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Introduction to Quantum Mechanics 3rd Edition Griffiths Solutions ManualFull Download: s-manuInstructors’ Solution ManualIntroduction to Quantum Mechanics, 3rd ed.David Griffiths, Darrell SchroeterReed CollegeAugust 3, 2018This sample only, Download all chapters at: AlibabaDownload.com

2Contents1 The Wave Function42 The Time-Independent Schrödinger Equation163 Formalism784 Quantum Mechanics in Three Dimensions1095 Identical Particles1686 Symmetries and Conservation Laws1977 Time-Independent Perturbation Theory2358 The Variational Principle3019 The WKB Approximation33310 Scattering35411 Quantum Dynamics37212 Afterword420A Linear Algebra427

3PrefaceThese are our own solutions to the problems in Introduction to Quantum Mechanics, 3rd ed. We havemade every effort to insure that they are clear and correct, but errors are bound to occur, and for thiswe apologize in advance. We would like to thank the many people who pointed out mistakes in the solution manual for the second edition, and encourage anyone who finds defects in this one to alert us ([email protected] or [email protected]). We especially thank Kenny Scott, Alain Thys, and Sergei Walter,who found many errors in the 2nd edition solution manual. We maintain a list of errata on the web fiths.html), and incorporate corrections in the manual itself fromtime to time. We also thank our students for many useful suggestions, and Neelaksh Sadhoo, who did much ofthe typesetting for the second edition.David Griffiths and Darrell Schroeter

4CHAPTER 1. THE WAVE FUNCTIONChapter 1The Wave FunctionProblem 1.1(a)hji2 212 441.hj 2 i 1 X 21 2j N (j) (14 ) (152 ) 3(162 ) 2(222 ) 2(242 ) 5(252 )N1416434(196 225 768 968 1152 3125) 459.571.1414j141516222425(b)σ2 σ j j hji14 21 715 21 616 21 522 21 124 21 325 21 4 1 X1 ( j)2 N (j) ( 7)2 ( 6)2 ( 5)2 · 3 (1)2 · 2 (3)2 · 2 (4)2 · 5N142601(49 36 75 2 18 80) 18.571.1414 18.571 4.309.(c)hj 2 i hji2 459.571 441 18.571.[Agrees with (b).]

CHAPTER 1. THE WAVE FUNCTION5Problem 1.2(a)hZ211x dx 2 hx2 h2hx i 0h2 σ hx i hxi 5222 2 5/2x5 hh2.5 0 2h4 22h h σ 0.2981h.3453 5(b)x Z 11 dx 1 (2 x)2 hx2 hP 1 x x x 1 1 x x .hx hxi σ 0.3333h 0.2981h 0.6315h; P 1 0.6315 x hxi σ 0.3333h 0.2981h 0.0352h.0.0352 0.393.Problem 1.3(a)Z 2Ae λ(x a) dx.1 Let u x a, du dx, u : . Z 1 Aer λu2du A rπλ A λ.π(b)Z 2xe λ(x a) dx Ahxi A λuue2ZZ du a hx2 i Ae λu2 r π a.du A 0 aλ2x2 e λ(x a) dx Z2u2 e λu du 2a AZ 2(u a)e λu du Z A Z1 A2λr 2ue λu du a2 π 0 a2λσ 2 hx2 i hxi2 a2 rπλ Z 1 a2 .2λ11 a2 ;2λ2λ1σ .2λ 2e λu du

6CHAPTER 1. THE WAVE FUNCTION(c)l(x)AxaProblem 1.4(a) A 21 2aZa A 22Z(b22(b x) dx A 2(b a) ar b a32 a2b A A . A 333bx dx 01a2 x33 a01 (b a)2 (b x)3 3(b) Aabx(c) At x a.(d)aZ A 2 Ψ dx 2a2P 0Z0aax dx A .3b22a P 1 if b a, XP 1/2 if b 2a. X(e) Z a Z b112x3 dx x(b x)dxx Ψ 2 dx A 2 2a 0(b a)2 a( 4 a b)2x1x3x43 12x b 2b b a24 0 (b a)2234 a 3 a2 (b a)2 2b4 8b4 /3 b4 2a2 b2 8a3 b/3 a424b(b a) 42 313b2a b2 2 a b a b (b3 3a2 b 2a3 ) . 224b(b a)334(b a)4Zhxi b)a

