Griffiths D J Introduction to Quantum Mechanics Solutions 2nd Ed Pearson s PDF

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Contents Preface 2 1 The Wave Function 3 2 Time-Independent Schrödinger Equation 14 3 Formalism 62 4 Quantum Mechanics in Three Dimensions 87 5 Identical Particles 132 6 Time-Independent Perturbation Theory 154 7 The Variational Principle 196 8 The WKB Approximation 219 9 Time-Dependent Perturbation...

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Contents Preface

2

1 The Wave Function

3

2 Time-Independent Schrödinger Equation

14

3 Formalism

62

4 Quantum Mechanics in Three Dimensions

87

5 Identical Particles

132

6 Time-Independent Perturbation Theory

154

7 The Variational Principle

196

8 The WKB Approximation

219

9 Time-Dependent Perturbation Theory

236

10 The Adiabatic Approximation

254

11 Scattering

268

12 Afterword

282

Appendix Linear Algebra

283

2nd Edition – 1st Edition Problem Correlation Grid

299

2

Preface These are my own solutions to the problems in Introduction to Quantum Mechanics, 2nd ed. I have made every effort to insure that they are clear and correct, but errors are bound to occur, and for this I apologize in advance. I would like to thank the many people who pointed out mistakes in the solution manual for the first edition, and encourage anyone who finds defects in this one to alert me (griffi[email protected]). I’ll maintain a list of errata on my web page (http://academic.reed.edu/physics/faculty/griffiths.html), and incorporate corrections in the manual itself from time to time. I also thank my students at Reed and at Smith for many useful suggestions, and above all Neelaksh Sadhoo, who did most of the typesetting. At the end of the manual there is a grid that correlates the problem numbers in the second edition with those in the first edition.

David Griffiths

c
2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 1. THE WAVE FUNCTION

3

Chapter 1

The Wave Function Problem 1.1 (a)
j2 = 212 = 441.
j 2  = =

 1  2 1  2 j N (j) = (14 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 ) N 14 1 6434 (196 + 225 + 768 + 968 + 1152 + 3125) = = 459.571. 14 14 j 14 15 16 22 24 25

(b)

σ2 = =

σ=

∆j = j −
j 14 − 21 = −7 15 − 21 = −6 16 − 21 = −5 22 − 21 = 1 24 − 21 = 3 25 − 21 = 4

 1  1  (∆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) = = 18.571. 14 14

√

18.571 = 4.309.

(c)
j 2  −
j2 = 459.571 − 441 = 18.571.

[Agrees with (b).]

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4

CHAPTER 1. THE WAVE FUNCTION

Problem 1.2 (a) 

h

1 1 x2 √ dx = √ 2 hx 2 h


x2  = 0

h2 σ =
x  −
x = − 5 2

2

2

(b)



h h2 2 5/2  = x .  5 5 0

 2 2h 4 2 h = h ⇒ σ = √ = 0.2981h. 3 45 3 5

 √ x+ 1 1 √ 1 √

√ dx = 1 − √ (2 x) = 1 − √ x+ − x− . 2 hx 2 h h x−

x+

P =1−



x−

x+ ≡
x + σ = 0.3333h + 0.2981h = 0.6315h; √

P =1−

0.6315 +

√

x− ≡
x − σ = 0.3333h − 0.2981h = 0.0352h.

0.0352 = 0.393.

Problem 1.3 (a)



∞

Ae−λ(x−a) dx. 2

1=

Let u ≡ x − a, du = dx, u : −∞ → ∞.

−∞





∞

−λu2

1=A

e

du = A

−∞

(b)


x = A

∞



π λ

⇒ A=

xe−λ(x−a) dx = A 2

−∞ ∞

 =A

ue

−λu

2




x  = A

∞

2

−∞ ∞

−λu2

e

  π du = A 0 + a = a. λ



x2 e−λ(x−a) dx 2

−∞ ∞

=A

2 −λu2

u e





∞

du + 2a

−∞

1 =A 2λ

(u + a)e−λu du

−∞



∞

du + a

−∞ 2



λ . π

π + 0 + a2 λ

σ 2 =
x2  −
x2 = a2 +



π λ

ue

−λu2



1 1 − a2 = ; 2λ 2λ

∞

du + a

−∞

= a2 +

2

−λu2

e

 du

−∞

1 . 2λ 1 σ=√ . 2λ

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2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 1. THE WAVE FUNCTION

5

(c) ρ(x) A

x

a

Problem 1.4 (a) |A|2 1= 2 a



a

|A|2





b

(b − x) dx = |A| 2 (b − a) a 

3 b−a 2 a 2b = |A| + = |A| ⇒ A= . 3 3 3 b 2

x dx +

2

2

0

(b)

1 a2



 a b  (b − x)3  x3  1 − +  3 0 (b − a)2 3 a

Ψ A

a

b

x

(c) At x = a. (d) 

a

|Ψ|2 dx =

P = 0

|A|2 a2



a

x2 dx = |A|2 0

a a = . 3 b



P = 1 if b = a,
P = 1/2 if b = 2a.


