   a ω    1   .  b  = 2    iλ   exp − ω (1 − cosω t )   To get back to the original state vectors,  x1  1  a + b  ψ =  =  x2  2  a − b   λ   cos  ω (1 − cosω t )  . =   λ   i sin  ω (1 − cosω t )   Therefore, the occupation probability for each of the four states at time t is  gE 00 z 10  2 Ps (t ) = x1 = cos 2  0 (1 − cosω t ) ω    gE 00 z 10  2 Pp(ml =0) (t ) = x 2 = sin 2  0 (1 − cosω t )  ω   Pp(ml =±1) (t ) = 0, where 00 z 10 is to be calculated using wave functions RnlYlm (θ ,ϕ ) for a particle in a central field and is a real number.

5065 An ion of a certain atom has L = 1 and S = 0 when it is in free space. The ion is implanted in a crystalline solid (at x = y = z = 0) and sees a local environment of 4 point charges as shown in the Fig. 5.16. One can show by applying the Wigner-Eckart theorem (DON’T TRY) that the effective perturbation to the Hamiltonian caused by this environment can be written as α H1 = 2 ( L2x − L2y ), 

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Where Lx and Ly are the x and y components of the orbital angular momentum operator, and α is a constant. In addition, a magnetic field is applied in βB L , where Lz is the z the z direction and causes a further perturbation H z =  z component of the angular momentum operator and β is a constant.

Fig. 5.16 a. Express the perturbation Hamiltionian H ′ = H1 + H2 in terms of L+ and L− , the “raising” and “lowering” operators for orbital angular momentum. b. Find the matrix of the perturbation Hamiltonian in the basis set using the three states 1, 0 , 1,1 and 1, − 1 . c. Find the energy levels of the ion as a function of B. Make a careful sketch of your result. d. When B = 0, what are the eigenfunctions describing the ion? (MIT) Sol: a. From the definitions L± = Lx ± iL y , we get [L+ , L− ] = −2i[Lx , L y ] = 2Lz , 1 1 1 L2x − L2y = ( L+ + L− )2 + ( L+ − L− )2 = ( L2+ + L2− ). 4 4 2 Thus the perturbation Hamiltonian is α βB H ′ = 2 ( L2x − L2y ) + Lz   α βB = 2 ( L2+ + L2− ) + 2 ( L+ L− − L− L+ ). 2 2

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391

b. Using the formula L± L , M L =  L( L + 1) − M L ( M L ± 1) L , M L ± 1 , we find the following non-vanishing elements L2− 1,1 = 2L− 1,0 = 2 2 1, −1 , L2+ 1, −1 = 2 2 1,1 , L+ L− 1,1 = 2 2 1,1 , L+ L− 1, 0 = 2 2 1,0 , L− L+ 1, 0 = 2 2 1,0 , L− L+ 1, − 1 = 2 2 1, −1

and hence the matrix

( L′, M ′ H ′ L, M ) as follows: L

L

1,1

1,0

 βB  H ′ = 1,0  0 1, −1  α

0

1,1

1, −1   0 . −β B  

α

0 0

c. The perturbation energy E is determined by the matrix equation  βB − E 0 α  −E 0 0   α − β −E 0 B 

 a      b  = 0,    c  

whose secular equation

βB − E 0 α −E 0 0 α 0 −β B − E gives the corrections E1 = − (β B)2 + α 2 , E2 = 0, E3 = (β B)2 + α 2 .

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The perturbation energy levels are shown in Fig. 5.17 as functions of B, where the dashed lines are the asymptotes.

Fig. 5.17 d. If B = 0, the energy levels are E1 = −α , E2 = 0, E3 = α , and the eigenstates are given, respectively, by a = −c ≠ 0, a = c = 0, a = c ≠ 0,

b = 0; b ≠ 0; b = 0.

Thus the corresponding energy eigenstates are  0   −l   a   a   b  = 1  0 ,  b  = l ,        2   c   0  1   c   (1) ( 2)  a   1   b  = 1  0 .    2   c   l  (3) In terms of the state vectors 1,1 , 1,0 , 1, −1 the wave functions are 1 1 1,1 + 1, −1 , 2 2 E2 = 0 = 1,0 ,

E1 = −α = −

E3 = α =

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1 1 1,1 + 1, −1 . 2 2

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393

5066 The spin-dependent part of the effective Hamiltonian for a positronium (bound state of electron and positron) in a magnetic field B may be written as Hspin = Aσ e ⋅ σ p + µB B(σ ez − σ pz ), where σ e and σ p are the Pauli spin matrices for the electron and the positron, and µB is the Bohr magneton. a. At zero magnetic field the singlet state lies 8 × 10 −4 eV below the triplet state. What is the value of A? b. Illustrate by a sketch the dependence of the energy of each of the four spin states on the magnetic field B, including both the weak and strong field cases. c. If the positronium atom is in its lowest energy state in a strong magnetic field and the field is instantaneously switched off, what is the probability of finding the atom in the singlet state? d. How would the result of (c) be changed if the field is switched off very slowly? (Wisconsin) Sol: a. When B = 0, the effective Hamiltonian has the expectation value H spin = A F ′mF′ σ e ⋅ σ p FmF , where F = se + s p . As

σ e ⋅σ p =

2 2 2 2 (F − se − s p ), 2

E = H spin = 2 A[F ( F + 1) − se (se + 1) − s p (s p + 1)]⋅δ F ′F ⋅δ mF′ mF . For the triplet state, F = 1, so EF =1 = 2 A (1 ⋅ 2 − 12 ⋅ 32 − 12 ⋅ 33 ) = A . For the singlet state, F = 0, so EF =0 = −3 A . Hence, EF =1 − EF =0 = 4 A = 8 × 10−4 eV , giving A = 2 × 10−4 eV.

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Problems and Solutions in Quantum Mechanics

b. We first transform from representation in coupling state vectors to that in non-coupling state vectors: 1 1 1 1 F = 1, mF = 1 = se = , mse = , s p = , msp = − , 2 2 2 2 F = 1, mF = 1 =

1  1 1 1  se = , mse = , s p = , 2 2 2 2

1 1 1  1 1 + se = , mse = − , s p = , msp = −  . 2 2 2 2 2  1 1 1 1 F = 1, mF = −1 = se = , mse = , s p = , msp = − , 2 2 2 2 msp = −

F = 0, mF = 0 =

1 1 1 1   se = , mse = , s p = , 2 2 2 2

msp = −

1 1 1 1 1  − se = , mse = − , s p = , msp =  . 2 2 2 2 2 

2µ B Then as µ B B(σ ez − σ pz ) = B (sez − s pz ) and

(sez − s pz ) F = 0, mF = 0 =

1 1 1 1 1 (sez − s pz ) , , , − 2 2 2 2 2 1 1 1 1  − ,− , ,  2 2 2 2 

=

  1 1  1 1 1 1  +  , , ,− 2  2 2  2 2 2 2  1 1 1 1 1 1 −− −  ,− , ,  2 2 2 2 2 2

=  F = 1, mF = 0 , (sez − s pz ) F = 1, mF = 0 =  F = 0,mF = 0 , (sez − s pz ) F = 1, mF = 1 = (sez − s pz ) F = 1,mF = −1 = 0, and using the results of (a), we have in the order of 1,1 , 1, −1 , 1,0 , 0,0 ,   H spin =    

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A 0 0 A 0 0 0

0

0 0 A 2µB B

  .  −3 A 

0 0 2µB B

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395

The secular equation A− E 0 0

0 A− E 0

0

0

0 0 A− E

0 0 2µB B

=0

2 µB B −3 A − E

then gives the spin-energy eigenvalues E1 = E2 = A, E3 = − A + 2 A2 + µ B2 B2 , E4 = − A − 2 A2 + µ B2 B2 . The variation of E with B is shown in Fig. 5.18. If the magnetic field B is weak we can consider the term µ B B(σ ez − σ pz ) as perturbation. The energy correction is zero in first order perturbation and is proportional to B2 in second order perturbation. When the magnetic field B is very strong, the energy correction is linear in B.

Fig. 5.18 c. When the positronium is placed in a strong magnetic field ( µ B B  A) , the lowest energy state, i.e., the eigenstate whose energy is approximately − A − 2 µ B B , is 1 1 1 mse = − , msp = = { 1,0 − 0,0 } , 2 2 2 where we have considered Aσ e ⋅ σ p as perturbation. If the magnetic field is switched off suddenly, the probability that the atom is in state F = 0, mF = 0 is 1 1 1 1 P = 0,0 , − , , 2 2 2 2

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2

=

1 1 0,0|1,0 − 0,0|0,0 2 2

2

1 = . 2

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Problems and Solutions in Quantum Mechanics

d. If the magnetic field is switched off very slowly, no transition occurs and the atom will remain in the state 12 , − 12 , 12 12 , and the energy of the system is E = − A.

5067 Positronium consists of an electron and a positron bound by their Coulomb attraction. a. What is the radius of the ground state? The binding energy of the ground state? b. The singlet and triplet ground states are split by their spin-spin interaction such that the singlet state lies about 10–3 volts below the triplet state. Explain the behavior of positronium in a magnetic field. Draw an energy level diagram to illustrate any dependence on the magnetic field. (Berkeley) Sol: a. The hydrogen atom has ground state radius and binding energy a0 =  2 µe 2 ≈ 0.53 Å , E1 = µe 4 2 2 ≈ 13.6 eV, where µ = mem p (me + m p ), the Coulomb potential V (r ) = −e 2 r1 − r2 having been used. The results may be applied to any hydrogen-like atom of nuclear charge Ze with the replacements V (r ) → V ′(r ) = − Ze 2 r1 − r2 ,

µ → µ ′ = m1m2 (m1 + m2 ), m1 , m2 being the mass of the nucleus and that of the orbiting electron. For positronium, µ ′ = me2 , Z = 1 , and so a0′ =  2 µ ′e 2 = a0 µ µ ′ = 2a0 ≈ 1 Å , 1 E1′ = µ ′e 4 2 2 = µ ′ E1 µ = E1 ≈ 6.8 eV. 2

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b. Choose 0, 0, S , Sz Hami­ltonian

397

as the eigenstate and take as the perturbation H ′ = Ase ⋅ s p +

eB (sez − s pz ), mc

where A = 10 2eV . Using the results of Problem 5066 we have the perturbation −3

energy matrix  A 2 4 0 0 0  2 0 A 4 0 0   eB Hmn ′ = 0 0 A 2 4 −   mc  eB 3  − − A 2 0 0  mc 4 (1,1)

(1, − 1)

(1, 0)

  (1,1)   (1, − 1) ,   (1, 0)  (0, 0)  

(0, 0)

from which we find the perturbation energies E1 = E2 = A 2 4, 2

2

2

2

 A 2   eB  E3 = − A 4 +  + 2   ,   mc   2  2

 Ah 2   eB  E4 = − A 4 −  + 2   .   mc   2  2

The dependence of the energy levels on B is shown in Fig. 5.19. The result shows that the energy levels of the hyperfine structure is further split in the presence of a weak magnetic field, whereas the hyperfine structure is destroyed in the presence of a strong magnetic field. These limiting situations have been discussed in Problem 5059.

Fig. 5.19

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Problems and Solutions in Quantum Mechanics

5068 Estimate the magnetic susceptibility of a He atom in its ground state. Is it paramagnetic or diamagnetic? (Chicago) Sol:

Suppose the He atom is in an external uniform magnetic field H = Heˆz (H being a constant). The vector potential A, defined by H = ∇ × A , can be taken to be A = 1 H (− yeˆx + xeˆy ) . The Hamiltonian of the system is then 2 2

2 1  e  Hˆ = ∑  Pi − A  − l J . H . c i =1 2m

As the helium atom is in the ground state, l J = 0. Thus 2 1  e  Hˆ = ∑  Pi − A  c i =1 2m

2

Where m and e are respectively the mass and charge of an electron. ∂2 E The magnetic susceptibility is given by χ = −4π ∂ H 2 |H =0 . For helium atom in its ground state, 2 1  2 e e e2 2  Hˆ = ∑ P − A ⋅ P − P ⋅ A + A , i i  i c c c 2  i =1 2m  2 ∂ 2 Hˆ e2 e2 2 2 = ( x 2 + y 2 ), = ( + ) x y ∑ ∂ H 2 i=1 4mc 2 2mc 2

and so

χ = −4π

∂2 E ∂H 2

H =0

∂2 = −4π He ground state Hˆ He ground state ∂H 2 e2 = −4π He ground state x 2 + y 2 He ground state 2mc 2 e2 = −4π x2 + y2 . 2mc 2

(

)

As x 2 + y 2 + z 2 = r 2 , we can take 1 2 x 2 = y 2 ≈ rHeg .s . , 3

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399

2 where rHeg . s . is the mean-square distance of each electron from the He nucleus in

the ground state, which gives

χ ≈−

2 2 4π e r H eg . s c2 3m

in Gaussian units. Note that as χ < 0, helium atom in its ground state is diamagnetic.

5069 An atom with no permanent magnetic moment is said to be diamagnetic. In this problem the object is to calculate the induced diamagnetic moment for a hydrogen atom in a weak magnetic field B ignoring the spins of the electron and proton. a. Write down the nonrelativistic Hamiltonian for a particle of mass m and charge q in a combined vector and scalar electromagnetic field. b. Using the fact that a satisfactory vector potential A for a uniform magnetic field is A = − 21 r × B , write down the steady-state Schrödinger equation for the hydrogen atom. Assume the proton to be infinitely massive so center of mass motion can be ignored. c. Treat the terms arising from the magnetic field as a perturbation and calculate the shift in the ground state energy due to the existence of the magnetic field. d. Calculate the induced diamagnetic moment per atom. (MIT) Sol: a. A particle of mass m and charge q in an electromagnetic field of potentials (φ , A ) has Hamiltonian H=

2

1  q   p − A  + qφ . 2m c

b. If the motion of the proton is neglected, the Schrödinger equation for a hydrogen atom in a uniform magnetic field B is  1 e e2  2 ( p + B × r ) −  ψ (r ) = Eψ (r ), 2c r   2me

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Problems and Solutions in Quantum Mechanics

where q = −e for an electron, or  1  2 e e e2 e2  2 ( ) p + p ⋅ B × r + B × r ⋅ p + B × r −    r ψ (r ) 2c 2c 4c 2   2me   = Eψ (r ). As p ⋅ B × r − B × r ⋅ p = −i∇⋅ (B × r ) = −i(∇ × B ) ⋅ r + iB ⋅∇ × r = 0 because B is uniform and ∇ × r = 0, we have p ⋅B × r = B × r ⋅p = B ⋅r × p = B ⋅L. Taking the direction of the uniform magnetic field B as the z direction, we have B = Beˆz , B ⋅ L = BLz = iB ∂ and ∂ϕ (B × r )2 = (− Byeˆx + Bxeˆy )2 = B 2 ( x 2 + y 2 ) = B 2r 2 sin 2 θ in spherical coordinates. The Schrödinger equation can then be written as   2 2 e 2 ieB ∂ e 2 B2 2 2  − ∇ − − + r sin θ  ψ (r ,θ ,ϕ )  2m r 2me c ∂ϕ 8me c 2  e = Eψ (r ,θ ,ϕ ) in spherical coordinates. c. Treat H′ =

eB e 2 B2 2 2 Lz + r sin θ 2me c 8me c 2

as a perturbation. The energy shift of the ground state is ∆E = 100 H ′ 100 with 100 = R10 (r )Y00 (θ ,ϕ ) = 2 where a =  2 . Thus m e

1 − ar e , π a3

3 ∞ π   e 2 B2 1  1 ∆E =  2π ∫ sinθ dθ ∫ r 2 dr   e −2r /a r 2 sin 2θ  ⋅ 2 0 0 π  a   8me c 1 ∞ e 2 B2 a 2e 2 B2 . = 3 ∫ r 4 e −2r /a dr ⋅ = 3me c 2 4me c 2 a 0

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401

Note that for the ground state, l = 0, ml = 0 , and the first term of H ′ makes no contribution to ∆E. d. The above energy shift is equal to the energy of a magnetic dipole in a magnetic field B, ∆E = − µ ⋅ B , if the dipole moment is

µ=−

e 2a 2 B. 4me c 2

This can be considered as the dipole moment induced by the field. The atom is diamgnetic as µ is antiparallel to B.

5070

∂2 E (H ) , where ∂H 2 H = 0 E(H) is the energy of the atom in a constant external magnetic field. The magnetic polarizability of an atom is defined by α H = −

a. Estimate the magnetic polarizability of the F = 0, 1s hyperfine ground state of a hydrogen atom. b. Estimate the magnetic polarizability of the ground state 1s2 of a helium. (CUSPEA) Sol: a. If the magnetic field H is very weak, the perturbation Hamiltonian is H′ = − µ ⋅ H . Taking the direction of H as that of the z-axis and letting the spins of the electron and proton be S and I respectively we have

µ ⋅ H =  −1 ( g e µB Sz + g p µ p I z )H 1 1 =  −1  ( g e µ B + g p µ p )(Sz + I z ) + ( g e µ B − g p µ p )(Sz − I z ) H . 2 2  The first order perturbation makes no contribution to α H as (cf. Problem 5066). F = 0, mF = 0 Sz ± I z F = 0,mF = 0 = 0 . We then consider the energy correction of second order perturbation for the ground state F = 0, 1s : (2)

E (H ) =

1

∑

m=−1

F = 1 − µ ⋅H F = 0

2

EF =0 − EF =1

where m is the quantum number of the projection on z axis of F.

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E (2) ( H ) =

1

∑

F = 1, m − µ ⋅ H F = 0,0

m=−1 1

= H 2  −2 ∑ F = 1, m m=−1

/ ( EF =0 − EF =1 )

2

( EF =0 − EF =1 )

1 g e µ B − g p µ p ( Sz − I z ) F = 0, 0 2

(

)

2

as ( Sz + I z ) 00 = 0 . Then as ( Sz − I z ) 0,0 =  1,0 , the matrix elements are all zero except for m = 0. Thus 1 E (2) ( H ) = F = 1, m = 0 ( g e µ B − g p µ p )(Sz − I z ) 2 F = 0,0 H 2  −2 ( EF =0 − EF =1 ). 2

As µ p  µ B , g e = 1, and the spectral line F = 1 → F = 0 has frequency 140 MHz, corresponding to EF =1 − EF =0 = 0.58 × 10−6 eV,

α (H ) =

2 µB2 = (5.8 × 10−9 eV Gs ) 2( EF =1 − EF =0 )

( 2 × 5.8 × 10

−7

eV )

= 2.9 × 10−11 eV Gs 2 . b. Consider a helium atom in a uniform magnetic field H. The vector potential is A = 1 H × r and it contributes e 2 A2 2mc 2 per electron to the perturbation 2 Hamiltonian that gives rise to the magnetic polarizabihty α ( H ) (Problem 5068). If the helium atom is in the ground state 1 s 2 , then L = S = J = 0. Taking the direction of H as the z direction, we have H′ = 2⋅

e 2 A2 e 2 H 2 2 = ( x + y 2 ), 2mc 2 4mc 2

the factor of 2 being added to account for the two electrons of helium atom. The energy correction is thus E( H ′ ) = ψ 0 =

e2H 2 2 (x + y2 )ψ 0 4mc 2

e2H 2 2 e2H 2 2 2 2 x + y = . r0 , 4mc 2 4mc 2 3

where r0 is the root-mean-square radius of helium atom in the ground state, as r02 = x 2 + y 2 + z 2 and x 2 = y 2 = z 2 . Since r0 = Bohr radius of hydrogen atom,

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a 2 = 0 , a0 being the zme 2 2

24/05/22 9:22 AM

Perturbation Theory

α (H ) = −

∂2 E ∂H 2

2

2

=−

1  e  2a0 e 2 2  a0  ⋅ ⋅  = −   2 2mc 3  2  6  2mc  e 2

=−

µB2 , 6 EI

H =0

403

e e2 is the Bohr magneton, EI = is the ionization potential 2mc 2ao of hydrogen. Thus

where µ B =

α (H ) = −

(0.6 × 10−8 )2 = −4.4 × 10−19 eV Gs 2 . 6 × 13.6

5071 A particle of mass m moves in a three-dimensional harmonic oscillator well. The Hamiltonian is H=

p2 1 2 + kr . 2m 2

a. Find the energy and orbital angular momentum of the ground state and the first three excited states.

b. If eight identical non-interacting (spin − 1 ) particles are placed in such 2 a harmonic potential, find the ground state energy for the eight-particle system. c. Assume that these particles have magnetic moment of magnitude µ . If a magnetic field B is applied, what is the approximate ground state energy of the eight-particle system as a function of B. Plot the magnetization − ∂E for the ground state as a function of B. ∂B

( )

(Columbia) Sol: a. A three-dimensional harmonic oscillator has energy levels 3  EN =  N +  ω ,  2 where ω =

k m

, N = 2nr + l , N = 0,1, 2, …, nr = 0,1, 2, …, l = N − 2nr .

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For the ground state, N = 0 and the energy is E0 = 3 ω , the orbit angular 2 momentum is L = 0. For the first exited state, N = 1, E1 = 5 ω , L =  . As l = 1 the level contains 2 three degenerate states. b. For spin − 1 particles, two can fill up a state. Thus when fully filled. The 2 ground state contains two particles and the first three excited states contain six particles. Thus the ground state energy of the eight-particle system is E0 = 2 × 3 ω + 6 × 5 ω = 18ω . 2 2 c. The Hamiltonian of the system is 8 8  p2 1  8 e Hˆ = ∑  i + kri 2  − ∑ µ i ⋅ B − ∑ Li ⋅ B 2  i=1 i =1  2m i =1 2mc 8 dV (ri ) 1 e2 A2 (ri ) + ∑ 2 2 2 L i ⋅ si , 2 2 mc 2 m c r dri i =1 i =1 i where V (ri ) = 1 kri 2 , A is the vector potential 1 B × r giving rise to B. 2 2 As the eight particles occupy two shells, all the shells are full and we have S = 0, L = 0, j = 0. 8

+∑

The wave functions of the system are the products of the following functions (excluding the radial parts):  1  Y00 (e1 )Y00 (e 2 ) 2 {α (1)β (2) − α (2)β (1)},   Y11 (e3 )Y11 (e 4 ) 1 {α (3)β (4) − α (4)β (3)},  2  1  Y10 (e5 )Y10 (e6 ) {α (5)β (6) − α (6)β (5)}, 2   1  Y1−1 (e7 )Y1−1 (e8 ) 2 {α (7)β (8) − α (8)β (7)},  where ei = ri ri . Note that the two space sub-wave functions in each are same. Then as the total space wave function is symmetric, the total spin wave function must be antisymmetric. As σ x α = β , σ y α = iβ , σ zα = α ,

σ x β = α , σ y β = −iα ,

σ z β = −β ,

we have 1 1 {α (1)β (2) − α (2)β (1)} = {β (1)β (2) − α (2)α (1)}, 2 2 1 1 σ 2x {α (1)β (2) − α (2)β (1)} = − {β (1)β (2) − α (2)α (1)}, 2 2 i 1 σ 1y {α (1)β (2) − α (2)β (1)} = {β (1)β (2) + α (2)α (1)}, 2 2 i 1 σ 2y {α (1)β (2) − α (2)β (1)} = − {β (1)β (2) + α (2)α (1)}, 2 2 1 1 σ 1z {α (1)β (2) − α (2)β (1)} = {α (1)β (2) + α (2)β (1)}, xxxxxxxxxxxxx_Chapter_05.indd  Page 404 24/05/22 2 2

σ 1x

9:22 AM

σ 1x σ 2x σ 1y σ 2y σ 1z σ 2z

1 1 {α (1)β (2) − α (2)β (1)} = {β (1)β (2) − α (2)α (1)}, 2 2 1 1 Theory {β (1)βPerturbation (2) − α (2)α (1)}, {α (1)β (2) − α (2)β (1)} = − 2 2 i 1 {α (1)β (2) − α (2)β (1)} = {β (1)β (2) + α (2)α (1)}, 2 2 i 1 {α (1)β (2) − α (2)β (1)} = − {β (1)β (2) + α (2)α (1)}, 2 2 1 1 {α (1)β (2) − α (2)β (1)} = {α (1)β (2) + α (2)β (1)}, 2 2 1 1 {α (1)β (2) − α (2)β (1)} = − {α (1)β (2) + α (2)β (1)}. 2 2

405

Their inner products with the bra 1 {α (1)β (2) − α (2)β (1)}+ will result in 2 σ 1x , σ 2 x , σ 1 y , σ 2 y , as well as σ 1z + σ 2 z being zero. Hence

σ 1z L1z + σ 2 z L2 z = −i

∂ ∂ ⋅σ 1z + ⋅σ 2 z ∂ϕ1 ∂ϕ 2

 ∂  ∂ 〈σ 2 z 〉  = −i   〈σ 1z 〉 + ∂ϕ 2  ∂ϕ1  = −i 

∂ 〈σ 1z + σ 2 z 〉 = 0. ∂ϕ

Thus the ground state energy is E = 18ω +

e2 8 ∑ (B × ri )2 8mc 2 i=1 8

= 18ω + e 2 B 2 8mc 2 ∑ ri 2 sin 2θ i , i =1

and the magnetization is −

∂ E −e 2 B 8 2 2 = ∑ ri sin θi = χ B, ∂ B 4mc 2 i=1

8 ∂E 2 2 e2 giving χ = − 4mc 2 ∑ i=1 ri sin θ i as the diamagnetic susceptibility. ∂ B as a function of B is shown in Fig. 5.20.

Fig. 5.20

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Problems and Solutions in Quantum Mechanics

5072 Suppose one has an electron in an S state of a hydrogen atom which is in a magnetic field along the z direction of strength Hz. At time t = 0 a magnetic field along the x direction is turned on; its strength increases uniformly from zero to H x at time t = T (H x  H z ) and then remains constant after t = T . Assume the nucleus has no spin and that 2

 Hx  Hx .  H   H z z Consider only the interaction between the electron spin and the magnetic 2 H field. Neglect all terms of order  x  or higher. If the electron has its spin in  Hz  the z direction at t = 0, find the state of the electron when t = T . Show that the state is an eigenstate of the Hamiltonian due to the combined magnetic fields H = (H x ,0, H z ) provided that T is sufficiently long. Explain what sufficiently long means in this case. (Berkeley) Sol:

teH x t (−e ) ˆ Treat the potential energy H ′ = − t µ ⋅ H x eˆ x = − s ⋅ ex Hx = s as a perTmc x T Tmc turbation. Before H x is turned on, the electron in the S state of a hydrogen has (−e ) 1 ⋅ H = e H , and − 12 with two spin states, 1 with energy E+ = − mc 2 z 2mc z 2 energy E− = − e H z . 2mc Use time-dependent perturbation theory, taking the wave function as

ψ (t ) = e − iE+t /

1 1 + a− e − iE−t / − T , 2 2

where a− is given by 1 T 1 ˆ 1 − i ( E+ − E− )t / − H′ e dt i ∫0 2 2 1 T eH x t 1 1  eH  = ∫ − sz exp  −i z t  dt  mc  i 0 mc T 2 2 eH x T  eH  t exp  −i z t  dt = ∫ 0  mc  2imcT

a− =

1 H   eH  imcH x   eH z   =  x  exp  −i z T  − T −1 , exp −i  mc  2eTH z2   mc   2  Hz 

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Perturbation Theory

407

where we have used sx | 1 〉 =  | − 1 〉. 2 2 2 Thus the spin state of the electron at time T is  eH  1  1  H x   eH  exp  −i z T  ψ (T ) = exp  −i z T  +  2mc  2  2  H z   mc   −

imcH x   eH z    exp  −i T  −1  2eTH z2   mc   

 eH  1 × exp  i z T  − .  2mc  2 mc If the time T is sufficiently long so that eT ≤ H x , we can neglected the second term of a− and obtain  eH   1 1 H x 1  . ψ (T ) = exp  −i z T   + −  2mc   2 2 H z 2  The Hamiltonian due to the combined magnetic field H = ( H x ,0, H z ) is eH eH Hˆ = − µ ⋅ B = x sx + z sz . mc mc e and consider Hψ (T ). As sx ± 12 = 2 ∓ 12 , sz ± 12 = ± 2 ± 12 mc have for T → ∞ ,

Let α =

we

 1 1 Hx 1   iα H z T  Hˆψ (T ) = α exp  − + −  ( H x sx + H z sz )  2  2 2 H z 2  =

1 1 H x2 1 α  iα H z T   exp  − H − +  x  2 2   2 2 Hz 2 + Hz

=

≈

1 1 1  − Hx −  2 2 2 

2  α H z  −iα H z T    1  H x   1 exp  1 +     2 2    2  H z   2  1 Hx 1 + − 2 Hz 2

α H z  iα H z T   1 1 H x 1  eH z exp  − + − = ψ (T ).   2 2   2 2 H z 2  2mc

The result shows that when T is sufficiently large ψ (T ) is an eigenstate of the Hamiltonian due to the combined magnetic field with eigenvalue eH z . 2mc

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Problems and Solutions in Quantum Mechanics

5073 An electron is in the n = 1 eigenstate of a one-dimensional infinite squarewell potential which extends from x = − a / 2 to x = a / 2. At t = 0 a uniform electric field E is applied in the x direction. It is left on for a time τ and then removed. Use time-dependent perturbation theory to calculate the probabilities P2 and P3 that the electron will be, respectively, in the n = 2 and n = 3 eigenstates at t > τ . Assume that τ is short in the sense that τ   , E1 − E2 where En is the energy of the eigenstate n. Specify any requirements on the parameters of the problem necessary for the validity of the approximations made in the application of time-dependent perturbation theory. (Columbia) Sol:

The electron in the n = 1 eigenstate of the potential well  0, x ≤ a / 2, V =  ∞ , otherwise has wave functions and corresponding energies

ψ n (x ) =

2 πn  a  sin  + x , a  a  2  

En =  2π 2n 2 / 2ma 2 ,

n = 1, 2, …

The uniform electric field Eeˆx has potential φ = − ∫ Edx = − Ex . The potential energy of the electron ( charge − e ) due to E , H ′ = eEx , is considered as a perturbation. We have Hn′2n1 = n2 H ′ n1 α

=

2 2 a    n2π  a   n1π  α sin   x +   eEx dx  x +   sin  ∫ − 2   a 2  a 2  a

=

a a  eE 2   (n1 − n2 )  πx+  a cos  ∫ −  2  a 2  a

a    (n + n )  − cos  1 2 π  x +    xdx  2   a  =

eE  a2 (−1)n1 −n 2 − 1  a  (n1 − n2 )2 π 2  −

=

 a2 [(−1)n1 +n2 − 1] 2 2 (n1 + n2 ) π 

4eEa n1n2 [(−1)n1 +n2 − 1], π 2 (n12 − n22 )2

π 2 2 2 1 ω n2n1 = ( En2 − En1 ) = (n2 − n1 ),  2ma 2 1 τ 1 1 Ck′k (t ) = ∫ H k′′k e iω k′k t dt = H k′′k (1 − e iω k′kτ ) . 0 xxxxxxxxxxxxx_Chapter_05.indd  iPage 408 ω k′k 

24/05/22 9:22 AM

=

eE 2   (n1 − n2 )  a  πx+  a cos  ∫ −  2  a 2  a

a    (n + n )  − cos  1 2 π  x +    xdx  2    a =

eE  a2 (−1)n1 −n 2 − 1 Perturbation Theory  a  (n1 − n2 )2 π 2  −

=

409

 a2 [(−1)n1 +n2 − 1] 2 2 (n1 + n2 ) π 

4eEa n1n2 [(−1)n1 +n2 − 1], π 2 (n12 − n22 )2

π 2 2 2 1 ω n2n1 = ( En2 − En1 ) = (n2 − n1 ),  2ma 2 1 τ 1 1 Ck′k (t ) = ∫ H k′′k e iω k′k t dt = H k′′k (1 − e iω k′kτ ) . 0 i  ω k′k For the transition 1 → 2, H 21 ′ = 2 H′ 1 = −

16eEa , ω 21 = 3π 2 /2ma 2 , 9π 2

and so the probability of finding the electron in the n = 2 state at t > τ is 1 P2 =| C21 (t )|2 = 2 2 H 21 ′ 2 (1 − e iω 21τ )(1 − e − iω 21τ )  ω 21 2

3

2  16a 2   eEm  3π 2    16 eEa  ≈ sin = τ τ     2  4ma 2   9π 2    2π   9π  

for τ 

 . E1 − E2

For the transition 1 → 3, H 31′ = 3 H ′ 1 = 0, and so 2

P3 = C31 (t ) = 0. The validity of the time-dependent perturbation theory requires the time τ during which the perturbation acts should be small. The perturbation potential itself should also be small.

5074 For a particle of mass m in a one-dimensional box of length l, the eigenfunctions and energies are 2 nπ x sin , 0 ≤ x ≤ l, l l

ψ n(x) =

2

En =

1  nπ     n = ±1, ± 2,… 2m  l 

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Problems and Solutions in Quantum Mechanics

Suppose the particle is originally in a state n and the box length is increased to a length of 2l (0 ≤ x ≤ 2l ) in a time t  h / En. Afterwards what is the probability that the particle will be found in an energy eigenstate with energy En? (MIT) Sol:

First consider the process in which the box length is increased from l to 2l. As t  h , it is reasonable to assume that the state of the particle in the box is unable En to respond to the change during such a short time. Therefore the wave function of the particle after the change is completed is   ψ (x ) =   

2 nπ x sin , 0 ≤ x ≤ l, l l 0, l ≤ x ≤ 2l .

On the other hand, the eigenstates and eigenvalues of the same particle in the one-dimensional box of length 2l would be, respectively,

φn′ ( x ) = En′ =

1 n′π x sin , (0 ≤ x ≤ 2l ), l 2l 2

1  n′π     , (n′ = ±1, ± 2,…). 2m  2l 

The energy En of the particle corresponds to the energy level En′ in the 2l box, where n = n′ , i.e., n′ = 2n . The corresponding eigenstate is then φ2n . Thus the l 2l probability amplitude is ∞

A = ∫ φ2n ( x )ψ ( x )dx = −∞

2 l 2 nπ x 1 sin dx = , ∫ 0 t l 2

and the probability of finding the particle in an eigenstate with energy En is 1 2 P= A = . 2

5075 A particle is initially in its ground state in a box with infinite walls at 0 and L. The wall of the box at x = L is suddenly moved to x = 2L . a. Calculate the probability that the particle will be found in the ground state of the expanded box. b. Find the state of the expanded box most likely to be occupied by the particle.

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411

c. Suppose the walls of the original box [0, L] are suddenly dissolved and that the particle was in the ground state. Construct the probability distribution for the momentum of the freed particle. (Berkeley) Sol: a. The wave function of the particle before the box expands is   ψ (x ) =   

2 πx sin , x ∈[0, L], L L 0 otherwise.

The wave function for the ground state of the system after the box has expanded is   φ1 ( x ) =   

1 πx sin , x ∈[0, 2 L], L 2L 0 otherwise.

The probability required is then P1 =

∫

∞

−∞

2

2 L πx πx 32 sin sin dx = 2 . ∫ 0 L 2L L 9π

2

φ1* ( x )ψ ( x )dx =

b. The probability that the particle is found in the first exited state of the expanded box is P2 = where

∫

∞

−∞

2

φ2* ( x )ψ ( x )dx =

  φ2 ( x ) =   

2

2 L 2 πx 1 sin dx = , L ∫0 L 2

1 πx sin , x ∈[0, 2 L], L L 0 otherwise.

For the particle to be found in a state n ≥ 3, the probability is Pn =

L

nπ x π x 2 sin sin dx L ∫0 L 2L

2

2

n  n  sin  − 1 π sin  + 1 π 2  2  2 = 2 − . π (n − 2) (n + 2) n  sin 2  + 1 π 2  (n 2 − 4)2 32 1 ≤ < . 25π 2 2

32 = 2 π

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L

nπ x π x 2 Pn = sin sin dx ∫ L 0 L 2L

2

2

412

n  n  sin  − 1 π sin  + 1 π 2  2  2 = 2 − . − 2) π (n Mechanics (n + 2) Problems and Solutions in Quantum n  sin 2  + 1 π 2  (n 2 − 4)2 32 1 ≤ < . 2 25π 2

32 = 2 π

Hence, the particle is most likely to occupy the first excited state of the expanded box. c. The wave function of the freed particle with a momentum p is 1 e ipx/ . 2π  The probability amplitude is then 1 2 πx sin dx e − ipx / L L 2π  1 L /π [1 + e − ipL/ ] = . 1 − ( pL / π )2 π L

Φ( p) = ∫

L

0

The probability distribution for the momentum is therefore | Φ( p)|2 =

2π  3 L pL    1 + cos  . ( π 2 − p 2 L2 )2   2

5076 A particle of mass M is in a one-dimensional harmonic oscillator potential V1 = 1 kx 2 . 2 a. It is initially in its ground state. The spring constant is suddenly doubled ( k → 2k ) so that the new potential is V2 = kx 2 . The particle’s energy is then measured. What is the probability for finding that particle in the ground state of the new potential V2? b. The spring constant is suddenly doubled as in part (a), so that V1 suddenly becomes V2, but the energy of the particle in the new potential V2 is not measured. Instead, after a time T has elapsed since the doubling of the spring constant, the spring constant is suddenly restored back to the original value. For what values of T would the initial ground state in V1 be restored with 100% certainty? (CUSPEA)

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413

Sol: a. The wave function of the system before k change is

ψ (x ) =

1 π

1/4

1

 Mω 0  − 2 Mω 0 x 2 / .   e  

Suppose that the particle is also in the ground state of the new potential well after k change. Then the new wave function is

ψ ′( x ) =

1/4

1

 Mω 1  − 2 Mω1x 2 / .   e  

1 π

The transition matrix element is

ψ ′ |ψ = ∫

1/2

1 M  1 (ω 0 + ω 1 )x 2  1/4   (ω 0ω 1 ) exp  − M  dx π     2 1/2

1 M π 1/4   (ω 0ω 1 ) ⋅ π   1 M (ω 0 + ω 1 ) 2  1/4 (ω ω ) = 1 1 0 . 2 (ω 0 + ω 1 )

=

When k changes into 2k , ω 0 changes into ω 1 = 2ω 0 , thus

ψ ′ |ψ

2

( 2ω 02 )1/2 (ω 1ω 0 )1/2 = 1 1 ( 2 + 1)ω 0 (ω + ω 1 ) 2 2 0 21/4 = = 2 ⋅ 21/4 ( 2 − 1). 1 ( 2 + 1) 2 =

Hence, the probability that the particle is in the state ψ ′( x ) is 5

24

(

)

2 −1 .

b. The quantum state is not destroyed as the energy is not measured. At t = 0, ψ ( x , 0) = ψ 0 ( x ), ψ n ( x ) being the eigenstates of V1. We expand

ψ ( x , 0) in the set of eigenstates of V2: ψ ( x , 0) = ψ m′ ψ 0 ψ m′ ( x ) . Here and below we shall use the convention that a repeated index implies summation over that index. Then

ψ ( x , t ) = e − iH 2t /ψ ( x , 0) = ψ m′ |ψ 0 ψ m′ ( x ) e − iEm′ t / ,

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Problems and Solutions in Quantum Mechanics

where H 2 is the Hamiltonian corresponding to V2. Since ψ 0 ( x ) has even parity, parity conservation gives m = 2n + 1,  0, (ψ m′ ( x )|ψ 0 ( x ) =   ≠ 0, m = 2n, and so

ψ ( x , τ ) = ψ 2m ( x ) ψ 2m ψ 0 e − iE2 mτ / . Hence ψ ( x ,τ ) = ψ 0 ( x ) can be expected only if E2mτ /  = 2 N π + c , where N is a natural number and c is a constant, for any m. As 1  E2m =  2m +  ω 1′ ,  2 we require 1   2m +  ω 1′τ = 2 N π + c. 2 Setting 1 c = ω 1′ , 2 we require 2mω 1′τ = 2 N π , or 2ω 1′τ = 2 N ′π i.e.,

τ=

πN′ , ω 1′

where N ′ = 0, 1, 2, … Thus only if τ = N ′π certainty.

M , will the state change into ψ 0 ( x ) with 100% 2k

5077 A particle which moves only in the x direction is confined between vertical walls at x = 0 and x = a . If the particle is in the ground state, what is the energy? Suppose the walls are suddenly separated to infinity; what is the

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probability that the particle has momentum of magnitude between p and p + dp ? What is the energy of such a particle? If this does not agree with the ground state energy, how do you account for energy conservation? (Chicago) Sol:

When the particle is confined between x = 0 and x = a in the ground state, its wave function is  2 πx  sin , 0 ≤ x ≤ a, ψ0 =  a a  0 otherwise,  and its energy is E=

π 22 . 2ma 2

When the walls are suddenly removed to infinity, the wave function of the particle cannot follow the change in such a short time but will remain in the original form. However, the Hamiltonian of the system is now changed and the original wave function is not an eigenstate of the new Hamiltonian. The original wave function is to be taken as the initial condition in solving the Schrödinger equation for the freed particle. The wave packet of the ground state in the original potential well will expand and become uniformly distributed in the whole space when t → ∞. Transforming the original wave packet to one in momentum ( p = k ), representation, we have

ψ ( p) =

a 2 1  π x  ikx sin   ⋅ e dx 2π  ∫0 a  a 

=−

aπ 1 + e ika .  (ka)2 − π 2

During the short time period of separating the walls the probability that the momentum is in the range p → p + dp is given by f ( p )dp = {|ψ ( p )|2 + |ψ (− p)|2 }dp  ka  cos 2   dp  2 aπ =8  [(ka)2 − π 2 ]2 a f (0)dp = |ψ (0)|2 dp = 4 3 dp π 

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if p ≠ 0; if p = 0.

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Because the new Hamiltonian is not time-dependent, we can calculate the average value of the energy using the original wave function:  ka  cos 2   2 2 2 ∞ p ∞ k  2 aπ E=∫ f ( p )dp = ∫ ⋅8 d(k ) 2 0 2m 0 2m  [(ka) − π 2 ]2 πy y 2cos 2  2 ∞  2  4 π 22 = = , dy 2ma 2 ma 2 ∫0 ( y 2 − 1)2 where y = ka . This means that the energy of the system is not changed during π the short time period of separating walls, which is to be expected as a p ψ 0 H before ψ 0 = ∫ ψ 0* ψ 0 dx , 0 2m  −iH after t  ψ (t ) H after ψ (t ) = ψ 0 exp(iH aftert / ) × H after exp  ψ    0 a

= ψ 0 H after ψ 0 = ∫ ψ 0* 0

p2 ψ 0 dx 2m

= ψ 0 H before ψ 0 . If the walls are separating to an infinite distance slowly or if the walls are not infinite high, there would be energy exchange between the particle and the walls. Consequently, the energy of the particle would change during the time of wall separation.

5078 A nucleus of charge Z has its atomic number suddenly changed to Z + 1 by

β -decay as shown in Fig. 5.21. What is the probability that a K-electron before the decay remains a K-electron around the new nucleus after the β − decay? Ignore all electron-electron interactions. (CUSPEA) Sol:

The wave function of a K-electron in an atom of nuclear charge Z is

ψ (r ) = NZ 3/2e − rZ /a .

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As

∫

∞

0

417

r 2 drN 2e −2r /a = 1, the probability that the K-electron remains in the original

orbit is 2

P = ψ Z +1 (r ) ψ Z (r )

=

(1 + Z1 )3 . (1 + 21Z )6

Fig. 5.21 5079 3

A tritium atom ( H ) can undergo spontaneous radioactive decay into a helium-3 ion ( 3 He + ) by emission of a beta particle. The departure of the electron is so fast that to the orbital electron the process appears as simply an instantaneous change in the nuclear charge from Z = 1 to Z = 2. Calculate the probability that the He ion will be left in its ground state. (Berkeley) Sol:

The wave function of the ground state of He + is +

ψ 1Hes =

1 π

 2   a

3/2

exp{−2r /a},

where a is the Bohr radius. Let the wave function of 3 H be ϕ (r ). As the process of β decay takes place very fast, during the time period in which the 3 H becomes 3 He + the wave function does not have time to change. Hence the probability that the 3 He + is in the ground state is +

P=

ψ 1Hes ϕ ϕϕ

2

2

.

Initially, the 3 H is in the ground state so that

ϕ (r ) =

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1  1   π  a

3/2

e − r /a .

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Therefore,  2 P = 4 2  a  =

27 36

∫

∞

0

3/2 ∞

{ }

3r ∫0 r exp − a dr 2

2

x 2e − x dx =

2

29 = 0.702. 36

5080 3

Tritium (hydrogen of mass 3 i.e. H ) is beta-radioactive and decays into a helium nucleus of mass 3 ( 3 He ) with the emission of an electron and a neutrino. Assume that the electron, originally bound to the tritium atom, was in its ground state and remains associated with the 3 He nucleus resulting from the decay, forming a 3 He + ion. a. Calculate the probability that the 3 He + ion is found in its 1s state. b. What is the probability that it is found in a 2p state? (MIT) Sol:

Neglect the small difference in reduced mass between the hydrogen atom and the helium atom systems. The radius of the ion 3 He + is a0 /2, where a0 is the Bohr radius, so the wave functions are 2 − r /a0 , e a03/2 + 2 −2 r /a0 ψ 1He = Y00 , s 3/2 e a  0   2

ψ 1Hs = Y00

+

ψ 2Hep = Y1m

1 2r − r /a0 e 3/2 2 6(a0 / 2) a0

(m = 1, 0, − 1). a. The amplitude of the probability that the ion He + is in the state 1s is

(

A = ∫ ψ 1He s

+

)

*

ψ 1Hs d 3 x =

∞

27/2 2 −3r /a0 16 2 . r e dr = a03 ∫0 27

Hence the probability is | A |2 = 2 ( 16 27 ) = 0.702 . 2

b. On account of the orthonormality of spherical harmonics, the probability that the ion 3 He + is in a state 2p is zero Y1m | Y00 = 0 .