CHAPTER 1. THE WAVE FUNCTION7Problem 1.5(a)Z2 Ψ dx 2 A 1 2 Ze 2λx2 dx 2 A 0e 2λx 2λ 0 A 2;λA λ.(b)Zhxi 2x Ψ dx A 2Z xe 2λ x dx 0.[Odd integrand.] 2hx i 2 A 2Z 2 2λxx e0 21 .dx 2λ(2λ)32λ2 (c)σ 2 hx2 i hxi2 1;2λ2 1σ .2λ Ψ( σ) 2 A 2 e 2λσ λe 2λ/2λ λe 2 0.2431λ. 2h.24h mx mProbability outside:Z 2 Ψ 2 dx 2 A 2σZ e 2λx dx 2λ σe 2λx 2λ e 2λσ e 2 0.2431.σProblem 1.6For integration by parts, the differentiation has to be with respect to the integration variable – in this case thedifferentiation is with respect to t, but the integration variable is x. It’s true that x 2 (x Ψ 2 ) Ψ x Ψ 2 x Ψ 2 , t t t tbut this does not allow us to perform the integration:Zab x Ψ 2 dx tZab b(x Ψ 2 )dx 6 (x Ψ 2 ) a . t

8CHAPTER 1. THE WAVE FUNCTIONProblem 1.7From Eq. 1.33, t Ψdhpidt Ψ xR i t Ψ Ψ x dx. But, noting that 2Ψ x t 2Ψ t xand using Eqs. 1.23-1.24: ii Ψ Ψ Ψi 2 Ψ i 2 Ψ Ψ VΨ VΨ Ψ Ψ t x x t2m x2 x x 2m x2 2 3 Ψ Ψi i Ψ Ψ Ψ 3 V Ψ Ψ (V Ψ)2m x x2 x x xThe first term integrates to zero, using integration by parts twice, and the second term can be simplified to Ψ V2 VV Ψ Ψ x Ψ V x Ψ x Ψ Ψ x . Sodhpi i dt Zi V V Ψ 2dx h i. x xQEDProblem 1.822 ΨSuppose Ψ satisfies the Schrödinger equation without V0 : i Ψ t 2m x2 V Ψ. We want to find the solution22 Ψ0 Ψ 0Ψ0 with V0 : i t 2m x2 (V V0 )Ψ0 .Claim: Ψ0 Ψe iV0 t/ .i iV t/ h 2 2 ΨiV0 Ψ iV0 t/ 00Proof: i Ψ i e i Ψ e VΨe iV0 t/ V0 Ψe iV0 t/ 2 t t 2m x22 Ψ0 2m x2 (V V0 )Ψ0 .QEDThis has no effect on the expectation value of a dynamical variable, since the extra phase factor, being independent of x, cancels out in Eq. 1.36.Problem 1.9(a)2Z 1 2 A 2amx2 / e21dx 2 A 02rπ A 2(2am/ )rπ ;2am A 2amπ 1/4.(b) Ψ iaΨ; t Ψ2amx Ψ; x 2Ψ2am x2 2 ΨΨ x x 2am 2 ΨPlug these into the Schrödinger equation, i Ψ t 2m x2 V Ψ: 22am2amx2V Ψ i ( ia)Ψ 1 Ψ2m 2amx2Ψ 2a2 mx2 Ψ, so a a 1 V (x) 2ma2 x2 . 2amx21 Ψ.

CHAPTER 1. THE WAVE FUNCTION9(c) Zx Ψ 2 dx 0.hxi [Odd integrand.] 2hx i 2 A 2 Z2 2amx2 / x e0hpi m1dx 2 A 22 (2am/ )r2π .2am4amdhxi 0.dt 2Z 2Ψhp i ΨΨdx 2 Ψ 2 dxi x x Z ZZ22am2amx2am2222 Ψ dx x Ψ dx Ψ 1 Ψ dx 2am 2am 22am 1 2am 1 hx i 2am 1 2am am . 4am2Z2 (d)σx2 hx i hxi σx 4am2σx σp r2q 4am ;4amσp2 hp2 i hpi2 am σp From Math Tables: π 3.141592653589793238462643 · · ·P (0) 0P (5) 3/25P (1) 2/25P (6) 3/25In general, P (j) Average: hji 125 [0(c) hj 2 i 125 [0P (2) 3/25P (7) 1/25P (3) 5/25P (8) 2/25125 [0Median: 13 are 4, 12 are 5, so median is 4.· 0 1 · 2 2 · 3 3 · 5 4 · 3 5 · 3 6 · 3 7 · 1 8 · 2 9 · 3] 2 6 15 12 15 18 7 16 27] 125 [0P (4) 3/25P (9) 3/25N (j)N .(b) Most probable: 3.am .am 2 . This is (just barely) consistent with the uncertainty principle.Problem 1.10(a) 11825 4.72. 12 · 2 22 · 3 32 · 5 42 · 3 52 · 3 62 · 3 72 · 1 82 · 2 92 · 3] 2 12 45 48 75 108 49 128 243] σ 2 hj 2 i hji2 28.4 4.722 28.4 22.2784 6.1216;71025 28.4. σ 6.1216 2.474.

