新视野大学英语视听说教程4第二版答案-求新视野大学英语视听说教程3第二版答案.（外研出版社,总主编：郑树棠,主编：王大伟）

发布时间：2017-08-03 所属栏目：圆周运动

一 : 求新视野大学英语视听说教程3第二版答案.（外研出版社,总主编：郑树棠,主编：王大伟）

求新视野大学英语视听说教程3第二版答案.（外研出版社,总主编：郑树棠,主编：王大伟）

如题!是视听说,不是读写.好人一生平安感激不尽呐!

新视野大学英语听说教程第三册答案求新视野大学英语视听说教程3第二版答案.（外研出版社,总主编：郑树棠,主编：王大伟）

求新视野大学英语视听说教程3第二版答案.（外研出版社,总主编：郑树棠,主编：王大伟）的参考答案

Unit 1

Part I

1.A 2.B 3.D 4.B 5.D

Part II

(1)for (2)with (3)opportunity (4)tuition (5)explore (6)encounter (7)adventure (8)As with any country, it is not advisable to carry large amounts of cash around with you (9)Traveler's checks are one of the safest and easiest ways to transport money, because you may have them replaced if they get lost or stolen (10)It is wise to bring about $100 with you in U.S. cash, so you will be able to manage upon your arrival in the States

Part III

1.C 2.A 3.B 4.D 5.A

Part IV

1.C 2.A 3.A 4.D 5.C 6.B 7.A 8.C 9.B 10.C

Unit2

Part I

1.C 2.C 3.B 4.C 5.D

Part II

(1)chemicals (2)atmosphere (3)particles (4)trapped (5)lasts (6)human-based (7)progressed(8)Even in Ancient Rome people complained about smoke put into the atmosphere (9)Air pollution can have serious consequences for the health of human beings (10)Cities with large numbers of automobiles or those that use great quantities of coal often suffer most severely from air pollution problems

Part III

1.A 2.C 3.D 4.C 5.B

Part IV

1.B 2.C 3.D 4.C 5.C 6.A 7.D 8.D 9.B 10.C

Unit3

Part I

1.B 2.C 3.B 4.B 5.D

Part II

(1) values(2) purpose (3) true (4) Nowhere (5) equality (6) races (7) laboratory (8) Although Edwards specifically mentions young men, young women also compete in organized sports without regard to their race or economic background (9) Many Americans believe that learning how to win in sports helps develop the habits necessary to compete successfully in later life (10) the competitive ethic taught in sports must be learned and cultivated in youth for the future success of American business and military efforts

Part III

1.C 2.A 3.B 4.D 5.D

Part IV

1.D 2.B 3.A 4.D 5.D 6.C 7.A 8.B 9.B 10.A

Unit4

Part I

1.D 2.A 3.D 4.C 5.C

Part II

(1)at (2) wheel (3) tone (4) expression (5) Honey (6) divorce (7) speed (8) I don't want you to try to talk me out of it because I don't love you any more (9) I want the bank accounts, and all the credit cards, too (10) No, I've got everything I need

Part III

1.C 2.A 3.B 4.D 5.A

Part IV

1.A 2.D 3.B 4.C 5.A 6.D 7.A 8.B 9.C 10.D

Unit5

Part I

1.C 2.A 3.C 4.D 5.D

Part II

(1)hang (2) remains (3) symbol (4) performed (5) what (6) that (7) represent (8) By the year 1600, some Germans began bringing evergreen trees into their homes. They put fruit, nuts and sweets on the trees (9) They say he did this to show how wonderful the stars had appeared to him as he traveled one night (10) The Christmas tree tradition spread to many parts of the world. Today, some form of Christmas tree is part of most Christmas celebrations

Part III

1.D 2.A 3.B 4.C 5.D

Part IV

1.C 2.A 3.B 4.C 5.C 6.A 7.B 8.D 9.A 10.C

Unit6

Part I

1C 2B 3C 4D 5C

Part II

(1) ice (2) Examples (3) parking (4) response (5) provide (6) impression (7) ages (8) If this is the place where you would be working, they would automatically be worried that you will be late for work everyday (9) They might be historical questions regarding your previous employment or education (10) Make a statement, then support it by giving an example of a situation and how you handled it

Part III

1C 2B 3A 4C 5D

Part IV

1D 2A 3B 4C 5B 6C 7A 8A 9D10C

Unit7

Part I

1A 2D 3D 4C 5D

Part II

(1) resulting (2) demand (3) on (4) trends (5) elderly (6) trend (7) pet-related (8) Look at existing businesses and the products and services they offer, and then determine if there's a need for more of those products or services (9) Keep an open mind and continue to assess everything you read and hear from a business point of view (10) Dream, think, plan, and you'll be ready to transform that business idea into the business you've always wanted

Part III

1C 2B 3D 4A 5C

Part IV

1B 2D 3B 4C 5A 6B 7D 8C 9D 10D

Unit8

1C 2B 3A 4D 5D

Part II

(1) habit (2) crucial (3)among (4) decade (5) did (6) issues (7) adult (8) teens get turned off if you ask them a lot of direct questions (9) The first phone call generally takes about 45 minutes, with discussion about why the teen began smoking and the extent of the habit (10) Now 260 teens who have participated in the program are being followed

Part III

1A 2B 3C 4B 5D

Part IV

1C 2B 3A 4C 5B 6A 7B 8C 9D 10C

Unit9

Part I

1C 2D 3C 4B 5C

Part II

(1) tiny (2) identity (3) maintain (4) fuel (5) networks (6) insurance (7) external (8) Ticket prices include a number of fees, taxes (9) If airlines carry passengers without proper documents on an international flight, they are responsible for carrying them back to the originating country (10) While airlines as a whole earned 6% return on capital employed, airports earned 10%

Part III

1C 2A 3B 4D 5C

Part IV

1A 2B 3D 4C 5B 6D 7B 8A 9D 10B

Unit10

Part I

1B 2C 3B 4C 5D

Part II

(1) damage (2) symptoms (3) sharp (4) involved (5) fitness (6) mind (7) over (8) They also may remember things better by mentally connecting them to other meaningful things (9) Stress, anxiety, or depression can make a person more forgetful (10) much pain and suffering can be avoided if older people, their family, and their doctors recognize dementia as a disease

Part III

1B 2C 3B 4A 5D

Part IV

1B 2C 3A 4C 5D 6C 7B 8A 9B 10D

二 : 哪里有新编大学英语泛听教程第二册mp3下载我要高等教育出版社出版

哪里有新编大学英语泛听教程第二册 mp3下载

我要高等出版社出版的，如果有，也可发邮件给我，
邮箱是

google了半天，没找到~~

三 : 量子力学教程答案(第二版)

量子力学习题及解答

第一章量子理论基础

1．1 由黑体辐射公式导出维恩位移定律：能量密度极大值所对应的波长?m与温度T成反比，即

?m T=b（常量）

；并近似计算b的数值，准确到二位有效数字。［www.61k.com]

解根据普朗克的黑体辐射公式

8?hv3

vdv?c

hvdv，（1）ekT

以及 ?v?c，（2）?vdv???vd?，（3）有

?dv????

d?d??c????????

v(?)

??v(?)??c

?8?hc5?1?

hc,e

?kT

这里的??的物理意义是黑体内波长介于λ与λ+dλ之间的辐射能量密

度。

本题关注的是λ取何值时，??取得极大值，因此，就得要求?? 对λ的一阶导数为零，由此可求得相应的λ的值，记作?m。但要注意的是，还需要验证??对λ的二阶导数在?m处的取值是否小于零，如果小于零，那么前面求得的?m就是要求的，具体如下：

??'

8?hc1

?hc1?

???6?hc??5???0e?kT?1???kT1?e?hc??kT?

? ?5?hc?kT

?0 1?e?hc

?kT? 5(1?e

?hc?kT

)?hc?kT

如果令x=hc

?kT

，则上述方程为

5(1?e?x)?x

这是一个超越方程。首先，易知此方程有解：x=0，但经过验证，此解是平庸的；另外的一个解可以通过逐步近似法或者数值计算法获得：x=4.97，经过验证，此解正是所要求的，这样则有

?hcmT?

把x以及三个物理常量代入到上式便知

?mT?2.9?10?3m?K

这便是维恩位移定律。据此，我们知识物体温度升高的话，辐射的能量分布的峰值向较短波长方面移动，这样便会根据热物体（如遥远星体）

量子力学答案量子力学教程答案(第二版)

的发光颜色来判定温度的高低。［www.61k.com］

1．2 在0K附近，钠的价电子能量约为3eV，求其德布罗意波长。

解根据德布罗意波粒二象性的关系，可知

E=hv，

如果所考虑的粒子是非相对论性的电子（E动???2ec），那么

E?p2

2? e

如果我们考察的是相对性的光子，那么

E=pc

注意到本题所考虑的钠的价电子的动能仅为3eV，远远小于电子的质量与光速平方的乘积，即0.51?106

eV，因此利用非相对论性的电子的能量——动量关系式，这样，便有

hp ?

h2eE

?hc2?2ecE6

1.24?10?2?0.51?106

m ?0.71?10?9m?0.71nm

在这里，利用了

hc?1.24?10?6eV?m

以及

?2ec?0.51?106eV

最后，对

hc2?2

ecE

作一点讨论，从上式可以看出，当粒子的质量越大时，这个粒子的波长就越短，因而这个粒子的波动性较弱，而粒子性较强；同样的，当粒子的动能越大时，这个粒子的波长就越短，因而这个粒子的波动性较弱，而粒子性较强，由于宏观世界的物体质量普遍很大，因而波动性极弱，显现出来的都是粒子性，这种波粒二象性，从某种子意义来说，只有在微观世界才能显现。

1．3 氦原子的动能是E?3

kT（k为玻耳兹曼常数），求T=1K时，氦原子的德布罗意波长。

解根据

1k?K?10?3eV，

知本题的氦原子的动能为

3kT?3

k?K?1.5?10?322

eV, 显然远远小于?2核c这样，便有

hc2?2

核cE

量子力学答案量子力学教程答案(第二版)

1.24?10?6

2?3.7?109

?1.5?10?3

?0.37?10?9m

?0.37nm

这里，利用了

?29核c?4?931?106eV?3.7?10eV

最后，再对德布罗意波长与温度的关系作一点讨论，由某种粒子构成的温度为T的体系，其中粒子的平均动能的数量级为kT，这样，其相庆的德布罗意波长就为

2?c2

hc2?kc2

据此可知，当体系的温度越低，相应的德布罗意波长就越长，这时这种粒子的波动性就越明显，特别是当波长长到比粒子间的平均距离还长时，粒子间的相干性就尤为明显，因此这时就能用经典的描述粒子统计分布的玻耳兹曼分布，而必须用量子的描述粒子的统计分布——玻色分布或费米公布。[www.61k.com］

1．4 利用玻尔——索末菲的量子化条件，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轨道的可能半径。已知外磁场H=10T，玻尔磁子M?24

B?9?10J?T，试计算运能的量子化间隔△E，并与T=4K及T=100K的热运动能量相比较。

解玻尔——索末菲的量子化条件为

pdq?nh

其中q是微观粒子的一个广义坐标，p是与之相对应的广义动量，回路积

分是沿运动轨道积一圈，n是正整数。

（1）设一维谐振子的劲度常数为k，谐振子质量为μ，于是有

E?p212

2??2kx

这样，便有

p??2?(E?

kx2) 这里的正负号分别表示谐振子沿着正方向运动和沿着负方向运动，一正一负正好表示一个来回，运动了一圈。此外，根据

12kx2 可解出 x2E

???k

这表示谐振子的正负方向的最大位移。这样，根据玻尔——索末菲的量子化条件，有

x2?(E?1kx2)dx??

?x?(?)2?(E?1

kx2?2x)dx?nh

x2?(E?1kx2)dx??x?2?(E?1

kx2?

2x)dx?nh

x?1?2?(E?kx2)dx?nh为了积分上述方程的左边，作以下变量代换； ?x?

sin? 这样，便有

量子力学答案量子力学教程答案(第二版)

?22?Ecos2

?d??2E?n?

?sin???k???

2h ?

2Ecos??

2Ekcos?d??n

?2?

2E?

cos2?d??

h这时，令上式左边的积分为A，此外再构造一个积分

B??2?

??2E?

?d?

sin2这样，便有

A?B??2?

??2E?

d??2E???

（1）A?B??2??2E?

cos2?d?

??2?

??E

cos2?d(2?)

??2??E

?d?,

cos这里? =2θ，这样，就有

A?B???

dsin??0 根据式（1）和（2），便有

A?E?

这样，便有

h ? E?n2?h?k

??k,

其中h?

最后，对此解作一点讨论。(www.61k.com)首先，注意到谐振子的能量被量子化了；其次，这量子化的能量是等间隔分布的。

（2）当电子在均匀磁场中作圆周运动时，有

?q?B ?这时，玻尔——索末菲的量子化条件就为 p????qBR

qBRd(R?)?nh

? qBR2?2??nh ? qBR2?nh

2）

又因为动能耐E?2?

，所以，有

（

量子力学答案量子力学教程答案(第二版)

E?(qBR)2q2B2R2

2??

2? ?qBn?q?2??nB?2? ?nBNB,

其中，MB?q?

是玻尔磁子，这样，发现量子化的能量也是等间隔的，而且

?E?BMB

具体到本题，有

?E?10?9?10?24J?9?10?23J

根据动能与温度的关系式

kT 以及

1k?K?10?3eV?1.6?10?22J

可知，当温度T=4K时，

E?1.5?4?1.6?10?22J?9.6?10?22J

当温度T=100K时，

E?1.5?100?1.6?10?22J?2.4?10?20J

显然，两种情况下的热运动所对应的能量要大于前面的量子化的能量的间隔。[www.61k.com)

1．5 两个光子在一定条件下可以转化为正负电子对，如果两光子的能量相等，问要实现实种转化，光子的波长最大是多少？

解关于两个光子转化为正负电子对的动力学过程，如两个光子以怎样的概率转化为正负电子对的问题，严格来说，需要用到相对性量子

场论的知识去计算，修正当涉及到这个过程的运动学方面，如能量守恒，动量守恒等，我们不需要用那么高深的知识去计算，具休到本题，两个光子能量相等，因此当对心碰撞时，转化为正风电子对反需的能量最小，因而所对应的波长也就最长，而且，有

E?hv??ec2

此外，还有

E?pc?

于是，有

??ec2

???

