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Perturbation theory, second-order correction - When does the sum stop?

  1. Oct 4, 2014 #1

    DataGG

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    I've no idea if I should be posting this here or in the general forums.
    This is not really an exercise as much as an example. I'm not understanding something though:

    1. The problem statement, all variables and given/known data


    Using perturbation theory, find the exact expression for the energy given by the hamiltonian:

    $$\hat{H}=\hat{H_0} + \hat{H_p} = - \dfrac{\hbar}{2m}\dfrac{d^2}{dX^2} + \dfrac{1}{2}m\omega \hat{X}^2 + q\epsilon \hat{X}$$

    I'm only having problems with the second-order correction.

    The formula for the second-order correction is below. My question is: How do we find ##m## and ##n##? Is their inequality the only restriction? Can they both go to infinity? How do I find where the sum stops?

    2. Relevant equations

    $$E^{(2)}_n=\sum _{m \ne n} \dfrac{|<\phi _m |\hat{W}|\phi _n >^2 |}{E_n ^{(0)} -E_m^{(0)}} $$

    3. The attempt at a solution

    This is an example. The full exercise is explained in a book. However, in order to compreehend one of the steps, I need to find out how to find out where the sum stops.
     
    Last edited: Oct 4, 2014
  2. jcsd
  3. Oct 4, 2014 #2

    DataGG

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    Oh.. I think I can answer my own question.

    The sum stops when the numerator is zero. In the above example, it's gonna be zero everytime ##m## and ##n## differ more than 1. The sum is, however, supposed to be an infinite sum, correct?
     
  4. Oct 4, 2014 #3

    vela

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    ##n## is fixed, and the sum is over ##m##. You need to sum over all states except for the nth state. In this particular case, you're right that almost all the terms vanish, so you're only left with one or two terms.
     
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