喀兴林高等量子力学习题EX32

练习32.1为什么vsrbbb的完全性关系(30.11)式(将其中积分理解为取和)与lnnn21的完全性关系(32.8)式都等于1？根据(32.5)式，这两个基矢不是相差一个常数因子吗？（梁立欢）解比照vsrbbb的完全性关系，可展开为对称化基矢的叠加：2211212211212'''1''''''''cnnncnnncnnnccnnnccbbbcbbbllllvsrvsr可见lnnn21的完全性关系与vsrbbb一致，不因相差一个常数而改变，改变的只是展开式每项前的系数。32.2练习32.3从la和la作用于基矢.......21lnnn的公式（32.13）和（32.14）二式出发，重新证明它们的对易关系（32.15）式。（完成人：张伟）证明：由公式（32.13）和（32.14）121...21212121...1...1.........1.........lnnnlllllllllllnnnnnnnannnnnnna其中12111211211121.........212121121121......)1(...212121...1...1...11...1......1.........ln......)1(...ln...1...1...11......1...1.........llllnnnnnnnllllllllllllllllllllllnnnnnnnlllllllllllllllllllnnnnnnnnnnnannnnaannnnnnnnnnnnnnnnnnannnnaann其中即其中样的种情况得到的结论是一的前面，不难证明另一排在假设0............1...1...11.........ln.........ln212121121121lllllllllllllllllllllllllllllaaaaaannnnaannnnnnnnnnaannnnnnn的对易关系和由此得到从而有可见即类似的方法和证明过程可以证明0llllllaaaaaa的对易关系为：和lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllnnnnnnnllllllllllllllllllllllnnnnnnnlllllllllllllllllllaaaaaaaannnnnnnnnnnannnaannnnnnnnnannnaannaaaannnnnnaannnnnnnnnnaannnnnnnnnnnnnnnnnnannnnaannnnnnnnnnnnnnnnnnannnnaannllll综合之后有，时：当得：比较上两式的结果，可时当时有：从而，当从而有可见即其中即其中时有当类似的可以证明：1......1...1...............1...1...1......0............1...1...1.........ln.........ln...1...1...1...1...............ln......)1(...ln...1...1...1......1...1.........212121212121212121121121.........212121121121......)1(...21212112111211211121至此la和la的对易关系（32.15）式已经全部证毕。#练习32.4证明不论玻色子或费米子，有（吴汉成）sjrjsrrsjjsraaaaaaaa],[],[证明：la和ma的对易关系的离散形式如下：0lmmlaaaa，0lmmlaaaa，lmlmmlaaaa----（1）（一）对于玻色子1，并代入（1）式中，则显然有：0lmmlaaaa，0lmmlaaaa，lmlmmlaaaasjrjsrjsraaaaaaaaa],[],[],[srjjrsjjsraaaaaaaaaa)()(rsjasjrjsrjsraaaaaaaaa],[],[],[srjjrsjjsraaaaaaaaaa)()(sjrrjaaaaa)(sjra（二）对于费米子1，并入（1）式中，显然有：0lmmlaaaa，0lmmlaaaa，lmlmmlaaaasjrjsrjsraaaaaaaaa],[],[],[srjjrsjjsraaaaaaaaaa)()(（sjsjjsjrrjaaaaaaaa,）sjrjrsjsjsjraaaaaaaaaa)()(sjrsjrrsjaaaaaaa22rsjasjrjsrjsraaaaaaaaa],[],[],[srjjrsjjsraaaaaaaaaa)()(（rjjrjrjssjaaaaaaaa,）srjrjjrjsjsraaaaaaaaaa)()(srjsjrjsraaaaaaa22(sjjsaaaa)srjsjrsjraaaaaaa22srjjrsjraaaaaa)(2sjra综合上述，显然得证。#32.5证明：（何贤文）srjrsjjsrsrirsjjsraaaaaaaaaa],[],[证：由于00rssrrssraaaaaaaa得到rssrrssraaaaaaaasrjrsjrsjrjsrjsjrsjrsjsrsrjrsjrjsjrsjrsjsraaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa],[],[],[],[],[],[],[],[],[],[32.6证明：（肖钰裴）utsbbbutjsutsjjtsrbbbtjsrjtsrjsjbsjjbrrjaaabbgbbbbgbbaGaaabbgbbbbgbbaGabgbaGbgbaaGutstsrsr)()(21,)()(21,,,)2()2()2()2()2()2()1()1()1()1(证明：lllbblljjllbbllljljllbblllljjllbblllljbblljllbblllllbbljjlllbbljlllbbljlllbblabgbaaaaaabgbaaaaaabgbaaaaaabgbaaabgbaaabgbabgbaaaabgbaaabgbaaGabgbaGllllllllllllllllllll)1(''')1(''')1(''')1('')1('')1(')1('')1('')1('')1()1('')1('''''''''')()()(,,sjbsljllljbblljlllljbblljlllbbllllbbljjlllbbljabgbaaaaaabgbaaaaaabgbaabgbaabgbaaaabgbaaGsllllllllll)1('')1(''')1(')1('')1('')1('')1()()(,,'''''lmmlbbbbmlmlaabbgbbaaGlmlm)(!21'')2('''')2()()(!21))((!21))((!21))((!21)(!21)(!21,)(!21,'')2('''''')2('''''')2('''''')2('')2('''')2('''')2('''')2(''''''''''''''lmjmljmlmlbbbbmllmjmljlmmlmlbbbbmllmmjljlmmlmlbbbbmllmmljjlmmlmlbbbbmllmmlbbbbmlmljjlmmlbbbbmlmljlmmlbbbbmlmljaaaabbgbbaaaaaaaaaabbgbbaaaaaaaaaabbgbbaaaaaaaaaabbgbbaabbgbbaaaaaabbgbbaaaaabbgbbaaaGlmlmlmlmlmlmlmlmlmlmlmlmlmlmtsrbbbtjsrjtsraaabbgbbbbgbbtsr)()(21)2()2(utsbbbutjsutsjjaaabbgbbbbgbbaGuts)()(21,)2()2()2(#32.7练习32.8写出全同粒子系统的总轨道角动量zL、L和L的二次量子化形式。（谷巍）解：在离散本征值谱的情况下，全同粒子系统的一般算符G的二次量子化形式为lmmlmlmbbbblllbbaabbgbbaaabgbaGGGlmlmll)2(††l)1(†l)2()1(21对于全同粒子系统的总轨道角动量zL、L和L，可知它们是单体算符之和的形式，所以只需要保留上式右边的第一项。则有：lllbbllllbblllzlbblzablbaLablbaLablbaLllllll†††又因为zL、L和L有共同的本征矢量lm,且存在下面的关系：1,1mlmlmllmLlmmlmLz我们可取††,,,,lmzabalmbLLLB则全同粒子系统的总轨道角动量zL、L和L的二次量子化形式为：lmlmlmlmmlmllmlmzmllmzaamalmmlamalmlmlaL†††mllmmllmlmmllmmllmmllmmlaamlmlamlmlamlmlalmlmlaL†1,††11,1lmmllmlmmllmmllmmllmmlaamlmlamlmlamlmlalmlmlaL†1,††11,1#32.9利用zzLLLLL22，以及上题结果写出2L的二次量子化形式，并与（32.23）式比