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  • Introduction to Quantum Mechanics, 3rd ed.
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  • Introduction To Quantum Mechanics, 3rd Ed.

Introduction to Quantum Mechanics Third Edition Instructor's Solution Manual David Griffiths | Complete. Contents 1 The Wave Function 4 2 The Time-Independent Schr¨odinger Equation 16 3 Formalism 78 4 Quantum Mechanics in Three Dimensions 109 5 Identical Particles 168 6 Symmetries and Conse...

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  • Introduction to Quantum Mechanics, 3rd ed.
  • Introduction to Quantum Mechanics, 3rd ed.
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Instructors’ Solution Manual
Introduction to Quantum Mechanics, 3rd ed.

David Griffiths, Darrell Schroeter
Reed College

August 3, 2018

,2




Contents

1 The Wave Function 4

2 The Time-Independent Schrödinger 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-
fith@reed.edu or schroetd@reed.edu). 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 page
(http://academic.reed.edu/physics/faculty/griffiths.html), 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 1.1
(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) = = 459.571.
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 
σ2 = (∆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


σ= 18.571 = 4.309.

(c)

hj 2 i − hji2 = 459.571 − 441 = 18.571. [Agrees with (b).]

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