?ec2 1.24?10?6?0.51?106

m?2.4?10?12m

?2.4?10?3nm

尽管这是光子转化为电子的最大波长，但从数值上看，也是相当小的，我们知道，电子是自然界中最轻的有质量的粒子，如果是光子转化为像正反质子对之类的更大质量的粒子，那么所对应的光子的最大波长将会更小，这从某种意义上告诉我们，当涉及到粒子的衰变，产生，转化等问题，一般所需的能量是很大的。能量越大，粒子间的转化等现象就越丰富，这样，也许就能发现新粒子，这便是世界上在造越来越高能的加速器的原因：期待发现新现象，新粒子，新物理。

第二章波函数和薛定谔方程

量子力学答案量子力学教程答案(第二版)

?i?**(1) J1?(?1??1??1??1)2m

i?1ikr?1?ikr1?ikr?1ikr??? ?[e(e)?e(e)]r0?(r，t)??(r)f(t)2mr?rrr?rri???Et i?111111? ??(r)e ?[(?2?ik)?(?2?ik)]r02mrrrrrr?i? J?(???*??*??)?k??k? ?r?r2m203mrmriiii??Et?Et?Et?Et?????i?* ] J1与r同向。[www.61k.com)表示向外传播的球面波。 ?[?(r)e??（?(r)e?）??*(r)e??（?(r)e?）2m

??i?*?*? ?[?(r)??(r)??(r)??(r)]?i?**2m(2) J2?(?2??2??2??) 2m? 可见J与t无关。 i?1?ikr?1ikr1ikr?1?ikr? ?[e(e)?e(e)]r0 2mr?rrr?rr2.2 由下列定态波函数计算几率流密度：

1ikr1?ikri?111111? (1)?1?e (2)?2?e ?[(??ik)?(??ik)]r0rr2mrr2rrr2r?2表示向内(即向原点) 传从所得结果说明?1表示向外传播的球面波，?k??k?播的球面波。 ??2r0??3r??mrmr 解：J1和J2只有r分量 2.1证明在定态中，几率流与时间无关。证：对于定态，可令

???1??1??e??e?在球坐标中 ??r0 ?rr??rsin??? 可见，J2与r反向。表示向内(即向原点) 传播的球面波。

ikx补充：设?(x)?e，粒子的位置几率分布如何？这个波函数能否归

一化？ ??

量子力学答案量子力学教程答案(第二版)

?*?dx?dx??

? ∴波函数不能按?

(x)

dx?1方式归一化。（www.61k.com]

其相对位置几率分布函数为 ??2

?1表示粒子在空间各处出现的几率相同。

2.3 一粒子在一维势场

? U(x)??

?，x?0?0，

0?x?a ??

?，x?a中运动，求粒子的能级和对应的波函数。

解：U(x)与t无关，是定态问题。其定态S—方程

??2d2

2mdx

?(x)?U(x)?(x)?E?(x) 在各区域的具体形式为 Ⅰ

?20 ?d2

x?2mdx2

?1(x)?U(x)?1(x)?E?1(x)

①

Ⅱ

0?x?a ?

2mdx

?2(x)?E?2(x) ② Ⅲ

?a ??2d2

x2mdx2

?3(x)?U(x)?3(x)?E?3(x) ③由于(1)、(3)方程中，由于U(x)??，要等式成立，必须

?1(x)?0 ?2(x)?0 即粒子不能运动到势阱以外的地方去。

方程(2)可变为d2?2(x)dx2

?2mE

2?2(x)?0

令k2

2mE

2，得 d2?2(x)dx

?k2

?2(x)?0 其解为 ?2(x)?Asinkx?Bcoskx ④

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) ⑤

?2(a)??3(a) ⑥

⑤ ?B?0

⑥

?Asinka?0 ?A?0

?sinka?0

?ka?n? (n?1, 2, 3,

∴?2(x)?Asinn?

x 由归一化条件

(x)2

dx?1

得 A

sin2

xdx?1 0

：

量子力学答案量子力学教程答案(第二版)

由

sinm?n?ab

ax?sinaxdx?2

?A?

??2n?

2(x)?asiax

?k2

?2mE?2

?E?2?2n?2ma

( n ? 1,2,3,?)可见E是量子化的。［www.61k.com］对应于En的归一化的定态波函数为

?2sinn?

xe?i

Ent ??

, 0?x?an(x,t)??a ?a

0, x?a, x?a #

2.4. 证明（2.6-14）式中的归一化常数是A??

n?证：

??A?sin(x?a),x?an??a??

0, x?a（2.6-14）

由归一化，得

1??2

2ndx??A?sin2

(x?a)dx?A?2?

a1?a2[1?cosn?a

(x?a)]dx ?A?2a

A?2

2?a2

cos

(x?a)dxA?2a

?A?2

a?2?an?n?sina(x?a)

?a?A?2a

∴归一化常数A??

2.5 求一维谐振子处在激发态时几率最大的位置。

2 解：?(x)??2??x

2 ? xe2

?1(x)?2

1(x)?4?2?

?x2e??

?2?

?x2e??2x2

d ? 1( x ) 2 ? 3

dx ?[2x?2?2x3]e??2x2 令

d?1(x)

?0，得 8

量子力学答案量子力学教程答案(第二版)

x?0

x??1

x???

由?1(x)的表达式可知，x?0 ，

x???时，?1(x)?0。［www.61k.com］显然不是最大几率的位置。

而d2?1(x) ?2?3[(2?6?2x2)?2?2x(2x?2?2x3)]e

??2x2

dx23

?4?

[(1?5?2x2?2?4x4)]e??2x

2 d2?1(x)dx2??24?31?0x??

可见x??

是所求几率最大的位置。 #

2.6 在一维势场中运动的粒子，势能对原点对称：U(?x)?U(x)，证明粒子的定态波函数具有确定的宇称。

证：在一维势场中运动的粒子的定态S-方程为

?2d2

?2?dx

?(x)?U(x)?(x)?E?(x) ①

将式中的x以(?x)代换，得

??2d2

2?dx

?(?x)?U(?x)?(?x)?E?(?x) ②

利用U(?x)?U(x)，得

??2d2

2?dx

?(?x)?U(x)?(?x)?E?(?x) ③

比较①、③式可知，?(?x)和?(x)都是描写在同一势场作用下的粒子状态的波函数。由于它们描写的是同一个状态，因此?(?x)和?(x)

之间只能相差一个常数c。方程①、③可相互进行空间反演 (x??x)而得其对方，由①经x??x反演，可得③，

?(?x)?c?(x) ④

由③再经?x?x反演，可得①，反演步骤与上完全相同，即是完全等价的。

?(x)?c?(?x) ⑤

④乘 ⑤，得

?(x)?(?x)?c2

?(x)?(?x) 可见，c2

c??1

当c??1时，

?(?x)??(x)，??(x)具有偶宇称，当c??1时，

?(?x)???(x)，??(x)具有奇宇称，当势场满足 U(?x)?U(x)时，粒子的定态波函数具有确定的宇称。

2.7 一粒子在一维势阱中

U(x)????U0?0,x?a

?? 0,

x?a

运动，求束缚态(0?E?U0)的能级所满足的方程。

量子力学答案量子力学教程答案(第二版)

解法一：粒子所满足的S-方程为

??2 d2

2?dx

?(x)?U(x)?(x)?E?(x) 按势能U(x)的形式分区域的具体形式为

Ⅰ

：

??2d22?dx2

?1(x)?U0?1(x)?E?1

(x)???x?a ①

Ⅱ

：

2?dx

?2(x)?E?2(x)?a?x?a ②

Ⅲ

：

?d2?dx2

?3(x)?U0?3(x)?E?3(x)a?x?? ③

整理后，得

Ⅰ： ?2?(U0?E)

????2

?1?0 ④

Ⅱ：. ??2

??2? E

2?2?0 ⑤ Ⅲ：??3

??2?(U0?E)?2?3?0 ⑥ 令 k22?(U0?E)1??2 k2

2?2?E?2

则

Ⅰ： ????k2

1?1?0 ⑦ Ⅱ：. ????k2

2?2?0 ⑧ Ⅲ：??3

??k21?1?0 ⑨ 各方程的解为

?k1?Ae?1

x?Bek1

?2?Csink2x?Dcosk2x

? 3?Ee?k1x?Fe?k1x

由波函数的有限性，有

?1(??)有限 ?A?0

? 有限 ?E?0 3 ( ? )因此

?1?Bekx

??k 3?Fe

1x 由波函数的连续性，有

?1(?a)??2(?a),?Be?k1

a??Csink2a?Dcosk2a (10)

????(?a),?kk1

a1?(?a)21Be??k2Ccosk2a?k2Dsink2a (11)

?2(a)??3(a),?Csink2a?Dcosk2a?Fe

?k1a

(12)

??(a)??3?(a),?k2Ccosk2a?k2Dsink2a??k1Fe?k1

a2 (13)

整理(10)、(11)、(12)、(13)式，并合并成方程组，得

e?k1aB?sink2aC?cosk2aD?0?0

k1e?1aB?k2cosk2aC?k2sink2a D?0?00?sink2aC?cosk2aD?e

?k1a

F?0

0?k2cosk2aC?k2sink2aD?k1e?k1aF?0

解此方程即可得出B、C、D、F，进而得出波函数的具体形式，要

方程组有非零解，必须

量子力学答案量子力学教程答案(第二版)

e?k1asink2a?cosk2a0a

k1e?k1?k2cosk2a?k2sink2a0

0sink2acosk?k2a

k2cosk2a

?k?k2sink2ak1Be1a

?k2cosk2a?k2sink2a

0?e?k1a

sink2a

coska2a?e?k1?

k2cosk2a

?k2sink2ak?k1e1a

sink2a

?cosk2a0 ?ka1e?k1sink2a

cosk2a

?e?k1a?

kk?k2cos2a?k2sink2ak1e

?e?k1a[?k?k?k1k2e1acos2k2a?k22e1a

sink

2acosk2a? ?k1asin2k?k1k2e?k2a?k22e

1asink2acosk2a]? ?k?k?k1e1a[k1e1asink2acosk2a?k2e?k1acos2k2a? ?k?kk1e1asink2acosk2a?k2e?1asin2k2a] ?e

?2k1a

[?2k2

1k2cos2k2a?ksin2k2

2a?k1sin2k2a]

?e?2k1a[(k22

2?k1)sin

2k2a?2k1k2co2sk2a] ∵ e

?2k1a

∴(k22

2?k1)sin2k2a?2k1k2cos2k2a?0

即 (k2

2?k1)tg2k2a?2k1k2?0为所求束缚态能级所满足的方程。（www.61k.com)# 解法二：接（13）式

?Csink2a?Dcoskk22a?

kCcoskk

22a?Dsink2a 1k1

Csink22a?Dcosk2a??

kkCcoska?k

22Dsink2a 1k1

k2kcoska?sinkk2

22asink2a?cosk2ak1k1

2kcoskk2

2a?sink2a?(ksink2a?cosk2a)11?(k2kcoskk

2a?sink2a)(2ksink2a?cosk2a)11?(

k2kcoskk

2a?sink2a)(2sink2a?cosk2a)?01k1 (

k2kcoska?sinkk

22a)(2sink2a?cosk2a)?01k1

2k2k

k2sink2acosk2a?sin2k2a?2cos2k2a?sink2acosk21kka?0

11 (?1? k2

2ksin2k2k

2)2a? 2cos2k1k2a?0

1 (k2

22?k1)sin2k2a? 2k1k2cos2k2a?0

解法三：

(11)-(13)?2k?k2Dsink2a?k1e

(B?F)

(10)+(12)?2Dcosk?ka

2a?e1(B?F)

量子力学答案量子力学教程答案(第二版)

(11)?(13)

(10)?(12)

?k2tgk2a?k1 (a)

(11)+(13)?2k?ik2Ccosk2a??k1(F?B)e1a

(12)-(10)?2Csink?ik2a?(F?B)e

(11 ) ? (13 )

(12 ) ? (10 )

? k 2 ctgk 2 a ? ? k 1

令 ??k2a，??k2a，则

? tg??? (c)或

? ctg???? (d)

?2??2?(k2

2?U0a221?k2)?

(f)

合并(a)、

(b)： tg2k2k1k22a?

k2k2 利用tg2k2tgk2a2a?2

2?11?tgk2a

解法四：（最简方法-平移坐标轴法）

Ⅰ：?

2??1

???U0?1?E?1 （χ≤0） Ⅱ：??2

2???2

??E?2 （0＜χ＜2a） Ⅲ：??2

??3

??U0?3?E?3 （χ≥2a） ?

??1???2?(U0?E)?1?0?

?????2?E?

2?2?2?0

???

??2?(U0?E)3???2

?3?0??1???k22

1?1?0 (1) k1?2?(U0?E)?2????2??k22

2?2?0 (2) k22?2?E?束缚态0＜??

??3??k21?3?0 (3)E＜U0

?k1

xk1

1?Ae??Be??2?Csink2x?Dcosk2x

?3?Ee?k1

x?Fe?k1

?1(??)有限 ?B?0

??0

3(?)有限 ?E因此

??1?Aek1x ?3?Fe

?k1x

由波函数的连续性，有

量子力学答案量子力学教程答案(第二版)

?1(0)??2(0),?A?D (4)

?1?(0)???2(0),?k1A?k2C (5)

??(2a)??3?(2a),?k2Ccos2k2a?k2Dsin2k2ka22a??k1Fe?1

(6)

?2(2a)??3(2a),?Csin2k2a?Dcos2kk2a?Fe?21

a (7)

(7)代入(6)

Csin2kk22a?Dcos2k2a??kCcos2kk

2a?2Dsin2k2a 1k1

利用(4)、(5)，得

k1Asin2ka??Acos2kk

2k2a?Acos2k22a?Dsin2k2a2k1A[(

k1k?k2

k)sin2k2a?2cos2k2a]?021

?A?0

k1k?k2

)sin2k2a?2cos2k2a?02k1

两边乘上(?k1k2)即得

(k22

2?k1)sin2k2a?2k1k2cos2k2a?0

2.8分子间的范德瓦耳斯力所产生的势能可以近似表示为

???,

x?0 ， U(x)???U0

, 0?x?a，?b，

??U1, a?x??0,

b?x ，求束缚态的能级所满足的方程。(www.61k.com]

解：势能曲线如图示，分成四个区域求解。定态S-方程为

??2d2

2?dx

?(x)?U(x)?(x)?E?(x) 对各区域的具体形式为

Ⅰ：??2

2??1

???U(x)?1?E?1 (x?0) Ⅱ：??2

2???2??U0?2?E?2 (0?x?a) Ⅲ：??2

2???3

??U1?3?E?3 (a?x?b) Ⅳ：??2

??4??0?E?4 (b?x) 对于区域Ⅰ，U(x)??，粒子不可能到达此区域，故 ?1(x)?0

而 . ??2? (2

??U0?E)?2

?2?0 ① ??2? 3

??(U1?E)?2

?3?0 ②13

量子力学答案量子力学教程答案(第二版)

2?E

???2?4?0 ③ ?4

对于束缚态来说，有?U?E?0

?3(b)??4(b)?Csink2b?Dcosk2b?Fe⑨

?k3b

?(b)??4?(b)?Ck2sink2b?Dk2cosk2b??Fk3e ?3

∴ ????k2 k22? (U0?E)

21?2?0 1??2

④

???E)

3??k2

? k2

33?03?

