Problems solutions in quantum mechanics

This page intentionally left blank www.pdfgrip.com PROBLEMS AND SOLUTIONS IN QUANTUM MECHANICS This collection of solved problems corresponds to the standard topics covered in established undergraduate and graduate courses in quantum mechanics Completely up-to-date problems are also included on topics of current interest that are absent from the existing literature Solutions are presented in considerable detail, to enable students to follow each step The emphasis is on stressing the principles and methods used, allowing students to master new ways of thinking and problem-solving techniques The problems themselves are longer than those usually encountered in textbooks and consist of a number of questions based around a central theme, highlighting properties and concepts of interest For undergraduate and graduate students, as well as those involved in teaching quantum mechanics, the book can be used as a supplementary text or as an independent self-study tool Kyriakos Tamvakis studied at the University of Athens and gained his Ph.D at Brown University, Providence, Rhode Island, USA in 1978 Since then he has held several positions at CERN’s Theory Division in Geneva, Switzerland He has been Professor of Theoretical Physics at the University of Ioannina, Greece, since 1982 Professor Tamvakis has published 90 articles on theoretical high-energy physics in various journals and has written two textbooks in Greek, on quantum mechanics and on classical electrodynamics This book is based on more than 20 years’ experience of teaching the subject www.pdfgrip.com www.pdfgrip.com PROBLEMS AND SOLUTIONS IN QUANTUM MECHANICS KYRIAKOS TAMVAKIS University of Ioannina www.pdfgrip.com    Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge  , UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521840873 © K Tamvakis 2005 This publication is in copyright Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press First published in print format 2005 - - ---- eBook (EBL) --- eBook (EBL) - - ---- hardback --- hardback - - ---- paperback --- paperback Cambridge University Press has no responsibility for the persistence or accuracy of s for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate www.pdfgrip.com Contents 10 Preface Wave functions The free particle Simple potentials The harmonic oscillator Angular momentum Quantum behaviour General motion Many-particle systems Approximation methods Scattering Bibliography Index page vii 17 32 82 118 155 178 244 273 304 332 333 v www.pdfgrip.com www.pdfgrip.com Preface This collection of quantum mechanics problems has grown out of many years of teaching the subject to undergraduate and graduate students It is addressed to both student and teacher and is intended to be used as an auxiliary tool in class or in selfstudy The emphasis is on stressing the principles, physical concepts and methods rather than supplying information for immediate use The problems have been designed primarily for their educational value but they are also used to point out certain properties and concepts worthy of interest; an additional aim is to condition the student to the atmosphere of change that will be encountered in the course of a career They are usually long and consist