第三章习题解答
3.1 一维谐振子处在基态?(x)? (1)势能的平均值U???e??2x2i2??t2，求：
1??2x2； 2p2 (2)动能的平均值T?；
2? (3)动量的几率分布函数。
?11222?2??2x2解：(1) U???x???xedx ???22?1?1?111?2 ???2 ?222???2????2224??2??2???211?3?5???(2n?1)? ??? ?x2ne?axdx? 04a2n?1anp21?*2? (2) T???(x)p?(x)dx ???2?2? ? ? ? ? ? ?2??x?1??2?x2de(??)e2dx 2?dx?2?????22???(1??2x2)e??xdx
???2????22???x?[?edx??2?x2e??xdx]
?????2???22???[??2?3]
?2??2???22??22?2?? ?????2?4?4???2?2222221122221?? 4111???????? 244?(x)dx (3) c(p)???*p(x) 或 T?E?U? ? ??2??12??1??????12 e?22xei?Px?dx
?????? e1??2x22ei?Px?dx
? ? ?12??12??12????????? e1ipp2??2(x?2)2?222??2??dx dx e?p22?2?2?e??e??p22??22???? e1ip??2(x?2)22??p22?2?22???1???
动量几率分布函数为 ?(p)?c(p)?#
3.2.氢原子处在基态?(r,?,?)? (1)r的平均值；
e2 (2)势能?的平均值；
r (3)最可几半径； (4)动能的平均值；
(5)动量的几率分布函数。 解：(1)r??r?(r,?,?)d??221???e?p2?2?2
13?a0e?r/a0，求：
13?a0???00?2??0re?2r/a0r2sin? drd? d?
?4?3?2r/a0n! ?3?radr ?xne?axdx?n?1 0aa0043!3?a0 ?342a0?2???a???0?e2e2(2)U?(?)??3r?a0e2??3?a0
???00?2??01?2r/a02ersin? drd? d?r???00?2??0e?2r/a0rsin? drd? d?r dr
4e2??3a0??0e?2r/a04e21e2??3??2a0?2?a0??a???0?
(3)电子出现在r+dr球壳内出现的几率为 ?(r)dr???0?2?0[?(r,?,?)]2r2sin? drd? d??4?2r/a02erdr 3a04?2r/a02er 3a0d?(r)42 ?3(2?r)re?2r/a0
dra0a0d?(r)?0， ? r1?0, r2??, r3?a0 令 dr 当 r1?0, r2??时，?(r)?0为几率最小位置
?(r)?d2?(r)4842?2r/a0 ?(2?r?r)e32a0dr2a0a0d2?(r)
dr2??r?a08?2e?0 3a0 ∴ r?a0是最可几半径。
21?1???1??1??2??????2?2?(r2)?(sin?)? (4)Tp?2 ??rsin?????r??rsin2???2?2?2??2?2??1?r/a02?r/a02 T??e?(e)rsin? drd? d? 3???0002??a0?2?2??1?r/a01d2d?r/a02??e[r(e)]rsin? drd? d? 32???0002?dr?a0rdr4?21 ??(?3a02?a0??0r2?r/a0(2r?)e dr
a022a0a04?2?2 ? (2?)?42442?a02?a0??(r)?(r,?,?)d? (5) c(p)???*p1 c(p)?(2??)3/2??10?a3030?e?r/a0rdr?e02?i?prcos??sin? d??d?
02? ?2?(2??)3/2?a?0re2?r/a0dr?e0?i?prcos?? d(?cos?)
? ?2?(2??)3/2?a30??0r2e?r/a0dr?eipri?i?prcos??0i
2?(2??)3/2?pr????r/a0?prn!n?ax?re(e?e)drxedx? n?1?003ip?a?a0?11[?]
3/23ip1i1i22(2??)?a0(?p)(?p)a0?a0?14ip ? 2332a0?ip?a?(1?p)202a0?2 ?2? ? ?4303244a0?2222a??a0(a0p??)
(2a0?)3/2??(a0p??)2222
动量几率分布函数
358a0? ?(p)?c(p)?2 224?(a0p??)#
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
e? m2?n?m Je??? rsin? 证：电子的电流密度为
??i?**(?n?m??n Je??eJ??e?m??n?m??n?m) 2? ?在球极坐标中为
??1???1??e??e? ??er ?rr??rsin??????式中er、e?、e?为单位矢量
????1???i?1?*Je??eJ??e[?n?m(er?e??e?)?n?m2??rr??rsin???
??1???1?* ??n(e?e?e)?n?m]?mr???rr??rsin????ie???*?1?**??[er(?n?m?n???)?e(??n?m?mn?mn?m?n?m2??r?rr??
?1?1?1?*** ??n?n?m)?e?(?n?m?n?n?n?m)]?m?m??mr??rsin???rsin??? ??n?m中的r和?部分是实数。
?ie?e?m22?2?(?im?n?m?im?n?m)e? ???n?me? ∴ Je??2?rsin??rsin? 可见，Jer?Je??0
2 Je???e?m2?n?m
?rsin?#
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???2? (SI)? M?Mz??me??? (CGS)??2?c 原子磁矩与角动量之比为
?e? (SI)?M?2? z??
eLz?? (CGS)??2?c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?m2?n?mdS??(rsin?)2
?rsin?e?m2?rsin??n?mdS ?? ??? ??e?m? (2)氢原子的磁矩为 M??dM?? ???0?r2sin??n?mdrd? (dS?rdrd?)
?2?0?e?m???n?mr2sin? drd?
2??e?m2?2????n?mr2sin? drd?
002?e?m2???22? ??n?mrsin? drd?d?
2??0?0?0e?m ?? (SI)
2?e?m 在CGS单位制中 M???
2?c 原子磁矩与角动量之比为
MMMee z??? (SI) z?? (CGS) #
LzLz2?Lz2?c
量子力学 第三章习题与解答
