Quantum Mechanics - 2 Time‐Independent Schrödinger Equation 14 ...

EXAM ELABORATIONS Aug 27, 2025
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Introduction to Quantum Mechanics Solutions Manual

David J Griffith

Second Edition 1 / 4

Contents

Preface 2

  • The Wave Function 3
  • Time‐Independent Schrödinger Equation 14
  • Formalism 62
  • Quantum Mechanics in Three Dimensions 87
  • Identical Particles 132
  • Time‐Independent Perturbation Theory 154
  • The Variational Principle 196
  • The WKB Approximation 219
  • 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 / 4

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 3 / 4

N 14 N 14

CHAPTER 1. THE WAVE FUNCTION 3

Chapter 1

The Wave Function

Problem 1.1 (a)

·j× 2 = 21 2 = ·j 2 × = 1 Σ j 2 N (j) = 1 (14 2

) + (15

2

) + 3(16

2

) + 2(22

2

) + 2(24

2

) + 5(25

2 )

= 1

(196 + 225 + 768 + 968 + 1152 + 3125) =

6434 =

(b) 14 14

j ∆j = j − ·j× 14 15 16 22 24 25

14 − 21 = −7

15 − 21 = −6

16 − 21 = −5

22 − 21 = 1

24 − 21 = 3

25 − 21 = 4

σ 2 = 1 Σ (∆j) 2 N (j) = 1

(−7)

2

+ (−6)

2

+ (−5)

2

· 3 + (1)

2

· 2 + (3)

2

· 2 + (4)

2 · 5

= 1

(49 + 36 + 75 + 2 + 18 + 80) =

260 =

14 14

σ = √

18.571 =

(c)

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

  • 2

4.309.

18.571.

459.571.

  • / 4

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Category: EXAM ELABORATIONS
Added: Aug 27, 2025
Description:

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 Me...

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