2? (U1?2

⑤

??4??k2?2

44?0 k4??2?E/?2

⑥

各方程的解分别为

?k1

x2?Ae?Be?k1

?3?Csink2x?Dcosk2x

?4?Ee?k3

x?Fe?k3

由波函数的有限性，得 ?4(?)有限， ? E?0 ∴ ??kx

4?Fe3 由波函数及其一阶导数的连续，得 ?1(0)??2(0) ?B??A ∴ ?k2?A(e3x

?e?k3x)

?(a)?A(ek3x?e?k3x

2(a)??3)?Csink2a?Dcoks2a ⑦

??k3

(a)??3?(a)?Ak1(e3a

?e?k3a)?Ck2cosk2a?Dk2sink2a⑧

⑩

由⑦、⑧，得k1ek1a?e?k1aCcosk2a?Dcosk2a

kka?

Csink (11) 2e1a?e?k12a?Dcosk2a

由

⑨

、

⑩

得

(k2ck2b)C?(ok2sk2b)Ds?i(?k3sk2bn)C?(ik3c

k2b)Dn o

(k2kcoskb)C?(?k

22b?sink2cosk2b?sink2b)D?0

3k3

(12)

ek1a?e?k1a 令??k1

ek1

a?e

?k1a?k，则①式变为 2 (?sink2a?cosk2a)C?(?cosk2a?sink2a)D?0

联立(12)、(13)得，要此方程组有非零解，必须

k2 (

kcoksk22b?sink2b)(?ksink2b?coks2b)

(?3?0 sink?coks3

2a2a)(?coks2a?sink2a)

量子力学答案量子力学教程答案(第二版)

即 (?cosksinkk

22a?2a)(kcosk2b?sink2b)?(?sink2a?cosk2a)?

3 ?(?

ksink2b?cosk2b)?03

?k2kcoska?k2

2bcosk2sink2bsink2a??sink2bcosk2a?

3k3 ?sink2bsinkk22a??

ksinkbsinkk

22a?2ksink2bcosk2a)?33 ??cosk2bsink2a?cosk2bcosk2a?0 sinkk22(b?a)(??

k)?cosk)((?k

22(b?a?1)?03k3

tgk?a)?(1?k2?)(k22(bk??)

3k3

把?代入即得

tgkk2ek1a?e?k1ak2k1ek1a?e?k1a

2(b?a)?(1?kaa)(??k1a

)

3ek1?e?k1k3k2ek1a?e 此即为所要求的束缚态能级所满足的方程。[www.61k.com) #

附：从方程⑩之后也可以直接用行列式求解。见附页。

(ek1a?e?k1a)?sink2a?cosk2a

(ek1a?e?k1a)k2

?k2cosk2ak2sink2a0

0sink?k2bcosk2b?e3a

?00ka

2cosk2b?k2sink?k2bk3e3?k2cosk2ak2sink2a0

0?(ek1a

?e?k1a)sinkk2bcosk2b?e?3a?

k?k2cosk2b?k2sink2bk3e3a

?sink2a?cosk2a0

?k1(ek1a

?e?k1a)?sink?k2bcosk2b?e3a

k2cosk2b?k2sink2bk3e?k3a

?(ek1a?e?k1a（)?k?ka2?k2k3e3cosk2acosk2b?k2e3a

sink2a

cosk?k2?k2b?k2k3e3asink2asink2b?k2e3acosk2asink2b）

?k1(ek1b?e?k1b（)kk2k3e?3bsink2acosk2b?k2e?k3bcosk2a cosk2b?k?k3e3bcosk?kb2asink2b?k2e3sink2asink2b））

?(ek1a?e?k1a)[?k22k3cosk2(b?a)?k2sink?k ? ( e 2(b?a)]e3b

k 1a ? e ? k1 a )[ k ?k1k 3sin k 2( b

?a)?k1k2cosk2(b?a)]e3b?ek1a[?(k(k2k1?k3)k2cosk2(b?a)?2?k1k3)sink2(b?a)]e?3b

e?k1a[(kcosk2

1?k3)k22(b?a)?(k2?k1k3)sink2(b?a)]e?k3b?0

量子力学答案量子力学教程答案(第二版)

? [?(k1?k3)k2?(k2?k1k3)tgk2(b?a)]e?k3b

?[(k1?k3)k2?(k?k1k3)tgk2(b?a)]e [(k?k1k3)e

2k1a

?k3b

2k1a

∴a?

在界限外发现振子的几率为

?a0

?(k?k1k3)]tgk2(b?a)?(k1?k3)k2e

?(k1?k3)k2?0

此即为所求方程。[www.61k.com） #

补充练习题一

1、设 ?(x)?Ae?1

?2x2(?为常数)，求A = ？解：由归一化条件，有 1?A

e?d( x)?A2

e??

d(? x)

?A2

e?dy?A2

∴A?

2、求基态微观线性谐振子在经典界限外被发现的几率。

解：基态能量为E10?

?? 设基态的经典界限的位置为a，则有

E122

10?2??a?2

? ?

? a 0

?22

e ? x dx ?? ? e ??

x dx (? 0 ?

e ?

2 x 2

? ? ?

? ? a 0

)

?2?

?ae??

2dx (偶函数性质)

ae?(?x)d(? x)0?2

e?y2

利用

[??

?y2

dy??1

?y2

dy]

# [?

2?2

e?t

dt] （令y?

?t2/2

式中2??

edt为正态分布函数?(x)?

2??

e???

当x?

2时的值?(2)。查表得?(2)??0.92

∴????

[??0.92] ?2(1?0.92)?0.16 ∴在经典极限外发现振子的几率为0.16。 #

量子力学答案量子力学教程答案(第二版)

??23、试证明?(x)?e

(2?3x3?3?x)是线性谐振子的波函

把2?(x)代入①式左边，得

数，并求此波函数对应的能量。(www.61k.com] 证：线性谐振子的S-方程为

??2 d2 2?dx?(x)?1

??2x2?(x)?E?(x) ①

把?(x)代入上式，有

d?(x)d??1

2x2dx?dx[3e(2?3x3?3?x)]

???1

?2x23[??2x(2?3x3?3?x)?(6?3x2?3?)]e

3e?122x2

(?2?5x4?9?3x2?3?)d2?(x)d???1

2dx2

?dx?e2?x2(?2?5x4?9?3x2

?3?)??3? ?

???12?2?2x25432?1

?2x2?3???xe

(?2?x?9?x?3?)?e2(?8?5x3?18?3

x)???

?(?4x2?7?2)

2x22

3e(2?3x3?3?x)

?(?4x2?7?2)?(x)

左边???2d2?(x)122

2?dx2

??x?(x)2

?2 ?7?2??(x)??22??4x2?(x)?1

??2x2?(x)

?7??????22??(x)??22????4x2?(x)?1

??2x2?(x)

72???(x)?12??2x2?(x)?1222??x?(x) ?7

2???(x)右边?E?(x) 当E?72??时，左边 = 右边。 n = 3 ?(x)??d?12?2x2

dxe(2?3x3?3?x)，是线性谐振子的波

??。第三章量子力学中的力学量

函数，其对应的能量为

量子力学答案量子力学教程答案(第二版)

?2x23.1 一维谐振子处在基态?(x)

?2?i

?te?

，求：

(1)势能的平均值?122

??x2

；

(2)动能的平均值?p2

；

(3)动量的几率分布函数。［www.61k.com］解：(1) ?

2??2x2?1?2??2?

x2edx

?12??2??211112

?22?2?

?2??22?2

?4????? ?1

(2) ?

2???2?

????*(x)p?2

?(x)dx ?

?1?

2x22

22?

???e2

(??2ddx2)e2

?2xdx ???22?

??(1??2x2)e??x??

???2?2[??e??2x2

dx??2

??x222

????

e??xdx] ?

?222??[???2?2?3] ?

?22??22???22?2??

4???4??

?? 或 ?E??12???14???1

(3) c(p)???*

p(x)?(x)dx

2?i?

Px2??

??e2?x

?1?

?2x2?i2

Px2??

?1??

?1ipp2?2??

2???

2?2(x??2)2?22?? e

dx ?

1??

p22?2?

?2(x??2?

)222??e?

?1ip??

dx ?

1?p2p2 2??

2?2?2

??e

2??

动量几率分布函数为 p2

?(p)?c(p)2

??e

?2?2

量子力学答案量子力学教程答案(第二版)

?r/ae2e2

(2)?(?)??3

???

?2??

1?2r/a02ersin? drd? d? 3.2.氢原子处在基态?(r,?,?)?a3e0，求：

(1)r的平均值； 2

(2)势能?

的平均值； (3)最可几半径；

(4)动能的平均值；

(5)动量的几率分布函数。（www.61k.com] 解

(1)??

r(r,?,?)2

d??12?

??a3r2sin? drd? d?0

re?2r/a0 ?

a3r3a?2r/a0 0

?!3a34

?a0 0???2?

2?a?0??

r?a0

??e2

?2??

?2r/a0?a3

ersin? drd? d?

?4e2?2r/a0a3

r dr

e???4e21e2

a32

??0??2?a0

??a?0??

(3)电子出现在r+dr球壳内出现的几率为

?(r)dr??

[?(r,?,?)]2r2sin? drd? d??

4a3e?2r/a0r2

dr 0

?(r)?

4a3e?2r/a0r2

d?(r)?42

2r/a0dra3(2?r)re?

a0 令

d?(r)

?0， ? r1?0, r2??, r3?a0 当 r1?0, r2??时，?(r)?0为几率最小位置

d2?(r)4842?2r/a0

dr2?a3(2?r?2

r)e 0a0a0

：

量子力学答案量子力学教程答案(第二版)

d2?(r)

dr2

r?a0

??3e?2?0

2?(2??)3/2

?pr???r/a0?pr

re(e?e)dr ?3ip0

∴ r?a0是最可几半径。［www.61k.com］

1?22????? (4)T

p?112?2??] 1i1i(?p)2(?p)2?2?2?? ??a0?a0?3???0002??a0

14ip2 ??2???1?r/a01d2d?r/a02233??e[r(e)]rsin? drd? d?2a0?ip?a?(1?p)23

2??0?0?0?a0drr2dr02

a0?2

4?1

(?3

a02?a0

(2r?

r?r/a0

)e dr a0

44a0?2

2a??a0(a0p??)

a0a04?2?2

? (2?)?42

442?a02?a0

*??(r)?(r,?,?)d? (5) c(p)???p

(2a0?)3/2?

?(a0p??)

2222

动量几率分布函数

c(p)?

1(2??)3/2

3a0

e?r/a0r2dr?e

?prcos??

sin? d??d?

?prcos??

358a0?

?(p)?c(p)?2 224

?(a0p??)

2?(2??)

3/2

?r/a0

dr?e

d(?cos?)

3.3 证明氢原子中电子运动所产生的电流密度在球极坐标中的分量是 Jer?Je??0 Je??

2?(2??)

3/2

2?r/a0

???prcos?dreipr

e? m2

n?m

? rsin?

证：电子的电流密度为

量子力学答案量子力学教程答案(第二版)

J??i?**

e??eJ??e2?

(?n?m??n?m??n?m??n?m)

?在球极坐标中为

??e??1???1?

r?e? 式中e???

?rr???e?

rsin???

r、e?、e?为单位矢量

J??eJ???ei???1???1?*e? 2?[?n?m(er?r?re????e?rsin???

)?n?m

? ? *

??1???1?

n?m(er?r?re????e?

rsin???

)?n?m]

ie?2?[e??**

??1?*r(?n?m?r?n?m??n?m?r?n?m)?e?(?n?mr??

?n?m ??*1??mr??n?m?e??(1rsin???n?m???*n?m?1rsin??*

?n)n?m

?n?m)]

??n?m中的r和?部分是实数。[www.61k.com)

∴ J?

e??

i?e2?rsi?n

(?i22?

n?m?in?m)e?

??e?m?rsin?

2?n?me?

可见，Jer?Je??0

Je?me???

?rsin?

n?m #

3.4 由上题可知，氢原子中的电流可以看作是由许多圆周电流组成的。

(1)求一圆周电流的磁矩。 (2)证明氢原子磁矩为

??me?

M?M?

?2?(SI)z??

?me????2?c

(CGS) 原子磁矩与角动量之比为

??e

Mz??2?

(SI)L??

????e2?c

(CGS)这个比值称为回转磁比率。

解：(1) 一圆周电流的磁矩为 dM?iA?Je?dS?A （i为圆周电流，A为圆周所围面积）

e?mrsin?2

n?mdS??(rsin?)2?

e?m?

?rsin?2

n?mdS ??

e?m

?rsin?2

n?mdrd?

(dS?rdrd?)

(2)氢原子的磁矩为

量子力学答案量子力学教程答案(第二版)

M??dM?

???

e?m

?n?mrsin? drd?

??e?m??2?

?2??2

0?02n?mrsin? drd?

??e?m2???2?

?0?0?022

n?mrsin? drd?d? ??e?m

(SI)

在CGS单位制中 M???e?m

2?c

原子磁矩与角动量之比为

MzL?M??e

(SI) zLz2?

MzL??e

2?c

(CGS) # z 一刚性转子转动惯量为I，它的能量的经典表示式是H?L23.52I

，L

为角动量，求与此对应的量子体系在下列情况下的定态能量及波函数： (1) 转子绕一固定轴转动： (2) 转子绕一固定点转动：

解：(1)设该固定轴沿Z轴方向，则有 L2?L2

哈米顿算符 H??1?22

2IL?2Z??d2Id?

2 其本征方程为 (H

?与t无关，属定态问题) ?2d2? 2Id?

2?(?)?E?(?)

d?(?)d?2??2IE?

?(?) 令 m2

2IE

，则 d2?(?)2

?m?(?)?0 取其解为 ?(?)?Aeim? (m可正可负可为零) 由波函数的单值性，应有

?(??2?)??(?)?eim(??2?)?eim?

即 e

i2m?

?1 ∴m= 0，±1，±2，…

Em2?2

转子的定态能量为m?2I

(m= 0，±1，±2，…)

可见能量只能取一系列分立值，构成分立谱。［www.61k.com）定态波函数为

?m?Aeim?

A为归一化常数，由归一化条件 1?2?

0mm

???d??

d??A21

?A?

∴ 转子的归一化波函数为

?m?

12?

eim?

量子力学答案量子力学教程答案(第二版)

综上所述，除m=0外，能级是二重简并的。（www.61k.com］

[1?cos2kx?coskx] 2Ai2kxikx

?e?i2kx)??e?ikx)] ?[1?2(e2(e ? (2)取固定点为坐标原点，则转子的哈米顿算符为 H

??12I

L?2 H

?与t无关，属定态问题，其本征方程为 12I

L?2

Y(?,?)?EY(?,?) (式中Y(?,?)设为H

?的本征函数，E为其本征值) L

?2Y(?,?)?2IEY(?,?) 令 2IE???2

，则有

?2Y(?,?)???2Y(?,?) 此即为角动量L

的本征方程，其本征值为 L2???2??(??1)?2

(??0, 1, 2, ?) 其波函数为球谐函数Y?m(?,?)?Nm

?mP?(cos?)eim? ∴ 转子的定态能量为 E?(??1)?

可见，能量是分立的，且是(2??1)重简并的。

3.6 设t=0时，粒子的状态为

?(x)?A[sin2

kx?1coskx]

求此时粒子的平均动量和平均动能。

解：?(x)?A[sin2

kx?11

coskx]?A[1(1?cos2kx)?coskx]

A2??2[ei0x?1ei2kx?1?i2kx1ikx1?ikx1

22e?2e?2e]?

