Quantum physics eisberg-resnick

Contents
Preface 1 The Wave Function 2 Time-Independent Schrödinger Equation 3 Formalism 4 Quantum Mechanics in Three Dimensions 5 Identical Particles 6 Time-Independent Perturbation Theory 7 The Variational Principle 8 The WKB Approximation 9 Time-Dependent Perturbation Theory 10 The Adiabatic Approximation 11 Scattering 12 Afterword Appendix Linear Algebra 2nd Edition – 1st Edition ProblemCorrelation Grid 2 3 14 62 87 132 154 196 219 236 254 268 282 283 299

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 forthe first edition, and encourage anyone who finds defects in this one to alert me (griffith@reed.edu). 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 thetypesetting. 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, withoutpermission in writing from the publisher.

CHAPTER 1. THE WAVE FUNCTION

3

Chapter 1

The Wave Function
Problem 1.1
(a) j
2

= 212 = 441. 1 N j 2 N (j) = 1 (142 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 ) 14

j2 = =

1 6434 (196 + 225 + 768 + 968 + 1152 + 3125) = = 459.571. 14 14 j 14 15 16 22 24 25 ∆j = j − j 14 − 21 = −7 15 − 21 = −6 16 − 21 = −5 22 − 21 = 1 24 − 21 = 3 25− 21 = 4

(b)

σ2 = =

1 N

(∆j)2 N (j) =

1 (−7)2 + (−6)2 + (−5)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5 14

1 260 (49 + 36 + 75 + 2 + 18 + 80) = = 18.571. 14 14

σ= (c)

√

18.571 = 4.309.

j2 − j

2

= 459.571 − 441 = 18.571.

[Agrees with (b).]

c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyrightlaws 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.

4

CHAPTER 1. THE WAVE FUNCTION

Problem 1.2
(a)
h

x2 =
0

1 1 x2 √ dx = √ 2 hx 2 h h2 − 5 h 3
2

2 5/2 x 5

h

=
0

h2 . 5

σ 2 = x2 − x (b) P =1−
x+ x−

2

=

=

2h 4 2 h ⇒ σ = √ = 0.2981h. 45 3 5

√ 11 √ dx = 1 − √ (2 x) 2 hx 2 h

x+ x−

1 √ √ =1− √ x+ − x− . h x− ≡ x − σ = 0.3333h − 0.2981h = 0.0352h.

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

P =1−

0.6315 +

0.0352 = 0.393.

Problem 1.3
(a)
∞

1=
−∞

Ae−λ(x−a) dx.
2

Let u ≡ x − a, du = dx, u : −∞ → ∞.

∞

1=A
−∞

e−λu du = A
2

π λ

⇒ A=

λ . π

(b)
∞

x =A =A

xe−λ(x−a) dx = A
2

∞ −∞(u + a)e−λu du
2 2

−∞ ∞ −∞

ue−λu du + a
2

∞ −∞

e−λu du = A 0 + a

π λ

= a.

x2 = A =A

∞

x2 e−λ(x−a) dx
2

−∞ ∞ −∞

u2 e−λu du + 2a
2

∞ −∞

ue−λu du + a2
2

∞ −∞

e−λu du
2

=A

1 2λ

π + 0 + a2 λ
2

π λ

= a2 +

1 . 2λ 1 σ=√ . 2λ

σ 2 = x2 − x

= a2 +

1 1 − a2 = ; 2λ 2λ

c 2005 Pearson Education, Inc., Upper Saddle River, NJ. Allrights 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 (c)
ρ(x) A

5

a

x

Problem 1.4
(a) 1= |A|2 a2
a

x2 dx +
0

|A|2 (b − a)
2 a

b

(b − x)2 dx = |A|2 3 . b

1 a2

x3 3...