外国教材量子力学概论2ndedition课后练习题含答案

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量子力学(第二版)答案 苏汝铿 第二章课后答案2.31-2#3

量子力学(第二版)答案 苏汝铿 第二章课后答案2.31-2#3
则有

x 0
2 x
x 0
eik0 y sin Reik0 y sin Te
ik y y
ik0 cos eik0 y sin Rik0 cos eik0 y sin Tik xe
ik y y
由于势能与 y 无关,则有 k y k0 sin ,上式变为
2 Ze / x, x 0 V x 的结果相比较. , x 0
解:根据维理定律
x 1 1 E V x Ze2 dx 2 2 x 2
如果当 x 0 时, x 不趋于零,上述积分会发散, E 会趋近于负无穷大 .这是不可能的 ,所以 我们得到 0 0 .这样我们就可以用 Laplace 变换来解决这个问题. 势能为 V x 一维薛定谔方程为
2
2
k02 sin 2
ii 在 x 0 区域中,波函数的形式和 i 中一样
在 0 x t 区域中, E V 0 ,薛定谔方程变为

2
/ 2m 2 0
在 y 方向上势能是不变的,我们有 exp ik x exp kx ,其中 k k0 sin ,
C
4ik
eikc e c 1 ik / e c 1 ik /
2 2
因为在 I 中已经取入射波的形式为 eikx ,所以透射系数 T CC * ,则有
T
k
2

2 2

4k 2 2 sinh 2 c 4k 2 2
2
当 c
1 ,即 V0 E
2
s
2
1 1 s 1 s ds

量子力学答案(第二版)苏汝铿第五章课后答案5.1-5#7

量子力学答案(第二版)苏汝铿第五章课后答案5.1-5#7
与视核为点电荷时电子的势能之差为
r r0 r r0
2 1 r2 3 Ze Ze 3 , ' H eV r r 2 r 2 r 0 0 r 0,
2
r r0 r r0
将其视为微扰。类氢离子中 1s 轨道电子波函数为
2
D
l 0 , m
2
l|m c o s | 0 / E 0
l E
由于
cos Y00
1 Y10 3
根据球谐函数的正交性可知,能量二级修正中只有 l 1, m 0 有贡献。
所以
E0 D 1 0 | c o s
2
2
| 00 E 0/ E
2
1
2
/ 2I ,
l 0,1, 2...
对确定的 l , m 0, 1, 2,..., l ,即能级的简并度为 2l 1 。 定理:某能级 En 非简并时, H 和宇称算符 具有共同本征矢 n 。 因而,
n r n n r n n r n n r n
07QMEx5.1-5.3 如果类氢原子的核不是点电荷,而是半径为 r0 ,电荷分布的小球,计算这种效应对类
5.1
氢原子基态能量的一级修正。 解: 由电磁学知球形电荷分布的静电势为
Ze 3 1 r 2 , r0 2 2 r02 V (r ) Ze , r
Z 1s R10Y00 a0
3/ 2
2e
Zr a0
1 4
2 Zr a0
1s 能级的一级修正为
E1s 1s H 1s
'
1

量子力学教程(二版)习题答案

量子力学教程(二版)习题答案

第一章 绪论1.1.由黑体辐射公式导出维恩位移定律:C m b b T m 03109.2 ,⋅⨯==-λ。

证明:由普朗克黑体辐射公式:ννπνρννd e ch d kT h 11833-=, 及λνc =、λλνd cd 2-=得1185-=kThc ehc λλλπρ,令kT hcx λ=,再由0=λρλd d ,得λ.所满足的超越方程为 15-=x xe xe用图解法求得97.4=x ,即得97.4=kThcm λ,将数据代入求得C m 109.2 ,03⋅⨯==-b b T m λ 1.2.在0K 附近,钠的价电子能量约为3eV,求de Broglie 波长.解:010A 7.09m 1009.72=⨯≈==-mEh p h λ #1.3. 氦原子的动能为kT E 23=,求K T 1=时氦原子的de Broglie 波长。

解:010A 63.12m 1063.1232=⨯≈===-mkTh mE h p h λ 其中kg 1066.1003.427-⨯⨯=m ,123K J 1038.1--⋅⨯=k #1.4利用玻尔—索末菲量子化条件,求: (1)一维谐振子的能量。

