Instructors’ Solution Manual Introduction to Quantum ...

1 Introduction to Quantum mechanics 3rd Edition Griffiths Solutions Manual Full Download: Instructors' Solution Manual Introduction to Quantum mechanics , 3rd ed. David Griffiths, Darrell Schroeter Reed College August 3, 2018. This sample only, Download all chapters at: 2. Contents 1 The Wave Function 4. 2 The Time-Independent Schro dinger Equation 16. 3 Formalism 78. 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. 3. Preface These are our own solutions to the problems in Introduction to Quantum mechanics , 3rd ed. We have made every effort to insure that they are clear and correct, but errors are bound to occur, and for this we apologize in advance. We would like to thank the many people who pointed out mistakes in the solu- tion Manual for the second edition, and encourage anyone who finds defects in this one to alert us (grif- or We especially thank Kenny Scott, Alain Thys, and Sergei Walter, who found many errors in the 2nd edition Solution Manual .)

2 We maintain a list of errata on the web page ( ), and incorporate corrections in the Manual itself from time to time. We also thank our students for many useful suggestions, and Neelaksh Sadhoo, who did much of the typesetting for the second edition. David Griffiths and Darrell Schroeter 4 CHAPTER 1. THE WAVE FUNCTION. Chapter 1. The Wave Function Problem (a). hji2 = 212 = 441. 1 X 2 1 2. hj 2 i = (14 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 ).. j N (j) =. N 14. 1 6434. = (196 + 225 + 768 + 968 + 1152 + 3125) = = 14 14. j j = j hji 14 14 21 = 7. 15 15 21 = 6. (b) 16 16 21 = 5. 22 22 21 = 1. 24 24 21 = 3. 25 25 21 = 4. 1 X 1 . 2 = ( j)2 N (j) = ( 7)2 + ( 6)2 + ( 5)2 3 + (1)2 2 + (3)2 2 + (4)2 5.. N 14. 1 260. = (49 + 36 + 75 + 2 + 18 + 80) = = 14 14.. = = (c). hj 2 i hji2 = 441 = [Agrees with (b).]. CHAPTER 1. THE WAVE FUNCTION 5. Problem (a). h h h2. Z . 2 2 1 1 2 5/2. hx i = x dx = x = . 0 2 hx 2 h 5 0 5. 2. 2 2 h2 2 h 4 2 2h = hx i hxi = = h = = 5 3 45 3 5.

3 (b). x+ x+. 1 . Z. 1 1. P =1 dx = 1 (2 x) =1 x+ x . x 2 hx 2 h x h x+ hxi + = + = ; x hxi = = . P =1 + = Problem (a). Z . 2. 1= Ae (x a) dx. Let u x a, du = dx, u : .. Z r r u2 . 1=A e du = A A= .. (b). Z Z . 2 2. hxi = A xe (x a) dx = A (u + a)e u du .. Z Z r . u 2. u2 . =A ue du + a e du = A 0 + a = a.. Z . 2. hx2 i = A x2 e (x a) dx . Z Z Z . 2 2 2. =A u2 e u du + 2a ue u du + a2 e u du . r r . 1 1. =A + 0 + a2 = a2 + . 2 2 . 1 1 1. 2 = hx2 i hxi2 = a2 + a2 = ; = . 2 2 2 . 6 CHAPTER 1. THE WAVE FUNCTION. (c). l(x). A. a x Problem (a). ( a b). a b |A|2 |A|2 x3 (b x)3. Z Z . 2 2 2 1 1. 1= 2 (b x) dx = |A|. x dx +. 2 + . a 0(b a) a a2 3 0 (b a)2 3 a r 2 a b a 2b 3. = |A| + = |A| A= . 3 3 3 b (b). ^. A. a b x (c) At x = a. (d). a a |A|2. Z Z . 2 2 a 2a P = 1 if b = a, X. P = | | dx = 2 x dx = |A| = . 0 a 0 3 b P = 1/2 if b = 2a. X. (e). Z Z a Z b . 1 1. hxi = x| |2 dx = |A|2 2 x3 dx + x(b x) 2. dx a 0 (b a)2 a ( 4 a b). 2. x3 x4.. 3 1 x 1 2x = + b 2b +.

