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《量子力学概论》系美国David J.Griffiths教授所著《Introduction to Quantum Mechanics》,其内容包含我国大学量子力学最主要的内容。该书的特色是强调量子力学的实验基础和基本概念,讲解直接从薛定谔方程开始,同时力图体现现代物理学内容,把问题扩展到多个前沿的研究领域,如统计物理、固体物理、粒子物理等;在写法上,作者从务实的角度出发,着重于交互式的写作,采用对话式的语言,叙述简明,文笔流畅。力图改变量子力学难于理解、难于接受的教学状况。该书内容分理论和应用两部分。理论部分包括:波函数、定态薛定谔方程、形式理论、三维空间中的量子力学和全同粒子;应用部分包括:不含时微扰理论、变分原理、WKB近似、含时微扰理论、绝热近似、散射和后记。为使读者更好的理解量子力学,书后还提供了附录线性代数。 大卫·格里菲斯,是里德学院(俄勒冈州珀特兰市)的物理学教授。在哈佛大学获得粒子物理博士学位后,曾在几所大学和学院任教,1978年进入里德学院。他专长于电动力学和量子力学,还有基本粒子物理,并在这三个领域都著有教科书。 本电子书为该教材的习题解答(英文原版)。
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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
2
nd
Edition – 1
st
Edition Problem Correlation Grid 299
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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
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.
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CHAPTER 1. THE WAVE FUNCTION 3
Chapter 1
The Wave Function
Problem 1.1
(a)
j
2
=21
2
= 441.
j
2
=
1
N
j
2
N(j)=
1
14
(14
2
) + (15
2
) + 3(16
2
) + 2(22
2
) + 2(24
2
) + 5(25
2
)
=
1
14
(196 + 225 + 768 + 968 + 1152 + 3125) =
6434
14
=
459.571.
(b)
j ∆j = j −j
14 14 − 21 = −7
15 15 − 21 = −6
16 16 − 21 = −5
22 22 − 21 = 1
24 24 − 21 = 3
25 25 − 21 = 4
σ
2
=
1
N
(∆j)
2
N(j)=
1
14
(−7)
2
+(−6)
2
+(−5)
2
· 3 + (1)
2
· 2 + (3)
2
· 2 + (4)
2
· 5
=
1
14
(49+36+75+2+18+80)=
260
14
=
18.571.
σ =
√
18.571 = 4.309.
(c)
j
2
−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 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.
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4 CHAPTER 1. THE WAVE FUNCTION
Problem 1.2
(a)
x
2
=
h
0
x
2
1
2
√
hx
dx =
1
2
√
h
2
5
x
5/2
h
0
=
h
2
5
.
σ
2
= x
2
−x
2
=
h
2
5
−
h
3
2
=
4
45
h
2
⇒ σ =
2h
3
√
5
=0.2981h.
(b)
P =1−
x
+
x
−
1
2
√
hx
dx =1−
1
2
√
h
(2
√
x)
x
+
x
−
=1−
1
√
h
√
x
+
−
√
x
−
.
x
+
≡x + σ =0.3333h +0.2981h =0.6315h; x
−
≡x−σ =0.3333h − 0.2981h =0.0352h.
P =1−
√
0.6315 +
√
0.0352 = 0.393.
Problem 1.3
(a)
1=
∞
−∞
Ae
−λ(x−a)
2
dx. Let u ≡ x − a, du = dx, u : −∞ → ∞.
1=A
∞
−∞
e
−λu
2
du = A
π
λ
⇒
A =
λ
π
.
(b)
x = A
∞
−∞
xe
−λ(x−a)
2
dx = A
∞
−∞
(u + a)e
−λu
2
du
= A
∞
−∞
ue
−λu
2
du + a
∞
−∞
e
−λu
2
du
= A
0+a
π
λ
=
a.
x
2
= A
∞
−∞
x
2
e
−λ(x−a)
2
dx
= A
∞
−∞
u
2
e
−λu
2
du +2a
∞
−∞
ue
−λu
2
du + a
2
∞
−∞
e
−λu
2
du
= A
1
2λ
π
λ
+0+a
2
π
λ
=
a
2
+
1
2λ
.
σ
2
= x
2
−x
2
= a
2
+
1
2λ
− a
2
=
1
2λ
;
σ =
1
√
2λ
.
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.
课后答案网 www.khdaw.com
CHAPTER 1. THE WAVE FUNCTION 5
(c)
A
x
a
ρ(x)
Problem 1.4
(a)
1=
|A|
2
a
2
a
0
x
2
dx +
|A|
2
(b − a)
2
b
a
(b − x)
2
dx = |A|
2
1
a
2
x
3
3
a
0
+
1
(b − a)
2
−
(b − x)
3
3
b
a
= |A|
2
a
3
+
b − a
3
= |A|
2
b
3
⇒
A =
3
b
.
(b)
x
a
A
b
Ψ
(c) At x = a.
(d)
P =
a
0
|Ψ|
2
dx =
|A|
2
a
2
a
0
x
2
dx = |A|
2
a
3
=
a
b
.
P =1 if b = a,
P =1/2if b =2a.
(e)
x =
x|Ψ|
2
dx = |A|
2
1
a
2
a
0
x
3
dx +
1
(b − a)
2
b
a
x(b − x)
2
dx
=
3
b
1
a
2
x
4
4
a
0
+
1
(b − a)
2
b
2
x
2
2
− 2b
x
3
3
+
x
4
4
b
a
=
3
4b(b − a)
2
a
2
(b − a)
2
+2b
4
− 8b
4
/3+b
4
− 2a
2
b
2
+8a
3
b/3 − a
4
=
3
4b(b − a)
2
b
4
3
− a
2
b
2
+
2
3
a
3
b
=
1
4(b − a)
2
(b
3
− 3a
2
b +2a
3
)=
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.
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