(e)   a   b 1 1 2 x|Ψ|2 dx = |A|2 2 x3 dx + x(b − x) dx a 0 (b − a)2 a   4 a  b  2 3 1 x  x3 x4  1 2x = b − 2b + + b a2 4 0 (b − a)2 2 3 4 a  2  3 = a (b − a)2 + 2b4 − 8b4 /3 + b4 − 2a2 b2 + 8a3 b/3 − a4 2 4b(b − a)  4  3 b 2a + b 2 3 1 2 2 = (b3 − 3a2 b + 2a3 ) = −a b + a b = . 2 2 4b(b − a) 3 3 4(b − a) 4 


x =

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6

CHAPTER 1. THE WAVE FUNCTION

Problem 1.5 (a) 



∞

|Ψ| dx = 2|A| 2

1=

2

−2λx

e

 2

dx = 2|A|

0

∞ e−2λx  |A|2 = ; −2λ 0 λ

A=

√

λ.

(b) 
x =

 x|Ψ| dx = |A| 2


x  = 2|A| 2

2

2

∞

∞

xe−2λ|x| dx = 0.

[Odd integrand.]

−∞

2 −2λx

x e 0

1 2 = dx = 2λ . 3 (2λ) 2λ2

(c) σ 2 =
x2  −
x2 =

1 ; 2λ2

√

1 σ=√ . 2λ

|Ψ(±σ)|2 = |A|2 e−2λσ = λe−2λ/

λ

2λ

√

= λe−

2

= 0.2431λ.

|Ψ| 2

.24λ

−σ

+σ

x

Probability outside: 

∞

2

 |Ψ|2 dx = 2|A|2

σ

∞

e−2λx dx = 2λ



σ

∞ √ e−2λx  = e−2λσ = e− 2 = 0.2431.  −2λ σ

Problem 1.6 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|Ψ|2 ) = |Ψ| + x |Ψ|2 = x |Ψ|2 , ∂t ∂t ∂t ∂t but this does not allow us to perform the integration:  a

b

∂ x |Ψ|2 dx = ∂t

 a

b

b ∂ (x|Ψ|2 )dx = (x|Ψ|2 )a . ∂t

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CHAPTER 1. THE WAVE FUNCTION

Problem 1.7 From Eq. 1.33, ∂ ∂t



dp dt

∗ ∂Ψ

Ψ



∂x



= −i


∂ ∂t



7

Ψ∗ ∂Ψ ∂x dx. But, noting that

∂2Ψ ∂x∂t

=

∂2Ψ ∂t∂x

and using Eqs. 1.23-1.24:

 

∂Ψ∗ ∂Ψ ∂Ψ i
∂ 2 Ψ∗ i
∂ 2 Ψ i i ∗ ∂ ∗ ∂Ψ ∗ ∂ = +Ψ = − +Ψ + VΨ − VΨ ∂t ∂x ∂x ∂t 2m ∂x2
∂x ∂x 2m ∂x2




3 2 ∗ i
∂ Ψ ∂ Ψ ∂Ψ ∂Ψ ∂ i = Ψ∗ 3 − + V Ψ∗ − Ψ∗ (V Ψ) 2 2m ∂x ∂x ∂x
∂x ∂x

The first term integrates to zero, using integration by parts twice, and the second term can be simplified to ∗ ∂Ψ ∗ ∂V 2 ∂V V Ψ∗ ∂Ψ ∂x − Ψ V ∂x − Ψ ∂x Ψ = −|Ψ| ∂x . So d
p = −i
dt

  ∂V i ∂V −|Ψ|2 dx =
− .
∂x ∂x

QED

Problem 1.8
∂ Ψ Suppose Ψ satisfies the Schr¨ odinger equation without V0 : i
∂Ψ ∂t = − 2m ∂x2 + V Ψ. We want to find the solution 2 2
∂ Ψ0 0 Ψ0 with V0 : i
∂Ψ ∂t = − 2m ∂x2 + (V + V0 )Ψ0 . 2

2

Claim: Ψ0 = Ψe−iV0 t/
.