2?? 可见，动量pn的可能值0 2k? ?2k? k? ?k?

动能p2n

的可能值

02k2?22k2?2k2?2k2?2

??2?2? 对

应

的

几

率

应A24A216A216A2A2

(1616

)?2??

(12 11118888

)?A2?? 上述的A为归一化常数，可由归一化条件，得

1???A2A2 A2

n?(n

4?4?16)?2???2?2??

∴ A?1/?

∴ 动量p的平均值为

为

为23

量子力学答案量子力学教程答案(第二版)

??pn?n

0?2k??A216?2???2k??A216?2???k??A216?2???k??A2

?16

?2???0

?p2

p2n2????n

n2?

?0?2k2?21k2?21

??8?2?2??8?2

?5k2?2

# 3.7 一维运动粒子的状态是 ?Axe??x

?(x)??, 当x?0

0, 当x?0

其中??0，求：

(1)粒子动量的几率分布函数； (2)粒子的平均动量。[www.61k.com)

解：(1)先求归一化常数，由 1??

???

?2?x

(x)dx0

Axe

4?3

∴A?2?3/2

?(x)?2?3/2xe?2?x

(x?0)

?(x)?0 (x?0)

c(p)??

1?ikx??

2??

e?(x)dx?(

1??

)1/2

?2?3/2??xe?(??ik)x2???(x)dx

?(2?3)1/2[?xe?(??ik)x?1??(??ik)x

2????ik0???ik

???edx ?(2?32??)1/2x(??ik)2

??(2?32??)1/21

(??ip

)

动量几率分布函数为

? (p)?c(p)2

?2?31??

2?3?3

(?2?

(?2?2?p2)

(2) ?

(x)p

i???

??(x)dx??4?3xe??xd??x

(e)dx ??i?4?3

x(1??x)e?2?xdx

??i?4?3

(x??x2)e?2?xdx

??i?4?3

14?

?14?

)

?0 #

3.8.在一维无限深势阱中运动的粒子，势阱的宽度为a，如果粒子的状态由波函数

?(x)?Ax(a?x)

量子力学答案量子力学教程答案(第二版)

描写，A为归一化常数，求粒子的几率分布和能量的平均值。[www.61k.com）

解：由波函数?(x)的形式可知一维无限深势阱的分布如图示。粒子能量的本征函数和本征值为

?2n?

?(x)?

?sinx, 0?x?a?aa

0, x?0, x?a En2?2?2

n?2?a

(n?1， 2， 3， ?) 动量的几率分布函数为?(E)?C2

n Cn?

*??

?(x)?(x)dx??a

sin

x?(x)dx 先把?(x)归一化，由归一化条件，

1???

(x)2

dx??a

A2x2(2

a?x)dx?A

x2(a2?2ax?x2)dx

(a2x2?2ax3?x4)dx

?A2

(a3?a52?a55

5)?A2a30

∴A?30

∴ Ca230n?

n??0a?a5

ax?x(a?x)dx

?2an?a2n?a3

[a?0xaxdx??0

xaxdx]

?2a2n?a3n?a2n?a3[?n?xcosax?n2?2ax?n?xcosax

2a2

n2?2xn?ax?2an3?3

cosn?ax]0

4n3?

[1?(?1)n

] ∴ ?(E)?C2240n2

n?n6?6[1?(?1)]

?960

?，n?1， 3， 5， ??n6?6

0，n?2, 4, 6, ? ???

???(x)H?(x)dx???p0?(x)2?

?(x)dx ??a30?2d2

0a5x(x?a)?[?2?dx

2x(x?a)]dx ?30?2a30?2 a3a3

?a5?0x(x?a)dx??a5(2

?3)

?5?2

3.9.设氢原子处于状态 ?(r,?,?)?

12R(r)Y32110(?,?)?2

R21(r)Y1?1(?,?)25

量子力学答案量子力学教程答案(第二版)

求氢原子能量、角动量平方及角动量Z分量的可能值，这些可能值出现的几率和这些力学量的平均值。[www.61k.com］

解：在此能量中，氢原子能量有确定值 2

E?e2

s2??

?es2?2

(n?2)

角动量平方有确定值为

L2??(??1)?2?2?2 (??1) 角动量Z分量的可能值为 LZ1?0LZ2??? 其相应的几率分别为 14， 34

其平均值为

133

Z?4?0???4??4

3.10一粒子在硬壁球形空腔中运动，势能为 U(r)??

??, r

?a;?0, r ?

a 求粒子的能级和定态函数。

解：据题意，在r?a的区域，U(r)??，所以粒子不可能运动到这一区域，即在这区域粒子的波函数

??0 (r?a)

由于在r?a的区域内，U(r)?0。只求角动量为零的情况，即??0，

这时在各个方向发现粒子的几率是相同的。即粒子的几率分布与角度?、?无关，是各向同性的，因此，粒子的波函数只与r有关，而与?、?

无关。设为?(r)，则粒子的能量的本征方程为

??21d2d?

2?rdr(rdr

)?E?

令 U(r)?rE?, k

2 ?2?E

2，得 d2udr

?k2

u?0 其通解为

u(r)?Acoskr?Bsinkr

???(r)?AB

rcoskr?r

sinkr

波函数的有限性条件知， ?(0)?有限，则

A = 0 ∴ ?(r)?

sinkr 由波函数的连续性条件，有

?(a)?0 ?B

sinka?0 ∵B?0 ∴ka?n?

(n?1,2,?) k?n?

∴ En2?22?

n?2?a2

?(r)?Bn?

rsin

r 其中B为归一化，由归一化条件得

量子力学答案量子力学教程答案(第二版)

d??

(r)r2sin? dr

?4???

B2sin2n?

rdr?2? aB2

∴ B?

12? a

∴ 归一化的波函数 1sinn?

ar ?(r)?2? ar

3.11. 求第3.6题中粒子位置和动量的测不准关系(?x)2?(?p)2?? 解： p?0

?2? ?

5224

k? x???2212

??Ax[sinkx?2coskx]dx?0

x2???22212

??Ax[sinkx?2

coskx]dx??

(?x)2?(?p)2?(x2?x2)?(p2?p2

)??

3.12 粒子处于状态 2

?(x)?(

11/2

2??2

)

i?px?x04?

式中?为常量。［www.61k.com)当粒子的动量平均值，并计算测不准关系

(?x)2?(?p)2??

解：①先把?(x)归一化，由归一化条件，得 x2 1??

1?2?

? (

x??

2??

dx?

???

x2?

)

2?2??(

1/2

2??2

) ∴?2?12? /

∴ 是归一化的

?(x)?ip?

0x?2

x2]

② 动量平均值为

???

d?? i

p?i?

?*(?i?0x?x2i p0x?x2dx)?dx??i????e?2(?

p0?? x)e?2

??dx

??i???

i ??x??( ?p0

?? x)e2

dx ?p2

??x??

edx?i? ???

xe ??x2

?p0

③ (?x)2?(?p)2?? ??

??x2

?*x?dx?? ??

dx （奇被积函

数）

量子力学答案量子力学教程答案(第二版)

??x2

dx??1

xe??x

2??

?12?

??1

p2???2??

?p0x??x2

p0x??x2??*

? dx???2???

??dx

e dx

??2

(??p0?

)?i2??p???x222

?20???xedx??????x2e??x dx

2 ??2(??p0?)?0?(??2?2)1?2

2??(2

?2?p0) (?x)2?x2

?x2?12?

(?p)2?p2

?p2

2?2?p2)?p2

200?

(?x)2?(?p)2

12???2122??4

? #

3.13利用测不准关系估计氢原子的基态能量。（www.61k.com）

解：设氢原子基态的最概然半径为R，则原子半径的不确定范围可近似取为

?r?R

由测不准关系

(?r)2

?(?p)2

??2

得 (?p)2

??2

对于氢原子，基态波函数为偶宇称，而动量算符p?

为奇宇称，所以

p?0

又有 (?p)2?p2

?p2

所以 p?(?p)?4R2

可近似取 p??22

E?P2e2

能量平均值为 s2??r

e22

作为数量级估算可近似取 sesr?

则有 E?e2s2?R

?R 基态能量应取E的极小值，由

?Ee2?R???2

s?R3?R

2?0 R??2

得?e2

代入E，得到基态能量为 Esmin??

?e2?

补充练习题二

量子力学答案量子力学教程答案(第二版)

1．试以基态氢原子为例证明：?不是T

?或U?的本征函数，而是T??U?的本征函数。（www.61k.com］

解：?112

100?)3/22e?r/a0 (1?? e

4?(

a2)

0a0?

T????21?22?1?1?2?r2[?r(r?r)?sin?(sin???)?sin2???2]2U

???esr

T???21?2??100

100??2?r2?r(r?r

)

???2112?(a)3/2?1?2??r/a0

r2?r

(re)

0?r ???2113/22?(a)(12ae?r/a0???2(12

2?)2?)?100

00a0r2?a0a0r ?常数??100

?100

不是T

?的本征函数2

U??100??esr

?100 可见，?100不是U

?的本征函数

而 (T??U?)????2113/212?100

2?(a)(r/a0es2?)e??100

0a0a0r

r ???21?2?2

2?a2

?100??100??100

0?a0r?a0r ???21

2?a2

?100

可见，?100

是(T

??U?)的本征函数。

2．证明：L?6?，L???的氢原子中的电子，在??45?和 135?的

方向上被发现的几率最大。

解： ?W2

?m(?,?)d??Y?md? ∴ W2?m(?,?)??m

6?，L???的电子，其??2, m??1

? Y)??

sin?cos? ei?21(?,?

Y2?1(?,?)??

si?nco?s e?i?8?

∴W?,?)?Y2

152?1(?m?8?sin2?cos2??1532?

sin22? 当??45?和 135?时

W15

2?1?32?为最大值。即在??45?，??135?方向发现电子的

几率最大。

量子力学答案量子力学教程答案(第二版)

在其它方向发现电子的几率密度均在0～

32?

之间。（www.61k.com)

3．试证明：处于1s，2p和3d态的氢原子的电子在离原子核的距离分别为a0、4a0和9a0的球壳内被发现的几率最大(a0为第一玻尔轨道半径 )。证：①对1s态，n?1, ??0, R13/2?r/a10?(

a)e

Wr2R2

110(r)?10(r)?(

a)34r2e?2r/a0

10?(1)34(2r?2

ar2)e?2r/a0

?ra00

令

?W10

?0 ?r1?0, r2??, r3?a0 易见，当?r1?0, r2??时，W10?0不是最大值。

W10(a0)?

4ae?2

为最大值，所以处于1s态的电子在 r?a0处被发0

现的几率最大。

②对2p态的电子n?2, ??1, R13/2

ra21?(

02a)e?r/2

3a0

?(143r2?r/a21(r)?rR21

2a)2

re03a0

211r??r?24a5

(4?)r/a00

ae0 令

?W21

?0 ?r1?0, r2??, r3?4a0 易见，当?r1?0, r2??时，W21?0为最小值。

?2W2118rr2??r2?24a5

(12?r/a00

a?2)e 0a0

?2W21

1?r2

24a5?16a20(12?32?16)e?4

??8a3

e?4?0 r?4a0

030

∴ r?4a0为几率最大位置，即在r?4a0的球壳内发现球态的电子

的几率最大。 ③对

于

态

的

电

子 n?3, ??2, R232?(

)3/21(ra)2e?r/3a0a 00

W(r)?r2R2

113a3232?0

72r6e?r/ a81?15

32?852r

3a0

?r812?15a7

r(6?0

3a)e?2r/0 令

?W32

?0 ?r1?0, r2??, r3?9a0 易见，当?r1?0, r2??时，W32?0为几率最小位置。 ?2W6 32162

4r52r?2r/3a0?r2?812?15a7

(15r??2)e 0

a09a030

量子力学答案量子力学教程答案(第二版)

?2W32?181?15aa4

36a2?81a2

002(90)(15??)e?6

?r2

r?9a0

0a9a2

165a3e?6

?00

∴ r?9a0为几率最大位置，即在r?9a0的球壳内发现球态的电子的几率最大。［www.61k.com］

4. 当无磁场时，在金属中的电子的势能可近似视为

U(x)???0, x?0 (在金属内部

)?U0, x?0 (在金属外部

)

其中 U0?0，求电子在均匀场外电场作用下穿过金属表面的透射系

数。

解：设电场强度为?，方向沿χ轴负向，则总势能为

V(x)??e? x (x?0)， V（x）?U0?e? x （ x?0）势能曲线如图所示。则透射系数为

D?exp[

?2?x1

?x2?(U2

0?e? x?E)dx] 式中E为电子能量。x1?0，x2由下式确定

p?2?(U0?e? x?E)?0 ∴ xU0?E

2?e?

令 x?U0

?Ee?

si2

n?，则有

x2?(U?e? x?E)dx?

2?(UU0?E

0?E)?e?

2sin2? d?

?2U2?

0?Ee?2?(U?E)(?cos3?03)

0 ?

2U0?E

3e?

2?(U0?E)

∴透射系数D?exp[

?2U0?E3?e?

2?(U0?E)] 5．指出下列算符哪个是线性的，说明其理由。

① 4x2

d22

dx2； ② ?? ； ③ ?

K?1

2 解：①4x2ddx2是线性算符 ? 4xd2222(c?c2d2d1u12u2)?4x(c1u1)?4x(c2 dx2dx2dx2

u2) 2 ? cdd

1?4x2dx2u1?c2?4x2dx

2u2

②?? 2

不是线性算符

? [c2u222

1u1?c2u2]2?c11?2c1c2u1u2?c2u2

?c1[u22

1]?c2[u2]

③

是线性算符

K?1

量子力学答案量子力学教程答案(第二版)

NNNN

?c1u

?c2u2?2K?1

?c1u1?K?1

?c2u2?c1K?1

?u1?cK?1

?u2

K?1

6．指出下列算符哪个是厄米算符，说明其理由。［www.61k.com]

ddx， i ddx， 4 d2

解： ??

d??d???*dx

? dx??*?-?????dx?*? dx

当 x???，??0，??0

? ??

?d???*ddx? dx?????dx?*? dx????d

??(dx?)*? dx

???d

??(dx?)*? dx

不是厄米算符

???*id

? dx?i?*? ??d??dx

-??i???dx?*? dx ??i????(ddx?)*? dx???d

??(idx

?)*? dx

?id

是厄米算符

???*4d2dx

2? dx?4?*d?dx ?

?d?*d?-??4???dxdx dx ??4??

d?*d?d?*

?d2

?*??dxdx dx?4dx

??4???dx2? dx

??4?

?*? ?

??dx2

dx??