of a number of related questions around a central theme Solutions are presented in sufficient detail to enable the reader to follow every step The degree of difficulty presented by the problems varies This approach requires an investment of time, effort and concentration by the student and aims at making him or her fit to deal with analogous problems in different situations Although problems and exercises are without exception useful, a collection of solved problems can be truly advantageous to the prospective student only if it is treated as a learning tool towards mastering ways of thinking and techniques to be used in addressing new problems rather than a solutions manual The problems cover most of the subjects that are traditionally covered in undergraduate and graduate courses In addition to this, the collection includes a number of problems corresponding to recent developments as well as topics that are normally encountered at a more advanced level vii www.pdfgrip.com www.pdfgrip.com 320 Problems and Solutions in Quantum Mechanics we obtain δ =− 2mg k h¯ ∞ dr r e−µr j (kr ) = − 2mg k 2h¯ ∞ d x x e−µx/k j (x) For s-waves, we get ∞ mg d x e−µx/k (1 − cos 2x) 2 k h¯ mg =− + (µ/2k)2 µk¯h δ0 = − This is negative for our repulsive potential Problem 10.9 (a) Prove the identity (±) G (±) (E) = G (±) (E) (E) + V G with −1 G (±) , (E) ≡ (E ± i − H0 ) G (±) (E) ≡ (E ± i − H )−1 (b) From the equation |ψk(±) = |k + G (±) (E)V |k derive the operator scattering equation (the Schwinger–Lipmann equation) (±) |ψk(±) = |k + G (±) (E)V |ψk (c) Establish the orthonormality property ψk(±) |ψk(±) = k |k = δ(k − k ) (d) If k = k, prove the relation ψk(−) |V |k = k |V |ψk(+) (e) Introduce the operator T (E) ≡ V + V G (+) (E)V and show that T (E) − T † (E) = −2πi V δ(E − H ) V www.pdfgrip.com 10 Scattering 321 Solution −1 (a) Multiplying by G −1 from the right, we arrive at the from the left and G identity −1 +V G −1 = G ↔ E − H0 = E − H + V (b) We have |k = (1 + GV )−1 |ψk This, substituted back, gives |ψk = |k + GV (1 + GV )−1 |ψk This would be the Schwinger–Lipmann equation if GV (1 + GV )−1 = G V which is equivalent to GV = G V (1 + GV ) or GV = G (1 + V G)V This, thanks to the identity proved in (a), is always true (c) We have ψk |ψk = ψk | [1 + G(E)V ] |k = ψk |k + ψk |(E − H )−1 V |k = ψk |k + ψk |(E − E )−1 V |k = ψk |k − ψk |V (E − H0 )−1 |k = ψk |k − ψk |V G (E )|k = ψk | − V G (E ) |k = k |k = δ(k − k ) (d) ψk(−) |V |k = k | + G (−) (E)V † V |k = k | + V G (+) (E) V |k = k |V + G (+) (E)V |k = k |V |ψk(+) (e) 1 − V E +i − H E −i − H −2i V = −2πi V δ(E − H )V =V (E − H )2 + T (E) − T † (E) = V www.pdfgrip.com 322 Problems and Solutions in Quantum Mechanics e k′ k θ −e Fig 45 Scattering from a dipole Problem 10.10 Consider an electric dipole consisting of two electric charges e and −e at a mutual distance 2a Consider also a particle of charge e and mass m with an incident wave vector k perpendicular to the direction of the dipole; see Fig 45 (a) Calculate the scattering amplitude in the Born approximation Find the directions at which the differential cross section is maximal (b) Consider a different system with a target consisting of two arbitrary charges q1 and q2 similarly placed Calculate again the scattering amplitude and the