4 B a2 4 0 (b a)2 2 3 4 a 3. a2 (b a)2 + 2b4 8b4 /3 + b4 2a2 b2 + 8a3 b/3 a4.. = 2. 4b(b a). 4 . 3 b 2 2 2 3 1 2a + b = 2. a b + a b = 2. (b3 3a2 b + 2a3 ) = . 4b(b a) 3 3 4(b a) 4. CHAPTER 1. THE WAVE FUNCTION 7. Problem (a).. e 2 x |A|2 . Z Z . 2 2 2 x 2. 1= | | dx = 2|A| e dx = 2|A| = ; A= . 0 2 0 . (b). Z Z . hxi = 2. x| | dx = |A| 2. xe 2 |x| dx = 0. [Odd integrand.].. Z . 2 2 2 2 x 2 1. hx i = 2|A| x e dx = 2 = . 0 (2 )3 2 2. (c). 1 1 . 2 = hx2 i hxi2 = ; = . | ( )|2 = |A|2 e 2 = e 2 / 2 . = e 2. = . 2 2 2 . |^| 2. h .24h x <m +m Probability outside: . e 2 x . Z Z . 2 | |2 dx = 2|A|2 e 2 x dx = 2 = e 2 = e 2. = 2 . Problem For integration by parts, the differentiation has to be with respect to the integration variable in this case the differentiation is with respect to t, but the integration variable is x. It's true that x 2 . (x| |2 ) = | | + x | |2 = x | |2 , t t t t but this does not allow us to perform the integration: Z b Z b b x | |2 dx = (x| |2 )dx 6= (x| |2 ) a.

5 A t a t 8 CHAPTER 1. THE WAVE FUNCTION. Problem dhpi 2 2 .. R . From Eq. , dt = i~ t x dx. But, noting that x t = t x and using Eqs. : i~ 2 i~ 2 .. i i = + = + V + V . t x t x x t 2m x2 ~ x x 2m x2 ~. 3 2 .. i~ i . = 3 + V (V ). 2m x x2 x ~ x x The first term integrates to zero, using integration by parts twice, and the second term can be simplified to V V 2 V. x V x x = | | x . So Z. dhpi i V V. = i~ | |2 dx = h i. QED. dt ~ x x Problem 2 2. Suppose satisfies the Schro dinger equation without V0 : i~ ~ . t = 2m x2 + V . We want to find the Solution 2 2. 0 . 0 with V0 : i~ t = 2m ~. x2 + (V + V0 ) 0 . 0. Claim: 0 = e iV0 t/~ . iV t/~ h ~2 2 i Proof: i~ . t 0. = i~ iV0 t/~. t e + i~ iV0. ~ e 0. = 2m x 2 + V e iV0 t/~ + V0 e iV0 t/~. 2 2. ~ 0. = 2m x2 + (V + V0 ) 0 . QED. This has no effect on the expectation value of a dynamical variable, since the extra phase factor, being inde- pendent of x, cancels out in Eq. Problem (a). Z r r 1/4. 2 2amx2 /~ 21 ~ 2am 1 = 2|A| e dx = 2|A| = |A|2 ; A=.

