Solutions Griffiths Quântica
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Solutions Griffiths Quântica


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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 \u2013 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
e\ufb00ort 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 \ufb01rst edition,
and encourage anyone who \ufb01nds defects in this one to alert me (gri\ufb03th@reed.edu). I\u2019ll maintain a list of errata
on my web page (http://academic.reed.edu/physics/faculty/gri\ufb03ths.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 \ufb01rst edition.
David Gri\ufb03ths
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)
\u3008j\u30092 = 212 = 441.
\u3008j2\u3009 = 1
N
\u2211
j2N(j) =
1
14
[
(142) + (152) + 3(162) + 2(222) + 2(242) + 5(252)
]
=
1
14
(196 + 225 + 768 + 968 + 1152 + 3125) =
6434
14
= 459.571.
(b)
j \u2206j = j \u2212 \u3008j\u3009
14 14\u2212 21 = \u22127
15 15\u2212 21 = \u22126
16 16\u2212 21 = \u22125
22 22\u2212 21 = 1
24 24\u2212 21 = 3
25 25\u2212 21 = 4
\u3c32 =
1
N
\u2211
(\u2206j)2N(j) =
1
14
[
(\u22127)2 + (\u22126)2 + (\u22125)2 · 3 + (1)2 · 2 + (3)2 · 2 + (4)2 · 5]
=
1
14
(49 + 36 + 75 + 2 + 18 + 80) =
260
14
= 18.571.
\u3c3 =
\u221a
18.571 = 4.309.
(c)
\u3008j2\u3009 \u2212 \u3008j\u30092 = 459.571\u2212 441 = 18.571. [Agrees with (b).]
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.
4 CHAPTER 1. THE WAVE FUNCTION
Problem 1.2
(a)
\u3008x2\u3009 =
\u222b h
0
x2
1
2
\u221a
hx
dx =
1
2
\u221a
h
(
2
5
x5/2
)\u2223\u2223\u2223\u2223h
0
=
h2
5
.
\u3c32 = \u3008x2\u3009 \u2212 \u3008x\u30092 = h
2
5
\u2212
(
h
3
)2
=
4
45
h2 \u21d2 \u3c3 = 2h
3
\u221a
5
= 0.2981h.
(b)
P = 1\u2212
\u222b x+
x\u2212
1
2
\u221a
hx
dx = 1\u2212 1
2
\u221a
h
(2
\u221a
x)
\u2223\u2223\u2223\u2223x+
x\u2212
= 1\u2212 1\u221a
h
(\u221a
x+ \u2212\u221ax\u2212
)
.
x+ \u2261 \u3008x\u3009+ \u3c3 = 0.3333h+ 0.2981h = 0.6315h; x\u2212 \u2261 \u3008x\u3009 \u2212 \u3c3 = 0.3333h\u2212 0.2981h = 0.0352h.
P = 1\u2212
\u221a
0.6315 +
\u221a
0.0352 = 0.393.
Problem 1.3
(a)
1 =
\u222b \u221e
\u2212\u221e
Ae\u2212\u3bb(x\u2212a)
2
dx. Let u \u2261 x\u2212 a, du = dx, u : \u2212\u221e\u2192\u221e.
1 = A
\u222b \u221e
\u2212\u221e
e\u2212\u3bbu
2
du = A
\u221a
\u3c0
\u3bb
\u21d2 A =
\u221a
\u3bb
\u3c0
.
(b)
\u3008x\u3009 = A
\u222b \u221e
\u2212\u221e
xe\u2212\u3bb(x\u2212a)
2
dx = A
\u222b \u221e
\u2212\u221e
(u+ a)e\u2212\u3bbu
2
du
= A
[\u222b \u221e
\u2212\u221e
ue\u2212\u3bbu
2
du+ a
\u222b \u221e
\u2212\u221e
e\u2212\u3bbu
2
du
]
= A
(
0 + a
\u221a
\u3c0
\u3bb
)
= a.
