2015中科院量子力学真题解析

1、中国2015 年招收攻读院学位入学试题参考科目名称：量子力学一、一个质量为m的粒子在一维盒子（0 x L)里y(0) =y(L)1)求系统的能级运动,波函数y(x)满足y(0) =y(L);2)第一激发态写成归一化动量本征态的组合形式,并给出当= 0时,组合系数满足条件p2-h2 d 2y2mE2mE：1)H =, Hy(x) = Ey(x);即= Eyy(x) +y(x) = 0(令k =）2m2m dx2h2h2得y(x) + k2y(x) = 0,取y(x) = Aeikx + Be-ikxA + B = 0A = -By(0) =y(L) = 0 cos kL = 1 kL = 2np

2、(A + B) cos kL + (A - B)i sin kL = 0 y(0) =y(L)(A - B)ik = ik(A - B) cos kL + ik(A + B)i sin kLsin kL = 02n2h2p2ikx-ikx E =(n = 0, 1, 2.);y(x) = A(e- e)mL2+2)利用归一化条件d(x - x ) = -yk *(x )yk (x)dk+-ikxikxikx-ikx= A(e- e)(e- e)dk2-+i( x-x)k-ik ( x+x)i( x+x)ki( x-x)k= A(e2- e- e+ e)dk-= A22pd(+ x )kdk +

3、2pd(x - x )p112nd(x - x ) = 4pA2d(x - x ) A =y (x) =(e- e)(k =ikx-ikx, n = 0, 1, 2.)4pn4pL又Q归一化动量本征态j(k ) = 1eik x (- k +)2pj(k = 2p +j(k = - 2p11212y(x) =(eik x - e-ik x ) =)1114pLL= 1 2ph + 1 -2ph = 02L2L二、一个三维简谐振子受到微扰H = l(xyz + x2 y + xy2 )的作用,求系统基态能量,并且精确到l2量级.：体系波函数y(0) (x, y, z) =y(0) (x)y(0)

4、 ( y)y(0) (z)n1n2n3n1n2n3= (n + 3)hw; (n = n + n + n ; n ,n ,n = 0,1, 2, 3.)E(0)n1n2n31231232=000 H 000 ,由波函数对称性得E(1) =000 H 000 = 0E(1)000000H = n n nl(xyz + x2 y + xy2 ) 000n1n2n3 ,0001 2 3= l 111 xyz 0003+010 x2 y 000 + 210 x2 y 000 + 120 xy2100 xy20001000 11111221= l 111(111 + 010010 + 210)2210a

5、32a2a2a2a2211 12 11 120 + 100100 + 120a 22a212a2a231111111= l( )2 +a322a322a32a322a3| H |2n1n2n3n1n2n3 ,000E(2)=E(0)- E(0)n1n2n3 000n1n2n3000l2a3l2a3l 13ll121222|( )2 |2|2|2|a32a32a32=+E(0) - E(0)E(0) - E(0)E(0) - E(0)E(0) - E(0)E(0) - E(0)000111000010000210000100000120l2l2l2l2l2 1 ()3 +h 1 ()3 +h 1

6、 ()3 +h 1 ()3 +h 1 (h=)3-3hw 8 mw-hw 8 mw-3hw 4 mw-hw 8 mw-3hw 4 mw= - l2+= -() ()11l2h2h1223hw mw2481224m3w311l2h23=hw-224m3w4E000u三、两个自旋粒子分别为ur r ur rur urrururH = 3(s1 n)(s1 n) - s1 s2，记s = s1 + s2，证明,设两个粒子相互作用为：rr r)2h24-1)2)H , s = 022r 2r2ururur urur urur ur3h 23h2s：(1)s = (s + s )2 = s 2 + s

7、2 + 2s s =+ 2s1 s s s =-121212212224rr选相对取向n为直角坐标系z轴；则n = (0, 0,1), 记sz = s1z + s2 zh2sz = (s1z + s2 z ) = s1z + s2 z + 2s1z s2 z =+ 2s1z s2 z22222s 2h2 s1z s2 z =-z24r2rrrs 2h2s23h23s 2s23s 2s32)H = 3(-) -+=-;H , s = -, s =sz , s2zzz242422222rrrrrrrQsz , s = sz , sx i + sy j + sz k = sz , sx i +sz

8、, sy j +sz , sz k = 0rrrrs 2, s = s s , s +s , ss = 0;H , s = 0zzzzz四、一个质量为m的粒子在势场V (x) = ; x 0 中运动.Bx; x 01) 用变分法求基态能量可以选取下列哪个作为近似波函数，说明理由-a) e a，b)xe a，c)1- e a (a为变分参数）2)用所选的波函数求基态的能量。- x a：1)由势场V (x)得波函数y(0) =y(x ) = 0;所以选b,y(x) = xe-h2d 20 y* (x)( 2m dx2 + Bx)y(x)dx2) =|y(x) |2 dx0-2 xa342!|y(x) |2 dx = x2e a dx =2003( )a- 2 x3!3Ba4y* (x)Bxy(x)dx = B x3e a dx = B=24800( )a-h2-h22md 2-y* (x)(y(x)dx =2xe a (-e aa +xe a )dxm dx2a200-h2- 2 x- 2 xh2a8m21=-xe a dx +x2e a dx =2ma2a00-h2d