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1 the Problem 2 the Interaction Picture Time dependent perturbation theory1 D. E. Soper2 University of Oregon 11 May 2012 I offer here some background for Chapter 5 of J. J. Sakurai, Modern Quantum Mechanics. 1 The problem Let the hamiltonian for a system of interest have the form H(t) = H0 + V (t) : (1) Here H0 is time-independent. We assume that we know the eigenvectors and eigenvalues of H0. The interaction hamiltonian V can be time independent or time dependent. We wish to solve the Schr¨odingerequation d i (t) = (H0 + V (t)) (t) ; (2) dt working order by order in powers of V . 2 The interaction picture In the first section of these notes, we have been working in the usual Schr¨odinger picture. Now let us switch to the \interaction picture" defined by iH0t I(t) = e S(t) : (3) One easily derives (see the derivation below for UI (t; t0)) that d i I(t) = VI(t) I(t) ; (4) dt 1Copyright, 2012, D. E. Soper [email protected] 1 where iH0t −iH0t VI(t) = e VS(t)e : (5) Note that VI(t) is time dependent even if VS(t) is time independent (unless V commutes with H0). Note also that VI(t) at one time generally does not commute with VI(t) at another time. Finally, note that if, for some time interval, VS(t) vanishes, then I(t) does not evolve during that time interval. That is, essentially, the point of working in the interaction picture. To express the evolution in the interaction picture using a starting time t0 at which initial conditions are specified, it is convenient to write iH0t I(t) = e S(t) iH0t = e US(t; t0) S(t0) (6) iH0t −iH0t = e US(t; t0)e I(t0) : Here US(t; t0) is the Schr¨odingerpicture evolution operator, equal to exp(−iH(t − t0)) if the hamiltonian is time independent. When the hamiltonian depends on time, US(t; t0) is more complicated, but all that we need is the differential equation that it obeys, d i U (t; t ) = H(t) U (t; t ) : (7) dt S 0 S 0 Thus we have I(t) = UI(t; t0) I(t0) ; (8) where iH0t −iH0t0 UI(t; t0) = e US(t; t0)e : (9) We have d i U (t; t ) = eiH0t[−H + H (t)]U (t; t )e−iH0t0 dt I 0 0 S S 0 iH0t −iH0t0 (10) = e VS(t)US(t; t0)e iH0t −iH0t iH0t −iH0t0 = [e VS(t)e ] e US(t; t0)e : Thus d i U (t; t ) = V (t) U (t; t ) (11) dt I 0 I I 0 with the initial condition UI(t0; t0) = 1 : (12) 2 3 Perturbative solution We can write an integral equation for UI(t; t0), Z t UI(t; t0) = 1 − i dτ VI(τ) UI(τ; t0) : (13) t0 The reader should check that if UI obeys this integral equation then it obeys both the differential equation and the boundary condition. We can think of this as follows. The operator UI(t; t0) tells us that the system evolves from time t0 to time t. The system can evolve in two ways. First, it may be that nothing happens: the \1" term. Second, it may be that the system evolves up to some time τ, then an interaction happens at time τ, then nothing happens from time τ to time t. We integrate over possible times τ. It is easy to solve this by iteration starting at UI(t; t0) = 1 so as to generate a solution that is a power series in V . We write 1 X (n) UI (t; t0) = UI (t; t0) ; (14) n=0 (n) where UI (t; t0) is proportional to exactly n powers of VI. Then (0) UI (t; t0) = 1 : (15) The integral equation gives Z t (n+1) (n) UI (t; t0) = −i dτ VI(τ) UI (t; t0) : (16) t0 (0) Substituting UI = 1, we get Z t (1) UI (t; t0) = −i dτ1 VI(τ1) : (17) t0 Then we get Z t Z τ2 (2) 2 UI (t; t0) = (−i) dτ2 dτ1 VI(τ2) VI(τ1) : (18) t0 t0 3 Continuing in this manner, we get Z t Z τ3 Z τ2 (n) n UI (t; t0) = (−i) dτn ··· dτ2 dτ1 VI(τn) ··· VI(τ2) VI(τ1) : (19) t0 t0 t0 Another way to write the integration is Z t Z t Z t (n) n UI (t; t0) = (−i) dτn ··· dτ2 dτ1 θ(τ1 < τ2 ··· < τn−1 < τn) t0 t0 t0 (20) × VI(τn) ··· VI(τ2) VI(τ1) : That is, we integrate over the times τj in the interval from t0 to t subject to the restriction that τ1 < τ2 ··· < τn−1 < τn : (21) 4 Shorthand