A4224?G??cD???kL?5???1??3?196L??LL? ?1?G??cD??B???2k2L2?5???1??3?/?1??49L??LL??L? ?在L很大的情况下
2(?x)2(?p)2?5222?kL49
当L??时，过渡到自由粒子的情况
(?x)2(?p)2??
?i?1?x2??(x)????p0x?4?2??2??2??exp???? 3.12. 粒子处于状态
式中?为常量。求粒子的动量平均值，并计算侧不准关系(?p)21/2(?x)2??
[解] 令?(x)?A?(x) 其中A为归一化常数，由归一化条件
*????(x)?(x)dx?A?2*????(x)?(x)dx?
?x2??exp?2?dx2??2?????2??
?A2??A??*A22??
?2???12?A?2??
2?2???1/2?i?ix2??d?x2???(x)dx?p???(x)pexp??p0x?2???i??exp?p0x?2?dx2???2??4???dx?4????????iA2x??x2??(?i?)??p?2?2?exp??2?2?dx????02??2?????
A2?A22??2?*?p02???p0
?x2?x???(x)x?(x)dx?xexp??2?dx?0??2??2????2??
?A2?x2?x???(x)x?(x)dx?xexp??2?dx2?????2???2??
2?*2?2A2?A2??2??2????2xexp??????x2?2?2???2??x2???????????exp???2?2??dx??? ?A22?2??
??2
p2????2?(x)dx?A2?????*(x)p???(x)*p?2?(x)dx
?A2????p??(x)2dx
2?A?2??2???i?ddxexp??i2??px2?0x?4?2??dx 222??A??ix??ix22??2???????p0?2?2???exp???px??04?2??dx
??2A2?p20132??2????22???4?42??????
?????p2?2?0?4?2???
(?x)2?(x?x)2?x2??2
?p)2?(p?p)2?p2?p2?p2?2?2(20?4?2?p0?4?2
(?x)2(?p)2??24
3.12 利用测不准关系估计氢原子的基态能量
[解] 氢原子的基态波函数
r?(r,?,?)?1?a0?a3e0
设电子的质量为?，相对核的位置为r。平均能量
p2e2E?s2??r
由于基本波函数?(r)为中心对称的，因此
?p?0
1）（
(?p)2?(?p?p)2?p2
电子相对核的距离r在其数量级内，偏差?r不会大于r，即
?r?r?r?r
(?r)2?r2
最大偏差 (?r)2?r2?r2
?r)22测不准关系 (?p)??2(4??2
(?p)2??2?2取等号则有：
(?r)2?r2 E?1?2e2s2?r2?r
dEe2sdr??2??r3?r2?0 ??2r?e2?a0s
?E?2e22s?2es?2?2?e2?a2?a?2?2?2??4min?2Es002?a0a0?es2?a0 或
min??2?2第四章 态和力学量的表象
4.1 求动量表象中角动量Lx的矩阵元和L2x的矩阵元
1i??(r??r[解] 动量算符P??的本征函数
P)?P?(2??)3/2e?
方法（一） 在动量表象中
(L*?x)P?P?????P?(r)L?x??P??(r)d?
???*?P?(r)(y?P?z?P??z?y)?P??(r)d?
??i??*P?(r?)??????p?P?p????zyy???P??(r)d?z????P
??i???????p*???z?P?py???P?(r)?P??(r)d?y???Pz????
（2）
（3）
（4）
?????????????p??i??pz?py?p???????P?Pyz??
???????(p)??(p?p) p（方法二） 在动量表象中，动量的本征函数 p????????(Lx)p?p?????(p?p?)Lx?(p?p)dpxdpydpz
??????????????(p?p)??i??pz?py???(p?p)dpxdpydpz???pz??????py? ?????????????(p??i??pz?py?p)??????p?pyz??
???2?????????????(p???p?)Lx?(p?p)dpxdpydpz
(L2x)p?p??????????????????dp??dp?????(p?p)??i??p??p???(p???p?)dp?zyxyz??????pz??????py?
2???????2???????p??p?p?p?zy????p?pyz?? ?????????(p???2?p?p?p)zy??py??pz??
22???????2???(p????pz?py?p)??p?pyz??
4.2 求一维无限势阱中粒子的坐标和动量在能量表象中的矩阵元
[解] 设势阱的势为
2??U(x)???0n2?2?2En?22?a能量
0?x?ax?a,x?a n?1,2,?
能量的本征波函数
?n(x)?2n?sinxaa 0?x?a
?(x)?0 x?0,x?a
在能量表象中
2am?n?xmn???(x)x?n(x)dx??xsinxsinxdx0a0aa
a*m1a?m?nm?n???x?cos?x?cos?x?dxa0?aa?
当m?n xnn?1a?2n?x1?cos?a?0?a?x?dx?
1?112n?a??x2?xsinx?a?22n?a2n?21?12?a?2?n???a??x??cosa?a??2n???2?0
a2n??sinxdx?0a??0
aa?1?1?1?1???a??22??2n???
?a/2
当m?n
1?1m?nam?nxmn??xsin?x?xsin?xa?(m?n)?a(m?n)?a
?a(m?n)??a0sinm?na?xdx?a(m?n)???m?nsin?xdx?a?0
aa?am?nam?n???cos?x?cos?x?2222(m?n)?a(m?n)?a??0
?(?1)m?n?1??(m?a??(?1)n)?22m?n?1?(m?an)?22
?(?1)m?n?1a??(m4?mnan)?2222
d??*Pmn???m(x)???i???n(x)dx0dx??
??2i?am?dn?sinx?sinxdx?0aadxa
??i?2n?a2?a0sinm?n?xcosxdxaa
??i?当m?n
n?a?m?nm?n?sin?x?sin?x?dx2?0?a?aa?