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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

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).

An electron in the n=3,l=0,m=0state of hydrogen decays by a sequence of (electric dipole) transitions to the ground state.

(a) What decay routes are open to it? Specify them in the following way:

|300|nlm|n'l'm'|100.

(b) If you had a bottle full of atoms in this state, what fraction of them would decay via each route?

(c) What is the lifetime of this state? Hint: Once it’s made the first transition, it’s no longer in the state |300\rangle∣300⟩, so only the first step in each sequence is relevant in computing the lifetime.

For the examples inProblem 11.24(c) and (d), calculate cm(t)to first order. Check the normalization condition:

m|cmt|2=1,

and comment on any discrepancy. Suppose you wanted to calculate the probability of remaining in the original state ψN ; would you do better to use |cNt|2,or1-mN|cmt|2 ?

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

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free