notation There is a convenient shorthand for writing our result. We could write Z t Z t (n) n UI (t; t0) = (−i) dξn ··· dξ1 t0 t0 1 X × θ(ξ < ξ ··· < ξ < ξ ) (22) n! π(1) π(2) π(n−1) π(n) π2Sn × VI(ξπ(n)) ··· VI(ξπ(2)) VI(ξπ(1)) : This may seem perverse. I have just made up new variable names ξj and matched them up to the τj in according to the n! permutations of n objects. The n! terms each gives the same integral. It is important that I always put the operators V (τ) that have the later times to the left in the product of operators. I can write this as Z t Z t (n) n UI (t; t0) = (−i) dξn ··· dξ1 t0 t0 1 X × θ(ξ < ξ ··· < ξ < ξ ) (23) n! π(1) π(2) π(n−1) π(n) π2Sn × TVI(ξn) ··· VI(ξ2) VI(ξ1) : 4 Here the notation \T " instructs me to put the operators V (τ) that have the later times to the left in the product of operators. Then I can use X θ(ξπ(1) < ξπ(1) ··· < ξπ(n−1) < ξπ(n)) = 1 : (24) π2Sn This gives Z t Z t (n) n 1 UI (t; t0) = (−i) dξn ··· dξ1 TVI(ξn) ··· VI(ξ2) VI(ξ1) : (25) n! t0 t0 With the \T " notation, the whole series can be written as Z t UI(t; t0) = T exp −i dτ VI(τ) : (26) t0 This is a nice, compact expression. What it means is that one is supposed to expand the exponential and time-order the operators. That is, what it means is Eq. (20). 5 Two level system As an example, let us consider a two level system subject to a perturbing interaction that links the two levels and oscillates with frequency !, 0 γ V (t) = e−i!t : (27) S γ 0 Then the perturbing potential in the interaction picture is ei!1t 0 0 γ e−i!1t 0 V (t) = e−i!t : (28) I 0 ei!2t γ 0 0 e−i!2t where !2 is the unperturbed energy of the upper level, !1 is the unperturbed energy of the lower level. When we multiply the matrices, we obtain 0 γe−2i!~2t V (t) = ; (29) I γe−2i!~1t 0 where 1 !~2 = [! − (!2 − !1)] ; 2 (30) 1 !~ = [! − (! − ! )] 1 2 1 2 5 We take the initial time to be t0 = 0. With this perturbation, the evolution operator evaluated to first order is Z t UI(t; 0) = 1 − i dτ VI (τ) + ··· 0 t Z 0 e−2i!~2t = 1 − iγ dτ −2i!~1t + ··· 0 e 0 0 i [e−2i!~2t − 1] = 1 − iγ 2~!2 + ··· i [e2i!~1t − 1] 0 2~!1 0 i e−i!~2t[e−i!~2t − ei!~2t] = 1 − iγ 2~!2 + ··· i e−i!~1t[e−i!~1t − ei!~1t] 0 2~!1 1 −i!~2t 0 e sin(~!2t) !~2 = 1 − iγ 1 −i!~1t + ··· : e sin(~!1t) 0 !~1 (31) If the initial condition is 0 (0) = (32) I 1 then at later times I(t) = UI (t; 0) I(0) is given by sin(~! t) 0 2 −i!~2t 1 I(t) = − iγ e + ··· : (33) 1 !~2 0 This appears to be singular for! ~2 ! 0, but note that the factor sin(~!2t) vanishes for! ~2 ! 0 so that the singularity is cancelled. The probability to find the system in the excited state after time t is 2 2 2 2 sin (~!2t) jC2(t; !~2)j ≡ j 2 I (t) j = γ 2 : (34) !~2 2 This solution applies when γ is small. If we do not have γ =!~2 1 then first order perturbation theory is not accurate and we need the exact solution for the two level system with a harmonic perturbation, which is analyzed in Sakurai. 6 Behavior integrated over energy Let us consider a slightly generalized problem. Suppose that we have a perturbation −i!t VS(t) = e V0 (35) 6 and suppose that the perturbation can take the system from a starting state 1 with unperturbed energy !1 at time t0 = 0 to a range of states j that have unperturbed energies !j. Then the first order transition amplitude to get from state 1 to state j is Z t (1) j UI (t; 0) 1 = − i dτ j VI (τ) 1 0 Z t iH0τ −i!τ −iH0τ = − i dτ j e V0e e 1 (36) 0 Z t −i(!+!1−!j )τ = − i dτ e j V0 1 0 Let us define 1 !~ = [! − (! − ! )] j 2 j 1 (37) γj = j V0 1 Then Z t (1) −2i!~j τ j UI (t; 0) 1 = − iγj dτ e 0 γj = e−i!~j t[e−i!~j t − ei!~j t] (38) 2~!j γ j −i!~j t = − i e sin(~!jt) !~j Thus the probability to get to state j , evaluated in lowest order pertur- bation theory, is 2 2 2 sin (~!jt) j j I (t) j = jγjj 2 : (39) !~j The result in this form may be just what we want.
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