 2 2 

∂Ψ −iV0 t/

∂ Ψ −iV0 t/
0 Proof: i
∂Ψ + i
Ψ − iV
0 e−iV0 t/
= − 2m + V0 Ψe−iV0 t/
∂t = i
∂t e ∂x2 + V Ψ e 2


∂ Ψ0 = − 2m ∂x2 + (V + V0 )Ψ0 . 2

QED

This 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)  2

∞

1 = 2|A|

−2amx2 /


e

21

dx = 2|A|

0

2



π = |A|2 (2am/
)



π
; 2am

 A=

2am π


1/4 .

(b) ∂Ψ = −iaΨ; ∂t

∂Ψ 2amx =− Ψ; ∂x


∂2Ψ 2am =− ∂x2




∂Ψ Ψ+x ∂x



2am =−




2amx2 1−


 Ψ.


∂ Ψ Plug these into the Schr¨ odinger equation, i
∂Ψ ∂t = − 2m ∂x2 + V Ψ: 2

2

  
2 2am 2amx2 V Ψ = i
(−ia)Ψ + − 1− Ψ 2m

  2amx2 =
a −
a 1 − Ψ = 2a2 mx2 Ψ, so


V (x) = 2ma2 x2 .

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8

CHAPTER 1. THE WAVE FUNCTION (c) 
x =

∞

x|Ψ|2 dx = 0.

[Odd integrand.]

−∞



∞


x  = 2|A| 2

2

2 −2amx2 /


x e 0


p = m

1 dx = 2|A| 2 2 (2am/
)



2

π

= . 2am 4am

d
x = 0. dt 

2 
∂ ∂2Ψ 2 Ψ∗ 2 dx
p  = Ψ Ψdx = −
i ∂x ∂x       2 2am 2amx 2am 2 ∗ 2 2 2 = −
Ψ − 1− Ψ dx = 2am
|Ψ| dx − x |Ψ| dx


      2am 2 2am
1 = 2am
1 −
x  = 2am
1 − = 2am
= am
.

4am 2 

∗

2

(d) σx2


=⇒ σx = =
x  −
x = 4am 2

 σx σp =



2


4am

√


; 4am

σp2 =
p2  −
p2 = am
=⇒ σp =

√

am
.

am
=
2 . This is (just barely) consistent with the uncertainty principle.

Problem 1.10 From Math Tables: π = 3.141592653589793238462643 · · · (a)

P (0) = 0 P (5) = 3/25

P (1) = 2/25 P (6) = 3/25

In general, P (j) =

=

1 25 [0

(c)
j 2  = =

1 25 [0

P (3) = 5/25 P (8) = 2/25

1 25 [0

Median: 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] = 1 25 [0

P (4) = 3/25 P (9) = 3/25

N (j) N .

(b) Most probable: 3. Average:
j =

P (2) = 3/25 P (7) = 1/25

118 25

= 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 =
j 2  −
j2 = 28.4 − 4.722 = 28.4 − 22.2784 = 6.1216;

710 25

= 28.4. √ σ = 6.1216 = 2.474.

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CHAPTER 1. THE WAVE FUNCTION

9

Problem 1.11 (a) Constant for 0 ≤ θ ≤ π, otherwise zero. In view of Eq. 1.16, the constant is 1/π.  1/π, if 0 ≤ θ ≤ π, ρ(θ) = 0, otherwise.

ρ(θ) 1/π −π/2

π

0

θ 3π/2

(b) 
θ =

θρ(θ) dθ =

1
θ  = π



π

1 π



1 θ dθ = π

2

π

θdθ = 0



2

0

σ 2 =
θ2  −
θ2 =

1 π



π π θ2  = 2 0 2

[of course].

π θ3  π2 . = 3 0 3

π2 π2 π2 − = ; 3 4 12

π σ= √ . 2 3

(c) 1
sin θ = π


cos θ =

1 π



π

sin θ dθ =

2 1 1 π (− cos θ)|0 = (1 − (−1)) = . π π π

cos θ dθ =

1 π (sin θ)|0 = 0. π

0



1
cos θ = π

π

0



2

π

1 cos θ dθ = π



2

0

π

(1/2)dθ = 0

1 . 2

[Because sin2 θ + cos2 θ = 1, and the integrals of sin2 and cos2 are equal (over suitable intervals), one can replace them by 1/2 in such cases.]

Problem 1.12 (a) x = r cos θ ⇒ dx = −r sin θ dθ. The probability that the needle lies in range dθ is ρ(θ)dθ = probability that it’s in the range dx is ρ(x)dx =

1 π dθ,

so the

1 dx dx 1 dx  = √ . = 2 π r sin θ π r 1 − (x/r) π r2 − x2

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10

CHAPTER 1. THE WAVE FUNCTION ρ(x)

-2r



√ 1 , π r 2 −x2

∴ ρ(x) =

Total:

0,

r

√ 1 dx −r π r 2 −x2

(b)
x =

1 π



2 π


x2  =

r

−r

 0

x√

r

√

=

r2

2 π

-r

r

if − r < x < r, otherwise.

r 0

√ 1 dx r 2 −x2

1 dx = 0 − x2

2 π

=

x

2r

[Note: We want the magnitude of dx here.] sin−1

 x r r 0

=

2 π

sin−1 (1) =

2 π

·

π 2

= 1.


