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Confused about perturbation theory with path integrals

  1. Jun 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi, I am just trying to wrap my head around using path integrals and there are a few things that are confusing me. Specifically, I have seen examples in which you can use it to calculate the ground state shift in energy levels of a harmonic oscillator but I don't see how you can find the shift for any other energy level. I am aware you don't need to use a path integral for this but I am hoping to get a better understanding of them by doing this.

    2. Relevant equations
    I know that you can calculate the energy shift in the ground state using the partition function by using

    [itex] E_0=-lim_{\beta\rightarrow\infty}\frac{1}{\beta} log K_E[J] |_{J=0}[/itex]

    I know if I am using the Hamiltonian
    [itex]\mathcal{H}=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^2\hat{q}^2+\gamma\hat{q}^4[/itex]

    then to first order you can just write ## K_E[J] ## as

    [itex] K_E[J]=K_E^0[J]-\gamma \int d\tau (\frac{\delta}{\delta J(\tau)})^4K^0_E[J]|_{J=0}[/itex]

    where

    [itex] K_E^0[J]=\frac{C}{2}\int d\tau d\tau'J(\tau)G(\tau,\tau')J(\tau')[/itex]

    and G is the green's function.

    3. The attempt at a solution

    So you can evaluate this all and work it out for the ground state which I don't have a problem with, but what confuses me is how this works for anything that isn't the ground state. Clearly the ##\beta\rightarrow\infty## limit is always going to put you there.

    I understand that you probably need to use

    [itex] \langle 0 |\hat{a}e^{-\beta H} \hat{a}^\dagger |0\rangle [/itex]

    In some capacity but I don't really see where to proceed from there. Do you have to worry about commuting ##\hat{a}^\dagger## through the exponential?

    I don't really know where to look for guidance with this sort of thing so any help would be appreciate. Thank you very much!
     
  2. jcsd
  3. Jun 27, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jun 27, 2015 #3
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