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Probability of perturbed harmonic oscillator

  • Thread starter 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.
 

Answers and Replies

  • #2
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First step, you will need to find the first order expansion using perturbation theory. Use the perturbed potential:

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

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

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

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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