(

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)

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419

5081 A beam of excited hydrogen atoms in the 2s state passes between the plates of a capacitor in which a uniform electric field E exists over a distance l. The hydrogen atoms have velocity υ along the x-axis and the E field is directed along the z-axis, as shown in Fig. 5.22.

Fig. 5.22 All the n = 2 states of hydrogen are degenerate in the absence of the E field, but certain of them mix when the field is present. a. Which of the n = 2 states are connected in first order via the perturbation? b. Find the linear combination of n = 2 states which removes the degeneracy as much as possible. c. For a system which starts out in the 2s state at t = 0 , express the l wave function at time t ≤ . v d. Find the probability that the emergent beam contains hydrogen in the various n = 2 states. (MIT) Sol:

Consider the potential energy eEz of the electron ( charge − e ) of a hydrogen atom in the external electric field Eeˆz as a perturbation. As the n = 2 states are degenerate, we calculate 2 ′m′ H ′ 2m , where H ′ = eEz, and  = 0, m = 0. It is known that only the following matrix elements are non-vanishing: 2,  + 1, m z 2,  ,m = −

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( − m + 1)( + m + 1) 2,  + 1 r 2,  , (2 + 1)(2l + 3)

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with 3a n n2 −  2 . 2 Thus all the matrix elements are zero except 2,  + 1| r | 2,  =

210 H ′ 200 = −3eEa . a. The 2s and 2p states are connected via the perturbation in first order since for the H ′ matrix only elements with ∆ = ±1 are nonzero. b. The perturbation Hamiltonian is  0 −3eEa  H′ =   , 0  −3eEa whose secular equation −λ −3eEa =0 −3eEa −λ gives eigenvalues ±3eEa , the corresponding eigenstate vectors being

1  1 . 2  ∓1

The degeneracy of the state n = 2 is now removed. c. As t = 0, just before the atoms enter the electric field,

ψ (0) =

1 ( + + − ), 2

where + =

1  1 , 2  1

− =

1  1 2  −1

are the state vectors obtained in (b). At time 0 < t ≤ l when the atoms are subject to the electric field, v 1 i 3eEat / |ψ (t )〉 = + + e − i 3eEat / − ) (e 2  cos(3eEat /  )   3eEat   3eEat  =  2s + i sin   2p ,  = cos  i eEat  sin(3 / )      where 2s =

  1 ( + + − ) =  10  , 2  

2p =

  1 ( + − − ) =  10  . 2  

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d. For t ≥ l , we find from (c) the probabilities v 2 3eEat 200 ψ (t ) = cos 2 ,  2 3eEat 210 ψ (t ) = sin 2 . 

5082 a. Consider a particle of mass m moving in a time-dependent potential V ( x , t ) in one dimension. Write down the Schrödinger equations appropriate for two reference systems ( x , t ) and ( x ′ , t ) moving with respect to each other with velocity υ (i.e. x = x ′ + υt ).

Fig. 5.23 b. Imaging that a particle sits in a one-dimensional well (Fig. 5.23) such that the well generates a potential of the form mω 2 x 2 /2. At t = 0 the well is instantly given a kick and moves to the right with velocity υ (see Fig. 5.24). In other words, assume that V ( x , t ) has the form    V(x, t) =   

1 mω 2 x 2 , for t < 0, 2 1 mω 2 x ′ 2 , for t > 0. 2

If for t < 0 the particle is in the ground state as viewed from the ( x , t ) coordinate system, what is the probability that for t > 0 it will be in the ground state as viewed from the ( x ′ , t ) system? (Columbia)

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Fig. 5.24 Sol: a. Both ( x , t ) and ( x ′ , t ) are inertial systems, and so the Schrödinger equations are: for the ( x , t ) system, ∂  2 d 2   − 2m dx 2 + V ( x , t )ψ ( x , t ) = i ∂t ψ ( x , t ),   for the ( x ′ ,t ) system, ∂  2 d 2   − 2m dx ′ 2 + V ′( x ′ , t )ψ ( x ′ , t ) = i ∂t ψ ( x ′ , t ),   where V ′( x ′ , t ) = V ′( x − υt , t ) = V ( x , t ). b. This problem is the same as Problem 6052 for the following reason. Consider an observer at rest in the ( x ′ , t ) system. At t < 0 he sees the particle as sitting in the ground state of the potential well V. At t = 0, the potential well V instantly requires a velocity υ to the right along the x direction. The situation is the same as if V remains stationary while the particle acquires a velocity −υ along the −x direction. It is required to find the probability that the particle remains in the ground state. Problem 6052 deals with an Al nucleus which by emitting a y to the right acquires a uniform velocity to the left. The physics involved is exactly the same as the present problem and we can just make use of the results there.

5083 If the baryon number is conserved, the transition n ↔ n known as “neutron oscillation” is forbidden. The experimental limit on the time scale of such oscillations in free space and zero magnetic field is τ n− n ≥ 3 × 10 6 sec. Since neutrons occur abundantly in stable nuclei, one would naively think it possible to obtain a much better limit on τ n− n . The object of this problem is to under-

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423

stand why the limit is so poor. Let H0 be the Hamiltonian of the world in the absence of any interaction which mixes n and n. Then H0 n = mn c 2 n and H0 n = mn c 2 n for states at rest. Let H′ be the interaction which turns n into n and vice versa: H ′ n = ε n and H ′ n = ε n , where ε is real and H′ does not flip spin. a. Start with a neutron at t = 0 and calculate the probability that it will be observed to be an antineutron at time t. When the probability is first equal to 50%, call that time τ n− n . In this way convert the experimental limit on τ n− n into a limit on ε . Note mn c 2 = 940 MeV . b. Now reconsider the problem in the presence of the earth’s magnetic 1   field  B0 ≥ gauss  . The magnetic moment of the neutron is   2

µn ≈ −6 × 10 −18 MeV/gauss. The magnetic moment of the antineutron is opposite. Begin with a neutron at t = 0 and calculate the probability it will be observed to be an antineutron at time t. (Hint: work to lowest order in small quantities.) Ignore possible radiative transitions. c. Nuclei with spin have non-vanishing magnetic fields. Explain briefly and qualitatively, in sight of part (b), how neutrons in such nuclei can be so stable while τ n− n is only bounded by τ n− n ≥ 3 × 10 6 sec. d. Nuclei with zero spin have vanishing average magnetic field. Explain briefly why neutron oscillation in such nuclei is also suppressed. (MIT) Sol: a. To find the eigenstates of the Hamiltonian H = H 0 + H ′ we introduce in the neutron-antineutron representation the state vectors  0 1 n ∼ , n ∼ . 1  0

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As n H 0 + H ′ n = mn c 2 , n H0 + H ′ n = ε we have energy eigenvalue equation  m c2 − E ε n   mn c 2 − E ε

  a    = 0.   b 

Solving the equation, we get  a 1  1 E+ = mn c 2 + ε ,   = ,  b + 2  1  a 1  1 E− = mn c 2 − ε ,   = .  b − 2  −1 At t = 0, the system is in the neutron state and so 1 1 n = E+ + E− , 2 2 where 1  1 1  1 , E− ∼ .   2  1 2  −1

E+ ∼

At time t, the state of the system becomes 1 − iE+t / 1 − iE−t / ψ,t = e E+ + e E− . 2 2 In the neutron-antineutron representation, we thus have  εt  cos  − imn c 2t/  ψ ,t = e εt   −i sin 

  ,  

and hence 2 εt εt n − ie − imnc t / sin n .   The probability that at time t the particle is observed as an antineutron is therefore 2

ψ , t = e − imnc t / cos

2

P(t ) = ie − imnc t / sin

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2

εt = sin 2 (ε t / ). 

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425

1 τ n−n is defined as the time at which P = , i.e., 2 1 π  . τ n−n = arcsin = ε 2 4ε Then as Tn−n ≥ 3 × 106 s , the experimental limit on ε is ε ≤

ε≤

h = 1.7 × 10−28 MeV . 8 × 3 × 106

h = 1.7 × 10−28 MeV 8 × 3 × 106

b. Noting that H ′ does not change the spin, after introducing the magnetic field one can take the neutron-antineutron representation 1  0  0  0  0 1  0  0 n ↑∼   , n ↑∼   , n ↓∼   , n ↓∼    0  0 1  0  0  0  0  1  and calculate the matrix elements of H ′ + µ B0. Thus one obtains the perturbation Hamiltonian  − µn B0  ε   0   0

ε

0

− µn B0

0

0

µn B0

0

ε

  0  ε   µn B0  0

with µn ≈ −6 × 10−18 MeV/Gs, µn ≈ 6 × 10−18 MeV/Gs . This gives rise to two eigenequations:  − µ B − E (1) n 0   ε  µ B − E (1)  n 0  ε

ε µn B0 − E (1) ε − µn B0 − E (1)

    a ↑  = 0,   b ↑      a ↓  = 0,   b ↓ 

where we have used the relation µn = − µn . Solving the two equations, we obtain E±(1) = ± λ = ±  2 + ( µn B0 )2 ,

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and hence   a ↑ 1  =   2λ   b ↑ + 

 λ + µn B0 λ − µn B0   a ↑  ,  = 1  2λ  − λ − µ0 B0 λ + µn B0   b ↑  − 

 ,  

  a ↑ 1    = 2λ   b ↑ + 

 λ − µn B0 λ + µn B0   a ↑  ,  = 1  2λ  − λ + µ0 B0 λ − µn B0   b ↑  − 

 .  

As t = 0, the system is in the neutron state n ↑∼

λ − µn B0  a ↑ λ + µn B0  a ↑   +   , 2λ  b ↑  2λ  b ↑  + −

n ↓∼

λ − µn B0  a ↑ λ − µn B0  a ↑   +   . 2λ  b ↑  2λ  b ↑  + −

At time t, the states of the system are 2 1 (↑) ∼ e − imnc t / 2λ  (λ − µ B )e − iλt / + (λ + µ B )e iλt / n 0 n 0 × λ  2 2 − i t /  (λ − ( µn B0 ) (e − e i λt /  )  2

(↓) ∼ e − imnc t /

 ,  

1 2λ

 (λ + µ B)e − iλt / + (λ − µ B )e iλt / n n 0 × λ  2 2 − i t /  (λ − ( µn B0 ) (e − e i λt /  ) 

 .  

Therefore the probability of n ↑→ n ↑ is Pn↑→n ↑ (t ) = =

ε 2 2 λt sin λ2  ε 2 + ( µn B0 )2 t ε2 , sin 2 2 ε + ( µn B0 )  2

and that of n ↓→ n ↓ is Pn↓→n ↓ (t ) = =

ε 2 2 λt sin λ2  ε 2 + ( µn B0 )2 t ε2 2 . sin ε 2 + ( µn B0 )2 

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Finally, if the neutron is not polarized the probability of n → n is 1 1 P(t ) = Pn↑→n ↑ (t ) + Pn↓→n ↓ (t ) 2 2 =

ε 2 + ( µn B0 )2 t ε2 2 , sin ε 2 + ( µn B0 )2 

which means that the polarization of the neutron has no effect on the transition probability. As µn B0  ε , 2

 1.65 × 10−28  P(t ) ≤  ≈ 0.3 × 10−20 ,  6 × 10−18 × 1/ 2  which shows that the transition probability is extremely small. c. If nuclear spin is not zero, the magnetic field inside a nucleus is strong, much larger than 0.5 Gs. Then the result of (b) shows that Pn→n  10−20 , which explains why the neutron is stable inside a nucleus. d. If nuclear spin is zero, then the average magnetic field in the nucleus is zero. Generally this means that the magnetic field outside the nucleus is zero while that inside the nucleus may not be zero, but may even be very large, with the result that Pn→n is very small. Besides, even if the magnetic field inside the nucleus averaged over a long period of time is zero, it may not be zero at every instant. So long as magnetic field exists inside the nucleus, Pn→n becomes very small. Neutron oscillation is again suppressed.

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Part VI

Scattering Theory & Quantum Transitions

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6 6001 Derive the quantum mechanical expression for the s-wave cross section for scattering from a hard sphere of radius R. (MIT) Sol:

The effect of the rigid sphere is equal to that of the potential ∞ (r < R ), V (r ) =  0 (r > R ). Let the radial wave functíon of the s-wave be R0 (r ) = χ 0 (r ) / r . Then the Schroödinger equation can be written as

χ 0′′(r ) + k 2 χ 0 (r ) = 0

(r > R ),

with

χ 0 (r ) = 0 (r < R ), k =

1 2mE . 

The solution for r > R is χ 0 (r ) = sin(kr + δ 0 ). The continuity of the wave function at r = R gives sin(kR + δ 0 ) = 0, which requires δ 0 = nπ − kR, or sin δ 0 = (−1)n+1 sin kR (n = 0, 1, 2, …) . Therefore the total cross-section of the s-wave is 4π 4π σ t = 2 sin 2δ 0 = 2 sin 2 kR. k k 2 For low energies, k → 0, sin kR ≈ kR , and so σ t = 4π R . For high energies, k → ∞

and σ t ≈ 0.

431

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6002 The range of the potential between two hydrogen atoms is approximately 4Å. For a gas in thermal equilibrium, obtain a numerical estimate of the temperature below which the atom-atom scattering is essentially s-wave. (MIT) Sol: The problem concerns atom-atom scatterings inside a gas. If mainly s partial waves are involved, the uncertainty principle requires µυ r a ≤  ,where µ = 12 m p is the reduced mass of the two atoms, v r = v 1 − v 2 is the relative velocity between the two atoms, of velocities v 1 , v 2 , a = 4 Å. When thermal equilibrium is reached,

υ = 0,

1 3 m p υ 2 = kT , 2 2

k being Boltzmann’s constant and T the absolute temperature. The meansquare value of the relative speed υ r is

υr2 = ( v 1 − v 2 )2 = υ12 + υ 22 − 2 v 1 ⋅ v 2 = 2 υ 2 =

6kT , mp

since on average v 1 ⋅ v 2 = 0, υ12 = υ 22 = υ 2 . Thus

µaυr ≈

m pa 6kT ≤ , 2 mp

i.e., T≤

2

2

2 2  c  1 2 × (6.58 × 10−16 )2  3 × 1010  1 ×   =  4 × 10−8  8.62 × 10−5 3m pc 2  a  k 3 × 938 × 106 = 2o K.

Hence, under normal temperatures the scattering of other partial waves must also be taken into account.

6003 A nonrelativistic particle of mass m and energy E scatters quantummechanically in a central potential V ( r ) given by 2

V (r ) =

2  λ  U ( r ), U ( r ) = −2  ,  cosh λ r  2m

where λ is a parameter. This particle potential has the property that the cross section σ ( E ) grows larger and larger as E → 0, diverging for E = 0 .

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Clearly, for E very small the cross section is dominated by the s-wave ( l = 0) contribution. Thus for small E one needs to compute only the l = 0 partial wave amplitude. In connection with this, to save you mathematical efforts, you are told mathematically that the equation d 2φ + Aφ = U ( r )φ , dr 2 where A is a positive constant, has a general solution

φ = α ( λ tanh λ r − ik )e ikr + β ( λ tanh λ r + ik )e − ikr , −x

x

where k = A and α and β are integration constants. Recall tanh x = e x −e− x e +e

and compute σ ( E ) for E → 0 . (CUS) Sol: The s partial wave function is spherically symmetric, its equation being −

2 2 1 d  2 d   U (r )φ (r ) = Eφ (r ) . ⋅ 2 ⋅ r φ (r ) +  2m 2m r dr  dr

With R(r ) = φ (r )r , the above becomes R′′(r ) +

2 2m   E − U (r ) R(r ) = 0, 2    2m 

i.e., R′′(r ) +

2mE R(r ) = U (r )R(r ). 2

The solution is R(r ) = α (λ tanh λ r − ik )e ikr + β (λ tanh λ r + ik )e − ikr , where k=

2mE . 2

ikr Consider r → 0. As φ (0) is finite, R → 0. Then as tanh λ r → λ r , e → 1, we have

for r → 0 R(r ) ≈ α (λ 2r − ik ) + β (λ 2r + ik ) → α (−ik ) + β (ik ) = 0,

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giving α = β . Consider r → ∞ . As tanh λ r → 1, we have for r → ∞ R(r ) → α (λ − ik )e ikr + β (λ + ik )e − ikr = α [(λ − ik )e ikr + (λ + ik )e − ikr ] = α λ 2 + k 2 (e ikr −iα1 + e − ikr +iα1 ) = α λ 2 + k 2 ⋅ 2cos(kr − α 1 ) π = 2α λ 2 + k 2 sin(kr + − α 1 )  sin(kr + δ 0 ), 2 where

δ0 =

π − α1 , 2

and α 1 is defined by tan α 1 = k / λ , or α 1 = tan −1k / λ . Thus the total cross section for scattering is 4π 4π σ t = 2 sin 2 δ 0 = 2 cos 2α 1. k k For low energies, E → 0, k → 0, α 1 → 0 , and so

σt =

4π 2π  2 . = k2 mE

6004 A particle of mass m is interacting in three dimensions with a spherically symmetric potential of the form V (r ) = −Cδ (| r | −a). In other words, the potential is a delta function that vanishes unless the particle is precisely a distance “a” from the center of the potential. Here C is a positive constant. a. Find the minimum value of C for which there is bound state. b. Consider a scattering experiment in which the particle is incident on the potential with a low velocity. In the limit of small incident velocities, what is the scattering cross section? What is the angular distribution? (Princeton) Sol: a. Suppose the eigenfunction of a bound state of the single-particle system has the form

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ψ (r ) = R(r )Ylm (θ , ϕ ). Then the radial function R(r ) satisfies l(l + 1)  2 2mC  R = 0, R′′ + R′ + (ik )2 + 2 δ (| r | −a) −  r 2  r 



(1)

where k = −2mE /  2 . Note E < 0 for a bound state. If r ≠ a , the equation is an imaginary-variable spherical Bessel equation. For r < a it has the solution that is finite at r = 0 R(r ) = Ak jl (ikr ), where j is spherical Bessel function of the first kind of order l. For r > a it has the solution that is finite for r → ∞ R(r ) = Bk hl(1) (ikr ), where hl(1) is spherical Bessel function of the third kind, (spherical Hankel function) of order l. The wave function is continuous at r = a . Thus Ak j (ika) = Bk h(1) (ika). Integrating Eq. (1) from a − ε to a + ε , where ε is a small positive number, and then letting ε → 0, we have R′(a + 0) − R′(a − 0) = −C ′R(a), (2)

where

C′ =

2mC . 2

Suppose there is at least a bound state. Consider the ground state l = 0, for which    R(r ) =    

Aj0 (ikr ) = A

sin(ikr ) , r < a, ikr

Bh0(1) (ikr ) = B

(−1)e − kr , r > a. kr

Differentiating R(r ) and letting r → a , we have B e − ka  1 ⋅  k +  , k a  a A  k cosh(ka) sinh(ka)  . R′(a − 0) =  − k a a 2  R′(a + 0) =

Substituting these in Eq. (2) gives

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aC ′ =

2ka . 1 − e −2 ka

As for x ≥ 0, x ≥ 1 − e − x , we have aC ′ ≥ 1 and Cmin ′ = 1/ a , or  Cmin =

2 . 2ma

This is the minimum value of C for which there is a bound state. b. We use the method of partial waves. When the particle is incident on the potential with a low velocity, only the  = 0 partial wave is important, for which the radial wave equation 2 2mC   R′′ + R′ +  k 2 + 2 δ (r − a) R = 0 . r    On setting R(r ) = χ 0 (r ) / r it becomes 2mC χ 0′′+  k 2 + 2 δ (r − a) χ 0 = 0,   



(3)

which has solutions finite at r → 0 and r → ∞  A sin kr , r < a, χ 0 (r ) =   sin(kr + δ 0 ), r > a. As χ 0 (r ) is continuous at r = a , we require A sin ka = sin(ka + δ 0 ) . Integrating Eq. (3) from a − ε to a + ε gives

χ 0′ (a + ε ) − χ 0′ (a − ε ) = −

2mC χ 0 (a). 2

Substituting in the expressions for χ 0 (r ), it becomes ka ka 2mCa − =− 2 . tan(ka + δ 0 ) tan ka  For k → 0 , the above becomes ka 2mCa −1 = − 2 , tan δ 0  or tan δ 0 =

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i.e., sin δ 0 =

ka  2maC  k 2a 2 +  1 −   2 

2

≈

437

ka .  2maC   1 −  2 

Hence, the total scattering cross section is

σt =

4π 4π a 2 2 . δ ≈ 2 2 sin 0 k  2maC   1 −  2 

Note that for low velocities only s-waves (l = 0) need be considered and the differential cross section is simply

σ (θ ) =

1 2maC  2 sin 2 δ 0 = a  1 −   k2 2 

−2

,

which is independent of the angles. Thus the angular distribution is isotropic.

6005 a. Find the s-wave phase shift, as a function of wave number k, for a spherically symmetric potential which is infinitely repulsive inside of a radius r0, and vanishes outside of r0. b. For k → 0 discuss the behavior of the phase shifts in the higher partial waves. (Wisconsin) Sol: a. This is a typical scattering problem that can be readily solved by the method of partial waves. The potential can be expressed as  ∞ , r < r0 , V (r ) =  0, r > r0 .  The radial wave function for the  partial wave is  0, r < r0 , R (kr ) =  (2)  j (kr )cos δ  − n (kr )sin δ  , r > r0 . Here j and n are spherical Bessel function and spherical Neumann function of order . These functions have the asymptotic forms

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1 →∞ j ( x ) x  → sin( x − π / 2), x 1 →∞ n ( x ) x  → − cos( x − π / 2). x Hence, for r > r0 we have π   →∞ R (kr ) kr → sin  kr − +δ .   2 The phase shift δ  can be determined by the continuity of the wave function at r = r0 . Writing kr0 = x , the continuity condition R ( x ) = j ( x )cos δ  − n ( x )sin δ  = 0 gives tan δ  =

j ( x ) . n ( x )

In the low-energy limit x → 0, the functions have the asymptotic forms x , (2 + 1)!! (2 − 1)!! →0 n ( x ) x →− , x +1 →0 j ( x ) x →

so that tan δ  =

j  ( x ) x →0 x 2 +1 . → − n ( x ) [(2l − 1)!!]2 (2 + 1)

Thus the s-wave ( = 0) phase shift is tan δ 0 = − x = − kr0 . It gives a finite contribution to the scattering and the corresponding total cross section is 4π 4π σ t = 2 sin 2 δ 0 ≈ 2 δ 02 ≈ 4π r02 . k k The scattering is spherically symmetric, and the total cross section is four 2 times the classical value π r0 . b. Consider the low-energy limit k → 0 . As tan δ l ≈ −

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δ  falls off very rapidly as  increases. All the phase shifts vanish as k → 0, except for the l = 0 partial wave. Hence, s-waves predominate in low-energy scattering. Physically, particles with higher partial waves are farther away from the force center so the effect of the force on such particles is smaller, causing | δ  | to be smaller.

6006 A particle of mass m is scattered by the central potential V (r ) = −

2 1 , 2 2 ma cosh ( r / a )

where a is a constant. Given that the equation d2 y 2 y=0 + k2 y + 2 dx cosh2x has the solutions y = e ± ikx (tanh x ∓ ik ), calculate the s-wave contribution to the total scattering cross section at energy E. (MIT) Sol: Letting χ 0 (r ) = rR(r ) we have for the radial part of the Schrödinger equation for s-waves (l = 0) d 2 χ 0 (r ) 2m  1 2  E χ 0 (r ) = 0. + + 2 2  2 2 dr ma cosh (r / a)    With x = r / a, y( x ) = χ 0 (r ) and k =

2mE 2

, the above becomes

d 2 y( x ) 2 2 2 y( x ) = 0 . + k a y( x ) + dx 2 cosh 2( x ) This equation has solutions y = e ± iakx (tanh x ∓ iak ) . For R finite at r = 0 we require y ( 0 ) = 0 . The solution that satisfies this condition has the form y( x ) = e iakx (tanh x − iak ) + e − iakx (tanh x + iak ) = 2cos(akx )tanh x + 2ak sin(akx ), or r χ 0 (r ) = 2cos(kr ) tanh   + 2ak sin(kr ).  a

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Thus 1 dχ0 = χ 0 dr

r 1 r ak 2 cos(kr ) − k sin(kr )tanh   + cos(kr )sech 2    a a  a r   ak sin(kr ) + cos(kr )tanh    a

→∞ r →

ak 2 cos(kr ) − k sin(kr ) ak cot(kr ) − 1 =k . cot(kr ) + ak ak sin(kr ) + cos(kr )

On the other hand if we write χ 0 in the form

χ 0 (r ) = sin(kr + δ 0 ), then as 1 dχ0 cot(kr )cot δ 0 − 1 , = k cot (kr + δ 0 ) = k χ 0 dr cot(kr ) + cot δ 0 we have to put cot δ 0 = ak , or sin 2δ 0 =

1 . 1 + a2k 2

Hence, the s-wave contribution to the total scattering cross section is

σt =

2π  2 4π 4π 1 2 δ = 2 = 2 2 2 sin 0 k k 1+ a k mE

1 . 2a 2mE 1+ 2

6007 A spinless particle of mass m, energy E scatters through angle θ in an attractive square-well potential V ( r ):  −V0 , 0 < r < a , V (r ) =  r > a.  0,

V0 > 0,

a. Establish a relation among the parameters V0 , a, m and universal constants which guarantees that the cross section vanishes at zero energy E = 0. This will involve a definite but transcendental equation, which you must derive but need not solve numerically. For parameters

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441

meeting the above condition, the differential cross section, as E → 0, will behave like ∂σ E → E λ F(cosθ ). →0 ∂Ω b. What is the numerical value of the exponent λ ? c. The angular distribution function F(cosθ ) is a polynomial in cosθ . What is the highest power of cosθ in this polynomial? (Princeton) Sol: a. When the energy is near zero, only the partial wave with l = 0 is important. Writing the radial wave function as R(r ) = χ (r ) / r , then χ (r ) must satisfy the equations  2mE r > a,  χ ′′ + 2 χ = 0,     χ ′′ + 2m ( E + V ) χ = 0, 0 < r < a 0  2 with k =

2mE 2

,K=

2m ( E + V0 ) 2

. The above has solutions

χ (r ) = sin(kr + δ 0 ), r > a, χ (r ) = A sin( Kr ), 0 < r < a. As both χ (r ) and χ ′(r ) are continuous at r = a , we require sin(ka + δ 0 ) = A sin( Ka), k cos(ka + δ 0 ) = K A cos( Ka), or K tan(ka + δ 0 ) = k tan( Ka), and hence k  δ 0 = tan −1  tan( Ka) − ka . K  For E → 0 k → 0, K → k0 ≡

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2mV0 , 2

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Problems and Solutions in Quantum Mechanics

and so  tan(k0a)  δ0 → k  − a .   k0 For the total cross section to be zero at E = 0, we require 2

 tan(k0a)  4π 2 δ 0 → 4π a 2  − 1 = 0, 2 sin k   k0a or tan(k0a) = k0a, i.e.  2mV0  2mV0 tan  a = a,     which is the transcendental equation that the parameters V0 , a, m and universal constant  must satisfy. b & c When k → 0, the partial wave with l = 0 is still very important for the differential cross-section, although its contribution also goes to zero. Expanding tan( Ka) as a Taylor series in k, we have

(

)

tan( Ka) = tan a k 2 + k02 = tan(k0a) +

ak 2 +. 2k0 cos 2(k0a)

2

Neglecting terms of orders higher than k , we have k  δ 0 = tan −1  tan( Ka) − ka K   2    k  k 2a k ≈ tan −1   1 − 2   tan(k0a) +   − ka 2 2k0 cos (k0a)    k0  2k0   k  k3 k 3a ≈ tan −1  tan(k0a) − 3 tan(k0a) + 2 2  − ka 2k0 2k0 cos (k0a)   k0   k 3a k 3a ≈ tan −1 ka − 2 + 2 2  − ka 2k0 2k0 cos (k0a)   k 3a k 3a k 3a 3 ≈ 2 2 − 2− . 2k0 cos (k0a) 2k0 3 Hence,

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443

dσ 1 ≈ sin 2δ 0 dΩ k 2 ≈ k 4 F (k0 , a) 2

 2m  = E 2  2  F (k0 , a).   Thus the differential cross section per unit solid angle is approximately iso2 tropic and proportional to E for E → 0 . To find the contribution of partial wave with  =1, consider its radial wave equations 1 d  2 d   2 2 R  +  K − 2  R = 0 (r < a), r r 2 dr  dr   r  1 d  2 d   2 2 R  +  k − 2  R = 0 (r > a). r r 2 dr  dr   r  The solutions have the form of the first-order spherical Bessel function sin ρ cos ρ j1 ( p ) = 2 − ρ , or ρ  sin( Kr ) cos( Kr ) − , 0 < r < a,  ( Kr )2 Kr  R1 =   A  sin(kr + δ 1 ) − cos(kr + δ 1 )  , r > a.    (kr )2 kr   The continuity of R1 and the first derivative of r 2 R at r = a gives sin Ka cos K a  sin(ka + δ 1 ) cos(ka + δ 1 )  − = A − 2 , ( Ka)2 Ka ka  (ka)  sin Ka = A sin(ka + δ 1 ). Taking the ratios we have k 2 [1 − Ka cot( Ka)] = K 2 [1 − ka cot(ka + δ 1 )] , or  k 2 k 2a cot( Ka)  tan(ka + δ 1 ) = ka 1 + 2 − + O( k 4 )  k0  k0  = ka + O ( k 3 ) ,

as  1 k2  + O( k 4 )  . K = k 2 + k02  k0 1 + 2 2 k 0   Hence, 1 δ 1 = tan −1  ka + O(k 3 ) − ka = − (ka)3 + O(k 3 ) = O(k 3 ) . 3

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∂σ Thus its contribution to ∂Ω , 9 sin 2 δ 1 cos 2θ , k2 is also proportional to k 4 . Similarly, for l = 2,    R2 =    

1  3 cos( Kr )  3  ( Kr )3 − Kr  sin( Kr ) − ( Kr )2 ,  

0 < r < a,

1 3 cos(kr + δ 2 )  3 , r > a.  (kr )3 − kr  sin(kr + δ 2 ) − (kr )2  

The continuity of R and (r 3 R2 )′ at r = a then gives k 2 [tan(ka + δ 2 ) − ka] [3 − (ka)2 ]tan(ka + δ 2 ) − 3ka = . K 2 [tan( Ka) − Ka] [3 − ( Ka)2 ]tan( Ka) − 3Ka Let y = tan(ka + δ 2 ) − ka. The above becomes 1 y(−1 + O(k ))  3  (ka)3 − ka  y − 1 = bK 2 (1 + O ⋅ (k 2 )) ,   where b=

a . 2 cos 2(k0a) − k0

Therefore 1 3 1 1 − + 2 + O( k ) 3 (ka) ka bK 1 = 2 3  (ka) (ka)3  + + O( k 4 )  1− (ka)3  3 3bK 2 

y=

=

(ka)3  (ka)2 (ka)3  − + O( k 4 )  1+ 3  3 3bK 2 

=

(ka)3 (ka)5 + + O(k 6 ), 3 9

and

δ 2 = tan −1( y + ka) − ka ≈ y = O(k 3 ).

dσ Thus the contribution of partial waves with l = 2 to dΩ is also proportional to k 4 . This is true for all  for E → 0.

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445

Hence ∂σ 1 = | f (θ )|2 = 2 ∂Ω k

∞

∑(2l + 1)e l sin δ l Pl (cosθ ) iδ

2

l =0

E → k 4 F (cosθ )  E 2 F (cosθ ), →0 and the exponent of E is λ = 2. The highest power of cosθ in the angular distribution function is also 2 since the waves consist mainly of  = 0 and l = 1 partial waves.

6008 1. The shell potential for the three-dimensional Schrödinger equation is V ( r ) = αδ ( r − r0 ) . a. Find the s-state ( l = 0) wave function for E > 0. Include an expression that determines the phase shift δ . With k = 2mE show that in the limit k → 0, δ → Ak, where A is a constant (called the scattering length). Solve for A in terms of α and r0. b. How many bound states can exist for l = 0 and how does their existence depend on α ? (Graphical proof is acceptable) c. What is the scattering length A when a bound state appears at E = 0? Describe the behavior of A as α changes from repulsive (α > 0) to attractive, and then when α becomes sufficiently negative to bind. Is the range of A distinctive for each range of α ? Sketch A as a function of α . (MIT) Sol: a. The radial part of the Schrödinger equation for  = 0 is −

2 1 ∂  2 ∂  ψ  + V (r )ψ = Eψ . r 2m r 2 ∂r  ∂r 

With ψ = µ / r , V (r ) = αδ (r − r0 ) it becomes −

2 µ ′′ + αδ (r − r0 )µ = E µ , 2m

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i. e.,

µ ′′ − βδ (r − r0 )µ = − k 2 µ ,

where

β=

2mα , 2

k=

(1)

2mE . 2

The equation has solution for which µ = 0 at r = 0 and µ = finite for r → ∞  sin kr , r < r0 , µ= a sin(kr + δ ), r > r0 .  Integrating Eq. (1) from r0 − ∈ to r0 + ε and letting ε → 0 give

µ ′(r0 + ) − µ ′(r0 − ) = βµ(r0 ). The continuity of µ at r = r0 and this condition give sin kr0 = a sin(kr0 + δ ) ,  β  k sin kr0 = a cos(kr0 + δ ) − cos kr0 Hence, a 2[sin 2(kr0 + δ ) + cos 2(kr0 + δ )] = a 2 = 1 + tan(kr0 + δ ) =



β β2 sin 2kr0 + 2 sin 2kr0 , k k

tan kr0 , β 1 + tan kr0 k

(2)

which determine a and the phase shift δ . In the limiting case of k → 0, the above equation becomes kr0 + tan δ kr0 ≈ , 1 − kr0 tan δ 1 + β r0 or tan δ ≈ −

β r02 k, 1 + β r0

neglecting O(k 2 ). Then, as k → 0, we have tan δ → 0 and so

δ ≈−

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r0 k = Ak , 1 1+ β r0

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where A=

−r0 2 1+ 2mα r0

is the scattering length. b. For bound states, E < 0 and Eq. (1) can be written as

µ ′′ − βδ (r − r0 )µ = k 2 µ with

β=

2mα , 2

k2 = −

2mE . 2

The solution in which µ = 0 at r = 0 and µ = finite for r → ∞ is  sinh kr , r < r0 , µ =  − kr ae , r > r0 .  The continuity conditions, as in (a), give  sinh kr = ae − kr0 ,  0 µ= − kr − kro  β ae = −ake 0 − k cosh kr0 . Eliminating a we have (β + k )sinh kr0 = − k cosh kr0 , or c −2 kr0 = 1 +

2kr0 . β r0

Fig. 6.1 For bound states E < 0. Between E = −∞ and E = 0, or between 2kr 0 = ∞ and 2kr0 = 0 , there is one intersection between the curves (I) y = e −2 kro and

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2kr0 1 (II) y = 1 + β r if − 1 < β ro < 0, as shown in Fig. 6.1. Thus if this condition is 0 satisfied there will be one bound state with  = 0.. This condition requires −1
a ) find the “matching equation” at a positive energy, which determines the energy dependence of the  = 0 phase shift δ 0 . From this show that at high energies,

δ (k ) →

maV0 , and obtain this result from the Born approximation. 2k

(Wisconsin) Sol: Let χ = rR. For the  = 0 partial wave, the Schrödinger equation becomes  V  χ ′′ + k′ 2 χ = 0, k′ 2 = k 2  1 + 0  ,  E mE 2 χ ′′ + k 2 χ = 0, k 2 = 2 , 

r < a, r > a.

The solutions are  sin(k′r ), r < a, χ= A sin(kr + δ 0 ), r > a.  The continuity condition (ln χ )′

r =a−

= (ln χ )′

r =a+

gives an equation for determining δ 0: k′ tan(ka + δ 0 ) = k tan(k′a). As 2mE 2  V  k′ 2 = k 2  1 + 0  and k = 2 ,    E when k → ∞ , k′ → k . Hence  k δ 0 = arctan  tan(k′a) − ka → (k′ − k )a as k → ∞ .  k′  Thus, letting k → ∞ we obtain  k′ 2 − k 2  kV maV δ0 →  a≈ 0 a= 2 0 .   k′ + k  2E k The Born approximation expression for the phase shift for l = 0 is 2mk ∞ 2mkV0 V (r ) j02 (kr )r 2dr = 2 ∫0  2 1 mV   = 2 02  ka − sin(2ka) , k  2 

δ0 ≈ −

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∫

a

0

sin 2kr dr k2

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453

whence

δ0 →

mV0a  2k

as k → ∞ , in agreement with the partial-wave calculation.

6013 Calculate the scattering cross section for a low energy particle from a potential given by V = −V0 for r < a , V = 0 for r > a. Compare this with the Born approximation result. (Columbia) Sol:

The radial Schrödinger equation can be written in the form l(l + 1)   χ l′′ ( r ) +  k 2 − 2  χ l (r ) = 0, r   l(l + 1)   χ1′′ (r ) +  k′ 2 − 2  χ l (r ) = 0, r  

r > a, r < a,

where χ = rR(r ), k2 =

2mE , 2

k′ 2 =

2m( E + V0 ) . 2

Scattering at low energies is dominated by the s partial wave, for which  = 0, and the above become

χ l′′ (r ) + k 2 χ l (r ) = 0,

r > a,

χ l′′ ( r ) + k′ χ l (r ) = 0,

r < a,

2

whose solutions are  A sin(k′r ) , r < a, χ l (r ) =  sin(kr + δ 0 ) , r > a.  The continuity condition (ln χ l )′ | − = (ln χ l )′ | + gives r =a r =a k tan(k′a) = k′ tan(ka + δ 0 ) , or k  δ 0 = arctan  tan(k′a) − ka .  k′ 

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For low energies k → 0, k′ → k0 =

2mV0 , 2

and the above becomes  tan(k0a)  δ 0 ≈ ka  − 1 .  k0a  The total scattering cross section is then 2

σ≈

 tan(k0a)  4π 4π 2 δ ≈ 2 δ 02 = 4π a 2  − 1 . 2 sin 0 k k  k0a 

If k0a  1, 2

 k a (k a)3  16π a6m 2V02 σ ≈ 4π a  0 + 0 − 1 = . 9 4  k0a 3k0a  2

In the Born approximation, m e − ik ⋅rV (r )e ik ′⋅r d 3r , 2π  2 ∫

f (θ ) = −

where k ′ , k are respectively the wave vectors of the incident and scattered waves. Let q = k − k ′ , with | k ′ | = | k | = k for elastic scattering. Then q = 2k sin θ2 , where

θ is the scattering angle. Thus π m ∞ V (r )r 2dr ∫ e − iqr cosθ ′ 2π sin θ ′dθ ′ f (θ ) = − 2 ∫0 0 2π  2m ∞ sin(qr ) 2 2mV a = − 2 ∫ V (r ) r dr = 2 0 ∫ r sin(qr )dr 0  qr q 0 2mV = 2 30 [sin(qa) − qa cos(qa)]. q Hence

σ (θ ) =| f (θ )|2 =

4m 2V02 [sin(qa) − qa cos(qa)]2 .  4q6

For low energies k → 0, q → 0, sin(qa) ≈ qa −

1 1 (qa)3 , cos(qa) ≈ 1 − (qa)2 , 3! 2!

and hence

σ (θ ) ≈

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4m 2V02a6 . 9 4

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The total cross section for scattering at low energies is then

σ = ∫σ (θ ) dΩ =

16π m 2V02a6 . 9 4

Therefore at low energies for which k → 0, k a  1, the two methods give the same result.

6014 In scattering from a potential V (r ), the wave function may be written as an incident plane wave plus an outgoing scattered wave: ψ = e ikz + υ (r ). Derive a differential equation for υ (r ) in the first Born approximation. (Wisconsin) Sol:

Two methods may be used for this problem. Method 1: For a particle of mass m in a central field V (r ), the Schrödinger equation can be written as (∇ 2 + k 2 )ψ = Uψ , where U=

2m V, 2

k=

2mE . 2

Define Green’s function G(r − r ′ ) by (∇ 2 + k 2 )G(r − r ′ ) = −4πδ (r − r ′ ). This is satisfied by the function G(r − r ′ ) =

exp(ik | r − r ′ |) , | r − r′ |

and the Schrödinger equation is satisfied by 1 ψ (r ) = ψ 0 (r ) − G(r − r ′ )U (r ′ )ψ (r ′ )d 3r ′ . 4π ∫ As the incident wave is a plane wave e ikz ′ , we replace U (r ′ )ψ (r ′ ) by U (r ′ )e ikz ′ in the first Born approximation:

ψ (r ) = e ikz −

1 4π

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∫

exp(ik | r − r ′ |) U (r ′ )e ikz ′d 3r ′ . | r − r′ |

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Hence, the scattered wave is 1 4π

υ (r ) = −

∫

exp(ik | r − r ′ |) U (r ′ )e ikz ′d 3r ′. | r − r′ |

2 2 Applying the operator (∇ + k ) to the two sides of the equation, we get

(∇ 2 + k 2 )v(r ) = −

1 exp(ik(r − r ′ )) (∇ 2 + k 2 ) U (r ′ )e ikz ′ d 3r ′ 4π ∫ | r − r′ |

= ∫δ (r − r ′ )U (r ′ )e ikz ′ d 3r ′ = U (r )e ikz . Hence, the differential equation for υ (r ) is (∇ 2 + k 2 )υ (r ) = U (r )e ikz . Method 2: Writing the radial Schrödinger equation as (∇ 2 + k 2 )ψ = Uψ , where k2 =

2mE 2m , U = 2 V, 2 

and substituting in ψ = e ikz + υ (r ), we get (∇ 2 + k 2 )e ikz + (∇ 2 + k 2 )υ (r ) = U[e ikz + υ (r )], or (∇ 2 + k 2 )υ (r ) =

2m V[e ikz + υ (r )], 2

as (∇ 2 + k 2 )e ikz = 0. In the first Born approximation, e ikz + υ (r ) ≈ e ikz , and so the differential equation for υ (r ) is (∇ 2 + k 2 ) υ (r ) ≈

2m ikz Ve . 2

6015 In the quantum theory of scattering from a fixed potential, we get the following expression for the asymptotic form of the wave function

ψ (r ) r→∞ → e ikz + f (θ ,ϕ )

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e ikr . r

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457

a. If the entire Hamiltonian is rotationally invariant, give the argument that the scattering amplitude f should be independent of the angle ϕ . b. Why cannot this argument be extended (considering rotation about any axis) to conclude that f should be independent of θ as well? c. Reconsider part(b) in the case where the incident energy approaches zero. d. What is the formula for the scattering cross section in terms of f? e. What is the formula for the first Born approximation for f? (Be sure to define all quantities introduced. You need not worry about simple dimensionless factors like 2 or π ). f. Under what conditions is the Born approximation valid? (Berkeley) Sol: ∧

a. The incident wave e ikz = e ikr cosθ is the eigenstate of lz , third component of the angular momentum L, with eigenvalue m = 0. If the Hamiltonian is rotationally invariant, the angular momentum is conserved and the outgoing wave is ∧ still the eigenstate of lz with eigenvalue m = 0, that is, ∧

lz f (θ , ϕ ) = mf (θ , ϕ ) = 0.

 ∂ Since lˆz = , =this means that ∂ f (θ , ϕ ) / ∂ϕ = 0. i ∂ϕ ∧

b. As the asymptotic form of the wave function ψ (r ) is not an eigenfunction of L2, we cannot extend the above argument to conclude that f is independent of θ . c. When the energy E → 0, i.e., k → 0, the incident wave consists mainly only of the l = 0 partial wave; other partial waves have very small amplitudes and can be neglected. Under such conditions, the rotational invariance of H results in ∧ the conservation of L2. Then the outgoing wave must also be the eigenstate of ∧ L2 with eigenvalue l = 0 (approximately). As ∧ ∂ 1 ∂2   1 ∂  L2 = −  2  ,  sin θ  + 2  ∂θ sin 2θ ∂ϕ   sin θ ∂θ

we have

1 d  df (θ )  sin θ =0.  sin θ dθ  dθ 

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As f (θ ) must be a wave function with all the appropriate properties, this means df (θ ) / dθ = 0. d. The differential scattering cross section is given by dσ = | f (θ , ϕ )|2 . dΩ e. In the first Born approximation, for scattering from a central field V (r ′ ), f is given by m V (r ′ )exp(−iq ⋅ r ′ )d 3r ′ 2π  2 ∫ 2m ∞ = − 2 ∫ r ′V (r ′ )sin(qr ′ )dr ′ , q 0

f (θ , ϕ ) = −

where q = k − k 0 , k and k 0 being respectively the momenta of the particle before and after scattering. f. The validity of Born approximation requires that the interaction potential is small compared with the energy of the incident particle.

6016 Consider a particle of mass m which scatters off a potential V ( x ) in one dimension. a. Show that GE ( x ) =

1 2π

∫

∞

dk

−∞

e ikx , 2k2 + iε E− 2m

with ε positive infinitesimal, is the free-particle Green’s function for the time-independent Schrödinger equation with energy E and outgoingwave boundary conditions. b. Write down an integral equation for the energy eigenfunction corresponding to an incident wave traveling in the positive x direction. Using this equation find the reflection probability in the first Born approximation for the potential V0 , | x | < a / 2, V(x) =  0, | x | > a / 2.

