作业六参考答案
1、[证明] 由题意得，谐振子波函数为?n(x)????e?2nn!22x/2Hn(?x)
?n?1(x)?x?n(x)? ? ? ? ?令???x??e?2n?1(n?1)!?22x/2Hn?1(?x)
1???xe?2nn!122x/2Hn(?x)?x/2????e?2nn!22x/2??xHn(?x)??1???e?2nn!???e?2nn!22??Hn(?)?
22x/2?1?1?H(?)?nH(?)n?1n?1???2?nH(?)??n?1n?12?2(n?1)!?e??2x2/2?(n?1)??2????Hn?1(?)?n?1?2(n?1)!????1?nn?1?(x)??(x)??n?1??2n?12?由x?n(x)??1?nn?1?(x)??(x)??得 n?1n?1??22??1?n?1nx?n?1(x)???(x)??n(x)? ??2n?22? ?1?n?1n?2x?n?1(x)???n(x)??n?2(x)???22?x2?n(x)??1?nn?1x?(x)?x?(x)??n?1??2n?12??1n?1?n?1??1?nn?2?1n?n?1? ???n?2(x)??n(x)???n(x)??n?2(x)??由厄????22?22??2?2?????1? ?n(n?1)?n?2(x)?(2n?1)?n(x)?(n?1)(n?2)?n?2(x)?2??2??米多项式的正交归一性：
?*??n(x)?m(x)dx??mn
???Hm???Hn(?)e??d??2nn!??mn，得到谐振子波函数的正交归一性：
2?? 9
???1?nn?1**x???(x)x?n(x)dx???(x)?(x)dx??(x)?(x)dx??0 nn?1nn?1????2??2?????*n?11*V???2x2???2??n(x)x2?n(x)dx22????????2?*** ?n(n?1)?(x)?(x)dx?(2n?1)?(x)?(x)dx?(n?1)(n?2)?(x)?(x)dx??nn?2nnnn?2???4?2???????????2?1 ?(2n?1)?(n?)?En24?2222、[证明] 由题意知，谐振子波函数为?n(x)????e?2nn!22x/2Hn(?x)；
?????2nHn?1???。我们计算得 而厄米多项式满足求导公式：Hnd?n(x)?dx ? ????e?2nn!???e?2nn!???e?2nn!22x/2d?2???xH?x?H(?x)??nn??dx???2?d??xH?x??H(?x)???n?nd?x????2???xHn??x??2n?Hn?1(?x)???22x/222x/2 ???2x?n(x)?2n?1?n?1(x)2n
?n?n?1 ?????n?1(x)??n?1(x)???2n?n?1(x)22???n?n?1 ????n?1(x)??n?1(x)?2?2?由上面等式得：
??1(x)???(n?1)2?n?2(x)?n2?n(x)??n????1(x)???(n?1)2?n(x)?(n?2)2?n?2(x)??n??
9
?nd?d2?nn?1d???(x)??(x)??n?1n?1dx22dx2dx??? ???n2??(n?1)2?n?2(x)?n2?n(x)?? ??(n?1)2??(n?1)2?n(x)?(n?2)2?n?2(x)?2??
?2?? ??n(n?1)?n?2(x)??2n?1??n(x)?(n?1)(n?2)?n?2(x)?由平均值公式A??*n*???A?d?得：
??*?n?(x)*nd????n(x)dx???(x)??ip???(x)p??n(x)dx??idx????????d?n(x)dxdx??n?*?n?1* ??i???n(x)?n?1(x)dx??n(x)?n?1(x)dx??0??22??????
22??d??(x)dx??(x)?T???(x)T?(x)dx?n2?n?2?dx??????*n*n??d2???(x)2?n(x)dx?2???dx*n**?n(n?1)??n(x)?n?2(x)dx??2n?1???n(x)?n(x)dx
2?2???????*?(n?1)(n?2)??n(x)?n?2(x)dx????22?11?En ?n??2n?1???????4?22??222? ????2????????2x223、[解] 由题意知，波函数为?n(x)??nHn??x?，其中?????e?2n!??在上题中我们求得：
12。
x?n(x)???n?1?nn?1n?1??(x)??(x)?(x)???(x)??(x) 及 ???? n?1n?1nn?1n?1??222??2??x?p?0 x2?n(x)?1?n(n?1)?n?2(x)?(2n?1)?n(x)?(n?1)(n?2)?n?2(x)?2?? 2??2????(x)??n?n(n?1)?n?2(x)?(2n?1)?n(x)?(n?1)(n?2)?n?2(x)?
2 9
?x?212?2?2n?1?、p222??22(2n?1)
???x??????p????x?x??x2?x2??1?2n?12(2n?1)2 ? ?x??p??n??p?p?2p2?p2??2??1?? 2?2d????V(x)，则能量本征方程为： 4、[解] 根据题意得：H2mdx2d2?(x)?1?22??m?x?q?x?(x)?E?(x) ??2mdx22??将势函数配方得：
2?d2?(x)1q??q2?2?2???m??x??(x)??E??(x) 22?2?2mdx2m??2m????q2?2q?令X?x?及E??E?得 222m?m?d2?(X)1??m?2X2?(x)?E??(x)， 22mdX2这是以为X变量的一维线性谐振子能量本征方程。其解为：
222?????2X22?n(X)??nHn(X) ??m??e?2n!??????n?12?En12 q2?2q?将X?x?及E??E?代入上式得所求的能量本征值及其对应的本征函数： 222m?m?????2?x?q??n(x)??n?e?2n!??q2?2En???n?12??2m?2?12m?2?22Hn(x?q?m?2)
??(x)?0 ?x?0??25、[解] 能量本征方程为: ? 122????(x)???x?(x)?E?(x) x?0???2?2? 9
??0, x?0根据题意知，其解为：?(x)??， ??2x22Hn(?x) x?0??Nne而对应于能级En???n???1?? n?0,1,2,2?。接下来由边界条件确定n的取值范围。
n由边界条件?(0)?0得Hn(0)?0，而Hn(?x)?(?1)Hn(x)要求n?2m?1 ?m?0,1,2,?。
因此所求能级为：E2m?1???2m???3?? m?0,1,2,2?
?0, x?0?对应的波函数为：?(x)?? ???2x22eH(?x) x?02m?1?22m?1(2m?1)!??6、[解] 由题意得：?(x,0)??0(x?x0)?????e?2?x?x0?22，E0?1? 2展开?(x,0)???C?nn?0n(x)，得
Cn???(x)?(x,0)dx???*n?2nn!?2????e??2x22Hn(?x)?(x,0)dx1 ?2nn!?1 ?n2n!?由母函数公式：e???????e???22Hn(?)e?????0?2d? ????x, ?0??x0?
Hn(?)e??n?0?2??2??0???02??d????s2?2s?Hn(?)ns得 n!??1???1?222????2???0??0???2???0??0???2???0??0???2H(?)sn??n?s?2s?2?2?2?n???Hn(?)ed?????s?ed???eed???n!n!n?0??????n?0???1?? ????e122?????2???0?2s????0?s?2??d??e??12??0??0s4??12???04?e???????s?0?2??2d??e12??0??0s4? ??e???1??n?0??0s?n!n12??0snn????0e4n?0n!212????2???0??0??0?snsnn2?????Hn(?)ed?????0e4 n?0n!??n?0n! 9
物理学相关 作业六参考答案