directions of maximal scattering Solution (a) The potential created by the dipole is V (r) = − e2 e2 + |r + a| |r − a| We have placed the charge −e at −a and the charge e at a The scattering amplitude in the Born approximation is d 3r ir·q 1 4π m 4π me2 + − k |V |k = − e 2 (2π ) |r + a| |r − a| h¯ h¯ d 3r ir·q 4π me2 −e−iq·a + eiq·a e =− (2π) r h¯ f k (k ) = − We have denoted q ≡ k − k The integral involved is d 3r eir·q ∞ = 2π r dr r −1 (d cos θ) eiqr cos θ = 4π q2 Thus, the scattering amplitude is f k (k ) = − 4ime2 2me2 −iq·a iq·a −e = − + e h¯ q h¯ www.pdfgrip.com sin q · a q2 10 Scattering 323 Taking k = k zˆ and a = a xˆ , we have k = k sin θ xˆ + k cos θ zˆ , q = 2k (1 − cos θ) and q · a = −ka sin θ Thus f k (k ) = 4m e4 sin2 (ka sin θ ) (¯h k)4 (1 − cos θ)2 The cross section becomes maximal when 2a sin θ = λ π (2n + 1) = (2n + 1) k (b) The scattering amplitude will be 4π me d 3r ir·q e q2 e−iq·a + q1 eiq·a (2π)3 r h¯ 2me = − 2 q2 e−iq·a + q1 eiq·a h¯ q me (q1 + q2 ) cos(ka sin θ) + i(q1 − q2 ) sin(ka sin θ ) =− (¯h k)2 − cos θ f k (k ) = − The differential cross section is m e2 (q1 + q2 )2 cos2 (ka sin θ) + (q1 − q2 )2 sin2 (ka sin θ) (¯h k)4 (1 − cos θ)2 2 2 m e q1 + q2 + 2q1 q2 cos(2ka sin θ) = (¯h k)4 (1 − cos θ )2 f k (k ) = For q1 q2 > 0, maximal cross section is achieved at 2a sin θ = n 2π = nλ k while if q1 q2 < it is achieved at 2a sin θ = (2n + 1) π λ = (2n + 1) k Problem 10.11 Consider the scattering of particles of mass m from an attractive potential that has a constant strength −V0 within a sphere of radius R but vanishes elsewhere Calculate the differential and the total cross section for k R (i.e small energies) Solution The phase shift is given by the formula tan δ = j (k R) − α j (k R) n (k R) − α n (k R) www.pdfgrip.com 324 Problems and Solutions in Quantum Mechanics where α is the logarithmic derivative8 at r = R In the limit k R small-argument behaviour of the spherical Bessel functions, we get tan δ ≈ − (2 + 1) (k R)2 [(2 + 1)!!]2 +1 1, using the α R− α R+ +1 These phase shifts are small; the scattering will be dominated by small values of The differential cross section is dσ ∼ |δ0 + 3δ1 P1 (cos θ ) + · · ·|2 ∼ C0 + C1 cos θ + · · · d k where R α02 δ02 = R k2 (1 + Rα0 )2 α0 R(−1 + Rα1 ) 6δ0 δ1 C1 = = 2R (k R)2 k (1 + Rα0 )(2 + Rα1 ) C0 = and (−1 + Rα1 )(1 + Rα0 ) C1 = 2(k R)2 C0 Rα0 (2 + Rα1 ) It is clear that C1 /C0 in the limit of low energies The total cross section will be σ = d (C0 + C1 cos θ + · · ·) ≈ 4πC0 = 4π R = 4π R α0 R + α0 R tan q R 1− qR Problem 10.12 The scattering amplitude of a particle of mass m in a potential V (r ) can be written as f k (ˆr) = − 4mπ k V ψk(+) h¯ where ψk(+) is the scattering wave function, which satisfies the integral equation9 (+) ψk(+) = k + G (+) (E) V ψk Write down the Born expansion of the scattering amplitude Using the optical The logarithmic derivative at r = R is α = j (q R)/ j (q R) The wave number q is defined as q ≡ 2m(E + V0 )/¯h The Green’s function operator is defined as G (+) (E) = E +i − H www.pdfgrip.com 10 Scattering 325 theorem calculate the total cross section for the potential V (r ) = g e−µ r to the lowest non-trivial order Solution The scattering amplitude in the forward direction is 4π m k V ψk(+) h¯ 4π m k|V |k + k|V G (+) =− (E)V |k + · · · h¯ Its imaginary part will be f k (k) = − 4π m + ··· Im [ k|V |k ] + Im k|V G (+) (E)V |k h¯ 1 2π m − k V V k =− E + i − H0 E − i − H0 i¯h Im [ f k (k)] = − 2π m (−2i) k V V k (E − H0 )2 + i¯h 4π m 4π m k|V δ(E − H )V |k = k|V d q |q q|δ(E − H0 )V |k = h¯ h¯ 4π m d q k|V |q q|V |k δ (E(k) − E(q)) = h¯ 4π m = d q | k|V |q |2 δ (E(k) − E(q)) h¯ 8π m = d q | k|V |q |2 δ(q − k ) h¯ Let us now calculate the potential matrix element, first setting Q ≡ k − q We have d 3r −iQ·r−µr k|V |q = g e (2π)3 ∞ g2 −µr = dr r e (d cos θ) e−i Qr cos θ 4π −1 ∞ ig = drr e−(µ+i Q)r − e−(µ−i Q)r 4π Q ∞ ig ∂ dr e−(µ+i Q)r − e−(µ−i Q)r =− 4π Q ∂µ 1 ig ∂ − =− 4π Q ∂µ µ + i Q µ − i Q 1 µg ig = − = 2 2 4π Q (µ + i Q) (µ − i Q) π (µ + Q )2 =− www.pdfgrip.com 326 Problems and Solutions in Quantum Mechanics For q = k, we may write Q = 2k (1 − cos θ) Thus, the imaginary part of the scattering amplitude is 8g µ2 m π δ(q − k ) d 3q )4 (µ + Q h¯ 1 8g µ2 m π ∞ dq qδ(q − k) (d cos θ) = (µ + Q )4 h¯ −1 8g kµ2 m π 1 = (d cos θ) (µ + 2k − 2k cos θ)4 h¯ −1 g µ2 m π 1 = (d cos θ) 2 (−µ /2k − + cos θ)4 2k h¯ −1 Im [ f k (k)] = = 3µ2 (µ2 + 4k ) + 16k (µ2 + 4k )3 g m kπ h¯ µ4 16 The total cross section is σ = 3µ2 (µ2 + 4k ) + 16k (µ2 + 4k )3 64π g m 3¯h µ4 Problem 10.13 Consider a one-dimensional potential that vanishes beyond some point, i.e V (x) = for |x| ≥ a > (a) The scattering wave functions satisfy the following integral equation10 ψk(+) (x) = φk (x) + √ ∞ −∞ (+) d x G (+) (x − x )V (x )ψk (x ) where φk (x) = eikx / 2π Determine the Green’s function G (+) (x − x ) Show that it can be written as the position matrix element of an operator (b) Show that in the asymptotic region |x| a, we can write eik|x| ψk (x) ∼ φk (x) + √ f (k, k ) 2π Determine the scattering amplitude f (k, k ) in terms of V and ψ (c) What is the connection of the scattering amplitude f (k, k ) to the reflection and transmission coefficients, familiar from standard one-dimensional problems? Prove the relation Re [ f (k, k)] = − 12 | f (k, k)|2 + | f (k, −k)|2 10 The wave number corresponds to the energy in the standard way, E = h¯ k /2m www.pdfgrip.com 10 Scattering 327 (d) Consider the exactly soluble problem for which V (x) = gδ(x) Calculate the scattering amplitude in the Born approximation and compare with the known exact answer Show that, in the attractive case, the Born approximation is valid only when the energy is large in comparison to the bound-state energy Solution (a) Acting on both sides of the integral equation with the operator H0 − E, we obtain (+) [H0 (x) − E] ψk (x) ∞ = [H0 (x) − E] φk (x) + −∞ (+) d x [H0 (x) − E] G (+) (x − x )V (x )ψk (x ) or −V (x)ψk(+) (x) = ∞ −∞ (+) d x [H0 (x) − E] G (+) (x − x )V (x )ψk (x ) which implies that (+) [E + i − H0 (x)] G (x − x ) = δ(x − x ) We have added the term +i to the energy in order to fix the boundary conditions at infinity Fourier transforming, we obtain dq iq(x−x ) h¯ q e E +i − 2π 2m dq iq(x−x ) e 2π ˜ (+) (q) = G and ˜ (+) (q) = G h¯ k h¯ q − +i 2m 2m −1 Thus finally we get G (+) (x −x)= = ∞ −∞ 2im h¯ dq iq(x−x ) h¯ k h¯ q e − +i 2π 2m 2m (x − x ) eik(x−x ) − 2k (x − x) −1 eik(x −x) (−2k) =− It is easy to see that the position matrix elements of the operator G (+) (E) = E + i − H0 give exactly the Green’s function