(2)在均匀磁场中作圆周运动的电子的轨道半径。

已知外磁场T 10=B ,玻尔磁子123T J 10923.0--⋅⨯=B μ,求动能的量子化间隔E ∆,并与K 4=T 及K 100=T 的热运动能量相比较。

解:(1)方法1:谐振子的能量222212q p E μωμ+=可以化为()12222222=⎪⎪⎭⎫ ⎝⎛+μωμE q Ep的平面运动,轨道为椭圆,两半轴分别为22,2μωμEb E a ==,相空间面积为,2,1,0,2=====⎰n nh EEab pdq νωππ所以,能量 ,2,1,0,==n nh E ν方法2:一维谐振子的运动方程为02=+''q q ω,其解为()ϕω+=t A q sin速度为 ()ϕωω+='t A q c o s ,动量为()ϕωμωμ+='=t A q p cos ,则相积分为 ()()nh TA dt t A dt t A pdq T T==++=+=⎰⎰⎰2)cos 1(2cos 220220222μωϕωμωϕωμω, ,2,1,0=nνμωnh Tnh A E ===222, ,2,1,0=n(2)设磁场垂直于电子运动方向,受洛仑兹力作用作匀速圆周运动。

量子力学(二)习题参考答案

量子力学(二)习题参考答案

2µ (U1 − E ) h2 2µ E h2
ψ 2 '' ( x) + k 2ψ 2 ( x ) = 0, k =
西华师大物理与电子信息学院
4
四川省精品课程——量子力学补充习题参考答案
ψ 3'' ( x) − β 2ψ 3 ( x) = 0, β =
其解分别为:
2µ (U 2 − E ) h2
ψ 1 ( x) = A1eα x + B1e −α x ψ 2 ( x) = C sin(kx + δ ) ψ 3 ( x ) = A2e β x + B2 e− β x
2
2

而透射系数

2) 、当 E<U0 时,有ψ 2 '' ( x ) − k3 2ψ 2 ( x ) = 0 , k3 = 其解为:ψ 2 ( x ) = Ce
− k3 x
+ De k3 x = Ce − k3 x (ψ 2 有限条件)

以下可以重复前面的求解过程。 不过, 为了简单我们亦可以在前面得到的结果⑤中做代 换 k2 =i k3 ,得到
由(18)式, (16) 、 (17)变成 或由 (19) 式, (16) 、 (17) 变成
(20)或(21)式就是讲义上习题 2.7 的结果。 a) 将 δ = 0 代入ψ 2 ( x) 中有:ψ 2 ( x) = C sin kx 由连续性条件:ψ 2 ( a) = ψ 3 ( a ) → C sin( ka ) = B2 e − β a
ψ m (ϕ ) =
除了 m=0 的态之外, E m 圴是二重简并的。 5、梯形式——— U ( x ) =
0, x < 0 U 0 , x > 0

苏汝铿量子力学(第二版)课后习题(含答案)---第一章1.7-1.8#04(延边大学)三年级

苏汝铿量子力学(第二版)课后习题(含答案)---第一章1.7-1.8#04(延边大学)三年级

1.7 一个德布罗意波在k空间的表示220()4()a k k C k --=求:(ⅰ)(,)x t ψ和2(,)x t ψ,在时刻t 这是否是个高斯波包? (ⅱ)波包的宽度()x t ∆; (ⅲ)2(,)x t dx ψ∞-∞⎰是否依赖于t ?解:(ⅰ) 由于已知德布罗意波在k 空间的表示220()4()a k k C k --=因此对该一维波包有 ()121(,)()(2)i kx t x t C k e dk ωψπ-=⎰(1)将()k ω在0k 附近展开并略去高阶项有 20001()()()()2g k k v k k k k ωωβ≈+-+- (2) 其中 0()g k d v dk ω= ,022()k d dkωβ= 将(2)式代入(1)式有 20001()[()()]212(,)()()g i k t i kx v k k k k e x t C k edk ωβψαπ-∞-----∞=⎰(3)当220()414()(2)a k k C k eπ--=代入(3)式可得:22200001()()[()()]4212(,)()g a i k t k k i kx v k k k k e x t e dk ωβψαπ-∞------=⎰积分上式可得0022()2()(,)]2(1)g i k tik x x v t x t e e i t ωαψβα--=+则222242()(,)]1g x v t x t tαψβα-=+故在时刻t 这是个高斯波包 (ⅱ) 波包宽度()x t ∆≈(ⅲ) 由222242()(,)]1g x v t x t t αψβα-=+易知2(,)x t dx ψ∞-∞⎰依赖于t1.8将平面波和波包的讨论推广到三维情况,求群速度。

