对各区域的具体形式为Ⅰ：??22??1???U(x)?1?E?1 (x ? 0 ) Ⅱ ：
2??22????U0?2?E?2 (0?2 ? x ? a ) Ⅲ：?2?2????U1?3?E?3 (a?x?b) ?3 Ⅳ：??2????0?E?4 (b?x) 对于区域Ⅰ，U(x)??，粒子不可能到达此区域，?4故 ?1(x)?0
??? 而 . ?22? (U0?E)?2?2????0 ① ?32? (U1?E)?2????3?0 ② ?42?E?2?4?0
③ 对于束缚态来说，有?U?E?0
???k1? ∴ ?2???k4?⑤ ?4222?0 k1?222? (U0?E)?2???k3? ④ ?323?0 k3?22? (U1?E)?2
4?0 k4??2?E/? ⑥
2?2?Ae 各方程的解分别为 ?3k1x?Be?k1x?Csink2x?Dcosk2x 由波函数的有限性，得
?k3x?4?Ee?Fe
?k3x?4(?)有限， ? E?0 ∴ ?4?Fe?k3x 由波函数及其一阶导数的连续，得 ?1(0)??2(0) ?B??A∴ ? ?2(a)??3(a)?A(ek3x2?A(ek3x?e
?k3x)
⑦
?e?k3a?k3x)?Csink2a?Dcosk2acosk2a?Dk2sink2a ⑧
?k3b?(a)??3?(a)?Ak1(e?3k3a?e)?Ck2 ?3(b)??4(b)?Csink2b?Dcosk2b?Fe
3 ⑨
?(b)??4?(b)?Ck2sink2b?Dk2cosk2b??Fk3e?kb ⑩ ?3由⑦、⑧，得
k1ek2ek1ak1a?e?e?k1a?k1a?Ccoks2a?Dcoks2aCsikn2a?Dcoks2a (11)由 ⑨、⑩得
(k2coks2b)C?(k2sikn2b)D?(?k3sikn2b)C?(k3coks2b)D
(k2k3cosk2b?sink2b)C?(?k2k3cosk2b?sink2b)D?0 (12)
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令??eek1ak1a?e?e?k1a?k1a?k1k2，则①式变为
?2?Aek1x?Be?k1x(?sink2a?cosk2a)C?(?cosk2a?sink2a)D?0 ?3?Csink2x?Dcosk2x
? 联
立
(12)
、
(13)
得
，
要
此
方
程
组
4?Ee非
?k3x?Fe?k3x有零解，必须
(k2k3(?sink2a?cosk2a)cosk2b?sink2b)(?sink2b?cosk2b)?0 k3(?cosk2a?sink2a)k2k3cosk2b?sink2b)?(?sink2a?cosk2a)?k2即 (?cosk2a?sink2a)( ?(? ?k2k3k2k3sink2b?cosk2b)?0k2k3k2k3k2kk3
cosk2bcosk2a?sink2bsink2a??sink2bcosk2a?sink2bsink2a?k2k3sink2bcosk2a)?
?sink2bsink2a?? ??cosk2bsink2a?cosk2bcosk2a?0 sink2(b?a)(?? tgk (b?a)?(1?)?cosk2(b?a)((?k2k3?1)?0k232?)(k2k3??)