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459

For what values of E do you expect this to be a good approximation? (Buffalo) Sol: a. To solve the one-dimensional time-independent Schrödinger equation  2 d2   2m dx 2 + E  ψ = Vψ , we define a Green’s function GE ( x ) by  2 d2   2m dx 2 + E  GE ( x ) = δ ( x ). Expressing GE ( x ) and δ ( x ) as Fourier integrals 1 2π 1 δ (x ) = 2π

GE ( x ) =

∫

∞

∫

∞

−∞

f (k )e ikx dk ,

e ikx dk

−∞

and substituting these into the equation for GE ( x ), we obtain   2k2   − 2m + E  f (k ) = 1, or f (k ) =

1 . 2k2 E− 2m

As the singularity of f (k ) is on the path of integration and the Fourier integral can be understood as an integral in the complex k-plane, we can add iε , where ε is a small positive number, to the denominator of f (k ). We can then let ε → 0 after the integration. Consider GE (k ) =

1 2π

∫

∞

dk

−∞

e ikx .  2k2 + iε E− 2m

The intergral is singular where ( E + iε ) −

2k2 = 0, 2m

i.e., at k = ± k1 ,

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where

2m( E + iε ) . 

k1 =

When x > 0, the integral becomes a contour integral on the upper half-plane with a singularity at k1 with residue a1 = −

me ik1z . π  2 k1

Cauchy’s integral formula then gives GE ( x ) = 2π ia1 = −i

m ik1x e ( x > 0).  2 k1

As

ε → 0, k1 →

2mE . 

This is the value of k1 to be used in the expression for GE ( x ). Similarly, when x < 0, we can carry out the integration along a contour on the lower half-plane and get G E ( x ) = −i Here k1 =

2mE 

m − ik1x e  2 k1

( x < 0).

also. Thus the free-particle Green’s function GE ( x ) represents

the outgoing wave whether x > 0 or x < 0. b. The solution of the stationary Schrödinger equation satisfies the integral equation ψ E ( x ) = ψ 0 (x) + GE ( x )*[V ( x )ψ E ( x )] ∞

= ψ 0 ( x ) + ∫ GE ( x − ξ )V (ξ )ψ E (ξ )dξ , −∞

where ψ 0 ( x ) is a solution of the equation  2 d 2   2m dx 2 + E  ψ ( x ) = 0 . In the first-order Born approximation we replace ψ 0 and ψ E on the right side of the integral equation by the incident wave function and get ∞

ψ E ( x ) = e ikx + ∫ GE ( x − ξ )V (ξ )e ikξ dξ = e ikx −∞

m ik ( x −ξ ) e V (ξ )e ikξ dξ 2k ∞ m + ∫ (−i ) 2 e − ik ( x −ξ )V (ξ )e ikξ dξ . x k x

+ ∫ ( −i ) −∞

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461

For reflection we require ψ E ( x ) for x → −∞ . For x → −∞ , m ik ( x −ξ ) e V (ξ )e ikξ dξ = 0, 2k ∞ a/2 m − ikx 2ikξ m − ikx 2ikξ ∫x (−i )  2 k e e V (ξ )dξ = ∫− a/2(−i )  2 k e V0e dξ mV = −i 2 02 sin(ka)e − ikx . k x

∫

−∞

( −i )

Hence, the reflection probability is R = |ψ E (−∞ )|2 |ψ 0 |2 =

m 2V02 2 sin ka . 4k4

When the energy is high, ∞

| ∫ GE ( x − ξ )V (ξ )e ikξ dξ |  | e ikx |, −∞

and replacing ψ ( x ) by e ikx is a good approximation.

6017 Calculate, using the Born approximation, the differential and total cross sections for the scattering from the potential that is given by V0 exp(−α 2r2). Sol:

The differential cross section is given by

dσ 2 = f (θ ) dΩ

where f (θ) is given by f (θ ) =

−2m  2q

∫

∞

0

r sin(qr )V (r ) dr

Plugging in the above given potential, we get f (θ ) = to evaluate the integral

∫

= − (1/ 4 )

∞

0

−2m  2q

∫

∞

0

sin(qr )re −α

r sin(qr )V0e −α

2 2

r

2 2

r

dr 2 2

r (e iqr − e − iqr )e −α r dr −∞ 2i

dr = (1/ 2 ) ∫

∞

{

∂ − q2 /4α 2 ∞ − (α r +iq/2α )2 − (α r −iq/2α )2 e +e ) dr ∫−∞ (e ∂q

}

2

−q π q 4α 2 ∂ − q2 /4α 2  π  = − (1/ 4 )  e  2 α  = 4α 3 e ∂q

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Problems and Solutions in Quantum Mechanics − q2

− π mV0 4α 2 So plugging this integral in f (θ), we get f (θ ) = e 2 2α 3 The relation between momentum transfer and the scattering angle in the Born approximation is q = 2k sin(θ/2). In this approximation, we get dσ π m 2V0 2 2 = f (θ ) = e dΩ 4  4α 6

−2 k 2

α2

sin 2 (θ /2)

And the total cross section is given by σ (θ ) = ∫ d Ω

π m 2V0 2 = 4  4α 6

∫

2π

0

+1

dφ ∫ d cos θ e

dσ dΩ

−2 k 2

α2

(1−cosθ )

−1

So the total cross section is given by σ (θ ) =

2 2 π 2m 2V0 2 (1 − e −2 k /a ) 4 6 2 α

6018 A particle of energy E scatters off a repulsive spherical potential V ( r ) = V0 for r < a   0 for r > a. 2

 tanh( ka )  − 1 In the low energy limit the total cross section is given by 4π a 2   ka  2m 2 where k = 2 (V0 − E ) > 0 . In the limit V0 → ∞ , find the ratio of σ to the  classical scattering cross section off a sphere of radius a. Sol:

 tanh(ka)  σ = 4π a 2  − 1   ka

2

In the low energy limit, as the potential V0 → ∞ and E is negligibly small 2m k 2 = 2 (V0 − E ) tends to infinity and when ka → ∞, tanh(ka) →1.  2  1  So σ = 4π a 2  − 1 and approximating on the limit, lim σ = 4π a 2. ka→∞  ka  The classical scattering cross section is given by σ c = π a 2.



The ratio of these cross section is

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σ = 4. σc

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463

6019 A nucleon is scattered elastically from a heavy nucleus. The effect of the heavy nucleus can be represented by a fixed potential −V0, V (r ) =  0,

r < R, r > R,

where V0 is a positive constant. Calculate the deferential cross section to the lowest order in V0 . (Berkeley)

Sol: Let µ be the reduced moves of the nucleon and the nucleus, q = k − k ′ where k ′ , k are respectively the wave vectors of the nucleon before and after the scattering. In the Born approximation, as in Problem 6013, we have 2

σ (θ ) = f (θ ) = =

4µ 2 4q2

∫

∞

0

r ′V (r ′ ) sin qr ′dr ′

2

4 µ 2V02 (sin qR − qR cos qR )2 ,  4q6

where q = 2k sin(θ /2) , k = k = k ′ .

6020 A particle of mass m, charge e, momentum p scatters in the electrostatic potential produced by a spherically symmetric distribution of charge. You are given the quantity ∫r 2 ρ d 3 x = A, ρ ( r )d 3 x being the charge in a volume element d 3 x . Supposing that ρ vanishes rapidly as r → ∞ and that ∫ ρd 3 x = 0 ; working in the first Born approximation, compute the differential cross secdσ tion for forward scattering. (That is θ = 0 , where θ is the scattering angle.) dΩ (Princeton) Sol:

In the first Born approximation, we have dσ m 2e 2 = dΩ 4π 2  4

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∫ U (r ) exp(iq ⋅ r )d x 3

2

,

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where

θ 2p θ q = k − k ′ , q = 2k sin = sin , k ′ and k 2  2 being the wave vectors of the particle before and after the scattering, U (r ) is the electrostatic Coulomb potential and satisfies the Poission equation ∇ 2U = −4πρ(r ). Let F (q ) = ∫ ρ (r ) exp(iq ⋅ r )d 3 x , where F (q ) is the Fourier transform of ρ(r ). Using the Poisson equation we have 2π

∫U (r ) exp(iq ⋅ r )d x = q 3

2

F (q ) .

Hence dσ m 2e 2 (4π )2 4m 2e 2 2 2 = 2 4 ⋅ 4 F ( q ) = 4 4 F (q ) . dΩ 4π  q q For forward scattering, θ is small and so q is small also. Then F (q ) = ∫ ρ(r ) exp(iq ⋅ r ) d 3 x

since as ∫ ρd 3 x = 0,

∫

π

0

1 = ∫ ρ(r ) 1 + iq ⋅ r + (iq ⋅ r )2 +  d 3 x 2!   1 = ∫ ρ(r )d 3 x − ∫ ρ(r )(q ⋅ r )2 d 3 x +  2! Aq 2 1 2 ≈ − q ∫ ρ(r )r 2d 3 x = − , 6 6 cos 2n+1 θ ⋅ sinθ dθ = 0, the lowest order term for θ → 0 is

q2 1 2 3 ρ ( r ) ( i q ⋅ r ) d x ≈ − ρ(r )r 2d 3 x . 2! ∫ 6 ∫ Hence dσ A 2m 2 e 2 . = dΩ θ = 0 9 4

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465

6021 Use Born approximation to find, up to a multiplicative constant, the differential scattering cross section for a particle of mass m moving in a repulsive potential V = Ae − r

2 / a2

.

(Berkeley) Sol:

In Born approximation we have (Problem 6013) 2m ∞ f (θ ) = − 2 ∫ rV (r ) sin(qr ) dr , q 0 where q = 2k sin(θ / 2), k being the momentum of incident particle. As f (θ ) = − =

2 2 2mA ∞ − r 2 /a2 mA ∞ sin(qr )dr = − 2 ∫ re − r /a sin(qr )dr re 2 ∫ q 0  q −∞

mAa 2 2 2 q

=−

∫

∞

−∞

(e − r

2 /a 2

)′ sin(qr )dr = −

mAa 2 2 2

∫

∞

e−r

2 /a 2

−∞

cos(qr )dr

mAa3 ∞ − ( r /a )2  r r e cos  qa  d    a  a 2 2 ∫−∞

mAa3 ∞ − r 2 e cos (qar )dr 2 2 ∫−∞ 2   iqa  2   − q2a2 /4 mAa3 ∞    iqa   =− dr + exp  −  r + exp  −  r −    e  2 ∫−∞   4 2   2        =−

=−

2 2 mAa3 π e − q a /4 , 2 2 2

σ (θ ) = f (θ ) =

m 2 A 2a6 − q2a2 /2 πe . 4 4

6022 A nonrelativistic particle is scattered by a square-well potential  −V0 , V (r ) =   0,

r < R , (V0 > 0) r > R.

a. Assuming the bombarding energy is sufficiently high, calculate the scattering cross section in the first Born approximation (normalization is not essential), and sketch the shape of the angular distribution, indicating angular units.

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b. How can this result be used to measure R? c. Assuming the validity of the Born approximation, if the particle is a proton and R = 5 × 10 −13 cm , roughly how high must the energy be in order for the scattering to be sensitive to R? (Wisconsin) Sol: a. Using Born approximation we have (Problem 6013) −1 ∞ f (θ ) ∝ ∫ rV (r ) sin(qr ) dr q 0 V R V = 0 ∫ r sin(qr )dr = 30 (sin qR − qR cos qR ). q 0 q Hence 2

dσ  sin x − x cos x  ∝  , dΩ  x3 where x = qR = 2kR sin θ2 . The angular distribution is shown in Fig. 6.4

Fig. 6.4 b. The first zero of

dσ dΩ

occurs at x for which x = tan x, whose solution is x ≈ 1.43π .

This gives R=

1.43π . θ 2k sin 1 2

By measuring the minimum angle θ1 for which dσ = 0, R can be determined. dΩ c. In order that R may be determined from the zero points of ddσ Ω , we require that the maximum value of x, 2kR , is larger than 1.43π , or

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E≥

2

 2  1.43π  (1.43π )2  2  c    =   2m p  2 R  8 mpc 2  R 

467

2

(1.43π )2 (6.58 × 10−22 )2  3 × 1010  = × ×  5 × 10 −13  8 938

2

= 4.2 MeV.

6023

Consider the potential V ( r ) = ∑V0 a 3δ (3) ( r − ri ), where ri are the position i

vectors of a cube of length a centred at the origin and V0 is a constant. If V0 a 2 1, the scattering is significantly nonisotropic. The  angular momentum at which only s-wave scattering, which is isotropic, is important must satisfy a ⋅ k ≤  , i.e., ka ≤ 1. When ka  1, the scattering begins to be significantly non-isotropic. This is in agreement with the result given by the first Born approximation, b. The wave function to the first order is

ψ (r ) = eikz −

1 e ik|r −r ′| 2m V (r ′ ) e ikz ′ dV ′. 4π ∫ | r − r ′ |  2

Fig. 6.5

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471

Hence ∆ψ (1) (0) V0 e ikr ′ 2m − r ′/a+ikz ′ e dV ′ = ψ (0) (0) 4π ∫ r ′  2 =

mV0 r ′e ikr ′−r ′/a+ikr ′ cosθ ′ sinθ ′dθ ′dr ′ 2 ∫

=

2m | V0 | a 2 1 + 4 k 2a 2 2m | V0 | a 2 . = 2 2 2 2  (4 k a + 1)  1 + 4 k 2a 2

The criterion for the validity of the first Born approximation is then 2m V0 a 2  2 1 + 4 k 2a 2

1 .

In the low-k, limit, ka  1, the above becomes 2m V0 a 2 2  1, or V0  . 2  2ma 2 In the high-k limit, ka  1, the criterion becomes m V0 a 2k  1, or V 0  . 2 k ma Since in this case k  1a the restriction on | V0 | is less than for the low-k limit.

6026 −1

For an interaction V ( r ) = β r exp( −α r ) find the differential scattering cross section in Born approximation. What are the conditions for validity? Suggest one or more physical applications of this model. (Berkeley) Sol: In Born approximation, we first calculate (Problem 6013) 2m ∞ f (θ ) = − 2 ∫ r ′V (r ′ ) sin qr ′ dr ′ q 0 =−

2mβ ∞ −α r −2mβ , e sin qr dr = 2 2 2 ∫ 0 q  (q + α 2 )

where q = 2k sin θ2 and m is the mass of the particle, and then the differential cross section 2

σ (θ ) = f (θ ) =

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4m 2 β 2 .  4 (q 2 + α 2 )2

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The derivation is based on the assumption that the interaction potential can be treated as a perturbation, so that the wave function of the scattered particle can be written as

ψ (r ) = ψ 0 (r ) + ψ 1 (r ),

where

ψ1  ψ 0 ,

ψ 0 (r ) = e , ikz

ψ 1 (r ) ≈ −

m e ik|r −r ′| V ( r ′ ) ψ 0 ( r ′ ) d 3r ′ . 2π  2 ∫ | r − r ′ |

Specifically, we shall consider two cases, taking a as the extent of space where the potential is appreciable. i. The potential is sufficiently weak or the potential is sufficiently localized. As

ψ1 ≤

m 2π  2

∫

V (r ′ ) ψ 0 ( r ′ ) d 3r ′ r − r′

2 a 4π r dr m V ψ0 ∫ 2 0 2π  r 2 2 ≈m V a ψ0 / ,

≈

for ψ 1  ψ 0 we require m V a2  1, 2 i.e., V 

2  . or a  2 ma mV

This means that where the potential is weak enough or where the field is sufficiently localized, the Born approximation is valid. Note that the condition does not involve the velocity of the incident particle, so that as long as the interaction potential satisfies this condition the Born approximation is valid for an incident particle of any energy. ii. High energy scattering with ka  1. The Born approximation assumes ψ 0 = e ikz and a ψ 1 that satisfies ∇ 2ψ 1 + k 2ψ 1 =

2m ikz Ve . 2

Let ψ 1 = e ikz f (θ ,ϕ ) . The above becomes im ∂f = − 2 V, ∂z k

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473

and so

ψ 1 = e ikz f = − Then as

ψ1 

im ikz e Vdz .  2k ∫

m V aψ0 ,  2k

for ψ 1  ψ 0 = 1 we require V 

 2 k υ = , ma a

where υ is the speed of the incident particle, υ = m = mk . Thus as long as the energy of the incident particle is large enough, the Born approximation is valid. p

From the above results we see that if a potential field can be treated as a small perturbation for incident particles of low energy, it can always be so treated for incident particles of high energy. The reverse, however, is not true. In the present problem, the range of interaction can be taken to be a ≈ α1 , so that V (a)  βa . The conditions then become i.

|β |

α 2 , m

ii. | β |  υ =

2 k m

, where k =

2mE 2

.

The given potential was used by Yukawa to represent the interaction between two nuclei and explain the short range of the strong nuclear force.

6027 Consider the scattering of a 1 keV proton by a hydrogen atom. a. What do you expect the angular distribution to look like? (Sketch a graph and comment on its shape). b. Estimate the total cross section. Give a numerical answer in cm2, m2 or barns = 10 −24 cm2, and a reason for you answer. (Wisconsin) Sol: The problem is equivalent to the scattering of a particle of reduced mass µ ≈ 12 m p = 470 MeV, energy Er = 0.5 keV by a potential which, on account of

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electron shielding, can be roughly represented by interaction given by the Bohr radius 0.53 Å. As a 2 µc 2 Er ka = c =

e2 r

e − r /a , where a is the range of

0.53 × 10−8 × 2 × 470 × 106 × 0.5 × 103 = 1.84 × 102  1, 10 −16 6.58 × 10 × 3 × 10

for Born approximation to be valid we require (Problem 6026) V ∼

e2 υ  , a a

i.e.,

e2 υ  . c c

Since LHS =

1 = 7.3 × 10−3 , 137 2 Er 2 × 0.5 × 103 = = 1.5 × 10−3 , µc 2 470 × 106

RHS =

the condition is not strictly satisfied. But in view of the roughness of the estimates, we still make use of the Born approximation. a. When the proton collides with the hydrogen atom, it experiences a repulsive Coulomb interaction with the nucleus, as well as an attractive one with the orbital electron having the appearance of a cloud of charge density e ρ(r ). The potential energy is then V (r ) =

e 2 2 ρ(r ′ ) −e ∫ dr ′. r − r′ r

Using Born approximation and the formula e iq⋅r 4π ∫ r dr = q2 , we obtain f (θ ) = − =−

 µe 2 iq⋅r  1 ρ(r ′ ) e  −∫ dr ′  dr r r 2π  2 ∫ r | − | ′   2 µe 2 [1 − F (θ )] ,  2q 2

where

µ=

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mp 2

θ , q = 2k sin , 2

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Scattering Theory & Quantum Transitions

F (θ ) = =

q2 4π

∫∫

475

eiq⋅r ρ(r ′ ) dr ′dr | r − r′|

q 2 iq⋅r ′ e iq⋅r e ρ r d ( ) r ′ ′ ∫ r dr 4π ∫

= ∫e iq⋅r ρ(r )dr . For the ground state of the hydrogen atom, we have 1 2 ρ(r ) = ψ 100 = 3 e −2r /a , πa and so F (θ ) =

2r iq⋅r − 1 e a dr 3 ∫ πa −2

 a 2q 2  = 1 + .  4  Hence f (θ ) = −

1 1 µe 2   1− ⋅ . 2 2 k 2 sin 2(θ / 2)  (1 + a 2 k 2 sin 2θ / 2)2 

Taking into account the identical nature of the two colliding particles (two 2 protons), we have for the singlet state: σ s = f (θ ) + f (π − θ ) , the triplet 2

state: σ A = f (θ ) − f (π − θ ) . Hence, the scattering cross section (not considering polarization) is 1 3 σ = σ s + σ A. 4 4 Some special cases are considered below, i.

θ ≈ 0: µ 2e 4  1 − F (θ ) 1 − F (π − θ )  σs = 4 4  2 + 4  k  sin θ / 2 cos 2 θ / 2  ≈

1 µ 2e 4  2 2   2a k +  44k 4  cos 2 θ / 2 

≈

µ 2e 4 4  m p  2 a = a , 4  2me 

2

2

2

use having been made of the approximation for x ≈ 0 1/ (1 + x )2 ≈ 1 − 2 x ,

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as well as the expression a=

2  2  mp  = . me e 2 µe 2  2me 

Similarly we obtain 2

 mp  2 σA = a .  2me  ii. θ ≈ π : A similar calculation gives 2

 mp  2 σS =σA =  a.  2me  iii. a 2 k 2 sin 2 θ2 = 10 or θ ≈ 0.07π : For 0.07π ≤ θ ≤ 0.93π , we have

σ (θ ) = =

2

2 4 1 µ 2e 4  1 1 1 1  3 µe   + + −    4 4  2 2 4 4  k  sin θ / 2 cos θ / 2  4 4  4 k 4  sin 2 θ / 2 cos 2 θ / 2 

2

µ 2e 4  3cos 2 θ + 1   4 k 4  sin 4 θ 

 3cos 2 θ + 1  = σ0 ,  sin 4 θ  where σ 0 =

µ2e4 4 k4

. The angular distribution is shown in Fig. 6.6.

Fig. 6.6 d. As f (θ ) → ∞ for θ → 0, θ → π , to estimate the total scattering cross section, consider the total cross section for large scattering angles (0.07π ≤ θ ≤ 0.93π ) and for small scattering angles:

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σ t large = 2πσ 0 ∫

0.93π

0.07 π

 3cos 2 θ + 1   sin 4 θ  sinθ dθ

4 − 3sin 2 θ dθ 0.07 π sin3 θ θ cot θ    = 2πσ 0  − ln  tan  − 2   2 sin θ   = 155πσ 0 , = 2πσ 0 ∫

σ t small > 2π ∫

477

0.93π

0.93π 0.07 π

0.07 π

0

σ (0.07π )sin θ dθ × 2

= 2π × 1703σ 0 [1 − cos(0.07π )] × 2

= 164πσ 0 .

6028 The study of the scattering of high-energy electrons from nuclei has yield­ed much interesting information about the charge distributions in nuclei and nucleons. We shall here consider a simple version of the theory, in which the “electron” is assumed to have zero spin. We also assume that the nucleus, of charge Ze , remains fixed in space (i.e., its mass is assumed infinite). Let ρ ( x ) denote the charge density in the nucleus. The charge distribution is assumed to be spherically symmetric, but otherwise arbitrary. Let fe (p i , p f ) , where pi is the initial, and p f is the final momentum, be the scattering amplitude in the first Born approximation for the scattering of an electron from a point nucleus of charge Ze. Let f (pi , p f ) be the scattering amplitude, also in the first Born approximation, for the scattering of an electron from a real nucleus of the same charge. Let q = pi − p f denote the momentum transfer. The quantity F defined by f (pi , p f ) = F ( q2 )fe (pi , p f ) is called the form factor: it is easily seen that F in fact depends on pi and p f only through the quantity q2 . a. The form factor F ( q2 ) and the Fourier transform of the charge density

ρ ( x ) are related in a very simple manner: state and derive this relation-

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Problems and Solutions in Quantum Mechanics

ship within the framework of the nonrelativistic Schrödinger theory. The assumption that the electrons are “nonrelativistic” is here made so that the problem will appear as simple as possible, but if you think about the matter it will probably be clear that the assumption is irrelevant: the same result applies in the “relativistic” case of the actual experiments. It is also the case that the neglect of the electron spin does not affect the essence of what we are here concerned with. b. The graph in Fig. 6.7 shows some experimental results pertaining to the form factor for the proton, and we shall regard our theory as applicable to these data. On the basis of the data shown, compute the root-meansquare (charge) radius of the proton. Hint: Note that there is a simple relationship between the root-mean2 2 square radius and derivative of F(q2 ) with respect to q at q = 0 . Find this relationship, and then compute. (Berkeley)

Fig. 6.7 Sol: a. In the nonrelativistic Schrödinger theory the first Born approximation gives the scattering amplitude of an electron (charge −e) due to a central force field as 2m ∞ f (pi , p f ) = − 2 ∫ r ′V (r ′ ) sin(qr ′ )dr ′ , q 0

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479

where q is the magnitude of the momentum transfer in the scattering. For a point nucleus, the scattering potential energy is Ze 2 . r For a real nucleus of charge density ρ(r ) , the scattering potential energy is Ve (r ) = −

V (r ) = − e ∫

2 ∞ p(r ′ )r ′ dr ′ ρ(r ′ )dr ′ = −4π e ∫ , 0 r − r′ r − r′

which satisfies Poisson’s equation ∇ 2V (r ) =

1 d2 (rV ) = +4π e ρ(r ). (1) r dr 2

Consider the integral in the expression for f: ∞  ∞ 1  iqr  1 iqr ∫0 rVe dr = iq e  rV − iq (rV )′  0   −

1 q2

∫

∞

0

(rV )′′e iqr dr .

By a method due to Wentzel, the first term can be made to vanish and so 2m ∞ f (pi , p f ) = − 2 ∫ rV (r )sin(qr )dr q 0 =− =



2m  1  ∞ − (rV )′′ sin qr dr  2q  q 2  ∫0

2m 4π e ∞ r ρ(r )sin (qr ) dr , (2)  2q q 2 ∫0

use having been made of Eq. (1). In the case of a point-charge nucleus, only the region near r = 0 makes appreciable contribution to the integral and so ∞

∞

0

0

4π ∫ r ρ(r )sin(qr )dr ≈ 4π q ∫ r 2 ρ(r )dr = qZe. Hence for a point nucleus, f e ( pi , p f ) =

2mZe 2 .  2q 2

For an extended nucleus we can then write 4π ∞ 2 sin(qr ) dr . f ( pi , p f ) = f e ( pi , p f ) ⋅ r ρ (r ) ∫ qr Ze 0 By definition the form factor is

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Problems and Solutions in Quantum Mechanics

F (q 2 ) =



4π Ze

∫

∞

0

or F (q 2 ) =

r 2 ρ (r )

sin(qr ) dr , (3) qr

1 ρ(r )e − q⋅r dr . ∫ Ze

This is the required relation with the Fourier transform of the charge density. b. Differentiating Eq. (3) with respect to q we have dF 4π = dq Ze

∫

 r cos(qr ) sin(qr )  r 2 ρ (r )  −  dr , q 2r   qr

∞

0

and hence dF dF dq 1 4π = = ⋅ 2 2 d(q ) dq d(q ) 2q Ze

∫

 r cos ( qr ) sin(qr )  r 2 ρ (r )  −  dr . q 2r   qr

∞

0

dF To find d(q 2 ) q 2 = 0 we first calculate  r cos(qr ) sin(qr )  − lim   2 q→0 q 3r   qr   1 1 2 3  r ⋅ 1 − 2 (qr )  qr − 6 (qr )  = lim  −  q→0 q 2r q 3r     2 r  1  = lim  − r 2  = − . q→0  3  3 Then

dF d (q 2 )

q2 =0

1 1 ∞ 1 = − ⋅ ∫ r 2 ρ(r ) ⋅ 4π r 2dr = − r 2 . 6 Ze 0 6

2 2 From Fig. 6.7 can be found the slope of the curve F (q ) at q = 0, whence the

root-mean-square charge radius of the nucleus:  dF 〈r 2 〉 =  −6 2  d (q )

1/2

 . q2 =0  

6029 In the early 1920’s, Ramsauer (and independently Townsend) discovered that the scattering cross-section for electrons with an energy of  0.4 eV was very

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481

2 much smaller than geometrical (π a , with a the radius of the atom) for scat-

tering by argon atoms in gaseous form. It was also found that the cross section for 6-volt electrons was 3.5 times as great as the geometrical cross section and that the scattering was approximately isotropic. What is the origin of the “anomalous” cross sections? (What is the maximum possible cross section for scattering of low-energy electrons (with wavelength λ  a)? (Princeton) Sol: If the attractive potential is strong enough, at a certain energy the partial wave with  = 0 has exactly a half-cycle more of oscillation inside the atomic potential. Then it has a phase shift of δ = π and so contributes nothing to f (θ ) and hence the cross section. At low energies, the wavelength of the electron is large compared with a so the higher-  partial waves’ contribution is also negligible. This accounts for the Ramsauer-Townsend effect that the scattering cross section is very small at a certain low energy. For low-energy electrons, the maximum possible cross section for scattering is four times the geometrical cross section. It should be noted that a raregas atom, which consists entirely of closed shells, is relatively small, and the combined force of nucleus and orbital electrons exerted on an incident electron is strong and sharply defined as to range.

6030 Let f (ω ) be the scattering amplitude for forward scattering of light at an individual scattering center in an optical medium. If the amplitude for incoming and outgoing light waves are denoted by Ain (ω ) and Aout (ω ) respectively, one has Aout (ω ) = f (ω ) Ain (ω ) . Suppose the Fourier transform Ain ( x − t ) =

1 2π

∫

+∞

−∞

e iω ( x −t ) Ain (ω )dω

vanishes for x − t > 0 . a. Use the causality condition (no propagation faster than the speed of light c = 1) to show that f (ω ) is an analytic function in the half-plane lm ω > 0 .

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Problems and Solutions in Quantum Mechanics

b. Use the analyticity of f (ω ) and the reality of Ain (ω ) and Aout (ω ), and assume that f (ω ) is bounded at infinity to derive the dispersion relation Re [ f (ω + iε ) − f (0)] =

∞ 2ω 2 Im f (ω ′ + iε ) P ∫ dω ′ , 0 ω ′(ω ′ 2 − ω 2 ) π

with ε arbitrarily small and positive. (Chicago) Sol: a. Ain ( x − t ) = 0 for t < x means A out ( x − t ) = 0 for t < x . Then Ain (ω ) = out

1 2π

∫

0

e − iωτ Ain (τ ) dτ

−∞

out

is a regular function when Im ω > 0, since when τ < 0 the factor exp(Im ωτ ) of the integrand converges. As Aout (ω ) = f (ω )Ain (ω ) , f (ω ) is also analytic when Im ω > 0 . b. For ω → ∞ , 0 ≤ arg ω ≤ π , we have f (ω ) < M , some positive number. Assume that f (0) is finite (if not we can choose another point at which f is f (ω ) − f (0) is sufficiently small at infinity, and so finite). Then χ (ω ) = ω 1 +∞ χ (ω ′ + i 0) χ (ω ) = dω ′ , Imω > 0 . 2π i ∫−∞ ω ′ − ω When ω is a real number, using 1 P = + iπδ (ω ′ − ω ), ω ′ − ω − i0 ω ′ − ω we get P +∞ Imχ (ω ′ + i 0) dω ′. π ∫−∞ ω ′ − ω Ain (ω ) being a real number means that Ain* (−ω * ) = Ain (ω ). Hence Re χ (ω ) =

out

out

out

f * (ω * ) = f (−ω ), and so lm f (ω + i 0) = − Im f (−ω + i 0) , and Re [ f (ω + i 0) − f (0)] =

∞ Imf (ω ′ + i 0) 2ω 2 P∫ dω ′ , 0 π ω ′(ω ′ 2 − ω 2 )

where P denotes the principal value of the integral.

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483

6031 2 2

A spin-one-half projectile of mass m and energy E = 2 mk scatters off an infinitely heavy spin-one-half target. The interaction Hamiltonian is Hint = Aσ 1 ⋅σ 2

e − µr r

( µ > 0),

where σ 1 and σ 2 are the Pauli spin operators of the projectile and target respectively. Compute the differential scattering cross section ddσΩ in lowest order Born approximation, averaging over initial and summing over final states of spin polarization. Express ddσΩ as a function of k and the scattering angle θ . (Princeton) Sol: Suppose that the projectile is incident on the target along the z-axis, i.e., k 0 = ke z . In lowest order Born approximation, the scattering amplitude is m e − µr ′ 3 i ( k − k )⋅r e 0 Aσ 1 ⋅ σ 2 d r′ 2 ∫ 2π  r′ −m iq⋅r e − µr ′ 3 = e A σ ⋅ σ d r′ 1 2 2π  2 ∫ r′ ∞ π −m = 2 Aσ 1 ⋅ σ 2 ∫ e − µr ′r ′dr ′ ∫ e iqr ′ cosθ sin θ dθ 0 0  2m 1 = − 2 Aσ 1 ⋅ σ 2 2 2 , µ +q  where q = k 0 − k, q = 2k sin θ . Denote the total spin of the system by S. Then 2 f (θ ) = −

S = 12 (σ 1 + σ 2 ) and 2 1 σ 1 ⋅ σ 2 = (σ 2 − σ 12 − σ 22 ) = [4 S(S + 1) − 3 − 3] 2 2 2  = [4 S(S + 1) − 6]. 2 2 For S = 0, σ 1 ⋅ σ 2 = −3 and f0 (θ ) =

6 Am dσ 0 (6 Am)2 , = 2 2 2. 2 2 µ + q dΩ ( µ + q )

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Problems and Solutions in Quantum Mechanics

For S = 1, σ 1 ⋅ σ 2 =  2 and f1 (θ ) = −

2 Am dσ 1 (2 Am)2 , = 2 2 2. 2 2 µ + q dΩ ( µ + q )

1 If the initial states of spin of the projectile and target are   P = α P ,  0

1  0 T = α T

respectively, then the initial state of spin of the system is Θ11 = α Pα T , the scat­ ikr

tered wave function is f1 (θ ) e r Θ11, and the corresponding differential scattering cross section is given by dσ  1 1 1 1  2 ;   = f1 (θ ) , dΩ  2 2 2 2  dσ  1 1 1 1  dσ  1 1 1 1 ; , − = ; −     dΩ 2 2 2 2 dΩ 2 2 2 2 dσ  1 1 1 1 = ; − , −  = 0.  dΩ 2 2 2 2 Noting that the triplet state vectors are Θ11 = α Pα T , Θ1,−1 = β P βT , Θ10 =

1 (α P βT + β Pα T ) , 2

the singlet state vector is Θ00 =

1 (α P βT − β Pα T ) 2

and that

α P βT =

1 (Θ10 + Θ00 ), etc., 2

we can obtain the remaining differential cross sections: dσ  1 1 1 1 1 2  , − ; , −  = f1 (θ ) − f0 (θ ) , 2 4 dΩ  2 2 2 dσ  1 1 1 1 1 2  , − ; −  = f1 (θ ) − f0 (θ ) , dΩ 2 2 22 4 dσ  1 1 1 1  dσ  1 1 1 1  , − ;  =  , − ; − , −  = 0; dΩ 2 2 22 dΩ 2 2 2 2 dσ  1 1 1 1 1 2 ; , −  = f1 (θ ) − f0 (θ ) , − dΩ  2 2 2 2 4 1 1 1 dσ  1 1 2 ; − −  = f1 (θ ) + f0 (θ ) , dΩ  2 2 2 2 4 dσ  1 1 1 1  dσ  1 1 1 1 ; ; − , −  = 0; − = − dΩ  2 2 2 2  dΩ  2 2 2 2 dσ  1 1 1 1 2  − , − ; − , −  = f1 (θ ) , 2 2 2 dΩ  2 dσ  1 1 1 1  dσ  1 1 1 1  − , − ; ,  =  − , − ; −  2 2 2 2 2 2 dΩ 2 dΩ 2 1 24/05/22 dσ  1 1 1 xxxxxxxxxxxxx_Chapter_06.indd  Page 484 =  − , − ; , −  = 0.

10:20 AM

dσ  1 1 1 1  dσ  1 1 1 1  , − ; =  , − ; − , −  = 0; dΩ  2 2 2 2  dΩ  2 2 2 2 1 1 dσ  1 1 1 2 ; , −  = f1 (θ ) − f0 (θ ) , − dΩ  2 2 2 2 4 1Scattering 1  1 Theory & Quantum dσ  1 1 2 Transitions ; − −  = f1 (θ ) + f0 (θ ) , dΩ  2 2 2 2 4 dσ  1 1 1 1  dσ  1 1 1 1 ; ; − , −  = 0;  −  =  − dΩ 2 2 2 2 dΩ 2 2 2 2 dσ  1 1 1 1 2  − , − ; − , −  = f1 (θ ) , dΩ 2 2 2 2 dσ  1 1 1 1  dσ  1 1 1 1  − , − ; ,  =  − , − ; −  dΩ 2 2 2 2 dΩ 2 2 2 2 =

485

1 1 1 dσ  1  − , − ; , −  = 0. dΩ  2 2 2 2

Averaging over the initial states (i) and summing over the final states (f) of spin polarization, we obtain dσ 1 dσ (i ) (i ) ( f ) ( f ) = SPz STz ; SPz STz ∑ ∑ d Ω 4 s( i ) , s( i ) s( f ) , s( f ) d Ω

(

Pz

Tz Pz

)

Tz

12 A 2m 2 1 . = 3 f12 (θ ) + f02 (θ ) = 2 4 µ + 4 k 2 sin 2 θ2

6032 Calculate in the Born approximation the differential scattering cross section for neutron-neutron scattering, assuming that the interaction potential responsible for scattering vanishes for the triplet spin state and is equal to − µr for the singlet spin state. [Evaluate the cross section for an V ( r ) = V0 e r unpolarized (random spin orientation) initial state.] (Berkeley) Sol:

The Born approximation gives (Problem 6013) 2m ∞ f s (θ ) = − 2 ∫ rV (r )sin qr dr q 0 2m ∞ = − 2 ∫ V0e − µr sin qr dr q 0 q 2mV =− 2 0 2 , q = 2k sin(θ / 2),  q q + µ2 where k is the wave vector of the relative motion of the neutrons, m = mn /2 is the reduced mass.

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As the spin wave function of the spin singlet state is antisymmetric, its spatial wave function must be symmetric. Thus

σ s = f (θ ) + f (π − θ )

2

=

 4m 2V02  1 1 +    4  µ 2 + 4 k 2 sin 2 θ2 µ 2 + 4 k 2 cos 2 θ2 

=

16m 2V02 ( µ 2 + 2k 2 )2 .  4 ( µ 2 + 4 k 2 sin 2 θ2 )2 ( µ 2 + 4 k 2 cos 2 θ2 )2

2

Because the neutrons are initially unpolarized, the scattering cross section is 1 3 1 σ (θ ) = σ s + σ t = σ s 4 4 4 4m 2V02 ( µ 2 + 2k 2 )2 . = 4 2  ( µ + 4 k 2 sin 2 θ2 )2 ( µ 2 + 4 k 2 cos 2 θ2 )2

6033 The scattering of low-energy neutrons on protons is spin dependent. When the neutron-proton system is in the singlet spin state the cross section is

σ 1 = 78 × 10 −24 cm2, whereas in the triplet spin state the cross section is σ 3 = 2 × 10 −24 cm2. Let f1 and f3 be the corresponding scattering amplitudes. Express your answers below in terms of f1 and f3 . a. What is the total scattering cross section for unpolarized neutrons on unpolarized protons? b. Suppose a neutron which initially has its spin up scatters from a proton which initially has its spin down. What is the probability that the neutron and proton flip their spins? (Assume s-wave scattering only.) c. The H2 molecule exists in two forms: ortho-hydrogen for which the total spin of the protons is 1 and para-hydrogen for which the total spin of the protons is 0. Suppose now a very low energy neutron ( λn  d , the average separation between the protons in the molecule) scatters from such molecules. What is the ratio of the cross section for scattering unpolarized neutrons from unpolarized ortho-hydrogen to that for scattering them from para-hydrogen? (Berkeley)

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Sol: a. The triplet and singlet spin states of a neutron-proton system can respectively be expressed as 1 χ13 = α nα p , χ 03 = (α n β p + α p βn ), χ −31 = βn β p , 2 1 (α n β p − α p βn ), χ −1 1 = 2 ∧ 1 1 with α =   , β =   . If we define an operator f by  0  0 ∧

f =

3 1 1 f3 + f1 + ( f3 − f1 )(σ n ⋅ σ p ), 4 4 4

then as

σ n ⋅ σ p = σ nxσ px + σ nyσ py + σ nzσ pz , σ x α = β , σ yα = i β , σ zα = α , σ x β = α , σ y β = −iα , σ z β = − β , we have ∧

f χ13 = f3 χ13 ,

∧

∧

f χ 03 = f3 χ 03

∧

f χ −31 = f3 χ −31 ,

f χ −1 1 = f1 χ −1 1 , ∧

i.e., the eigenvalues of f for the triplet and singlet spin states are f3 and f1 respectively. Similarly, if we define 3 2 1 2 1 2 f3 + f1 + ( f3 − f12 )(σ n ⋅ σ p ) 4 4 4 ∧ 2 2 2 then f has eigenvalues f3 and f1 for the triplet and singlet states, and we ∧

f

2

=

∧

can express the total cross section for the scattering as σ t = 4π f 2 . Assume the spin state of the incident neutron is  e − iα cosβ   iα .  e sinβ  Note here (2β , 2α ) are the polar angles of the spin direction of the neutron. 1 If the state of the polarized proton is   , then the cross section is  0 +

+

 e − iα cos β   1 ∧ 2  1  e − iα cos β  σ t = 4π  iα    f  0  iα  . sin β  n  e sin β  n  0 p pe

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As +

 1  1  0 (σ n ⋅ σ p )  0 p p   0  0 1  = (1 0) p σ nx   + iσ ny   + σ nz     0 p   l p   1 p  = σ nz , +

 e − iα cos β   e − iα cos β   iα  σ nz  iα   e sin β  n  e sin β  n − iα

0  e 1 = (e iα cos β e − iα sin β )n  − 0 1  e iα 

cos β   sin β  n

 e − iα cos β  = (e iα cos β e − iα sin β )n  − iα  sin β  n  −e = cos β − sin β = cos 2β , 2

2

we have

σ t = π {3 f32 + f12 − ( f32 − f12 )cos 2β } cos 2β  1 3 =  σ 3 + σ 1 − (σ 3 − σ 1 ) . 4 4  4 Since the incident neutrons are unpolarized, cos 2β = 0 and so 3 1 σ t = σ 3 + σ 1. 4 4 Because the direction of the z-axis is arbitrary, the total scattering cross section of unpolarized protons is as the same as that of polarized protons. b. The state vector before interaction is  1   0  0  1 . n p We can expand this in terms of the wave functions of the singlet and triplet states:  1   0 1  1  1  0  0  1   0  1 = 2  2  0  1 +  1  0  n p n p n p    +

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1   1   0  0  1  −  2  0 n  1 p  1 n  0

  p   .

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The scattered wave is then e ikr 1  1  1   0  0  1  +    fs r 2 2  0 n  1  p  1 n  0 p   + f1 =

1  1  0  0  1   −   2  0 n  1 p  1 n  0 p   

e ikr r

 f + f  1  0 f − f  0  1  3 1 + 3 1     .      2  0 n  1  p 2  1  n  0 p   

Hence the probability that the neutron and proton both flip their spins is ( f3 − f1 )2 1 ( f3 − f1 )2 = . ( f3 + f1 )2 + ( f3 − f1 )2 2 f32 + f12 c. Let ^

^

^

F = f1 + f 2 = with S=

3 f3 + f1 1 + ( f3 − f1 )σ n ⋅ S 2 2

(

) (

)

1 σ p + σ p2 = s p1 + s p2 . 2 1

As

σ xσ y = iσ z , σ yσ z = iσ x , σ zσ x = iσ y , σ iσ j + σ jσ i = 2δ ij , we have (σ n ⋅ S)2 = S 2 − σ n ⋅ S, and hence 1 ^ F 2 = {(3 f3 + f1 )2 + (5 f32 − 2 f1 f3 − 3 f12 )σ n ⋅ S 4 + ( f3 − f1 )2 S 2 }. For para-hydrogen, S = 0 and so

σ P = π (3 f3 + f1 )2 . In this case, as there is no preferred direction the cross section is independent of the polarization of the incident neutrons.

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1 1 For ortho-hydrogen, S 2 = 1(1 + 1) = 2 . Taking the proton states as     ,  0 p1  0 p2 using 1 σ n ⋅ S = (σ n ⋅ σ p1 + σ n ⋅ σ p2 ) 2 and following the calculation in (a) we have

σ ⋅ S = cos 2β . Hence

σ 0 = π {(3 f3 + f1 )2 + (5 f32 − 2 f1 f3 − 3 f12 )cos 2β + 2( f3 − f1 )2 }, where 2β is the angle between S and σ rι. If the neutrons are unpolarized, cos 2β = 0 and so

σ 0 = π (3 f3 + f1 )2 + 2( f3 − f1 )2  . This result is independent of the polarization of the hydrogen. The ratio we require is: σ0 = 1 + 2( f3 − f1 )2 / ( f1 + 3 f3 )2 . σP

6034 Consider a hypothetical neutron-neutron scattering at zero energy. The interaction potential is  σ ⋅ σ V , r ≤ a, V (r ) =  1 2 0 r > a,  0,

where σ 1 and σ 2 are the Pauli spin matrices of the two neutrons. Compute the total scattering cross section. Both the incident and target neutrons are unpolarized. (CUS) Sol:

Consider the problem in the coupling representation. Let 1 1 S = s1 + s 2 = σ 1 + σ 2 . 2 2

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Then 1 σ 1 ⋅ σ 2 = (4 S 2 − σ 12 − σ 22 ) 2 1 = [ 4 S(S + 1) − 3 − 3] 2 = 2S(S + 1) − 3, where S = 1 or 0. It is noted that an eigenstate of S is also an eigenstate of V(r). For zero-energy scattering we need to consider only the s partial wave, which is symmetrical. The Pauli principle then requires the spin wave function to be antisymmetric. Thus we have S = 0 and  −3V0 , r < a, V= r > a.  0, For s-waves, the wave equation for r < a is d 2u 2 + k0 u = 0, dr 2 where u(r ) = rψ , ψ being the radial wave function, k02 = 6mV0 / h 2, and the solution is u(r ) = A sin(k0r ). For r > a , the wave equation is d 2u 2 + k1 u = 0, dr 2

where k12 = 2m E , and the solution is u(r ) = sin(k1r + δ 0 ). 2 The continuity of u and u′ at r = a gives k1 tan(k0a) = k0 tan(k1a + δ 0 ). For E → 0, k1 → 0 and k1 tan(k1a) + tan δ 0 tan(k0a) = k0 1 − tan(k1a)tan δ 0 → k1a + tan δ 0 , i.e.,  tan(k0a)  δ 0 ≈ k1a  − 1 .   k0a For collisions of identical particles,

σ (θ ) = f (θ ) + f (π −θ ) 1 = 2 k1

∑ 2(2l + 1)e

l =0,2,4

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2 2 iδ t

sin δ l Pl (cosθ ) .

