Time-Dependent Perturbation of a 1D Infinite Square Well

  • #1

Homework Statement


At t < 0 we have an unperturbed infinite square well. At 0 < t < T, a small perturbation is added to the potential: V(x) + V'(x), where V'(x) is the perturbation. At t > T, the perturbation is removed. Suppose the system is initially in the tenth excited state if the unperturbed potential.

What are the possible results of energy measurements at t = T/2, 3T/2, and infinity. Explain how you would calculate the probability of each result. Indicate whether or not these probabilities should depend on time.

Homework Equations



ψn = A sin(nπx/a)
φ(x,t) = c(t)φ(x)e-iEt/ħ
E = (nπħ)2/2ma2.

The Attempt at a Solution


Since the system begins only in the tenth excited state of the unperturbed potential, there is no time dependence present via the wavefunction. And because the perturbation itself has no time dependence I believe that in all cases the probabilities should not depend on time for any t > 0. As t approaches infinity, the perturbation is long since removed, and I think the energy should be the same as it would be for any stationary wave function in the square well: E = (10πħ)2/2ma2.

What I'm not certain of is exactly how to determine the possible results of the energy measurements of the system for t = T/2 and 3T/2, or what these results would be. I am also not sure if my time dependence reasoning above is correct.
 

Answers and Replies

  • #2
DrClaude
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As t approaches infinity, the perturbation is long since removed, and I think the energy should be the same as it would be for any stationary wave function in the square well: E = (10πħ)2/2ma2.
The perturbation is sudden: that corresponds to a kick. Do you really think this wouldn't change the energy of the system?

An any case, nothing beats doing the calculation. Apply TDPT and see what you get.
 
  • #3
The perturbation is sudden: that corresponds to a kick. Do you really think this wouldn't change the energy of the system?

An any case, nothing beats doing the calculation. Apply TDPT and see what you get.
Hmm. I guess that makes sense about the sudden perturbation removal causing an energy change. I'll give it go with applying TDPT for the first two cases t= T/2 and 3T/2. When the perturbation is removed, can the precise energy really be calculated? Given that we don't really know exactly how much the kick perturbs the system. I figure it would raise the energy, but I don't know what I would use in the TDPT to represent the kick.

Thank you for your help.
 

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