把?代入即得 tgk2(b?a)?(1?k2ek3ek1ak1a?e?e?k1a?k1a)(k2k3?k1ek2ek1ak1a?e?e?k1a?k1a) 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。
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(e(ek1a?e?k1a)?sink2a?k2cosk2asink2bk2cosk2b?k2cosk2a?cosk2ak2sink2acosk2b?k2sink2bk2sink2acosk2b?k2sink2b00?ek3e0?ek3e?k3a?k3a?k3a?k3ak1a?e00?k1a)k2?00?(ek1a?e?k1a)sink2bk2cosk2b??sink2a ? k 1(ek1ak1a?cosk2acosk2b?k2sink2b20?ek3e?k3a?k3a?k3a?e?k1a)?sink2bk2cosk2b?k3a (e ??e?k1a（)?k2k3e?k3acosk2acosk2b?k2e2?k3asink2a
c o s k2b?k2k3e ? k 1(ek1bsink2asink2b?k2e?k3bcosk2asink2b）?k3b?e?k1b（)k2k3esink2acosk2b?k2e?k3bcosk2a c o sk2b?k3e?k3bcosk2asink2b?k2e2sink2asink2b））?(ek1a?e?k1ak1a)[?k2k3cosk2(b?a)?k2sink2(b?a)]e?k1a?k3b?k3b?k3b?k3b ?(e ?ek1a?e)[k1k3sink2(b?a)?k1k2cosk2(b?a)]e2[?(k1?k3)k2cosk2(b?a)?(k2?k1k3)sink2(b?a)]e?k1a2
e?0[(k1?k3)k2cosk2(b?a)?(k2?k1k3)sink2(b?a)]e? [?(k1?k3)k2?(k2?k1k3)tgk2(b?a)]e22?k3b?k3b
?[(k1?k3)k2?(k2?k1k3)tgk2(b?a)]e [(k22?02k1a?k1k3)e2k1a?(k22 此即为所求方
?k1k3)]tgk2(b?a)?(k1?k3)k2e ? (k 1?k3)k2?0程。
3.1 一维谐振子处在基态?(x)???e??x222i??t2，求：(1)势能的平均值U?12??x；(2)动
22能的平均值T?解
p22?； (3)动量的几率分布函数。
：
(1) U?12??2x2?12??2??????xe2??x22dx?12??2???212?22???12??212?2?1413
??2?????14??
??0x2ne?ax2dx?1?3?5???(2n?1)2n?1?aan (2)
T?p22???12?2??????(x)dx?(x)p*2??1?2??22????e122??x2(??2ddx22)e122??x2dx??2?2????[?e??2???x22dx??2????xe2??x22dx]???2??14???[????2??2?3]
??2??2?2???24?14?2?14?24???
???? 或
T?E?U? (3)
12??????c(p)???(x)?(x)dx*p?12????????2 e122??x2e?i?Pxdx ?
12???????? e??x2122e?i?Pxdx?12???????? e12ip2p??(x?)?2222??2??dx
?12??1????p22e222??2???? e12ip2??(x?)22??dx?12???p222???p22e2??22??p????e2??2 动量几率分布函数为 ?(p)?c(p)2?1???e??
3.2.氢原子处在基态?(r,?,?)?1?a 解(1)
30e?r/a0，求： (1)r的平均值；(2)势能?e2r的平均值；(3)最
可几半径； (4)动能的平均值；(5)动量的几率分布函数。
：
r??r?(r,?,?)d??dx?n!an?121?a30???00?2??0re?2r/a0rsin? drd? d?2?4a30??0ra3?2r/a0dr
?
?0xen?ax
?4a303!?2????a??0?4?32a0
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(2)U?(?e2re2)???e23?a02??0???00?2??1r0e?2r/a0rsin? drd? d?2??
?a4ea04ea330???00e?2r/a0rsin? drd? d?
2????0e1?2r/a0r dre22??30?2????a??0?出
2??a0 (3)
电
子
现
22在r+dr球壳内
2出现的几率
?2r/a02为
?(r)dr???0?2?0[?(r,?,?)]rsin? drd? d??4a03e?2r/a0rdr ?(r)?4a03er
d?(r)dr?4a03(2?2a0r)re?2r/a0 令
d?(r)dr?0， ? r1?0, r2??, r3?a0 当
r1?0, r2??时，?(r)?0为几率最小位置
d?(r)dr22 ?4a30(2?8a0r?4a20r)e2?2r/a0
d?(r)dr2r?a02??8a30e?2?0 ∴
r?a0是最可几半径。
?? (4)T
12??p2???22?1??1??1??22???2?(r)?(sin?)?22?? r??r?rsin?????sin????2T???22?2???002???2??10?a130e?r/a0?(e2?r/a0)rsin? drd? d?2???2????00?0?a01a03e?r/a01dr2dr2[r2ddr(e?r/a0)]rsin? drd? d?
2 ??4?2302?a(???0(2r?ra0)e?r/a0 dr?4?2402?a(2a042?a042)??2202?a
(5) c(p)?*??p?(r)?(r,?,?)d?
? c(p)?1(2??)3/2??10?a30e?r/a0rdr?e02??i?prcos?sin? d??2?0d?
15