Time Dependent Perturbation Theory - Lecture Notes | PHY 4605, Study notes of Physics

Download Time Dependent Perturbation Theory - Lecture Notes | PHY 4605 and more Study notes Physics in PDF only on Docsity! 5 Time-dependent Perturbation Theory I Consider time-dependent perturbation in Hamiltonian H = H0 + V̂ (t) (1) with H0 constant in time and exactly soluble as before, H0|n〉 = En|n〉, 〈n|n′〉 = δnn′. Recall H0 time-ind. =⇒ time evolution of soln to ih̄∂|ψ〉/∂t = H0|ψ〉 is |ψ(t)〉 = ∑n cn|n〉 exp(−ih̄Ent/h̄). With full t-dependent H , write solution with time-dependent coefficients |ψ(t)〉 = ∑ n cn(t)e −ih̄Ent/h̄|n〉 (2) and plug in: (H0 + V̂ (t))|ψ〉 = ih̄∂|ψ〉 ∂t (3) ∑ n ( En + V̂ ) cn(t)e −ih̄Ent/h̄|n〉 = ∑ n  ih̄ dcn dt + Encn(t)   e−ih̄Ent/h̄|n〉 Inner product with 〈m|: ih̄ dcm dt = ∑ n 〈m|V̂ (t)|n〉cn(t)ei(Em−En)t/h̄, (4) i.e. set of coupled diff. eq. for cm(t). Perturbation turned on at t=0 Large class of interesting problems can be defined by assuming system evolves according to H0 until t = 0, at which time perturbation V̂ (t) is turned on. Assume system is in eigenstate |n〉 at t = 0, then initial conditions are cm(t = 0) = δm,n (5) Now look at t > 0 but very small, such that still have cn ' 1 and cm ¿ cn, m 6= n. Then can drop all terms except m = n on rhs of (4), find 1 ih̄ dcm dt = 〈m|V̂ (t)|n〉ei(Em−En)t/h̄, (6) which can be directly integrated: cm ' 1ih̄ ∫ t 0 dt′〈m|V̂ (t′)|n〉ei(Em−En)t′/h̄, m 6= n (7) 5.1 Transition probabilities Can derive some quite general 1st order results for transition probabilities which go under name of Fermi golden rule–useful for calculations in wide variety of physical situations, back-of-envelope estimates! 1st consider situation when perturbation is oscillatory: V̂ (t) = V̂0 cos ωt (8) & we want to consider transitions induced between two eigenstates |i〉 and |f〉 of H0, i.e. at t = 0 state is |i〉.1 Plugging into (7) and using cos ωt = (eiωt + e−iωt)/2, and2 h̄ω0 ≡ Ef − Ei, (9) find after performing t-integral cf(t) = − 1 2h̄ 〈f |V̂0|i〉   ei(ω0+ω)t − 1 ω0 + ω + ei(ω0−ω)t − 1 ω0 − ω   (10) If ω ' ω0 the external frequency nearly matches the energy of the transi- tion, so we can neglect the 1st term,3 giving cf(t) ' − 1 2h̄ 〈f |V̂0|i〉e i(ω0−ω)t − 1 ω0 − ω (11) Probability atom is in state |f〉 at time t is |cf(t)|2. Using identity |eiθ − 1|2 = 2(1− cos θ) = 4 sin2 θ/2, find 1We will neglect coupling to other states, so formally we are solving the two-level system. 2Note h̄ω0 > 0 means atom has absorbed photon, h̄ω0 < 0 that atom has emitted photon. 3One has to be a little careful about this argument, in the sense that we are assuming that we can make ω sufficiently close to ω0 such that the 2nd term becomes arbitrarily large with respect to the first. Still, for fixed t and V (t), we must remember that we can’t have cf (t) > 1. We are assuming that the time can be chosen sufficiently small such that our perturbation expansion still works, even arbitrarily close to ω0. 2 Angular dependence for linear polarization Hyperfine radiation has characteristic polarization dependence. Sup- pose ~² is in x-z plane for simplicity, S · ~² = Ŝex sin θ + Ŝez cos θ. Redo previous calculation, noting that Ŝez has no nonzero matrix elements with 〈f | and |i〉, since both 〈f | and |i〉 are eigenstates of Ŝz and 〈f |i〉 = 0. So only contribution is Ŝx component, which now enters with add’l factor sin2 θ. P = g2e2B20 32m2c2 sin2(ω − ω0)t/2 (ω − ω0)2 sin 2 θ (~² = sin θx̂ + cos θẑ) (22) Physical interpretation of P → 0 when θ → 0: if B ‖ ẑ, system invariant under rotations about ẑ =⇒ L̂z is conserved. So transitions with M = ±1 are not allowed. Circular polarization Now shine circularly polarized light on H-atom., B = B0(̂i cos ωt + ĵ sin ωt) (23) i.e. B rotates in x-y plane w/ angular freq. ω. In addition to matrix elts of Ŝx and Ŝz which we’ve calculated, we’ll need 〈f |Ŝy|i〉 = 〈0, 0|Ŝey|1M〉 (24) = (〈↑↓|−〈↓↑|√ 2 ) ( Ŝe+−Ŝe− 2i )    |↑↑〉 |↑↓〉+|↓↑〉√ 2 |↓↓〉 (25) = −ih̄2 (〈↑↓|−〈↓↑|√ 2 )    −|↓↑〉 |↑↑〉−|↓↓〉√ 2 |↑↓〉 (26) = h̄ 2 √ 2    −i 0 −i (27) 5 So for M = ±1, prob. ampl. is (recall h̄ω0 ≡ Ef − Ei < 0 for the emission process we’re calculating here!): cf = 1 ih̄ ∫ t 0 dt′〈f |V̂ |i〉eiω0t′, where 〈f |V̂ |i〉 = geB0 2mc ( 〈f |Ŝx|i〉 cos ωt′ + 〈f |Ŝy|i〉 sin ωt′ ) = geB0 2mc · −Mh̄ 2 √ 2 · (cos ωt′ + iM sin ωt′) (28) and we find cf = iM 4 √ 2 geB0 2mc ∫ t 0 dt′ei(ω0+Mω)t ′ . (29) Now the exponent in (29) can be small only if ω0 +Mω ' 0, or M = +1, since ω0 < 0, take ω > 0 always. For this case we get large transition amplitude. But if M = −1, exponent is large and integrand oscillates rapidly =⇒ cf → 0. “photon” interpretation: circularly polarized light has angular momentum z-component M = +1 if propagation is along ẑ and phases are chosen as in (23). Angular momentum conservation means that atom initially in state