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).

Short Answer

Expert verified

cao-ca-o=-ihHab0oeigtcbtdt

But

cbt1.

Step by step solution

01

Given data

Given that

ca=-ihHabe-igtCb

And

cb=-ihHabe-igtCb

02

Solving the case of a time independent perturbation

ca=-ihHabe-igtCbAndcb=-ihHabe-igtCb

Differentiating with respect to t :

Cb=-ihHabiω0eiω0tCa-ihHabeiω0t-ihHabeiω0tcborCb=iω0cb-1h2Hab2cb.Letα2=1h2Hab2.Thencb-iω0Cb+α2cb=0

or

This is a linear differential equation with constant coefficients, so it can be solved by a function of the form :

λ2-iω0λ+α2=0;λ=12iω0±--ω02-4α2=i2ω0±ω,whereω=-ω02-4α2

The general solution is therefore

cbt=Aeiω0+ωt/2+Beiω0+ωt/2=eiω0t/2Aeiω0t/2+Beω0t/2,Or

cbt=eω0t/2Ccosω0t/2+Dsinω0t/2.Butcb0=0,soC=0,andhence

cb=t=Deω0t/2sinω0t/2

Then

cb=Diω02eω0t/2sinω0t/2+ω2eω0t/2cosω0t/2=ω2Deω0t/2cosω0t/2+iω0ωsinω0t/2=-ihHbaeω0t/2ca

ca=ihHbaω2e-ω0t/2Dcosω0t/2+iω0ωsinω0t/2.Butca0=1soihHbaω2

Conclusion:

cat=e-iω0t/2cosiω0t/2+iω0ωsiniω0t/2,cbt=2Habihωeiω0t/2siniω0t/2

Where

ω=ω02+4Hab2h2ca2+cb2=cos2ωt/2+ω02ω2sin2ωt/2+4Hab2h2ω2sin2ωt/2=cos2ωt/2+1ω2ω02+4Hab2h2sin2ωt/2

cos2ωt/2+sin2ωt/2=1

[In light of the Comment you might question the initial conditions. If the perturbation includes a factor θ(t) , are we sure this doesn’t alter and That is, are we sure and are continuous at a step function potential? The answer is “yes”, for if we integrate Eq. 9.13 from to ,

ca=-ihHabe-iω0tcb1,ca=-ihHabe-iω0tcacao-ca-o=-ihHab0oe-iω0tcbtdt

Butcbt1,

So the intergral goes to zero as o0,and hencecb-o=cao.cb

The same goes for of course

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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).

In Equation 9.31 assumed that the atom is so small (in comparison to the wavelength of light) that spatial variations in the field can be ignored. The true electric field would be E(r,t)=E0cos(krωt).

If the atom is centered at the origin, thenkr1 over the relevant volume,|k|=2π/λ sokr~r/λ1) and that's why we could afford to drop this term. Suppose we keep the first-order correction:

E(r,t)=E0[cos(ωt)+(kr)sin(ωt)].

The first term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of k.rlead to even more "forbidden" transitions, associated with higher multipole moments).

(a) Obtain the spontaneous emission rate for forbidden transitions (don't bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer:role="math" localid="1659008133999" Rba=q2ω5πϵ0c5|a|(n^r)(k^r)|b|2.

(b) Show that for a one-dimensional oscillator the forbidden transitions go from leveln to levelrole="math" localid="1659008239387" n-2 and the transition rate (suitably averaged over n^andk^) isR=q2ω3n(n1)15πϵ0m2c5.

(Note: Hereω is the frequency of the photon, not the oscillator.) Find the ratio of the "forbidden" rate to the "allowed" rate, and comment on the terminology.

(c) Show that the2S1S transition in hydrogen is not possible even by a "forbidden" transition. (As it turns out, this is true for all the higher multipoles as well; the dominant decay is in fact by two-photon emission, and the lifetime it is about a tenth of a second

Prove the commutation relation in Equation 9.74. Hint: First show that

[L2,z]=2ih(xLy-yLx-ihz)

Use this, and the fact that localid="1657963185161" r.L=r.(r×p)=0, to demonstrate that

[L2,[L2,z]]=2h2(zL2+L2z)

The generalization from z to r is trivial.

As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody radiation is the source). Show that at room temperature (T = 300 K) thermal stimulation dominates for frequencies well below 5×1012Hz , whereas spontaneous emission dominates for frequencies well above . Which mechanism dominates for visible light?

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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