6 0 2 (2am/~) 2am ~. (b). 2 2amx2.. 2amx 2am 2am = ia ; = ; = +x = 1 . t x ~ x2 ~ x ~ ~. 2 2. Plug these into the Schro dinger equation, i~ ~ . t = 2m x2 + V : ~2 2amx2.. 2am V = i~( ia) + 1 . 2m ~ ~. 2amx2.. = ~a ~a 1 = 2a2 mx2 , so V (x) = 2ma2 x2 . ~. CHAPTER 1. THE WAVE FUNCTION 9. (c). Z . hxi = x| |2 dx = 0. [Odd integrand.].. Z r 2 2 2 2amx2 /~ 2 1 ~ ~. hx i = 2|A| x e dx = 2|A| 2 = . 0 2 (2am/~) 2am 4am dhxi hpi = m = 0. dt 2. 2 . Z Z. ~ . 2. hp i = dx = ~2 2 dx i x x 2. Z Z Z . 2 2am 2amx 2 2am 2 2. = ~ 1 dx = 2am~ | | dx x | | dx ~ ~ ~.. 2am 2 2am ~ 1. = 2am~ 1 hx i = 2am~ 1 = 2am~ = am~. ~ ~ 4am 2. (d). r ~ ~ . x2 2. = hx i hxi = 2. = x = ; p2 = hp2 i hpi2 = am~ = p = am~. 4am 4am q ~.. x p = 4am am~ = ~2 . This is (just barely) consistent with the uncertainty principle. Problem From Math Tables: = . P (0) = 0 P (1) = 2/25 P (2) = 3/25 P (3) = 5/25 P (4) = 3/25. (a). P (5) = 3/25 P (6) = 3/25 P (7) = 1/25 P (8) = 2/25 P (9) = 3/25.

7 N (j). In general, P (j) = N . (b) Most probable: 3. Median: 13 are 4, 12 are 5, so median is 4. 1. Average: hji = 25 [0 0 + 1 2 + 2 3 + 3 5 + 4 3 + 5 3 + 6 3 + 7 1 + 8 2 + 9 3]. 1 118. = 25 [0 + 2 + 6 + 15 + 12 + 15 + 18 + 7 + 16 + 27] = 25 = 1. (c) hj 2 i = 25 [0 + 12 2 + 22 3 + 32 5 + 42 3 + 52 3 + 62 3 + 72 1 + 82 2 + 92 3]. 1 710. = 25 [0 + 2 + 12 + 45 + 48 + 75 + 108 + 49 + 128 + 243] = 25 = . 2 = hj 2 i hji2 = = = ; = = 10 CHAPTER 1. THE WAVE FUNCTION. Problem (a). r 1 2. mv 2 + V = E v(x) = (E V (x)) . 2 m (b). b b Z r Z. 1 m 1. T = q dx = p dx. a 2. E 1 2 k a (2E/k) x2. m 2 kx p Turning points: v = 0 E = V = 12 kb2 b = 2E/k; a = b. r Z b r b r m 1 m 1 x m T =2 dx = 2 sin =2 sin 1 (1). k 0 2. b x 2 k b 0 k r r m m =2 = . k 2 k 1 1 (x). (x) = pmq 2 1. = b2 x 2 . E 2. k m 2 kx 1. Z b Z b 2 1 2 . (x) dx = dx = = 1. X. a 0 b2 x 2 2. x -b b (c) hxi = 0. b x2 2 b x2. Z Z. 1. hx2 i = . dx = dx b b2 x2 0 b2 x2. 2.. xp 2 b2 x b b2 b2 b2 E.

8 = b x2 + sin 1 = sin 1 (1) = = = . 2 2 b 0 2 2 k r p p b E. x = hx2 i hxi2 = hx2 i = = . 2 k Problem (a). dt |dt/dp| dp = (p) dp =. T T. where dt is now the time it spends with momentum in the range dp (dt is intrinsically positive, but dp/dt = F = kx runs negative hence the absolute value). Now s . p2 p2.. 1 2 2. + kx = E x = E , 2m 2 k 2m CHAPTER 1. THE WAVE FUNCTION 11. so 1 1 1. (p) = r = p = p , pm 2 p2 2mE p2 c2 p2. k k k E 2m . where c 2mE. This is the same as (x) (Problem (b)), with c in place of b (and, of course, p in place of x). c2 c . (b) From Problem (c), hpi = 0, hp2 i = , p = = mE . 2 2. E . r r m E 1 1. (c) x p = mE = E = . If E 2 ~ , then x p 2 ~, which is precisely the Heisenberg k k . uncertainty principle! x@t_D := snapshots = Table@x@ p RandomReal@jDD, 8j, 10 000<D. Problem 8- , , - , - , - , - , - , 9987 , x@t_D := Cos@tD , - , , , - , <. snapshots = Table@x@ p RandomReal@jDD, 8j, 10 000<D. Histogram@snapshots, 100, "PDF", PlotRange 80, 2<D.

