4 Solved Problems on Time-dependent Theory - Homework 5 | PHYSICS 137B, Assignments of Quantum Mechanics

Download 4 Solved Problems on Time-dependent Theory - Homework 5 | PHYSICS 137B and more Assignments Quantum Mechanics in PDF only on Docsity! Physics 137B, Fall 2007, Moore Problem Set 5 Solutions 1. Our unperturbed Hamiltonian is H0 = gµBB h̄ Sz, with eigenstates |↓〉 , E↓ = − gµBB 2 ≡ − , |↑〉 , E↓ = + gµBB 2 = + . At time t = 0 we introduce the perturbation H ′ = gµBB ′ h̄ Sx, whose eigenstates we will label as |1〉 = 1√ 2 (|↓〉 − |↑〉) , E1 = − gµBB ′ 2 , |2〉 = 1√ 2 (|↓〉+ |↑〉) , E2 = + gµBB ′ 2 . (a) We start in the state |ψ(0)〉 = |↓〉 at time t = 0. According to the formalism of time-dependent perturbation theory, we can expand our state |ψ(t)〉 for t > 0 as |ψ(t)〉 = c↓(t)e−iE↓t/h̄ |↓〉+ c↑(t)e−iE↑t/h̄ |↑〉 . To first order, the time-dependent coefficients c↓(t) and c↑(t) are given by c↓(t) = 1 + (ih̄) −1 ∫ t 0 〈↓ |H ′| ↓〉 dt′ = 1 , 1 c↑(t) = (ih̄) −1 ∫ t 0 〈↑ |H ′| ↓〉 eiωt′dt′ = 1 ih̄ gµBB ′ 2 ∫ t 0 eiωt ′ dt′ = 1 ih̄ gµBB ′ 2 1 iω ( eiωt − 1 ) = −iB ′ B eiωt/2 sin ωt 2 , where ω = (E↑ − E↓)/h̄ = 2/h̄ is the Bohr frequency. Thus our time-evolved state can be written to first order as |ψ(t)〉 = e+it/h̄ |↓〉+ ( −iB ′ B eiωt/2 sin ωt 2 ) e−it/h̄ |↑〉 = e+iωt/2 |↓〉 − iB ′ B sin ωt 2 |↑〉 . The transition probability is given by P↑↓(t) = | 〈↑ |ψ(t)〉 |2 = ( B′ B )2 sin2 ωt 2 , exactly as you calculated in the previous homework assignment. To calculate the expectation value of H0, we need 〈ψ(t)|H0|ψ(t)〉 = ( e−iωt/2 〈↓|+ iB ′ B sin ωt 2 〈↑| )( e−iωt/2(−) |↓〉+ iB ′ B sin ωt 2 (+) |↑〉 ) = − [ 1− ( B′ B )2 sin2 ωt 2 ] . This is of order (B′/B)2, whereas |ψ(t)〉 is only normalized to order B′/B. So we should divide by 〈ψ(t)|ψ(t)〉 = 1 + ( B′ B )2 sin2 ωt 2 . Expanding the denominator to leading order we obtain the expectation value 〈H0〉 = 〈ψ(t)|H0|ψ(t)〉 〈ψ(t)|ψ(t)〉 = − [ 1− 2 ( B′ B )2 sin2 ωt 2 ] . 2 2. (a) The 1D harmonic oscillator has energies En = h̄ω0(n+ 1 2 ). The number of states with energies less than or equal to E = En is N(E) = n+ 1 = E h̄ω0 + 1 2 . For large n (i.e. for E  h̄ω0), we treat this function as a smooth function of the energy E. Then the density of states is ρ(E) = dN dE = 1 h̄ω0 . (b) A 2D harmonic oscillator is equivalent to two independent 1D harmonic oscil- lators, so the eigenstates are labeled by pairs of non-negative integers (nx, ny), with energy Enx,ny = h̄ω0(nx + 1 2 ) + h̄ω0(ny + 1 2 ) = h̄ω0(nx + ny + 1) . Let n = nx + ny. There are n + 1 energy eigenstates with energy equal to E = h̄ω0(n+ 1), namely (nx, ny) = (n, 0), (n− 1, 1), . . . , (1, n− 1), (0, n) . Thus the number of states with energy less than or equal to E is N(E) = n∑ k=0 (k + 1) = 1 2 (n+ 1)(n+ 2) = 1 2 ( E h̄ω0 )( E h̄ω0 + 1 ) ≈ 1 2 E2 (h̄ω0)2 . where the last approximation is valid for E  h̄ω0. The density of states is therefore ρ(E) = dN dE = E (h̄ω0)2 . (Note: Many of you inserted an extra factor of 2 for spin. For a 2D harmonic oscillator the particles are generally assumed to be spin 0, unless you are explicitly told otherwise.) 