?1a?(m?n)?(m?n)?xcosx?cosa?0?aa??x?dx?a1?a2(m?n)?ax(m?n)???[cosx?sinx]22a?(m?n)?a(m?n)?a0?a?a2(m?n)?ax(m?n)? ?[cosx?sinx]? 22a(m?n)?a(m?n)?0????a11?2(?1)m?n?1???2?(m?n)2??(m?n)???4mnm?n(?1)?12222?(m?n)a??a0?un(x)dx??i??pmn??um(x)p*2m?dn?sinx?sinxdxaadxa2n??am?n???i2?sinx?cosxdx0aaan??a?(m?n)?(m?n)???i2??sinx?sinaaa0?
?x?dx??x? ?0an???a(m?n)?a(m?n)??i2?cosx?cosa(m?n)?aa?(m?n)??11?m?n??1]?(m?n)(m?n)?(?1)??i2mn??(?1)m?n?12(m?n2)a?in??aa2?
???? ?sinmucosnud?u?cos(m?n)ucos(m?n)u??C 2(m?n)2(m?n)15 求线性谐振子哈密顿量在动量表象中的矩阵元。
2211??122222??????x?? 解：Hp???x 22?22??x2? Hpp????*p(x)H?p(x)dx
?pxp?x1?2?2122?? ?e(????x)edx 2?2??2??x2(p??p)x(p??p)x???2i121212?? ??(p?)edx???xedx ??????2??2??22??iiii2(p??p)x??p?21212?? ??(p??p)???()edx 2???2?22??i?p?i2p?212?2? ??(p??p)???()2?2i?p?2??1?????ei(p??p)x?dx
2p2122? ??(p??p)?????(p??p) 2?2?2?p2p2122? ??(p??p)?????(p??p) 22?2?p16 求连续性方程的矩阵表示
解：连续性方程为
???????J ?t?i?(???*??*??) ∴ J?2??i???(???*??*??) 而 ??J?2? ? ? ∴ i?
i?(??2?*??*?2?) 2?1??*??*T??) (?Ti???????T??*) ?(?*T?t?(?*?)????T??*) i??(?*T?t 写成矩阵形式为
?????T???(???)???T?t
????(??T??)*?T?T*?0i?(???)???T?ti???的作用，17 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??H21??a，H11??H22??b；a、b都是实数。用微扰公式求能量微扰矩阵元为H12至二级修正值。
解：由微扰公式得
(1)? En?Hnn(2)??' Enm?Hmn2(0)(0)En?Em
(1)(1)??b ??b 得 E01?H11E02?H22 E(2)01??m'?1Hm2E01?E0m'a2? E01?E02a2? E02?E01 E(2)02??m?1Hm2E02?E0m ∴ 能量的二级修正值为
a2 E1?E01?b?
E01?E02a2 E2?E02?b?
E02?E0118 计算氢原子由第一激发态到基态的自发发射几率。
解： Amk34es2?mk?2?r mk33?c 由选择定则????1，知2s?1s是禁戒的 故只需计算2p?1s的几率 ?21?E2?E1 ?13?es4 ?3(1?)?
48?32??2222 而 r21?x21?y21?z21
?es4 2p有三个状态，即 ?210, ?211, ?21?1
? (1)先计算z的矩阵元 z?rcos*(r)R10(r)r3dr???1*mcos? Y00d? (z)21m,100??R210? ?f?Y1*m ?f1313 Y00d?
?m0
?(z)21,1000?13f
(z)21,1100?0 (z)21?1,100?0
?cos?? (2)计算x的矩阵元 x?rsin (x)21m,100?rsin?(ei??e?i?) 21?*3*i??i?R(r)R(r)rdr?Ysin? (e?e)Y00d? 21101m??02 ?12*f?Y1m(?Y11 ?Y1?1)d? ?2316f(??m1??m?1)
? Y11??Y00?14?3sin? ei? 8?
Y1?1?3sin? e?i? 8?
?(x)21,1000?0 (x)211,100??16f
(x)21?1,100?16f
1rsin?(ei??e?i?) 2i?sin??(3)计算y的矩阵元 y?rsin (y)21m,100 ?1?*i???R21(r)R10(r)r3dr??Y1*?e?i?) Y00d? msin?(e2i012f?(??m1??m?1) 2i31i6f(??m1??m?1)
? ?(y)21,1000?0
(y)211,100?i6i6f
f
(y)21?1,100?? ?r2p?1s2f2f212?(2??2??f)?f2
663(4)计算f
*(r)R10(r)r3dr? f??R210?256816a0
313/2) ?(2a013/2?4?2a0r?()?redr
0a3a002114！?25525627 ??a0?a0?a0445336a08162152 f?9a0
322 3 A2p?1s34es2?21?2?r21 3?c34es23?es432152?()?9a0 ?333?c8?328?3e14?22s ?7?103(?) 23?c? es28?e10 ?7?6s3?1.91?109s?1
3?c ??1?5.23?10?10s?0.52?10?9s A2119 求线性谐振子偶极跃迁的选择定则
?22 解： Amk?rmk?xmk
*x?kdx xmk???m