2?)*? dx?4d2

2是厄米算符

7、下列函数哪些是算符d2

dx2的本征函数，其本征值是什么？

①x2

， ② ex， ③sinx， ④3cosx， ⑤sinx?cosx

解：①d2dx

2(x2

)?2 ∴ x2

不是d2dx

2的本征函数。

② d2dx

?ex e不是d2 ∴x

2的本征函数，其对应的本征值为1。

③d2dx

(sinx)?d

dx(cosx)??sinx 可见，sinx是d2

∴dx2

的本征函数，其对应的本征值为－1。

量子力学答案量子力学教程答案(第二版)

④d2dx

(3cosx)?d

dx(?3sinx)??3cosx?(3cosx) ∴ 3cosx 是d2

2的本征函数，其对应的本征值为－1。（www.61k.com]

d2 ⑤dx2(sinx?cosx)?d

dx(cosx?sinx）??sinx?cosx

??(sinx?cosx)

∴ sin

x?cosx是d2

2的本征函数，其对应的本征值为－1。

8、试求算符F

???ieix

的本征函数。解：F

?的本征方程为 F

???F? 即 ?ieix

?F?

?iFeixdx??d(Feix

dx)?d(?Feixddx

) ln???Feix

dx?lnc ??ce?Fe?ix

（F

?是F的本征值）

9、如果把坐标原点取在一维无限深势阱的中心，求阱中粒子的波函数和能级的表达式。

解： U(x)???0,x?a?2

???

?,x?

2 方程（分区域）：

Ⅰ：U(x)?? (x??a2

)

Ⅲ：U(x)?? (x?a2

)

Ⅱ：??2d2?II

2?dx2

?E?II

d2?IIdx2?2?E

?2?II?0 令 k2

?2?E?2

d2?IIdx

?k2

?II?0 ?II?Asin(

kx??) ??(?aa 标准条件：?I)??II(?)?2?a2

?(2)??aIIIII(2)

∴ Asin(

?kx??)?0 ∴

?I(x)?0 ?III(x)?033

∴

量子力学答案量子力学教程答案(第二版)

∵ A?0 ∴ sin(?kx??)?0

取 ??ka2?0，即 ??ka

∴ ?sink(x?a

II(x)?A2

)

Asinka?0 ? sinka?0

∴ ka?n? (n?1, 2, ?)

?Asin?n(x?a),xa

∴ 粒子的波函数为 ?(x)????

?2?a??

0, x?2

?22

粒子的能级为E?2n2?2k2?k?

2?a

(n?1, 2, 3, ?) 由归一化条件，得

(x)2

d??A2?

a/2

sin2

n????a/2a(x?a

)dx ?A2

?a/21?a/22[1?cos2n?a(x?a2)]dx ?A2?a2

a/22?A??a/2cos2n?a(x?a2

)dx a

aA2?A2?asin2n?(a2

22n?ax?2)a

a22A ∴ A?2

∴ 粒子的归一化波函数为

??2?n(x?aa

?(x)???

aa2),x?2

???

0, x?a2

10、证明：处于1s、2p和3d态的氢原子中的电子，当它处于距原子核的距离分别为a0、4a0、9a0的球壳处的几率最

（a0为第一玻尔轨道半径）。(www.61k.com］证：1s:

?(r)2

10dr?R10r2dr ?(1)3

?4e?2r/a0a?r2dr 0

?110(r)?(

a)3

?4r2e?2r/a0 0

d?10?4(1)3?(2r?2

r2)e?2r/a0draa

00 ?8(131

a)?(1?r)re?2r/a0

0a0

量子力学答案量子力学教程答案(第二版)

令

d?10

?0，则得 dr

r21?0 r22?4a0

r?0 r?a d2?21

?0 ∴ r22?4a0为最大几率位置。[www.61k.com］

111102

d?10dr2

?8(132?rr?2a)?[(1?r)?(1?)er/a0

]0a0a0a08(

?)3?(1?4r?2r

?2r/a0a2)e] 0a0a0

d2 ?10

dr2?0 ∴r11?0为几率最小处。

r11?0

d2?10

dr2

?0 ∴r11?a0为几率最大处。

r11?a0

2p: ?21(r)dr?R2

21r2dr

?(123r?r/a022a)?a2e?rdr

030 ?r)?(13r2?r/a21(0

2a)?2e

03a0

d?21113?r/a0

dr?24a5

?(4?r)re 0

d2?2118r22?r/a0

dr2?24a5(1?r?2

）re] 0a0a0

令

d?21

?0，则得 drr22?4a0

当 0?r?4a0时，

d2?10

?0 ∴r?0为几率最小位置。2r

3d: ?32(r)?R2

?86

?3a032

98415a7

re 0

d?2r32dr?898415a7

(5?2r0

3a)r5e?3a0

0令

d?32

?0，得 r31?0, r32?9a0

同理可知 r31?0为几率最小处。

r32?9a0为几率最大处。

11、求一维谐振子处在第一激发态时几率最大的位置。

122

解：??

??x1(x)??2?xe2

2 ?2

2?32??2

1(x)?1(x)?xe

d?1?4?3

(x??2x3)e??2x2

量子力学答案量子力学教程答案(第二版)

4?2(1??x2)xe??

d2?31dx2

?4?(1?5?2x2?2?4x4)e??2x2

令

d?1

?0，得 x1?

1?0，x2??2??????x0

d2?1

dx2

?0， ∴ x1?0为几率最小处。（www.61k.com)

x1?0

dx2

?0， ∴ x1

2??

x2??x0为几率最大处。 2??

6．设氢原子处在?(r,?,?)?1

的态（a0为第一玻尔

轨道半径），求

①r的平均值；

②势能?e2

的平均值。

解：①???13

?2r0?2?0?a3redrsin? d?d? 0

?0?0

a0?a3

?3?2?1?()3?(a0)?4? 0

22 ?

a0 ②?

e2sr??e2

?1??2r

0?a3?4?0

?0readr ??e2saa

a3?4?(0)?(0)

022e2

??sa

12、粒子在势能为

?U1, 当x?0 U??

?0, 当0?x?a

??U2

, 当x?a

的场中运动。证明对于能量E?U1?U2的状态，其能量由下式决定： ka?n??sin?1?k2?U?

?k1

2?U

（其中k?

2?E

）证：方程

Ⅰ：??2d2?I

2?dx

?U1?I?E?II (x?0) 36

量子力学答案量子力学教程答案(第二版)

Ⅱ：?

2??2

Ⅲ：?

d2?II

?0?II?E?II (?0x?A) 2

dxd2?III

?U2?III?E?III (x?0) 2

?I(0)??II(0)?C1?Asin? ①

? ??s I(0)??II(0)??C1?kAco?

②

令

2?(U1?E)

2?E

, k?

, 则得

Ⅰ：d2?I

dx2

??2?I?0 Ⅱ： d2?II

dx2

?k2?II?0 d2 Ⅲ： ?IIIdx

??2

?III?0 其通解为

??x

I?C1e?Dx1e??

?II?Asin(kx??) ??xIII?C2e?D?x2e?

利用标准条件，由有限性知

x ? ?, ?I 0,D1?0 x ? ??, ?III ?0,C2?0 ∴ ?I?C1e

?II?Asin(

kx??) ???x

III?D2e

由连续性知

??2?(U2?E)

, ?II(a)??III(a)?Asin(kx??)?D??x2e ③

???(a)?kAcos(kx??)???D?xII(a)??III

2e? ④

由①、②，得

tg??k

⑤

由③、④，得

tg(ka??)??k

⑥

而tg(ka??)?

tgka?tg?

1?tgka?tg?

把⑤、⑥代入，得tgka?tg?1?tgka?tg???k

整理，得 ?tgka??

?tg?

1?k?

tg?

量子力学答案量子力学教程答案(第二版)

tg? tg(n??ka)?

?1?

tg?

令 tg??

?tg? tg(n??ka)?

tg(???) 1?

tg?

∴ n??ka????

ka?n????? 由sinx?

tgx

?tg2

，得

sin??k?k?(k?22

??k2?U2? sin???k??k1?(k2

?)?k2?U1ka?n??sin?1?k2?U?sin?1

12?U2

###

13、设波函数?(x)?sinx，求[(ddx)x]2??[xd

]?? 解：原式?[(dddx)xdx)x]

??[xddx][xd

?[(

ddx)x][sinx?xcosx]??[xd

][xcosx]?

?(sinx?xx)?x(cosx?cosx?x)?x(x?x)

?sinx?2xcosx

14、说明：如果算符A?和B?都是厄米的，那么 (A?+B?)也是厄米的证： ??*1(A??B?)?2d????*1A??*2d????1B??2d? ??

?2(A??1)*d???

?2(B??1

)*d? ???

[(A??B?)?1

]*d? ∴ A

?+B?也是厄米的。(www.61k.com]

15、问下列算符是否是厄米算符：

①x

?p? ②1x 2

(x?p?x?p?xx?) 解：①??*1(x

?p?????*

x)?2d1x?(p?x?2)d? ??（x

??1)*p?x?2d???(p?xx??1)*?2d? 因为 ?p

xx???p?x ∴ x

?p?x 不是厄米算符。 38

量子力学答案量子力学教程答案(第二版)

②

??*

111[2

?p?x?p?xx?)]?2d??2??*1(x??p1*

x)?2d??2

??1(?pxx?)?2d? ?12?(?p)*?1

xx??12d??2?(x?p?x?1)*?2d? ??[12(x?p?*x?p?xx?))?1]?2d? ??[1

(?pxx??x?p?x)?*1]?2d? ∴ 12

(x?p?x?p?xx?)是厄米算符。[www.61k.com） ## 16、如果算符?

?、??满足关系式??????????1，求证 ①?

???2???2???2?? ②?

???3???3???3??2 证： ① ?

???2???2???(1???2??)???2?? ??

?2??????????2?? ???2???(1?????)???2?? ?2?

? ②?

???3???3???(2?????2??)?????3?? ?2?

?2???2???????3?? ?2??2???2(1?????)???3?? ?3??2 17、求 L?xP?x?P?xL?x??

L?yP?x?P?xL?y?? L?zP?x?P?xL?z??

解： L?xP?x?P?xL?x?(?yP?z?z?P?y)P?x?P?x(?yP?z?z?P?y) ??yP?zP?x?z?P?yP?x?P?x?yP?z?P?xz?P?y) ??yP?zP?x?z?P?yP?x??yP?zP?x?z?P?yP?x)

= 0

L?yP?x?P?xL?y?(z?P?x?x?P?z)P?x?P?x(z?P?x?x?P?z)

?z?P?2x?x?P?zP?x?P?xz?P?z?P?xx?P?z) ?z

?P?2x

?x?P?z

P?x

?z?P?2x

?P?x

x?P?z

) ??(x?P?x?P?xx?)P?z ??i?P?z

L?zP?x?P?xL?z?(x?P?y??yP?x)P?x?P?x(x?P?y??yP?x) ?x?P?yP?x?y?P?2x?P?xx?P?y?P?x?yP?x ?x?P?xP?y??yP?2x?P?xx?P?y??yP?2x ?(x?P?x?P?xx?)P?y ?i?P?y 18、 L?xx??x?L?x

?? L?yx??x?L?y?? L?zx??x?L?z??

量子力学答案量子力学教程答案(第二版)

解： L?xx??x?L?x?(y?P?z?z?P?y)x??x?(y?P?z?z?P?y) ??yP?zx??z?P?yx??x?y?P?z?x?z?P?y ??yP?zx??z?P?yx??y?P?zx??z?P?yx?

= 0

L?yx??x?L?y?(z?P?x?x?P?z)x??x?(z?P?x?x?P?z) ?z?P?xx??x?P?zx??x?z?P??x?2xP?z

?z?(P?xx??x?P?x

) ??i?z

? L?zx??x?L?z?(x?P?y?y?P?x)x??x?(x?P?y?y?P?x

) ?x

?2P??y?P?x??x?2y

P?y

?y?x?P?x

??y(x?P?x?P?xx?)

??i??y

第四章态和力学量的表象

4.1.求在动量表象中角动量Lx的矩阵元和L2x的矩阵元。[www.61k.com］

i??

解：(L1x)p?p?(

2??)3?

e??p??r

(yp?zpp?rz??y)e?d? ?(1i??i?3??p??

r2??

)?ep??r

(ypz?zpy)e?d?

i??i??

?(12??)3?e??p??r(?i?)(p??z?p?r

?p?py)ed? y?pz

i???

?(?i?)(p??13z?p?p?）?r?p?py)()y?pz2??

?ed?

?i?(p??

??y?p?pz)?(p?p?)

z?py

(L2*x)p?p????p?(x?)L2

x??p

d? i??

i??

?(13??p??r

2??

)?e(yp

?e?p?rz?zp?y)2d?

i??

?(12??)3?

e??p??r

(yp?z?zp?y)(yp?z?zp?y)e?p?rd?

i??i??

?(13??p?r2??)?e(yp?p??z?zp?y)(i?)(y?p?r

?p?pz)ed?

z?py

i??

?(i?)(p??13?p?ry?p?pz)(?)?

e?p??r

(yp?z?zp?y)e?d?

z?py2?

i??

???2

(p??213?p?p?）

?ry?p?pz?p)()?ed?

zy2??

???2

(p?y

?p?p?2?(?p??z)p?) z?py

量子力学答案量子力学教程答案(第二版)

4.2 求能量表象中，一维无限深势阱的坐标与动量的矩阵元。[www.61k.com］

解：基矢：u2n(x)?asinn?

ax 能量：E?2?2n2

n?2?a

对

角

元

：

x2mmm??

axsin2?a

axdx?2 当时，m?n xa?mn?a?0(sin

ax)?x?(sinn?

)dx

1aa?0x??cos(m?n)?ax?cos(m?n)?

a??

dx?1?a2(m?n)aa??[?ax(m?n)?

(m?n)2?2

cosax?(m?n)?sinax]0a

?[a2(m?n)??(m?n)2?2

cosax(m?n)?

ax?(m?n)?sinax]? 0???a??

?11

??2(?1)m?n?1??(m?n)

2?(m?n)2???

4mnm??2(m2?n2)

?(?1)n

?1?

p2mn

??u*

m(x)p

?un(x)dx??i??a

asinm?ax?ddxsinn?0

xdx??i2n??am?n?

2?0sinax?cosaxdx

??in??a?(m?n)?(m?n)a0??

sin?2?ax?sin

ax?

dxa

?in???a(m?n)?a(m?n)?a2??(m?n)?

cosax?(m?n)?cos

ax??? 0

n??a?a2?

?1?(m?n)?1?(m?n)?

?(?1)m?n

?1?

]??(?1)m?n?1?

i2mn?

(m2

?n2)a

4.3 求在动量表象中线性谐振子的能量本征函数。解：定态薛定谔方程为

?12

p22??2?2ddp2C(p,t)?2?C(p,t)?EC(p,t) 2 即 ?12??2?2ddp

2C(p,t)?(E?p22?)C(p,t)?0 41

量子力学答案量子力学教程答案(第二版)

两边乘以2

，得

?1d21dp2C(p,t)?(2E???p2

???)C(p,t)?0

???

令??

???

p?? p, ??

???

C(p,t)?(???2)C(p,t)?0 跟课本P.39(2.7-4)式比较可知，线性谐振子的能量本征值和本征函数为En?(n?2)??

C(p,t)?Nne

?2pHn(?p)e

?i?

