Download 5 Solved Problems of Harmonic Oscillator - Homework 3 | PHYSICS 137B and more Assignments Quantum Mechanics in PDF only on Docsity! Physics 137B, Fall 2007, Moore Problem Set 3 Solutions (Revised 10/7/2007) 1. (a) In the usual basis, the Hamiltonian is H = Bh̄ 2 ( 1 0 0 −1 ) . The initial state is |ψ(0)〉 = ( 1 0 ) , so the time-evolved state is |ψ(t)〉 = e−iHt/h̄ |ψ(0)〉 = e−iBt/2 ( 1 0 ) . (b) Now the Hamiltonian is H = Bh̄ 2 ( 0 1 1 0 ) , whose eigenstates are |+〉 = 1√ 2 ( 1 1 ) and |−〉 = 1√ 2 ( 1 −1 ) , with eigenvalues E± = ±Bh̄2 . Since we can expand the initial state as |ψ(0)〉 = 1√ 2 (|+〉+ |−〉), its time evolution is |ψ(t)〉 = e−iHt/h̄ |ψ(0)〉 = 1√ 2 (e−iBt/2 |+〉+ e+iBt/2 |−〉) = ( cos Bt 2 −i sin Bt 2 ) . 1 The expectation value of Sz for this state is therefore 〈Sz〉 = 〈ψ(t)|Sz|ψ(t)〉 = ( cos Bt 2 +i sin Bt 2 ) h̄ 2 ( 1 0 0 −1 ) ( cos Bt 2 −i sin Bt 2 ) = h̄ 2 cos2 Bt 2 + ( − h̄ 2 ) sin2 Bt 2 = h̄ 2 cosBt . Remark: Once again, don’t forget to complex-conjugate your coefficients when you go from a ket to a bra. Many of you forgot to do this, and ended up with 〈Sz〉 = h̄/2 instead of the answer above. 2. The unperturbed energy eigenstates of the infinite square well potential are ψ(0)n (x) = √ 2 L sin nπx L , with energies E(0)n = n2π2h̄2 2mL2 . The first-order energy shifts for the perturbation H ′ = Aδ(x− L/2) are E(1)n = 〈ψ(0)n |H ′|ψ(0)n 〉 = ∫ L 0 dx 2 L sin2 nπx L Aδ(x− L/2) = 2A L sin2 nπ 2 = { 2A/L n odd 0 n even . By the same type of calculation we find that | 〈ψ(0)n |H ′|ψ(0)m 〉 |2 = ∣∣∣∣2AL sin nπ2 sin mπ2 ∣∣∣∣2 = { (2A/L)2 n,m odd 0 otherwise 2 4. (a) |`− s| ≤ j ≤ `+ s, so j = 3/2, 5/2, 7/2 or 9/2. (b) There are 4 states with m = 3/2, one for each value of j. (c) L · S = 1 2 (J2 − L2 − S2) = 1 2 ( j(j + 1)− `(`+ 1)− s(s+ 1) ) h̄2 = h̄2 2 ( j(j + 1)− 63 4 ) j 3/2 5/2 7/2 9/2 L · S −6h̄2 −7 2 h̄2 0 +9 2 h̄2 5. The two lowest eigenfunctions for the one-particle harmonic oscillator are ψ0(x) = (mω πh̄ )1/4 exp ( −mω 2x2 2h̄ ) , ψ1(x) = (mω πh̄ )1/4 (mω 2h̄ )1/2 x exp ( −mω 2x2 2h̄ ) . (a) If the two spin-half particles are allowed to occupy different spin states, the lowest energy state is the antisymmetrized tensor product state |φ〉 = 1√ 2 [ |ψ0 ↑〉 ⊗ |ψ0 ↓〉 − |ψ0 ↓〉 ⊗ |ψ0 ↑〉 ] , φ(x1, x2) = 1√ 2 ψ0(x1)ψ0(x2) [ χ↑ ⊗ χ↓ − χ↓ ⊗ χ↑ ] . This state has energy 1 2 h̄ω + 1 2 h̄ω = h̄ω. Plugging in the form of ψ0(x) given above you can easily check that it’s an eigenfunction of the Hamiltonian. 5 (b) If the two particles must both be spin up, the lowest energy state is |φ〉 = 1√ 2 [ |ψ0 ↑〉 ⊗ |ψ1 ↑〉 − |ψ1 ↑〉 ⊗ |ψ0 ↑〉 ] , φ(x1, x2) = 1√ 2 [ ψ0(x1)ψ1(x2)− ψ0(x2)ψ1(x1) ] χ↑ ⊗ χ↑ . This state has energy 1 2 h̄ω + 3 2 h̄ω = 2h̄ω. Since the wavefunction is now antisymmetric in x1 and x2, the wavefunction vanishes at x1 = x2. Under an interaction potential Λδ(x1 − x2), the first-order energy shift would vanish. 6