\u3008x2\u3009 = A
\u222b \u221e
\u2212\u221e
x2e\u2212\u3bb(x\u2212a)
2
dx
= A
{\u222b \u221e
\u2212\u221e
u2e\u2212\u3bbu
2
du+ 2a
\u222b \u221e
\u2212\u221e
ue\u2212\u3bbu
2
du+ a2
\u222b \u221e
\u2212\u221e
e\u2212\u3bbu
2
du
}
= A
[
1
2\u3bb
\u221a
\u3c0
\u3bb
+ 0 + a2
\u221a
\u3c0
\u3bb
]
= a2 +
1
2\u3bb
.
\u3c32 = \u3008x2\u3009 \u2212 \u3008x\u30092 = a2 + 1
2\u3bb
\u2212 a2 = 1
2\u3bb
; \u3c3 =
1\u221a
2\u3bb
.
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 5
(c)
A
xa
\u3c1(x)
Problem 1.4
(a)
1 =
|A|2
a2
\u222b a
0
x2dx+
|A|2
(b\u2212 a)2
\u222b b
a
(b\u2212 x)2dx = |A|2
{
1
a2
(
x3
3
)\u2223\u2223\u2223\u2223a
0
+
1
(b\u2212 a)2
(
\u2212 (b\u2212 x)
3
3
)\u2223\u2223\u2223\u2223b
a
}
= |A|2
[
a
3
+
b\u2212 a
3
]
= |A|2 b
3
\u21d2 A =
\u221a
3
b
.
(b)
xa
A
b
\u3a8
(c) At x = a.
(d)
P =
\u222b a
0
|\u3a8|2dx = |A|
2
a2
\u222b a
0
x2dx = |A|2 a
3
=
a
b
.
{
P = 1 if b = a, \ufffd
P = 1/2 if b = 2a. \ufffd
(e)
\u3008x\u3009 =
\u222b
x|\u3a8|2dx = |A|2
{
1
a2
\u222b a
0
x3dx+
1
(b\u2212 a)2
\u222b b
a
x(b\u2212 x)2dx
}
=
3
b
{
1
a2
(
x4
4
)\u2223\u2223\u2223\u2223a
0
+
1
(b\u2212 a)2
(
b2
x2
2
\u2212 2bx
3
3
+
x4
4
)\u2223\u2223\u2223\u2223b
a
}
=
3
4b(b\u2212 a)2
[
a2(b\u2212 a)2 + 2b4 \u2212 8b4/3 + b4 \u2212 2a2b2 + 8a3b/3\u2212 a4]
=
3
4b(b\u2212 a)2
(
b4
3
\u2212 a2b2 + 2
3
a3b
)
=
1
4(b\u2212 a)2 (b
3 \u2212 3a2b+ 2a3) = 2a+ b
4
.
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.
6 CHAPTER 1. THE WAVE FUNCTION
Problem 1.5
(a)
1 =
\u222b
|\u3a8|2dx = 2|A|2
\u222b \u221e
0
e\u22122\u3bbxdx = 2|A|2
(
e\u22122\u3bbx
\u22122\u3bb
)\u2223\u2223\u2223\u2223\u221e
0
=
|A|2
\u3bb
; A =
\u221a
\u3bb.
(b)
\u3008x\u3009 =
\u222b
x|\u3a8|2dx = |A|2
\u222b \u221e
\u2212\u221e
xe\u22122\u3bb|x|dx = 0. [Odd integrand.]
\u3008x2\u3009 = 2|A|2
\u222b \u221e
0
x2e\u22122\u3bbxdx = 2\u3bb
[
2
(2\u3bb)3
]
=
1
2\u3bb2
.
(c)
\u3c32 = \u3008x2\u3009 \u2212 \u3008x\u30092 = 1
2\u3bb2
; \u3c3 =
1\u221a
2\u3bb
. |\u3a8(±\u3c3)|2 = |A|2e\u22122\u3bb\u3c3 = \u3bbe\u22122\u3bb/
\u221a
2\u3bb = \u3bbe\u2212
\u221a
2 = 0.2431\u3bb.
|\u3a8|2
\u3bb
\u3c3\u2212\u3c3 +
x
.24\u3bb
Probability outside:
2
\u222b \u221e
\u3c3
|\u3a8|2dx = 2|A|2
\u222b \u221e
\u3c3
e\u22122\u3bbxdx = 2\u3bb
(
e\u22122\u3bbx
\u22122\u3bb
)\u2223\u2223\u2223\u2223\u221e
\u3c3
= e\u22122\u3bb\u3c3 = e\u2212
\u221a
2 = 0.2431.
