第四章习题解答
4.1.求在动量表象中角动量Lx的矩阵元和L2x的矩阵元。
p?r13??p??r??z?zp?y)ed? 解：(Lx)p?p?()e(yp2???i??i??p?r13??p??r? ?()e(ypz?zpy)ed?
2???i??i??13??p??r???p?r ?()?e(?i?)(pz?py)ed?
2???py?pzi??i???r??13?（p?p?） ?(?i?)(pz?py)()ed?
?py?pz2??????? ?i?(py?pz)?(p?p?)
?pz?py*?2?(x)L??d? )?? (L2?xp?ppxp?p?r13??p??r2??? ?()e(ypz?zpy)ed?
2???i??i??p?r13??p??r?z?zp?y)(yp?z?zp?y)e?d? ?()?e(yp2??i??i??13??p?r???p?r?z?zp?y)(i?)(py ?()?e(yp?pz)ed?
2???pz?pyi??i??i???p?r??13??p??r?z?zp?y)e?d? ?(i?)(py?pz)()?e(yp?pz?py2???r??213?（p?p?）2 ???(py?pz)()ed?
?pz?py2?????2?? ???2(py?pz)?(p?p?)
?pz?py#
4.2 求能量表象中，一维无限深势阱的坐标与动量的矩阵元。
2n?解：基矢：un(x)?sinx
aa?2?2n2 能量：En? 22?aa2m?a1uxdx? ?ucosnudu?2cosnu?sinnu?c 对角元：xmm??xsin20aa2nn2am?n?x)?x?(sin)dx 当时，m?n xmn??(sin0aaai???i??i???1a?(m?n)?(m?n)?xcosx?cosa?0?aa??x?dx?a1?a2(m?n)?ax(m?n)???[cosx?sinx]22a?(m?n)?a(m?n)?a0?a?a2(m?n)?ax(m?n)? ?[cosx?sinx]? 22a(m?n)?a(m?n)?0????a11?2(?1)m?n?1???2?(m?n)2??(m?n)??4mnm?n(?1)?12222?(m?n)a2m?dn?*?un(x)dx??i??sinpmn??um(x)px?sinxdx0aadxa2n??am?n???i2?sinx?cosxdx0aaan??a?(m?n)?(m?n)????i2??sinx?sinx?dxaaa0???a??
n???a(m?n)?a(m?n)??i2?cosx?cosa(m?n)?aa?(m?n)??11?m?n??1]?(m?n)(m?n)?(?1)??i2mn??(?1)m?n?12(m?n2)a?in??aa2??x? ?0a
????
?sinmucosnudu??#
cos(m?n)ucos(m?n)u??C 2(m?n)2(m?n)4.3 求在动量表象中线性谐振子的能量本征函数。
解：定态薛定谔方程为
21p222d ????C(p,t)?C(p,t)?EC(p,t) 222?dp21p222d 即 ????C(p,t)?(E?)C(p,t)?0 222?dp2 两边乘以，得
??1 ?1d22Ep2C(p,t)?(?)C(p,t)?0 2?????dp???令?? ??1???p?? p, ??1???
2E ??d2 2C(p,t)?(???2)C(p,t)?0
d? 跟课本P.39(2.7-4)式比较可知，线性谐振子的能量本征值和本征函数为
En?(n?12)?? 1i??2p2?EntC(p,t)?Nne2Hn(?p)e?式中Nn为归一化因子，即
? Nn?(1/2n)1/2
?2n!#
4.4.求线性谐振子哈密顿量在动量表象中的矩阵元。
121?2?2122?????x?? 解：H?p???2x2 22?22??x2??(x)dx H???*(x)Hpp?pp?pxp?x1?2?2122?? ?e(????x)edx 2?2??2??x2(p??p)x(p??p)x???2i121212?? ??(p?)edx???xedx ??????2??2??22??iiii2(p??p)x??p?21212??? ??(p?p)???()edx 2???2?22??i?p?2p?212?2? ??(p??p)???()2?2i?p?2i??1?????ei(p??p)x?dx
2p2122? ??(p??p)?????(p??p) 2?2?p?22p2122? ??(p??p)?????(p??p) 22?2?p#
?和L?的矩阵分别为 ?2和L?的共同表象中，算符L4.5 设已知在LZxy?0?i0??010???2???? Lx??i0?i? ?101? Ly?22??0i?0????010? 求它们的本征值和归一化的本征函数。最后将矩阵Lx和Ly对角化。
解：Lx的久期方程为
??
?2???0?2???0???3??2??0
?202 ??1?0，?2??，?3???
?的本征值为0，?，?? ∴Lx?的本征方程 Lx?a1??010??a1????????101a?? ?a2? ???2?2??a?????010??a3??3??a1????的本征函数L?2和L?共同表象中的矩阵 其中???a2?设为LZx?a??3? 当?1?0时，有
?010??a1??0???????? ?101??a2???0?
2???????010??a3??0??a2??0?????? ?a1?a3???0? ?a3??a1，a2?0
2???0??a2????a1??? ∴ ?0??0?
??a??1? 由归一化条件
?a1???2?** 1??0?0?(a1,0,?a1)?0??2a1
??a??1? 取 a1?12
?1???2???的本征值0 。 ?对应于L ?0??0x????1???2?? 当?2??时，有
?a1??010??a1????????101a?? ?a2? ???2?2??a?????010??a3??3??1?a2???2??a??a2?2a1?1??1??? ?(a1?a3)???a2???a2?2a3
?2????a?a1?1??a3???3a2???2??a1??? ∴ ????2a1?
??a1???? 由归一化条件
?a1???2*** 1?(a1,2a1,a1)?2a1??4a1
??a1????1 取 a1?
2?1????2??1??的本征值? ∴归一化的?????对应于Lx?2??1????2? 当?2???时，有
?a1??010??a1???????? ?101??a2?????a2?
2??a?????010??a3??3?
量子力学周世勋习题解答第四章
