Solve Equation 9.13 to second order in perturbation theory, for the general case ca(0)=a,cb(0)=bca=-ihH'abe-0tcb,cb=-ihH'bae-0tca

(9.13).

Short Answer

Expert verified

The second order perturbation theory for the equation is

cα(2)(t)=α-ibh0tHabc(tc)e-iω0tcdtc-αh20tHabc(tc)e-iω0tc0tcHbac(tcc)eiω0tccdtccdtccb(2)(t)=b-iah0tHabc(t¢)eiω0tcdtc-bh20tHabc(t¢)eiω0tcdtc0tcHbac(tcc)e-iω0tccdtccdtc

Step by step solution

01

Concept.

Consider a time dependent potential, we can solve the Schrodinger equation for this potential in a two state system if we split the Hamiltonian into a time independent part H°and time dependent part H', that is:

H=H°+H1

This is done in the section 9.1 to get the solution.

02

Solving the given equation to second order perturbation theory

Wehave,Zeroorder:cα(0)(t)=α,cb(0)(t)=bThecoefficientsaresolutionsofthecouplesODEs,thatis:Theequation9.13isc.a=-iHab'e-iω0tcb,c.b=-iHab'e-iω0tcαFirstorder;c.a=-iHab'e-0tbcα(1)(t)=α-ib0tHab'(t)e-0tdt'c.a=-iHab'e-0tαcb(1)(t)=b-0tHab'(t)e-0tdt'Secondorder:c.a=-iHab'e-0tb-0tHab'(t)e-0tdt'cα(2)(t)=α-ibHab'(t)e-0tdt'-α20tHab'(t)e-0tdt'0tHba'(t)e-0tdt'dt'Togetcbjustswitchab(whichentailsalsochangingthesignofω0)cb(2)(t)=b-iaHba'(t')e0tdt'-b20tHab'(t')e0tdt'0tHba'(t')e-0tdt'dt'

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Most popular questions from this chapter

The half-life of (t1/2)an excited state is the time it would take for half the atoms in a large sample to make a transition. Find the relation betweenrole="math" localid="1658300900358" t1/2andT(the “life time” of the state).

Solve Equation 9.13 for the case of a time-independent perturbation, assumingthatandcheck that

. Comment: Ostensibly, this system oscillates between “” Doesn’t this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this are not, and never were, Eigen states of the Hamiltonian—a measurement of the energy never yields. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end,are Eigen states of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T —this doesn’t affect the calculations, but it allows for a more sensible interpretation of the result.

ca=-ihHabeigtcb,cb=-ihHbaeigtca …(9.13).

The first term in Equation 9.25 comes from the eiωt/2, and the second from e-iωt/2.. Thus dropping the first term is formally equivalent to writing H^=(V/2)e-iωt, which is to say,

cbl-ihvba0tcos(ωt')eiω0t'dt'=-iVba2h0tej(ω0+ω)t'+ej(ω0-ω)t'dt'=--iVba2hej(ω0+ω)t'-1ω0+ω+ej(ω0-ω)t'-1ω0-ω(9.25).Hba'=Vba2e-iωt,Hab'=Vab2eiωt(9.29).

(The latter is required to make the Hamiltonian matrix hermitian—or, if you prefer, to pick out the dominant term in the formula analogous to Equation 9.25 forca(t). ) Rabi noticed that if you make this so-called rotating wave approximation at the beginning of the calculation, Equation 9.13 can be solved exactly, with no need for perturbation theory, and no assumption about the strength of the field.

c.a=-ihHab'e-iω0tcb,c.b=-ihHba'e-iω0tca,

(a) Solve Equation 9.13 in the rotating wave approximation (Equation 9.29), for the usual initial conditions: ca(0)=1,cb(0)=0. Express your results (ca(t)andcb(t))in terms of the Rabi flopping frequency,

ωr=12(ω-ω0)2+(Vab/h)2 (9.30).

(b) Determine the transition probability,Pab(t), and show that it never exceeds 1. Confirm that.

ca(t)2+cb(t)2=1.

(c) Check that Pab(t)reduces to the perturbation theory result (Equation 9.28) when the perturbation is “small,” and state precisely what small means in this context, as a constraint on V.

Pab(t)=cb(t)2Vab2hsin2ω0-ωt/2ω0-ω2(9.28)

(d) At what time does the system first return to its initial state?


We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption?

Develop time-dependent perturbation theory for a multi-level system, starting with the generalization of Equations 9.1 and 9.2:

H^0ψn=Enψn,   ψnψm=δnm (9.79)

At time t = 0 we turn on a perturbation H'(t)so that the total Hamiltonian is

H^=H^0+H^'(t)(9.80).

(a) Generalize Equation 9.6 to read

Ψ(t)=ca(t)ψaeiEat/+cb(t)ψbeiEbt/(9.81).

and show that

c˙m=incnHmn'ei(EmEn)t/ (9.82).

Where

Hmn'ψm|H^'|ψn (9.83).

(b) If the system starts out in the state ψN, show that (in first-order perturbation theory)

cN(t)1i0tHNN'(t')dt'(9.84).

and

cm(t)i0tHmN'(t')ei(EmEN)t'/dt',   (mN)(9.85).

(c) For example, supposeH^'is constant (except that it was turned on at t = 0 , and switched off again at some later time . Find the probability of transition from state N to state M (MN),as a function of T. Answer:

4|HMN'|2sin2[(ENEM)T/2](ENEM)2 (9.86).

(d) Now supposeH^'is a sinusoidal function of timeH^'=Vcos(ωt): Making the usual assumptions, show that transitions occur only to states with energy EM=EN±, and the transition probability is

PNM=|VMN|2sin2[(ENEM±ω)T/2](ENEM±ω)2 (9.87).

(e) Suppose a multi-level system is immersed in incoherent electromagnetic radiation. Using Section 9.2.3 as a guide, show that the transition rate for stimulated emission is given by the same formula (Equation 9.47) as for a two-level system.

Rba=π3ϵ02||2ρ(ω0)Rb (9.47).

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