358a0? ?(p)?c(p)?2 224?(a0p??)26 设t=0时，粒子的状态为
?(x)?A[si2nkx?1s] 2cokx求此时粒子的平均动量和平均动能。
1解：?(x)?A[sin2kx?1 [12coskx]?A2(1?cos2kx)?2coskx]A[1?cos2kx?coskx] 2Ai2kxikx?e?i2kx)?1?e?ikx)] ?[1?12(e2(e2 ? ?A2??i0x1i2kx1?i2kx1ikx1?ikx1 [e?2e?2e?2e?2e]?22??2k? ?2k? k? ?k? 可见，动量pn的可能值为0 2pn2k2?22k2?2k2?2k2?2 动能的可能值为0 2???2?2? 对应的几率?n应为
A2A2A2A2A2( )?2?? 416161616 11111( ) ? A2?? 28888 上述的A为归一化常数，可由归一化条件，得
A2A2A2 1???n?(?4?)?2????2??
4162n ∴ A?1/?? ∴ 动量p的平均值为
p??pn?nnA2A2A2A2?0?2k???2???2k???2???k???2???k???2???0161616162pnp2 T????n
2?n2?
2k2?21k2?21 ?0???2???2
?82?85k2?2 ?
8?7 设氢原子处于状态 ?(r,?,?)?13R21(r)Y10(?,?)?R21(r)Y1?1(?,?) 22求氢原子能量、角动量平方及角动量Z分量的可能值，这些可能值出现的几率
和这些力学量的平均值。
解：在此能量中，氢原子能量有确定值 E2???es22?n22???es28?2 (n?2)
角动量平方有确定值为
L2??(??1)?2?2?2 (??1) 角动量Z分量的可能值为 LZ1?0LZ2??? 其相应的几率分别为 其平均值为 LZ?133?0?????? 44413， 44???ieixd的本征函数。 8 试求算符Fdx?的本征方程为 解：F???F? F即 ?ieixd?F?dxdd)?d(?Feix) dxdx
d???iFeixdx??d(Feixd?lncdxln???Feix?ix?是F的本征值） ??ce?Fe（Fdd)x]2??[x]?? dxdxdddd解：原式?[()x][()x]??[x][x]?
dxdxdxdxddx]? ?[()x][sinx?xcosx]??[x][xcosdxdx9 设波函数?(x)?sinx，求[(n?xx)?x(coxs?coxs?x)?x(x?x) ?(sixx?2xcosx ?sin?和B?都是厄米的，那么 10 证明：如果算符A?+B?)也是厄米的 (A??B??d???*B?)?d???*A 证： ??1*(A2?12?1??2d? ??)*d???(B ???2(A1?2??1)*d? ??B?)?]*d? ???2[(A1?+B?也是厄米的。 ∴ A?P??L?11 求 Lxx?Pxx??
?P??? Lyx?PxLy?? ?P??L? Lzx?Pxz??
?P??L??)P??P?(y??z?) ??z?z?P?P?P解： Lxx?Pxx?(yPyxxzy?P??P?P??y?z?) ?P??z?P ?yzx?zyx?PxPx?Py?P??P?P???P??P?P??P ?yzx?zyx?yPzx?zyx) = 0
?P??L???x?)P??P?(z??x?) ?P?P?P Lyx?Pxy?(zxzxx?Pxz?2?x?P??z??P?x?P??z) ?P ?zxzx?Px?PzxP?2?x?P??P?2?P?x?P??z) ?P ?zxzx?zxxP??P?x?P??z ??(xxx)P? ??i?Pz?P??L???P?(x??y?) ????x)P?P?P Lzx?Pxz?(xPy?yPxxyx?P??y?2?P?x? ?P????xy?P ?xPyxxxPy?Px?P???2?P?x?2 ?P??y?y?P ?xxy?yPxxPx??P?x?P?? ?(xxx)Py ? ?i?Py?x?x???x??? L???????L12 求Lz?xLz?? x?xLx?? Lyxy?x?)x??z?) ?????z?z??x?(y?P?P?P 解： Lx?xLx?(yPyzy?x?x? ?P??P????z?x?z?P ?yz?zy?xyPy?x?x?x?P??P???zx??z??P ?yz?zy?yPy = 0
?x??x?)x??x?) ????P??x?(z?P?P?P Ly?xLy?(zxzxz?x??x? ???zx??x?z?2P?P?P ?zx?xPxz?x???x) ?(P ?zx?xP? ??i?z?x?)x??y?) ?????y?y?P??x?(x?P?P Lz?xLz?(xPxyx??y?x?2P?P??2????x ?xyx?xPy?yxP??P?x?(x?P? ?yxx)
? ??i?y13 求在动量表象中角动量Lx的矩阵元和L2x的矩阵元。 解：(Lx)p?pp?r13??p??r??z?zp?y)ed? ?()e(yp2???i??i??i??i??p?r13??p??r)?e(ypz?zpy)e?d? ?(2??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??? ?i?(py?????pz)?(p?p?) ?pz?py*?2?(x)L??d? )?? (L2??xpp?pxpp?r13??p??r2??z?zp?y)ed? ?()e(yp2???p?r13??p??r?z?zp?y)(yp?z?zp?y)e?d? ?()?e(yp2??i??i??i??i??
13??p?r???p?r?z?zp?y)(i?)(py?()?e(yp?pz)ed?
2???pz?py
p?r??13??p??r????(i?)(py?pz)()e(ypz?zpy)ed?
?pz?py2????r??213?（p?p?） ???(py?pz)()?ed?
?pz?py2??2i???i??i??i??i?? ???2(py??2???pz)?(p?p?) ?pz?py14 求能量表象中，一维无限深势阱的坐标与动量的矩阵元。
解：基矢：un(x)?2n?sinx aa?2?2n2 能量：En?
2?a22m?axsin2xdx? 0aa21u ?ucosnudu?2cosnu?sinnu?c nn2am?n?x)?x?(sin)dx 当时，m?n xmn??(sina0aa
对角元：xmm??a