10CHAPTER 1. THE WAVE FUNCTIONProblem 1.11(a)1mv 2 V E2r v(x) 2(E V (x)) .m(b)ZbT ar1q2mE 122 kx dx mkZab1pdx.(2E/k) x2pTurning points: v 0 E V 12 kb2 b 2E/k; a b.rr Z br b1mmm 1 x 2T 2dx 2sinsin 1 (1)22k 0kbk0b xr rm πm 2 π.k 2k1ρ(x) πZpmq 2kbρ(x) dx aE m2πbZ 0122 kx 11 π b2 x 2 .ρ(x)12 π 1. Xdx π 2b2 x 2-bbx(c) hxi 0.Zx2x22 b dx dxπ 0b2 x2b2 x2 b x b2xp 2b2b2b2 πb2E b x2 sin 1 sin 1 (1) .π22bππ 22k0hx2 i 1πZb ppbσx hx2 i hxi2 hx2 i 2rE.kProblem 1.12(a)dt dt/dp dp TTwhere dt is now the time it spends with momentum in the range dp (dt is intrinsically positive, butdp/dt F kx runs negative—hence the absolute value). Nows p21 22p2 kx E x E ,2m 2k2mρ(p) dp

CHAPTER 1. THE WAVE FUNCTION11soρ(p) πpmkk1r 2kE p2 1πp2mE p2 1πpc2 p2,2m where c 2mE. This is the same as ρ(x) (Problem 1.11(b)), with c in place of b (and, of course, p inplace of x).(b) From Problem 1.11(c), hpi 0, hp2 i E mE kuncertainty principle!rr(c) σx σp c2c, σp mE .22mEE . If E kω12 ω,then σx σp 12 ,which is precisely the [email protected] D : [email protected] [email protected]@ p [email protected], 8j, 10 000 DProblem 1.138- 0.977661, 0.371147, - 0.0389023, - 0.0257086, - 0.968682, - 0.681951, - 0.874704,[email protected] D : [email protected] 0.584236, - 0.974667, 0.10798, 0.996311, - 0.767106, 0.926913 Ü9987á,snapshots [email protected]@ p [email protected], 8j, 10 000 [email protected], 100, "PDF", PlotRange Æ 80, 2 D8- 0.977661, 0.371147, - 0.0389023, - 0.0257086, - 0.968682, - 0.681951, - 0.874704,2.0Ü9987á,0.584236, - 0.974667, 0.10798, 0.996311, - 0.767106, 0.926913 [email protected], 100, "PDF", PlotRange Æ 80, 2 [email protected] D : p1 - [email protected] [email protected], 8x, - 1, 1 , PlotRange - 80, 2 D-0.50.0 [email protected] D : p1 - [email protected] [email protected], 8x, - 1, 1 , PlotRange - 80, 2 D1.0

1.51.00.512CHAPTER 1. THE WAVE [email protected] D : 1 - [email protected] [email protected], 8x, - 1, 1 , PlotRange - 80, 2 snapshots, 100, "PDF", PlotRange Æ 80, 2 D,[email protected] [email protected], 8x, - 1, 1 , PlotRange - 80, 2 DD2.01.51.00.5-0.50.00.51.0Problem 1.14(a) Pab (t) Rba Ψ(x, t) 2 dx,sodPabdt Rb Ψ 2 dx.a tBut (Eq. 1.25): Ψ 2 i Ψ Ψ Ψ J(x, t).Ψ t x 2m x x xdPab dtZab bJ(x, t)dx [J(x, t)] a J(a, t) J(b, t). xQEDProbability is dimensionless, so J has the dimensions 1/time, and units seconds 1 .(b) Here Ψ(x, t) f (x)e iat , where f (x) Ae amxdfand Ψ Ψ x f dx too, so J(x, t) 0.2/ df iat df iat, so Ψ Ψ f dx, x f edx e