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Considering only the s partial wave, we have the differential cross section 4 4 σ (θ ) = 2 sin 2 δ 0 ≈ 2 δ 02 k1 k1 and the total cross section 2

 tan(k0a)  σ t = 4πσ = 16π a 2  − 1 .   k0a As the incident and target neutrons are unpolarized the probability that S = 0 (i.e., opposite spins) is 1 . Hence 4 2

 tan(k0a)  σ t = 4π a  − 1 .   k0a 2

6035 A beam of spin − particles of mass m is scattered from a target consisting of 1 2

heavy nuclei, also of spin 1/2. The interaction of a test particle with a nucleus 3 is cs1 ⋅ s 2δ (x1 − x 2 ) , where c is a small constant, s1 and s 2 are the test particle and nuclear spins respectively, and x1 and x 2 are their respective positions. a. Calculate the differential scattering cross section, averaging over the initial spin states. What is the total cross section? b. If the incident test particles all have spin up along the z-axis but the nuclear spins are oriented at random, what is the probability that after scattering the test particles still have spin up along the z-axis? (Princeton) Sol: a. As the nuclear target, being heavy, acts as a fixed scattering center, the center-of-mass and laboratory frames coincide. Then the equation of relative motion is  2 2  (3)  − 2m ∇ r + cs1 ⋅ s2δ (r )ψ (r ) = Eψ (r ).  

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As c is a small constant, we can employ the Born approximation m f (θ ) = − e i(k − k ′)⋅r ′cs1 ⋅ s2δ (3) (r ′ )d 3r ′ 2π  2 ∫ cm =− s1 ⋅ s 2 , 2π  2 (3) using the property of δ (r ′ ). The differential scattering cross section is

σ (θ ) =| f (θ )|2 =

c 2m 2 | s1 ⋅ s 2|2 , (2π  2 )2

where 1 s1 ⋅ s 2 = S2 − s12 − s 22  2 1 1 3 1 3 =  S(S + 1) − ⋅ − ⋅   2 , 2 2 2 2 2 S = s1 + s 2 . For the state of total spin S = 0, 2

σ (θ ) =

c 2m 2 1  3  2 (3mc )2 . ⋅ −   = 2 2 (8π )2 (2π  ) 2  2 

For S =1,

σ 1 (θ ) =

2

c 2m 2 1 1 2 (mc )2 ⋅  = . 2 2 (2π  ) 2 2 (8π )2

Averaging over the initial spin states, the differential scattering cross section is 3 3(mc )2 1 σ t (θ ) = σ 0 (θ ) + σ 1 (θ ) = . 4 4 (8π )2 Alternative solution: Let | α p 〉 denote the initial spin state of the incident particle. The spin of the target is unpolarized, so its state is a “mixture” of α and β states (not “superposition”), c1 (t ) α N + c2 (t ) β N . Here c1 (t ) and c2 (t ) have no fixed phase difference, and so the mean-square values are each 1/2. In the coupling representation, the initial spin state is a mixture of the states c (t ) 1 c1 (t ) χ , 2 χ 0 + χ 00 , 2

(

)

where | χ 〉 = | α p 〉 | α N 〉,

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( (

χ 00

) )

1 α p βN + β p α N , 2 1 = α p βN − β p α N . 2

χ 01 =

After scattering, the spin state of the system, with part of the asymptotic spatial states, becomes c (t ) ψ (S ) f = c1 (t ) f1 (θ ) χ11 + 2 [ f1 (θ ) χ 01 2 + f0 (θ ) χ 00  . Taking the dot product of the different χ states with ψ f , we obtain the corresponding probability amplitudes, which are then added together to give the total cross section 1 2 2 σ t (θ ) = c1 (t ) σ 1 (θ ) + c2 (t ) σ 1 (θ ) 2 1 2 + c2 (t ) σ 0 (θ ). 2 As c1 (t ) , c2 (t ) each has the mean-square value 12 , we have 1 1 1 σ t (θ ) = σ 1 (θ ) + σ 1 (θ ) + σ 0 (θ ) 2 4 4 3 1 = σ 1 (θ ) + σ 0 (θ ), 4 4 same as that obtained before. b. After scattering the two irrelevant spin states of the system are 1 χ11 and χ 01 + χ 00 . 2

(

)

Taking the dot product of the two states with ψ f , we obtain the corresponding scattering amplitudes. Hence the scattering cross section is 1 2 2 2 σ (t ) = c1 (t ) σ 1 (θ ) + c2 (t ) [ f1 (θ ) + f0 (θ )] . 4 Averaging over the ensemble, we have 1 1 m 2c 2 2 σ (t ) = σ 1 (θ ) + f1 (θ ) + f 2 (θ ) = . 2 8 (8π )2 The probability that the test particles still have spin up is P=

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σ (t ) 1 = . σt 3

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6036 a. Two identical particles of spin

1 2

and mass m interact through a

screened Coulomb potential V ( r ) = e 2 exp( − λ r ) / r , where 1/ λ is the screening length. Consider a scattering experiment in which each particle has kinetic energy E in the center-of-mass frame. Assume that E is large. The incoming spins are oriented at random. Calculate (in the

center-of-mass frame) the scattering cross section ∂σ for observation ∂Ω

of a particle emerging at an angle θ relative to the axis of the incoming particles as shown in Fig. 6.8.

Fig. 6.8 b. Assuming that the outgoing particles are observed at an angle θ relative to the beam axis, what is the probability that after the scattering event the two particles are in a state of total spin one? What is the probability that, if one particle has spin up along the z-axis, the other particle also has spin up along the z-axis? c. How large must the energy be for your approximations to be valid? Suppose that, instead, the energy is much less than this. In the limit of low energies, what is the probability that after being scattered the two particles are left in a state of S = 1? (Princeton) Sol:

In the CM frame, the motions of the two particles are symmetric. The interaction is equivalent to a potential centered at the midpoint of the line joining the particles. V ( ρ ) = e 2 exp(−2λρ ) / 2 ρ ,

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where ρ = 2r , r being the separation of the two particles. As the energy E is large, we can use the first Born approximation m f (θ ) = − e − iq⋅ρV ( ρ )d 3 ρ 2π  2 ∫ 2m ∞ e 2 =− 2 ∫ exp(−2λρ )sin(q ρ )d ρ q 0 2 =−

me 2 ,  2 q 2 + (2λ )2 

where q is the momentum transfer during the scattering, q = 2k sin(θ / 2). Considering the symmetry of the wave function of the two-identical-particle system, we have 2

for S = 0,

σ s (θ ) = f (θ ) + f (π − θ ) ,

for S = 1,

σ a (θ ) = f (θ ) − f (π − θ ) .

2

Thus 2

 1  me 2   1 1 σ s (θ ) =  2   2 2 θ + 2 2θ 2  2 4     k sin 2 + λ k cos 2 + λ 

2

2

2  1  me 2   k 2 + 2λ 2  , =  2  4     (k 2 sin 2 θ + λ 2 )(k 2 cos 2 θ + λ 2 )  2   2 2

2  1  me 2   k 2 cosθ  . σ a (θ ) =  2   2 2 2 2 4     (k sin 2 θ + λ )(k cos 2 θ + λ )    2 2

Hence, the total cross section is 1 3 σt = σs + σa 4 4 2

1  me 2  (k 2 + 2λ 2 )2 + 3(k 2 cosθ )2 =  2  2 . 16     k 2 2 θ + λ 2 k 2 2 θ + λ 2  sin cos ( ) 2 2  

(

)

a. If the incident particle is unpolarized, then the probability that after the scattering the two particles are in a state of total spin one is 3 σa 3(k 2 cosθ )2 4 = 2 . (k + 2λ 2 )2 + 3(k 2 cosθ )2 σt

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The probability that after the scattering both particles have spin up along the z-axis is 1 σa (k 2 cosθ )2 4 = 2 . (k + 2λ 2 )2 + 3(k 2 cosθ )2 σt b. The  = 0 partial wave has the symmetry f (θ ) = f (π − θ ). It makes no contribution to the S = 1 state, while it is the main contributor to the S = 0 state. Therefore the ratio of the scattering cross section of the S = 1 state to that of the S = 0 state is equal to the ratio of the scattering cross section of the  = 1 partial wave to that of the  = 0 partial wave. It tends to zero in the low energy limit.

6037 An electron (mass m) of momentum p scatters through angle θ in a spin2 dependent (and parity-violating) potential V = e − µr ( A + Bσ ⋅ r ), where

µ(> 0), A, B are constants and σ x , σ y , σ z are the usual Pauli spin matrices. Let ∂σ i ∂Ω

be the differential scattering cross section, summed over final spin states

but for definite initial spin state, labeled by the index i, of the incident electron. In particular, quantizing spin along the line of flight of the incident electron, we may consider alternatively: incident spin “up” (i =↑) or “down” (i =↓) . Compute ∂σ ↑ and ∂σ ↓ as functions of p and θ in lowest order Born ∂Ω ∂Ω approximation. (Princeton) Sol:

Let the incident direction of the electron be along the x-axis. In a diagonal representation of σ z, the spin wave function of the incident electron may be expressed as

ψ+ =

1  1  , or 2  l 

ψ− =

1  l  2  − l  .

Let n be the unit vector along the r direction, i.e., n = (sin θ cosϕ , sin θ sin ϕ , cosθ ). Then

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 0 − i  0 1 σ ⋅n =  sin θ cosϕ +  sin θ sin ϕ   i 0   1 0  cosθ sinθ e − iϕ   1 0 = + cos θ  . iϕ  0 − 1  sin θ e − cosθ  First consider ψ + ; (σ ⋅ n )ψ + = =

− iϕ 1  cosθ + sinθ e  2  sinθ e iϕ − cosθ

  

1 1 (cosθ + sin θ e − iϕ )α + (sin θ e iϕ − cosθ )β , 2 2

 0 1 where α =   , β =   are the eigenstates of σ z in Pauli’s representation. In first 1  0 Born approximation the scattering amplitude (including spin) is given by m f (θ ) = − e − iq⋅r'V (r′ )ψ + d 3 x ′ , 2 ∫ 2π  1 2p where q = (p f − p ), | q | = q = sin(θ /2) . This can be written as   m f (θ ) = − e − iqr ′ cosθ ′e − µr ′ 2 2π  2 ∫ × [ Aψ + + r ′B(σ ⋅ n )ψ + ] d 3 x ′ = I1 (θ )α + I 2 (θ )β , where 2 m 1 e − iqr ′ cosθ ′e − µr ′ 2 ∫ 2π  2 × [ A + r ′B(cosθ ′ + sin θ ′e − iϕ ′ )]r ′ 2 sin θ ′dr ′dθ ′dϕ ′ , m B − iqr ′ cosθ ′ − µr ′2 I 2 (θ ) = − e e (sin θ ′e iϕ ′ − cosθ ′ )r ′ 3 2π  2 2 ∫ × sin θ ′dr ′dθ ′dϕ ′.

I1 (θ ) = −

As

∫

2π

0

e ± iϕ ′dϕ ′ = 0,

we have m π ∞ − iqr ′ cosθ ′ − µr ′2 1 e e 2π  2 ∫0 ∫0 2 × ( A + r ′B cosθ ′ )r ′ 2 sin θ ′dθ ′dr ′ ,

I1 (θ ) = −

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499

or, integrating over θ ′, 2m A ∞ − µr ′2 2mi ⋅ r ′e sin(qr ′ )dr ′ − 2 2 ∫ 0 q 2 q 3 ∞ 2 − µr ′2 r′ e cos(qr ′ )dr ′ × 2 ∫0 2mi B ∞ − µr ′2 sin(qr ′ )dr ′. + 2 2⋅ ⋅ r ′e q 2 ∫0

I1 (θ ) = −

As ∞

f ( µ , q ) = ∫ e − µr ′ cos(qr ′ )dr ′ = 2

0

 q2  1 π exp  −  ,  4µ  2 µ

we have ∞ 2  q 2   −2q  ∂f 1 π = − ∫ r ′e − µr ′ sin(qr ′ )dr ′ = exp  −   , 0  4 µ   4 µ  ∂q 2 µ

or

∫

∞

0

2

r ′e − µr ′ sin(qr ′ )dr ′ =

 q2  π q ⋅ exp  −  ,  4µ  µ 4µ

as well as ∞ 2 ∂f π = − ∫ r ′ 2e − µr ′ cos(qr ′ )dr ′ = 0 ∂µ µ

 1   q2  q2 π  q2  ×  −  exp  −  + 2 exp  −  ,  4µ   4 µ  8µ µ  4µ  or

∫

∞

0

2

r ′ 2 e − µr ′ cos qr ′dr ′ =

 q2 q2  1 π exp  − − 2  4 µ 8µ  4µ µ ×

 q2  π exp  −  .  4µ  µ

Thus I1 (θ ) = −

=

 q2  2m A π q ⋅ ⋅ exp  −  2  4µ   q 2 µ 4µ

−

 q2   1 2mi B π q2  exp ⋅ − −  4 µ   4 µ 8 µ 2   2q 2 µ

+

 q2  q π 2mi B exp ⋅ ⋅  − 4 µ  ⋅ 4 µ  2q 2 2 µ  q2  π m  Bq  ⋅ 2  − A + i  exp  −  .  4µ  2 µ 2 µ  2µ 

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Hence, ∂σ ↑ ∂Ω

Similarly we obtain

 q 2   2 q 2 B2  π m2 exp  − 2 µ   A + 4 µ 2  . 8 µ 3 4  q2  mA π I 2 (θ ) = − 2 exp  −   4µ  2 µ 2 µ 2

= I1 (θ ) =

−

imBq 4 2 µ 2

 q2  π exp  −  ,  4µ  2µ

and hence ∂σ ↓ ∂Ω

= | I 2 (θ )|2 =

 q2   µm 2 q 2 B2  exp  −   A2 + . 3 4  2µ   4 µ 2  8µ 

The same results are found for ψ − .

6038 A spinless charged particle P1 is bound in a spherically symmetric state whose wave function is ψ 1 (r ) = (π a)−3/2 e − r

2

/2 a2

. If a spinless, nonrelativistic projectile P2

interacts with P1 via the contact potential V (r − r ′ ) = V0 b 3δ 3 (r − r ′ ) , calculate, in first Born approximation, the amplitude for the elastic scattering of P2 from the above bound state of P1 (without worrying about the overall normalization). Assuming P1 is sufficiently massive that its recoil energy is negligible, sketch

the shape of the angular distribution dσ of the scattered projectiles. How dΩ

does this shape change with bombarding energy, and how can it be used to determine the size of the P1 bound state? What determines the minimum energy of P2 necessary to measure this size? (Wisconsin) Sol:

Because P1 is very heavy and so can be considered as fixed, the Schrödinger equation of P2 is  2 2  3  − 2m ∇ + ∫dr ′ρ1 (r ′ )V0b δ (r − r ′ )ψ (r ) = Eψ (r ),   or  2 2  3  − 2m ∇ + V0b ρ1 (r ) ψ (r ) = Eψ (r ),

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501

2

where p1 (r ) = ψ 1 (r ) is the probability density of the particle P1 at r and m is the mass of P2. Then Born approximation gives f (θ ) = −

2m ∞ r ′V0b3 ρ1 (r ′ )sin(qr ′ )dr ′ | .  2q ∫0

Thus 1  b f (θ ) ∝   q  a

3

 r ′2  r exp ′ ∫0  − a 2  sin(qr ′ )dr ′ ∞

 1  ∝ b3 exp  − (qa)2  ,  4  and hence dσ 2  1  = f (θ ) = σ 0 exp  − (qa)2  dΩ 2   θ   = σ 0 exp  −2(ka)2 sin 2  , 2 

Fig. 6.9

p where k = , σ 0 = σ (θ = 0) . dσ vs. θ for different energies are shown in the  dΩ Fig. 6.9.

dσ When the incident energy is increased, dΩ will more rapidly decrease with increasing θ . As dσ θ ln = −2k 2a 2 sin 2 + c , dΩ 2 d σ where c is a constant, a plot of ln dΩ against sin 2(θ / 2) will give a straight line p 2 2 2 2 with slope −2k a = −2(  ) a , which can be used to determine the size a of the dσ P1 bound state. The expression for dΩ does not appear to impose any restriction on the incident energy, except where the validity of Born approximation, on the basis of which the expression was derived, is concerned. In fact (Problem 6026), the validity of Born approximation requires

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Problems and Solutions in Quantum Mechanics 3

 2k  b  V ~ V0   ,  a ma i.e., kmin 

mb3V0 .  2a 2

6039 a. State the electric-dipole selection rules for atomic states connected by emission or absorption of a photon. b. Interpret the selection rule in terms of photon orbital angular momentum, spin, helicity and parity. c. Make a semi-classical estimate of the lifetime of the 2p state of hydrogen, using the Bohr model and the classical formula P=

2 q2 2 υ 3 c3

(c.g.s. units)

for the power radiated by a particle of charge q and acceleration υ . Express your result in terms of e ,  , c , a and ω , where a is the Bohr radius and ω is the angular velocity in the circular orbit. d. Using the answer from (c), what is the width of the 2p state in electron volts? (Berkeley) Sol: a. The selection rules are ∆l = ±1, ∆m = ±1, 0. b. A photon has orbital angular momentum 0, spin 1, helicity ±1 , and negative parity. The conservation of angular momentum requires ∆l = 0, ±1,

∆m = ±1,0,

while the conservation of parity requires (−1)l = −(−1)l ′ ,

i.e.,

l ≠ l′.

Therefore, ∆l = ±1, ∆m = ±1,0 .

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503

c. Classically, the power radiated by an electron of acceleration v is P=

2 e2 | v |2 . 3 c3

An electron in a circular orbit of radius a has acceleration v = ω 2a, where ω 2a is given by e2 = mω 2a . 2 a For the 2p state, n = 2, l = 1, so the average radius of the electron orbit is 1 a = [3n 2 − l(l + 1)] = 5a0 , 2 where a0 is the Bohr radius. This can also be obtained by a direct integration

∫R r ⋅ r dr = 5a , a= ∫R ⋅ r dr 2 21

(

)

2

2 21

2

0

r where R21 ∝ r exp − 2ao . Thus the power radiated is P=

2 e6 . 3 (5a0 )4 m 2c 3

In a transition to the ground state, the energy difference is e 2  1 1  3e 2 ∆E = E2 − E1 = −  2 −  = . 2a  2 1  8a0 Hence the lifetime of the 2p state is 2

3 2 3  75  a m c T = ∆E / P =   0 4  4 e 2

2 3  75   mc  a =   2  0  4  e  c 2

2

−8 3 1  75    (0.53 x 10 ) =   × = 2.2 × 10−8 s .   4   2.82 × 10−13  3 × 1010

d. The width of the 2p state is Γ =  / T = 3.0 × 10−8 eV .

The neutral K-meson states K

o

6040 o and K can be expressed in terms of states

KL , KS :

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Problems and Solutions in Quantum Mechanics

1 ( KL + KS ), 2 1 = ( KL − KS ), 2

Ko = Ko

1 1 where K L and K S are states with definite lifetimes τ L = γ L and τ S = γ S

and distinct rest energies mL c 2 ≠ mS c 2 . At time t = 0 , a meson is produced in the state ψ (t = 0) = K o . Let the probability of finding the system in state K o at time t be Po (t ) and that of finding the system in state K o at time t be Po (t ) . Find an expression for P0 (t ) − P0 (t ) in terms of γ L , γ S , mL c 2 and mS c 2 . Neglect CP violation. (Columbia) Sol:

Suppose the K meson is a metastable state of width Γ at energy ε 0 . In the region of energy i E = E0 − Γ , 2 its wave function may be expressed as 1  i     K L exp  −i  mL c 2 /  − γ L  t  2   2    i   + K S exp  −i(mS c 2 /  − γ S )t  2   1   1  K L exp(−imL c 2t /  )exp  − γ Lt  =  2  2 

ψ (t ) =

+ K S e − imsc t /e −γ st /2  . 2

The probability of its being in the K o state at time t is P0 (t ) = 〈 K 0 ψ (t )

2

1 − imLc 2t / −γ Lt /2 − imS c 2t / −γ st /2 2 e e +e e 4 1 = {e −γ Lt + e −γ st + 2e − (γ L +γ S )t /2 cos[(mL − mS )c 2t / ]}, 4 =

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505

and the probability of its being in the K o state is P0 (t ) = 〈 K 0 ψ (t )

2

1 = {e −γ L t + e −γ S t − 2e − (γ L +γ S )t /2 cos[(mL − mS )c 2t / ]}. 4 Thus

P0 (t ) − P0 (t ) = e − (γ L +γ S )/2 cos[(mL − mS ) c 2t / ] .

6041 The energy levels of the four lowest states of an atom are Eo = −14.0 eV, E1 = −9.0 eV, E2 = −7.0 eV, E3 = −5.5 eV.

Fig. 6.10 and the transition rates Aij (Einstein’s A coefficients) for the i → j transitions are A10 = 3.0 × 10 8 s −1 , A20 = 1.2 × 10 8 s −1 , A30 = 4.5 × 107 s −1 , A21 = 8.0 × 107 s −1 , A31 = 0, A32 = 1.0 × 107 s −1 . Imagine a vessel containing a substantial number of atoms in the level E2. a. Find the ratio of the energy emitted per unit time for the E2 → E0 transition to that for the E2 → E1 transition. b. Calculate the radiative lifetime of the E2 level. (Wisconsin)

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Problems and Solutions in Quantum Mechanics

Sol: a. The energy emitted per unit time is given by ( Ei − E j )Aij , and so the ratio is E2 − E0 A20 7 12 ⋅ = ⋅ = 5.25. E2 − E1 A21 2 8 b. An atom at level E2 can transit to E0 or E1 through spontaneous transition. So the decrease in the number of such atoms in the time period dt is dN 2 = −( A20 + A21 )N 2dt . i.e., dN 2 = −( A20 + A21 )dt , N2 or, by integration,

N 2 = N 20 exp [ −( A20 + A21 )t ] ,

where N 20 is the number of atoms at energy level E2 at time t = 0. The mean lifetime is then 1 τ= N 20 =

∫

∞

t (−dN 2 ) = ( A20 + A21 )

t =0

∫

t exp [ −( A20 + A21 )t ] dt

∞

0

1 = 5.0 × 10−9 s . A20 + A21

6042 The energy difference between the 3p and 3s levels in Na is 2.1eV. Spin-orbit coupling splits the 3p level, resulting in two emission lines differing by 6Å. Find the splitting of the 3p level. Sol:

The fine structure splitting of Na in ground and excited states is as follows: 3 2p3/2 3p 3 2p1/2 2.1eV

λ1

λ2 3 2s1/2

3s

Fig. 6.11

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507

The transition 3 2p3/2 to 3 2p1/2 produces a photon of wavelength λ2 and the corresponding photon energy is as follows: E2 =

12, 400 eV. λ2 ( Å )

The transition 3 2p1/2 to 3 2s1/2 produces a photon of wavelength λ1 and 12, 400 corresponding photon energy is E1 = eV. λ1 ( Å) The separation between 2p3/2 and 2p1/2 is given by λ −λ  1 1 ∆E = E2 − E1 = 12, 400  −  eV = 12, 400  1 2  eV  λ2 λ1   λ1 λ2  We are given ∆λ = λ1 − λ2 = 6 Å, and we can take λ1 = λ2 = λ as a first order approxi­mation and use the transition corresponding to 3p → 3s and to obtain  12, 400  λ = (12, 400 / 2.1)Å. Therefore λ1 λ2 = λ 2 =    2.1 

2

Thus, evaluating:   6 = 2 meV. ∆E = 12, 400  2  (12, 400 / 2.1)  The fine structure splitting is of the order of 10–3eV.

 

6043 A hydrogen atom (with spinless electron and proton) in its ground state is placed between the plates of a condenser and subjected to a uniform weak electric field E = E0 e − Γt θ (t ), where θ (t ) is the step function: θ (t ) = 0 for t < 0 and θ (t ) = 1 for t > 0 . Find the first order probability for the atom to be in any of the n = 2 states after a long time. Some hydrogenic wave functions in spherical coordinates are ψ 100 = ψ 200 =

1

πa

3 0

e − r /a0 , ψ 210 =

1 32π a

3 0

e − r /2a0

r cosθ , a0

 r  − r /2a0 ,  1 − 2a  e 8π a 0

ψ 21±1 = ∓

1

3 0

r − r /2a0 sinθ e ± iφ . e 64π a a0 1

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3 0

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Problems and Solutions in Quantum Mechanics

A useful integral is

∫

∞

0

x ne − ax dx =

n! . an+1

(Wisconsin) Sol:

Take the direction of the electric field as the z direction. Then the perturbation Hamiltonian is Hˆ ′ = er ⋅ E(t ) = ezE(t ) . The non-vanishing matrix elements of H ′ are those between states of opposite parities. Thus P(1s → 2s ) = 0 . Consider P(1s → 2 p ) . The 2 p state is three-fold degenerate, i.e., m = 1, 0, − 1 .

2 p, m , with

For m′ | z | m′′ not to vanish, the rule is ∆m = 0 . Thus 2p, 1 H ′ 1s , 0 = 2p, −1 H ′ 1s , 0 = 0, and finding the probability of the transition 1s → 2 p is reduced to finding that of 1s , 0 → 2 p, 0 . As * 2 p , 0 H ′ 1s , 0 = eE(t ) ∫ψ 210 r cosθψ 100dr

=

eE(t ) 4 2π a04

∞

π

2π

0

0

0

∫ ∫∫

 3  exp  − r / a0   2 

× r 4 cos 2 θ sinθ dr dθ dϕ =

2 4! 27 2a0e eE(t ) = 2 E(t ), ⋅ ⋅ π 5 35 3  3  4 2π a04  2a  0

the probability amplitude is 1 ∞ 2p, 0 H ′ 1s , 0 e iω 21 t dt i ∫−∞ ∞ 1 27 2a0e iω = E0 ∫ e − Γt e 21 t dt 5 0 i 3 27 2a0eE0 1 = , ⋅ 35 i Γ − iω 21

C2 p 0, 1s 0 =

1 where ω 21 =  ( E2 − E1 ). Hence the probability of the transition 1s , 0 → 2 p, 0 is 2

P(1s → 2 p) = C2 p 0,1s 0 =

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215 a02e 2 E02 . 310  2 (Γ 2 + ω 212 )

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Scattering Theory & Quantum Transitions

509

Note that 1 e 2  1 1  3e 2 ω 21 = ( E2 − E1 ) = .  − =  2a0  22 1  8a0 

6044 A diatomic molecule with equally massive atoms, each with mass M, separated by D is electrically polarized, rotating about an axis perpendicular to D and running through the center of mass of the molecule. a. Express the energy of the rotational state of the molecule with angular momentum quantum number J in terms of its mechanical properties. b. What is the selection rule for electric dipole radiation emission from the molecule in one of its rotational states? (DERIVE ANSWER) c. Determine the frequency of the electric dipole radiation emitted from the rotating molecule as a function of J. (Express answer as a function of J , M , D and any universal constants that may enter). (Buffalo) Sol:

∧ 2 2 a. As H = 21I J , where J is the total angular momentum operator, I = 12 MD is

the moment of inertia of the molecule about the rotating axis, the energy of the rotating state of quantum number J is 1 EJ = J ( J + 1) 2 . MD 2 b. The eigenfunctions of the rotational states are the spherical harmonic functions Yjm . Take the z-axis along the electric field and consider the requirements for j ′′m′′ | cosθ | j ′m′ ≠ 0 . As J ′′m′′ | cosθ | J ′m′ =

( J ′ + 1 − m′ )( J ′ + 1 + m′ ) δ J ′′,J ′+1δ m ′′m ′ (2 J ′ + 1)(2 J ′ + 3) +

( J ′ + m′ )( J ′ − m′ ) δ J ′′,J ′-1δ m ′′m ′ , (2 J ′ + 1)(2 J ′ − 1)

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we require ∆J = j ′′ − j ′ = ±1, ∆m = m′′ − m′ = 0. c. For the transition from energy level J to J − 1 , we have ω = E J − E J −1 =

1 1 2 J 2 2 2 ( 1) ( 1) , J J +  − J − J  = MD 2 MD 2 MD 2

giving

ω=

2J . MD 2

6045 a. Find the energies above the bottom of the potential well of the ground state and first two excited states of a particle of mass m bound in a deep one-dimensional square-well potential of width l as shown in Fig. 6.12. Sketch the corresponding wave functions.

Fig. 6.12 b. Calculate the matrix elements for electric dipole transitions from the first two excited states to the ground state, and explain any qualitative differences. [You need not carry out all the integration] c. Give the general selection rule for electric dipole transition between any two states in the system. (Wisconsin) Sol: a. The energy levels of the system are

π 2  2n 2 , where n =1, 2, … . 2ml 2 The wave functions of even parity are given by E=

ψ n+ ( x ) =

nπ x 2 cos , l l

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where n = an odd integer .

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511

The wave functions of odd parity are

ψ n− ( x ) =

nπ x 2 sin , l l

where n = an even integer .

The ground state and the first two excited states are respectively  2π 2 , 2ml 2

n = 1,

E1 =

n = 2,

E2 = 4 E1 ,

n = 3,

E3 = 9 E1 ,ψ 3+ ( x ) =

ψ 1+ ( x ) =

2  πx , cos   l  l

ψ 2− ( x ) =

2  2π x  sin  ,  l  l

2  3π x  cos  .  l  l

These wave functions are sketched in Fig. 6.13.

Fig. 6.13 b. The Einstein coefficient for electric dipole transition is Akk ′ =

4e 2ω kk3 ′ 2 rkk ′ . 3 3c

The matrix element for the transition of an electric dipole from the first excited state to the ground state is x

21

=∫

l 2

ψ 1∗ ( x ) xψ 2 ( x ) dx

−l 2

=

πx 2 l2 2π x x cos sin dx . l ∫− l 2 l l

The matrix element for the transition of an electric dipole from the second excited state to the ground state is x

l /2

31

= ∫ ψ 1* ( x )xψ 3 ( x )dx − l /2

2 l /2 3π x πx cos = ∫ x cos dx . l − l /2 l l The second matrix element x 31 is zero because the integrand is an odd function. Thus the second excited state cannot transit to the ground state by electric dipole transition. There is however no such restriction on electric dipole transition from the first excited state.

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c. The matrix element for electric dipole transition from a state k to a state k′ of the system is l /2

x

kk ′

= ∫ ψ k* ′ ( x )ψ k ( x ) ⋅ xdx . − l /2

If the initial and final states have the same parity, the integrand is an odd function and x kk ′ vanishes. Thus the general selection rule for electric dipole transition is that any such transition between states of the same parity is forbidden.

6046 Consider a particle in a one-dimensional infinite potential well. Let the origin be at the center as shown in Fig. 6.14.

Fig. 6.14 a. What are the allowed energies? b. What are the allowed wave functions? c. For what class of solutions will a perturbing potential ∆V ( x ) = kx have no first order effect on the energy? d. If transitions between states can occur by dipole radiation, what are the selection rules? (Wisconsin) Sol: a. The allowed energies are En =

 2π 2n 2 ⋅ 8ma 2

(n = 1, 2…)

b. There are two classes of allowed wave functions, one of even parity,

ψ n+ ( x ) =

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nπ x 1 cos , a 2a

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513

where n is an odd integer, and one of odd parity,

ψ n− ( x ) =

nπ x 1 sin , a 2a

where n is an even integer. c. First order perturbation gives the energy as En = En(0) + ∆V

nn

= En(0) + kx

nn

.

As ∆V = kx is an odd function, the diagonal matrix elements are all zero. This means that, as long as the wave function has a definite parity (whatever it is), there is no energy correction of the first order. Only for states of mixed parities will there be energy correction arising from first order perturbation. d. Electric dipole transitions are determined by x x

kk ′

kk ′

. Since

= k′ | x | k ∝ k′ | a + + a | k ≈ k + 1 δ k ′ , k+1 + kδ k ′ , k−1 ,

the selection rule is ∆k = ±1 .

6047 A particle of charge q moving in one dimension is initially bound to a delta function potential at the origin. From time t = 0 to t = τ it is exposed to a constant electric field ε 0 in the x direction as shown in Fig. 6.15. The object of this problem is to find the probability that for t > τ the particle will be found in an unbound state with energy between Ek and Ek + dEk .

Fig. 6.15 a. Find the normalized bound-state energy eigenfunction corresponding to the delta function potential V ( x ) = − Aδ ( x ) . b. Assume that the unbound states may be approximated by free particle states with periodic boundary conditions in a box of length L. Find the normalized wave function of wave vector k ,ψ k ( x ) , the density of states

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as a function of k , D( k ), and the density of states as a function of free-particle energy, D( Ek ). c. Assume that the electric field may be treated as a perturbation. Write ∧

down the perturbation term in the Hamiltonian, H1 , and find the matrix ∧

∧

element of H1 between the initial state and the final state, 0 | H1 | k . d. The probability of a transition between an initially occupied state I and a final state F due to a weak perturbation Hamiltonian Hˆ 1 (t ) is given by PI→F (t ) =

1 2

∫

t −∞

2

iw t ′ F Hˆ1(t ′ ) I e F I dt ′ ,

where ω F I = ( EF − EI )  . Find an expression for the probability P( Ek )dEk that the particle will be in an unbound state with energy between Ek and Ek + dEk for t > τ . (MIT) Sol: a. The energy eigenfunction satisfies the Schrödinger equation −

 2 d 2ψ − Aδ (x )ψ = Eψ , 2m dx 2

where E < 0 , or d 2ψ − k 2ψ + A0δ (x )ψ = 0, dx 2 with k=

2m E , 2

A0 =

2mA . 2

Integrating this equation from −ε to +ε , ε being an arbitrary small positive number and then letting ε → 0, we get

ψ ′(+0) − ψ ′(−0) = −A0ψ (0). We also have from the continuity of the wave function at x = 0

ψ (+0) = ψ (−0) = ψ (0).

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515

Thus

ψ ′(+0) ψ ′(−0) − = −A0 . ψ (+0) ψ (−0) Solving the Schrödinger equation for x ≠ 0 we get ψ ( x ) = Ce − k x . Then as

ψ ( x ) = Ce − kx for x > 0 and ψ ( x ) = Ce kx for x < 0 we have ψ ′(+0) ψ ′(−0) − = −2k . ψ (+0) ψ (−0) Hence k=

A0 mA = 2. 2 

The energy level for the bound state is E=−

 2k 2 mA2 =− 2 , 2m 2

and the corresponding normalized eigenfunction is

ψ (x ) =

mA  mA  exp  − 2 x  .    2

b. If the unbound states can be approximately represented by a plane wave e ikx in a one-dimensional box of length L with periodic boundary conditions, we have   L   L exp ik  −   = exp  ik  ,    2 2   which gives e ikL = 1 , or kL = 2nπ .

(n = 0, ± 1, ± 2, …)

Hence k=

2nπ = kn , L

say.

Thus the normalized plane wave function for wave vector k is

ψ k (x ) =

1 ikx 1   2nπ   e = exp i  x  . L L   L 

Note that the state of energy Ek is two-fold degenerate when k ≠ 0, so the number of states with momenta between p and p + dp is Ldp 1 = D(k )dk = D( Ek )dEk . 2 2π  As k, p and Ek are related by k=

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p p2  2k 2 = , Ek = ,  2m 2m

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Problems and Solutions in Quantum Mechanics

we have D( k ) =

L , 2π

D ( Ek ) =

L m . π  2 Ek

c. Treating ε 0 as a perturbation, the perturbation Hamiltonian is Hˆ 1 = −qε 0 x . Its matrix element between the initial and final states is ∞

∧

k H 1 0 = ∫ ψ k* (−qε 0 x )ψ dx −∞

=−

qε 0 mA +∞ x exp ( −ikx − k0 x ) dx L  2 ∫−∞

=−

qε 0 mA d +∞ i exp ( −ikx − k0 x ) dx L  2 dk ∫−∞

=−

qε 0 mA d  0 i exp(−ikx + k0 x )dx L  2 dk  ∫−∞

∞ + ∫ exp(−ikx − k0 x )dx  0 

=− =

qε 0 mA 1 (−4ikk0 ) 2 2 2 2 [k + k0 ] L 

4iqε 0  mA    L  2 

3/2

k  2  mA  2   k +  2     

2

.

d. The perturbation Hamiltonian is  0, (−∞ < t < 0)  Hˆ 1 =  −qε 0 x , (0 ≤ t ≤ τ )  0. (τ < t < +∞ )  The transition probability at t > τ is PI→F (t ) = =

∧ 1 k H1 0 2  ∧ 1 k | H1 | 0 2 

2

∫

τ

0

2

dt ′ exp(iw F I t ′ )

(

sin 2 ω F Iτ 2

(ω

FI

2

)

2

2

).

As EF =

 2k 2 , 2m

EI = −

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mA2 , 2 2

1 ω FI = ( EF − EI ), 

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517

we have  τ  2  mA  2   sin   k +  2       .  4m  2

sin 2(ω F Iτ 2) = 2 (ω F I 2)2    2  mA  2     k +  2        4m  Hence the probability required is P( Ek )dEk = PI→F (t )D( Ek )dEk =

(16qε 0 )2 m3k03 2mEk 6 2mE π  7  k02 + 2 k      τ  2 2mEk   × sin 2   k0 + 2   dEk    4m 

=

 mA  (16qε 0 )2 m3  2    

3

 mA 2 2mE  π   2  + 2 k      

6

2mEk

7

 τ  mA  2 2mEk   × sin 2   2  + 2   dEk .     4m  

6048 Consider a two-level atom with internal states 1 and 2 of energy separation E2 − E1 = ω 21. It is initially in its ground state 1 and is exposed to electromagnetic radiation described by E = Em (e iwt + e − iwt ) . a. If ω = ω 12 , calculate the probability that the atom will be in the state 2 at a later time t. b. If ω is only approximately equal to ω 12, what qualitative difference will this make? Calculate the same probability for this case as you did in part (a). (Buffalo) Sol: Let H 0 1 = E1 1 = ω 1 1 , H 0 2 = E 2 2 = ω 2 2 .

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The Hamiltonian can be written as H = H0 + H ′, where H ′ = −ex ⋅ Em (e iω t + e − iω t ) is the perturbation arising from the presence of the electromagnetic radiation. The time-dependent Schrödinger equation ∂ t =H t ∂t is to be solved with the initial condition t = 0 = 1 . Suppose the solution is i

t = c1 (t )e − iω1t 1 + c2 (t )e − iω 2 k 2 with c1 (0) = 1, c2 (0) = 0 . Substitution in the Schrödinger equation gives i (c1e

− iω1t

1 − ic1ω 1e − iω1t 1

+ c2e − iω 2t 2 − ic2ω 2e − iω 2t 2 ) = c1e − iω1t ω 1 1 + c 2e − iω 2t ω 2 2 + c1e − iω1 t H ′ 1 + c2e − iω 2 t H ′ 2 . As i

∂ − iω1t (e |1〉) = ω 1e − iω1t 1 , ∂t

etc.,

the above simplifies to ic1e − iω t 1 + ic2e − iω 2t 2 = c1e − iω1t H ′ 1 + c2e − iω 2 t H ′ 2 . Multiplying the above equation by 1 and by 2 , we obtain ic1 = c1 1 H ′ 1 + c2e − i (ω 2 −ω1 )t 1 H ′ 2 , ic2 = c1e i (ω 2 −ω1 )t 2 H ′ 1 + c2 2 H ′ 2 . Writing i − ex ⋅ Em j = aij , or i H ′ j = (e iω t + e − iω t )aij , the above equations become icl = c1 (e iω t + e − iω t )a11 + c2e − iω 21t (e iω t + e − iω t )a12 , ic2 = c1e iω 21t (e iω t + e − iω t )a21 + c2 (e iω t + e − iω t )a22 ,

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where ω 21 = ω 2 − ω 1 . If Em is small, the fast-oscillation terms can be neglected and the equations written as c1 = −ia12c2e i (ω −ω 21 )t , c2 = −ia21c1e i (ω 21 −ω )t . Eliminating c1 from the above we find  c2 − i(ω 21 − ω )c2 + a12a21c2 = 0. As t = 0 = 1 , we have the initial conditions c1 (0) = 1, c 2 (0) = 0, c2 (0) = −ia21c1 (0) = −ia21 . a. For ω = ω 21 the above becomes  c2 + Ω2c2 = 0, 2

where Ω2 = a12a21 = a12 . The solution is c2 = Ae iΩt + Be − iΩt . The boundary conditions for c2 give A = −B = − Hence c2 (t ) = −i

a21 . 2Ω

a21 sin Ωt , Ω

and the probability that the atom will be in the state 2 at time t is | c2 (t )|2 = sin 2 Ωt . b. For ω ≈ ω 21, try a solution c2  e iλt . Substitution gives c2 = Ae iλ +t + Be iλ −t where

λ± =

1 [(ω 21 − ω ) ± Λ ] 2

with Λ = [(ω 21 − ω )2 + 4Ω2 ]1/2 . The boundary conditions for c2 thus give 2ia21 Λ  i  exp  (ω 21 − ω ) t  sin  t  . 2  Λ 2  Hence the probability is c2 (t ) = −

2

4 a21 Λ  c2 (t ) = sin 2  t  . 2 2  Λ 2

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6049 In HCl a number of absorption lines with wave numbers (in cm–1) 83.03, 103.73, 124.30, 145.03, 165.51 and 185.86 have been observed. Are these vibrational or rotational transition? If the former, what is the characteristic frequency? If the latter, what J values do they correspond to, and what is the moment of inertia of HCl? In that case, estimate the separation between the nuclei. (Chicago) Sol:

Diatomic molecular energy levels are given to a first approximation by 1 J ( J + 1) 2  WnJ = −U 0 +  n +  ω + ,  2 2 MR 2 where n is an integer, M and R are the reduced mass and separation of the two atoms. On the right-hand side, the terms are energies associated with, respectively, the electronic structure, nuclear vibrational motion, and rotation of the molecule. As Wn+1, J − Wn , J = ω , the vibrational lines have only one frequency, and so the lines are not due to vibrational transitions. The rotational energy levels EJ =

2 J ( J + 1), 2I

where I = MR 2 , and the selection rule ∆J = 1 give the wave number for the line arising from J + 1 → J as v ≡

 2 ( J + 1) 1 1 . = ∆E J = λ hc hc I

Hence I=

2 . hc ∆v

For J → J − 1, the energy of the spectral line is hcv =

2 J , I

which is proportional to J. The spacing of the neighboring lines ∆v =

2 2 ∆J = I I

is a constant. For the given lines we have

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521

v = λ1 (cm −1 ) Transition J → J − 1 ∆ ( λ1 ) (cm −1 )

83.03

4 → 3 20.70

103.73

5 → 4 20.57

124.30

6 → 5 20.73

145.03

7 → 6 20.48

165.51

8 → 7 20.35

185.86

9→ 8

( ∆ ) = 20.57(cm 1 λ

−1

).

The moment of inertia of the HCl molecule is therefore I=

=

2  1 hc ∆    λ 6.63 × 10−34 (2π )2 × 3.0 × 108 × 20.57 × 102

= 2.72 × 10−47 kg ⋅ m 2 . As M −1 = mH−1 + mCl−1 , the nuclear separation is 1/2

 m + mCl   R =  H  I  mHmCl  

1/2

(1 + 35)   = × 2.72 × 10−47  −27  1 × 35 × 1.67 × 10  −10 = 1.29 × 10 m = 1.29 Å.

6050 An arbitrary quantum mechanical system is initially in the ground state 0 . At −t T t = 0 , a perturbation of the form H ′(t ) = H0 e is applied. Show that at large

times the probability that the system is in state 1 is given by 0 H0 1 2

2

2     + ( ∆ε ) T

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,

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Problems and Solutions in Quantum Mechanics

where ∆ε is the difference in energy of states 0 and 1 . Be specific about what assumption, if any, were made arriving at your conclusion. (Columbia) Sol:

In the first order perturbation method the transition probability amplitude is given by 1 t iω t ′ ck ′k (t ) = ∫ H k′ ′k e k ′k dt ′ , 0 i where H k′ ′k = k′ H ′ k . Then 1 ∞ iω10 t − t T 1 H 0 0 dt e e i ∫0 1 1 = 0,  0

(α = apositive constant),

is applied to the system. The vector E0 is perpendicular to one of the surfaces of the box. Calculate to lowest order in E0 the probability that the charged particle, in the ground state at t = 0 , is excited to the first excited state by the

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523

time t = ∞. (If you wish, you can leave your result in terms of properly defined dimensionless definite integrals.) (Berkeley) Sol:

Replace the cubical box by the potential 0, 0 < x < 2b, 0 < y < 2b, 0 < z < 2b, V (x , y , z ) =  ∞ , otherwise. The zero order wave function of the particle in the box is lπ x mπ y nπ z 1 sin sin sin ≡ lmn . 3 b 2b 2b 2b

ψ lmn ( x , y , z ) =

Then the ground state is 111 , the first excited states are 211 , 121 , 112 . Let E be in the x direction, i.e., E0 = E0 e x . Then H ′ = −eE0 xe −αt . Calculate the matrix elements for the transitions between the ground and the first excited states: 1 2b πx πx 32b dx = − 2 . x sin sin b 9π b ∫0 2b 111 x 121 = 111 x 112 = 0.

