d 2 2 A = [A, d 2 A = 1 2 [[A, H ], dt 2 Visualizing IVR. [Chem. Phys. Lett. 320, 553 (2000)] 5.74 RWF Lecture #

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1 MIT Department of Chemistry 5.74, Spring 004: Introductory Quantum Mechanics II Instructor: Prof. Robert Field 5.74 RWF Lecture # Visualizing IVR. [Chem. Phys. Lett. 30, 553 (000)] Readings: Chapter and , The Spectra and Dynamics of Diatomic Molecules, H. Lefebvre- Brion and R. Field, nd Ed., Academic Ps, 004. Last time: Looking for causes of dynamics d dt 1 A = [A, H] ih d A = 1 [[A, H ], H] dt ih discussed 3 mode, onance model (H O) d derived many operator relationships for Ν j found that the cause of dt Ν Ν j is transfer rate operators # of quanta changed d e.g. N = Ω Ω dt ih k 11, H () = (Ω + Ω ) 1 + (Ω + Ω ) 1 3 aa k 1, aa 1

2 5.74 RWF Lecture #15 15 The Two-Level Problem Again E 0 δe 0 H = + + Ω+ Ω 0 E 0 δe E = E A + E B δe = E A E B 0 0 Ω= 0 V Ω We are Ω = = 0 0 V * 0 going to let V be real Suppose our pluck at t = 0 is ψ A Ψ(0) = A = cos θ + sin θ verify by plugging in + =cosθ A + sinθ B = sinθ A + cosθ B () Ψ 0 = cos θ A + cosθ sin B + sin = (cos θ + sin θ ) A. θ θ A cos θ sin θ B checks Now put in t-dependence: Ψ () 0 = cos θ e ie + t h + sinθ e ie t h in the +, basis and E ± = E ± =[δe + V e ie + t h Ψ () t = cosθ (cosθ A + sinθ B ) sinθ e ie t h ( sinθ A + cosθ B ) θ e ie + t h θ e ie t h = (cos + sin A ie + t h ie t h +cos θ sinθ ( e e ) B ]1 θ +δe 1 cos θ = sin = ) δe 1

3 5.74 RWF Lecture # Let us ask several questions for this maximally simple problem that will give us insights into the relationships between observable properties and the fundamental control parameters in H eff. 1. H E is independent of t and E = E A if Ψ () 0 = A. This is going to tell us how to measure the energy of a zero-order state.. What is H diag t = E diag () t t Ω Ω ()? t H = Ω + Ω t = E And what do E diag (t) and E (t) tell us? Evaluate H. We can evaluate this in either the A, B, or +, basis set. In the A,B basis set simplify 4 ρ()= t [cos θ + sin 4 θ + cos θ sin θ ( cosω + t)] A A +[cos θ sin θ ( cosω + t)] B B 3 θ θ 1 e iω + t) 3 θ θ 1 e +iω + t ) +iω + t +[cos sin ( + cosθ sin 3 θ ( e 1 )] A B iω + t +[cos sin ( + cosθ sin 3 θ ( e 1 )] B A V ρ()= t 1 ( V + δe ) ( 1 cosω + t) A A V + ( V + δe ) ( 1 cosω + t) B B VE δ iv sin ω + t + ] 1 [ A B V + δe VE δ iv sinω + t + ] 1 B A [ V + δe + ( V + δe ) ( 1 cosω t) + + ( V + δe ) (1 cosω t)

4 5.74 RWF Lecture # in the +, basis ρ()= t cos θ sin θ sin θ cosθ e iω + t + sin θ cosθ e +iω + t + +δe δe = δe 1 iω [e + t + e +iω + + t + ] 4 Want H t = Trace(ρH ) = ρ jk H kj jk, = ρ 11 H 11 + ρ 1 H 1 + ρ 1 H + ρ H 1 = 0 in +, basis because H 1 = H 1 = 0 in the +, basis +δe δe H t = E E + + E ± = E ± δe H t = E + = E + δe = E A! So we have our first really important ult. If you know your pluck perfectly, then, for any H, H t = E pluck independent of time. To determine the energy of the plucked zero-order state, simply measure the intensity weighted average energy in the absorption spectrum.

5 5.74 RWF Lecture # Now let us divide H into H diag and H = Ω + Ω E diag ()= t H diag t E ()= t Ω+ Ω Ω t ()must t be independent of t, but E diag and E are certainly time of course E = E diag ()+ t E dependent. For our set. Ψ() 0 = A pluck of a two-level system, we want to exps both H and ρ in the A, B Basis ρ()= t 1 V ) A ( V + δe ) ( 1 cosω + t A V ) B VE δ iv sinω + t + ] 1 A B [ V + δe VE δ iv sinω + t + ] 1 B A [ V + δe + ( V + δe ) ( 1 cosω + t B + ( V + δe ) ( 1 cosω t) + + ( V + δe ) ( 1 cosω t)

6 5.74 RWF Lecture # Thus H diag t diag AA diag diag AA BB BB = Trace ρh = ρ H + ρ H = ρ AA E E +δ E + ρ BB E = E δ E (ρ AA + ρ )E + (ρ AA ρ )δe A B BB BB V δ E = E V + δe (1 cosω + t) + δe at t = 0 H diag = E + δe = E A H t = ρ AB H BA + ρ BA H AB V δ E = V + δe ( 1 cosω + t) H diag t at t = 0 H = 0 t + H t = E + δe = E A In a more general situation, when there are several onances, we have a polyad H eff that is larger than, we need help sorting out which onance is most important for the post-pluck dynamics of each conceivable pluck. See HLB-RWF Chapter and Chem. Phys. Lett. 30, 553 (000).

7 5.74 RWF Lecture # Let E ()= t Ω j + Ω j j sum over onances we still have E E diag ()+ E () = t, j t j two useful quantities: E, j = lim T T 1 T 0 dt( Ω k + Ω k and fractional importance, for a specific pluck, of the j-th onance ) f j = E, j E What do we do with E, j, E, j, and f j? We attempt to correlate what we see in the spectrum to the various onance terms that could cause the observed effects. We look for a pluck which is maximally sensitive to one particular onance term. In order to do this, we need a working model for H eff. We then use insights guided by the onance operator dynamics. See Figure The survival probability of the plucked state is directly observable because it is the F. T. of the pluck-related piece of the spectrum. The dynamical featu in the survival probability encode Fourier components and amplitudes associated with each of the dynamically important onance operators. We want to know what causes each of the dynamical featu pent in any do-able experiment. Strategy: use computed simple dynamical quantities to sharpen up corpondences with observable dynamical featu.

8 5.74 RWF Lecture # Figu removed due to copyright reasons.