没有合适的资源?快使用搜索试试~ 我知道了~
首页量子力学概论 习题解答 (英文版)
作者格里菲斯
这本电子书是《量子力学概论》的习题解答(英文版)。该教材由美国里德学院的物理学教授大卫·格里菲斯所著,原著为《Introduction to Quantum Mechanics》。这本教材包含了我国大学量子力学的主要内容。该书着重强调量子力学的实验基础和基本概念,并从薛定谔方程开始讲解。同时,该书也试图体现现代物理学的内容,并将问题扩展到其他前沿的研究领域,如统计物理、固体物理和粒子物理等。在写作风格上,作者采用了务实的角度,着重于互动式写作,使用了对话式的语言,使叙述简明流畅。目的是改变传统教学中难以理解和接受的量子力学状况。 这本电子书分为理论和应用两个部分。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等内容。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、散射和后记等内容。为了帮助读者更好地理解量子力学,书的最后还提供了附录,介绍了线性代数的相关知识。 大卫·格里菲斯是里德学院的物理学教授。他在哈佛大学获得了粒子物理的博士学位,之后在几所大学和学院任教。格里菲斯教授的专长领域包括电动力学、量子力学和基本粒子物理,并在这三个领域都撰写了教科书。 这本电子书为《量子力学概论》的习题解答(英文版),目录包括了前言、波函数、定态薛定谔方程、形式理论、三维空间中的量子力学、全同粒子、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。
资源详情
资源评论
资源推荐
16 CHAPTER 2. THE TIME-INDEPENDENT SCHR
¨
ODINGER EQUATION
x
2
=
2
a
a
0
x
2
sin
2
nπ
a
x
dx =
2
a
a
nπ
3
nπ
0
y
2
sin
2
ydy
=
2a
2
(nπ)
3
y
3
6
−
y
3
4
−
1
8
sin 2y −
y cos 2y
4
nπ
0
=
2a
2
(nπ)
3
(nπ)
3
6
−
nπ cos(2nπ)
4
=
a
2
1
3
−
1
2(nπ)
2
.
p = m
dx
dt
=
0. (Note :Eq. 1.33 is much faster than Eq. 1.35.)
p
2
=
ψ
∗
n
i
d
dx
2
ψ
n
dx = −
2
ψ
∗
n
d
2
ψ
n
dx
2
dx
=(−
2
)
−
2mE
n
2
ψ
∗
n
ψ
n
dx =2mE
n
=
nπ
a
2
.
σ
2
x
= x
2
−x
2
= a
2
1
3
−
1
2(nπ)
2
−
1
4
=
a
2
4
1
3
−
2
(nπ)
2
;
σ
x
=
a
2
1
3
−
2
(nπ)
2
.
σ
2
p
= p
2
−p
2
=
nπ
a
2
; σ
p
=
nπ
a
.
∴ σ
x
σ
p
=
2
(nπ)
2
3
− 2.
The product σ
x
σ
p
is smallest for n =1; in that case, σ
x
σ
p
=
2
π
2
3
− 2=(1.136)/2 > /2.
Problem 2.5
(a)
|Ψ|
2
=Ψ
2
Ψ=|A|
2
(ψ
∗
1
+ ψ
∗
2
)(ψ
1
+ ψ
2
)=|A|
2
[ψ
∗
1
ψ
1
+ ψ
∗
1
ψ
2
+ ψ
∗
2
ψ
1
+ ψ
∗
2
ψ
2
].
1=
|Ψ|
2
dx = |A|
2
[|ψ
1
|
2
+ ψ
∗
1
ψ
2
+ ψ
∗
2
ψ
1
+ |ψ
2
|
2
]dx =2|A|
2
⇒ A =1/
√
2.
(b)
Ψ(x, t)=
1
√
2
ψ
1
e
−iE
1
t/
+ ψ
2
e
−iE
2
t/
(but
E
n
= n
2
ω)
=
1
√
2
2
a
sin
π
a
x
e
−iωt
+ sin
2π
a
x
e
−i4ωt
=
1
√
a
e
−iωt
sin
π
a
x
+ sin
2π
a
x
e
−3iωt
.
|Ψ(x, t)|
2
=
1
a
sin
2
π
a
x
+ sin
π
a
x
sin
2π
a
x
e
−3iωt
+ e
3iωt
+ sin
2
2π
a
x
=
1
a
sin
2
π
a
x
+ sin
2
2π
a
x
+ 2 sin
π
a
x
sin
2π
a
x
cos(3ωt)
.