9 8- , , - , - , - , - , - , 9987 , , - , , , - , <. Histogram@snapshots, 100, "PDF", PlotRange 80, 2<D. 1. r@x_D :=. p 1 - x2. Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<D. 1. r@x_D :=. p 1 - x2 Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<D. 12 CHAPTER 1. THE WAVE FUNCTION. 1. r@x_D :=. p 1 - x2. Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<D. 2 Show@Histogram@snapshots, 100, "PDF", PlotRange 80, 2<D, Plot@ r@xD, 8x, - 1, 1<, PlotRange -> 80, 2<DD. Problem Rb dPab Rb . (a) Pab (t) = a | (x, t)|2 dx, so dt = a t | |2 dx. But (Eq. ): | |2 .. i~ . = = J(x, t). t x 2m x x x Z b dPab b = J(x, t)dx = [J(x, t)]|a = J(a, t) J(b, t). QED. dt a x Probability is dimensionless, so J has the dimensions 1/time, and units seconds 1 . 2 . iat df iat df (b) Here (x, t) = f (x)e iat , where f (x) Ae amx /~. , so . x = f e dx e = f dx , df and . x = f dx too, so J(x, t) = 0. CHAPTER 1. THE WAVE FUNCTION 13. Problem Use Eqs. [ ] and [ ], and integration by parts: Z Z Z.

10 D 1 2. 1 2 dx = ( 1 2 ) dx = 2 + 1 dx dt t t t Z . i~ 2 1 i~ 2 2.. i i = + V 1 2 + 1 V 2 dx 2m x2 ~ 2m x2 ~. Z 2 2.. i~ 1 2. = 2 1 dx 2m x2 x2. " #. Z Z . i~ 1 1 2 2 1 2. = 2 dx 1 + dx = 0. QED. 2m x x x x x x Problem (a). a a 3 a x5. Z Z . 2 2 2 2 2 4 2x 4 2 2 4 2.. 1 = |A| a x dx = 2|A| a 2a x + x dx = 2|A| a x 2a +. a 0 3 5 0. r 2 1 16 5 2 15. = 2|A|2 a5 1 + = a |A| , so A = . 3 5 15 16a5. (b). Z a hxi = x| |2 dx = 0. (Odd integrand.). a (c). Z a d 2. ~ 2. a2 x2 a x2 dx = 0.. hpi = A (Odd integrand.). i a |dx {z }. 2x Since we only know hxi at t = 0 we cannot calculate dhxi/dt directly. (d). Z a 2. Z a hx2 i = A2 x2 a2 x2 dx = 2A2 a4 x2 2a2 x4 + x6 dx . a 0. 3 5. a x7.. 15 4x 2x 15 7 1 2 1. =2 a 2a + = a +. 16a5 3 5 7 0 8a5 3 5 7. 2 35 42 + 15 a2 8 a2.. 15a =.. = = . 8 5. 3 7 8 7 7. 14 CHAPTER 1. THE WAVE FUNCTION. (e). Z a d2 2. Z a hp2 i = A2 ~2 a 2 x2 a x2 dx = 2A2 ~2 2 a2 x2 dx . 2. a |dx {z } 0. 2. a x3 15~2 3 a3 15~2 2 5 ~2.. 15 2 2.