111 x 211 =

Hence the transition probability is 1 P= 2 

∫

∞

0

2

 i ∆Et  211 H ′ 111 exp  dt ,   

where ∆E =

π 2  2 2 2 2 2 2 2 3π 2  2 . (2 + 1 + 1 − 1 − 1 − 1 ) = 8mb 2 8mb 2

Thus  32beE0  P=  9π 2 

2

∫

∞

0

∆Et   exp  −α t + i  dt   

2

2

2  32beE0  = .  9π 2  α 2  2 + ( ∆E )2

6052 27

An Al nucleus is bound in a one-dimensional harmonic oscillator potential with natural frequency ω . Label the states as ψ mα = ψ m ( x )φα , where ψ m ( x ), m = 0, 1, 2,… , an eigenstate of the harmonic oscillator potential, describes

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Problems and Solutions in Quantum Mechanics

center-of-mass motion and φα ( x ) , α = 0, 1, 2,… , is the wave function specifying the intrinsic nuclear state. Suppose the nucleus is initially in the state ψ 0 ( x )φ1 and then decays to the ground state φ0 by emitting a photon in the − x direction. Assume that the nuclear excitation energy is much greater than the harmonic oscillator excitation energy: E * = ( Eα =1 − Eα =0 )  ω . a. What is the wave function for the nucleus after the photon has been emitted? b. Write an expression for the relative probability P1 / P0 , where Pn is the probability for the nucleus to be left in the state ψ n 0 = ψ n ( x )φ0 . c. Estimate numerically P1 / P0 with E * = 840 keV and ω = 1 keV . (MIT) Sol: a. The Galilean transformation x ′ = x − υt ,

t′ = t

transforms a wave function ψ ( x , t ) by  Mυ 2 Mυ  ψ ( x , t ) = exp  i t′ + i x ′ ψ ( x ′, t ′ ) ,  2   where υ is the velocity of frame L′ with respect to frame L and is taken to be in the x direction, and M is the mass of the particle. By emitting a photon of energy E * in the −x direction the nucleus acquires * a velocity υ = E in the x direction. At the same time it decays to the ground Mc state φ0. Thus initially (t ′ = t = 0 ) the nucleus has a velocity υ and is, in its own frame of reference L′ , in the ground state φ0 . Hence after emitting the photon the nucleus is initially in the state ψ given by (t = 0, x = x′ )  Mυ  x ψ ( x )φ0 ψ ( x , 0 ) = exp  i    0 in the observer’s frame L. b. The probability that the nucleus is in the state ψ no = ψ n ( x )φo is

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Pn = ψ n 0 ψ ( x , 0 )

525

2

 Mv  x φ ψ (x ) = ψ n ( x )φ0 exp  i  h  0 0  Mv  = n exp  i x 0  h 

2

2

,

where n = ψ n ( x ) . Using the creation and destructron operators a + , a, we can express x=

 (a+ + a ) , 2 Mω

and as e A+ B = e Ae Be −[ A ,B] 2 ([ A, B ] commutes with A, B ) we have  Mυ Pn = n exp  i  

  a + + a ) 0 ( 2 Mω 

 Mυ = n exp  i  

 Mυ  + a exp  i 2 Mω   

 Mυ 2  × exp  − | 0〉  4 ω 

2

  a 2 Mω 

2

m

 Mυ 2   Mυ 2  i  i   Mυ 2  ∞  2ω   2ω  = exp  − ∑ m!  2ω  m ,l =0 l!

 Mυ 2  i   Mυ 2   2ω  = exp  −  2ω  n!

=

1  12 Mυ 2  n !  ω 

n

2

l

n (a + )m al | 0〉

2

n

n!

 1 Mv 2  exp  − 2 .  ω 

Then P1 Mυ 2 E *2 0.84 2 = ≈ = 2 P0 2ω 2 Mc ω 2 × 27 × 931.5 × 10−3 ≈ 1.4 × 10−2.

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6053 Consider the situation which arises when a negative muon is captured by an aluminum atom (atomic number Z = 13). After the muon gets inside the “electron cloud” it forms a hydrogen-like muonic atom with the aluminum nucleus. The mass of the muon is 105.7 MeV . a. Compute the wavelength (in Å) of the photon emitted when this muonic atom decays from the 3d state. (Slide rule accuracy; neglect nuclear motion) b. Compute the mean life of the above muonic atom in the 3d state, taking into account the fact that the mean life of a hydrogen atom in the 3d state is 1.6 × 10 −8 sec . (Berkeley) Sol: a. For spontaneous transitions from the 3d state, the largest probability is for 3d → 2 p . In nonrelativistic approximation, the photon energy is given by hυ =

mµ Z 2e 4  1 1   −  2  2  2 2 32  2

mµ c 2 Z 2  e 2   5  =   2  c   36  =

2

105.7 × 132  1   5  −2     = 6.61 × 10 MeV. 2 137 36

The corresponding wavelength is

λ=

c hc 4.135 × 10−15 × 3 × 1010 = = = 1.88 × 10−9 cm . v hv 6.61 × 104

b. The transition probability per unit time is A ∝ ω 3 | rkk ′ |2 . For hydrogen-like atoms, as 1 rkk ′ ∝ , ω ∝ mZ 2 , Z

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and so

A ∝ m3 Z 4 ,

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527

the mean life of the muonic atom in the 3d state is 3

m  T A  Tµ =  0  T0 =  0  04  A  mµ  Z 3

1  0.51  = × 4 × 1.6 × 10−8 = 6.3 × 10−20 s .   105.7  13

6054 A particle of mass M, charge e, and spin zero moves in an attractive potential

(

)

k x 2 + y 2 + z 2 . Neglect relativistic effects. a. Find the three lowest energy levels E0 , E1 , E2 ; in each case state the degeneracy. b. Suppose the particle is perturbed by a small constant magnetic field of magnitude B in the z direction. Considering only states with unper­ turbed energy E2 , find the perturbations to the energy. c. Suppose a small perturbing potential Ax cos ω t  causes transitions among the various states in (a). Using a convenient basis for degenerate states, specify in detail the allowed transitions, neglecting effects proportional to  A2  or higher powers of A. d. In (c), suppose the particle is in the ground state at time t = 0 . Find the probability the energy is  E1  at time t. e. For the unperturbed Hamiltonian, what are the constants of the motion? (Berkeley) Sol: a. The Schrödinger equation for the particle in a rectangular coordinate system,   2  ∂2  ∂2 ∂2  + + + k ( x 2 + y 2 + z 2 ) ψ ( x , y , z ) − 2 2 2   2m  ∂ x ∂ y ∂ z   = Eψ ( x , y , z ) ,

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can be reduced to three equations of the harmonic oscillator type and the energy of the particle can be written as a sum 3 EN = El + Em + En = ω + ( l + m + n ) ω , 2 where

ω = 2k / M ,

N = l + m + n = 0, 1, 2, … .

Therefore, 3 3 E0 = ω =  2k / M , 2 2 no degeneracy; 5 5 E1 = ω =  2k / M , 2 2

three-fold degeneracy  (ψ 100 , ψ 010 , ψ 001 ) ;

7 7 E2 = ω =  2k / M , 2 2

six-fold degeneracy (ψ 200 , ψ 020 , ψ 002 , ψ 110 , ψ 101 , ψ 011 ) . In spherical coordinates the wave function is

ψ nlm ( r , θ , ϕ ) = Rnr l ( r ) Ylm (θ , ϕ ) ,

where 1/2

Rnr l ( r ) = α

3/2

 2l +2−n r ( 2l + 2nr + 1)!!  2 2 (α r )l e −α r /2  2   π nr ![( 2l + 1)!!] 

× F ( −nr , l + 3 / 2, α 2r 2 ) ,  0, 2,… N , l = N − 2nr =   1, 3,… N ,

( N = even ) , ( N = odd ) ,

N being related to the energy by 3  EN =  N +  ω , N = 0, 1, 2, … ,  2 and the degeneracy is  f N = 12 ( N + 1)( N + 2 ) . b. For a weak magnetic field B in the z direction, the perturbation Hamiltonian is eB ∧ H′ = − Lz . 2 Mc

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Then in spherical coordinates we have Enlm = Enl − ∧

eB m , 2 Mc

where  m  is the eigenvalue of  Lz . Thus the different degenerate states of E2 have perturbed energies: E200 = E20 eB , Mc eB E221 = E22 − , 2 Mc E220 = E22 , eB E22−1 = E22 + , 2 Mc eB E22−2 = E22 + . Mc E222 = E22 −

It is seen that the degeneracy is partially destroyed. c. At time t = 0, H ′ = Ax cos (ω t ). Consider the three-dimensional harmonic oscillator in the rectangular coordinate system. The first order perturbation gives, with l being the quantum number for the component oscillator along the x-axis, l ′m′n′ H ′ ( x , t ) lmn = δ m ′mδ n ′n l ′ H ′ ( x , t ) l

= A cos (ω t )δ m ′mδ n ′n l ′ x l   l +1 l = Aα −1 cos (ω t )  δ l ′ ,l+1 + δ l ′ ,l −1  2   2 × δ m ′mδ n ′n ,

where α =

M 2 k/ M = (2kM /  2 )1/4 . Thus the allowed transitions are 

those between states for which ∆m = ∆n = 0,

∆l = ±1 .

d. Between the states E0  and  E1 , the selection rules allow only the transition

ψ 000 → ψ 100. The probability is P10 =

1 2

∫

t

0

2

H10′ e iω ′ t ′ dt ′ =

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A2 2α 2  2

∫

t

0

2

cos (ω t ′ ) e xω ′ t ′ dt ′ ,

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where

ω′ = t

∫ cos (ω t ′ )e

2k , M iω ′ t ′

0

H10′ = 100 H ′ 000 , 1 t iω t ′ − iω t ′ iω ′ t ′ (e + e )e dt ′ 2 ∫0 1  e i(ω ′+ω ) t − 1 e i(ω ′−ω ) t − 1  =  + . 2i  ω ′ + ω ω −ω′ 

dt ′ =

In the microscope world, ω  and  ω ′  are usually very large. Only when

ω ∼ ω ′  will the above integral make a significant contribution to the integral. Hence P10 ≈

2 A 2 sin [(ω ′ − ω )t /2 ] , 2 8α 2h 2 [(ω ′ − ω ) /2 ]

or, when t is large enough, P10 ≈

A2π t δ (ω ′ − ω ) . 4α 2  2

Note that when t is large, 2

 sin ( x ′ − x )t /2   ≈ 2π tδ ( x ′ − x ) .   ( x ′ − x ) /2  e. The energy, angular momentum, third component of the angular momentum, and parity are constants of the motion.

6055 a. Suppose the state of a certain harmonic oscillator with angular frequency ω  is given by the wave function ∞

ψ = N∑ n=0

αn ψ n ( x ) e − nω t n!

α = x0

mω iφ e , 2

2

N = e −|α| /2 .

Calculate the average position of the oscillator, x , in this state and show that the time dependence of  x  is that of a classical oscillator with amplitude  x 0  and phase  φ .

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531

b. The Hamiltonian for a one-dimensional harmonic oscillator in a laser electromagnetic field is given by H=

ep p2 1 1 + E0 sin ω t − eE0 x cos ω t + mω 02 x 2 , 2 2 2m 2mω

where ω 0 , m and e are the angular frequency, mass and charge of the oscillator, and ω is the angular frequency of the radiation. Assume the laser is turned on at t = 0 with the oscillator in its ground state ψ 0. Treat the electromagnetic interaction as a perturbation in first order, and find the probability for any time t > 0 that the oscillator will be found in one of its excited states ψ n . Useful information: The normalized oscillator wave functions ψ n ( x ) have the property that  d 2 n  ψ n−1 ,  ψ n =  x + mω dx mω  d 2 (n + 1)  ψ n+1 .  ψ n =  x − mω dx mω

(Wisconsin) Sol: a. Adding up the two equations for ψ n  given in the question, we have 2 mω

(

nψ n−1 + n + 1ψ n+1 ,

x n = h / 2mω

(

n n −1 + n +1 n +1 .

2 xψ n = or

)

)

Hence

α *n ∞ α k i(n−k )ω t nx k ∑ e n ! k =0 k ! n=0 ∞

x = ψ x ψ = N 2∑

α *n ∞ α k i(n−k )ω t  e n  k k − 1 ∑ 2mω n ! k =0 k ! n=0 ∞

= N 2∑

+ k + 1 k + 1 

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Problems and Solutions in Quantum Mechanics

α *n ∞ α k i(n−k )ω t ∑ e n ! k =0 k ! n=0 ∞

= N 2∑ ∞

= N 2∑ n=0

 2mω

 α *n ⋅   n !

(

kδ k ,n+1 + k + 1δ k ,n−1

)

α n+1 n + 1e − iω t (n + 1)!

 α *(n+1) α n ⋅ n + 1e iω t  (n + 1)! n ! 

+ = N2

 ∞ | α |2n (α e −iωt + α *eiωt ) ∑ 2mω n=0 n ! 2

= N 2e|α|

 (α e −iωt + α *eiωt ) 2mω

1 = x 0  e i(φ −ω t ) + e − i(φ −ω t )  2 = x 0 cos (φ − ω t ) .

Thus  x  is the same as that for a classical oscillator of amplitude x 0  and initial phase φ . b. Initially the oscillator is in the state ψ 0 = 0 . Writing ω 0  for  ω , the given equations for ψ n  give

( n n −1 + n +1 n +1 ), / 2 ( n +1 n +1 − n n −1 ),

x n = h / 2mω 0 pˆ n = i hmω 0

where  pˆ = −i d . It follows that dx x 0 = h / 2mω 0 1 , pˆ 0 = i hmω 0 / 2 1 . The perturbation Hamiltonian is ep 1 Hˆ ′ = E0 sin ω t − eE0 x cos ω t . 2 2mω As  n |1 1 = δ n ,1 , H n′ 0 = 0 for n ≠ 1 . Hence  Pn0 = 0  for  n > 1. Consider  H10′ . We have ∧

ep 1 H10′ = 1 E0 sin ω t − eE0 x cos ω t 0 2mω 2 =

eE0  i 2  mω

 mω 0  sin ω t − cos ω t  2 2mω 0 

=

eE0 2

 ω0   i sin ω t − cos ω t  , ω

 2mω 0

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and hence the probability that the oscillator transits to state ψ 1 at time t 1 P10 = 2 

  ω0  iω ot ′  i sin ω t ′ − cos ω t ′  e dt ′ 2mω 0  ω

eE0 ∫0 2 t

 ω 0 iω t ′ − iω t ′ 1 iω t ′ − iω ot ′  iω t ′ ∫0  2ω (e − e ) − 2 (e + e ) e dt ′

=

e 2 E02 8mω 0 

=

 8ω 02 1 e 2 E02  1 cos 2ω t  −  2+ 2 8mω 0   2ω (ω 0 − ω 2 )2 2ω 2  +

2

t

2

e 2 E02  cos (ω 0 + ω ) t cos (ω 0 − ω ) t  −  . 4mω   (ω 0 + ω )2 (ω 0 − ω )2 

6056 Suppose that, because of a small parity-violating force, the 22 S1/2  level of the hydrogen atom has a small P-wave admixture 1 1    ψ  n = 2, j =  = ψ s  n = 2, j = , l = 0     2 2 1   + εψ p  n = 2, j = , l = 1 .   2

What first-order radiative decay will de-excite this state? What is the form of the decay matrix element? What does it become if  ε → 0 , and why? (Wisconsin) Sol:

The first-order radiative decay is related to electric dipole transition. It causes the 1 state to decay to ψ n = 1, j = 2 , which is two-fold degenerate, corresponding to  m j = ± 12 , l = 0. The matrix element for such an electric dipole transition is given

(

)

by

1 1   H12′ = ψ  n = 1, j =  − er ψ  n = 2, j =    2 2 1 1    = ε ψ  n = 1, j =  − er ψ p  n = 2, j = , l = 1     2 2 because of the selection rule ∆l = ±1  for the transition. Using non-coupling representation basic vectors to denote the relevant coupling representation basic vectors we have the following final and initial states:

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 1 1 1  ψ  n = 1, j = , m j =  = 100   ,    0 2 2  0 1 1  ψ  n = 1, j = , m j = −  = 100   ,    1 2 2  1 1 1 1  210   ψ p  n = 2, j = , l = 1, m j =  = −   0 2 2 3 +

 0 2 211   ,  1 3

 1 1 2 1  21, − 1   ψ p  n = 2, j = , l = 1, m j = −  = −   0 2 2 3 +

 0 1 210   .  1 3

Hence the non-vanishing matrix elements of  H12′  are 1 1   ε ψ  n = 1, j = m j =  − er ψ p  m j =    2 2 1 1 eε 100 r 210 = eε 100 z 210 e z 3 3 e eε A ez , = 100 r 200 e z = 3 3 where  A = 100 r 200 , e z  is a unit vector along the z direction, =

1 1 1   ε ψ  n = 1, j = , m j = −  − er ψ p  m j =     2 2 2 =−

2 eε 100 r 211 3

eε A 2 eε 100 xe x + ye y 211 = − ex + ie y , 3 3 1 1   ε ψ  n = 1, j = m j =  − er ψ p  m j = −    2 2

(

=−

)

eε A ex − ie y , 3 1 1 1   ε ψ  n = 1, j = , m j = −  − er ψ p  m j = −    2 2 2 =−

=−

(

)

eε A 1 eε 100 r 210 = ex . 3 3

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In the above we have used the relations  1 (1 0)   = 1,  0

 0 (1 0)   = 0,  1

etc.,

r = xe x + ye y + ze z = r sin θ cosϕ e x + r sin θ sin ϕ e y + r cosθ e z , 100 z 210 = 100 r 200 l = 0, m = 0 cos θ l = 1, m = 0 , etc., and the selection rules for the z-component of er, ∆m = 0 for the x-, y-components of er. ∆m = ±1 Thus if the parity of the  2 2 S1/2  state is destroyed by the inclusion of the  ε  term the electric dipole radiation would cause transition from the  2 2 S1/2  state to the ground 2 state 1 S1/2 , the probability of such de-excitation being ∝ ε 2 . If ε = 0 electric

dipole radiation does not allow de-excitation from the state 1 1     ψ s  n = 2, j = , l = 0  to the state ψ  n = 1, j = , l = 0     2 2 because the perturbation H ′  is a polar vector whose the matrix elements are nonzero only when ∆l = ±1 .

6057 a. The part of the Hamiltonian describing the hyperfine interaction between the electron and proton in atomic hydrogen is given by 8π H′ = − µ ⋅ µ δ 3 (r ), 3 e p eg

where  µi = 2mi iic Si  is the magnetic moment and  Si = 12 ri  is the spin of particle i (the σ ′s  are Pauli matrices). Calculate the hyperfine splitting between the 1s 3S1  and  1s 1S0  states of atomic hydrogen. Which state has the lower energy? Explain why physically. b. The vector potential of the radiation field emitted in a transition between the states in part (a) has the general form that, as  r → ∞ ,

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  ω ω ωe e  ˆ ˆ A = −i x +i n× L +i n × σ e +…   2 c 2me c 2me c  c  ×

e

i ωc r −iω t

,

r where  nˆ  is a unit vector along the direction of propagation of the radiation and  ⋅  denotes the matrix element for this transition. Show explicitly for each of the three terms whether or not  ⋅  is nonzero. What is the character of the radiation emitted in the transition? (Wisconsin) Sol:

Let the spatial wave function of ls state be ψ 0 ( r ) , the spin singlet state be  χ 00 , and the spin triplet state be  χ1 M ( M = 0, ± 1). a. The perturbation method for degenerate states is to be used, the perturbation Hamiltonian being H′ =

2 B 2π e g e g p S ⋅ S pδ 3 ( r ) = (Se + S p )2 ⋅ 2 e 3 mem pc 2

− S 2e − S 2p  δ 3 ( r ) = when  B = 2π 3

e 2 ge g p me m p c 2

B  2 3 2 3 S − h δ (r ), 2  2 

ψ 0 ( r ) χ 00  and  and  S = Se + S p , with  Se2 = Sp2 = 12 ⋅ 32  . If 

ψ 0 ( r ) χ1 M  are chosen to be the basis vectors, then H ′  is a diagonal matrix. 1 For the 1s S0  energy level,  S = 0  and we have

∆E1 = ψ 0 χ 00 H ′ ψ 0 χ 00

3 2. = − B 2 ψ 0 ( r = 0 ) 4

3 For the 1s S1 energy level, S 2 = 1(1 + 1)  2 and we have

1 2 ∆E2 = ψ 0 χ1 M H ′ ψ 0 χ1 M = B 2 ψ 0 ( r = 0 ) . 4 As

ψ 0 (r ) =

1 2 − r /a e , 4π a3/2

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2

ψ 0 ( r = 0) =

1 . π a3

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Hence the hyperfine splitting is 1 B 2 . π a3 The above calculation shows that the singlet state  (1 S0 )  has lower energy. The reason is as follows. The intensity of the field produced by a magnetic dipole decreases rapidly with increasing distance. So for the magnetic dipole interaction between the electron and proton we have to consider the case when they are very close. When µe  is parallel to µ p , the energy of the ∆E = ∆E2 − ∆E1 =

magnetic interaction is lower ( as E = − µ ⋅ B )  than when they are antiparallel. Since when µe  and  µ p  are parallel,  Se  and  S p  are antiparallel. The gap in singlet state has the lower energy. b. For the transition from the triplet state to the singlet state, due to the vector potential A, as the terms for xˆ  and  Lˆ  do not contain spin operators, we have x = ψ 0 χ 00 xˆ ψ 0 χ1 M = ψ 0 xˆ ψ 0 χ 00 | χ1 M = 0, L = ψ 0 χ 00 Lˆ ψ 0 χ1 M = 0,

σ e = ψ 0 χ 00 σ e ψ 0 χ1 M = χ 00 σ e χ1 M . As 1   1   0  1   0  −  , 2   0 e  1 p  0 p  1 e   1  1 χ11 =     ,  0 e  0 p

χ 00 =

 1   0  1   1   0     +      , 2   0 e  1 p  0 p  1 e   0  0 χ1,−1 =     ,  1 e  1 p

χ10 =

and if we take the z-axis parallel to n the z component of  σ e  will contribute nothing to  n × σ e , thus we have effectively

χ 00 σ e χ11 = − χ 00 σ e χ1,−1 =

 1 1 1 i ( 0 1)σ e   = − ex − e y , 0   2 2 2

 0 1 1 i (1 0 )σ e   = e x − e y , 1   2 2 2

 0  1 1 1 χ 00 σ e χ10 = (1 0 )σ e   − ( 0 1)σ e   = e Z ,  1  0 2 2

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where e x , e y , e z  are unit vectors along the x-, y-, z-axis respectively. Hence

σ e ≠ 0 . Note that the direction of A is parallel to  n × σ e . It is similar to the vector potential of magnetic dipole radiation, so the radiation emitted in the transition has the character of magnetic dipole radiation.

6058 Protons (magnetic moment  µ ) are in a magnetic field of the form Bx = B0 cos ω t ,

By = B0 sin ω t ,

Bz = constant,

B0  Bz .

At  t = 0  all the protons are polarized in the +z direction. a. What value of  ω  gives resonant transitions? b. What is the probability for a proton at time t to have spin in the  −z  direction? (Assume  B0  Bz )

(Princeton) Sol: a. As  B0  Bz , B′ = Bx xˆ + By yˆ  may be considered as a perturbation. Then the gives the energy unperturbed Hamiltonian (spin part)  H 0 = − µ Bzσ z   1  0 with spins along + z and − z difference of the two states    and       0 1 directions as 2 µ Bz . Hence resonant transition occurs at angular frequency

ω = 2 µ Bz /  . b. As

H = − µσ ⋅ B

(

= − µ σ x Bx + σ y By + σ z Bz

)

Bx − iBy   Bz = −µ  ,  Bx + iBy − Bz  the Schrödinger equation can be written as i

 Bz B0e − iω t   a ∂  a = − µ  iω t   , ∂t  b  − Bz   b   B0e

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539

where a and b are the probability amplitudes of the electron with its spin oriented along +z  and  −z  directions respectively. Letting a=e

− i ω2 t

f,

iω t

b = e 2 g,

one obtains the equations for f and g :

ω ∂f f + i + µ Bz f + µ B0g = 0, (1) 2 ∂t ∂g ω µ B0 f + i −  g − µ Bz g = 0 . (2) ∂t 2 Taking the time derivative of Eq. (2) and substituting in the expressions of ∂ f ∂g from (1) and (2), we obtain , ∂t ∂t 

∂2 g + Ω2g = 0, (3) ∂t 2 where Ω2 =

2 1  2 2  ω  . µ B + + µ B   0 z    2 2 

Initially the protons are polarized in the +z  direction. Hence  f = 1, g = 0  at t = 0. Then the solution of (3) is  g = A sinΩt , where A is a constant. Assume f = B sin Ωt + C cos Ωt  and substitute these in (1). Supposing f = i at  t = 0, we have C = i,

B=−

1  ω  + µ Bz  ,   Ω  2

A=−

µ B0 . ω

Hence 1  ω    + µ Bz  sin Ωt + i cos Ωt , Ω 2 − µ B0 g= sin Ωt . Ω f =−

Thus the probability for the protons to have spin in the −z  direction at time t is 2

 µB  P = b |2 = g |2 =  0  sin 2 Ωt  Ω  with Ω=

2

1  ω 1  ω   µ 2 B02 +  + µ Bz  ≈  + µ Bz       2  2

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as

Bz  B0 .

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6059 A piece of paraffin is placed in a uniform magnetic field H0 . The sample contains many hydrogen nuclei. The spins of these nuclei are relatively free from interaction with their environment and, to first approximation, interact only with the applied magnetic field. a. Give an expression for the numbers of protons in the various magnetic substates at a temperature T. b. A radio-frequency coil is to be introduced in order to observe resonance absorption produced by an oscillating magnetic field. What should be the direction of the oscillating field relative to the steady magnetic field H 0  and why?

c. At what frequency will resonance absorption be observed? Give the units of all quantities appearing in your expression so that the frequency will be given in megacycles per second. d. In terms of the transition mechanism of the proton spins, explain why the absorption of energy from the radio-frequency field does not disappear after an initial pulse, but in fact continues at a steady rate. What happens to the absorption rate as the strength of the oscillating field is increased to very large values? Explain. (CUS) Sol: a. As the spins of the hydrogen nuclei are assumed to interact only with the external field, the interaction Hamiltonian is ∧

∧

H = − m ⋅ H0 = −g µ N s z H 0 , taking the z-axis in the direction of H0. Then there are two states  sz = 12  and sz = − 12  with respective energies 1 1 E1/2 = − g µ N H 0 , E−1/2 = g µ N H 0 , 2 2 where g = 5.6  is a constant and µ N  the nuclear magneton  µ N = e / 2m pc . The condition for statistical equilibrium at temperature T gives the probabilities for a nucleus to be in the two states as

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541

for  sz = 12 : 1  P = exp  g µ N H 0 /kT  2 

 1  exp  2 g µ N H 0 /kT 

 1  + exp  − g µ N H 0 /kT   ,  2 

1 for sz = − 2 :

 1  P = exp  − g µ N H 0 /kT   2 

 1  exp  2 g µ N H 0 /kT 

 1  + exp  − g µ N H 0 /kT   ,  2  which are also the proportions of protons in the two states. ˆ  must be perpendicular to  H ˆ ,  say along the b. The oscillating magnetic field  H 1 0 x direction. This is because only if the spin part of the Hamiltonian has the form Hˆ = − m ⋅ H = −g µ s H − g µ s H N z

N x

0

1

∧

∧

will the matrix elements  sz = 12 H sz = − 12  and  sz = − 12 H sz = 12  be nonvanishing and transitions between the spin states occur, since  1 0   0 1 1 1 s z − = (1 0 )  = 0,  0 − 1  1  2 2 2 0 1 1 1 s x − = (1 0 )  1 2 2 2

1  0 1 = . 0   1  2

c. Resonance absorption occurs only when the oscillating frequency satisfies the condition ω = E−1/2 − E1/2 , or

ω = g µN H 0 /  . With g = 5.6,  = 1.054 × 10

µN =

−34

Js,

e e me 9.274 × 10−28 J Gs −1 , = = 2m pc 2me c m p 1836

and ω  in megacycles per second is given by

ω=

5.6 9.274 × 10−28 × × 10−6 H 0 = 2.7 × 10−2 H 0 . 1836 1.054 × 10−34

where H 0  is in gauss.

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d. Spin interactions between the protons tend to maintain a thermal equilibrium, so that even if the external field vanishes the magnetic interaction between a proton and the magnetic field caused by other protons still exists and the transitions take place. When the external magnetic field is very strong, the absorption rate saturates.

6060 An electron is bound in the ground state by a potential  β − , V= x ∞ ,

x > 0, x < 0,

which is independent of y and z. Find the cross section for absorption of planewave light of energy ω h , direction k, polarization  ε . Specify the final state of the electron. Take β 2m  ω  mc 2 . 2

(Wisconsin) Sol: As initially the electron moves freely in the y, z directions, its initial state is given by

(

 i p y y + pz z ψ i ( r ) = ϕ ( x ) exp   

)  , 

with   2 d 2ϕ β − ϕ = Eϕ , −  2m dx 2 x ϕ = 0, 

x > 0, x < 0.

The equation for ϕ  is the same as the radial equation of a hydrogen atom with l = 0. Hence the energy states are En = −

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mβ 2 1 , 2 2 n 2

n = 1, 2,... .

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543

Thus the ground (initial) state for x motion has energy and wave function E1 = −

ϕ1 ( x ) =

mβ 2 , 2 2

2 x − x /a e , x > 0, a3/2

where a= The condition  ω 

mβ 2 2

2 . mβ

≈ E1  means that the photon energy is much higher than

the mean binding energy of the electron and can therefore liberate it from the potential well. However, as  ω  mc 2 , this energy is much lower than the electron rest mass and cannot produce electron pairs. So the electron initial state is

(

)

ψ i ( r ) ≡ r | i = Cϕ1 ( x ) exp i k(ye ) y + kz(e ) z  , where  k(ye ) = p y  , kz(e ) = pz   are the wave numbers of the electron in the y, z 2 if the initial-state wave function is directions respectively, C =  1  = 1   L  L normalized to one electron in a 2-dimensional box of sides L in the y – z plane. The final state of the electron is that of a free electron moving in direction  k (fe ) (direction of observation):  1 ψ f (r ) ≡ r | f =    L

3/2

(

)

exp ik (fe ) ⋅ r ,

where L3  is the 3-dimensional box used for normalization. The perturbation Hamiltonian is H ′ = H − H0 = 

e A ⋅ pˆ . mc

{

}{

1 e 1 2 (pˆ + A )2 + V − pˆ + V 2m c 2m

}

where A is the vector potential of the photon field and the charge of the electron is –e. In the above we have chosen the gauge ∇ ⋅ A = 0  and omitted terms of orders higher than  A2 . Let the corresponding initial electric field be E = Eε sin (ω t − k i ⋅ r + δ 0 ) ,

{

}

where k i  is the wave vector of the incident photon and  ε = ε x , ε y , ε z ,  is a unit vector in the direction of E. The vector potential can be taken to be cE A = ε cos (ω t − k i ⋅ r + δ 0 ) ω

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as  E = − 1c ∂A, and the perturbation Hamiltonian is then ∂t e ˆ = −ie exp i (ω t − k ⋅ r + δ )  A ⋅P H′  i 0   mc 2mω

{

}

+ exp  −i (ω t − k i ⋅ r + δ 0 )  Eε∇.

In photon absorption,  E f > Ei  and we need to consider only the second term (the first term is for photo-emission); so the perturbation Hamiltonian is − i e H′ = exp  −i (ω t − k i ⋅ r + δ 0 )  Eε∇ . 2mω For a plane electromagnetic wave, 1 H = k × E, k and so the Poynting vector is S=

c c 2 E k. (E × H) = 4π 4π k

Averaging over time we have S=

cE 2 . 8π

Hence the number of incident photons crossing unit area per unit time is n=

S cE 2 = . ω 8π ω

The differential absorption cross section for photoelectric effect is given by dσ ω i→ f = , dΩ f n where ω i→ j  is the number of electrons in solid angle  dΩ f  which transit from the initial state to final states f near energy E f  in unit time. First order perturbation theory gives the transition probability per unit time as 2 2π ω i→ f d Ω f = ρ E f W f i dΩ f , 

( )

( )

where  ρ E f  is the density of the final states per unit energy range. For nonrelativistic electrons,

ρ=

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mk(fe ) L3 8π 3h 2

,

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545

(e )

where  k f  is the wave number of the electrons in the final states of energies near  E f , −iiee − exp  − W i ( − k ⋅ r + δ )  Eε ⋅∇ i Wfifi = = ff H H ′′ ii = = ff exp  −i − k ii ⋅ r + δ 00  Eε ⋅∇ i 22m ω mω + + ∞ ∞ − +∞ −iiee e −− iiδδ 00 +∞ 3/2 = dx dy dz dz ⋅⋅ L L−−3/2 = e ∫0 dx ∫ ∫ dy mω 0 ω 22m − ∞ − ∞

× ik((ee )) ⋅ r + ik ⋅ r  ( Eε ⋅∇ ) Cϕ ( x ) exp  − × exp  −ik ff ⋅ r + ikii ⋅ r  Eε ⋅∇ Cϕ11 ( x ) × exp + kkzz((ee )) zz  exp ii kk((yyee )) yy + ×   −iieCEe eCEe −− iiδδ 00 −−3/2   1 11  − L 3/2 ∫ dx dx dy dy dz dz εε xx  1 − = L =  −  mω 22m ω   xx aa  + iiεε yy kk((yyee )) + + iiεε zz kkzz((ee )) ϕ −ik ((ee )) ⋅ r + ik ⋅ r  exp  − + ϕ11 (( xx )) exp  ik ff ⋅ r + ik ii ⋅ r  (e ) × exp ii kk((yyee )) yy + × exp + kkzz(e ) zz   − iδ 0 2 44π (i ) π 2 aaeEe eEe − iδ 0 = ε xx kkxx(( ff )) − − kkxx(i ) = 22 ε f ( ) 5/2  (ii ))  f ( ) 5/2 ( mω ωL L 11 + ia k − k  m  + ia kxx − kxx  ( ff ) (e ) (e ) (i ) − − εε yy kk(yye ) − − εε zz kkzz(e ) δδ ((− − kkyy( ) + + kk(yyi )

(

)

}

(

(

+ k((ee)) )δ

(

)

}

{ ( )

f − k(( f )) + k((ii )) + k((ee ))

+ kyy )δ − kzz + kzz + kzz

)

) ..

Hence the differential absorption cross section is 8π ak f e 2 2 dσ ε − ε y k(ye ) − ε z kz(e )  = 2 2 2  x dΩ f mω c(1 + a ∆ )

(

)(

)

f f × δ ky(i ) + ky( e ) − ky( ) δ kz(i ) + kz(e ) − kz( ) .

In above calculation, we have made the change of symbols

{

}

{

}

f f f k (fe ) → k f ≡ kx( ) , ky( ) , kz( ) , k i → k (i ) ≡ kx(i ) , k(yi ) , kz(i ) , f kx( ) − kx(i ) = ∆

and used

(

)

( × exp iy ( k  L = δ (k + k 2π

2 δ k(yi ) + k(ye ) − ky( f )  = 1   2π

∫

L /2

− L /2

(i ) y

(i ) y

xxxxxxxxxxxxx_Chapter_06.indd  Page 545

) + k − k( ) )  dy  − k( ) ) ,

δ k(yi ) + k(ye ) − ky( f )

(e ) y

(e ) y

y

y

f

f

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546

Problems and Solutions in Quantum Mechanics

and

(

)

(

)

2 δ kz(i ) + kz(e ) − kz( f )  = L δ kz(i ) + kz(e ) − kz( f ) .   2π

Note that the two  δ -functions express momentum conservation in the directions of y and z. Also, energy conservation requires  2 k 2f 2m

= E1 +

(

2

2

 2 k (ye ) + kz( e ) 2

2

 k (ye ) + kz( e ) 2

= − E1 +

(

2m

) + ω

2m

) + ω .

The δ  functions mean that the y and z components of  k f  are fixed, and so is the x component of  k f . The physical reason why the  δ -functions appear in the expression for the differential absorption cross section is that when an electron which has definite momenta in the y and z directions collide with an incident photon which also has a definite momentum, energy and momentum conservation laws require the scattering direction of the electron in the final state to be fixed. To find the total absorption cross section, we note that 1 δ (α x ) = δ ( x ) α and so  k(yi ) + k(ye )  1  f δ k(y ) − k(yi ) − k(ye ) = δ  sin θ f sin ϕ f −  , kf  kf    1  kz(i ) + kz(e )   (f) (i ) (e ) δ k k k δ cos θ − − = −  . z z z f  kf  kf  

(

)

(

)

Then the total absorption cross section is

σa = ∫

dσ dΩ f dΩ f

(

)

(i ) (e ) (e )   1 ε z k f sin θ f cos ϕ f − kx − ε y ky − ε z kz  = ∫ 2 2  mω c k f  1 + a 2 k f sin θ f cos ϕ f − kx(i )  

8π ae 2 k f

(

2

)

 k(yi ) + k(ye )   kz(i ) + kz(e)  × δ  sin θ f sin ϕ f − δ cos θ −    f kf   kf   × sin θ f dθ f dϕ f 8π ae 2 = mω ck f

∫

2π

0

(

)

 ε k sin θ cos ϕ − k(i ) − ε k(e ) − ε k(e )  f f x y y z z  x f  2 (i ) 2   1 a k sin θ cos ϕ k + − f f f x  

(

(

 k(yi ) + k(ye )  d sin ϕ f δ  sin ϕ f −  k f sin θ f  cos ϕ f  × sin θ f

xxxxxxxxxxxxx_Chapter_06.indd  Page 546

)

2

)

24/05/22 10:20 AM 2

=

8π ae 2 k f

∫

mω c

(

)

(i ) (e ) (e )   1  ε z k f sin θ f cos ϕ f − kx − ε y k y − ε z kz  2  k 2f  1 + a 2 k f sin θ f cos ϕ f − kx(i )  

(

)

 k(yi ) + k(ye )   kz(i ) + kz(e)  × δ  sin θ f sin ϕ f − δ θ cos −    f kf   kf  

Scattering Theory & Quantum Transitions

× sin θ f dθ f dϕ f 8π ae 2 = mω ck f

∫

2π

0

(

)

 ε k sin θ cos ϕ − k(i ) − ε k(e ) − ε k(e )  f f x y y z z  x f  2 (i ) 2   + − 1 a k sin θ cos ϕ k f f f x  

(

(

 k(yi ) + k(ye )  d sin ϕ f δ  sin ϕ f −  k f sin θ f  cos ϕ f  × sin θ f

(

547

2

)

)

)

2

(i ) (e ) (e )   8π ae 2  ε x k f sin θ f cos ϕ f − kx − ε y k y − ε z kz  = ⋅ . 2  mω ck f  1 + a 2 k f sin θ f cos ϕ f − kx(i )   1 × , sin θ f ⋅ cos ϕ f

(

)

where k(i ) + kz(e ) cos θ f = z , kf cos ϕ f = sin ϕ f =

sin θ f =

(

k 2f sin 2θ f − k(yi ) + k(ye )

)

k f sinθ f

)

2

,

kf

2

k f sinθ f k(yi ) + k(ye )

(

k 2f − kz(i ) + kz(e )

,

.

Finally we get

σa =

8π ae 2 mω c

( − (k

1

) ( ) − (k 2

k 2f − k(yi ) + k(ye ) − kz(i ) + kz(e )

 × ε x  k 2f 

(i ) y

+ k(ye )

2

(i ) z

+ kz(e )

)

)

  ε y k(ye ) + ε z kz(e )  ×  1 + a 2  k 2 − k (i ) + k ( e ) 2 − k (i ) + k ( e ) y y z z  f 

(

) (

2

2

 − kx(i )   2

)

2

   2  .  − kx(i )    

6061 A system of two distinguishable spin- 1   particles is described by the 2 Hamiltonian

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548

Problems and Solutions in Quantum Mechanics

H=−

 2 2  2 2 1 m1m2 ∇1 − ∇2 + ω 2 (r1 − r2 )2 + gσ 1 ⋅ σ 2 2m1 2m2 2 m1 + m2

with  g  ω . a. What are the energy levels of the system? Give the explicit form of the wave functions for the two lowest energy levels (you need not specify the normalization). b. The system is in its ground state at time t → −∞ . An external timedependent potential is applied which has the form  z −z  V ( t ) =  V1 +V2 1 2  f ( t ) s1 ⋅xˆ  L  with  f (t ) = 0  as  t → ∞. Derive a set of coupled equations for the probability amplitudes Cn (t ) = n |ψ (t ) , where  n  denotes an eigen­ state of  H 0  and  ψ (t )  is the time-dependent wave function. c. Calculate  Cn ( ∞ )  for the case 0, t < 0 f (t ) =  1, 0 < t < τ

with 

gτ 

t >τ,

and

 1  and  V2  very small. Work through first order in  V2  and specify

clearly the quantum numbers for the states. (MIT) Sol: a. Let r = r1 − r2 , m1r1 + m 2r2 , m1 + m 2 1 S = s1 + s 2 = ( s1 + s 2 ) . 2 R=

Then the Hamiltonian of the system can be reduced to H=−

2 2 2 2 µ 2 2 ∇R − ∇T + ω r + 2g 2M 2µ 2

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3   S ( S + 1) − 2  ,

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Scattering Theory & Quantum Transitions

549

mm

where  M = m1 + m2 , µ = m11+m22 ,  S is the total spin. Note that in the expression for H, the first term is due to the motion of the system as a whole, the second and third terms together are the Hamiltonian of a spinless isotropic harmonic oscillator, and the last term is due to the spins of the particles. Hence the energy of the system is P2  3 +  n +  ω + 2g  2M 2 n = 0, 1, 2,… .

En =

3   S ( S + 1) − 2  ,

For energy levels of the internal motion, we shall omit the first term on the right-hand side of the above. For the ground state of the internal motion, n = 0, S = 0 and 3 E00 = ω − 3g . 2 Write the wave function as ψ 0 = 0 α 00 . For the first excited state, we similarly have 3 E01 = ω + g, 2

ψ 1 = 0 α 10 , α 1,±.1

Note that  0  is the wave function of the harmonic oscillator ground state and

α SM is the coupled spin wave function. b. Let 2 2 µ 2 2 ∇ r + ω r + 2g 2µ 2

H0 = −

3   S ( S + 1) − 2  ,

H 0 n = En n ,

[H

0

+ V (t )]ψ (t ) = i

Expanding ψ (t ) :

∂ψ . ∂t

ψ (t ) = ∑ Cn (t ) exp ( −iEnt /  ) n , n

and substituting it in the last equation we obtain

[ H0 + V (t )]

∑ Cne

−i

n

En t 

n = i

∑ n

+ i

Cne

∑ n

xxxxxxxxxxxxx_Chapter_06.indd  Page 549

−i

En t 

n

 −iEn  − i e Cn    

En t 

n .

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550

Problems and Solutions in Quantum Mechanics

Multiplying both sides by  m e

∑C

m

m

i

Emt 

 and summing over m we get

Em + ∑∑ Cne m

i

( Em − En )t 

m V (t ) n

n

 −iEm  , = i∑ C m + i∑ Cm     m m or

iC n (t ) = ∑ n V (t ) m exp  −i ( Em − En ) t /   Cm (t ) , m

which is the required set of coupled equations. c. Denote the initial state by  000α 00 , the final state by  nlmα SM . As ∧

σ 1 ⋅ x = sin θ cos ϕσ 1x + sin θ sin ϕσ 1 y + cos θσ 1z ; 1 (α1β 2 − β1α 2 ) , 2 α 11 = α 1α 2 , α 1,−1 = β1β 2 ,

α 00 =

1 (α 1β 2 + α 2β1 ) ; 2 σ 1 yα 1 = iβ1 , σ 1 zα 1 = α 1 ;

α 10 =

σ 1xα 1 = β1 ,

σ 1 y β1 = −iα 1 ,

σ 1x β1 = α 1 ,

σ 1z β1 = − β1 ;

we have

σ 1 ⋅ xα 00 = ∧

Therefore, as Y11 =

3 8π

1 sin θ e iϕα 1,−1 − sin θ e − iϕα 11 + 2cos θα 10 . 2

(

sin θ e iϕ , Y1,−1 =

)

3 8π

sin θ e − iϕ , Y10 =

nlmαSM σ 1 × x α 00 000 = nlmα SM ∧

− =

3 4π

cos θ , we have

4π Y11α1, −1 3

4π 4π Y1, −1α11 + Y10α10 000 3 3

1 δ n0δ l1δ S1 δ m0δ M0 - δ m1δ M, −1 3 −δ m1δ M1 ) ≡ 0,

(

since   = 0  for  n = 0 . Thus to first order perturbation the first term V1 f (t )σ 1 ⋅ x  in V(t) makes no contribution to the transition. Consider next ∧

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Scattering Theory & Quantum Transitions

551

nlmα sM r cos θσ 1 × x α 00 000 ∧

 4π 4π = nlmα SM r  − Y21α1,-1 − Y2, −1α11 15 15  +

1 16π 1  Y20α10 + α10  000 3 5 3 

1  2 δ m0δ M0 − δ m1δ M, −1 = λ1δ l2δ S1  15 3 5 1  1 − δ m, −1δ M1  + λ1δ l0δ S1δ m0δ M0 , 15  3 where ∞

∞

0

0

λ1 = ∫ Rnl ⋅ R00 ⋅ r 3 dr = ∫ Rn 2 R00r 3 dr .