Ent式中Nn为归一化因子，即 N?

n?(

?1/22n

)1/2 #

4.4.求线性谐振子哈密顿量在动量表象中的矩阵元。（www.61k.com）

解：H??122

2?p?2?122??1222??x??2??x2?2

??x H*pp

???

?p(x)H

??p

(x)dx 1?ipx?22

?p?x2???e?(??122?

2??x2?2

??x)edx

???22?(i?p?)21??

(p??p)x121?2?(p??p)x2?????edx?2??2??

???xedx

?p?2(p??p)x2??(p??p)?12??21?2?????(?i)2??p

?2e?

?p?2?(p??p)?1?2

22??

)x2?2??(i)?p?2

???

(p??pdx

?p212

22?2??(p??p)?2????p?2

?(p??p) ?p212

22?2??(p??p)?2????p

2?(p??p) #

4.5 设已知在L?2和L?Z的共同表象中，算符L?x和L?y的矩阵分别为? L??

010??x?2?101?

??010??

?0?L?2??

i0?

?y2?i0?i?

?0i0??

量子力学答案量子力学教程答案(第二版)

求它们的本征值和归一化的本征函数。［www.61k.com）最后将矩阵Lx和Ly对角化。解：Lx的久期方程为

?a2??0??????

?a1?a3???0? ?a3??a1，a2?0

?20

?2???2?0???3??2??0

??1?0，?2??，?3???

∴L?x的本征值为0，?，?? L?x

的本征方程 ?010???a1? ?

???a1?

???2?101???a2????a2?010????a???

3??a?3?? 其中???a1??a?

?的本征函数?2?设为Lx

L?2和L?Z共同表象中的矩阵 ?a?3?

当?1?0时，有

?010??a1 ??

2?101?????a???0?

?2????010????a3??0? ???0??

2??a2????0??

?a1? ∴ ???

0??0?

???a1??

?a1? 1???**

??20?0?(a1,0,?a1)?0??2a?1

??a1??

取 a1?

??1??2??

?0???0???对应于L?x

的本征值0 。 ??

?1?2?

?当?2??时，有

?010???a1? ?

?2?101???a1?

?a?????a?2?2?

??010????a3????a3??

由归一化条件

量子力学答案量子力学教程答案(第二版)

??1

a??22

?1?a??a2?2a1

?2(a?????1?a???

1?a3)2???a2?2a3 ??1

???a3???

a?a1

2a??32??

??a1??

∴ ?????2a1?

??a?1??

由归一化条件

??(a?a1?

? 1***

21,2a1,a1)??2a1??4a1??a?1??

取 a1

1?2

??1??2? ∴归一化的?????1?

??的本征值? ?2?对应于Lx

???1??2??

当?2???时，有

?010?? ?

?2?101??a1?

???a1?????a2?????a2?

?010????a??3??a?3?

??1??2a?1????a??a ?12??2a1

??2(a??1???

1?a3)?????a?2???a2??2a3 ??1???a?3???a?a1?2a2???3?

??a1??

∴ ???????2a1?

??a?1??

由归一化条件

? 1?(a***

?a1?

?21,?2a1,a1)???2a1??4a1??a?1??

取 a1

1?2

量子力学答案量子力学教程答案(第二版)

?1????2??1??的本征值?? ∴归一化的???????对应于Lx

???1????000??2??11??

1 ????0

121??????

121

?2??1???2?? 由以上结果可知，从L?2和L?Z的共同表象变到L?x表象的变换矩阵为 ??111??22?

S??2

?01?1??

2? ????1

112??222??

∴对角化的矩阵为L??

x?SLxS ?

?10?

1?22? L?????12112????1???

010?11

?x2?

2??101???2

???2??121??

??0??010????

?12

???1

1212???2???

2?? 2?22????12?

?121?1??2???

??122

?00??000? ??

??2?0

200?????0???00?2???

?00

????

按照与上同样的方法可得

L?y的本征值为0，?，?? L?y

的归一化的本征函数为 ??1?

?2??

?0???0?

??1???2??

??1??2????

?????i??

???1???2??

12??2????1??2?

????i????2??????1?2??

量子力学答案量子力学教程答案(第二版)

?2和L?的共同表象变到L?表象的变换矩阵为从LZy

i???????T??*)?(?*T?

?11

1??222??1i

??? S??

0i???S?

??12???12?2

?12

?12???2

???2???1?2利用S可使L

对角化

?00 L??

?0??y?SLy

S??0?0?? ?00????

4.6求连续性方程的矩阵表示解：连续性方程为

????J? ∴ J??t

?i?

2?(???*??*??) 而 ??J??i?

??(???*??*??) ?i?

2?(??2?*??*?2?) ?1?i?

(?T?*??*T??) 01?2??i?1??2?i2

1?2

??2??

∴ ?t

i??(?*?)?t?(?*T????T??*) 写成矩阵形式为

?(???)?????T??? ?t

??Ti??

(???)?????T??(??T??)*??*?0

第五章微扰理论

5.1 如果类氢原子的核不是点电荷，而是半径为r0、电荷均匀分布的小球，计算这种效应对类氢原子基态能量的一级修正。（www.61k.com］

解：这种分布只对r?r0的区域有影响，对r?r0的区域无影响。据题意知

H???U(r)?U0

(r) 其中U0(r)是不考虑这种效应的势能分布，即 2 U（r）??

ze4??0r

U(r)为考虑这种效应后的势能分布，在r?r0区域，

U(r)??Ze2

4??r

0在r?r0区域，U(r)可由下式得出，

量子力学答案量子力学教程答案(第二版)

U(r)??e

Ed r

??1Ze43Ze E???4??0r2?4?r3?03?r?4??3r, (r?r0)0r0

??Ze ?

4??2

(r?r0)0r

U(r)??e

Edr?e??

rEdr

Ze2

4??3

0r0?

Ze2

rdr4??

?1r0r2dr

??Ze28??r3(r0?r2)?Ze2??Ze22(3r2?r2

004??8??3

0)0r00r0

(r?r0)

? H???U(r)?U(r)???Ze23(3r2Ze220?r)?(r?r0)0?8??r4??r ?000?

0 ( r ? r 0) 由于r(0)

0很小，所以H

????H????22?

?U0(r)，可视为一种微扰，

由它引起的一级修正为（基态?(0)

1?(Z31/2?ar0?a3)e） 0

E(1)

???(0)*

1H???(0)1d? ?

?r0

?a3[?Ze28??3

(3r2

Ze22?ar00

0?r)?]e0r0

4??4?r2dr 0r?2Z

∴r??aar0

0，故e?1。(www.61k.com)

∴ E

(1)1

??Z4e2

(3r2r2

?r4

Z4e2

2??33

)dr?r

0a0r0

??a3

rd ??Z4e25

r50Z4e222??3r3(r0?)?3r 0a0052??0

0a0 ?Z4e22

10??3r0

0a0

?2Z4e2s2

5a3

r0 0

5.2 转动惯量为I、电偶极矩为D?

的空间转子处在均匀电场在??

中，如

果电场较小，用微扰法求转子基态能量的二级修正。解：取??

的正方向为Z轴正方向建立坐标系，则转子的哈米顿算符为

??L?2??122I?D???2I

L??D?cos? 取H

?(0)?12IL?2, H????D?cos?，则 H

??H?(0)?H?? 由于电场较小，又把H

??视为微扰，用微扰法求得此问题。 47

量子力学答案量子力学教程答案(第二版)

H?(0)的本征值为E(())?

?12I?(??1)?2 本征函数为 ?(0)

??Y?m(?,?)

?(0)的基态能量为E（0）0

?0，为非简并情况。［www.61k.com）根据定态非简并微扰论可知 2 E(2)

?? H??0

(0)

?E?

H?*(0)*?0????H???(0)

0d???Y?m(?D?cos?)Y00sin? d? d?

??D??

(cos? Y00)sin? d? d? ??D??

Y*4??m

104?

sin? d? d?

??D?3?Y*?0 Y10sin? d? d?

??D?3

??1

(2)'

D2?2?2I2

120

??'

H??0

E(0)(0)

???

3?(??1)?

0?E??

D?2

I #

5.3 设一体系未受微扰作用时有两个能级：E01及E02，

现在受到微扰H??的作用，微扰矩阵元为H12

??H21??a，H11??H22??b；a、b都是实数。用微扰公式求能量至二级修正值。解：由微扰公式得

E(1)

n?H?nn

2 E(2)

mnn

??'

H?m

(0)

得 E(1)(1)

01?H11??b E02?H?22

?b 2

(2)Hm

1?a01

E01?E0m

E?E 0102 E

(2)'

102

??a2

E02?E0m

EE 02?01

∴ 能量的二级修正值为

Ea2

1?E01?b?E

01?E02 Ea2

2?E02?b?E?E

0201

5.4设在t?0时，氢原子处于基态，以后受到单色光的照射而电离。设单色光的电场可以近似地表示为?sin? t，?及? 均为零；电离电子的波函数近似地以平面波表示。求这单色光的最小频率和在时刻t跃迁到电离态的几率。

解：①当电离后的电子动能为零时，这时对应的单色光的频率最小，其值为

???e4

smin?hvmin?E??E1?2?2

v??e4s13.6?1.6?10?19min

2?2h

?6.62?10?34

?3.3?1015

Hz ②t?0时，氢原子处于基态，其波函数为

量子力学答案量子力学教程答案(第二版)

r/a0

i??

在t时刻， ?13/2m?(?p?r

2??

微扰 H

???(t)?e???r?sin?t?e??r?

i? t

(e?e?i? t) ?F?(ei? t?e?i? t) 其中F??e???r?2i

在t时刻跃迁到电离态的几率为

k?m?am(t)

a?1tm(t)i??0

H?mk

ei?mkt?

dt? ?Fmk??t0

(ei(?mk

??)t?

i?ei(?mk??)t?)dt? ??F)tmkei(?mk???1ei(?mk??)t?1

?[??]

mk???mk??

对于吸收跃迁情况，上式起主要作用的第二项，故不考虑第一项，?mk??)t

aFi(mke?1

m(t)???

mk??

??)tW?1)(ei(?mk??)t?1)

k?m?am(t)2

Fmk(ei(?mk?

(?2

mk??)

2 ?

4Fmksin2(?mk??)t

(?2

mk??)

其中F?i?3/2

?mk??

??d??(1mFk

2??

取电子电离后的动量方向为取??、?

Z方向，

p所在平面为xoz面，则有

???r?

??xx??yy??zz

?(?sin?)(rsin?cos?)?(?cos?)( ?? rsin?sin?cos??? cos?rcos

F/2

mk?(

12?ae??p rcos?

)3(? rsin?30

2i?F11

mk?(2??

)3/2e?a30

2i ?

??2?

p rcos?0

(?rsin?sin?cos??? rcos?cos?)e?r/a0r2sin?drd? d?

e??2?i

12?

)

3/2

a3e??p rcos?(?cos? r3

cos?sin?)?r/a00

2i?0?0?0edrd? d?

e?cos??3?r??p rcos?2??

)

3/2

a32??0re/a0

dr[?0e?cos?sin? d?0

量子力学答案量子力学教程答案(第二版)

e?cos?i2??

?2a?

3?r/a0

p r?p rp r????p r?2?

[(e?e)?22(e??e?)]driprpr

iiii

*???d? (H???e?(t)rcos?) ?,100??210H H210100

?为Z轴方向)

?RYe?(t)rcos? RYd? (取方向1000?2110

e?cos?16p

i2??2a32

0ia0

?(1p

2?2)3a0?

16pe?cos?(a7/20?)

8?(a2

??)

∴ W4Fmksin21(?mk??)t

k?m?

mk??)

21 ?

128pe?cos?a0

?sin(?mk??)t

?2(a2226?2

0p??)(mk??)

5.5基态氢原子处于平行板电场中，若电场是均匀的且随时间按指数下降，即

????0,

当t?0???t/?

0e,

当t?0(?为大于零的参数) 求经过长时间后氢原子处在2p态的几率。［www.61k.com）

解：对于2p态，??1，m可取0, ?1三值，其相应的状态为 ?210 ?211 ?21?1

氢原子处在2p态的几率也就是从?100跃迁到?210 、 ?211 、?21?1的几率之和。

由 a1tm(t)?i ?

?0H?mkei?mkt?

dt? ?e?(t)??

21rR10dr?

Y*10Y00cos?sin?d? d? ?e?(t)f?

Y*10

Y10sin?d? d?

e?(t)f

R*21(r)R10(r)r3dr?

2566

?(12

)3/2

a?(1)3/2?4?2ar02a0

0a0

redr

?116a4?4！?25a5?256

a0035

H?*21,0100???21H0???10d0

??1e?(t)f ?e?(t)2566a2

0?243e?(t)a0

H?211

,100?e?(t)?

211rcos??100d?

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四 : 新视野大学英语视听说第二版第二册答案

新视野大学英语视听说教程2(第二版)答案(带unit test)

Unit1

Listening skills BBDCA Listening in Task1 FTFTF Listening in Task2 BBDAC Task3 classical，peaceful，relaxing，Jazz，sadness，heavy metal，energy，sporting events，physical labor，road accidents Let's talk Task 1 Good Morning to All，success，musical talents，without，second part，replaced，legal action，real owners Further listening and speaking Task 1 special，joke，talent，proud，loud，joy，honesty，dancer，talk，wondered，capture，fan Task2 BAADC Task 3 TFTTT Unit 1 test 1-5 CCDCB 1.favorite band ,2.Not anymore ，3.no longer ，

4.a big fan ，5.collected ，6.the ones ，7.Going crazy ，8.Maybe to you ，9.pressure ，

10.fall in love ，11.get it ，12.from time to time ，13.Go on ，14.music video 1-5BCDDA 1-5CDABA 6-10 CDCBC

Unit2

Listening skills: Making inferences ADBCB Listening in Task 1 director，sound effects， good story， think， sad，violent， light-hearted movies， mystery movie，in the future Task 2 DCDAB Task3 film, directors, actors, important, the United Kingdom,viewers,40 million, Olympic, World Cup, time, artistic value, depth, Best Picture, favor, science fiction Let's talk Task 1 ABCDD Further listening and speaking Task 1 memorizing lines，remember one line，I hear the guns roar，a loud boom，forgot his line Task 2 ACCAD Task 3 FTFTF Unit 2 test 1-5 CBABD 1 .a very good 2.starring 3.best-seller 4.a new classic 5.violent

6.somewhat 7.before 8.the calmness 9. death 10.thought 1-5 DBDCA 1-5 CCBAB 6-10 CCBAC Unit3

Listening skills: Identifying people's identity BCDAA Listening in Task 1 go out，flaming red，crush，chicken，likes，guts，turn him down，keeping everything，tell her Task 2early，late teens，Double-dating，Group dating，marry，Adult dating，hardly，Blind date，never Task 3 (3) An announcement about the wedding was published in the newspaper. (2) Wedding invitations were sent out to relatives and friends. (1) The minister greeted the guests in the church. (8) The minister talked about the meaning of marriage. (4) The bride and the bridegroom exchanged vows. (7) The minister prayed for the couple. (6) The minister declared the couple husband and wife. (9) The reception was over. (5) The minister signed the certificate. Let's talk Task 1 BBCDA Further listening and speaking Task 1 BDCA Task 2 the young man's plans，a nice house，God will provide for us，Don't worry, sir，play the role of God Task 3FFTTF Unit 3 test 1-5BDCDC 1.He's engaged 2.even 3.fell in love with 4.swept him off his feet 5.popped the question 6.getting married 7.lonely