解:对于三维平面波和波包,也可将波包视为由若干个平面波叠家而来,则有 ()321(,)()(2)i k r t r t C k e dk ωψπ⋅-=⎰由于()()i C k C k e α= 令 k r t ϕωα=⋅-+()C k 在点0k k = 周围宽度为k ∆的一个小区域内有一个明显的峰值,只有当相位ϕ在小区域内基本上保持不变时,ψ才有最大值。

量子力学答案(第二版)苏汝铿第五章课后答案5.16-5#7 @

量子力学答案(第二版)苏汝铿第五章课后答案5.16-5#7 @
A 2 p 1s A 200,100 A 210,100 A 211,100 A 21-1,100
批注 [JL3]: 对于固定初末态(即具有固定的 m 与 m )的跃迁,不需要求和。

2 3
8
2 me10 3 c3 6

8
2 me10 3 c3 6

8
me10 28 me10 c3 6 37 c3 6
The transition coefficient is
...
氢原子的初态(k 态)的波函数是: 100 ,末态( k ' 态)的波函数是 21m : 1s 态
100
1
a3
1
e

r a
(1)
r
2s 态
200
211
r ( 2 )e 2 a 3 a 32a
r ( )e 2 a sin e i 8 a 3 a r ( )e 2 a sin e i 8 a 3 a r ( )e 2 a cos 32a 3 a
0
i
t
(ez ) k 'k (ez ) k 'k
t o

e
1 [ i ( ' k ) ]t
k

dt t
(7)

0
i
e
t [ i ( ' k ) t ]t
k

i (k ' k )

0
i [(k ' k ) ]

1 t 0
(ez ) k 'k
| r k 'k
|
2
|
x | | y | | z

量子力学答案(第二版)苏汝铿第五章课后答案5.13-5#11

量子力学答案(第二版)苏汝铿第五章课后答案5.13-5#11

0
2 2 2 0 0 0 0 0 2 2 2
5.14 一根长度为 d 质量均匀分布的棒可绕其中心在一平面内转动,棒的质量为 M ,在棒的 两端分别有电荷+Q 和-Q。 (i)写出体系的哈密顿量,本征函数和本征值; (ii)如果在转动平面内存在一电场强度为 的弱电场,准确到一级修正,他的本征函数和 能量如何变化? (iii)如果这个电场很强,求基态的近似波函数和相应的能量值。 解: (i)该系统的哈密顿量为 H 式中 I
0
1
m1
n
n H' m Em 0 En 0
n H' m
1 2
2
dE cos e
0
i m n
d
1 1 dE 2 m n 1,0 m n 1,0 2 2 1 dE m n 1,0 m n 1,0 2
式中用了 k
0
0
0
取到 的一阶
B 0 C
0
的完备性

k
0 0
kLeabharlann k 1(ii)根据已给的条件
3 P2 1 H 0 i m 2 xi 2 , H ' x3 2 i 1 2m
可看出相应的 A
m
P3
2
, 它使 H ' i A, H 0 x3
计算 xi 在基态的平均值 xi
i 1, 2,3 至 的最低阶,并将这个结果和精确解相比较。
0
解: (i)设系统非微扰的本征态及对应的能量分别为 k 即 H0 k
0
, Ek 0
Ek 0 k

量子力学课后习题答案

量子力学课后习题答案

Wnl (r)dr Rnl2 (r)r 2dr
例如:对于基态 n 1, l 0
W10 (r) R102 (r)r 2

4 a03
r e2 2r / a0
求最可几半径
R e 2 r / a0
10
a03 / 2
dW10 (r) 4 (2r 2 r 2 )e2r / a0
x)

k
2
2
(
x)

0
其解为 2 (x) Asin kx B cos kx
根据波函数的标准条件确定系数A、B,由连续性条件,得
2 (0) 1(0) B 0
2 (a) 3 (a) Asin ka 0
A0
sin ka 0
ka n
(n 1, 2, 3,)
[1 r
eikr
r
(1 r
eikr )