For the three-dimensional harmonic oscillator, n = l + 2nr = 2 (1 + nr ) = even. We have 1 ∞ iω n 0t e nlmα SM V (t ) 000α 00 dt i ∫0 1 τ z  = ∫ e inω t nlmα SM  V1 + V2  σ 1 ⋅ x 000α 00  i 0 L 1 = (einωτ − 1) nlmα SM VL2 r cos θσ 1 ⋅ x α 00 000 , nω 

Cn ( ∞ ) =

∧

∧

and

C2 k−1 ( ∞ ) = 0, C2 k−1 ( ∞ ) = 0, 1V V 1 i 2 kωτ C2 k((e∞ C2 k ( ∞ ) = ) = − 1) 2 (λe1i 2 kωτ − 1) 2 λ1 2kω L 2kω L 1  21  2 δδmm0δ,1δMM0 ,−−1 δ m ,1δ M , −1 × δ S1  δ m×0δδMS10 − 5 15  3 15 3 5 −

1  1 δ m , −1δ−M 1  , δ m , −1δ M 1  , 15   15 ∞

∞

0

0

3 = ∫dr r. 3 R( 2 k )2 R00 dr . where k = 1, 2,where … , lk==2,1, λ2,1 … = ∫ , lr = R2, 1 00 2R ( 2 k )λ

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Part VII

Many-Particle Systems

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7 7001 In one dimension, a particle of mass m is attracted to the origin by a linear force − kx . Its Schrödinger equation has eigenfunctions  1  Ψ n (ξ ) = Hn (ξ )exp  − ξ 2  ,  2  where  mk  ξ = 2    

1/4

x

and Hn is the Hermite polynomial of order n. The eigenvalues are 1/2

1  k En =  n +  ω , where ω =   .   m 2 Consider two non-interacting distinguishable particles (i = 1, 2), each of mass m, each attracted to the origin by a force − kx i . Write down expressions for the eigenfunctions, eigenvalues, and degeneracies for the two-particle system using each of the following coordinate systems: a. single-particle coordinates x1 and x2,

(

b. relative ( x = x 2 − x1 ) and center-of-mass X =

x1 + x 2 2

) coordinates. (MIT)

Sol: a. For the single-particle system, the Hamiltonian is 2 d 2 1 2 + kx 2m dx 2 2  2  d 2 =− −α 4x 2   2m  dx 2

H=−

555

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556

Problems and Solutions in Quantum Mechanics

. The Schrödinger equation can be written as with α 4 = mk 2 d 2ψ + (λ − ξ 2 )ψ = 0, dξ 2 where

ξ = α x, λ =

2 m E.  k

The eigenfunctions are  1  ψ n (ξ ) = Hn (ξ )exp  − ξ 2 ,  2  with eigenvalues 1  En =  n +  ω ,  2 where

ω=

k . m

Using the single-particle coordinates x1 and x2, we can write the Hamiltonian for the two-particle system as  2 ∂2  2 ∂2 1 1 − + mω 2 x12 + mω 2 x 22 2 2 2m ∂ x1 2m ∂ x 2 2 2 = H1 + H 2 .

H=−

The energy eigenfunctions can be obtained as the common eigenfunctions of {H1 , H 2 }, i.e., ψ ( x1 , x 2 ) = ψ ( x1 )ψ ( x 2 ), the corresponding energy being E = E1 + E2. Thus  1  ψnm ( x1 , x 2 ) = Hn (α x1 )Hm (α x 2 ) exp − α 2 ( x12 + x 22 ) ,  2  (N ) Enm = (n + m + 1)ω = ( N + 1)ω ,

where 1/4

 mk  α = 2  ,  

1/2

 k ω =  ,  m

N = n + m.

(N )  is equal to the number of nonThe degeneracy of the energy level Enm negative integer pairs  (n, m)  which satisfy the condition n + m = N , i.e.,

f ( N ) = N + 1.

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Many-Particle Systems

b. Using the relative and center-of-mass coordinates x = x 2 − x1  and  X = we can write the Hamiltonian for the system as H=−

557 x1 + x 2 2

,

 2 ∂2  2 ∂2 1 1 − + Mω 2 X 2 + µω 2 x 2 , 2m ∂ X 2 2 µ ∂ x 2 2 2

where M = 2m, µ = 12 m, ω = ( mk ) . As in (a) we have 1/2

1 ψ nm ( X , x ) = Hn (α X )Hm (β x ) exp  − (α 2 X 2 + β 2x 2 ) ,  2  (N ) = (n + m + 1)ω = ( N + 1)ω , Enm f ( N ) = N + 1, where  Mω  α =    

1/2

1/2

 µω  β =  .   

,

7002 Consider three identical noninteracting particles confined to an infinite square well of width L. Find the energy of the ground state, first excited state, and π 2 2 second exited state in terms of ε1, where ε1 = if the particles are as follows: 2mL2 i. Bosons ii. Spin-1/2 fermions Sol: i. For bosons, there is no restriction, so any number of particles can be placed in a single state. The energy-level diagram for bosons are shown in the following figure: 9 ε1 4 ε1  so the ground state energy is 3 ε1

ε1 For the first excited state, the energy-level diagram is given by 9 ε1 4 ε1  First excited state energy is 1 × 4 ε1 + 2 × ε1 = 6 ε1

ε1

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Problems and Solutions in Quantum Mechanics

For the second excited state, the energy-level diagram is given by 9 ε1 4 ε1  Second excited state energy is 2 × 4 ε1 + 1 × ε1 = 9 ε1

ε1 ii. For spin-1/2 fermions, there is a restriction concerning the Pauli’s exclusion principle; therefore, the maximum number of fermions in a state is given by 2s + 1, where s is the spin of the fermion. For spin-1/2 fermions, the maximum number is 2(1/2) + 1 = 2. Therefore, for a state, only two fermions can be placed. The ground state of the fermion is given by 9 ε1 4 ε1  Ground state energy is 2 × ε1 + 4 × ε1= 6 ε1

ε1 The first excited state is given by 9 ε1 4 ε1  First excited state energy is 2 × 4 ε1 + 1 × ε1 = 9 ε1

ε1 The second excited state is given by 9 ε1 4 ε1  Second excited state energy is 1 × 9 ε1 + 2 × ε1 = 11 ε1

ε1

7003 a. Write down the Hamiltonian and Schrödinger equation for a onedimensional harmonic oscillator. 2

b. If xe − v x  is a solution find v and then give the energy E1 and the expectation values for  x , x 2 , p 2 , px .

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Many-Particle Systems

559

c. Show that for two equal particles in a single one-dimensional harmonic oscillator well the ground state may be written either as

φ0 ( m, x1 ) × φ0 ( m, x 2 ) or x + x2   m   φ0  2 m , 1  φ0  ,( x1 − x 2 ) ,  2  2 where φ0 ( m, x )  is the ground state solution for a single particle of mass m. (Wisconsin) Sol: a. The Hamiltonian for a one-dimensional harmonic oscillator is H=

p2 1 + mω 2 x 2. 2m 2

The stationary Schrödinger equation is  2 d 2 1 2 2  − 2m dx 2 + 2 mω x  ψ (x) = Eψ (x). 2

b. Substitution of ψ = xe − vx  in the Schrödinger equation gives 1  2 2 2 2 − vx 2  − 2m (−2ν )(3 − 2ν x ) + 2 mω x  xe    3 2 1 2 2ν 2  2  − vx 2 = v +  mω 2 − x  xe 2 m    m 2

= E1 xe − vx . Equating the coefficients gives  1 2 2ν 2  mω 2 − = 0,  2 m  2  E = 3 ν , 1  m  whose solution is

xxxxxxxxxxxxx_Chapter_07.indd  Page 559

 mω ,  ν=  2   E1 = 3 ω .  2

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560

Problems and Solutions in Quantum Mechanics

From the symmetry we know  x = 0 . The Virial theorem gives 1 1 1 3 p 2 = mω 2 x 2 = ⋅ ω . 2m 2 2 2 Therefore 3 p 2 = mω , 2

x2 =

3  . 2 mω

To find  px  we first normalize the wave function: ∞

A2 ∫ x 2e −2ν x dx = 1, 2

−∞

giving A2 = 4ν Then consider

2ν . π

 ∞ d ψ ( xψ )dx i ∫−∞ dx ∞ 2 d 2  = A2 ∫ xe −ν x ( x 2e −ν x )dx −∞ i dx 2  2 ∞ = A ∫ 2( x 2 − ν x 4 )e −2ν x dx −∞ i i =− . 2

px = −

c. The Schrödinger equation for the two-particle system is 1  2 2 2 2 2 2   − 2m (∇1 + ∇ 2 ) + 2 mω ( x1 + x 2 )ψ ( x1 , x 2 ) = Eψ ( x1 , x 2 ).   Suppose ψ ( x1 , x 2 ) = φ ( x1 )φ ( x 2 ). The variables can be separated:  2 2 1 2 2  − 2m ∇i + 2 mω xi  φ ( xi ) = Eiφ ( xi ),

i = 1, 2,

with E = E1 + E2 . Hence the system can be considered as consisting of two identical harmonic oscillators without coupling. The ground state is then ψ 0 ( x1 , x 2 ) = φ0 (m, x1 )φ0 (m, x 2 ). On the other hand, introducing the Jacobi coordinates 1 R = ( x1 + x 2 ), r = x1 − x 2 , 2

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the Schrödinger equation becomes 1 2  1  2  1 2 2 2 2  − 2m  2 ∇ R + 2∇ r  + 2 mω  2 R + 2 r   ψ ( R , r ) = Eψ ( R , r ) .   Writing ψ ( R , r ) = φ ( R ) ϕ (r), it can also be separated in the variables to  2 2 2 2  − 4m ∇ R + mω R  φ ( R ) = ERφ ( R ),  2 2 1 2 2  − m ∇ r + 4 mω r  ϕ (r ) = Erϕ (r ), where ER + Er = E , with ER, Er respectively describing the motion of the center of mass and the relative motion. Therefore the wave function of the ground state can be written as m  ψ 0 ( x1 , x 2 ) = φ0 (2m, R )φ0  , r 2  x + x2  m   = φ0  2m, 1  φ0  , x1 − x 2 .  2  2

7004 Consider two particles of masses m1 ≠ m2  interacting via the Hamiltonian H=

p12 p2 1 1 1 + 2 + m1ω 2 x12 + m2ω 2 x 22 + K ( x1 − x 2 )2. 2m1 2m2 2 2 2

a. Find the exact solutions. b. Sketch the spectrum in the weak coupling limit K  µω 2, where µ  is the reduced mass. (Berkeley) Sol: a. Let R = (m1 x1 + m2 x 2 ) (m1 + m2 ),

r = x1 − x 2 .

We have d ∂R d ∂r d = + dx1 ∂ x1 dR ∂ x1 dr =

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m1 d d + , m1 + m2 dR dr

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and so 2

d 2  m1  d 2 m1 d2 d2 +2 + 2, = 2 2  dx1  m1 + m2  dR m1 + m2 drdR dr and similarly 2

d 2  m2  d 2 m2 d2 d2 − + = 2 . 2 2 dx 2  m1 + m2  dR m1 + m2 dRdr dr 2 We also have x12 = R 2 + 2

m2 m22 Rr + r2, m1 + m2 (m1 + m2 )2

x 22 = R 2 − 2

m1 m12 Rr + r2. (m1 + m2 )2 m1 + m2

Hence 2 d2  2 m1 + m2 d 2 − 2 2(m1 + m2 ) dR 2 m1m2 dr 2 1 1 m1m2 2 2 1 2 + (m1 + m2 )ω 2 R 2 + ω r + Kr . 2 2 m1 + m2 2

H=−

Let M = m1 + m2 ,

µ=

m1m2 . m1 + m2

The equation of motion becomes  −2 d 2 2 d 2 µ  K  2 2 1 2 2 ω ω r ψ ( R , r ) M R + − + 1 +  2 2 µ ω 2  2 2 µ dr 2   2 M dR = Eψ ( R , r ). It can be reduced to two independent oscillator equations with energy states and wave functions 1 K  1  E = Elm = El + Em =  l +  ω +  m +  ω 1 + ,  2   2 µω2 1 ψ lm ( R , r ) = ψ l ( R )ψ m (r ) = N l N m exp  − α 12 R 2  H l (α 1 R )  2  1 × exp  − α 22r 2  Hm (α 2r ),  2  where  α1  Nl =   π 2l l ! 

1/2

,

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1/2

 Mω  , α1 =    

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Many-Particle Systems

 α2  Nm =   π 2m m ! 

1/2

 µω  α2 =    

,

1/2

 K   1 + µω 2 

563

1/4

,

and Hm are Hermite polynomials. b. For  K  µω 2 , if we can take  K   1 + µω 2 

1/2

≈ 1,

we have Elm ≈ (l + m + 1)ω = ( N + 1)ω ,

N ≡ l + m = 0, 1, 2, ….

This means that the degeneracy of the Nth energy level is N + 1: N ... ... N =3

l = 3, m = 0;

l = 2, m = 1;

l = 1, m = 2;

N =2

l = 2, m = 0;

l = 1, m = 1;

l = 0, m = 2.

N =1

l = 1, m = 0;

l = 0, m = 1.

N =0

l = m = 0.

l = 0, m = 3.

If K is not so small and we have to take 1+

K K =1+ + 2 µω 2 µω 2

then the energy levels with the same m will move upward by  K  1  +  ,  m +  ω  2 2  2 µω  and so the degeneracy is destroyed.

7005 Consider two identical spin-1/2 fermions confined to an infinite square well of width L. What is the spatial wave function of the system such that the total 5π 2  2 energy ε = if the spin state is singlet? mL2 Sol:

The total wave function of the fermionic system is antisymmetric, so the spatial part of wave function must be symmetric in nature.

ψ fermion = ψ fermion × ψ fermion total spin space

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The given energy is 10ε1, where ε1 is the ground state energy of infinite square well. 10π 2  2 ε= = 10ε1 2mL2



Distributing the two fermions in the energy level,



9ε1

3



4ε 1

2



ε1

1

sym ψ space =



sym ψ space =

1 [ψ1 (1)ψ3 (2) + ψ3 (1)ψ1 (2)] 2

 π x   3π x1  1 2   π x1   3π x 2   sin   + sin  2  sin   . sin       L   L  L L 2L

7006 Consider a system defined by the Schrödinger eigenvalue equation   2 2 k (∇1 + ∇ 22 ) + | r1 − r2 |2 ψ (r1 , r2 ) = Eψ (r1 , r 2 ) . − 2  2m  a. List all symmetries of this Schrödinger equation. b. Indicate all constants of motion. c. Indicate the form of the ground-state wave function. You may assume that the ground-state wave function of one-­dimensional harmonic oscillator is a Gaussian. (Berkeley) Sol: a. The Schrödinger equation has symmetries with respect to time translation, space inversion, translation of the whole system, and the exchange of r1 and r2, as well as symmetry with respect to the Galilean transformation. b. Let  r = r1 − r2 , R = 12 (r1 + r2 ) . The Schrödinger equation can then be written as  2 2 2 2 k 2  ∇ R − ∇ r + r ψ (R , r ) = Eψ (R , r ) . − m 2   4m

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565

This equation can be separated into two, one for the motion of a particle of mass 2m at the center of mass and one for the motion of a harmonic oscillator of mass m 2  relative to the second particle. The motion of the center of mass is a free motion, so that PR2 , Px , Py , Pz , ER , L2R , Lx , L y , Lz  are all constants of the motion. Of the relative motion, Er , L2r , Lz , as well as the parity of the wave function, are constants of the motion. c. The ground-state wave function has the form ψ (R , r ) = φ (R )ϕ (r ) . φ (r )  is the wave function of a harmonic oscillator of mass  m2 :  1  ϕ (r ) ∼ exp  − α 2r 2   2  with

α2 =

mk . 2 2

φ (R )  is the wave function of a free particle of mass 2m: φ (R ) ∼ exp(−ip ⋅ R  ) with | p | = 4mER ,

1 2k ER = E −  . 2 m

7007 Two identical bosons, each of mass m, move in the one-dimensional harmonic oscillator potential V = 21 mω 2 x 2 . They also interact with each other via the potential 2

Vint ( x1 , x 2 ) = α e − β ( x1− x2 ) , where  β  is a positive parameter. Compute the ground state energy of the system to first order in the interaction strength parameter α . (Berkeley) Sol:

Being bosons the two particles can simultaneously stay in the ground state. Taking Vint as perturbation, the unperturbed wave function of the ground state for the system is

ψ 0 ( x1 , x 2 ) = φ0 ( x1 )φ0 ( x 2 ) =

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α0 1 mω exp  − α 02 ( x12 + x 22 )  , α 0 = .  π  2 

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The perturbation energy to first order in α  is then ∞

∆E = ∫ ∫ ψ 0* ( x1 , x 2 )Vint ( x1 , x 2 )ψ 0 ( x1 , x 2 )dx1 dx 2 −∞

α 02α ∞ exp[−α 02 ( x12 + x 22 ) − β ( x1 − x 2 )2 ]dx1 dx 2 π ∫ ∫−∞ αα = 2 0 1/2 , (α 0 + 2β )

=

where the integration has been facilitated by the transformation x1 + x 2 = y1 , 2

x1 − x 2 = y 2. 2

The ground state energy of the system is therefore E = ω +

1/2

α 0α  mω  α = . 2 1/2  with  0    (α 0 + 2β )

7008 A one-dimensional square well of infinite depth and 1 Å width contains °

3 electrons. The potential well is described by  V = 0  for  0 ≤ x ≤ 1 A  and  V = +∞ °

for  x < 0  and  x > 1 A. For a temperature of  T = 0 K,  the average energy of the 3 electrons is  E = 12.4 eV  in the approximation that one neglects the Coulomb interaction between electrons. In the same approximation and for T = 0 K , what is the average energy for 4 electrons in this potential well? (Wisconsin) Sol:

For a one-dimensional potential well the energy levels are given by En = E1n 2 , where E1 is the ground state energy and  n = 1, 2, …  Pauli’s exclusion principle and the lowest energy principle require that two of the three electrons are in the energy level E1 and the third one is in the energy level E2. Thus  12 ⋅ 4 × 3 = 2 E1 + 4 E1, giving  E1 = 6 ⋅ 2 eV. For the case of four electrons, two are in E1 and the other two in E2, and so the average energy is 1 5 E = (2 E1 + 2 E2 ) = E1 = 15.5 eV. 4 2

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567

(Note: the correct value of E1 is 2

 π × 6.58 × 10−16 × 3 × 1010  π 22 1  π c  1 = =  2 2  6   2ma 2mc  a  1.02 × 10  10−8

2

= 37.7 eV.)

7009 Consider two electrons moving in a central potential well in which there are only three single-particle states ψ 1 , ψ 2  and  ψ 3 . a. Write down all of the wave functions ψ (r1 , r2 )  for the two-electron system. b. If now the electrons interact with a Hamiltonian δ H = V ′(r1 , r2 ) = V ′(r2 , r1 ), show that the following expression for the matrix element is correct:

ψ 13 | δ H |ψ 12 = ψ 3 (r1 )ψ 1(r2 )| V ′(r1 , r 2 )|ψ 2(r1 )ψ 1(r2 ) − ψ 1(r1 )ψ 3 (r2 )| V ′(r1 , r2 )|ψ 2(r1 )ψ 1(r2 ) . (Buffalo) Sol: a. The wave functions for a fermion system are antisymmetric for interchange of particles, so the possible wave functions for the system are 1 (ψ 1 (r1 )ψ 2 (r2 ) − ψ 1 (r2 )ψ 2 (r1 )), 2 1 ψ 13 = (ψ 1 (r1 )ψ 3 (r2 ) − ψ 1 (r2 )ψ 3 (r1 )), 2 1 ψ 23 = (ψ 2 (r1 )ψ 3 (r2 ) − ψ 2 (r2 )ψ 3 (r1 )). 2

ψ 12 =

b. We can write

ψ 13 | δ H |ψ 12 =

1 ψ 1 (r1 )ψ 3 (r2 )| V ′(r1 , r 2 )ψ 1 (r1 )ψ 2 (r2 ) 2 1 − ψ 1 (r1 )ψ 3 (r2 )| V ′(r1 , r 2 )ψ 2 (r1 )ψ 1 (r 2 ) 2 1 − ψ 1 (r2 )ψ 3 (r1 )| V ′(r1 , r 2 )ψ 1 (r1 )ψ 2 (r2 ) 2 1 + ψ 1 (r2 )ψ 3 (r1 )| V ′(r1 , r 2 )|ψ 2 (r1 )ψ 1 (r2 ) . 2

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Problems and Solutions in Quantum Mechanics

Since the particles are identical, r1 and r2 may be interchanged in each term. Do this and as V ′(r1 , r 2 ) = V ′(r 2 , r 1 ), we again obtain the same expression, showing its correctness.

7010 Two identical nonrelativistic fermions of mass m, spin 1/2 are in a onedimensional square well of length L, with V infinitely large and repulsive outside the well. The fermions are subject to a repulsive inter-particle potential V ( x1 − x 2 ) , which may be treated as a perturbation. Classify the three lowest-energy states in terms of the states of the individual particles and state the spin of each. Calculate (in first-order perturbation theory) the energies of second- and third-lowest states; leave your result in the form of an integral. Neglect spin-dependent forces throughout. (Berkeley) Sol:

For the unperturbed potential  0, x ∈[0, L], V (x ) =   ∞ , otherwise, the single-particle spatial wave functions are  2 nπx sin , x ∈[0 , L]  ψn (x ) =  L L 0, otherwise,  where n is an integer.

 a The spin wave function of a single particle has the form    .  b As we do not consider spin-dependent forces, the wave function of the two particles can be written as the product of a space part and a spin part. The spin part  χ J ( M ) ≡ χ JM  is chosen as the eigenstate of  S = s1 + s2  and  Sz = s1z + s2 z ,  i.e., S 2 χ JM = J (J + 1) χ JM , S J χ JM = M χ JM . J = 0  gives the spin singlet state, which is antisymmetric for interchange of the two particles.  J = 1  gives the spin triplet state, which is symmetric for interchange of the two particles. Symmetrizing and antisymmetrizing the space wave functions of the two-particle system we have

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Many-Particle Systems A ψ nm ( x1 , x 2 ) =

569

1 [ψ n ( x1 )ψ m ( x 2 ) − ψ n ( x 2 )ψ m ( x1 )], 2

 1 [ψ n ( x1 )ψ m ( x 2 ) + ψ n ( x 2 )ψ m ( x1 )], n ≠ m,  ψ ( x1 , x 2 ) =  2  ψ n ( x1 )ψ n ( x 2 ), n = m.  s nm

The corresponding energies are E=

π 22 2 (n + m 2 ), n, m = 1, 2, . 2mL2

The total wave functions, which are antisymmetric, can be written as A s ψ nm ( x1 , x 2 )χ JM , s A ψ nm ( x1 , x 2 )χ JM .

The three lowest energy states are the following. i. Ground state, n = m = 1. The space wave function is symmetric, so the spin state must be singlet. Thus

ψ 0 = ψ 11s ( x1 , x 2 )χ 00. ii. First excited states,  n = 1, m = 2.  ψ A ( x , x )χ , M = 0, ±1,  12 1 2 1 M ψ1 =  s  ψ 12 ( x1 , x 2 )χ 00 . The degeneracy is 4. iii. Second excited state,  n = 2, m = 2 . The space wave function is symmetric, the spin state is singlet:

ψ 2 = ψ 22s ( x1 , x 2 )χ 00 , which is nondegenerate. Because the perturbation Hamiltonian is independent of spin, the calculation of perturbation of the first excited state can be treated as a nondegenerate case. The perturbation energies of second and third lowest energy states are given by ∆E1A = ∫ dx1dx 2 |ψ 12A ( x1 , x 2 )|2 V ( x1 − x 2 ), ∆E1s = ∫ dx1dx 2 |ψ 12s ( x1 , x 2 )|2 V ( x1 − x 2 ), ∆E2 = ∫ dx1dx 2 |ψ 22s ( x1 , x 2 )|2 V ( x1 − x 2 ).

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7011 Consider two particles of masses m1 and m2 interacting via the Hamiltonian p12 p2 1 1 1 + 2 + m1ω x12 + m2ω x 22 + k ( x1 − x 2 )2 . Find the exact solutions to 2m1 2m2 2 2 2 the energy and the wave function? H=

Sol: We let R =

m1 x1 + m2 x 2 , r = x1 − x 2 m1 + m2

d ∂R d ∂r d m1 d d We then have dx = ∂x dR + ∂x dr = m + m dR + dr 1 1 1 1 2 2

2

2

2

 d  d 2  m1  2m1 d 2 d2 + 2   = 2  +  dx1  dR  m1 + m2  m1 + m2 drdR dr  d  d 2  m2  2m2 d 2 d2 And similarly,  + 2  = 2  −  dx 2  dR  m1 + m2  m1 + m2 drdR dr 2

 m1  2 2m2 rR In addition, x12 = R 2 +   r + m m + m  1 2 1 + m2 2

 m1  2 2m1 rR x 22 = R 2 +   r − m1 + m2  m1 + m2  Therefore, the Hamiltonian becomes  2 d 2 1   2 d 2 1  K  2 2 2 + ω µ ωr  H = − M R 1 + − + +    2 2 2  2 M dR 2   2µ dr 2  µω   where M = m1 + m2 and µ =

m1m2 m1 + m2

We thus have two independent harmonic oscillators and we can write  1  1 K E pq =  p +  ω + q +  ω 1 + , where p, q = 0, 1, 2, 3 . . .  2  2 µω 2

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2

571

2 2

and ψ pq ( R , r ) = ψ p ( R )ψq (r ) = N p N qe −α1 R /2 H p (α1 R )e −α2 r /2 H q (α 2r ), 1/2

1/4 1/2  α    α µω  K  Mω 2 1 = N  and α1 = where N p =  , α2 = 1 +  .  , q  q p   µω 2    π 2 q!   π 2 p! 

7012 Consider a system of two spin half particles in a state with total spin quantum number S = 0. Find the eigenvalue of the spin Hamiltonian H = AS1 × S2 , where A is a positive constant in this state. Sol: The total spin angular momentum S of this two spin-half system is S = S1 + S2 S 2 = S12 + S22 + 2S1 .S2 Hence, H =

A 2 2 2 (S − S1 − S2 ) 2

Let the simultaneous eigenkets of S 2 , Sz , S12 ,and S22 be | sms . Then, H | sms = =

A 2 2 2 (S − S1 − S2 )| sms 2 A (0 − 3 / 4 − 3 / 4 ) 2 −3 2  = A 4 2

The eigenvalue of the spin Hamiltonian H is

−3 2 A . 4

7013 a. A 2-fermion system has a wave function ψ (1, 2). What condition must it satisfy if the particles are identical? b. How does this imply the elementary statement of the Pauli exclusion principle that no two electrons in an atom can have identical quantum numbers?

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c. The first excited state of Mg has the configuration (3s, 3p) for the valence electrons. In the LS-coupling limit, which values of L and S are possible? What is the form of the spatial part of their wave functions using the single-particle functions ψ s (r )  and  φ p (r )? Which will have the lowest energy, and Why? (Berkeley) Sol: a. ψ (1, 2) must satisfy the condition of antisymmetry for interchange of the two particles: Pˆ ψ (1, 2) = ψ (2, 1) = −ψ (1, 2) . 12

b. In an atom, if there are two electrons having identical quantum numbers then ψ (1, 2) = ψ (2, 1). The antisymmetry condition above then gives ψ (1, 2) = 0, implying that such a state does not exist. c. The electron configuration (3s, 3p) correspond to l1 = 0, l2 = 1, s1 = s2 = 1/2 . Hence L = 1,

S = 0,1 .

ψ (1, 2) = φ (1, 2)χ S (1, 2), L S

L S

where  1 φ0 (1, 2) =   =    1 φ1 (1, 2) = 

1 (φs (r1 )φ p (r2 ) + φs (r2 )φ p (r1 )) 2 1 (1 + Pˆ12 )φs (r1 )φ p (r2 ), 2 1 (1 − Pˆ12 )φs (r1 )φ p (r 2). 2

The lowest energy state is Ψ11 (1, 2), i.e. the state of S = 1. Because the spatial is antisymmetric for the interchange 1 ↔ 2,  the part of the state S = 1   probability that the two electrons get close together is small, and so the Coulomb repulsive energy is small, resulting in a lower total energy.

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573

7014 Two particles, each of mass M, are bound in a one-dimensional harmonic oscillator potential  V = 21 kx 2  and interact with each other through an attractive harmonic force  F12 = − K ( x1 − x 2 ). You may take K to be small. a. What are the energies of the three lowest states of this system? b. If the particles are identical and spinless, which of the states of (a) are allowed? c. If the particles are identical and have spin 1/2, what is the total spin of each of the states of (a)? (Wisconsin) Sol:

The Hamiltonian of the system is ∧

H = Let  ξ =

1 2

K −  2  ∂2 ∂2  1 + k( x12 + x 22 ) + ( x1 − x 2 )2 . + 2 2 M  ∂ x1 ∂ x 22  2 2

( x1 + x 2 ) , η = ∧

1 2

H =− =−

∧

( x1 − x 2 )  and write  H as

∂2  1  2  ∂2 + k(ξ 2 + η 2 ) + Kη 2 + 2 M  ∂ξ 2 ∂η 2  2 ∂2  1 2 1  2  ∂2 + kξ + (k + 2 K )η 2 . + 2 M  ∂ξ 2 ∂η 2  2 2

The system can be considered as consisting of two independent harmonic oscillators of angular frequencies ω 1  and  ω 2  given by

ω1 =

k , M

ω2 =

k + 2K . M

The total energy is therefore 1 1   Enm =  n +  ω 1 +  m +  ω 2 ,    2 2 and the corresponding eigenstate is | nm = ψ nm = ϕ n( k ) (ξ )ϕ m( k+2 K ) (η ), (k)

where  n,m = 0, 1, 2, ...,  and  ϕ n  is the nth eigenstate of a harmonic oscillator of spring constant k.

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Problems and Solutions in Quantum Mechanics

a. The energies of the three lowest states of the system are 1 E00 = (ω 1 + ω 2 ), 2 1 E10 = (ω 1 + ω 2 ) + ω 1 , 2 1 E01 = (ω 1 + ω 2 ) + ω 2 . 2 b. If the particles are identical and spinless, the wave function must be symmetric with respect to the interchange of the two particles. Thus the states |00 , |10 are allowed, while the state |01  is not allowed. c. If the particles are identical with spin 1/2, the total wave function, including both spatial and spin, must be antisymmetric with respect to an interchange of the two particles. As the spin function for total spin  S = 0  is antisymmetric and that for  S = 1  is symmetric, we have S = 0 for |00 , S = 0 for |10 , S = 1 for |01 .

7015 A particular one-dimensional potential well has the following bound-state single-particle energy eigenfunctions: ψ a ( x ), ψ b ( x ), ψ c ( x ), where  Ea < Eb < Ec  . Two non-interacting particles are placed in the well. For each of the cases (a), (b), (c) listed below write down: The two lowest total energies available to the two-particle system, the degeneracy of each of the two energy levels, the possible two-particle wave functions associated with each of the levels. (Use ψ  to express the spatial part and a ket | S , ms  to express the spin part. S is the total spin.) a. Two distinguishable spin- 21  particles. b. Two identical spin- 21  particles. c. Two identical spin-0 particles. (MIT)

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Sol:

575

As the two particles, each of mass M, have no mutual interaction, their spatial wave functions separately satisfy the Schrödinger equation:   2 ∂2  + V ( x1 )ψ i ( x1 ) = Eiψ i ( x1 ), − 2 ∂ M x 2   1   2 ∂2   − 2 ∂ 2 + V ( x 2 )ψ j ( x 2 ) = E jψ j ( x 2 ). M x   2 (i , j = a, b, c…) The equations may also be combined to give   2 ∂2   2 ∂2  − 2 M ∂ x 2 − 2 M ∂ x 2 + V ( x1 ) + V ( x 2 ) ψ i ( x1 )ψ j ( x 2 )   1 2 = ( Ei + E j )ψ i ( x1 )ψ j ( x 2 ) . Consider the two lowest energy levels (i) and (ii). a. Two distinguishable spin- 12 particles. i. Total energy  = Ea + Ea , degeneracy  = 4 . The wave functions are  ψ ( x )ψ ( x )|0, 0 ,  a 1 a 2   ψ a ( x1 )ψ a ( x 2 )|1, m . (m = 0, ±1) ii. Total energy  = Ea + Eb , degeneracy = 8. The wave functions are  ψ ( x )ψ ( x )|0, 0 ,  a 1 a 2   ψ a ( x1 )ψ a ( x 2 )|1,m ,

 ψ ( x )ψ ( x )|0, 0 ,  a 1 b 2   ψ a ( x1 )ψ b ( x 2 )|1,m .

(m = 0, ±1)

b. Two identical spin-1/2 particles. i. total energy = Ea + Ea , degeneracy = 1. The wave function is ψ a ( x1 )ψ a ( x 2 )|0,0 ψ a ( x1 )ψ a ( x 2 )|0,0 . ii. total energy  = Ea + Eb , degeneracy  = 4. The wave functions are  1 [ψ a ( x1 )ψ b ( x 2 ) + ψ b ( x1 )ψ a ( x 2 )]|0, 0 ,   2   1 [ψ a ( x1 )ψ b ( x 2 ) − ψ b ( x1 )ψ a ( x 2 )]|1, m . (m = 0, ±1)  2 c. Two identical spin-0 particles. i. Total energy = Ea + Ea , degeneracy = 1. The wave function is ψ a ( x1 )ψ a ( x 2 )|0,0

ψ a ( x1 )ψ a ( x 2 )|0, 0 .

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Problems and Solutions in Quantum Mechanics

ii. Total energy  = Ea + Ea , degeneracy = 1. The wave function is 1 [ψ a ( x1 )ψ b ( x 2 ) + ψ b ( x1 )ψ a ( x 2 )|0, 0 . 2

7016 Two electrons move in a central field. Consider the electrostatic interaction e 2 | r1 − r2 |  between the electrons as a perturbation. a. Find the first order energy shifts for the states (terms) of the (1s )(2 s ) configuration. (Express your answers in terms of unperturbed quantities and matrix elements of the interaction e 2 | r1 − r2 |) b. Discuss the symmetry of the two-particle wave function for the states in part (a).

Fig. 7.1 c. Suppose that, at time t = 0, one electron is found to be in the 1s unperturbed state with spin up and the other electron in the 2s unperturbed state with spin down as shown in Fig. 7.1. At what time t will the occupation of the states be reversed? (Berkeley) Sol: a. The zero order wave function of the two electrons has the forms φ + (r1 , r 2 ) χ 00 (s1z , s 2z ),

φ − ( r1 , r2 ) χ1Ms (s1z , s 2z ),

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577

where

φε =

1 [u1s (1)υ 2 s (2) + ε u1s (2)υ 2 s (1)], ε = ±1, 2

being the normalized symmetric  (+ )  and antisymmetric  (− )  wave functions,

χ 0  and  χ1  denote the singlet and triplet spin states respectively. Denoting u1s (1)υ 2 s (2) by 1,2 and u1s (2)υ 2 s (1) by  2,1 , we can write the above as

φε = ( 1,2 + ε 2,1

)

2.

Because the perturbation Hamiltonian is independent of spin, we need not consider  χ . Thus ∆Eε = ∫ d 3r1d 3r2φε*

e2 φε | r1 − r2 |

1 e2 = ( 1, 2| +ε 2,1|) (|1, 2 + ε |2, 1 ) 2 | r1 − r2 | 1 = [ 1, 2| A | 1, 2 + 2,1| A |2,1 2 + ε 1, 2 | A |2, 1 + ε 2,1 | A |1, 2 ] = K + ε J , where  A = e 2 | r1 − r2 |, K = 1,2 A 1,2 = 2,1 A 2,1   is the direct integral, J = 1,2 A 2,1 = 2,1 A 1,2  is the exchange integral. b. The singlet state  χ 0  is antisymmetric for interchange of spins. The triplet state χ1  is symmetric for interchange of spins. Similarly, φ+  is symmetric for the interchange of r1 and r2, and φ−  is antisymmetric for the interchange. Hence the total wave function is always antisymmetric for interchange of the electrons. c. The initial state of the system is 1 [ 1s 2s ↑↓ − 2s1s ↓↑ ] 2 1  = ( 1, 2 + 2,1 )( ↑↓ − ↑↓ ) 2 2 + ( 1, 2 − 2,1 )( ↑↓ + ↑↓ )

ψ (t = 0) =

=

1 (ψ + χ 00 + ψ − χ10 ), 2

and so the wave function at t is 1 ψ (t ) = (ψ + χ 00e − iE+t / + ψ − χ10e − iE−t / ). 2

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When e − iE−t //e − iE+t / = −1, the wave function becomes 1 ψ (tn ) = e − iE+tn / ⋅ (ψ + χ 00 − ψ − χ10 ) 2 1  2,1 ↓↑ − 1,2 ↓↑  , = e − E +tn / ⋅  2 which shows that at this time the ls electron has spin down and the 2s electron has spin up i.e., the spins are reversed. As −1 = e i(2n+1)π , n = 0, 1, 2, this happens at times  (2n + 1)π  t = (2n + 1)π ⋅ = . E+ − E− 2J

7017 a. Show that the parity operator commutes with the orbital angular momentum operator. What is the parity quantum number of the spherical harmonic Ylm (θ ,ϕ ) ? b. Show for a one-dimensional harmonic oscillator in state  En = ( n + 21 ) ω that  ∆x 2

n

∆p 2

= ( n + 21 )  2 . 2

n

c. Consider the rotation of a hydrogen molecule H2. What are its rotational energy levels? How does the identity of the two nuclei affect this spectrum? What type of radiative transition occurs between these levels? Remember that the proton is a fermion. 2 d. Show that (n ⋅σ ) = 1 , where n is a unit vector in an arbitrary direction and σ  are the Pauli spin matrices. (Berkeley)

Sol: a. Applying the parity operator P  and the orbital angular momentum operator ∧

L=r×p

to an arbitrary wave function  f (r ), we have ^

^

P Lf (r ) = P (r × p ) f (r) = (− r ) × (− p ) f (− r ) ^

= r × pf (− r ) = LP f (r ).

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579

Hence  P^  and L commute, i.e., ^

[P , L]= 0 . As ^

PYlm (θ , φ ) = Ylm (π − θ , π + φ ) = (−1)l Ylm (θ , φ ), l

the parity quantum number of Ylm (θ , ϕ )  is  (−1) . b. For a one-dimensional harmonic oscillator, we can use the Fock representation  (a + a + ), 2mω

x=

mω + (a − a), 2

p=i

where a, a +  are annihilation and creation operators having the properties a n = n n −1 , a+ n = n + 1 n + 1 . Using these operators we have nxn = =

 n a + a+ n 2mω  2mω

(

)

n n − 1 + n + 1 n n + 1 = 0,

 n (a + a + )(a + a + ) n 2mω  = n n a + a+ n − 1 + n + 1 n a + a+ n + 1 2mω   n(n − 1) n n − 2 + n n n = 2mω  +(n + 1) n n + (n + 1)(n + 2) n n + 2 

n x2 n =

(

=

 ( 2n + 1), 2mω

and similarly n p n = 0, n p2 n =

mω (2n + 1) . 2

As ∆x 2 ∆p 2

n

n

= (x − x = p2

)2

n 2

n

= x2

− p n=

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n

− x

2 n

=

mω (2n + 1), 2

 (2n + 1), 2mω

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Problems and Solutions in Quantum Mechanics

we find ∆x 2 n ⋅ ∆p 2

2

1  = 2  n +  . n  2

c. The rotational energy levels of a hydrogen molecule are given by Er =  2 K ( K + 1) 2 I 0 , where  I 0 = MR02  is the moment of inertia of the molecule about the rotating axis, which is perpendicular to the line connecting the two nuclei, K is the angular momentum quantum number, K = 0, 1, 2, . Since the spin of a proton is  / 2, the total wave function of the molecule is antisymmetric for interchange of the two protons. When the two protons are interchanged, the wave function for the motion of the center of mass and the wave function for the atomic vibration are not changed; only the wave function for rotation is altered: YKM K (θ , ϕ ) → YKM K (π − θ , π + ϕ ) = (−1)K YKM K (θ , ϕ ) . If K is even, (−1)K YKM K (θ , ϕ ) = YKM K (θ , ϕ )  and the spin wave function  χ 0  must be antisymmetric, i.e., χ 0  is a spin singlet state; If K is odd, and the spin wave function χ1  must be (−1)K YKM K (θ , ϕ ) = −YKM K (θ , ϕ )   symmetric, i.e., χ1  is a spin triplet state. The hydrogen molecule in the former case is called a para-hydrogen, and in the latter case is called an ortho-hydrogen. There is no inter-conversion between para-hydrogen and ortho-hydrogen. Transitions can take place between rotational energy levels with ∆K = 2,4,6,...  within each type. Electric quadruple transitions may also occur between these levels. d.

2

  (n ⋅σ ) =  ∑ni σ i  = ∑nin j σ i σ j   i ij 2

=

1 ∑nin j {σi , σ j } = ∑nin j δij = ∑nini = 1. 2 i ,i i,j i

In the above i, j refer to x, y, z, and  {σ i ,σ j } ≡ σ iσ j + σ jσ i = 2δ ij .

7018 Two particles of mass m are placed in a rectangular box of sides  a > b > c  in the lowest energy state of the system compatible with the conditions below. The particles interact with each other according to the potential V = Aδ (r1 − r2 ). Using first order perturbation theory to calculate the energy of the system under the following conditions:

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581

a. Particles not identical. b. Identical particles of spin zero. c. Identical particles of spin one-half with spins parallel. (Berkeley) Sol: a. The unperturbed system can be treated as consisting of two separate singleparticle systems and the wave function as a product of two single-particle wave functions:

ψ (r1 , r2 ) = ψ (r1 )ψ (r2 ) . The lowest energy state wave function is thus   8 sin π x1 sin π x 2 sin π y1 sin π y 2 sin π z1 sin π z 2 ,  abc a a b b c c ψ 0 (r1 , r2 ) =  < < < < < < = x a y b z c i for 0 , 0 , 0 , ( 1,2) i i i   0, otherwise,  corresponding to an energy E0 =

2π2  1 1 1   + +  . m  a2 b2 c 2 

First order perturbation theory gives an energy correction ∆E = ∫ψ 0* (r1 , r2 )Aδ (r1 − r2 )ψ 0 (r1 , r2 )d 3r1d 3r2 = ∫ A |ψ 0 (r1 , r1 )|2 d 3r1 = ∫

c

0

 8  A 0  abc 

b

∫∫ 0

a

2

4

πy πz  27 A  πx ×  sin 1 sin 1 sin 1  dx1dy1dz1 = ,  a b c  8abc and hence E′ =

 2 π 2  1 1 1  27 A .  + + + m  a 2 b 2 c 2  8abc

b. For a system of spin-0 particles, the total wave function must be symmetric for interchange of a pair of particles. Hence the lowest energy state is ψ s (r1 , r2 ) = ψ 0 (r1 , r2 ), which is the same as that in (a). The energy to first order perturbation is also Es′ =

 2 π 2  1 1 1  27 A .  + + + m  a 2 b 2 c 2  8abc

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Problems and Solutions in Quantum Mechanics

c. For a system of spin- 12  particles the total wave function must be antisymmetric. As the spins are parallel, the spin wave function is symmetric and so the spatial wave function must be antisymmetric. As  a12 < b12 < c12 ,  the lowest energy state is ψ A (r1 , r2 ) =

1 [ψ 211 (r1 )ψ 111 (r2 ) − ψ 211 (r2 )ψ 111 (r1 )], 2

where ψ 111 (r )  and  ψ 211 (r )  are the ground and first excited single-particle states respectively. The unperturbed energy is EA0 =

π 2  5 1 1 + + .  m  2a 2 b 2 c 2 

First order perturbation theory gives ∆E = ∫ ψ *A (r1 , r2 )Aδ(r1 − r2 )ψ A (r1 , r2 )d 3r1d 3r2 = 0 . Therefore EA′ =

2π2  5 1 1  2 + 2 + 2  . m 2a b c

7019 A porphyrin ring is a molecule which is present in chlorophyll, hemoglobin, and other important compounds. Some aspects of the physics of its molecular properties can be described by representing it as a one-dimensional circular 

path of radius  r = 4 A, along which 18 electrons are constrained to move. a. Write down the normalized one-particle energy eigenfunctions of the system, assuming that the electrons do not interact with each other. b. How many electrons are there in each level in the ground state of the molecule? c. What is the lowest electronic excitation energy of the molecule? What is the corresponding wavelength (give a numerical value) at which the molecule can absorb radiation? (Berkeley) Sol: a. Denote the angular coordinate of an electron by  θ . The Schrödinger equation of an electron

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Many-Particle Systems

−

583

 2 ∂2 ψ (θ ) = Eψ (θ ) 2mr 2 ∂θ 2

has solution

ψ (θ ) =

1 ikθ . e 2π

The single-valuedness of ψ (θ ), ψ (θ ) = ψ (θ + 2π ), requires  k = 0, ± 1, ± 2. The energy levels are given by E=

2 2 k . 2mr 2

b. Let 0,1,2, … denote the energy levels  E0 , E1 , E2 ,…  respectively. In accordance with Pauli’s exclusion principle, E0 can accommodate two electrons of opposite spins, while Ek , k ≠ 0 , which is 2-fold degenerate with respect to ± | k |, can accommodate four electrons. Thus the electron configuration of the ground state of the system is 0214 24 34 4 4 . c. The electron configuration of the first excited state is 0214 24 34 4 351. The energy difference between E4 and E5 is ∆E = E5 − E4 =

9 2 2 (52 − 4 2 ) = 2 2mr 2mr 2

and the corresponding absorption wavelength is

λ=

ch 8π 2  mc  2 8π 2 42 = × = 5800 Å ,   r = ∆E 9 h 9 0.0242

where  mch = 0.0242 Å  is the Compton wavelength of electron.