x|G (+) (E)|x = = dq x|q q|G (+) (E)|x dq i(x−x )q e 2π h¯ q E +i − 2m www.pdfgrip.com −1 im ik|x−x | e h¯ k 328 Problems and Solutions in Quantum Mechanics (b) In the asymptotic region we can make the approximation |x − x | = (x − x )2 = x + (x )2 − 2x x ≈ |x| x = |x| − x ≈ |x| − x x x − 2x x Thus, defining k ≡ k|x|/x, we approximate the Green’s function as follows: G (+) (x − x ) = − Then we have ψk(+) (x) im ik|x−x | im e ≈ − eik|x|−ik x h¯ k h¯ k √ i 2πm ik|x| ikx e ≈√ e − h¯ k 2π d x e−ik x V (x )ψk(+) (x ) The scattering amplitude is √ mi 2π f (k, k ) = − h¯ k (c) In the far positive region, x d x e−ik x V (x )ψk(+) (x ) a, the wave function is eikx eikx + f (k, k) √ √ 2π 2π −a, it is In the far negative region, x e−ikx eikx + f (k, −k) √ √ 2π 2π From these expressions we can recognize immediately (1) the incident wave and current eikx √ , 2π Ji = h¯ k m(2π ) (2) the reflected wave and current e−ikx f (k, −k) √ , 2π Jr = − h¯ k | f (k, −k)|2 m(2π ) (3) the transmitted wave and current e−ikx [1 + f (k, k)] √ , 2π Jt = h¯ k |1 + f (k, k)|2 m(2π ) Thus, the reflection and transmission coefficients are T = Jt /Ji = |1 + f (k, k)|2 , R = |Jr |/Ji = | f (k, −k)|2 www.pdfgrip.com 10 Scattering 329 Probability conservation dictates that R+T =1 which is equivalent to |1 + f (k, k)|2 + | f (k, k)|2 = =⇒ | f (k, −k)|2 + | f (k, k)|2 = −2 Re[ f (k, k)] (d) The exact answer for the scattering amplitude of the delta function potential is f = −ig ig + h¯ k/m and the bound-state energy is Eb = − mg 2¯h On the other hand, the Born approximation to the scattering amplitude gives f B ≈ −i gm h¯ k Notice that the exact answer can be rewritten as (g < 0) f = −i √ i − E/|E b | |E b |, this is approximated by In the high-energy limit E mg |E b | = −i = f B E h¯ k f ≈i which coincides with the Born approximation result Problem 10.14 The neutron–proton scattering amplitude is of the form † f = χf a + 4¯h −2 b Sn · Sp χi where χi and χf are the initial and final states Calculate the cross section for neutron–proton scattering when the initial and final proton spins are not measured Solution The operator appearing in the amplitude is Sn · Sp = Sp(+) Sn(−) + Sp(−) Sn(+) + 2Spz Snz www.pdfgrip.com 330 Problems and Solutions in Quantum Mechanics The cross section corresponding to the scattering of neutrons from protons when the target spin is not measured will be m ,m m σm nn = f m pp, m nn mp, mp In particular, we have σ+ ≡ σ++ = m ,+ mp, mp f m pp, + and σ− ≡ σ+− = m ,− f m pp, + mp, mp The first relevant amplitude is m ,+ f m pp, + = m p m n = a + 4¯h −2 b Sn · Sp mn = mp = a + 2bm p δm p m p and leads to the cross section σ+ = a + 2bm p mp, mp a + 2bm p δm p m p = mp = | a + b | + | a − b | = 2|a|2 + 2|b|2 2 Similarly, we have for the other amplitude m ,− f m pp, + = m p m n = a + 4¯h −2 b Sn · Sp m n = − 12 m p = 2bδm p , 1/2 δm p ,−1/2 and for the corresponding cross section σ− = mp, mp The polarization 4|b|2 δm p , 1/2 δm p ,−1/2 = 4|b|2 will be = |a|2 − |b|2 σ+ − σ− = σ+ + σ− |a|2 + 3|b|2 Problem 10.15 Consider the scattering of a particle by a distribution of scattering centres Each scatterer is located at a point ri and scatters with a given potential V0 (|r − ri |) Write down the scattering amplitude in the Born approximation (a) Consider the case of a cube of side a with the scatterers placed at its eight vertices (b) Do the same for an infinite cubic lattice of lattice spacing a www.pdfgrip.com 10 Scattering 331 Solution The Born-approximation scattering amplitude is (q ≡ k − k ) f k (ˆr) = − m 2π¯h V0 (|r − ri |) d 3r eiq·r i m =− 2π¯h e d ρe iq·ri iq·ρ i √ m 2π ˜ V0 (ρ) = − V0 (q) h¯ eiq·ri i where V˜ is the Fourier transform of the given potential V0 (a) In the case of the cube, the sum is cos(aqx /2) cos(aq y /2) cos(aqz /2) The cross section will be qy a qx a qz a 64m (2π) ˜ dσ |V0 | cos2 = cos2 cos2 d 2 h¯ Maximal value is achieved when all qi = 2n i π/a (b) In the case of an infinite lattice, the sum is ∞ ∞ eiaqx n x n x =0 ∞ eiaq y n y n y =0 eiaqz n z n z =0 = (1 − eiaqx )−1 (1 − eiaq y )−1 (1 − eiaqz )−1 qy a qz a qx a sin sin = −8i e−i(qx +q y +qz )a/2 sin 2 The cross section will be dσ 64m(2π) ˜ qx a q y a qz a −1 |V0 | sin2 = sin sin d 2 h¯ Maximal (infinite) cross section corresponds to any of the momentum transfers qx = 2n x π , a qy = 2n y π , a qz = 2n z π a www.pdfgrip.com (n x , n y , n z = 1, 2, ) −1 Bibliography G Baym Lectures in Quantum Mechanics, New York: W A Benjamin, 1969 J S Bell Speakable and Unspeakable in Quantum Mechanics, Cambridge University Press, 1993 A Capri Problems and Solutions in Non-relativistic Quantum Mechanics, World Scientific, 2001 C Cohen-Tannoudji et al Quantum Mechanics, vols I and II, Wiley, 1977 F Constantinescu and E Magyari Problems in Quantum Mechanics, Pergamon Press, 1971 P A M Dirac Quantum Mechanics, 4th edn, London: Oxford University Press, 1958 S Flăugge Practical Quantum Mechanics, Springer-Verlag, 1971 S Gasiorowitcz Quantum Physics, New York: Wiley, 1996 I I Goldman and V D Krivchenkov Problems in Quantum Mechanics, New York: Dover Publications, 1993 K Gottfried Quantum Mechanics, vol 1, New York: W A Benjamin, 1966 W Greiner Quantum Mechanics: An Introduction, Springer-Verlag, 1989 L Landau and E M Lifshitz Quantum Mechanics, Reading MA: Addison-Wesley, 1965 F Mandl Quantum Mechanics, London: Butterworths Scientific Publications, 1957 A Messiah Quantum Mechanics, vols I and II, North Holland, 1970 E Merzbacher Quantum Mechanics, Wiley, 1970 J J Sakurai Modern Quantum Mechanics, Reading MA: Addison-Wesley, 1995 L Schiff Quantum Mechanics, New York, MacGraw-Hill, 1968 G L Squires Problems in Quantum Mechanics, Cambridge University Press, 1995 Yung-Kuo Lim (ed.) Problems and Solutions on Quantum Mechanics, World Scientific, 1998 332 www.pdfgrip.com Index addition of angular momenta, 125, 129, 146, 150, 153 angular momentum, 118 anharmonic perturbation, 275 Bell’s inequality, 157, 158, 161 Born approximation, 319, 322, 324, 330 central potential, 183, 206, 217 charged harmonic oscillator, 84 classical action, 297 coherent states, 100 Coulomb interaction screened at short distances, 273, 277 cross section, 306, 309, 330 helium atom, 249 hydrogen atom in a magnetic field, 118, 122, 131 hydrogen atom in an electric field, 279 hyperfine splitting interaction, 262, 264, 269 identical particles, 244, 247, 248 infinite square well, 42 infinite square well with a delta function, 55 infinite square well with time-dependent electric field, 287 instantaneous momentum transfer to a harmonic