1 r
eikr
r
(1 r
eikr )]er
i1 1 11 1 1

2
[ r
(
r2
ik
) r

r
(
r2
ik
r )]er

k
r2
er
J1与er 同向。 1 表示向外传播的球面波。
习题
(2)
J2

i
2
(
2
* 2
2*
解:U (x)与t 无关,是定态问题
薛定谔方程为

2
2
d2 dx2

(x) U (x) (x)

E (x)
在各区域的具体形式为:
x0
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Introduction to Quantum MechanicsOverviewQuantum Mechanics is a branch of Physics that describes the behavior of matter and energy at a microscopic level. This discipline has had a significant impact on modern science and technology, and its principles have been applied to the development of various fields, such as computing, cryptography and medicine. The study of Quantum Mechanics requires a basic understanding of the principles of Mathematics and Physics. The m of this document is to provide an introduction to Quantum Mechanics and to provide a set of practice exercises with answers that will allow students to test their knowledge and understanding of the subject.Fundamental PrinciplesThe fundamental principles of Quantum Mechanics are based on the concept of a wave-particle duality, which means that particles can behave as both waves and particles simultaneously. The behavior of particles at the microscopic level is probabilistic, and it is described by a wave function. A wave function is a complex function that describes the probability of finding a particle at a givenlocation. The square of the amplitude of the wave function gives the probability density of finding the particle at that point in space. The wave function can be used to calculate various physical quantities, such as the position, momentum and energy of a particle.Operators and ObservablesIn Quantum Mechanics, physical quantities are represented by operators. An operator is a mathematical function that acts on a wave function and generates a new wave function as a result. Operators are used to represent physical observables, such as the position, momentum and energy of a particle. The eigenvalues of an operator correspond to the possible results of a measurement of the corresponding observable. The eigenvectors of an operator correspond to the possible states of a particle. The state of a particle is described by a linear combination of its eigenvectors, which is called a superposition.Schrödinger EquationThe Schrödinger Equation is a mathematical equation that describes the time evolution of a wave function. It is based on the principle of conservation of energy, and it representsthe motion of a quantum system in terms of its wave function. The equation is given by:$$\\hat{H}\\Psi=E\\Psi$$where $\\hat{H}$ is the Hamiltonian operator, $\\Psi$ is the wave function, and E is the energy of the system. The Schrödinger Equation is the foundation of Quantum Mechanics, and it is used to calculate various physical properties of a particle, such as its energy and momentum.Practice Exercises1.Calculate the wave function for a particle that isin a 1D box of length L.–Answer: The wave function for a particle in a 1D box is given by:$$\\Psi(x)=\\sqrt{\\frac{2}{L}}\\sin{\\frac{n\\pi x}{L}}$$where n is a positive integer.2.Derive the time-dependent Schrödinger Equation.–Answer: The time-dependent SchrödingerEquation is given by:$$i\\hbar\\frac{\\partial\\Psi}{\\partialt}=\\hat{H}\\Psi$$3.Calculate the momentum operator for a particle in1D.–Answer: The momentum operator for a particle in 1D is given by:$$\\hat{p_x}=-i\\hbar\\frac{\\partial}{\\partial x}$$4.What is the uncertnty principle?–Answer: The uncertnty principle is afundamental principle of Quantum Mechanics thatstates that the position and momentum of a particlecannot be measured simultaneously with arbitraryprecision. Mathematically, it is given by: $$\\Delta x\\Delta p_x\\geq\\frac{\\hbar}{2}$$5.Calculate the energy of a particle in a 1D box oflength L with quantum number n.–Answer: The energy of a particle in a 1D box is given by:$$E_n=\\frac{n^2\\pi^2\\hbar^2}{2mL^2}$$ConclusionQuantum Mechanics is a fascinating and challenging fieldof study that has provided a deeper understanding of the behavior of matter and energy at the microscopic level. Theprinciples of Quantum Mechanics have been applied to various fields of study, including computing, cryptography and medicine, and they have contributed to significant advances in these fields. The practice exercises provided in this document are intended as a tool for students to test their knowledge and understanding of Quantum Mechanics. By solving these exercises, students will gn a deeper understanding of the fundamental principles of Quantum Mechanics and strengthen their problem-solving skills in this exciting field of study.。

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