2? ????0?E?4 Ⅳ：?4 (b?x) 2? 对于区域Ⅰ，U(x)??，粒子不可能到达此区域，故 ?1(x)?0
??2? (U0?E)??2?0 ① 而 . ?2?2
2? (U1?E) ????3?0 ② ?3 ?2 2?E???2?4?0 ③ ?4 ??U?E?0 对于束缚态来说，有
2? (U0?E) ?? ∴ ?2 ④ ?k12?2?0 k12?2?
2? (U1?E) 2?? ?3 ⑤ ?k3?3?0 k32?2 ?22?? ?k4 ?4 ⑥ ?4?0 k4??2?E/?2 各方程的解分别为 ?2? Aek1x?Be?k1x sinkx?Dcoskx ?3?C22 ?k3x?4?Ee?Fe?k3x 由波函数的有限性，得 ?4(?)有限， ?E?0 ??Fe?k3x ∴4 由波函数及其一阶导数的连续，得 ?(0) ?B??A ?1(0)?2 ?A(ek3x?e?k3x) ∴ ?2 kx?kx ?2(a)??3(a)?A(e3?e3)?Csink2a?Dcosk2a ⑦ k3a?k3a?(a)?? ?3?(a)?Ak(e?e)?Ck2cosk2a?Dk2sink2a ⑧ 31
?kb ?3(b)? ?4(b)?Csink2b?Dcosk2b?Fe3 ⑨ ?(b)? ?4?(b)?Ck2sink2b?Dk2cosk2b??Fk3e?k3b ⑩ ?3 k1ek1a?e?k1aCcosk2a?Dcosk2a由⑦、⑧，得 (11) ? k1a?k1ak2e?eCsink2a?Dcosk2a
由 ⑨、⑩得(k2cosk2b)C?(k2sink2b)D?(?k3sink2b)C?(k3cosk2b)D k k (2cosk2b?sink2b)C?(?2cosk2b?sink2b)D?0 (12) k3 k3 1a?e?k1ak1ek 令??ka，则①式变为 (?sink2a?cosk2a)C?(?cosk2a?sink2a)D?0 ?e 1?e?k1ak2
联立(12)、(13)得，要此方程组有非零解，必须 k2k( cosk2b?sink2b)(?2sink2b?cosk2b) k3?0 k3
(?sink2a?cosk2a)(?cosk2a?sink2a)
(?coska?sinka)(k2coskb?sinkb)?(?sinka?coska)?即222222k3
k
?(?2sink2b?cosk2b)?0 k3 k2k2 ?coskbcoska?sink2bsink2a??sink2bcosk2a?22
k3k3
kk ?sink2bsink2a??2sink2bsink2a?2sink2bcosk2a)? k3k3 ??cosk2bsink2a?cosk2bcosk2a?0 kk sink2(b?a)(??2)?cosk2(b?a)((?2?1)?0 k3k3 k2k2 tgk(b?a)?(1??)(??) 2k3k3
把?代入即得 k2ek1a?e?k1ak2k1ek1a?e?k1a tgk2(b?a)?(1?)(?) k1a?k1ak1a?k1a k3e?ek3k2e?e 此即为所要求的束缚态能级所满足的方程。 # 附：从方程⑩之后也可以直接用行列式求解。见附页。
(ek1a?e?k1a)?sink2a?cosk2a0 (ek1a?e?k1a)k?k2cosk2ak2sink2a02?0?k3a 0sink2bcosk2b?e 0k2cosk2b?k2sink2bk3e?k3a ?k2cosk2ak2sink2a0
ka?kacosk2b?e?k3a? 0?(e1?e1)sink2b k2cosk2b?k2sink2bk3e?k3a
?sink2a?cosk2a0
?k1(ek1a?e?k1a)?sink2bcosk2b?e?k3a
k2cosk2b?k2sink2bk3e?k3a
ka?ka2?k3a)?k2k3e?k3acosk2acosk2b?k2esink2a ?(e1?e1（ 2?k3a cosk2b?k2k3e?k3asink2asink2b?k2ecosk2asink2b）
?k3b?k3bk1b?k1b ?k(e?e（)kkesinkacoskb?kecosk2a123222
cosk2b?k3e?k3bcosk2asink2b?k2e?k3bsink2asink2b））
2?(ek1a?e?k1a)[?k2k3cosk2(b?a)?k2sink2(b?a)]e?k3b
2 ?e [?(k1?k3)k2cosk2(b?a)?(k2?k1k3)sink2(b?a)]e?k3b 2 e?k1a[(k1?k3)k2cosk2(b?a)?(k2 ?k1k3)sink2(b?a)]e?k3b?0
2? [?(k1?k3)k2?(k2?k1k3)tgk2(b?a)]e?k3b 2 ?[(k1?k3)k2?(k2?k1k3)tgk2(b?a)]e?k3b?0 2k1a2k1a22 ?(k2?k1k3)]tgk2(b?a)?(k1?k3)k2e [(k2?k1k3)e ?(k1?k3)k2?0 此即为所求方程。 第三章 力学量的算符表示
?2x2i???t? 22e3.1 一维谐振子处在基态?(x)?，求： ? 122U???x； (1)势能的平均值 2
p2 (2)动能的平均值； T?
k1a ?(ek1a?e?k1a)[k1k3sink2(b?a)?k1k2cosk2(b?a)]e?k3b
(3)动量的几率分布函数。
?11222?2??2x2解：(1) xedx U???x?????22?