CHAPTER 1. THE WAVE FUNCTION13Problem 1.15Use Eqs. [1.23] and [1.24], and integration by parts:ddtZ Z Ψ1 Ψ2(Ψ 1 Ψ2 ) dx Ψ2 Ψ 1dx t t t Z ii i 2 Ψ 1i 2 Ψ2 VΨΨ ΨVΨdx22112m x2 2m x2 Z 2 2 Ψ1i Ψ2Ψ2 Ψ1dx 2m x2 x2"#Z Z i Ψ 1 Ψ 1 Ψ2 Ψ 1 Ψ2 Ψ2 Ψ2dx Ψ1dx 0. QED2m x x x x x x Ψ 1 Ψ2 dx ZProblem 1.16(a)aa 3x542x1 A a x dx 2 A a 2a x x dx 2 A a x 2a35 a0r 2 116 5 215 2 A 2 a5 1 a A , so A .3 51516a52Z2 2 22Z42 24 (b)Zahxi x Ψ 2 dx 0.(Odd integrand.) a(c)hpi 2AiZaa2 x2 a d 2 a x2 dx 0. dx {z}(Odd integrand.) 2xSince we only know hxi at t 0 we cannot calculate dhxi/dt directly.(d)hx2 i A2Za a 2x2 a2 x2 dx 2A2Za a4 x2 2a2 x4 x6 dx0 a3515x72x4x 2a 2a 16a5357 0 2 35 42 1515aa2 8 · 88 73 ·5 ·7 15 7 1 2 1a 8a53 5 7a2.7a20

14CHAPTER 1. THE WAVE FUNCTION(e)hp2 i A2 2Zaa 2 x2 a d2 2 a x2 dx 2A2 2 22 dx {z}Za a2 x2 dx0 215 2 4· 16a5 x32a x 3a0 15 2 3 a315 2 25 2 a ·. 4a534a2 32 a2(f )rσx pσp phx2 i hxi2 1 2aa .77(g)rhp2 i hpi2 5 2 2 a2r5 .2a(h)aσx σp ·7r5 2ar5 14r 10 .X7 22Problem 1.17(a) Eq. 1.24 now reads Ψ t2 i Ψi 2m x2 V Ψ , and Eq. 1.25 picks up an extra term: ii2Γ 2 Ψ 2 · · · Ψ 2 (V V ) · · · Ψ 2 (V0 iΓ V0 iΓ) · · · Ψ , t R2Γ 2Γ2and Eq. 1.27 becomes dPQEDdt Ψ dx P .(b)dP2Γ2Γ dt ln P t constant P (t) P (0)e 2Γt/ , so τ .P 2ΓProblem 1.18 hh2 d T .3mkB d23mkB T(a) Electrons (m 9.1 10 31 kg):T (6.6 10 34 )2 1.3 105 K.3(9.1 10 31 )(1.4 10 23 )(3 10 10 )2Silicon nuclei (m 28mp 28(1.7 10 27 ) 4.8 10 26 kg):T (6.6 10 34 )2 2.4 K.3(4.8 10 26 )(1.4 10 23 )(3 10 10 )2

Introduction to Quantum Mechanics 3rd Edition Griffiths Solutions ManualFull Download: s-manuCHAPTER 1. THE WAVE FUNCTION15(b) P V N kB T ; volume occupied by one molecule (N 1, V d3 ) d (kB T /P )1/3 .h2T 3mkB PkB T 2/3 T5/3h2 P 2/31 T 5/33m kkB Bh23m 3/5For helium (m 4mp 6.8 10 27 kg) at 1 atm 1.0 105 N/m2 :1T (1.4 10 23 ) (6.6 10 34 )23(6.8 10 27 ) 3/5(1.0 105 )2/5 2.8 K.For atomic hydrogen (m mp 1.7 10 27 kg) with d 0.01 m:T (6.6 10 34 )2 6.2 10 14 K.3(1.7 10 27 )(1.4 10 23 )(10 2 )2At 3 K it is definitely in the classical regime.This sample only, Download all chapters at: AlibabaDownload.comP 2/5 .

4 Quantum Mechanics in Three Dimensions 109 5 Identical Particles 168 6 Symmetries and Conservation Laws 197 7 Time-Independent Perturbation Theory 235 8 The Variational Principle 301 9 The WKB Approximation 333 10 Scattering 354 11 Quantum Dynamics 372 12 Afterword 420 A Linear Algebra 427