oscillator, 94 isotropic harmonic oscillator, 185, 281 linear potential, 74 delta function near a wall, 50 delta function potential, 45, 66, 70, 315 delta-shell potential, 190, 192, 198, 306 density matrix, 171 deuteron, 264, 267 dilatations, 212 double delta-shell potential, 311 double-well potential, 296 driven harmonic oscillator, 112 magnetic dipole moment, 137, 139, 261, 264 magnetic flux quantization, 242 many-particle systems, 244 neutrino oscillations, 161 neutron interferometer, 164, 174 neutron–proton scattering, 329 number–phase uncertainty relation, 104 Einstein–Podolsky–Rosen paradox, 157 electric dipole moment, 207, 255, 260, 279, 281 electric quadrupole moment, 207, 260, 267 electromagnetic transitions, 290, 293 energy–time uncertainty relation, 22 entanglement, 157, 175 expanding square well, 38 exponential decay, 171 free particle, 17 Gaussian wave function, 25, 27 hard sphere, scattering from, 308 harmonic oscillator, 82 harmonic oscillator dispersion, 90 harmonic oscillator with a delta function, 91 heavy quark–antiquark bound states, 203 one-dimensional scattering, 326 one-dimensional well, arbitrary shape, 63 operator scattering equation, 320, 324 optical theorem, 304, 309, 324 oscillating magnetic field, 141, 164 parabolic well, 64 partial scattering amplitude, 306 particle in a gravitational field, 239 particle in a magnetic field, 132, 134, 135, 146, 219, 223, 230, 237 particle in an electric field, 132, 134, 208, 215, 255, 281, 301 periodic system, phase difference due to magnetic flux, 222 phase shift, 306, 308, 318, 323 plane-wave superposition, potential step with a delta function, 47 333 www.pdfgrip.com 334 Index propagator, 8–10, 110, 208, 242, 296 proton–deuteron ion, 265 step potential, 39, 41 Stern–Gerlach analyzer, 157, 166 system of three spinors, 151 quantum behaviour, 155 quantum measurement, 157, 166 quantum Zeno effect, 156, 168 radial square well, 178, 194, 323 rotations, 124, 126 scattering, 304 scattering amplitude, 304 scattering integral equation, 311 Schwinger–Lipmann equation, 320 sinusoidal perturbation, 274 spin, 122, 139, 141, 144, 146, 148, 150, 153, 164, 230 spin–orbit interaction, 267 square barrier with a delta function, 73 square well, 52, 247 teleportation, 177 Thomas–Reiche–Kuhn sum rule, 196, 198 three-dimensional reflection, 32 three-dimensional refraction, 32 time-dependent uniform electric field, 285, 286 triangular well, 64 two-delta-function potential, 59, 61 two-dimensional harmonic oscillator, 187, 189 two-state system, 155, 171 variational method, 299, 303 vector operator, 126, 128 virial theorem, 181 wave function, WKB approximation, 77, 78, 79, 301 www.pdfgrip.com ... x which implies that k sin θ = k sin θ = q sin θ www.pdfgrip.com 34 Problems and Solutions in Quantum Mechanics or, equivalently, sin θ k = sin θ q θ = θ, (b) The continuity of the wave function... only if it is treated as a learning tool towards mastering ways of thinking and techniques to be used in addressing new problems rather than a solutions manual The problems cover most of the subjects... continuity equation www.pdfgrip.com h¯ tα x mk0 24 Problems and Solutions in Quantum Mechanics (e) We obtain2 h¯ α + h¯ k02 h¯ tα h¯ k0 t 1+ = + m 2α m p2 t = x2 t The corresponding uncertainties