1 Rabi oscillations. Physical Chemistry V Solution II 8 March 2013

Transcription

1 Physcal Chemstry V Soluton II 8 March Rab oscllatons a The key to ths part of the exercse s correctly substtutng c = b e ωt. You wll need the followng equatons: b = c e ωt 1 db dc = dt dt ωc e ωt. Keepng n mnd that the cosne can be expressed n terms of complex exponentals cosωt = 1 e ωt + e ωt. 3 Settng b 1 = c 1 and ω 1 = 0 as requested, Equaton 1.96 becomes dc 1 dt = V 1 eωt + e ωt c e ωt = V e ωt c V 1 c 4 where we have dropped the e ωt term due to the rotatng wave approxmaton mentoned n the problem. Carryng out the same procedure on Equaton 1.97 from the lecture notes gves dc dt ωc dc dt ωc e ωt = V 1 dc dt eωt + e ωt c 1 + ω c e ωt, = V 1 eωt + 1c 1 + ω c and nally = V 1 c 1 + ω ωc. 5 b In the resonant case ω = ω the problem reduces to a par of symmetrc equatons dc 1 dt dc dt = V 1 c 6 = V 1 c 1. 7 Derentatng Equaton 7 gves d c dt = V 1 dc 1 dt. 8 1 / 8

2 Physcal Chemstry V Soluton II 8 March 013 Insertng Equaton 6 and takng nto account the equalty V 1 = V 1 d c dt = V1 c 9 we see that the tme behavour of c s descrbed by a harmonc oscllaton, whose general soluton reads V1 c t = A sn t + ϕ. 10 Snce the derental equaton par s symmetrc, c 1 must harmoncally oscllate as well. Consderng normalzaton c 1 + c = 1 we end up wth where ϕ s determned by the ntal condtons. V1 c 1 t = cos t + ϕ 11 V1 c t = sn t + ϕ, 1 c c t = sn V1 t + ϕ = 1 cosv 1t + ϕ. 13 The characterstc oscllaton frequency s the Rab frequency V 1. Addtonal Informaton: Physcal meanng and applcatons. In ther work I. Gerhardt et al. were able to montor Rab oscllatons between the ground and the excted state of dbenzantanthrene DBATT [1]. Rab cycles were recorded durng the passage of a long several ns pulse on resonance and the subsequent decay of the excted state populaton by spontaneous emsson to vbratonal levels of the electronc ground state see Fg.1 left. Ths emsson was detected as the Stokes uorescence sgnal whch, to a very good approxmaton, s proportonal to the populaton of the excted state. Ths means that durng the Rab oscllaton the excted state populaton can be montored by measurng the ntensty of the Stokes uorescence sgnal. The result of these studes s presented n Fgure 1. The dashed lne s the exctaton pulse energy. The red lne s the measured, oscllatng Stokes uorescence whch corresponds to an oscllatng excted state populaton. The oscllaton cycles are observed only durng the passage of the laser pulse,.e. whle the lght-matter nteracton s domnated by absorpton and stmulated emsson. The spontaneous emsson channel s used only to record the excted state populaton. After the passage of the laser pulse, the excted state only relaxes through the spontaneous emsson channel. Ths s typcally an exponental decay of an excted state, whch can also be seen n the measured curve. / 8

3 Physcal Chemstry V Soluton II 8 March 013 Fgure 1: Left: Level scheme of a sngle DBATT. Rght: Raw Stokes-shfted uorescence of a sngle molecule red curve and a theoretcal t black curve as a functon of delay wth respect to the optcal pulse. The blue dashed lne dsplays the measured exctaton pulse ntensty [1]. Two oscllaton cycles are observed at ths partcular pulse length ndcated n Fgure 1. One Rab cycle s completed after 3 ns. [1] I. Gerhardt, G. Wrgge, G. Zumofen, J. Hwang, A. Renn, and V. Sandoghdar, Phys. Rev. A 79, Ladder operator formalsm a ˆX can be regarded as the real part of â, ˆP as ts magnary part. Thus ˆX = mω â + â 14 mω ˆP = â â 15 b [â, â ] = ââ â â = mω = mω mω = 1 ˆX + ˆP m ω mω [ ˆX, ˆP ] ˆX ˆP m ω mω [ ˆX, ˆP ] 16 3 / 8

4 Physcal Chemstry V Soluton II 8 March 013 c â â = mω = 1 ω = 1 ω ˆX + ˆP m ω + mω ˆX H ω mω [ ˆX, ˆP ] + ˆP m + ω [ ˆX, ˆP ] The Schrödnger equaton cann generalbe wrtten lke 17 H ψ = ψ 18 t where ψ s not necessarly an egenstate and can also be tme-dependent. The tmendependent case s concerned wth the egenstates of the Hamltonan: H φ = E φ 19 As can be seen from Equaton 17, the tme-ndependent Schrödnger equaton for our case reads lke: where φ n are egenstates wth energy E E n. ω â â + 1 φ = E φ 0 d We want to know the energy of state â n,. e. calculate Hâ n. Our goal s to commute H and â so that we can reduce the problem to the known egenvalue relaton of n. Hâ n = ω commutator = ω = ω = â ω â ââ + â = âh ω n = âe n ω n = E n ωâ n = E n 1 â n n ââ â [â, â ]â + â ââ â â + â n â â n n = E n 1 n / 8