1?1?111?2 ???2?222???2????2224??2??2??
? 11?3?5???(2n?1)?2n?ax2 ??? xe dx?0 4a2n?1an
p21?*?2?(x)dx (2) T???(x)p
2?2??? 112??2x2???2x2?1d2 ? e2(??)e2dx 2??dx ?2?22??22?
??(1??2x2)e??xdx ?? ?2? ?22??22???2x2 ??[edx??2x2e??xdx] ?????2?
??22?? ??[??2?3]
2????????
222???????22 ??????2?4?4???2?
1 ??? 4
111 或 T?E?U????????? 244
* (3) c(p)??p(x)?(x)dx ?2??2?? ??2??1??????12 e?2x2ei?Px?dx 2??12ip2p2 ??(x?)??1?2?2?2?2?2? edx ??2??? p21ip???2(x?2)2?1?22?? ?e2?? e2dx ??2???
p2p2 ??1?2?2?2212?2?2?e??e ?2?????? 动量几率分布函数为 p2?1222 (p)?c(p)? ?e?? ???
#
1 3.2. 氢原子处在基态?(r,?,?)?e?r/a0，求： 3?a0
e2 (1)r的平均值； (2)势能?的平均值； (3)最可几半径； (4)动能的平均值； r (5) 动量的几率分布函数。 ??? 1???? e1??2x22ei?Px?dx ??1?2???2r/a022
解：(1)r?r?(r,?,?)d??rersin? drd? d? 3000?a0
?4?3?2r/a0n!
?3radr xne?axdx?n?1 0 aa00 43!3??a0 34 2a0?2?? ?a??0??
e2e2?2??1?2r/a02(2)U?(?)??3ersin? drd? d? r?a0000r
e2?2???2r/a0??3ersin? drd? d? 000?a0
2? 4e ??3e?2r/a0r dra00
4e21e2??3??2 a0?2?a0? ?a??0??
(3)电子出现在r+dr球壳内出现的几率为 ?2?4?2r/a0222 ?erdr ?(r)dr?[?(r,?,?)]rsin? drd? d? 300a0
4?2r/a02r ?(r)?3e
???????????????a0d?(r)42?3(2?r)re?2r/a0 dra0a0d?(r)?0， ? r1?0, r2??, r3?a0 令 dr 当 r1?0, r2??时，?(r)?0为几率最小位置 d2?(r)4842?2r/a0 ?(2?r?r)e232 a0dra0a0
d2?(r)8?2??e?0 3dr2r?a0a0
∴ r ?a0是最可几半径。 12?22 ?1???1??1??????2?2?(r2)?(sin?)? (4)T?p? ??rsin?????r??rsin2???2? 2?2? ?2?2??1?r/a02?r/a02 T??e?(e)rsin? drd? d? 3 2?000?a0
?2?2??1?r/a01d2d?r/a02?? e[r(e)]rsin? drd? d? 32?000?a0drr2dr ?241?r2?r/a0 ??(?(2r?)e dr 3 0a02?a0a0 22a0a04?2?2 ? (2?)?42???????00
*??(r)?(r,?,?)d? (5) c(p )??pi ???prcos?2?11?r/a02?erdresin? d?d? c(p )?003(2??)3/20?a0
i???prcos?2?2?r/a0 redre? d(?cos?) ?03/230 (2??)?a0 ?i?prcos??2??? ? r2e?r/a0dre3/230ipr?a0 (2??)0ii ?pr?2????r/a0?prn!re(e?e?)dr xne?axdx?n?1 ?03ip0a(2??)3/2?a0
2??11
?[?] i21i2 (2??)3/2?a3ip10(?p)(?p) a0?a0?44 a0?414ip? ? 2222233332a0??a0(a0p??)2a0?ip?a?(1?p)2 02a0?2
3/2(2a?)?0 ?2222?(a0p??)
35 8a0?2动量几率分布函数 ?(p)?c(p)?2 224 ?(a0p??)#
3.3 证明氢原子中电子运动所产生的电流密度在球极坐标中的分量是 Jer?Je??0 2?a442?a????????? Je??e? m2?n?m ? rsin?