7020 A large number N of spinless fermions of mass m are placed in a onedimensional oscillator well, with a repulsive δ -function potential between each pair: V=

k N 2 λ ∑xi + 2 ∑δ ( xi − x j ), 2 i =1 i≠ j

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k , λ > 0.

24/05/22 10:03 AM

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Problems and Solutions in Quantum Mechanics

a. In terms of normalized single-particle harmonic oscillator functions ψ n ( x ), obtain the normalized wave functions and energies for the three lowest energies. What are the degeneracies of these levels? b. Compute the expectation value of  ∑ i =1 x i2  for each of these states. For partial credit, you may do parts (a) and (b) with  λ = 0. N

(Berkeley) Sol: a. Treat the δ -function potential as perturbation. As for a system of fermions, the total wave function is antisymmetric, the zero-order wave function for the system can be written as the determinant

ψ n1 ( x1 ) ψ n1 ( x 2 ) … ψ n1 ( x N ) ψ n1nN ( x1x N ) =

=

ψ n2 ( x1 ) ψ n2 ( x 2 ) … ψ n2 ( x N ) 1 N!    ψ nN ( x1 ) ψ nN ( x 2 ) … ψ nN ( x N ) 1 ∑δ p P[ψ n1 ( x1 )ψ nN (x N )], N! P

where ni label the states from ground state up, P denotes permutation of xi and δ P = +1, −1  for even and odd permutations respectively. On account of the δ -function, the matrix elements of the perturbation Hamiltonian are all zero. Thus the energy levels are N N  E(n1 ,n2 ,,nN ) = n1 nN | H | n1 ...nN = ω  + ∑ni  ,  2 i=1  where  ω = k m . i. For the ground state:  n1 nN  are respectively  0, 1, …, N − 1, the energy is  N N ( N − 1)  ω 2 E(0,…, N −1) = ω  +  = 2 N , 2 2 and the wave function is

ψ 0,1,…,N −1 ( x1 x N ) =

1 ∑δ P P[ψ 0 (x1 )ψ N −1 (x N )] . N! P

ii. For the first excited state: n1 nN  are respectively  0, 1, …, N − 2, N , the energy is 1 E(0,…, N −2, N ) = ω ( N 2 + 2), 2

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585

and the wave function is 1 ∑δ P P[ψ 0 (x1 )ψ N −2 (x N −1 )ψ N (x N )]. N! P

ψ 0,1N −2,N ( x1 x N ) =

iii. For the second excited state: n1 nN  are respectively either 0, 1, ... N − 2, N + 1, or 0, 1, … N − 3, N − 1, N . The energy is E(0,…N −2, N +1) = E(0,1,…N −3, N −1, N ) =

ω 2 ( N + 4), 2

and the corresponding wave functions are 1 ψ 0,…N −2, N +1 ( x1 x N ) = ∑δ P P[ψ 0 (x1 ) N! P

ψ N −2 ( x N −1 )ψ N +1 ( x N )], 1 ∑δ P P[ψ 0 (x1 ) N! P

ψ 0,…N −3, N −1, N ( x1 x N ) =

ψ N −3 ( x N −2 )ψ N −1 ( x N −1 )ψ N ( x N )]. It can be seen that the ground and first excited states are nondegenerate, where the second excited state is two-fold degenerate. b. For stationary states,

∑x ∂ V (x x

2T =

i

i

i

1

N

) .

As

λ

∑x ∂ ∑ 2 δ ( x − x ) k

k

∑x i

i

k

i

i≠ j

= 0,

j

 ∂ k x 2j  = k ∑  ∂ xi  2 j 

∑x i

2 i

,

we have 2 T =k

N

∑x i

2 i

.

The virial theorem T = V ( x1  x N ) , or E=2 T , then gives N

∑x i =1

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2 i

=

1 E. mω 2

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Problems and Solutions in Quantum Mechanics

Hence N  N 2 00 ∑xxi22 00 = N = N 2 ,, ∑ 22m ω i =1 i mω i =1 N  N 2 11 ∑xxi22 11 = = 2mω (( N N2 + + 2), 2), ∑ i i =1 2mω i =1 N N 2 i2 i =1 i i =1

N  N 2 2 22 ∑xx 22 = 2 =  (( N + 4), 4), ′ ∑ = 2′ ∑xxii2 22′′ = N2 + ∑ 22m ω i =1 mω i =1

where  0 , 1 , 2 , and  2′  are the ground state, the first excited state and the two second excited states respectively.

7021 What is the energy difference in eV between the two lowest rotational levels of the HD molecule? The HD (D is a deuteron) distance is 0.75 Å. (Berkeley) Sol:

The rotational energy levels are given by EJ =

2 J ( J + 1). 2I

Thus for the two lowest levels, ∆E10 =

2 J ( J + 1) 2I

J =1 −

2 2 J ( J + 1) = . 2I I J =0

As the mass mD of the deuteron is approximately twice that of the hydrogen nucleus mp, we have 2m p ⋅ m p 2 2 r = mpr 2 , I = µr 2 = 2m p + m p 3 and hence ∆E10 =

3 (c )2 1 1.5 2 = = 2 2 2 2 2 mpc r 938 × 106 3 mP r 2

 6.582 × 10−16 × 3 × 1010  = 1.11 × 10−2 eV. × −8    0.75 × 10

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587

7022 Consider the (homonuclear) molecule N214 . Use the fact that a nitrogen nucleus has spin  I = 1 in order to derive the result that the ratio of intensities of adjacent rotational lines in the molecule’s spectrum is 2:1. (Chicago) Sol:

In the adiabatic approximation, the wave function of N2 molecule whose center of mass is at rest can be expressed as the product of the electron wave function ψ e , the total nuclear spin wave function ψ s , the vibrational wave function ψ 0 , and the rotational wave function ψ I; that is, ψ = ψ eψ sψ 0ψ I . For the molecular rotational spectrum, the wave functions of the energy states involved in the transition have the same ψ e , ψ 0, but different ψ s , ψ I . For interchange of the nitrogen nuclei, we have ψ eψ 0 → ψ eψ 0 or − ψ eψ 0. The N nucleus is a boson as its spin is 1, so the total nuclear spin of the N2 molecule can only be 0, 1 or 2, making it a boson also. For the exchange operator  ^ P  between the N nuclei, we have  +ψ s for S = 0, 2, ∧  ψ I for I = even integer, ∧ P ψs =  Pψ I =   −ψ s for S = 1,  −ψ I for I = odd integer. As N2 obeys the Bose-Einstein statistics, the total wave function does not change on interchange of the two nitrogen nuclei. So for adjacent rotational energy levels with ∆I = 1, one must have  S = 0  or 2, the other  S = 1, and the ratio of their degeneracies is  [2 × 2 + 1 + 2 × 0 + 1] :  (2 × 1 + 1) = 2 :1. For the molecular rotational spectrum, the transition rule is  ∆J = 2. As S usually remains unchanged in optical transitions, two adjacent lines are formed by transitions from  I =  even to even and  I =  odd to odd. Since the energy difference between two adjacent rotational energy levels is very small compared with kT at room temperature, we can neglect the effect of any heat distribution. Therefore, the ratio of intensities of adjacent spectral lines is equal to the ratio of the degeneracy of I =  even rotational energy level to that of the adjacent I =  odd rotational energy level, which has been given above as 2:1.

7023 a. Assuming that two protons of H2+  molecule are fixed at their normal separation of 1.06 Å, sketch the potential energy of the electron along an axis passing through the protons. b. Sketch the electron wave functions for the two lowest states in H2+ , indicating roughly how they are related to hydrogenic wave functions. Which wave function corresponds to the ground state, and why?

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Problems and Solutions in Quantum Mechanics

c. What happens to the two lowest energy levels of  H2+  in the limit that the protons are moved far apart? (Wisconsin) Sol: a. As the protons are fixed, the potential energy of the system is that of the electron, apart from the potential energy due to the Coulomb interaction 2 between the nuclei  eR . Thus V =−

e2 e2 − , | r1 | | r2 |

where r1, r2 are as shown in Fig. 7.2. When the electron is on the line connecting the two protons, the potential energy is V =−

e2 e2 − , |x| |R− x|

where x is the distance from the proton on the left. V is shown in Fig. 7.3 as a function of x. The wave function must be symmetrical or antisymmetrical with respect to the interchange of the protons. Thus the wave functions of the two lowest states are 1 ψ± = [φ (r1 ) ± φ (r2 )], 2 where φ (r )  has the form of the wave function of the ground state hydrogen atom:

Fig. 7.2

Fig. 7.3

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589

Fig. 7.4 1  λ   π  a

φ (r ) =

3/2

e − λ r /a

where a is the Bohr radius and λ  is a constant. The shape of the two wave functions along the x-axis are sketched in Fig. 7.4. It can be seen that the probability that the electron is near the two nuclei is larger for ψ + . Hence ψ + corresponds to a lower V and is therefore the ground state wave function. The fact that  E+ < E−  can also be seen from E± = ψ ± | H |ψ ± = φ (r1 )| H |φ (r1 ) ± φ (r1 )| H |φ (r2 ) , since

φ (r1 )| H |φ (r1 ) = φ (r2 )| H |φ (r2 ) , φ (r1 )| H |φ (r2 ) = φ (r2 )| H |φ (r1 ) . and all the integrals are negative. c. As the protons are being moved apart, the overlap of the two bound states φ (r1 )  and φ (r2 ) becomes less and less, and so  φ (r1 )| H |φ (r2 )  and  φ (r2 )| H |φ (r1 ) → 0 . In other words, the two lowest energy levels will become the same, same as the ground state energy of a hydrogen atom.

7024 Write the Schrödinger equation for atomic helium, treating the nucleus as an infinitely heavy point charge. (Berkeley) Sol:

Treating the nucleus as an infinitely heavy point charge, we can neglect its motion, as well as the interaction between the nucleons inside the nucleus and the distribution of the nuclear charge. The Schrödinger equation is then  p12  p 22 2e 2 2e 2 e2 + − − +  2m 2m  ψ (R1 , R 2 ) = Eψ (R1 , R 2 ), R R | R − R | e e 1 2 1 2 

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where R1, R2 are as shown in Fig. 7.5.

Fig. 7.5 On the left side of the equation, the first and second terms are the kinetic energies of the electrons, the third and fourth terms correspond to the attractive potentials between the nucleus and the electrons, and the last term is the repulsive potential between the two electrons.

7025 The excited electronic configuration (1s )1(2 s )1  of the helium atom can exist as either a singlet or a triplet state. Tell which state has the lower energy and explain why. Give an expression which represents the energy separation between the singlet and triplet states in terms of the one-electron orbitals

ψ 1s (r )  and  ψ 2 s (r ). (MIT) Sol:

Electrons being fermions, the total wave function of a system of electrons must be antisymmetric for the interchange of any two electrons. As the spin triplet state of helium atom is symmetric, its spatial wave function must be antisymmetric. In this state the electrons have parallel spins so the probability for them to get close is small (Pauli’s principle), and consequently the repulsive energy, which is positive, is small. Whereas for the spin singlet state the reverse is true, i.e., the probability for the two electrons to get close is larger, so is the repulsive energy. Hence the triplet state has the lower energy. Consider the interaction between the electrons as perturbation. The perturbation Hamiltonian is e2 H′ = , r12 where  r12 = | r1 − r2 |. For the singlet state, using the one-electron wave functions ψ 1s ,ψ 2 s, we have

ψ=

1

and for the triplet state

ψ=

3

1 [ψ 1s (r1 )ψ 2 s (r2 ) + ψ 1s (r2 )ψ 2 s (r1 )]χ00 , 2 1 [ψ 1s (r1 )ψ 2 s (r2 ) − ψ 1s (r2 )ψ 2 s (r1 )]χ1m 2

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with  m = 1, 0, − 1. The energy separation between the states is ∆E = 1ψ | H ′ |1 ψ − 3ψ | H ′ |3 ψ . With ψ ns* = ψ ns , we have ∆E = 2 ∫

e2 [ψ 1s (r1 )ψ 2 s (r1 )ψ 1s (r2 )ψ 2 s (r2 )]dr1dr2. r12

7026 a. Suppose you have solved the Schrödinger equation for the singlyionized helium atom and found a set of eigenfunctions ψ N (r ). 1. How do the φN (r )  compare with the hydrogen atom wave functions? 2. If we include a spin part σ +  (or σ − ) for spin up (or spin down), how do you combine the φ's  and σ 's to form an eigenfunction of definite spin? b. Now consider the helium atom to have two electrons, but ignore the electromagnetic interactions between them. 1. Write down a typical two-electron wave function, in terms of the φ's  and σ 's, of definite spin. Do not choose the ground state. 2. What is the total spin in your example? 3. Demonstrate that your example is consistent with the Pauli exclusion principle. 4. Demonstrate that your example is antisymmetric with respect to electron interchange. (Buffalo) Sol: a. 1. Th  e Schrödinger equation for singly-charged He atom is the same as that 2 2 for H atom with  e → Ze , where Z is the charge of the He nucleus. Hence the wave functions for hydrogen-like ion are the same as those for H atom with the Bohr radius replaced: h- 2 h- 2 r0 = 2 → a = , µe µ Ze 2

µ  being the reduced mass of the system. For helium  Z = 2.

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2. As φ N  and  σ ±  belong to different spaces we can simply multiply them to form an eigenfunction of a definite spin. b. 1. and 2. A He atom, which has two electrons, may be represented by a wave function 1 φ (1)φ N (2)[σ + (1)σ − (2) − σ − (1)σ + (2)] 2 N if the total spin is zero, and by 1 [φ N 1 (1)φ N 2 (2) − φ N 2 (1)φ N 1 (2)]σ + (1)σ + (2) 2 if the total spin is 1. + − 3. If  σ = σ ,φ N 1 = φ N 2 , the wave functions vanish, in agreement with the Pauli exclusion principle.

4.  Denote the wave functions by ψ (1, 2). Interchanging particles 1 and 2 we have

ψ (2, 1) = −ψ (1, 2).

7027 Ignoring electron spin, the Hamiltonian for the two electrons of helium atom, whose positions relative to the nucleus are  ri (i = 1, 2) , can be written as 2  p2 2e 2  e2 , H = ∑ i − V V . + = | ri |  | r1 − r2 | i =1  2 m

a. Show that there are 8 orbital wave functions that are eigenfunctions of H − V  with one electron in the hydrogenic ground state and the others in the first excited state. b. Using symmetry arguments show that all the matrix elements of V among these 8 states can be expressed in terms of four of them. [Hint: It may be helpful to use linear combinations of l = 1  spherical harmonics proportional to x y z , and .] |r | |r | |r | c. Show that the variational principle leads to a determinantal equation for the energies of the 8 excited states if a linear combination of the

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8 eigenfunctions of  H − V  is used as a trial function. Express the energy splitting in terms of the four independent matrix elements of V. d. Discuss the degeneracies of the levels due to the Pauli principle. (Buffalo) Sol:

Treating V as perturbation, the zero-order wave function is a product of two eigenfunctions  n, l , m  of a hydrogen-like atom. Thus the 8 eigenfunctions for H 0 = H − V  with one electron in the hydrogen ground state can be written as | lm ± =

1  (100)1 (2lm)2 ± (2lm)1 (100)2  2

with  l = 0, 1, m = −l ,…l , where the subscripts 1 and 2 refer to the two electrons. The corresponding energies are Eb = E1 + E2 = −

µ(2e 2 )2  1 5 µe 4  1 + 2  = − 2 . 2 2 2 2

To take account of the perturbation we have to calculate the matrix elements l ′m′± | V | lm ± . As V is rotation-invariant and symmetric in the two electrons and  lm ±  are spatial rotation eigenstates, we have (100)1 (2l ′m′ )2 | V |(100)1 (2lm)2 = (2l ′m′ )1 (100)2 | V |(2lm)1 (100)2 = δ ll ′δ mm′ Al , (100)1 (2l ′m′ )2 | V |(2lm)1 (100)2 = (2l ′m′ )1 (100)2 | V |(100)1 (2lm)2 = δ ll ′δ mm′ Bl , and hence l ′m′+ | V | lm + = δ ll ′δ mm′ ( Al + Bl ), l ′m′+ | V | lm − = 0, l ′m′− | V | lm + = 0, l ′m′− | V | lm − = δ ll ′δ mm′ ( Al − Bl ). Because the wave functions were formed taking into account the symmetry with respect to the interchange of the two electrons, the perturbation matrix is diagonal, whence the four discrete energy levels follow: The first levels  1m +  have energy  Eb + A1 + B1, second levels  1m −  have energy the fourth level Eb + A1 − B1, the third level 00 +  has energy  Eb + A0 + B0 ,  

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00 −  has energy  Eb + A0 − B0 . Note that the levels 1m +  and  1m −  are each three-fold degenerate (m = ±1, 0). According to Pauli’s principle, we must also consider the spin wave functions. Neglecting spin-orbit coupling, the total spin wave functions are  χ 0 , antisymmetric, a singlet state;  χ1sz , symmetric, a triplet state. Since the total electron wave function must be antisymmetric for inter-change of the electrons, we must take combinations as follows, lm + χ 0 , lm − χ1sz . Hence, the degeneracies of the energy levels are Eb + A0 − B0 Eb + A0 + B0 Eb + A1 − B1 Eb + A1 + B1

: : : :

1× 3 = 3 1×1 = 1 3× 3=9 3 × 1 = 3.

7028 Describe approximate wave functions and energy levels of the lowest set of P-states  ( L = 1)  of the neutral helium atom, using as a starting basis the known wave functions for the hydrogen atom of nuclear charge Z:

ψ 1s = π −1/2 a −3/2 e − r / a , a = a0 Z , ψ 2 p ,ml = 0 = (32π )−1/2 a −5/2 re − r /2 a cosθ , etc. a. There are a total of 12 states (2 spin components  × 2  spin components × 3  orbital components) which you should classify according to the Russell-Saunders coupling scheme, giving all the appropriate quantum numbers. Be sure that the states are properly antisymmetrized. b. Give an estimate (to the nearest integer) for the values of “Z” to use for each of the two orbital wave functions. What energy above the ground state results? What mathematical process could be used to calculate the optimum Z values?

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c. Write down an integral which gives the separation between two subsets of these 12 states due to the Coulomb repulsion between the two electrons. Which states are lower in energy? d. Which of these P-states, if any, can decay to the atomic ground state by the emission of a single photon. (Electric dipole only) e. Do there exist any other excited states with L = 1 which can decay to any one of the P-states discussed above by emission of a single photon by electric dipole interaction? If so, give an example of such a state in the usual scheme of spectroscopic notation. (Berkeley) Sol: a. Since  L = l1 + l 2 , Lz = l1z + l2 z , L = 1  means that  l1 , l2 = 0, 1  or 1, 0, i.e. one electron is in ls state, the other in 2p state. For convenience, Dirac’s bra-ket notation is used to represent the states. The symmetrized and antisymmetrized spatial wave functions are

ψ1 = ψ2 = ψ3 = ψ4 = ψ5 = ψ6 =

1 2 1 2 1 2 1 2 1 2 1 2

( 1s

2 p,ml = 1 + 2 p,ml = 1 1s ,

)

( 1s

2 p,ml = 1 − 2 p,ml = 1 1s ,

( 1s

2 p, ml = 0 + 2 p, ml = 0 1s ,

( 1s

2 p, ml = 0 − 2 p, ml = 0 1s ,

( 1s

2 p, ml = −1 + 2 p, ml = −1 1s ,

( 1s

2 p, ml = −1 − 2 p, ml = −1 1s .

)

) )

) )

For the total wave functions to be antisymmetric, we must choose the products of the spin singlet state χ00   and the symmetric space wave functions ψ 1 , ψ 3 , ψ 5 , forming three singlet states  ψ i χ00 (i = 1, 3, 5); and the products of the spin triplet states χ11  and the antisymmetric space wave functions  ψ 2 , ψ 4 , ψ 6 , forming nine triplet states  ψ i χ11 (i = 2, 4, 6, m = 0, ± 1). To denote the twelve states in the coupling representation, we must combine the

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above antisymmetrized wave functions: The wave functions of the three singlet states are 1

| mJ = 1 = |ψ 1 χ00 ,

P1 :

| mJ = 0 = |ψ 3 χ00 , | mJ = −1 = |ψ 5 χ00 . The wave functions of the nine triplet states are 3

mJ = 2 = ψ 2 χ11 ,

P2 :

mJ = 1 =

1 ( ψ 2 χ10 + ψ 4 χ11 ) , 2

1 2 1 ψ 2 χ1, −1 + ψ 4 χ10 + ψ 6 χ11 , 6 3 6 1 mJ = −1 = ψ 4 χ1, −1 + ψ 6 χ10 , 2 mJ = 0 =

(

)

mJ = −2 = ψ 6 χ1, −1 . 1 ( ψ 2 χ10 − ψ 4 χ11 ) , 2 1 mJ = 0 = ψ 2 χ1,−1 − ψ 6 χ11 , 2 1 mJ = −1 = ψ 4 χ1,−1 − ψ 6 χ10 ). 2 1 3 P0 : mJ = 0 = ψ 2 χ1,−1 − ψ 4 χ10 + ψ 6 χ11 . 3 3

P1 : mJ = 1 =

(

)

(

)

b. As the electron cloud of the 2p orbits is mainly outside the electron cloud of the ls orbit, the value of Z of the  1s  wave function is 2 and that of the  2 p  wave function is 1. The energy levels of a hydrogen-like atom is given by E=−

mZ 2e 4 . 2  2n 2

Hence the energy of the 2p states above the ground state is 2

∆E = −

mc 2  e 2   1 22  − 2  c   22 1  2

0.51 × 106  1  15 × ×  137  2 4 = 51 eV. =

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The optimum Z can be obtained from shielding effect calculations using the given wave functions. c. Denote the two subsets of symmetric and antisymmetric spatial wave functions with a parameter  ε = ±1  and write 1 ψε = ( 1s 2 p + ε 2 p 1s ) . 2 The repulsive interaction between the electrons, H′ =

e2 , | r1 − r2 |

results in a splitting of the energy levels of the two sets of wave functions. As 1 ψ ε | H ′ |ψ ε = ( 1s 2 p + ε 2 p 1s )H ′( 1s 2 p + ε 2 p 1s ) 2 = 1s 2 p | H ′ |1s 2 p + ε 1s 2 p | H ′ | 2 p1s , the splitting is equal to twice the exchange integral in the second term of the right-hand side, i.e., K = ∫ψ 1*s (r1 )ψ 1s (r2 )ψ 2* p (r2 )ψ 2 p (r1 )

e2 dr1dr2. | r1 − r2 |

As  K > 0, the energy of the triplet state (ε = −1)  is lower than that of the singlet state. (This is to be expected since when the space wave function is antisymmetric, the two electrons having parallel spins tend to avoid each other.)

d. The selection rules for electric dipole radiation transition are ∆ L = 0, ± 1; ∆ S = 0; ∆ J = 0, ± ∆ L = 0, ±1; ∆ S = 0; ∆ J = 0, ± 1 and a change of parity. Hence the state that can transit to the ground state 1S0 is the 1P1 state. e. Such excited states do exist. For example, the 3P1 state of the electronic configuration 2p3p can transit to any of the above 3P2,1,0 states through electric dipole interaction.

7029 Justify, as well as you can, the following statement: “In the system of two ground state H atoms, there are three repulsive states and one attractive (bound) state.” (Wisconsin)

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Problems and Solutions in Quantum Mechanics

In the adiabatic approximation, when discussing the motion of the two electrons in the two H atoms we can treat the distance between the nuclei as fixed and consider only the wave functions of the motion of the two electrons. For total spin S = 1 , the total spin wave function is symmetric for interchange of the two electrons and so the total space wave function is antisymmetric. Pauli’s principle requires the electrons, which in this case have parallel spins, to get away from each other as far as possible. This means that the probability for the two electrons to come near each other is small and the states are repulsive states. As  S = 1  there are three such states. For total spin  S = 0, the space wave function is symmetric. The probability of the electrons being close together is rather large and so the state is an attractive one. As  S = 0, there is only one such state.

7030 In a simplified model for a deuteron the potential energy part of the Hamiltonian is V = Va ( r ) + Vb ( r )s n ⋅ s p. The spin operators for the two spin-1/2 particles are sn and sp; the masses are mn and mp; Va and Vb are functions of the particle separation r. a. The energy eigenvalue problem can be reduced to a one-dimensional problem in the one variable r. Write out this one-dimensional equation. b. Given that Va and Vb both are negative or zero, state (and explain) whether the ground state is singlet or triplet. (Princeton) Sol: a. In units where   = 1, we have for the singlet state  (S = 0)  of the deuteron, 1 1 1 1 1 3  3 s n ⋅ s p = (s n + s p )2 − s n2 − s 2p = −  × × 2 = − 2 2 2 2 2 2  4 and the potential energy 3 Vsinglet = Va (r ) − Vb (r ) . 4 The Hamiltonian is then H=−

1 2 1 2 3 ∇n − ∇ p + Va (r ) − Vb (r ), 2mn 2m p 4

whence the Hamiltonian describing the relative motion Hr = −

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where ∇ r2  is the Laplacian with respect to the relative position coordinate mm r =| rp − rn |, m = mnn+mpP  is the reduced mass of the two particles.

After separating out the angular variables from the Schrödinger equation, the energy eigenvalues are obtained from the one-dimensional equation satisfied by the radial wave function R(r): −

1 d2 3  l(l + 1)  (rR ) +  + Va (r ) − Vb (r ) (rR ) = E(rR ). 2 2mr dr 2 4  2mr 

Similarly, for the triplet state  (S = 1)  we have 1 Vtriplet = Va (r ) + Vb (r ), 4 and the corresponding one-dimensional equation −

1 d2 1  l(l + 1)  (rR ) +  + Va (r ) + Vb (r ) (rR ) = E( Rr ). 2 2 2mr dr 4  2mr 

b. We shall make use of the lemma: For a one-dimensional problem of energy eigenvalues, if the conditions are all the same except that two potential energies satisfy the inequality V ′( x ) ≥ V ( x ), (−∞ < x < ∞ ), then the corresponding energy levels satisfy the inequality En′ ≥ En. For the ground state,  l = 0. As Vb ≤ 0  for a stable deuteron,  Vsinglet ≥ Vtriplet  and so the triplet state is the ground state.

7031 a. The ground state of the hydrogen atom is split by the hyperfine interaction. Indicate the level diagram and show from first principles which state lies higher in energy. b. The ground state of the hydrogen molecule is split into total nuclear spin triplet and singlet states. Show from first principles which state lies higher in energy. (Chicago) Sol: a. The hyperfine interaction is one between the intrinsic magnetic moment  µ p  of the proton nucleus and the magnetic field Be arising from the external electron structure, and is represented by the Hamiltonian H hf = − lµ p ⋅ Be . For

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the ground state, the probability density for the electron is spherically symmetric and so Be can be considered to be in the same direction as  µe , the intrinsic magnetic moment of the electron. Then as

µe = −

eg p e se , µ p = s p , ( g p > 0) me c 2m pc

Be is antiparallel to se and − µ p ⋅Be  has the same sign as  s e ⋅ s p . Let  S = s e + s p  and consider the eigenstates of S2 and Sz. We have 1 2 2 2 S − se − s p 2 1 3 3  =  S(S + 1) 2 −  2 −  2  2 4 4  1 = [2S(S + 1) − 3] 2 . 4

se ⋅ s p =

As the spins of electron and proton are both  12  , we can have 0, singlet state, S= 1, triplet state, and correspondingly se ⋅ s p

 3 2 − 4  < 0, singlet state, =  1  2 > 0, triplet state.  4

The hyperfine interaction causes the ground state to split into two states, S = 0  and  S = 1  (respectively the singlet and triplet total spin states). As Hnf has the same sign as  s e ⋅ s p , the energy of the triplet states is higher. The diagram of the energy levels of the ground state is shown in Fig. 7.6. Physically, hyperfine splitting is caused by the interaction of the intrinsic magnetic moments of the electron and the proton. For the electron the intrinsic magnetic moment is antiparallel to its spin; while for the proton

Fig. 7.6

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the magnetic moment is parallel to its spin. For the spin triplet, the spins of the electron and the proton are parallel, and so their magnetic moments are antiparallel. For the spin singlet, the reverse is true. If the spatial wave functions are same, the Coulomb energy between the election and proton is higher for the triplet state. b. For the hydrogen molecule H2, as protons are fermions, the total wave function must be antisymmetric for interchange of the two protons. Then for the nuclear spin singlet, the rotation quantum number can only be L = 0, 2, 4 . . ., where L = 0  has the lowest energy; for the spin triplet, the rotation quantum number can only be L = 1, 3, 5, . . ., where L = 1  has the lowest energy. As the energy difference caused by difference of L is larger than that caused by difference of nuclear spins, the energy of the state  L = 1  (total nuclear spin S = 1) is higher than that of the state  L = 0  (total nuclear spin  S = 0). So, for the ground state splitting of H2, the nuclear spin triplet  (S = 1)  has the higher energy. Because the spatial wave functions of  L = 1  and  L = 0  states are antisymmetric and symmetric respectively, the probability for the protons to come close is larger in the latter case than in the former, and so the Coulomb interaction energy is higher (for the same principal quantum number n). However, the difference between the energies of  L = 1  and  L = 0  is larger for the rotational energy levels than for the Coulomb energy levels. So for the ground state splitting of hydrogen atom, the nuclear spin triplet (S = 1)  has the higher energy.

7032 The wave function for a system of two hydrogen atoms can be described approximately in terms of hydrogenic wave functions. a. Give the complete wave functions for the lowest states of the system for singlet and triplet spin configurations. Sketch the spatial part of each wave function along a line through the two atoms. b. Sketch the effective potential energy for the atoms in the two cases as functions of the internuclear separation. (Neglect rotation of the system.) Explain the physical origin of the main features of the curves, and of any differences between them. (Wisconsin)

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The Hamiltonian of the system of two hydrogen atoms can be written as H = Hn + H e , and correspondingly the total wave function is ψ = ψ nφ ,  consisting of a nuclear part ψ n  and an electron part φ , with  Rv (r )YIm (θ , ϕ )χ 0 , for I = even, (para − hydrogen), ψn =  R (r )YIm (θ , ϕ )χ1 , for I = odd, (ortho − hydrogen),  ν where v denotes vibration, I denotes rotation quantum numbers, and χ 0 , χ1 are nuclear spin singlet and triplet wave functions.

Fig. 7.7 a. The configuration of the system is shown in Fig. 7.7. The wave function of a single electron is taken to be approximately

ϕ (r ) =

1 π

 λ   a

3/2

e − λ r /a .

Note that when λ = 1, ϕ (r ) is the wave function of an electron in the ground state of a hydrogen atom. For two electrons, the lowest singlet state wave function is 1 φs = [ϕ (ra1 )ϕ (rb 2 ) + ϕ (ra 2 )ϕ (rb1 )]χ 0e , 2 and the lowest triplet state wave function is 1 φt = [ϕ (ra1 )ϕ (rb 2 ) − ϕ (ra 2 )ϕ (rb1 )]χ1e , 2 where  χ 0e  and  χ1e  are electron spin singlet and triplet wave functions. Taking the x-axis along ab with the origin at a, we can express the spatial parts of  φs  and  φt  by

φs = b(e − k|x1|e − k|R− x2| + e − k|x2|e − k|R− x1| ), φt = b(e − k|x1|e − k|R− x2| − e − k|x2|e − k|R− x1| ). Keeping one variable (say x2) fixed, we sketch the variation of the spatial wave functions with the other variable in Fig. 7.8. It is seen that if one electron gets

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close to a nucleus, the probability is large for the other electron to be close to the other nucleus.

Fig. 7.8 b.

 1 1 1 1 V = −  + + +  e2  ra1 ra 2 rb1 rb 2  +

e2 e2 + + V0 . r12 R

The effective potential energy, V ≡ φ | V |φ ,  for the ground state as a function of R /a  is shown in Fig. 7.9. It is seen that the potential energy vanishes when the neutral atoms are infinitely far apart: R → ∞ , V → 0. When  R → 0, the potential energy between the two hydrogen nuclei becomes infinitely large while that between the electrons and the nuclei is finite, similar to the electron potential energy of a He atom. Hence  R → 0,

Fig. 7.9 V → +∞ . As R decreases from a large value, the repulsive potential between the nuclei increases, at the same time the attractive potential between the electrons and the nuclei increases also, competing against each other. For the singlet state, the probability that the electrons are close to the adjacent nuclei is large, and so the potential has a minimum value. For the triplet state, the probability that the electrons are close to the nuclei is small, and so the decrease of the potential energy due to the attractive force between the electrons and

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nuclei, which is negative, has a small value, and the repulsive potential between the nuclei, which is positive, is the main part of the total potential. Therefore  V > 0 and no minimum occurs.

7033 a. Using hydrogen atom ground state wave functions (including the electron spin), write wave functions for the hydrogen molecule which satisfy the Pauli exclusion principle. Omit terms which place both electrons on the same nucleus. Classify the wave functions in terms of their total spin. b. Assuming that the only potential energy terms in the Hamiltonian arise from Coulomb forces, discuss qualitatively the energies of the above states at the normal internuclear separation in the molecule and in the limit of very large internuclear separation. c. What is meant by an “exchange force”? (Wisconsin) Sol: a. The configuration of a hydrogen molecule is as shown in Fig. 7.7. Denote the λ ground state wave function of hydrogen atom by  100  and let  ϕ (r ) = ( 100 ) , where λ  is a parameter to be determined. Then the singlet state (S = 0)  wave function of hydrogen molecule is 1 ψ1 = [ϕ (ra1 )ϕ (rb 2 ) + ϕ (ra 2 )ϕ (rb1 )]χ 00 , 2 and the triplet state  (S = 1)  wave functions are

ψ3 =

1 [ϕ (ra1 )ϕ (rb 2 ) − ϕ (ra 2 )ϕ (rb1 )]χ1 M 2

with  M = −1, 0, 1. b. The energy of a hydrogen atom is 2

EH 2

me 4 1 mc 2  e 2  1 =− 2 2 =− 2 n 2  c  n 2 2

0.511 × 106  1  1 ×  137  n 2 2 13.6 = − 2 eV. n =−

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Thus the sum of the energies of two separate ground-state hydrogen atoms is −2 × 13.6 = −27.2 eV. On the other hand, for the He atom which also contains two protons and two electrons, the ground state energy is mZ ′ 2e 4 5  = −2 × 13.6 ×  2 −   16  2 2 = −77.5 eV.

EHe = −2 ×

2

where the factor 2 is for the two electrons of He atom and Z ′ = 2 − 165  is the effective charge number of the He nucleus. i. For the singlet state, the probability for the two electrons to be close to each other is rather large (on account of the Pauli principle), which enhances the repulsive exchange potential energy between them. The probability that the two electrons are near to the two nuclei is also large which tends to increase the attractive exchange potential. Taking both into account the exchange interaction potential lowers the energy. It is easily seen that for the singlet state, −77.5 eV < E1 < −27.2 eV. For the triplet state, the spins are parallel and so the spatial wave function is antisymmetric. In this case the potential energy is increased by the exchange interaction and so E3 > −27.2 eV, which makes it difficult to form a bound state. ii. When the distance between the nuclei → ∞, H 2  reduces to two separate hydrogen atoms. Hence the energy → −27.2 eV. c. The symmetrization or antisymmetrization of the wave function causes a mean shift of the potential energy by ∆V = ∫∫ ϕ (ra1 )ϕ (rb 2 )Vϕ (ra 2 )ϕ (rb1 )dτ 1 dτ 2. This is said to be caused by an “exchange force”.

7034 Describe the low-lying states of the H2 molecule. Give a rough value for their excitation energies. Characterize the radiative transitions of the first two excited states to the ground state. (Wisconsin) Sol:

In an approximate treatment of hydrogen atom, the zero order wave function is taken to be the product of two ground state hydrogen-like wave functions, which have the form

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ϕ (r ) =

1  λ  − λr /ao ,  e π  a0 

where a0 is the Bohr radius, λ  is a parameter to be determined. The spin part of the electron wave function of the H2 molecule ground state (S = 0)  is antisymmetric which requires the spatial part to be symmetric. As the spins of the two electrons are antiparallel, they can get quite close to each other (Pauli’s principle). This means that the density of “electron cloud” is rather large in the region of space between the two nuclei. In this region, the attractive potential between the two electrons and the two nuclei is quite large and thus can form a bound state, with wave function 1 ψ= [ϕ (ra1 )ϕ (rb 2 ) + ϕ (ra 2 )ϕ (rb1 )]χ 00 , 2 where the variables are as shown in Fig. 7.7. If the spins of the electrons are parallel (S = 1), then the spatial wave function must be antisymmetric, the probability that the two electrons getting close is small, and no bound state occurs. Of the energy levels related to the electronic, vibrational and rotational motions of H2, the rotational levels have the smallest spacing between two adjacent levels. For simplicity, we shall only consider rotational energy levels with the electrons in the ground state initially and in the absence of vibration between the nuclei. With no loss of generality, we can take the molecule’s energy to be zero when there is no rotation. The rotational levels are given by E=

2 J ( J + 1), 2I

where I is the moment of inertia and J is the total angular momentum of the twonuclei system. When  J = even, the total spin of the two protons in H2 is S = 0  and para-hydrogen results; when  J = odd, the total spin of the two protons is  S = 1  and ortho-hydrogen results. Suppose the distance between the two protons is  R = 1.5 × 0.53 = 0.80 Å  (Fig. 7.9). As 2 2 1  c  = =   2 I µ R 2 µc 2  R  =

2

 6.582 × 10−16 × 3 × 1010  2 6   938 × 10  0.8 × 10−8

2

= 1.3 × 10−2 eV, the energies of the low-lying states are as follows. J 0 Para-hydrogen : −2 0 E(10 eV)

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4 26.0

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Many-Particle Systems

J Ortho-hydrogen : E(10−2 eV)

1 2.6

3 15.6

607

5 39.0

As the interactions between two atoms are spin-independent, para-hydrogen and ortho-hydrogen cannot transform to each other, hence the selection rule ∆J = even. In nature the ratio of the number of molecules of ortho-hydrogen to that of para-hydrogen is 3:1. This means that the spectral line for  J = 2 → J = 0  is weaker than for  J = 3 → J = 1.

7035 The density matrix for a collection of atoms of spin J is ρ . If these spins are subject to a randomly fluctuating magnetic field, it is found that the density matrix relaxes according to the following equation: ∂ρ 1 = [ Jop ⋅ ρ Jop − J( J + 1) p ]. ∂t T Prove that the relaxation equation implies the following: a. 1 ∂ ∂ J z = Tr( Jop ρ ) = − J z , T ∂t ∂t b. J( J + 1) ∂ 2 ∂ 3 J z = Tr( J z2 ρ ) = − J z2 + . T T ∂t ∂t [Hint: Raising and lowering operators are useful here] (Columbia) Sol:

From definition,  J z = Tr( ρ J z ). Thus the following:

∂  ∂ρ  1 J z = Tr  J z  = Tr( J op ⋅ ρ J op J z − J(J +1) ρ J z ]  ∂t  T ∂t 1 = Tr [J x ρ J x J z + J y ρ J y J z + J z ρ J z2 − J(J +1)ρ J z ]. T As Tr AB = Tr B A,

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J x J y − J y J x = iJ z , J y J z − J z J y = iJ x , J z J x − J x J z = iJ y , J 2 J z = J ( J + 1) J z , (using units in which  = 1) we have 1 ∂ J z = Tr[ ρ J x J z J x + ρ J y J z J y + ρ J z3 − pJ ( J + 1) J z ] ∂t T 1 = Tr{ ρ[( J x2 + J y2 + J z2 ) J z + iJ x J y − iJ y J x − J ( J + 1) J z ]} T 1 1 = − Tr( ρ J z ) = − J z . T T b. ∂ 2 1  ∂ρ 2  J z = Tr  Jz  ∂t T  ∂t  1 = Tr[ J x pJ x J z2 + J y ρ J y J z2 + J z ρ J z3 − J ( J + 1)ρ J z2 ] T 1 = Tr[ ρ J x J z2 J x + ρ J y J z2 J y + ρ J z4 − ρ J ( J + 1) J z2 ] T 1 = Tr[ p( J x J z J x J z + J y J z J y J z + iJ x J z J y − iJ y J z J x + J z4 ) T − J ( J + 1)ρ J z2 ] 1 = Tr{ ρ[ J x2 J z2 + J y2 J z2 + J z4 + iJx Jy J z − iJ y J x J z + iJ x J z J y T − iJ y J z J x − J ( J + 1) J z2 ]} =

1 Tr{ ρ[i(iJ z ) J z + iJ x ( J y J z ) + iJ x (−iJ x ) − iJ y ( J x J z ) T − iJ y (iJ y )]}

1 Tr{ ρ[− J z2 + J x2 + J y2 + i(iJ z ) J z ]} T 1 = Tr{ ρ[−3 J z2 + ( J x2 + J y2 + J z2 )]} T 3 J ( J + 1) , = − J z2 + T T =

since

1 Tr[ ρ J ( J + 1)] = J ( J + 1) = J ( J + 1). T

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7036 A molecule is made up of three identical atoms at the corners of an equilateral triangle as shown in Fig. 7.10. We consider its ion to be made by adding one electron with some amplitude on each site. Suppose the matrix element of the Hamiltonian for the electron on two adjacent sites i, j is  i H j = − a  for  i ≠ j .

Fig. 7.10 a. Calculate the energy splittings. b. Suppose an electric field in the z direction is applied, so that the potential energy for the electron on top is lowered by b with  b  a . Now calculate the levels. c. Suppose the electron is in the ground state. Suddenly the field is rotated by 120° and points toward site 2. Calculate the probability for the electron to remain in the ground state. (Princeton) Sol: a. Denote the basis vectors by 1 , 2 , 3  and let  i H i = E0 , i = 1, 2, 3. Then  E −a −a  0   H =  − a E0 − a  .  −a −a E  0   To diagonalize H, solve E0 − λ

−a

−a

E0 − λ

−a

−a

−a

E0 − λ

−a = 0.

The solution gives energy levels E1,2 = E0 + a  (two-fold degenerate) and E3 = E0 − 2a .

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b. The H matrix is now  E − b −a −a  0   E0 −a  . H =  −a  −a −a E0   Its diagonalization gives energy levels E1 = E0 + a, E2 = E0 −

a + b + (a − b)2 + 8a 2 , 2

E3 = E0 −

a + b − (a − b )2 + 8a 2 . 2

E2 has the lowest energy and thus corresponds to the ground state, with wave function 1 ψ0 = [( E0 − E2 − a) 1 ( E0 − E2 − a)2 + 2a 2 + a 2 + a 3 ]. c. After the rotation of the field the system has the same configuration as before but the sites are renamed: 1 → 2, 2 → 3, 3 → 1. Hence the new ground state is 1 ψ 0′ = [a 1 + ( E0 − E2 − a) 2 + a 3 ]. ( E0 − E2 − a)2 + 2a 2 Hence the probability for the electron to remain in the ground state is

ψ 0′ |ψ 0

2

2

 2a( E0 − E2 − a) + a 2  . = 2 2   ( E0 − E2 − a) + 2a 

7037 Consider three particles, each of mass m, moving in one dimension and bound to each other by harmonic forces, i.e., 1 V = [( x1 − x 2 )2 + ( x 2 − x 3 )2 + ( x 3 − x1 )2 ]. 2 a. Write the Schrödinger equation for the system. b. Transform to a center-of-mass coordinate system in which it is apparent that the wave functions and eigenenergies may be solved for exactly.

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611

c. Using (b) find the ground state energy if the particles are identical bosons. d. What is the ground state energy if the particles are identical spin −1/ 2 fermions? (Wisconsin) Sol: a. As 3

E=∑ i =1

pi2 +V , 2m

The Schrödinger equation is i

 2  ∂2 ∂ψ ∂2 ∂2  =− + + ψ ∂t 2m  ∂ x12 ∂ x 22 ∂ x 32  k + [( x1 − x 2 )2 + ( x 2 − x 3 )2 + ( x 3 − x1 )2 ]ψ . 2

b. Using the Jacobi coordinates  y1 = x1 − x 2 ,   y = x1 + x 2 − x , 3  2 2  x +x +x  y3 = 1 2 3 , 3  or  y1 y 2  x1 = y 3 + + , 2 3  y1 y 2   x 2 = y3 − + , 2 3  2   x 3 = y3 − 3 y 2 ,  we have k 3  V =  y12 + 2 y 22  ,  2 2 3

pi2

 2  1 ∂2

∑ 2m = − 2m  3 ∂ y i =1

2 3

+2

∂2 3 ∂2  + , ∂ y12 2 ∂ y 22 

and hence the stationary eigenequation

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Problems and Solutions in Quantum Mechanics

 2 ∂ 2ψ  2  ∂ 2 3 ∂ 2  − + 2 ψ 6m ∂ y32 2m  ∂ y12 2 ∂ y 22  k 3 2 y1 + 2 y 22 ψ . + 2 2

ETψ = −

{

}

Try

ψ = Y ( y3 )φ ( y1 , y 2 ). The equation is separated into two equations:   2 ∂2 Y  − = EcY , 2  6m ∂ y3  2 2 2  −   2 ∂ + 3 ∂  φ + k  3 y 2 + 2 y 2  φ = Eφ ,  1 2  2m  ∂ y12 2 ∂ y 22   2 2  where  Ec = E − ET  is the energy due to the motion of the center of mass. The first equation gives Y=

1 i e 2π

6mEc y3 

.