[odd integrand, even interval].

 x  r 2 2 x2 2 r2 x 2  = 2 r sin−1 (1) = r . dx = r − x2 + − sin−1  π 2 2 r π 2 2 r2 − x2 0

√ σ 2 =
x2  −
x2 = r2 /2 =⇒ σ = r/ 2. To get
x and
x2  from Problem 1.11(c), use x = r cos θ, so
x = r
cos θ = 0,
x2  = r2
cos2 θ = r2 /2.

Problem 1.13 Suppose the eye end lands a distance y up from a line (0 ≤ y < l), and let x be the projection along that same direction (−l ≤ x < l). The needle crosses the line above if y + x ≥ l (i.e. x ≥ l − y), and it crosses the line below if y + x < 0 (i.e. x < −y). So for a given value of y, the probability of crossing (using Problem 1.12) is 

−y

P (y) = −l

1 = π



−1

sin



l

1 ρ(x)dx + ρ(x)dx = π l−y



−y

−l

1 √ dx + 2 l − x2



l

l−y



1 √ dx 2 l − x2

 x −y  l   1  −1 x  = − sin−1 (y/l) + 2 sin−1 (1) − sin−1 (1 − y/l)  + sin  l −l l l−y π

sin−1 (y/l) sin−1 (1 − y/l) − . π π Now, all values of y are equally likely, so ρ(y) = 1/l, and hence the probability of crossing is        1 l l−y 1 l −1 y −1 P = π − sin π − 2 sin−1 (y/l) dy − sin dy = πl 0 l l πl 0 =1−

=

l   1 2 2 2  πl − 2 y sin−1 (y/l) + l 1 − (y/l)2  = 1 − [l sin−1 (1) − l] = 1 − 1 + = . πl πl π π 0

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CHAPTER 1. THE WAVE FUNCTION

11

Problem 1.14 (a) Pab (t) =

b a

|Ψ(x, t)2 dx,

so

dPab dt

=

b

∂ |Ψ|2 dx. a ∂t

But (Eq. 1.25):

  ∂|Ψ|2 ∂ i
∂Ψ∗ ∂ ∗ ∂Ψ = Ψ − Ψ = − J(x, t). ∂t ∂x 2m ∂x ∂x ∂t

∴

dPab =− dt



b

a

∂ b J(x, t)dx = − [J(x, t)]|a = J(a, t) − J(b, t). ∂x

QED

Probability 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−amx

2

/


∗

df −iat df iat , so Ψ ∂Ψ = f dx , ∂x = f e dx e

df and Ψ∗ ∂Ψ ∂x = f dx too, so J(x, t) = 0.

Problem 1.15 (a) Eq. 1.24 now reads

∂Ψ∗ ∂t

2

∗

i
∂ Ψ i ∗ ∗ = − 2m ∂x2 +
V Ψ , and Eq. 1.25 picks up an extra term:

∂ i i 2Γ 2 |Ψ|2 = · · · + |Ψ|2 (V ∗ − V ) = · · · + |Ψ|2 (V0 + iΓ − V0 + iΓ) = · · · − |Ψ| , ∂t


 2Γ ∞ 2Γ 2 and Eq. 1.27 becomes dP QED dt = −
−∞ |Ψ| dx = −
P . (b) dP 2Γ 2Γ
= − dt =⇒ ln P = − t + constant =⇒ P (t) = P (0)e−2Γt/
, so τ = . P

2Γ

Problem 1.16 Use Eqs. [1.23] and [1.24], and integration by parts: d dt



∞

−∞

Ψ∗1 Ψ2

  ∞ ∗ ∂Ψ1 ∂ ∗ ∗ ∂Ψ2 (Ψ1 Ψ2 ) dx = Ψ2 + Ψ1 dx ∂t ∂t −∞ ∂t −∞     ∞  −i
∂ 2 Ψ∗1 i
∂ 2 Ψ2 i i ∗ ∗ Ψ dx + + Ψ − V Ψ V Ψ 2 2 1 1 2m ∂x2
2m ∂x2
−∞   ∞ 2 ∗ 2 i
∂ Ψ1 ∗ ∂ Ψ2 − dx Ψ − Ψ 2 1 2m −∞ ∂x2 ∂x2   ∞ ∞  ∞  ∞  i
∂Ψ∗1  ∂Ψ∗1 ∂Ψ2 ∂...