8.divorced 9.different 1-5CADBA 1-5BCADD 6-10CACBD

Unit4

Listening skills Obey your thirst.，Drivers wanted.，She works while you rest.，Make yourself heard.，We lead. Others copy.，Good to the last drop.，Don't leave home without it.，No business too small, no problem too big. Listening in Task 1 shoes，funny，a lot of life，ad，extra spring，money，millions of dollars，for nothing，reach the top，hard work，focus on，forget Task 2 FTFFT Task 3 ABBCC Let's talk Task 1 TFTFT Further listening and speaking Task 1 banned tobacco ads，young people smoking cigarettes，ways around the law，new customers old ones，smoking on campus Task 2 ACBDD Task 3 FTFTF Unit 4 test 1-5BBCDC 1.We do

2.change the image 3.do some surveys 4.different age groups 5.all the time 6.your users

7.target 1-5CAAAB 1-5CDCDC 6-10CBDBC

Unit5

Listening skills ABBDC Listening in Task 1 (2) The couple was going to celebrate the wife's birthday. (5) The husband put the cat out before the taxi arrived. (3) The cat shot back into the house when the couple was going to the taxi. (1) The husband went back to chase the cat out. (4) The wife told the driver a lie. (6) The husband got into the car. Task 2 ADBCC Task 3CDBAC Let's talk Task 1 1-6: five dollars, chops, worst/lowest quality, barked, finest, money 7-12: own/have, followed, apartment house, third floor, scratched on, door 13-18: yelled at, stop, smartest/cleverest, lamb chops, looked at, his key Further listening and speaking Task 1 1-5: a dog, delighted/pleased, one of her friends, be close, interview 6-9: warn, smell, foretell/predict, a chicken Task 2 CBCDD Task 3 1.in the theater2. a ticket3. well-trained, intelligent, human 4. any problem, leave the theater, any other dog 5. let the dog in, almost empty Unit 5 test 1-5ABACA

1.how's 2.they seem to be 3.protects them from 4.all the time 5.scaring away 6.though

7.they're lazy/they are lazy 8.avoid being seen 9.much faster 10.like horses 11.in Greek 1-5BCDAC 1-5AACBC 6-10DBCAD

Unit6

Listening skills: Dialog 1: D, The first girl (The girl who wears the short skirt Dialog 2: B, No, she hasn’t. Listening in Task 1 1-4: looks funny, looks fine, out of fashion, good on her 5-7: out-dated, following fashion trends, no wonder Task 2 1-5: fashionable clothes, trends, copied, traditional center, are guarded 6-10: July, great numbers, a high price, starting point, only a part Task 3 1-5: DACBC Let's talk Task 1 1-5: different and daring, Teenagers, their legs, shorter and shorter, five inches 6-10: loose morals, never appear, look childlike, women’s liberation, traditional Further listening and speaking Task 1 1-5: TFTFF Task 2 1-7: nervous, afraid, first time, locker, blanket, come out, wore Task 3 1-4: new clothes, full of clothes, university tuition, clothes in fashion 5-7: the Middle Ages, part-time job, not too expensive Unit 6 test 1-5DDCAC 1.trying on 2.a business suit 3.in a blouse 4.cotton tights 5.a dress

6.a miniskirt 7.to an office 8.catch more eyes 9.suit 10.a typical boy 1-5ACBDC 1-5CBAAD 6-10BCDAC

Unit7

五 : 新视野大学英语视听说教程第二版

Unit1

Enjoy the colorful campus life!

II. Basic Listening Practice

Keys: 1.C 2.D 3. B 4.D 5.A

III. Listening In

Task 1:On the first day

Keys:

(1)the first day

(2)changing

(3)really good

(4)hard workers

(5)went over her head

(6)explained

(7)notes

(8)Wednesday

(9)participation

(10)education

Task 2:How to select elective courses?

Keys:

(1)extra training

(2)chemistry

(3)accounting

(4)many fields of study

(5)better potential

(6)business degree

(7)challenging

(8)how to learn

(9)better understanding

(10)narrow-minded

Task3: How to get straight A's?

Keys:1.B 2.C 3. D 4.A 5.D

IV. Speaking Out

MODEL1

MODEL2

MODEL3

V. Let?ˉs Talk

Task 1:Maintaining the quality or increasing the intake? Keys:

(1)quality

(2)young

(3)25

(4)difficult

(5)government

(6)quality

(7)cut

(8)extra-cautious

(9)afford

(10)experience

(11)more

(12)blame

(13)budget

(14)puzzled

(15)service

Task 2:What's your answer?

According to the interviewee,

it's difficult to strike a balance between maintaining the educational quality and making sure as many people as possible receive university education.

In the past years, China has been confronted with the same problem.

The university enrollment has been on the increase and some institutions of higher learning do not have enough teachers and teaching facilities,

thus affecting the educational quality.

Task 3:Let's group work!

VI. Further Listening and Speaking

Listening Task:

Task1: Problems with our educational system

Script:

Hi, everybody. My topic today is "Problems with our educational system".

I disagree on a lot of the ways that things have happened for a long time in our educational system.

It seems that educators just want to give standardized tests,

which focus only on academic performance and neglect students' abilities and interest in other areas.

I think there are a lot of people who are very intelligent,

but haven't had the opportunities they could have had if they had learned in a broader-minded educational system.

I feel that a lot of courses that students are required to take in high school are too academic, and, as a result,

many kids have lost their interest in learning.Educators often fail to recognize various kinds of intelligence.

They simply exert a lot of pressure on students to be as well-rounded as possible. I think being well-rounded isn't really possible.

And as a consequence, some students I believe to be intelligent can't get into good colleges if they,

you know, haven't scored well on the math section, even if they are brilliant writers. Another thing that disturbs me is that the so-called weak students are separated

from the rest of the school.

Some kids are kept in a separate class if their grades are lower than others'. And they're very aware of their social position, you know.

I think it causes them to act in a way that is not really positive.

They're just acting in a way they are expected. Often their grades go from bad to worse. And that's pretty sad.

I think that many of the kids in those classes are intelligent,

but they never actually realize their potential because of the way they are treated early on in their education.

Keys:

(1)standardized tests

(2)abilities and interest

(3)interest

(4)pressure

(5)well-rounded

(6)get into good colleges

(7)the rest of the school

(8)from bad to worse

Task 2: The final exam

Script

At a university, there were four sophomores taking a chemistry course.

They were doing so well on all the quizzes, midterms, labs, etc. that each had an "A" so far for the semester.

These four friends were so confident that on the weekend before the final, they decided to go up to the University of Virginia and party with some friends there.

They had a great time and didn't make it back to school until early Monday morning. Rather than taking the final then,

they decided to find their professor after the final and explain to him why they missed it.

They explained that they had planned to come back in time for the final exam, but, unfortunately, they had a flat tire on the way back and didn't have a spare. As a result, they missed the final.

The professor thought it over and then agreed they could make up the final the following day.

The guys were relieved and elated.

The next day, the professor placed them in separate rooms, handed each of them a paper, and told them to begin.

They looked at the first problem, worth five points.

It was a simple question on a chemical reaction.

"Cool," they all thought at the same time, each one in his separate room, "this is going to be easy."

Each finished the problem and then turned the page.

On the second page was a question worth 95 points: "Which of the tires was flat?"

Keys:

(1)course

(2)quizzes

(3)semester

(4)confident

(5)party

(6)make

(7)missed

(8)they had a flat tire on the way back and didn't have a spare

(9)placed them in separate rooms, handed each of them a paper

(10)On the second page was a question worth

Task3: Harvard University

Script

Harvard University is the oldest institute of higher learning in the United States. Founded 16 years after the arrival of the Pilgrims at Plymouth,

the university has grown from nine students with a single master to the present enrollment of more than 21,000 students,

including undergraduates and students in 10 graduate and professional schools. Over 14,000 people work at Harvard, including more than 2,100 faculty members. Harvard has produced eight American presidents and many Nobel Prize winners. During its early years, Harvard offered a classic academic course based on the model of English universities,

but consistent with the prevailing Puritan philosophy.

Although many of its early graduates became ministers in Puritan churches throughout New England,

the university was never formally affiliated with a specific religious group. Under President Pusey, Harvard started what was then the largest fundraising campaign in the history of American higher education.

It was an 82.5-million-dollar program for the university.

The program increased faculty salaries, broadened student aid, created new professorships,

and expanded Harvard's physical facilities.

NeiI L. Rudenstine took office as Harvard's 26th president in 1991.

As part of an overall effort to achieve greater coordination among the university's schools and faculties,

Rudenstine encouraged academic planning and identified some of Harvard's main intellectual priorities.

He also stressed the importance of the university's excellence in undergraduate education,

the significance of keeping Harvard's doors open to students from families of different economic backgrounds,

and the task of adapting the research university to an era of both rapid information growth and serious fund shortage.

Keys: 1.B 2.A 3. C 4.D 5.D

Viewing and speaking:

Task 1:University budget cuts

Script

Host: Well, David Lammy, the University Minister, joins me from Westminster. Now thanks for joining us this lunchtime,

Mr. Lammy. How do these cuts... tie in with your much trumpeted commitment to increasing higher education?

Interviewee: Well, I think it's important to remind viewers that we will spend well over 12 billion pounds on higher education this year and to also say that there will be more students at university next year than ever before in our history.

But what is important is that when they get to university, is that they have good facilities, good buildings,

that they have good contact with their lecturers and, for students from poorer backgrounds that they receive a grant.

And you know 40 percent of students who are going to university are in receipt of some grants. So we have to plan...

Host:But there's going to be less money next year. That's the bottom line, isn't it? Interviewee: Er, we are asking universities to make a one percent cut in their teaching grant.

That's 51 million out of a total budget of over five billion.

Look, I think there are families across the country preparing for Christmas spending a bit less and they're,

you know, it's a lot more than one percent that they're, they're feeling.

So I think this is reasonable to ask universities if we are to ensure that we can

continue to send more young people to university and we remain committed to that. Host:And briefly, what about these two-year degrees? This is a reduction in standards, isn't it?

Interviewee:Well, we, we, we... It's important in this country that we remember that students aren't just the classic 18- to 21-year-old undergraduates.

We want mature students. We want more parttime students and over the last few years we've been growing the number of foundation degree,

two-year degrees that ensure that transition into high university and high level skills. That's what we want to support and that's what we're indicating in the grant letter that we've sent to universities over the last few days.

Host:OK. David Lammy, thank you so much for joining us.

Interviewee:Thank you.

Keys:

(1)increasing

(2)12 / twelve

(3)more

(4)facilities

(5)lecturers

(6)poorer

(7)grant

(8)one

(9)budget

(10)less

(11)young

(12)classic

(13)part-time

(14)growing

(15)two-year

Task 2:Talk after viewing

Script

There will be more students at universities next year than ever before in our history. When they get to university, they have good facilities, good buildings, as well as good contact with their lecturers.

Students from poorer backgrounds receive a grant.

40 percent of students who are going to university will receive some sort of grant. Unit 1 test

Part I

Keys: 1.A 2.B 3.D 4.B 5.D

Part II

(1)for

(2)with

(3)opportunity

(4)tuition

(5)explore

(6)encounter

(7)adventure

(8)As with any country, it is not advisable to carry large amounts of cash around with you

(9)Traveler's checks are one of the safest and easiest ways to transport money, because you may have them replaced if they get lost or stolen

(10)It is wise to bring about $100 with you in U.S. cash, so you will be able to manage upon your arrival in the States

Part III

1.C 2.A 3.B 4.D 5.A

Part IV

1.C 2.A 3.A 4.D 5.C 6.B 7.A 8.C 9.B 10.C

Uint2

Our globe is in danger!

I.Lead in

Task 1:

sandstorm ; air pollution ; deforestation

water pollution ; melting polar ice cap ; light pollution

drought ; desertification ; littering

II. Basic Listening Practice

Keys: 1.B 2.D 3. A 4.A 5.C

III. Listening In

Task 1:We should have proper respect for nature!

Script:

Martha:Do you think most people in your culture respect nature?

Ed:I think so. Um... more now than before.

Martha:What do you think is the most serious environmental problem in the world today?

Ed: Today... I think damage to the ozone layer is a big problem, and another problem is pollution in big cities and things like that.

Martha: How do you learn about environmental problems?

Ed: Um... through school. A lot of clubs promote environmental safety, and some TV programs, too. They talk about environmental safety and stuff like that.

Martha: Do you think students should learn more about the environment at school? Ed: I think so. So, as they grow older, they can be more aware of all the problems that are going on, and also to prevent more problems from occurring.

Martha: If you could create a new law to help the environment, what would it be? Ed: A new law for the environment? Um... I'd probably say that when people throw away their cigarette butts, they have to throw them into the garbage bin, not just throw them everywhere because it's just littering and I hate that. So they should be fined if they throw them on the floor or on the ground.

Martha: That's a good idea. What do you personally do to help protect the environment?

Ed: I'm so against littering. I never litter. If I see somebody litter, I get really angry. So I always throw my trash into the garbage bin.

Keys:

(1)nature

(2)environmental problem

(3)pollution

(4)promote

(5)aware of

(6)from occurring

(7)law

(8)throw away

(9)fined

(10)trash

Task 2:River pollution

Script:

If you see dead fish floating on the river or notice that the water is discolored and smelly,

you know the river has been polluted, and there are four main possible causes for it. First, fertilizer. If large amounts of fertilizer or farm waste drain into a river, the concentrations of nitrate and phosphate in the water increase considerably.Algae use these substances to grow rapidly, turning the water green.

This massive growth of algae leads to pollution. When the algae die, they are broken down by the action of the bacteria, which quickly multiply, using up all the oxygen in the water and therefore causing the death of fish.

Second, industrial waste. Factories sometimes discharge chemical waste into rivers. Examples of such pollutants include cyanide, lead, copper, and mercury. These substances may enter the water in such high concentrations that fish and other animals are killed immediately. Sometimes the pollutants enter the food chain and accumulate until they reach toxic levels, eventually killing fish and other animals. Third, oil pollution. If oil enters a slow-moving river, it forms a rainbow-colored film over the entire surface, preventing oxygen from entering the water.

Fourth, warm water. Industry often uses water for cooling processes, sometimes discharging large quantities of warm water back into rivers. A higher temperature of the water lowers the level of dissolved oxygen and upsets the balance of life in the water.

Keys:1.D 2.A 3.B 4.C 5.D

Task3: Curbing carbon emissions

Script:

Although it is not an easy task, China is striving to fulfill the promise to cut its carbon dioxide emissions per unit of GDP by 40 to 45 percent in the next 10 years.

Zhang Guobao, Director of the National Energy Administration, said, "The government puts great emphasis on seeking harmonious development between cities and the environment, and is readjusting the energy structure by giving priority to the development of clean and low-carbon energies, including hydroelectric, nuclear, wind, and solar power."