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 2. THE TIME-INDEPENDENT SCHR
¨
ODINGER EQUATION 17
(c)
x =
x|Ψ(x, t)|
2
dx
=
1
a
a
0
x
sin
2
π
a
x
+ sin
2
2π
a
x
+ 2 sin
π
a
x
sin
2π
a
x
cos(3ωt)
dx
a
0
x sin
2
π
a
x
dx =
x
2
4
−
x sin
2π
a
x
4π/a
−
cos
2π
a
x
8(π/a)
2
a
0
=
a
2
4
=
a
0
x sin
2
2π
a
x
dx.
a
0
x sin
π
a
x
sin
2π
a
x
dx =
1
2
a
0
x
cos
π
a
x
− cos
3π
a
x
dx
=
1
2
a
2
π
2
cos
π
a
x
+
ax
π
sin
π
a
x
−
a
2
9π
2
cos
3π
a
x
−
ax
3π
sin
3π
a
x
a
0
=
1
2
a
2
π
2
cos(π) − cos(0)
−
a
2
9π
2
cos(3π) − cos(0)
= −
a
2
π
2
1 −
1
9
= −
8a
2
9π
2
.
∴ x =
1
a
a
2
4
+
a
2
4
−
16a
2
9π
2
cos(3ωt)
=
a
2
1 −
32
9π
2
cos(3ωt)
.
Amplitude:
32
9π
2
a
2
=0.3603(a/2);
angular frequency: 3ω =
3π
2
2ma
2
.
(d)
p = m
dx
dt
= m
a
2
−
32
9π
2
(−3ω) sin(3ωt)=
8
3a
sin(3ωt).
(e) You could get either E
1
= π
2
2
/2ma
2
or E
2
=2π
2
2
/ma
2
, with equal probability P
1
= P
2
=1/2.
So H =
1
2
(E
1
+ E
2
)=
5π
2
2
4ma
2
; it’s the average of E
1
and E
2
.
Problem 2.6
From Problem 2.5, we see that
Ψ(x, t)=
1
√
a
e
−iωt
sin
π
a
x
+ sin
2π
a
x
e
−3iωt
e
iφ
;
|Ψ(x, t)|
2
=
1
a
sin
2
π
a
x
+ sin
2
2π
a
x
+ 2 sin
π
a
x
sin
2π
a
x
cos(3ωt − φ)
;
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
18 CHAPTER 2. THE TIME-INDEPENDENT SCHR
¨
ODINGER EQUATION
and hence x =
a
2
1 −
32
9π
2
cos(3ωt − φ)
. This amounts physically to starting the clock at a different time
(i.e., shifting the t = 0 point).
If φ =
π
2
, so Ψ(x, 0) = A[ψ
1
(x)+iψ
2
(x)], then cos(3ωt − φ) = sin(3ωt); x starts at
a
2
.
If φ = π, so Ψ(x, 0) = A[ψ
1
(x) − ψ
2
(x)], then cos(3ωt − φ)=−cos(3ωt); x starts at
a
2
1+
32
9π
2
.
Problem 2.7
Ψ(x,0)
x
a
a
/2
Aa/2
(a)
1=A
2
a/2
0
x
2
dx + A
2
a
a/2
(a − x)
2
dx = A
2
x
3
3
a/2
0
−
(a − x)
3
3
a
a/2
=
A
2
3
a
3
8
+
a
3
8
=
A
2
a
3
12
⇒
A =
2
√
3
√
a
3
.
(b)
c
n
=
2
a
2
√
3
a
√
a
a/2
0
x sin
nπ
a
x
dx +
a
a/2
(a − x) sin
nπ
a
x
dx
=
2
√
6
a
2
a
nπ
2
sin
nπ
a
x
−
xa
nπ
cos
nπ
a
x
a/2
0
+ a
−
a
nπ
cos
nπ
a
x
a
a/2
−
a
nπ
2
sin
nπ
a
x
−
ax
nπ
cos
nπ
a
x
a
a/2
=
2
√
6
a
2
a
nπ
2
sin
nπ
2
−
✘
✘
✘
✘
✘
✘
✘
a
2
2nπ
cos
nπ
2
−
✟
✟
✟
✟
✟
a
2
nπ
cos nπ +
✟
✟
✟
✟
✟
✟
✟
a
2
nπ
cos
nπ
2
+
a
nπ
2
sin
nπ
2
+
✟
✟
✟
✟
✟
a
2
nπ
cos nπ −
✘
✘
✘
✘
✘
✘
✘
a
2
2nπ
cos
nπ
2
=
2
√
6
a
2
2
a
2
(nπ)
2
sin
nπ
2
=
4
√
6
(nπ)
2
sin
nπ
2
=
0,neven,
(−1)
(n−1)/2
4
√
6
(nπ)
2
,nodd.
So
Ψ(x, t)=
4
√
6
π
2
2
a
n=1,3,5,...
(−1)
(n−1)/2
1
n
2
sin
nπ
a
x
e
−E
n
t/
, where E
n
=
n
2
π
2
2
2ma
2
.
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 2. THE TIME-INDEPENDENT SCHR
¨
ODINGER EQUATION 19
(c)
P
1
= |c
1
|
2
=
16 · 6
π
4
= 0.9855.
(d)
H =
|c
n
|
2
E
n
=
96
π
4
π
2
2
2ma
2
1
1
+
1
3
2
+
1
5
2
+
1
7
2
+ ···
π
2
/8
=
48
2
π
2
ma
2
π
2
8
=
6
2
ma
2
.
