Download Solutions to Physics 325 Homework Problems #5 and #18 and more Assignments Quantum Mechanics in PDF only on Docsity! 1 Physics 325 Solution to Homework Problems #5 Winter 2004 1. Your textbook Prob. 6.13 The relativistic perturbation is H ′r = −p4/(8m3c2) and following the treatment in section 6.3.1 in your text, the first order correction to the energy can be found from: E(1)r = − 1 2mc2 [ (E(0))2 − 2E(0)〈V 〉 + 〈V 2〉 ] For the harmonic oscillator potential, E(0)n = (n + 1/2)h̄ω and V (x) = mω2x2/2. We must evaluate 〈V 〉 = 〈ψ(0)n |V (x)|ψ(0)n 〉 and 〈V 2〉 = 〈ψ(0)n |V (x)2|ψ(0)n 〉. Using equations 2.41, 2.52, and2.53 in yout text, we have: x2 = (a+ − a−)2 −2mω2 ; a+ψ(0)n = i √ (n+ 1)h̄ω ψ(0)n+1; and a−ψ (0) n = −i √ nh̄ω ψ (0) n−1 giving us: 〈V 〉 = 〈ψ(0)n |V (x)|ψ(0)n 〉 = − 1 4 〈ψ(0)n |(a2+ − a+a− − a−a+ + a2−)|ψ(0)n 〉 => 〈V 〉 = 1 4 〈ψ(0)n |(a+a− − a−a+)|ψ(0)n 〉 = h̄ω 4 (n+ n+ 1) = 1 2 (n+ 1 2 )h̄ω (We drop the a2± terms because they produce the states ψ (0) n±2 which are orthogonal to ψ (0) n .) Note, this result for 〈V 〉, which is 1/2E(0)n , is the result of the Virial Theorem that we used in the homework last week. Therefore, (E(0)n )2 − 2E(0)n 〈V 〉 = 0, leaving us with only the V 2 term. 〈V 2〉 = 〈ψ(0)n |V (x)2|ψ(0)n 〉 = m2ω4 4 〈ψ(0)n |x4|ψ(0)n 〉 = 1 16 〈ψ(0)n |(a+ − a−)4|ψ(0)n 〉 We expand (a+ − a−)4 keeping track of the order of the operators and keeping only the terms that raise twice and lower twice (ie that bring us back to state ψ(0)n ). We did this in class and found: 〈ψ(0)n |(a+ − a−)4|ψ(0)n 〉 = 〈ψ(0)n |a2+a2− + a2−a2+ + a+a−a+a− + a−a+a−a+ + a−a2+a− + a+a2−a+|ψ(0)n 〉 = 3(h̄ω)2 (2n2 + 2n + 1) Therefore E(1)n = − 1 2mc2 〈V 2〉 = −3(h̄ω) 2 32mc2 (2n2 + 2n + 1) 2. Your textbook Prob. 6.14 We use the fact that: [ Li, Lj ] = ih̄ijkLk, [ Si, Sj ] = ih̄ijkSk, and [ Li, Sj ] = 0 (a) [ L · S, L ] = [ LxSx + LySy + LzSz, L ] 2 = [ LxSx + LySy + LzSz, Lx ] x̂ + [ LxSx + LySy + LzSz, Ly ] ŷ + [ LxSx + LySy + LzSz, Lz ] ẑ Consider the first commutator above: [ LxSx + LySy + LzSz, Lx ] = Sx [ Lx, Lx ] + Sy [ Ly, Lx ] + Sz [ Lz, Lx ] = 0 + Sy(−ih̄Lz) + Sz(ih̄Ly) = ih̄(LySz − LzSy) = ih̄(L × S)x Therfore, this first term gives us ih̄(L × S)x x̂. Similarly, the second and third terms give us the ŷ and ẑ components of the cross product, leaving us with: [ L · S, L ] = ih̄ L × S (b) Just exchange L and S in part (a) to find: [ L · S, S ] = ih̄ S× L (c) [ L · S, J ] = [ L · S, L + S ] = ih̄ L × S + ih̄ S× L = 0 (d) [ L · S, L2 ] = Sx [ Lx, L 2 ] + Sy [ Ly, L 2 ] + Sz [ Lz, L 2 ] = 0 because [ Li, L 2 ] = 0 for i = x, y, z. (e) [ L · S, S2 ] = 0 for the same reason as part (d). (f) J2 = (L + S)2 = L2 + S2 + 2L · S [ L · S, J2 ] = [ L · S, L2 + S2 + 2L · S ] = 0 because from parts (d) and (e), L · S commutes with L2 and S2, and it commutes with itself. 3. Your textbook Prob. 6.16 From Bohr theory, the energy of the photon is E(0)3 − E (0) 2 = E (0) 1 (1/9 − 1/4) = −5E (0) 1 /36 = 1.890 eV (using E(0)1 = −13.6057 eV). Therefore ν = ∆E h = 1.89 eV × 1.602× 10−19 J/eV 2π × 1.0546−34 J − s = 4.569× 1014 Hz and λ = c ν = 3.0 × 108 m/s 4.569× 1014 Hz = 6.565× 10−7 m = 656.5 nm For n = 2, we can have l = 1 or l = 0, so that we can have j = 3/2 or j = 1/2. From equation 6.66 in your text, we then have for E(1)n,j :
