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
(0)??H??1?2??D???L?D?co?s取H2I2I(0)?2L?1?2????D?cos?，则 L, H2I(0)??H? HE?(())??视为微扰，用微扰法求得此问题。H??? 由于电场较小，又把H?H2的本征值为
?12I?(??1)?本征函数为 ?
（0）(0)??Y?m(?,?)
? H(0)的基态能量为E0?0，为非简并情况。根据定态非简并微扰论可知
E0
(2)?? ?E?*(0)?H??0(0)02(0)??E
?0??H?????H(0)0d???Y?m(?D?cos?)Y00sin? d? d?***??D??Y?m(cos? Y00)sin? d? d? ??D??Y?m 4?3Y1014?sin? d? d?
??D?3?Y?0 Y10sin? d? d???'*D?32??1
E(2)0??? H??0E(0)02(0)??E??? ?'D?2?2I23?(??1)?1??12??13?2D?I22 其中
Fmk???*m??d??(Fk12??)3/2?a30?e?i??p?r?(??e??r2i)e?r/a0d?
z（p） ?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?)
?? ?α θ ?r ??O yx ?? rsin?sin?cos??? cos?rcos?
Fmk?(12??)3/21e30?a2i?e?i?p rcos?(? rsin?sin?cos??? rcos?cos?)e?r/a0d? 31
Fmk?(?12??)3/21e
?r/a03?a02i?i?p rcos? ?0??0?2?0e(?rsin?sin?cos??? rcos?cos?)ersin?drd? d?2??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??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。 H12(1) 解：由微扰公式得
En??Hnn
En(2)??m2'?HmnEn(0)2(0)?Em 得
E(1)01??b E?H11(1)02??bE?H22(2)01??m'?1Hm2E01?E0m?aE01?E02
E(2)02??m'?1Hm2E02?E0ma2?a2E02?E01∴ 能量的二级修正值为E1?E01?b?a2E01?E02
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
??min?hvmin?E??E1??es2?42 vmin??es242?h?13.6?1.6?106.62?10?34?19?3.3?1015Hz
②t?0时，氢原子处于基态，其波函数为 ?k?1?a30e?r/a0 在t时刻，
?m?(12??)3/2e?i??p?r
??????e??re??ri? t?i? t??(t)?e??rsin?t????(ei? t?e?i? t)其中F(e?e) ?F微扰 H
2i2i在t时刻跃迁到电离态的几率为Wk?m?am(t)
2am(t)??i?t01t0?ei?mktdt?Hmki(?mk??)t??
?Fmki??(ei(?mk??)t??e)dt???Fmk?[ei(?mk??)t?1?mk???ei(?mk??)t?1?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，
32
am(t)?Fmke?2i(?mk??)t?1?mk???Fmk?22
Wk?m?am(t)(ei(?mk??)t?1)(ei(?mk??)t2?1)(?mk??)12??)3/2?4Fmk2sin2212(?mk??)t2?(?mk??))e?r/a0
其中Fmk?为Z方向，
???mF?kd??(*1?a30?e?i??p?r?(??e??r2id?取电子电离后的动量方向
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?
????z（p） ?? ?α θ ?r Fmk?(12??12??)3/21?a)3/230e?2iee?i?p rcos?(? rsin?sin?cos??? rcos?cos?)e?r/a0d?O Fmk?(?1x 3?a02i?i?p rcos??r/a0 ?0??0?2?0e(?rsin?sin?cos??? rcos?cos?)e?rsin?drd? d?2?(12??12??)3/21?a)3/230???2i00e?2?0e?i?p rcos?(?cos? rcos?sin?)e3?r/a0drd? d?
?(1e?cos?30?a2i2???0re3?r/a0dr[?e0??i?p rcos?cos?sin? d?
?e?cos?i2??2a?30??0re3?r/a0[??ipr(e?i?p ri?e?p r)??222pr(e?i?p ri?e?p r)]dr?e?cos?i2??2a3016pia0?(211a02?p?22??)316pe?cos?(a0?)20227/238?(ap??)
∴ Wk?m?4Fmksin2212(?mk??)t2?(?mk??) ?128pe?cos?a0?222275sin212(?mk??)t2?(ap??)220226(?mk??)
33
?(12??12??)3/21?a)3/230???2i00e??2?0e?i?p rcos?(?cos? rcos?sin?)e3?r/a0drd? d??(1e?cos?30?a2i2???0re3?r/a0dr[?e0??i?p rcos?cos?sin? d?
?e?cos?i2??2a?30??0re3?r/a0[??ipr(e?i?p ri?e?p r)??222pr(e?i?p ri?e?p r)]dr?e?cos?i2??2a3016pia0?(1a021?p?22 ??16pe?cos?(a0?)20227/23)38?(ap??)
∴ Wk?m?4Fmk2sin2212(?mk??)t2?(?mk??) ?128pe?cos?a0?222275sin212(?mk??)t2?(ap??)220226(?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??0e?t/?, 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。解：对于2p态，??1，m可取0, ?1三值，其相应的状态为?210 ?211 ?1i?21?1t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 am(t)??0?ei?mktdt? Hmk?10 H2?,100 ???*210???e?(t)rcos?) ???d? (HH100?*? (取方向为Z轴方向) RYe?(t)rcos? RYd?1000?2110 ?e?(t)??0R21rR10dr?1332?0??0Y10Y00cos?sin?d? d? (cos?Y00?*13Y10) ?e?(t)f
??02??0Y10*Y10sin?d? d?
?13e?(t)f f???0R21(r)R10(r)rdr?*3256816?32a0ra0
?(12a0)3/223a0?(1a0)3/2??0re4dr?1146a0?4！?2355a0?5256816a0
34
?,100? H210??*210???d??1e?(t)f ?e?(t)256a?1282e?(t)a H0010024338163??11 H2?e?(t),100?3?0?211rcos??*100d?
?e?(t)?R21rR10dr?0?32?0??0Y11cos?Y00sin? d? d?Y11**?e?(t)?R21rR10dr?02?0??130Y10sin? d? d? = 0
*???d???1,100??21H21100??1H?e?(t)?R21rR10dr?0?3?0?2?0Y1?1cos?Y00sin? d? d?*?e?(t)?R21rR10dr?0?3?0?2?0Y1?1*13知
Y10sin? d? d? = 0
由上述结果可，
W100?211?0，
W100?21?1?0
2∴W1s?2p?W100?210?W100?211?W100?21?1?W100?210?1?2?t0?,100eH210ti?21t?dt?
2 ?2?2(128243)(ea0?0)22?t2e?2?2i?21t???11
0ei?21t?e?t?/?dt?(128243)ea0?02222?21?222?2 当t??时，
?1s?2p?2?2(128243)ea0?02221?21?1其中
?2?21?1?(E2?E1)?? es2?34(1?14)?3? es8?34?3 es28?a0
5.6计算氢原子由第一激发态到基态的自发发射几率。 解： Amk4es?mk??rmk33?c232 由选择定则????1，知2s?1s是禁戒的 故只需计算
2p?1s的几率 ?21??22E2?E1?2?2?es2?43(1?14)?3?es8?34
而 r21(1)
?x21计
?y21算
?z21z
2p有三个状态，即 ?矩
阵
元
210, ?
211, ?
21?1
先的
z?rc?o s 35