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?1)??2C(p,t)?Nne?12?p22Hn(?p)e?i?Ent 式中Nn为归一化因子，即Nn?(??1/22n!n)1/2
4.4.求线性谐振子哈密顿量在动量表象中的矩阵元。
?? 解：H12???p212??x??i?22?2?222??xpx?12?2222??x
Hpp???(x)dx????(x)Hp*p12???e?(??22??x1??12ii??x)e?22p?xdx???22??2(ip?)2?2??1212121?i??e?(p??p)xdx?122??2?2??i???xe?2(p??p)xdx
?p?2?p?2?(p??p)???212???i????()e2i?p?22???2(p??p)xdx?2?p2?(p??p)???()22??p?22?1i?????e?(p??p)xdx?2??(p??p)????22??p??(p??p)
?0????12??01010??1? 0???和L?的矩阵分别为 L?2和L?的共同表象中，算符L4.5 设已知在LxyZxLy??02???i2??0?i0i0???i? 0?? 求它们的本征值和归一化的本征函数。最后将矩阵Lx和Ly对角化。 解
：
Lx的久期方程为
???20?2???20?2???0?32?的本征值为??????0??1?0，?2??，?3???∴Lx0，?，??
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?的本征方程 Lx?0???12??01010??a1??1??a2?0???a3??a1??a1???????的本征函数??a??a其中??2??2?设为Lx??????a3???a3??2和L?共同表象中的矩阵 LZ 当
?1?0时，有
?0???12??01010??a1??1??a2?0???a3??0???????0? ??0?????a2??0???????a1?a3???0? ?a3??a1，a2?0 ∴ ?2???0??a2???0?a1?????0? ??a?1??12 由归一化条件
1??0??0?a1???**?(a1,0,?a1)?0??2a1??a?1??2 取
a1?
?0?1??2??0?1???2?????的本征值0 。 ?对应于Lx?????0?? 当?2??时，有 ?12??01010??a1??1??a2?0???a3??a1????????a2? ??a???3??1?a2???2??a2?2a1?a1??1?????(a1?a3)???a2???a2?2a3 ∴ ? ??2????a?aa3?31???1??a2???2???a1?????2a1? ???a1??? 由归一化条件 1?(a1,*?a1???**2a1,a1)?2a1??4a1???a??1?2 取 a1?12 ∴归一化的
27
??????????????2?1???对应于Lx的本征值? 2?1??2?1 当
?2???时，有
?0???12??01010??a1??1??a2?0???a3??a1?????????a2? ???a??3??1???a1?2??a2??2a1??a1??1???? ∴ ???(a1?a3)????a2???a2??2a3?2????a?a1?1???a3???3??a2??2?????a1??????2a1? ???a??1??a1???*** 由归一化条件 1?(a1,?2a1,a1)??2a1??4a1???a??1?2 取 a1?12 ∴归一化的
????1??2?1???2??1??2??????对应于Lx的本征值?? ????12012121212??2?1???
2?1??2?1????表象的变换矩阵为 S???2和L?的共同表象变到L 由以上结果可知，从LZ?x??????SLxS ∴对角化的矩阵为L?x?????L??x?2????12121201212?1??2??01????12??1??0?2???0????1???0??????12012121212??2?1???2?1??2?1?101? 28
???????2?????01212011????0??1????2??1??????2??12012121212??2?1???2?1??2?1?0????02??0020??0??0???0??2???000?00??0? ?????的本征值为0，?，?的归一化的本征函数为 ??? L 按照与上同样的方法可得Lyy0?1??2??0??1??2???? ???????1???2???i???? ??2??1?????2??2和L?LZ12i21212?????1???2???i?????
2???1?????2?共
同
表
象
变
到
从
?1??2?S??0??1??2的
?Ly表象的变换矩阵为
i212???1????2??1???S??2????1????20?i2i21??2?1? ??2?1???2???对角化 ?利用S可使LyL?y?SLyS??0?0??00?00??0? ????4.6. 求连续性方程的矩阵表示
??i?????J ∴ J? 解：连续性方程为 (???*??*??) ?t2???而
?i???J???(???*??*??)2?*?i?2?(???*??*??)22 ∴
????T??*) i??(??)?(?*T????T??*) i??(?*T?t?t??i? 写成矩阵形式为
??t??t?????(??)??T???T?
?????*(??)??T??(?T?)?T?T*i??0 29
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。
???U(r)?U(r) 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 H0其中U0(r)是不考虑这种效应的势能分布，即
U（r）??ze24??0r区
U(r)为考虑这种效应后的势能分布，在r?r0区域，U(r)??Ze24??0r?
在
r?r0域，U(r)可由下式得出，
U(r)??e?Er rd1Ze4Ze?3???r?r, ( r?r0)3?4??r24?r334??r?00003 E??Ze? ( r ? r0)2?4??0r? U(r)??e?r0rEdr?e?Edrr0???Ze23004??r?r0rrdr?Ze4??20??1r2r0dr
??Ze238??0r0(r?r)?202Ze24??0r0??Ze238??0r0(3r0?r) (r?r0)
2222?ZeZe22(3r0?r)? (r?r0)??3?H??U(r)?U0(r)??8??0r04??0r 由于r0很小，所以
? 0 (r?r)0?2????H?H(0)???2??Za0??U0(r)，可视为一种微扰，由它引起的一级修正为（基态
2?(0)1?(Z33?a0)1/2re）
E(1)1???1?(0)*???(0)d?H1?2Za0r?Z330?a?r00[?Ze28??r300(3r?r)?202Ze2?2Za0r4??0r]e4?rdr2
∴r??a0，故e ∴
?1。
E(1)1??Ze42332??0a0r042320?r00(3rr?r)dr?2024Ze423??0a0?r00r
d??Ze42332??0a0r0(r?50r055)?Ze4232??0a0r ?20Ze10??0a0r?2Zes5a0342r0
2 30