??i?? ?p?re??r11?r/a0*?3/2???F?d??()e()ed? 其中Fmk?mk z（） p32i2???a0
取电子电离后的动量方向为Z方向， ??
取?、p所在平面为xoz面，则有 ?? ??r??xx??yy??zz ? θ r α ? ?(? sin?)(rsin?cos?)?(?cos?)(rcos?) ? ?? rsin?sin?cos??? cos?rcos? O y
x i?p rcos? 13/21eFmk?()e?(? rsin?sin?cos??? rcos?cos?)e?r/a0d?32i ??2?a0
13/21eFmk?( )32i2???a 0 i 2??p rcos??? e?(?rsin?sin?cos??? rcos?cos?)e?r/a0r2sin?drd? d?000
??的作用，微扰矩阵元为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
??H??a，H11??H22??b；a、b都是实数。用微扰公式求能量至二级修正值。 H12 21 解：由微扰公式得
(1)? E n?Hnn 2 ?Hmn(2)' En ? (0)(0)E?Emnm (1)(1)??b ??b 得 E01?H11E02?H22
2?Ha2 )m1(2'? E01? E01?E0mE01?E02m2
?Ha2m1(2)'? E02 ?E02?E0mE02?E01m
∴ 能量的二级修正值为
2a E1 ?E01?b? E?E0102
a2
E2?E02?b? E02?E01
?????????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 4?es ? ? min?hvmin?E??E1?22?
?es413.6?1.6?10?19 ??3.3?1015Hz vmin??3426.62?102?h
②t ?0时，氢原子处于基态，其波函数为 1 ?k?e?r/a0 3?a0
i?? 13/2?p?r 在t 时刻， ?m?( )e2???? ??e??ri? t??(t)?e??rsin?t?(e?e?i? t) 微扰 H2i ?i? t ?F(e?e?i? t) ??
e??r?? 其中 F 2i
在t时刻跃迁到电离态的几率为 2?a(t) Wk?m m
1t ?ei?mkt?dt? Hmk am(t)?i?0
tF i(???)t??ei(?mk??)t?)dt? ?mk(emk i?0 Fmkei(?mk??)t?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， Fmkei(?mk??)t?1 a m(t)? ??mk?? 2F(ei(?mk??)t?1)(ei(?mk??)t?1)2mk Wk?m?am(t)? 22?(???) mk 4Fmk2sin21(?mk??)t2 ? ?2(?mk??)2 ??i???p?re??r11? ??*F?)3/2e?()e?r/a0d? 其中Fmkm?kd??(z（） p32i2???a0
取电子电离后的动量方向为Z方向， ??取?、 p所在平面为xoz面，则有 ???? ??r ??xx??yy??zz ?(?sin?)(rsin?cos?)?(?cos?)(rcos?) ?? rsin?sin?cos??? cos?rcos? ???? α θ ?r O y Fmk?(12??)3/2ee32i??a01i?p rcos??x (? rsin?sin?cos??? rcos?cos?)e?r/a0d? 2?? i??2??p rcos? e?(?rsin?sin?cos??? rcos?cos?)e?r/a0r2sin?drd? d?000 i??2??p rcos?11e3/2?( )e?(?cos? r3cos?sin?)e?r/a0drd? d? 32i000?a0 2??i ?p rcos???13/21e?cos??r/a)2?r3e0dr[e?cos?sin? d? ?(0032i2???a0
iiii?p rp r?p rp r?e?cos??2 3?r/a0???????re[(e?e)?22(e?e)]dr 03 iprpri2??2a0?
e?cos?16p1 ? 23ia?i2??2a00(1?p)3 2a0?2
7/216pe?cos?(a?)0 ?? 22238?(a0p??)
2 4Fmksin212(?mk??)tW? ∴ k?m22?(?mk??) 21222275sin128pe?cos?a?02(?mk??)t ? 222262?(a0p??)(?mk??)
i??2??p rcos?11e3/2?( )e?(?cos? r3cos?sin?)e?r/a0drd? d? 32i000?a0 2??i ???p rcos?13/21e?cos?3?r/a0?)2?redr[ecos?sin? d? ?(0032i2???a0
iiii2?p rp r?p rp r?e?cos?? 3?r/a0???re[(e??e?)?22(e??e?)]dr 3iprpri2 ??2a0?0
e?cos?16p1 ?23ia?i2??2a00(1?p)3 2a0?2
7/216pe?cos?(a?)0 ?? 22238?(a0p??)
2 4Fmksin212(?mk??)tW? ∴ k?m22?(?mk??) 21222275sin128pe?cos?a?02(?mk??)t ? 222262?(a0p??)(?mk??)
Fmk?(1)3/2e32i?a01???????????????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
当t?0?0, ??? ?t/?当t?0(?为大于零的参数)??0e,
求经过长时间后氢原子处在2p态的几率。
解：对于2p态，??1，m可取0, ?1三值，其相应的状态为 ?211 ?21?1 ?210 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。 1t?ei?mkt?dt? Hmk 由 am(t)? i?0*???d? (H???e?(t)rcos?) ?,100??210H H 210100 ?*??RYe?(t)rcos? RYd? (取方向为Z轴方向) 21101000
?2??3* ?e?(t)R21rR10drY10Y00cos?sin?d? d? 000
1 (cos?Y00?Y10) 3
2??*1 ?e?(t)fY10Y10sin?d? d?003
1?e?(t)f 3? 256*3a0 f?R21(r)R10(r)rdr?0 8163
13/2213/2?4?2a0r?()?()redr 02a03a0a0
114！?255256 ??a?a0 05 6a438160
*???d??1e?(t)f ? H??H210100 210,1003
e?(t)2561282 ?a0?e?(t)a0 2433816
?* ? H211,100?e?(t)?211rcos??100d? 0 ?2??3* ?e?(t)RrRdrYcos?Y00sin? d? d? 211011 000?2?? 3*1Y11Y10sin? d? d? ?e?(t)R21rR10dr000 3 = 0 *????1,100??21 H21?1H?100d? ??2?*3?e?(t)RrRdrY1?1cos?Y00sin? d? d? 2110 000??2?* 13Y1?1Y10sin? d? d? ?e?(t)R21rR10dr000 3 = 0 ????????????????????????? 由上述结果可知，W100?211?0， W100?21?1?0 ∴ W1s?2p?W100?210?W100?211?W100?21?1 21ti?21t??,100edt? ?W100?210?2H210
?0 2t21282i?21t??t?/?2?2()(ea0?0)eedt?
0?243 2ti?21t? ?e?1 21282222)ea0?0 ?2( 1?2432??221 ? ?时， 当t? 212822221)ea0?0 ?1s?2p?2( 1?2432?21?2 ? ? es43? es43 es211 其中?21?(E2?E1)? (1?)??33 ?48?a02?8?
5.6计算氢原子由第一激发态到基态的自发发射几率。 34es2?mk?2 解： r Amk?mk3?c3
由选择定则????1，知2s?1s是禁戒的
故只需计算2p?1s的几率
E?E1 ?21?2 ??
?es413?es4 ?3(1?)? 3 ?22?2428?2 而 r21?x21?y21?z21 2p有三个状态，即 ?210, ?211, ?21?1 (1)先计算z的矩阵元 z?rcos? ? *3* (z)21m,100?R21(r)R10(r)rdr??1mcos? Y00d? 0
1 ?fY1* Y00d? m3
1 ?f?m0 3
1
f ?(z)210,100? 3 (z)211,100?0 (z) 21?1,100?0
????