5 Physcal Chemstry V Soluton II 8 March 013 e Have a look at the arbtrary state φ = â ψ. The norm φ φ of any state s postve by denton φ φ 0. From equaton 6 and 7 gven n the exercse sheet one can see that â = â and thus â ψ = âψ = ψ â. 3 We conclude âψ âψ = ψ â â ψ = ψ ˆn ψ 0 4 f From problem.d we see that â n must be proportonal to n 1,. e. â n = c n 1, c C 5 Takng the norm on both sdes leads to n â â n = c n 1 n 1 n ˆn n = c n 1 n 1 n n n = c n 1 n 1 n = c c = n 6 Applyng ths result to the ground state: â 0 = 0 as c = Probng and controllng vbratonal wave packets a Assume that the fs pulse creates a coherent superposton of states φ n q of the 1D harmonc oscllator wth coecents c n t, where q s the spatal vbratonal coordnate and n s the assocated quantum number. From the spatal representaton of ψ0, ψq, 0 = q ψ0 = n c n0φ n q, and the expresson of φ n q n Equaton we obtan ψq, 0 = e q / n c n 0 n n! H nq, 8 5 / 8

6 Physcal Chemstry V Soluton II 8 March 013 where H n q are Hermte polynomals. Replacng c n 0 wth 1/ n n! n the prevous equaton yelds ψq, 0 = e q / n 1 n n! H nq. 9 We easly recognze that the expresson contans the exponental generatng functon of Hermte polynomals H n q, expqw w = n H nqw n /n!, wth w = 1/. Hence ψq, 0 = 4 e π e 1 q b After the pulse rradaton the wave packet evolves accordng to the Hamltonan of the 1D harmonc oscllator to yeld ψq, t = n c n tφ n q = n e Ent/ c n 0φ n q, 31 where E n = ωn+1/ are the energy levels. A few smple steps lead us to the expresson ψq, t = e q / e ωt/ n H n q n! e ωt n. 3 Once more we take advantage of the exponental generatng functon to derve the more compact and elegant form: ψq, t = e ωt/ e ωt exp 4 e 1 q exp ωt. 33 The wave packet has a center of mass q 0 that moves back and forth as q 0 = e ωt,.e. the moton s perodc. c At tme t we just need to supermpose the result of Equaton 30 wth that of Equaton 33 for t = t. Recallng that Ψ t, φ = e φ ψ0 + ψ t we obtan Ψq; t, φ = q Ψ t, φ = e φ 4 e π e 1 q 1 + e ω t/ d The uorescence sgnal s proportonal to Ψ t, φ. e ω t exp 4 e 1 q exp ω t. 34 We smplfy Equaton 34 by assumng ω t 1. Under ths condton we can approxmate usng the power seres expanson e ω t 1 ω t. Wth ω t 1 ths becomes e ω t 1 and we can wrte Ψq; t, φ e φ 4 e π e 1 q e π e ω t e 1 q 1+ω t / 8

7 Physcal Chemstry V Soluton II 8 March 013 A few more algebrac operatons and droppng terms of the order of ω t lead to Ψq; t, φ = 4 e π e 1 q 1 e qω t + e φ. 36 From ths we get Ψ t, φ : Ψ t, φ = Ψ t, φψ t, φ e = 4 π e 1 q 1 e qω t + e φ e 4 π e 1 q 1 e qω t + e φ 37 e = π e q 1 e qω t + e φ e qω t + e φ 38 e = π e q e qω t+φ + e qω t+φ e = π e q 1 + cosqω t + φ 40 Ths means, that the uorescence s a snusodal functon of the delay tme t between the pulses and also a functon of the phase φ. Addtonal Informaton: Physcal meanng and applcatons. In contrast to the long some ns resonant exctaton n the case of Rab oscllaton, Brnks et al. appled two broad-band, femtosecond pulses separated by a tme t n order to generate vbratonal wave packets and to observe ther nterference []. They also examned how the phase aects the uorescence sgnal. Accordng to the result from exercse d the uorescence sgnal from the nterference of two vbratonal wave packets has to be a cosne functon. Moreover, the phase φ sgncantly determnes ths nterference. Ths s exactly what Brnks et al. have observed. Fgure shows the uorescence sgnal measured at derent delay tmes t for n-phase φ = 0 and n-antphase φ = π exctaton pulses. One can clearly see the oscllatng behavor as predcted by the cosne functon and also the nuence of the phase factor. In ths case, the n-phase and n-antphase exctaton pulses lead to an nverted molecular response. Vbratonal relaxaton, not consdered n the exercse, s supermposed on the oscllatons. [] Daan Brnks et al. Nature Letters 465, / 8

8 Physcal Chemstry V Soluton II 8 March 013 Fgure : Sngle-molecule uorescence ntensty as a functon of the tme delay between two n-phase φ = 0 and n-antphase φ = π exctaton pulses. Some molecules present uctuatons even for tme delays t as long as 10 fs and wth more than one frequency component []. 8 / 8