5 3. Consider the 1D harmonic oscillator, initially in the ground state, with a time- dependent perturbation H ′ = −qxE(t) added at t = 0, where E(t) = E0e−t/τ . The transition probabilities in first-order time-dependent perturbation theory are given by the usual formula: Pn0(t) = 1 h̄2 ∣∣∣∣∫ t 0 〈n|H ′(t′)|0〉 eiωn0t′dt′ ∣∣∣∣2 = q2E20 | 〈n|x|0〉 |2 h̄2 ∣∣∣∣∫ t 0 dt′ e−t ′/τeinωt ′ ∣∣∣∣2 , where ωn0 = (En − E0)/h̄ = nω. The matrix element 〈n|x|0〉 is most easily computed by introducing the raising and lowering operators a† and a, in terms of which x = √ h̄ 2mω (a+ a†) . Using the standard rules a |n〉 = √ n |n− 1〉 and a† |n〉 = √ n+ 1 |n+ 1〉, we have 〈n|x|0〉 = √ h̄ 2mω 〈n|a+ a†|0〉 = √ h̄ 2mω 〈n|1〉 = √ h̄ 2mω δn1 Therefore only transitions to the n = 1 state are allowed. The integral is easy to compute, but we must not forget the boundary term coming from the lower limit of integration. Setting n = 1, we have I = ∫ t 0 dt′ e−t ′/τeiωt ′ = ∫ t 0 dt′ einω−1/τ t′ = 1 iω − 1/τ [ e(iω−1/τ)t − 1 ] . |I|2 = 1 ω2 + 1/τ 2 [( e−t/τ cosωt− 1 )2 + ( e−t/τ sinωt )2] = 1 ω2 + 1/τ 2 [ e−2t/τ − 2e−t/τ cosωt+ 1 ] . Thus P10(t) = q2E20 2h̄mω 1 ω2 + 1/τ 2 [ e−2t/τ − 2e−t/τ cosωt+ 1 ] . 6 In the limit t→∞, this becomes P10(∞) = q2E20 2h̄mω 1 ω2 + 1/τ 2 . 4. To use the sudden approximation in the limit τ →∞, we again have to solve for the exact eigenfunctions for t > 0. If τ → ∞, our perturbation is just H ′ = −qE0x, so our total Hamiltonian is H = H0 +H ′ = p2 2m + 1 2 kx2 − qE0x = p2 2m + 1 2 k ( x− qE0 k )2 − q 2E20 2k = p′2 2m + 1 2 kx′2 − q 2E20 2k , where x′ = x − qE0/k. Up to an irrelevant offset q2E20/2k, this is just another 1D harmonic oscillator, but now centered around a = qE0/k rather than about 0. The eigenfunctions, as functions of x′, are exactly the same as the usual 1D harmonic oscillator eigenfunctions. Let us use the notation |n〉 to denote the original harmonic oscillator eigenstates centered around 0, and |n′〉 to denote the harmonic oscillator eigenstates centered around a. At t = 0, our initial state is |ψ(0)〉 = |0〉. Since the |n〉’s are no longer energy eigenstates for t > 0, we need to expand this in the new basis. Thus, if we define d′n = 〈n′|0〉, then our time-evolved state at time t > 0 is |ψ(t)〉 = ∞∑ n=0 d′ne −iE′nt/h̄ |n′〉 , where E ′n = En = h̄ω0(n+ 1 2 ), as usual. The probability of finding this system in an excited state is just 1 minus the probability of finding it in the ground state. Thus Pexcited = 1− P0 = 1− | 〈0′|ψ(t)〉 |2 = 1− |d′0|2 . It is now straightforward to calculate d′0 = 〈0′|0〉 = ∫ dxψ∗0(x− a)ψ0(x) 7