Problem 1.6
For integration by parts, the di\ufb00erentiation has to be with respect to the integration variable \u2013 in this case the
di\ufb00erentiation is with respect to t, but the integration variable is x. It\u2019s true that
\u2202
\u2202t
(x|\u3a8|2) = \u2202x
\u2202t
|\u3a8|2 + x \u2202
\u2202t
|\u3a8|2 = x \u2202
\u2202t
|\u3a8|2,
but this does not allow us to perform the integration:\u222b b
a
x
\u2202
\u2202t
|\u3a8|2dx =
\u222b b
a
\u2202
\u2202t
(x|\u3a8|2)dx \ufffd= (x|\u3a8|2)\u2223\u2223b
a
.
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 7
Problem 1.7
From Eq. 1.33, d\u3008p\u3009dt = \u2212i\ufffd
\u222b
\u2202
\u2202t
(
\u3a8\u2217 \u2202\u3a8\u2202x
)
dx. But, noting that \u2202
2\u3a8
\u2202x\u2202t =
\u22022\u3a8
\u2202t\u2202x and using Eqs. 1.23-1.24:
\u2202
\u2202t
(
\u3a8\u2217
\u2202\u3a8
\u2202x
)
=
\u2202\u3a8\u2217
\u2202t
\u2202\u3a8
\u2202x
+ \u3a8\u2217
\u2202
\u2202x
(
\u2202\u3a8
\u2202t
)
=
[
\u2212 i\ufffd
2m
\u22022\u3a8\u2217
\u2202x2
+
i
\ufffd
V\u3a8\u2217
]
\u2202\u3a8
\u2202x
+ \u3a8\u2217
\u2202
\u2202x
[
i\ufffd
2m
\u22022\u3a8
\u2202x2
\u2212 i
\ufffd
V\u3a8
]
=
i\ufffd
2m
[
\u3a8\u2217
\u22023\u3a8
\u2202x3
\u2212 \u2202
2\u3a8\u2217
\u2202x2
\u2202\u3a8
\u2202x
]
+
i
\ufffd
[
V\u3a8\u2217
\u2202\u3a8
\u2202x
\u2212\u3a8\u2217 \u2202
\u2202x
(V\u3a8)
]
The \ufb01rst term integrates to zero, using integration by parts twice, and the second term can be simpli\ufb01ed to
V\u3a8\u2217 \u2202\u3a8\u2202x \u2212\u3a8\u2217V \u2202\u3a8\u2202x \u2212\u3a8\u2217 \u2202V\u2202x \u3a8 = \u2212|\u3a8|2 \u2202V\u2202x . So
d\u3008p\u3009
dt
= \u2212i\ufffd
(
i
\ufffd
) \u222b
\u2212|\u3a8|2 \u2202V
\u2202x
dx = \u3008\u2212\u2202V
\u2202x
\u3009. QED
Problem 1.8
Suppose \u3a8 satis\ufb01es the Schro¨dinger equation without V0: i\ufffd\u2202\u3a8\u2202t = \u2212 \ufffd
2
2m
\u22022\u3a8
\u2202x2 + V\u3a8. We want to \ufb01nd the solution
\u3a80 with V0: i\ufffd\u2202\u3a80\u2202t = \u2212 \ufffd
2
2m
\u22022\u3a80
\u2202x2 + (V + V0)\u3a80.
Claim: \u3a80 = \u3a8e\u2212iV0t/\ufffd.
Proof: i\ufffd\u2202\u3a80\u2202t = i\ufffd
\u2202\u3a8
\u2202t e
\u2212iV0t/\ufffd + i\ufffd\u3a8
(\u2212 iV0
\ufffd
)
e\u2212iV0t/\ufffd =
[
\u2212 \ufffd22m \u2202
2\u3a8
\u2202x2 + V\u3a8
]
e\u2212iV0t/\ufffd + V0\u3a8e\u2212iV0t/\ufffd
= \u2212 \ufffd22m