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Introduction to Quantum Mechanics Solutions Manual David J Griffith Second Edition $30.99   Add to cart

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Introduction to Quantum Mechanics Solutions Manual David J Griffith Second Edition

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Introduction to Quantum Mechanics Solutions Manual David J Griffith Second Edition (Complete and Comprehensive Guide)

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  • June 24, 2023
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Introduction to
Quantum Mechanics
Solutions Manual



David J Griffith

Second Edition

,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 (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 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

, CHAPTER 1. THE WAVE FUNCTION 3




Chapter 1

The Wave Function

Problem 1.1
(a)

·j×2 = 212 = 441.

1 Σ 2 1
·j 2 × = j N (j) = (142 ) + (152 ) + 3(162 ) + 2(222 ) + 2(242 ) + 5(252 )
N 14
1 6434
= (196 + 225 + 768 + 968 + 1152 + 3125) = =
459.571.
14 14

j ∆j = j − ·j×
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 Σ 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)
·j 2 × − ·j× 2 = 459.571 − 441 = 18.571. [Agrees with (b).]

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