Solutions for Homework 7 - Quantum Mechanics | PHYS 325, Assignments of Quantum Mechanics

Download Solutions for Homework 7 - Quantum Mechanics | PHYS 325 and more Assignments Quantum Mechanics in PDF only on Docsity! 1 Physics 325 Solution to Homework Problems #7 Winter 2004 1. Your textbook Prob. 9.7 We start with H ′ab = Vabe iωt/2 and: ċa = 1 ih̄ H ′abe −iω0tcb ċb = 1 ih̄ H ′∗abe iω0tca with ca(0) = 1, cb(0) = 0 (a) Let ∆ω = ω0 − ω. Then: ċa = 1 2ih̄ Vabe −i∆ωtcb ċb = 1 2ih̄ V ∗abe i∆ωtca => c̈a = 1 2ih̄ Vab [ − i∆ωe−i∆ωt cb + e−i∆ωt ċb ] The first term in the brackets is just −i∆ωċa and we substitute for ċb from our original equation: c̈a = −i∆ω ċa + 1 2ih̄ Vabe −i∆ωt [ 1 2ih̄ V ∗ab e i∆ωt ca ] = −i∆ω ċa − 1 4h̄2 |Vab|2 ca We turn this second order differential equation for ca into an algebraic equation by looking for a solution of the form ca = eiqt. Using this form for ca in the equation above gives us: −q2eiqt = −i∆ω (iq)eiqt − 1 4h̄2 |Vab|2 eiqt => q2 + q∆ω − 1 4h̄2 |Vab|2 = 0 => q± = −∆ω ± √ (∆ω)2 + |Vab/h̄|2 2 = −∆ω 2 ± ωr where ωr is the Rabi flopping frequency defined in the text. We see that there are two possible values for q, making the most general solution to the differential equation: ca = Aeiq+t + Beiq−t We now apply the initial conditions: ca(0) = 0 => A + B = 1 ċa(0) = 1 2ih̄ Vabe −i∆ωtcb(0) = 0 => iq+A + iq−B = 0 => q+A + q−(1 − A) = 0 => A = −q− q+ − q− = ωr + ∆ω/2 2ωr Similarly, B = q+ q+ − q− = ωr − ∆ω/2 2ωr Therefore : ca(t) = e−i∆ωt/2 2ωr [ (ωr + ∆ω/2)eiωrt + (ωr − ∆ω/2)e−iωrt ] = e−i∆ωt/2 2ωr [ 2ωr cos(ωrt) + i∆ω sin(ωrt) ] or ca(t) = e−i∆ωt/2 [ cos(ωrt) + i∆ω 2ωr sin(ωrt) ] We can follow the same procedure for finding cb, but note that we will get the same second order differential equation, except that ∆ω → −∆ω. Therefore we can write: cb(t) = Ceis+t + Deis−t where s± = ∆ω 2 ± ωr 2 Applying the initial conditions: cb(0) = 0 gives C = −D and: ċb(0) = 1 2ih̄ V ∗abca(0) = 1 2ih̄ V ∗ab => is+C − is−C = 1 2ih̄ V ∗ab => C = −1 4h̄ωr V ∗ab => cb(t) = −1 4h̄ωr V ∗ab e i∆ωt/2 [ eiωrt − e−iωrt ] = 1 2ih̄ωr V ∗ab e i∆ωt/2 sin(ωrt) (b) Pa→b = |cb|2 = 1 4h̄2ω2r |Vab|2 sin2(ωrt) = |Vab|2 sin2(ωrt) h̄2(∆ω2 + |Vab/h̄|2) Therefore, Pa→b is maximum when ∆ω = 0 (”on resonance”) and sin2(ωrt) = 1, at which time Pa→b takes its maximum value of 1. |ca(t)|2 + |cb(t)|2 = [ cos(ωrt) + i∆ω 2ωr sin(ωrt) ][ cos(ωrt) − i∆ω 2ωr sin(ωrt) ] + Pa→b = cos2(ωrt) + (∆ω)2 4ω2r sin2(ωrt) + 1 4h̄2ω2r |Vab|2 sin2(ωrt) = cos2(ωrt) + sin2(ωrt) [ (∆ω)2 + |Vab/h̄|2 4ω2r ] = cos2(ωrt) + sin2(ωrt) = 1 (c) For |Vab/h̄|2 << (∆ω)2 = (ω0 − ω)2, we have ωr ≈ (ω0 − ω)/2, so that our exact solution goes to: Pa→b ≈ |Vab|2 sin2[(ω0 − ω)t/2] h̄2(ω0 − ω)2 which is the same as the first order perturbation result, Eqn. 9.28. Therefore, first order perturbation works when |Vab| << h̄|ω0 − ω|, ie the interaction energy (the energy that couples states a and b), is much less than the energy of the ”detuning” (the energy difference between the applied frequency, ω, and the resonant energy, ω0). This condition guarantees that Pa→b << 1, which is necessary for the perturbation to be considered small. (d) When ωrt = π ie t = π/ωr, we first return to the state where cb = 0. The minus sign in front of ca can be absorbed into the phase factor. 2. Your textbook Prob. 9.8 The rate for spontaneous emission is A, Einstein’s A coeficient, while the rate for stimulated emission is Bρ(ω0) where B is Einstein’s B coefficient and ρ(ω0) is the energy density of black-body photons at a temperature of T = 300K. Using Eqns. 9.52 and 9.54 in your text, we can compare the two rates: Bρ(ω0) A = h̄ π2c3 ω3 eh̄ω/kBT − 1 π2c3 ω3h̄ = 1 eh̄ω/kBT − 1 kBT ≈ .025 eV for T = 300K. For f = 5× 1012Hz, h̄ω = 6.6× 10−16eV − s× 2π× 5× 1012Hz ≈ .021eV. that is, kBT ≈ h̄ω for f ≈ 5 × 1012Hz. For f << 5 × 1012Hz, 1 eh̄ω/kBT − 1 ≈ 1 1 + h̄ω/kBT − 1 ≈ kBT h̄ω >> 1