With

φ = φ1 ( y1 )φ2 ( y 2 ) the second equation is separated into two equations   2 ∂2 φ 3 1  − + ky12φ1 = E1φ1 , 2 ∂ m y 4  1  2 2  − 3 ∂ φ2 + ky 2φ = E φ . 2 2 2 2  4m ∂ y 22  where  E = E1 + E2. These are equations for harmonic oscillators of masses m2 , 23m  and force constants 2k and 3k respectively, both having the same angular frequency ω = 3mk . Hence the total energy is 1  3k  1  3k  E = E1 + E2 =  n +   +l +   ,  2 m  2 m with n, l = 0, 1, 2, 3,….

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613

c. Let α 2 = mw  . The ground state wave functions of φ1 , φ2  are  1  φ10 ( y1 ) =    2π 

1/4

 1  α exp  − α 2 y12  ,  4 

 2  φ20 ( y 2 ) =    3π 

1/4

 1  α exp  − α 2 y 22  ,  3 

and so 1/4

 −α 2  1  φ0 ( y1 , y 2 ) = φ10 ( y1 )φ20 ( y 2 ) =  2  α exp  (3 y12 + 4 y22 ) ,  3π  12   where 3 y12 + 4 y 22 = 3( x1 − x 2 )2 + ( x1 + x 2 − 2 x 3 )2 = 4( x12 + x 22 + x 32 − x1 x 2 − x 2 x 3 − x 3 x1 ). As  y3 = 13 ( x1 + x 2 + x 3 )  it is obvious that the spatial wave function ψ 0  is symmetric for the interchange of any two of the particles, which is required as the bosons are identical. The ground state energy of the three bosons, excluding the translational energy of the center of mass, is 1 3k 1 3k 3k E0 =  +  = . 2 m 2 m m d. If the particles are identical spin-1/2 fermions, as spin is not involved in the expression for the Hamiltonian, the eigenfunction is a product of the spatial wave function and the spin wave function, and must be antisymmetric for interchange of particles. For the coordinate transformation in (b), we could have used  y1′ = x 2 − x 3 ,   y ′ = x 2 + x 3 − x , 1  2 2  x +x +x  y3′ = 1 2 3 3  and still obtain the same result. In this case the spatial eigenfunction is ψ ( y1′ , y 2′ , y3′ ) = φ1n ( y1′ )φ2l ( y 2′ )Y ( y3′ ) and the energy is E = (n + l + 1)

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3k . m

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Since  φ10 ( y1 )φ20 ( y 2 ) = φ10 ( y1′ )φ20 ( y 2′ ), the spatial wave function is symmetric for the interchange of two particles. However, for three spin-1/2 fermions it is not possible to construct a spin wave function which is antisymmetric. Hence this state cannot be formed for three spin-1/2 fermions and higher states are to be considered. Looking at the wave functions of a harmonic oscillator, we see that the exponential part of φ1n ( y1 )φ2l ( y 2 )  is the same as that of φ10 ( y1 )φ20 ( y 2 )  and is symmetric. Let Φ1 = φ11 ( y1 )φ20 ( y 2 ), Φ 2 = φ11 ( y1′ )φ20 ( y 2′ ). and construct the total wave function  0   1   l   1   l   0  + Φ2  Φ = Φ1       0 0 l  1  2 3  l  1  0  2  0  3  1   0   1  . − (Φ 2 + Φ1 )  0  1  l  2  0  3 As Φ1 = C( x1 − x 2 ), Φ 2 = C( x 2 − x 3 ), Φ1 + Φ 2 = C( x1 − x 3 ), where C is symmetric for interchange of the particles,  Φ  is antisymmetric as required for a system of identical fermions. Hence the ground state energy of the system, excluding the translational energy of the center of mass, is E0 = 2

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3k . m

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Part VIII

Miscellaneous Topics

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8 Express e

 0 a   −a 0   

8001 as a 2 × 2 matrix; a is a positive constant. (Berkeley)

Sol 1: Let S(a ) = e

 0 a   −a 0   

=e

 0 1  a   −1 0 

= e aA

with  0 1  . A≡  −1 0  As  0 1  0 1   1 0  A2 =  = − = −I ,     −1 0   −1 0   0 1  I being the unit matrix, we have d S(a) = AS(a), da d2 S(a) = A 2 S(a) = − S(a), da 2 and thus S ′′ (a) + S(a) = 0. The general solution is S(a) = c1e ia + c2e − ia, subject to boundary conditions S(0) = I , S ′ (0) = A Hence c1 + c2 = I ,  c1 − c2 = −iA,

617

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giving  I − iA ,  c1 =  2   c = I + iA .  2 2 Therefore I − iA ia I + iA −ia e + e 2 2 I iA = (e ia + e −ia ) + (e −ia − e ia ) = I cos a + A sin a 2 2  cos a sin a  = .  − sin a cos a 

S(a ) =

 0 1   1 0 . As A2 = −  Sol 2: Let A =  = − I , A3 = − A, A 4 = I , …   0 1  −1 0  ∞

∞ an An ∞ a 2 k (−1)k a 2 k+1 (−1)k =∑ I +∑ A n=0 n ! k=0 (2k )! k=0 (2k + 1)!

e aA = ∑

 cos a sin a  = cos aI + sin aA =  .  − sin a cos a 

8002 a. Sum the series y = 1+ 2 x + 3 x + 4 x 3 + , | x |< 1. 2

b. If f ( x ) = xe − x/ λ over the interval 0 < x < ∞ , find the mean and most probable values of x . f ( x ) is the probability density of x. ∞ dx c. Evaluate I = ∫ . 0 4 + x4 d. Find the eigenvalues and normalized eigenvectors of the matrix  1 2 4  2 3 0 .    5 0 3 Are the eigenvectors orthogonal? Comment on this. (Chicago)

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Miscellaneous Topics

619

Sol: a. As | x | < 1 , y − xy = 1 + x + x 2 + x 3 +  =

1 , (1 − x )

or y=

1 . (1 − x )2

b. The mean value of x is ∞

∞

x = ∫ x f ( x )dx / ∫ f ( x )dx 0

0

∞

= ∫ x ⋅ xe 0

=λ

− x /λ

∞

dx / ∫ xe − x /λ dx 0

Γ(3) = 2λ . Γ(2)

The probability density is an extremum when 1 f ′ ( x ) = e − x /λ − xe − x /λ = 0 , λ i.e. at x = λ or x → ∞. Note that λ > 0 if f ( x ) is to be finite in 0 < x < ∞ . As 1 f ′′ (λ ) = − e −1 < 0, f (λ ) = λ e −1 > lim f ( x ) = 0 , x →∞ λ the probability density is maximum at x = λ . Hence the most probable value of x is λ c. Consider the complex integral dz dz dz ∫c 4 + z 4 = ∫c1 4 + z 4 + ∫c2 4 + z 4 along the contour c = c1 + c2 as shown in Fig. 8.1.

Fig. 8.1 The integrand has singular points −1 + i , 1 + i inside the closed contour c. Hence the residue theorem gives

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dz

∫ 4 + z c

4

= 2π i[Res(1 + i ) + Res( −1 + i )]  1 + i −1 + i  π = 2π i  − − = .  16 16  4

Now let R → ∞ , we have

∫

c2

dz → 0. 4 + z2

Then as

∫

c1

0 ∞ dx ∞ dx dz dx , = + = 2∫ 0 4 + x4 4 + z 4 ∫−∞ 4 + x 4 ∫0 4 + x 4

we have

∫

∞

0

dx π = . 4 + x4 8

d. Let the eigenvalue be E and the eigenvector be  x1  X =  x2  .    x  3

Then  1 2 4   x1   x1   2 3 0   x  = E x  .    2  2  5 0 3   x   x  3 3 For non-vanishing X, we require E −1 − 2 − 4 −2 E−3 0 =0 . −5 0 E−3 The solution is E1 = 3, E2 = −3, E3 = 7. Substitution in the matrix equation gives the eigenvectors, which, after normalization, are  0 1   X1 = 2 5   −  1

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621

for E = E1 and  −6 1   X2 = 2 , 65    5

X3 =

 4 1   2 3 5    5

for E = E2 , E3 . Note that these eigenvectors are not orthogonal. Generally, only for a Hermition matrix are the eigenvectors corresponding to different eigenvalues orthogonal.

8003 Please indicate briefly (in one sentence) what contributions to physics are associated with the following pairs of names. (Where applicable write an appropriate equation.) a. Franck–Hertz b. Davisson–Germer c. Breit–Wigner d. Hartree–Fock e. Lee–Yang f. Dulong–Petit g. Cockroft–Walton h. Hahn–Strassmann i. Ramsauer–Townsend j. Thomas–Fermi (Berkeley) Sol: a. Franck and Hertz verified experimentally the existence of discrete energy ­levels of an atom. b. Davisson and Germer verified the wave properties of electrons by demonstrating their diffraction in a crystal.

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c. Breit and Wigner discovered the Breit–Wigner resonance formula in nuclear physics. d. Hartree and Fock developed a self-consistent field method for obtaining approximate many-electron wave functions. e. Lee and Yang proposed the non-conservation of parity in weak interactions. f. Dulong and Petit discovered that atomic heat is the same for all solids at high temperatures, being equal to 3R, R being the ideal gas constant. g. Cockroft and Walton effected the first artificial disintegration of an atomic nucleus. h. Hahn and Strassmann first demonstrated the fission of uranium by neutrons. i. Ramsauer and Townsend first observed the resonant transmission of low energy electrons through rare-gas atoms. j. Thomas and Fermi proposed an approximate statistical model for the structure of metals.

8004 Give estimates of magnitude order for the following quantities. a. The kinetic energy of a nucleon in a typical nucleus. b. The magnetic field in gauss required to give a Zeeman splitting in atomic hydrogen comparable to the Coulomb binding energy of the ground state. c. The occupation number n of the harmonic oscillator energy eigenstate that contributes most to the wave function of a classical onedimensional oscillator with mass m = 1gram,, period T = 1sec, amplitude x 0 = 1cm. d. The ratio of the hyperfine structure splitting to the binding energy in the 1s state of atomic hydrogen, expressed in terms of the fine structure constant α , the electron mass me , and the proton mass mp . (Berkeley)

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Miscellaneous Topics

Sol: a. The kinetic energy T =

623

p2 of a nucleon in a nucleus can be estimated using 2m

the approximation p ∼ ∆ p and the uncertainty principle ∆x ∆ p ∼ h. As h ∆x ∼ 10−12 cm, ∆p ∼ , ∆x T~

2

h2  1  1  hc    =   2m  ∆x  2mc 2  ∆x 

2

=

 4.1 × 10−15 × 3 × 1010  1  2 × 938 × 106  10−12

~

150 × 108 ~107 eV. 2000

2

b. The Zeeman splitting is given by ∆E ~ µ B ⋅ B, µ B being the Bohr magneton, and the Coulomb binding energy of a hydrogen atom is 13.6 eV. For the two to be comparable we require B≈

13.6 × 1.6 × 10−19 13.6 × 1.6 = × 1013 wbm −2 ~109 Gs. 9.3 × 10−32 9.3

c. The energy of a classical one-dimensional oscillator is m 2 E = (ω x 0 ) = 2π 2mx 02 / T 2 = 2π 2erg . 2 For nω = E, we require n=

π ×1 E 2π 2 π T = = = = 3 × 1027. ω ω  1.054 × 10−27

d. The energy shift due to hyperfine-structure splitting of a hydrogen atom in the ground state (in units where c =  = 1) is ∆E ∼ me2α 4 /m p where α is the fine-structure constant. The binding energy of the electron in the ground state is Ee = meα 2 /2. Hence  me  ∆E / E = 2α 2    mp 

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8005 Consider the density matrix ρ (t ) = ψ (t ) ψ (t ) corresponding to the state

ψ (t ) of the system. Show that it satisfies the time-evolution equation dρ i = [H , ρ ]. dt  Sol: Using the Schrödinger equation, H |ψ = i the density matrix as follows:

∂ |ψ and its conjugate, we express ∂t

dρ ∂ = | ψ (t ) ψ (t ) | + |ψ (t ) dt ∂t

∂ ψ (t ) | ∂t

dρ i = − ( Hψ (t ) ψ (t ) − ψ (t ) ψ (t )H dt 

)

dρ i = − ( H ρ − ρH ) dt  dρ i = − [H , ρ]. dt 

8006 Answer each of the following questions with a brief and, where possible, quantitative statement. Give your reasoning. a. A beam of neutral atoms passes through a Stern–Gerlach apparatus. Five equally-spaced lines are observed. What is the total angular momentum of the atom? b. What is the magnetic moment of an atom in the state 3 P0? (Disregard nuclear effects) c. Why are the noble gases chemically inert? d. Estimate the energy density of black body radiation in this room in erg cm−3. Assume the walls are black.

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e. In a hydrogen gas discharge both the spectral lines corresponding to the transitions 22 P1/2 → 12 S1/2 and 22 P3/2 → 12 S1/2 are observed. Estimate the ratios of their intensities. f. What is the cause for the existence of two independent term level schemes, the singlet and triplet systems, in atomic helium? (Chicago) Sol: a. When unpolarized neutral atoms of total angular momentum J pass through the Stern-Gerlach apparatus, the incident beam will split into 2 J + 1 lines. Thus 2 J + 1 = 5, giving J = 2. 3 b. An atom in the state P0 has total angular momentum J = 0. Hence its magnetic moment is equal to zero, if nuclear spin is neglected.

c. The molecules of noble gases consist of atoms with full-shell structures, which makes it very difficult for the atoms to gain or lose electrons. Hence noble gases are chemically inert. d. The energy density of black body radiation at room temperature T ≈ 300 K is 4 µ = σT 4 c 4 = × 5.7 × 10−5 × 3004 10 3 × 10 = 6 × 10−5 erg / cm3 . e.

I ( 22 P1/2 → 12 S1/2 )

I ( 2 P3/2 → 1 S1/2 ) 2

2

≈

2 J1 + 1 2J2 + 1

=

2 × 1/ 2 + 1 1 = . 2 × 3 / 2 +1 2

f. The helium atom contains two spin-1/2 electrons, whose total spin S = s1 + s2 can have two values S = 1 (triplet) and S = 0 (singlet). Transition between the two states is forbidden by the selection rule ∆S = 0 . As a result we have two independent term level schemes in atomic helium.

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8007 a. Derive the conditions for the validity of the WKB approximation for the one-dimensional time-independent Schrödinger equation, and show that the approximation must fail in the immediate neighborhood of a classical turning point. b. Explain, using perturbation theory, why the ground state energy of an atom always decreases when the atom is placed in an external electric field. (Berkeley) Sol: a. The WKB method starts from the Schrödinger equation  2 d 2   − 2m dx 2 + V ( x )ψ ( x ) = Eψ ( x ),   where it is assumed

ψ ( x ) = e is ( x )/ . Substitution in the Schrődinger equation gives 2



1  ds   1 d 2 s = E − V ( x ). (1)   + 2m  dx  i 2m dx 2 Expanding s as a series in powers of  i 2

   s = s0 + s1 +   s2 + , i i and substituting it in Eq. (1), we obtain 2



1 ′2  1 ′′   s0 + s0 + 2s0′ s1′ ) +   ( s1′2 + 2s0′ s2′ + s1′′ ) ( i 2m i 2m + = E − V ( x ).

(2)

If we impose the conditions

s0′′  s0′2 , (3)



2s0′ s1′  s0′2 , (4)



Eq. (2) can be approximated by 1 ′2 s0 = E − V ( x ), (5) 2m

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which is equivalent to setting 2s0′ s1′ + s0′′ = 0, 2s0′ s2′ + s1′2 + s1′′ = 0, …… (3) and (4) are the conditions for the validity of the WKB method. Integration of Eq. (5) gives s0 ( x ) = ± ∫

x

x

2m( E − V ( x ))dx = ± ∫ pdx ,

so that (3) can be written as  dp  1, (6) p 2 dx

i.e.,



d  1  1, dx  p 

or dλ  1, dx where

λ=

  = . p 2m( E − V ( x ))

Near a turning point V ( x ) ~ E , p → 0 and (6) is not satisfied. Hence the WKB method cannot be used near classical turning points. b. Consider an atom in an external electric field ε in the z direction. The perturbation Hamiltonian is H ′ = −ez, where z = ∑ i zi is the sum of the z coordinates of the electrons of the atom, and the energy correction is

′ ∆E0 = H 00′ + ∑ H on / ( E0 − En ) . 2

n≠0

As z is an odd operator and the parity of the ground state is definite, H 00′ = 0.. Futhermore E0 − En < 0. Hence ∆E0 < 0. This means that the energy of the ground state decreases when the atom is placed in an electric field.

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8008 Consider a particle of mass m in the ground state of an infinitely high box extending x = 0 to x = a. It is suddenly expanded from x = a to x = 2a . What is the probability of that particle in the ground state after this sudden expansion? Sol:

ψi =

πx 2 sin a a

Probability = 〈ψ i |ψ f 〉

2 πx sin 2a 2a

ψf = 2

Probability amplitude = ψ i |ψ f a

=∫ 0

2 πx 2 πx sin sin dx 2a a a 2a a

2 1 πx πx sin sin dx = × ∫ 2a a a 20 π  1   π  1 sin  1 −  x sin  1 +  x    2 2 a a  2 − =   π  1 a  π  1 2 1 −  2 1 +    a  2  a  2   a

=

2  πx a 3πx a  ×  sin × − sin  a  2a π a 3π 0

=

a 2 a [1 − 0 ] − [ −1 − 0] a π 3π

=

2 4 2 a  1 × 1+ = a π  3  π 3

{

}

2



 2 4  16 × 2 32 Probability =  × = = 2. 9π 2 9π  π 3 

8009 Set up the relevant equations with estimates of all missing parameters. The molecular bond (spring constant) of HCl is about 470 N/m. The moment of inertia is 2.3 × 10 −47 kg ⋅ m2 .

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a. At 300 K what is the probability that the molecule is in its lowest excited vibrational state? b. Of the molecules in the vibrational ground state what is the ratio of the number in the ground rotational state to the number in the first excited rotational state? (Wisconsin) Sol: a. The Hamiltonian for the vibrational motion of the system is pˆ 2 1 Hˆ v = + µω 2 xˆ 2 , 2µ 2 and the vibrational states are 1  Ev(n ) =  n +  ω , n = 0, 1, 2, …,  2 with ω = K / µ , K being the force constant and µ the reduced mass of the oscillating atoms. Statistically, the number of molecules in state E (n ) is proportional to exp(−nx ), ω where x = , k being Boltzmann’s constant and T is the absolute temperakT ture. Thus the probability that the molecule is in the first excited state is P1 =

1+ e

−x

e−x = e − x (1 − e − x ) , + e −2 x +…

As  1 × 35  µ= m ≈ m p = 1.67 × 10−27  kg,  1 + 35  p

27 −34 ω 1.054 × 10 × ( 470 /1.67 × 10 ) = x= kT 1.38 × 10−23 × 300

1/2

we have P1 ≈ e

−13.5

= 13.5,

= 1.37 × 10−6 .

b. The Hamiltonian for rotation is 1 2 Hˆ r = J , 2I and the energy states are Er( J ) =

2 J ( J + 1), J = 0, 1, 2, … . 2I

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Since the number of molecules in rotational state J is proportional to  E( J )  (2 J + 1)exp  − r  as the J state is (2 J + 1)-times degenerate mJ = − J , − J + 1, … J ,  kT 

(

(m

J

)

= − J , − J + 1, … J , we have  2  N ( J = 0) 1 . = exp   IkT  N ( J = 1) 3 As

(1.054 × 10 ) 2 = = 0.117, IkT 2.3 × 10−47 × 1.38 × 10−23 × 300 N ( J = 0) 0.117 = e /3 = 0.37. N ( J = 1) −34 2

8010 The potential curves for the ground electronic state (A) and an excited electronic state (B) of a diatomic molecule are shown in Fig. 8.2. Each electronic state has a series of vibrational levels which are labelled by the quantum ­number v . a. The energy differences between the two lowest vibrational levels are designated as ∆ A and ∆ B for the electronic states A and B respectively. Is ∆ A larger or smaller than ∆ B ? Why? b. Some molecules were initially at the lowest vibrational level of the electronic state B, followed by subsequent transitions to the various vibrational levels of the electronic state A through spontaneous emission of radiation. Which vibrational level of the electronic state A would be most favorably populated by these transitions? Explain your reasoning. (Wisconsin)

Fig. 8.2

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Miscellaneous Topics

Sol:

 ∂2 V  a. The force constant is K =  2   ∂r 

r =r0

631

where r0 is the equilibrium position. It

can be seen from Fig. 8.2 that K A > K B . The vibrational energy levels are given by 1 K  E (n ) =  n +  ω , ω = .   2 µ Hence. ∆A ≈ 

KA KB , ∆B ≈  , µ µ

and so ∆ A > ∆ B . b. Electrons move much faster than nuclei in vibration. When an electron transits to another state, the distance between the vibrating nuclei remains practically unchanged. Hence the probability of an electron to transit to the various levels is determined by the electrons’ initial distribution probability. As the molecules are initially on the ground state of vibrational levels, the probability that the electrons are at the equilibrium position r = r0 B is largest. Then from Fig. 8.2 we see that the vibrational level v ≈ 5 of A is most favorably occupied.

8011 Singlet positronium decays by emitting two photons which are polarized at right angles with respect to each other. An experiment is performed with photon detectors behind polarization analyzers, as shown in Fig. 8.3. Each analyzer has a preferred axis such that light polarized in that direction is transmitted perfectly, while light polarized in the perpendicular direction is absorbed completely. The analyzer axes are at right angles with respect to each other. When many events are observed, what is the ratio of the number of events in which both detectors record a photon to the number in which only one detector records a photon? (MIT)

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Fig. 8.3 Sol:

Suppose the positronium is initially at rest. Then the two photons will move in opposite directions to conserve momentum, and will reach the respective analyzers at the same time. Assume further that the detector solid angle is very smaller. Then the directions of those photons that reach the analyzers must be almost perpendicular to the latter. Hence the directions of polarization of these photons are parallel to the analyzers. Denote by θ the angle between one photon’s direction of polarizationn and the direction of transmission of the analyzer reached by it. The probability that it can pass through the analyzer is cos 2 θ . Consider the second photon produced in the same decay. As it is polarized at right angles with respect to the first one, the angle between its direction of polarization and the direction of transmission of the second analyzer, which is oriented at right angles to that of the first analyzer, is also θ . Hence the probability that, of the two detectors, only one records the passage of a photon is  1 P1 ∝ Ω   2π Ω = , 4

∫

2π

0

cos 2θ (1 − cos 2 θ ) dθ +

1 2π

 ∫ (1 − cos θ ) cos θ dθ  2π

2

2

0

where Ω is the solid angle subtended by the detector, and the probability that both detectors record the passage of photons is  1 P2 ∝ Ω   2π

∫

2π

0

 3Ω cos 2θ cos 2 θ dθ  = .  8

Hence the ratio of the number of events of both detectors recording to that of only one detector recording in a given time is P2 3 1 3 = ÷ = . P1 8 4 2

8012 A point source Q emits coherent light isotropically at two frequencies ω and

ω + ∆ω with equal power I joules/sec at each frequency. Two detectors A and

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B each with a (small) sensitive area s, capable of responding to individual photons are located at distances l A and lB from Q as shown in Fig. 8.4. In the following take ∆ ω ω  1 and assume the experiment is carried out in vacuum. a. Calculate the individual photon counting rates (photons/sec) at A and B as functions of time. Consider time scales  1 ω . b. If now the output pulses from A and B are put into a coincidence circuit of resolving time τ , what is the time-averaged coincidence counting rate? Assume that τ  1 ∆ω and recall that a coincidence circuit will produce an output pulse if the two input pulses arrive within a time τ of each other. (CUS)

Fig. 8.4 Sol: a. The wave function of a photon at A is  iω  lA −t  i (ω +∆ω ) lcA −t   ψ A ( lA , t ) = C1 e  c  + e ,   where C1 is real and, hence, the probability of finding a proton at A in unit time is PA = ψ 1*ψ 1     l = C12 2 + 2cos  ∆ω  A − t     c      ∆ω = 4C12 cos 2  (lA / c − t ) .  2  If there is only a single frequency, PA = C12. As each photon has energy ω , the number of photons arriving at A per second is s I . i 2 4π lA ω

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Hence C2 =

Is , 4π lA2 ω

and PA =

Is  ∆ω cos 2  (lA / c − t ) . π lA2 ω  2 

Similarly we have  ∆ω PB = 4C22 cos 2  (lA / c − t ) ,  2  where C22 =

Is . 4π lB2 ω

b. In a coincidence of resolving time τ , the time-averaged coincidence counting rate τ 1 T P = lim dt ∫ PA (t )PB (t + x )dx −τ T →∞ 2T ∫− T τ 1 T dt ∫ 4C 4 1 + cos ∆ω ( lA /c − t )  = lim ∫ −τ T →∞ 2T − T   l × 1 + cos ∆ω  B − t − x   dx   c  = lim

T →∞

1 2T

∫

T

−T

{

4C 4 ⋅ 1 + cos ( lA /c − t ) ∆ω 

}

    l × 2τ + 2τ cos  ∆ω  B − t    dt ,  c     where C4 =

I 2s 2 = C12C22 , 16π 2lB2lA2  2ω 2

as

∫

τ

−τ

cos  ∆ω ( lB / c − t − x )  dx =

1  l   2 sin(τ∆ω )cos  B − t  ∆ω    ∆ω c  

≈ 2τ cos  ∆ω ( lB /c − t )  .

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Hence P = 8τ C 4 lim

T →∞

1 2T

∫

T

−T

{1 + cos (l /c − t ) ∆ω } A

  l   × 1 + cos  B − t  ∆ω   dt    c   = 8τ C 4 lim

T →∞

1 2T

∫

T

−T

   lB   1 + cos  − t  ∆ω   c  

 l   l   l     + cos  A − t  ∆ω  + cos  A − t  ∆ω  cos  B − t  ∆ω   dt        c   c   c   1  ∆ω (lA − lB ) 1 + cos   c   2  l − l    l + l   + cos  A B − t  ∆ω  cos  A B  ∆ω      2 2 c c    

= 8τ C 4 lim

T →∞

1 2T

∫

T

−T

1  l + l   + cos  A B − 2t  ∆ω   dt  2  c  = 8τ C 4 lim

T →∞

1 2T

 2T  l − l   cos  A B  ∆ω   2T + 2  c   

 1  ∆ω ( lA − lB )   = 8τ C 4 1 + cos   c     2 =

τ I 2s 2  1  ∆ω 1 + cos  (lA − lB )  . 2 2 2 2 2  2π lBlA  ω  2  c 

8013 The state of an infinite potential well of width a is ψ = iψ 0 + (1+ i )ψ 1 + ( 2 − i )ψ 2 . Normalize the wave function and find the expectation value of energy. Sum of squares of coefficients = 1 Sol:

1 = A2 (1 + 2 + 5 ) = A2 8

A=

1 8

ε1 = ε ε2 = 4ε ε 3 = 9ε

ε=

π 2 2 2ma 2

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〈ε〉 = ε s Ps 27

1 2 5 54 ε = ε + × 4 ε + × 9ε = 8 8 8 84 =



27 ε 27 π 2  2 . = 4 8ma 2

8014 A spinless particle of mass m and charge q is constrained to move in a circle of radius R as shown in Fig. 8.5. Find its allowed energy levels (up to a common additive constant) for each of the following cases: a. The motion of the particle is nonrelativistic. b. There is a uniform magnetic field B perpendicular to the plane of the circle. c. The same magnetic flux which passed through the circle is now contained into a solenoid of radius b(b < R ). d. There is a very strong electric field F in the plane of the circle

( q | F | 

2

)

/ mR 2 .

e. F and B are zero, but the electron’s motion around the circle is extremely relativistic. f. The circle is replaced by an ellipse with the same perimeter but half the area. (CUS ) Sol: a. Let the momentum of the particle be p. The quantization condition p ⋅ 2π R = nh gives p=

n R

and hence E=

2

2 2 1  n  p2 n , =   = 2m 2m R 2mR 2

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where n = 0, ± 1, ± 2,. . . .

Fig. 8.5 b. Take coordinates with origin at the center of the circle and the z-axis along the direction of B. Then the vector potential at a point on the circle is 1 A = BReϕ . 2 The Schrödinger equation Hˆψ = Eψ , where 2

1  q  Hˆ =  p − A  , 2m c can be written as 2

1  i ∂ q 1  − − ⋅ BR ψ (ϕ ) = Eψ (ϕ ). 2m  R ∂ϕ c 2  Its solution is ψ (ϕ ) = Ce inϕ , The single-valuedness condition ψ (ϕ ) = ψ (ϕ + 2π ) demands n = 0, ± 1, ± 2, . . . . Substituting the solution in the equation gives E=

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2

1  n q   − BR  . 2m  R 2c 

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c. When the magnetic flux is confined to the inside of a solonoid of radius b enclosed by the circle, magnetic field is zero on the circular path. As ∇ × A = B = 0, A can be taken to be a constant which is equal to 12 BR when b → R. Then A=

1 Bπ R 2 φ = . 2 πR 2π R

As φ remains the same, the energy levels are the same as in (b). d. Take the x-axis parallel to F. Then F = F (cosϕ , − sinϕ ), dr = (0, Rdϕ ), and hence V = − ∫ qF ⋅ dr = qFR ∫ sinϕ dϕ = −qFR cosϕ . Thus the Hamiltonian is −2 1 d 2 Hˆ = − qFR cosϕ . 2m R 2 dϕ 2 Because the electric field F is very strong, the probability that the particle moves near ϕ ∼ 0 is large. Hence we can make the approximation 1 ϕ2 cosϕ = 1 − ϕ 2 + O (ϕ 4 ) ≈ 1 − 2 2 and obtain 2 d 2  1  Hˆ = − − qFR  1 − ϕ 2  , 2 2  2  2mR dϕ or 2 d 2 1 Hˆ + qFR = − + qFRϕ 2 , 2mR 2 dϕ 2 2 which has the form of the Hamiltonian of a harmonic oscillator of mass M = mR 2 and angular frequency ω given by Mω 2 = qFR , whose eigenvalues are 1  En + qFR =  n +  ω ,  2 or 1  En =  n +  ω − qFR ,  2 with

ω=

qFR qF = , n = 0, 1, 2, … . M mR

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Therefore 1 qF  En = −qFR +  n +   .   2 mR e. The quantization condition gives p ⋅ 2π R = nh, or p = n / R. If the particle is highly relativistic, nc E = pc = , n = 0, 1, 2, … . R f. The quantization condition gives p = n / R , and hence E = pc =

nc , R

same as for a circular orbit.

8015 Consider the density matrix ρ (t ) = |ψ (t ) >< ψ (t ) | corresponding to the state

|ψ (t ) > of the system. Show that it satisfies the time-evolution equation dρ i = − [H, ρ ] dt  Sol: Using the Schrodinger equation, H |ψ > = i express the density matrix as follows:

∂ |ψ > and its conjugate, we ∂t

dρ ∂ ∂ = | ψ (t ) >< ψ (t ) | +|ψ (t ) >< ψ (t ) | dt ∂t ∂t dρ i = − ( |Hψ (t ) >< ψ (t ) | −|ψ (t ) >< ψ (t ) H |) dt  dρ i = − ( Hρ − ρ H ) dt  dρ i = − [H, ρ] dt 

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8016 To find approximate eigenfunctions of the Hamiltonian H we can use trial functions of the form ψ = ∑ k =1akφk in the variational method (where the φk are n

given functions, and the ak are parameters to be varied). Show that one gets n solutions ψ α with energies ε α = ψ α | H |ψ α / ψ α ∣ψ α , where H is the Hamiltonian. We will order them so that ε1 ≤ ε 2 ≤ ε 3 . . . Show from the Hermitian properties of the Hamiltonian that the ψ α either automatically have or can be chosen to have the properties 〈ψ α |ψ β 〉 = δ αβ , 〈ψ α | H |ψ β 〉 = ε α δ αβ . From the fact that one can certainly find a linear combination of ψ 1 and ψ 2 which is orthogonal to ψ 1 , the exact ground state of H with eigenvalue E1, prove that

ε 2 ≥ E2 , where E2 is the exact energy of the first excited state. (Wisconsin) Sol: Suppose {φk } is the set of linearly independent functions. We may assume that 〈φi | φ j 〉 = δ ij , first using Schmidt’s orthogonalization process if necessary. Then n

H=

∑ a* a λ

ψ | H |ψ = ψ |ψ

i

i,j n

j

ij

∑ a* a δ i

i,j

= ∑ xi* λij x j = X + λˆ X ,

j ij

i,j

where xi =

ai

∑a j

Note that

2

, λij = φi | H | φ j = λ *ji .

j

n

∑x i =1

2 i

= 1.

ˆ , such that As λˆ is Hermitian, we can choose a rotational transformation X = pY ˆ = pˆ + λˆ pˆ = pˆ −1λˆ pˆ is a diagonal matrix with diagonal elements Λ ≤ Λ ≤ Λ . Λ 11

22

33

Then n

H = ∑ Λii yi , 2

i =1

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Miscellaneous Topics

where yi satisfy

∑

n i =1

641

2

yi = 1.

Applying the variational principle      2 2 0 = δ  H − α  ∑ yi − 1  = δ  ∑ ( Λii − α ) yi + α  ,   i  i   where α is the Lagrange multiplier, we get

∑(Λ i

or

− α ) yi δ yi = 0,

ii

( Λii − α ) yi

= 0,

(i = 1, 2, ⋅⋅⋅)

i.e., α = Λii or yi = 0. Hence the solutions of the variational equations are

α = Λii , y (ji ) = δ ij = δ ij , (i = 1, 2, ⋅⋅⋅, n). Thus we get n solutions ψ a , the α th solution yi(α ) = δ i(α ) corresponding to energy

ψ α | H |ψ α 2 = ∑ Λii yi(α ) = Λαα ψ α |ψ α i

εα = with ε1 ≤ ε 2 ≤ ε 3 ... .

For ψ α = ψ α [ X (Y )] , we have

ψ α |ψ β = ∑ ai(α ) *ai( β ) i

=

∑a

(α ) j

∑a

(α ) j

j

=

j

ψ α | H |ψ β = ∑ a

(α )* i

i,j

= =

(β ) j

∑a

(β ) j

j

2

j

2

* ⋅ ∑ xi(α ) xi( β ) i

2

 2 * ⋅ ∑ yi(α ) yi( β ) =  ∑ a(jα )  δ αβ ,  j  i

λa

(α ) j

∑a

(α ) j

j

∑a

(β ) ij j

∑a j

2

2

∑a

(β ) j

∑a

(β ) j

j

2

j

2

⋅ ∑ xi(α )* λij x (j β ) i,j

2

⋅ ∑ yi(α )* Λij y (j β ) i,j

 2 =  ∑ a(jα )  ε α δ αβ .  j  Then, by setting Ψα = ψ α /

∑

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2

j

a(jα ) , we have

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Problems and Solutions in Quantum Mechanics

Ψα | Ψ β = δ αβ , Ψα | H | Ψ β = ε α δ αβ ⋅ Let the exact wave functions of the ground state and the first excited state of H be Φ1 and Φ 2, their exact energies be E1 and E2 respectively. Then there must exist 2 2 two numbers µ1 and µ2 such that Φ1 = µ1Ψ1 + µ2 Ψ 2 , µ1 + µ2 = 1. From the orthogonality of Φ1 and Φ 2, we have Φ 2 = µ2* Ψ1 − µ1* Ψ 2, and hence 2

2

E1 = ε1 µ1 + ε 2 µ2 , E2 = ε1 µ2 + ε 2 µ1 = ( ε1 − ε 2 ) µ2 + ε 2 ≤ ε 2 . 2

2

2

8017 2 2

Find the value of the parameter λ in the trial function φ ( x ) = Ae − λ x , where A is a normalization constant, which would lead to the best approximation for the energy of the ground state of the one-particle Hamiltonian 2 d 2 H=− + bx 4 , where b is a constant. The following integrals may be 2m dx 2 useful:

∫ ∫

∞ −∞

∞ −∞

π ∞ 2 − ax 2 1 π , ∫ x e dx = , −∞ a 2 a3

2

e − ax dx = 2

x 4 e − ax dx =

3 π . 4 a5 (Wisconsin)

Sol:

2 2

Using the trial function φ = Ae − λ x , consider the integrals

∫

∞

−∞

∫

∞

−∞

∞

φ * ( x )φ ( x )dx = ∫ A2e −2 λ x dx = A2 2 2

−∞ ∞

φ * ( x )Hφ ( x )dx = ∫ A2e − λ −∞

2 2

x

π = 1, 2λ 2

 2 d 2 4  −λ2x2 dx  − 2m dx 2 + bx  e

2 ∞   2 2  = A 2 ∫  − ( 2λ 4 x 2 − λ 2 ) + bx 4  e −2 λ x dx −∞ m  

 2  1 = A 2  −  2λ 4 ⋅  m  2 

π

( 2λ )

2 3

− λ2

π  3  +b 2λ 2  4

π

( 2λ )

2 5

   

 1 2 π π 3  λ +b = A2  , 2 16λ 5  2 m 2

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643

and obtain 〈H 〉 =

∫ φ * Hφdx = 1   λ  ∫ φ *φdx 2  m 2

2

+b

3  . 8λ 4 

As 13 (a + b + c ) ≥ (abc )1/3 for positive numbers a, b, c, we have 1

1   2 λ 2  2 λ 2 3b  3   4 3b  3 . 〈H 〉 =  + + ≥ ⋅ 2  2m 2m 8λ 4  2  4m 2 8  Hence the best approximation for the energy of the ground state is 1

1

3  3  3  b 4  3 〈 H 〉 min =    2  . 4  4  m 

8018 Two electrons in a 10-potential well with a state ψ = state and expectation value of energy.

1 1 ψG + ψ FE . Find the 2 2

Ground state can be either ψ 12 or ψ 21. First excited state can be either ψ 13 or ψ 31. Sol: ⇒| ψ〉 =

πx1 πx 2  1 2 2 πx 2 2 2 πx1  a sin a sin a − a sin a sin a  + 2 1 2 3πx 2 2 3πx1 πx1 πx 2   sin a sin a − a sin a sin a  2 a

1 π 2 2 1 π 2 2 ψπψ = ⋅ (1 + 4 ) + (1 + 9 ) 2 2 2ma 2 2ma 2 15π 2  2 . = 4ma 2

8019 (Use nonrelativistic methods to solve this problem.) Most mesons can be described as bound quark-antiquark states ( qq ) . C ­ onsider the case of a meson made of a ( qq ) pair in an s-state. Let mq be the quark mass.

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Problems and Solutions in Quantum Mechanics

A + Br with r A < 0 and B > 0. You are asked to find a reasonable approximation to the ground state energy of this system in terms of A, B, mq and ћ. Unfortunately, Assume the potential binding the q to the q can be written as V =

for a class of trial functions appropriate to this problem, a cubic equation has to be solved. If this happens to you, and you do not want to spend your limited time trying to solve such a cubic equation, you may complete your solution for the case A = 0 (without loss of credit). Please express your final answer in terms of a numerical constant, which you should explicitly evaluate, multiplying a function of B, mq and  . (Berkeley) Sol: Method I Use for the trial function the wave function of a ground state hydrogen atom.

ψ ( r ) = e − r /a and calculate H = ψ | H |ψ / ψ | ψ ∞  2 1 ∂  2 ∂  = ∫ drr 2e − r /a  − r  2 0  2 µ r ∂r  ∂r  ∞

+ Ar −1 + Br  e − r /a / ∫ drr 2e −2r /a 0

=

3Ba  2 1 A + + , 2 2µ a2 a

mq where µ = is the reduced mass of the qq. system. Vary a to minimize H by  2 δH = 0, which gives letting δa 2 3 3 Ba − Aa − = 0. µ 2 1/3

 2 2  When A = 0, the solution is a =  . Hence the estimated ground state  3Bµ  energy is 3  36 B 2  2  Eg = H =   4  mq 

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1/3

1/3

 B2 2  = 2.48   .  mq 

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645

Method II Another estimate of the energy of the ground state can be obtained from the uncertainty principle. Consider H=

p2 A + + Br . 2µ r

As the principle requires    px x ≥ , p y y ≥ , pz z ≥ , 2 2 2 we take the equal sign for the ground state and obtain H=

2 2 2 A + + + + Br . 2 2 2 8µ x 8µ y 8µ z r

To minimize H, let ∂H = 0, ∂x i.e.

−  2 Ax Bx − + = 0. 4µx 2 r 3 r

As H is symmetric with respect to x, y, z, when it reaches the optimal value, we have x = y = z , or r = 3x , and the above equation becomes − Letting A = 0 we get

2 B A − + = 0. 3 2 4 µ x 3 3x 3

1  2  x = 36   4 µ B 

1/3

 9 2  ,  or  r =   4 µ B 

1/3

,

Hence 1

1/3

 9  2 B2  3   2 B2  3 2 + = = H= Br 2 3.30     . 8µx 2  2 mq   mq 

8020 An attractive potential well in one dimension satisfies +∞

V ( x ) < 0, ∫ V ( x )dx finite,  −∞

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∫

+∞ −∞

x 2V ( x )dx finite.

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Problems and Solutions in Quantum Mechanics

−βx a. Using trial wave functions of the form e has at least one bound state.

(∫

+∞ −∞

2

/2

, prove that the potential

b. Assuming further that the potential is quite weak V ( x )dx , ∫

+∞

−∞

(∫

+∞ −∞

V ( x )dx , ∫

x 2V ( x )dx are both ”small”), find the best upper bound (for the

energy) for this class of trial functions.

c. In a dimensionless statement, state precisely what is meant by ”small” in part (b). (Berkeley) Sol: a. The given trial function is the ground state wave function of a onedimensional harmonic oscillator. We shall use the normalized function β ψ (x ) =   π

where β = H=−

1/4

2

e − β x /2 ,

mω . The Hamiltonian can be written as 

2 d 2 1 1 1 + mω 2 x 2 + V ( x ) − mω 2 x 2 = H 0 + V ( x ) − mω 2 x 2 . 2 2m dx 2 2 2

As 1 ψ H 0 ψ = ω , 2 we have 1 1 H = 〈ψ | H |ψ 〉 = ω + ψ V ( x ) − mω 2 x 2 ψ 2 2 1 2 β + 〈ψ | V ( x )|ψ , = ω + 〈ψ | V ( x )|ψ 〉 = 4 4m and

δ H 2 1 = + 〈ψ | V ( x )|ψ 〉 − ψ x 2V ( x ) ψ . δβ 4m 2β Since when β → 0 1 1 β 〈ψ | V |ψ 〉 = 2β 2β π

∫

∞

−∞

Vdx → −∞

as V is negative,

ψ x 2V ψ =

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β π

∫

∞

−∞

x 2Vdx → 0,

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+∞ −∞

x 2V ( x )dx

Miscellaneous Topics

647

2 δH → > 0 . Therefore we have δ H → −∞ as β → 0. When β → ∞ , 4m δβ δβ δH = 0 at least for a certain positive β , say β 0 . Thus the trial function is suitδβ able and the energy for the corresponding state is E = H ( β0 ) = =−

 2   2β0 + 2β 0  − + ψ x 2V ( x ) ψ   4m  4m

 2β0 + 2β 0 ψ x 2V ( x ) ψ < 0. 4m

Therefore the system has at least one bound state. Note that we have used the δH = 0 , which gives, for β = β 0 , fact   δβ  β0  2  〈ψ | V ( x )|ψ 〉 = 2β 0  − + ψ x 2V ( x ) ψ  .  4m  b. c. Let

∫

∞

−∞

∞

V ( x )dx = A, ∫ x 2V ( x )dx = B. The requirement that A and B are small −∞

means that the potential V ( x ) can have large values only in the region of small | x |. Furthermore, for large | x |, V ( x ) must attenuate rapidly. This means that we can expand the integrals ∞

A1 = ∫ e − β x V ( x )dx  ∫ 2

−∞ ∞

∞

−∞

(1 − β x )V (x )dx = A − β B, 2

∞

B1 = ∫ x 2e − β x V ( x )dx  ∫ x 2V ( x )dx = B, 2

−∞

−∞

Then the minimization condition

δH = 0 gives δβ

2 β A1 + − B1 = 0, π 4m 2 πβ or 2 A 3 β + − B = 0, 4m 2 πβ 2 π i.e.

β=

A π 2 π 4 + + . 2 2 12mB 144m B 3B

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Problems and Solutions in Quantum Mechanics

Hence the bound state energy is estimated to be E=−

 2β β B + 2β 4m π 2

 π 2 A   −2 4 AB  π 4 4 = + + − +   . 2 2 2 144m B 3B   12m 36m 3π   12mB As A and B are both negative, E < 0. Hence 2

 π 2 π 4 A   2  E