Government authorities have closed small, coal-fired plants with a total capacity of 60.06 million kilowatts in the past four years. This year's target of closing 10 million kilowatts of capacity will be achieved by August.

"We have promised to the international community that 15 percent of our power will be generated from nonfossil sources by 2020," Director Zhang said. At present, non-fossil energy accounts for only 7.8 percent.

China is making efforts to increase the proportion of clean energy in its total energy consumption. Statistics show that China invested US$34.6 billion in clean energy last

year, exceeding the United States which invested US$18.6 billion. Thus, China has become the world leader in generating clean energy. Five years earlier, China's investment in clean energy was only US$2.5 billion.

However, China's carbon emission reduction target cannot be achieved easily. The shift to a low-carbon economy might be met at a cost to society. For instance, more than 400,000 people were laid off as a result of the shutdown of small coal-fired power plants in the past four years. Many studies indicate that the effort to curb greenhouse gas emissions may delay China's development, affect people's income, and lead to unemployment.

Keys:1.D 2.D 3.A 4.B 5.C

IV. Speaking Out

MODEL1

MODEL2

MODEL3

V. Let?ˉs Talk

Task 1:Disappointment over climate negotiations

Keys:

1) doesn't really actually commit anyone to doing anything,

2) the atmosphere simple can't take the kind of emissions we've been seeing in the business-as-usual scenario and there's no real commitment to change that.

3) There's no real commitment to put serious money on the table

4) It's not backed by action,

5) It does not constitute a... a deal.

6) It's a hollow shell

7) it lacks anything on emissions cuts

8) There's no guarantee that there will be new money, that the money will be real, that there's actually a commitment to get there, or that it will be channeled in new ways,

Task 2:What's your answer? Two speakers hold a negative attitude toward the world climate conference.

Task 3:Let's group work!

Developed countries should bear more responsibilities.

First, they caused most of the world pollution in their industrialization process.

Second, they are already economically advanced, with more funds to deal with pollution problems.

On the other hand, developing countries caused less pollution in the past, and now they need rapid development.

Therefore, an international agreement the Kyoto Protocol in 1997 stipulated common but differentiated responsibilities for developed and developing nations, which means that both should be responsible for environmental protection and cut carbon dioxide emissions, but the former must take more responsibility.

Recently, however, some wealthy countries have insisted that developing countries like China should shoulder more responsibilities.

VI. Further Listening and Speaking

Listening Task:

Task1: The environment and the development

Script:

Li:Hi, Professor Wang.

I'm Li Lin, a correspondent for the university newspaper.

The staff and students here are getting more and more interested in the relationship between the environment and economic development.

What do you think is the most serious environmental problem at present? What measures should we adopt to improve the environment and develop the economy at the same time?

Wang: There are many environmental problems: air pollution, water pollution, desertification, overfishing, destruction of natural habitats, acid rain, overconsumption of wild animals and plants, etc. But lying at the center of all those problems, as I see it, is the contradiction between economic growth and the environment.

Since the United Nations Earth Summit in 1992, growing number of people and governments have adopted the new idea of "sustainable development". This means today's economic growth should not wipe out the resources and options for future generations. Planning and development should ensure not only economic growth, but also social advancement and environmental health. In other words, some economic behavior must be restricted or controlled. Instilling principles of sustainable development into government planning, resource management and economic policy is the most important step China can take to solve its environmental problems.

China has already taken some remarkable steps to reduce damage to the environment. For instance, following the huge floods of 1998, the government banned logging in the upper reaches of the Yangtze River in order to protect forests and reduce the risk of floods.

Still, the basic contradiction between the environment and development persists. Much work is to be done before we can achieve the aim of a balance between economic growth and the environment.

Keys:

(1)water pollution

(2)overconsumption

(3)economic growth

(4)resources

(5)social advancement

(6)restricted

(7)government planning

(8)economic policy

(9)remarkable steps

(10)balance

Task 2: Thick cloud of pollution covering southern Asia

Script

A United Nations study says a thick cloud of pollution covering southern Asia threatens the lives of millions of people. Scientists say the pollution could increase lung disease and cause early death. The cloud is also damaging agriculture and affecting rainfall levels. It has affected many countries in southern Asia. The pollution cloud is three kilometers high. Scientists say it can move halfway around the world in a week.

The cloud is the result of forest fires, the burning of agricultural waste, and huge increases in the burning of fuels by vehicles, industries, and power stations. Pollution from millions of bad cooking stoves has made the problem worse. Many poor people burn fuels like wood and animal waste in such stoves.

Scientists say the cloud of pollution appears to cool the land and oceans by blocking sunlight.

They say it reduces the amount of sunlight reaching the Earth's surface by as much as 15 percent.

At the same time, heat inside the cloud warms the lower parts of the atmosphere. Harmful chemicals from the cloud are mixing with rainfall.

This acid rain damages crops and trees and threatens public health.

Scientists are concerned that the pollution will intensify during the next 30 years as the population of Asia increases to an estimated 5,000 million people.

Keys:1.D 2.B 3.A 4.C 5.C

Task3: Mountain regions face a number of dangers

Script

Mountain people around the world are in great danger of the negative effects of the worsening environment, according to a UN report.

As global warming and deforestation accelerate and technology makes wilder places more accessible, environmental and social pressures on the world's remotest regions increase.

The UN has found that many mountainous regions—inhabited by one out of five of the world's people—are barely recognizable when they are compared to what they were like 60 years ago.

This is mostly because forests were cut to make way for cattle grazing and agriculture. The authors of the UN study expect 98 percent of the mountain areas to experience severe climate change by 2055.

Biological losses are expected to be heavy.

The mountains of Europe, parts of California and the northwest Andes in South America are among the most threatened mountain areas in the world and should be given priority in conservation.

The UN is anxious to raise awareness of the problems facing mountain areas because they are inhabited by some of the most vulnerable people.

These people could lose their culture and their livelihood with even the smallest shifts in climate.

At the same time, many mountain regions are losing people. Thousands of villages in Europe are deserted most of the year.

In other areas like Nepal, people are drifting to the cities in search of work.

Keys:

(1)negative

(2)report

(3)warming

(4)pressures

(5)mountainous

(6)barely

(7)make way for

(8)expect 98 percent of the mountain areas to experience severe climate change by 2055

(9)The UN is anxious to raise awareness of the problems facing mountain areas

(10)Thousands of villages in Europe are deserted most of the year

Viewing and speaking:

Task 1:Santa's home town in danger

Script

Weather experts may have found a new problem caused by global warming, one which many people will pay attention to: There are signs that Santa's home in the North may be in trouble because of warmer temperatures.

The Finnish town of Rovaniemi on the Arctic Circle, which many Europeans say is the home of Santa Claus, has had its warmest winter in 40 years.

As a result, there has been much less snow than usual—meaning no snowmen, no snowballs, and possibly not enough snow for Santa to ride his sleigh on.

More important for local residents, it may mean fewer tourists, as well. Santa's wintry home town normally attracts thousands of visitors each year, and millions of dollars.

Anne Pelttari-Bergman, the town's tourist director, worries that the town could be in trouble if snow levels do not return to normal.

She explains: "Snow is really important for us, of course, for Santa Claus, for Christmas tourism, and also for our winter tourism because winter is our best season. It is really important for us."

Weather experts and town residents are hoping this warm winter is an one-time thing. Few people can imagine a holiday when even Santa does not have a white Christmas.

Keys:

(1)signs

(2)warmer temperatures

(3)warmest

(4)snow

(5)attracts

(6)trouble

(7)normal

(8)season

(9)one-time

(10)white

Task 2:Talk after viewing

Script

The image of Santa Claus flying in a sleigh pulled by reindeer and leaving toys and gifts for every child is known worldwide, and Santa Claus has become the most beloved of Christmas symbols.

But if the weather at Santa's home town were to become so warm that there was no more snow at Christmas, Santa Claus would lose his charm.

So snow is really important for Santa Claus, for his home town's Christmas tourism, and its winter tourism as winter is the best season in Santa's home town. Unit 2 test

Part I

Keys: 1.C 2.C 3.B 4.C 5.D

Part II

(1)chemicals

(2)atmosphere

(3)particles

(4)trapped

(5)lasts

(6)human-based

(7)progressed

(8)Even in Ancient Rome people complained about smoke put into the atmosphere

(9)Air pollution can have serious consequences for the health of human beings

(10)Cities with large numbers of automobiles or those that use great quantities of coal often suffer most severely from air pollution problems

Part III

1.A 2.C 3.D 4.C 5.B

Part IV

1.B 2.C 3.D 4.C 5.C 6.A 7.D 8.D 9.B 10.C

Uint 3

Culture makes me what I am

Task 1:On the first day

Collectivism Individualism Individualism Collectivism

II. Basic Listening Practice

Keys: 1.B 2.A 3. D 4.B 5.A

III. Listening In

Task 1:Competition in America

Keys:1.C 2.A 3. C 4.B 5.D

Task 2:How to select elective courses?

Keys:

(1)roots

(2)what

(3)moral

(4)diligence

(5)sin

(6)teachings

(7)centuries

(8)If it's worth doing at all, it's worth doing well

(9)In English a new word has been created to describe people who work compulsively

(10)Others hold that workaholics are valuable members of society because they are extremely productive

Task3: Cross-cultural tips on doing business

Keys:1.D 2.A 3. D 4.A 5.C

IV. Speaking Out

MODEL1

MODEL2

MODEL3

V. Let?ˉs Talk

Task 1:International Slavery Museum

Keys:

(1)transatlantic slave trade

(2)unknown lands

(3)dignity or payment

(1)rich

(2)cotton and sugar

(1)all over the world

(2)family history

(3)changed

200th anniversary

(1)banned

(2)taking part in

(3)slavery

(4)freed

Task 2:What's your answer?

VI. Further Listening and Speaking

Listening Task:

Task1: Problems with our educational system

Script:

To Americans, punctuality is a way of showing respect for other people's time.

Being more than 10 minutes late to an appointment usually calls for an apology and maybe an explanation.

People who are running late often call ahead to let others know of the delay.

Of course, the less formal the situation, the less important it is to be exactly on time. At informal get-togethers, for example, people often arrive as much as 30 minutes past the appointed time.

But they usually don't try that at work.American lifestyles show how much people respect the time of others.

When people plan an event, they often set the time days or weeks in advance. Once the time is fixed, it takes almost an emergency to change it.

If people want to come to your house for a friendly visit, they will usually call first to make sure it is convenient.

Only very close friends will just "drop in" unannounced.

Also, people hesitate to call others late at night for fear they might already be in bed. The time may vary, but most people think twice about calling after 10:00 p.m.

To outsiders, Americans seem tied to the clock. People in some Eastern cultures value relationships more than schedules.

In these societies, people don't try to control time, but to experience it.

Many Eastern cultures, for example, view time as a cycle.

The rhythm of nature—from the passing of seasons to the monthly cycle of the moon—shapes their view of events.

If they have wasted some time or let an opportunity pass by, they are not very worried, knowing that more time and opportunities will come in the next cycle. But Americans often want to jump at the first opportunity.

They are unwilling to stand by idly and give up the opportunity.

The early American hero Benjamin Franklin expressed that view of time like this: "Do you love life? Then do not waste time, for that is the stuff life is made of." Keys:1.C 2.B 3.A 4.C 5.D

Task 2: Our personal space

Script

Our personal space, that piece of the universe we occupy and call our own, is contained within an invisible boundary surrounding our body. As the owners of this area, we usually decide who may enter and who may not. When our space is invaded, we react in a variety of ways. We back up and retreat, stand our ground as our hands become moist from nervousness, or sometimes even react violently. Our response shows not only our unique personality, but also our cultural background.

For example, cultures that stress individualism such as England, the United States, Germany, and Australia, generally demand more space than collective cultures do, and tend to become aggressive when their space is invaded. This idea of space is quite different from the one found in the Mexican and Arab cultures. In Mexico, the physical distance between people when engaged in conversation is closer than what is usual north of the border. And for Middle Easterners, typical Arab conversations are at close range. Closeness cannot be avoided.

As is the case with most of our behavior, our use of space is directly linked to the value system of our culture. In some Asian cultures, for example, employees do not stand near their bosses; the extended distance demonstrates respect. Extra interpersonal distance is also part of the cultural experience of the people of Scotland and Sweden, for whom it reflects privacy. And in Germany, private space is sacred.

Keys:1.A 2.B 3.A 4.B 5.B

Task3:We don't know what to do with them.

Script

A Russian, a Cuban, an American businessman, and an American lawyer were passengers on a fast train speeding across the French countryside. As time wore on, they gradually became friendly with one another, introducing themselves and shaking hands. Eventually, the Russian took out a large bottle of vodka and poured each of his traveling companions a drink. Just as the American businessman was sipping the vodka and praising its fine quality, the Russian hurled the half-full bottle out of the open window.

"What did you do that for?" asked the startled American businessman.

"Vodka is plentiful in my country," said the Russian. "In fact, we have thousands and thousands of liters of it—far more than we need."

The American businessman shook his head and leaned back in his seat, obviously baffled by the Russian's reasoning.

A little later, the young Cuban passed around a box of fine Havana cigars. The men enjoyed this treat and made admiring remarks about the pleasure of smoking good Havana cigars. At that very moment the Cuban took a couple of puffs of his cigar and then tossed it out of the open window.

"I thought the Cuban economy was not good this year," the American businessman said. "Yet you threw that perfectly good cigar away. I find your actions quite puzzling."

"Cigars," the Cuban replied, "are a dime a dozen in Cuba. We have more of them

than we know what to do with."

The American businessman sat in silence for a moment. Then he got up, grabbed the lawyer, and threw him out of the window.

Keys:

1.The Russian hurled the half-full bottle of vodka out of the open window.

2.He answered, "Vodka is plentiful in my country. In fact, we have thousands and thousands of liters of it—far more than we need."

3.The businessman said, "I thought the Cuban economy was not good this year. Yet you threw that perfectly good cigar away. I find your actions quite puzzling."

4.He replied, "Cigars are a dime a dozen in Cuba. We have more of them than we know what to do with."

5.The American businessman sat in silence for a moment. Then he got up, grabbed the lawyer, and threw him out of the window. He did that probably because he thought there were too many lawyers in the United States.

Viewing and speaking:

Task 1:Reviving the image of tea

Keys:

(1)disappearing

(2)kicking

(3)tea shop

(4)image crisis

(5)update

(6)compete

(7)fast

(8)product designers

(9)contacts

(10)similar

(11)want

(12)challenge

(13)consumer

(14)traditional

Task 2:Talk after viewing

Script

As people become espresso fiends and cappuccino connoisseurs, the traditional British tea are being kicked off the menu by designer coffees.

Unit 3 test

Part I

Keys: 1.B 2.C 3.B 4.B 5.D

Part II

(1) values

(2) purpose

(3) true

(4) Nowhere

(5) equality

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新视野大学英语视听说教程4第二版答案-求新视野大学英语视听说教程3第二版答案.（外研出版社,总主编：郑树棠,主编：王大伟）

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