Probability of perturbed harmonic oscillator

phyzzydud
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An experimenter has carefully prepared a particle of mass m in the first excited state of a one dimensional harmonic oscillator. Suddenly he coughs and knocks the center of the potential a small distance, a, to one side. It takes him a time T to recover and when he has done so he immediately puts the center back where it was.
A) Find, to lowest order in a, the probability that the oscillator will now be in its ground state.
B) Find, to lowest order in a, the probability that the oscillator will now be in its second excited state.


I know that there are two sudden approximations that can be made, the first after he coughs, and the second after he moves the center back. I was thinking Hi = -hbar^2/2m * d^2 phi/dx^2 + 1/2 m w^2 x^2. For the first sudden approximation, the x^2 would be replaced by (x-a)^2, and then it would return to x^2 for the second approximation. I think I'm not sure where the a fits into the calculation, and how to find the answer to the lowest order in a.
 
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First step, you will need to find the first order expansion using perturbation theory. Use the perturbed potential:

V = \frac{1}{2}m\omega^2 (x-a)^2- \frac{1}{2}m\omega^2x^2

And expand the the wavefunction to first order in 'a' by finding,

|n'> = |n>+\sum_{k\ne n}|k>\frac{<k|V|n>}{E_n - E_k}

Once you have that you will have to make use of the fact that the change in the potential was sudden.
 
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To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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