Problem 2.8
(a)
Ψ(x, 0) =
A, 0 <x<a/2;
0, otherwise.
1=A
2
a/2
0
dx = A
2
(a/2) ⇒ A =
2
a
.
(b) From Eq. 2.37,
c
1
= A
2
a
a/2
0
sin
π
a
x
dx =
2
a
−
a
π
cos
π
a
x
a/2
0
= −
2
π
cos
π
2
− cos 0
=
2
π
.
P
1
= |c
1
|
2
= (2/π)
2
=0.4053.
Problem 2.9
ˆ
HΨ(x, 0) = −
2
2m
∂
2
∂x
2
[Ax(a − x)] = −A
2
2m
∂
∂x
(a − 2x)=A
2
m
.
Ψ(x, 0)
∗
ˆ
HΨ(x, 0) dx = A
2
2
m
a
0
x(a − x) dx = A
2
2
m
a
x
2
2
−
x
3
3
a
0
= A
2
2
m
a
3
2
−
a
3
3
=
30
a
5
2
m
a
3
6
=
5
2
ma
2
(same as Example 2.3).
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
20 CHAPTER 2. THE TIME-INDEPENDENT SCHR
¨
ODINGER EQUATION
Problem 2.10
(a) Using Eqs. 2.47 and 2.59,
a
+
ψ
0
=
1
√
2mω
−
d
dx
+ mωx
mω
π
1/4
e
−
mω
2
x
2
=
1
√
2mω
mω
π
1/4
−
−
mω
2
2x + mωx
e
−
mω
2
x
2
=
1
√
2mω
mω
π
1/4
2mωxe
−
mω
2
x
2
.
(a
+
)
2
ψ
0
=
1
2mω
mω
π
1/4
2mω
−
d
dx
+ mωx
xe
−
mω
2
x
2
=
1
mω
π
1/4
−
1 − x
mω
2
2x
+ mωx
2
e
−
mω
2
x
2
=
mω
π
1/4
2mω
x
2
− 1
e
−
mω
2
x
2
.
Therefore, from Eq. 2.67,
ψ
2
=
1
√
2
(a
+
)
2
ψ
0
=
1
√
2
mω
π
1/4
2mω
x
2
− 1
e
−
mω
2
x
2
.
(b)
ψψ
ψ
1
2
0
(c) Since ψ
0
and ψ
2
are even, whereas ψ
1
is odd,
ψ
∗
0
ψ
1
dx and
ψ
∗
2
ψ
1
dx vanish automatically. The only one
we need to check is
ψ
∗
2
ψ
0
dx:
ψ
∗
2
ψ
0
dx =
1
√
2
mω
π
∞
−∞
2mω
x
2
− 1
e
−
mω
x
2
dx
= −
mω
2π
∞
−∞
e
−
mω
x
2
dx −
2mω
∞
−∞
x
2
e
−
mω
x
2
dx
= −
mω
2π
π
mω
−
2mω
2mω
π
mω
=0.
Problem 2.11
(a) Note that ψ
0
is even, and ψ
1
is odd. In either case |ψ|
2
is even, so x =
x|ψ|
2
dx = 0. Therefore
p = mdx/dt =
0. (These results hold for any stationary state of the harmonic oscillator.)
From Eqs. 2.59 and 2.62, ψ
0
= αe
−ξ
2
/2
,ψ
1
=
√
2αξe
−ξ
2
/2
.So
n =0
:
x
2
= α
2
∞
−∞
x
2
e
−ξ
2
/2
dx = α
2
mω
3/2
∞
−∞
ξ
2
e
−ξ
2
dξ =
1
√
π
mω
√
π
2
=
2mω
.
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
剩余302页未读,继续阅读
wonringking
- 粉丝: 0
- 资源: 5
上传资源 快速赚钱
- 我的内容管理 收起
- 我的资源 快来上传第一个资源
- 我的收益 登录查看自己的收益
- 我的积分 登录查看自己的积分
- 我的C币 登录后查看C币余额
- 我的收藏
- 我的下载
- 下载帮助
会员权益专享
最新资源
- zigbee-cluster-library-specification
- JSBSim Reference Manual
- c++校园超市商品信息管理系统课程设计说明书(含源代码) (2).pdf
- 建筑供配电系统相关课件.pptx
- 企业管理规章制度及管理模式.doc
- vb打开摄像头.doc
- 云计算-可信计算中认证协议改进方案.pdf
- [详细完整版]单片机编程4.ppt
- c语言常用算法.pdf
- c++经典程序代码大全.pdf
- 单片机数字时钟资料.doc
- 11项目管理前沿1.0.pptx
- 基于ssm的“魅力”繁峙宣传网站的设计与实现论文.doc
- 智慧交通综合解决方案.pptx
- 建筑防潮设计-PowerPointPresentati.pptx
- SPC统计过程控制程序.pptx
资源上传下载、课程学习等过程中有任何疑问或建议,欢迎提出宝贵意见哦~我们会及时处理!
点击此处反馈
安全验证
文档复制为VIP权益,开通VIP直接复制
信息提交成功
评论0