Solutions to Problem Set 7 for Quantum Mechanics | PHYS 324, Assignments of Quantum Mechanics

Download Solutions to Problem Set 7 for Quantum Mechanics | PHYS 324 and more Assignments Quantum Mechanics in PDF only on Docsity! 1 Physics 324 Solutions to Selected Problems #7 Autumn 2003 1. Prob. 4.02 in your textbook: Let Φ(x, y, z) = X(x)Y (y)Z(z), then the time independent Schrodinger equation, after dividing both sides by X(x)Y (y)Z(z) becomes: 1 X(x) d2X(x) dx2 + 1 Y (y) d2Y (y) dy2 + 1 Z(z) d2Z(z) dz2 = 2M h̄2 ( V (x, y, z) − E ) a.) For x, (y, z) ≤ 0 or x, (y, z) ≥ a, we know that X(x), (Y (y), Z(z)) = 0, while for 0 ≤ x, y, z ≤ a, we have: 1 X(x) d2X(x) dx2 + 1 Y (y) d2Y (y) dy2 + 1 Z(z) d2Z(z) dz2 = −2M h̄2 Elmn where Elmn ≥ 0. This requires: 1 X(x) d2X(x) dx2 = −2M h̄2 Elmn − 1 Y (y) d2Y (y) dy2 − 1 Z(z) d2Z(z) dz2 = Cx = −2M h̄2 El where we relabeled the separation constant Cx = −2MEl/h̄2. Similarly, 1 Y (y) d2Y (y) dy2 = −2M h̄2 (Elmn − El) − 1 Z(z) d2Z(z) dz2 = Cy = −2M h̄2 Em and 1 Z(z) d2Z(z) dz2 = −2M h̄2 (Elmn − El − Em) = −2M h̄2 En where Elmn = El + Em + En Defining k2i = 2MEi/h̄ 2, we then have 3 equations: d2X(x) dx2 = −k2l X(x), d2Y (y) dy2 = −k2mY (y), d2Z(z) dz2 = −k2nZ(z) whose solutions are the same as the one dimensional infinite square well potential: X(x) = Asin(klx) + Bcos(klx); X(0) = X(a) = 0 => B = 0 and sin(kla) = 0 The normalization is the same as the one dimensional infinite square well problem: A = √ 2/a, as is the energy spectrum: kla = lπ, or El = h̄2k2l /2M = h̄ 2l2π2/(2Ma2) where l = 1, 2, 3.... The solutions for Y and Z are the same as for X : X(x) = √ 2 a sin(kla) with kl = lπ a and El = h̄2l2π2 2Ma2 Y (y) = √ 2 a sin(kma) with km = mπ a and Em = h̄2m2π2 2Ma2 Z(z) = √ 2 a sin(kna) with kn = nπ a and En = h̄2n2π2 2Ma2 b.) Note that Elmn = El + Em + En = h̄2π2(l2 + m2 + n2)/(2Ma2). Let’s define E0 = h̄2π2/(2Ma2), then the lowest energy, E1, occurs for l = m = n = 1: E1 = E0(12 + 12 + 12) = 3E0 with a degeneracy of 1 2 E2 = E0(22 + 12 + 12) = 6E0 with a degeneracy of 3 because (l, m, n) = (2, 1, 1), (1, 2, 1) and (1, 1, 2) give the same energy. Similarly, E3 = E0(22 + 22 + 12) = 9E0 with a degeneracy of 3 E4 = E0(32 + 12 + 12) = 11E0 with a degeneracy of 3 E5 = E0(22 + 22 + 22) = 12E0 with a degeneracy of 1 E6 = E0(12 + 22 + 32) = 14E0 with a degeneracy of 6 E14 has an energy of 27E0 and comes from both (l, m, n) = (1, 1, 5) and (3, 3, 3). Its degeneracy is 4. It is the first energy for which different distinct sets of (l, m, n) are degenerate. 2. Prob. 4.20 in your textbook: For simplicity, I will leave the hats off the operators for this problem. Recall that [ ri, pj ] = ih̄δij and that A [ B, C ] + [ A, C ] B = [ AB, C ] . a.) [ Lz, x ] = [ xpy − ypx, x ] = [ xpy, x ] − [ ypx, x ] = 0 − y [ px, x ] = y [ x, px ] = ih̄y [ Lz, y ] = [ xpy − ypx, y ] = [ xpy, y ] − [ ypx, y ] = x [ py, y ] − 0 = −x [ y, py ] = −ih̄x [ Lz, z ] = [ xpy − ypx, z ] = [ xpy, z ] − [ ypx, z ] = 0 − 0 = 0 [ Lz, px ] = [ xpy − ypx, px ] = [ xpy, px ] − [ ypx, px ] = py [ x, px ] − 0 = ih̄py [ Lz, py ] = [ xpy − ypx, py ] = [ xpy, py ] − [ ypx, py ] = 0 − px [ y, py ] = −ih̄px [ Lz, pz ] = [ xpy − ypx, pz ] = [ xpy, pz ] − [ ypx, pz ] = 0 − 0 = 0 b.) [ Lz, Lx ] = [ Lz, ypz − zpy ] = [ Lz, ypz ] − [ Lz, zpy ] = y [ Lz, pz ] + [ Lz, y ] pz − z [ Lz, py ] − [ Lz, z ] py = ih̄(0 − xpz + zpx − 0) = ih̄Ly c.) [ Lz, r 2 ] = [ Lz, x 2 + y2 + z2 ] = [ Lz, x 2 ] + [ Lz, y 2 ] + [ Lz, z 2 ] = x [ Lz, x ] + [ Lz, x ] x + y [ Lz, y ] + [ Lz, y ] y + z [ Lz, z ] + [ Lz, z ] z = ih̄(xy + yx − yx − xy + 0 + 0) = 0 [ Lz, p 2 ] = [ Lz, p 2 x + p 2 y + p 2 z ] = [ Lz, p 2 x ] + [ Lz, p 2 y ] + [ Lz, p 2 z ] = px [ Lz, px ] + [ Lz, px ] px + py [ Lz, py ] + [ Lz, py ] py + pz [ Lz, pz ] + [ Lz, pz ] pz