M = +1, finally in state M = 0, outgoing photon has M = +1, consistent with general conservation law Mi = Mf + Mphoton (emission) (30) and we have deduced that Mphoton is +1 for right- and−1 for left-circularly polarized radiation.6 We could redo the argument for an absorption pro- cess, and find Mi + Mphoton = Mf (absorption) (31) 6In this naive treatment of the electromagnetic field, there is no difference between the argument for stimulated and sponatneous emission. It is easiest to think of spontaneous emission, where the outgoing photon has Mphoton = +1, as stated. But the calculation also applies to stimulated emission, where now Mphoton has to be thought of as the difference in the z-components of the photon field in the final state (2 photons) minus the initial state (1 photon). So a process where the incoming photon has M = −1, but the outgoing photons have both M=0, during which process the atom makes the transition from M = +1 to M = 0 is also possible. See sec. 5.3. 6 5.3 Einstein argument relating absorption, stimulated & spon- taneous emission Figure 1: Three types of emission processes: a) absorption; b) stimulated emission; and c) spontaneous emission The calculations we have done for 2-level systems so far makes it plausible that an externally applied electromagnetic field can cause transitions be- tween states, and we have seen there is a possible crude interpretation in terms of “photons” even at this (“first-quantized”) level. But the simplest process one is taught about in, e.g. chemistry classes is in some sense the hardest to understand. Why does atom in excited state emit light spon- taneously? A single excited atom sitting in empty space has an infinite lifetime based on the quantum mechanics we have learned so far, because, the excited state is an eigenstate of the Hamiltonian! How can it decay, emitting light? Answer is actually beyond scope of course. What we think of as “vacuum” is in quantum electrodynamics a very active medium, continually being “polarized” by quantum fluctuations, i.e. particle-antiparticle pairs which live for a short time (short enough to satisfy Heisenberg’s uncertainty principle ∆E∆t ' h̄) and then decay. These processes can, in analogy to an externally applied classical field (stimulated emission), cause transitions in nearby atoms. Another way to think of it is to put an atom in a large box. The Hamiltonian now has to be solved together with the modes of 7 Number conservation for atoms in ground state (in equilibrium!): dN1 dt ∣∣∣∣∣∣ abs + dN1 dt ∣∣∣∣∣∣ spon + dN1 dt ∣∣∣∣∣∣ stim = 0 (45) so C = A(1− e−h̄ω0/kBT ) + Be−h̄ω0/kBT (46) Now draw several consequences from simple result: 1. As T → ∞, e−h̄ω0/kBT → 1 so we learn that B = C, rate per atom of stimulated emission equal to rate of absorption. This is the same result we derived from microscopic theory for a single atomic transition. 7 2. Insert into (46) to find B = A as well, so there’s only one overall const. 3. Summarize: If rate of absorption from given mode is nN1A, sponta- neous decay rate is N2A, stimulated rate is nN2A, and the net (sum of stimulated & spontaneous) decay rate to mode is N2(1 + n)A. 5.4 Continuum of final states: Fermi Golden Rule ? Note this whole calculation has been done thinking only about atoms interacting with monochromatic radiation, ang. freq. ω0. If we put atoms in a cavity at some temperature T , would expect a distribution of fre- quencies, e.g. blackbody. Let’s be more general and ask what happens if the radiation field has some distribution of frequencies to which the atoms might decay, characterized by a density of states (no. states/ energy in- terval) ρ(ω). So to get probability for transition due to one of the modes, integrate over final states assuming that probabilities for transitions from different modes add independently (incoherent perturbations): Pf = |〈f |V̂0|i〉|2 h̄2 ∫ ∞ −∞ dω ρ(ω) sin2[(ω − ω0)t/2] (ω − ω0)2 (47) 7Be amazed: Einstein was able to derive this result in a statistical fashion knowing almost zero about quantum theory! At the time, absorption and spontaneous emission had classical counterparts (think of the decay of an orbiting charged particle in a classical atom), but stimulated emission was a new idea. Einstein showed there was no detailed balance unless these processes were included. 10 Suppose now that the peak of sin2[(ω − ω0)t/2]/(ω − ω0)2 is much nar- rower than the spread of frequencies in ρ0. Then allowed to approximate ρ(ω) by ρ(ω0). Integral may now be made dimensionless and performed, ∫∞ −∞ dx sin 2(x/2)/x2 = π/2, so we get Pf ' π|〈f |V̂0|i〉| 2 2h̄2 ρ(ω0)t (48) Pf is the probability of a transition, so the rate of making transitions is Rf = dPf dt ' Pf t = π|〈f |V̂0|i〉|2 2h̄2 ρ(ω0) (49) This is one version of the Fermi golden rule. 11