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Particle on a ring with perturbation

  1. Jul 10, 2014 #1
    So I'm trying to solve old qualifying exam problems, one of which is a particle on a ring with a constant electric field perturbation. The un-perturbed problem is straightforward, and we then add a constant electric field in the x-direction (the ring lies in the xy-plane) of magnitude E. Therefore, we add [itex] -ReE \cos \phi [/itex] as a perturbing term to the Hamiltonian. R is the radius of the ring and phi is the angular coordinate on the ring.

    We then ask for the first (non-zero) energy correction to the original ground state. Because of the cosine term, the first order correction vanishes, I believe, so we need to look to second order or beyond.

    However, the second order correction involved a sum of terms of the form:

    [tex] \langle \psi_{k} | H_{1} | \psi_{0} \rangle =\alpha \int_{0}^{2 \pi} d \phi e^{ik \phi} \cos \phi [/tex]

    Unless I'm screwing up badly, I believe the above terms are also zero. So am I making a mistake here? Or am I correct and I need to look for third order corrections to get a non-zero term?
    Last edited: Jul 10, 2014
  2. jcsd
  3. Jul 10, 2014 #2
    I don't think this integral is zero for all ##k##. Write ##\cos \phi = (e^{i\phi} + e^{-i\phi})/2##.
  4. Jul 10, 2014 #3
    No? I agree it certainly isn't zero in general, but k must be an integer thanks to the periodic boundary conditions on the un-perturbed eigenfunctions. Won't it always vanish when k is an integer? Thanks a lot.
  5. Jul 10, 2014 #4
    Oh wait...it doesn't vanish for k=1, right? It will vanish for all other integral values of k?
  6. Jul 10, 2014 #5
    It doesnt vanish for k=-1 as well but negative integers are out of play anyway.
  7. Jul 10, 2014 #6
    Awesome, thank you. So the point is that the infinite sum which is, in general, required for the 2nd order perturbation correction, simply contains one term?
  8. Jul 11, 2014 #7
    No, you have to consider negative integers too; ##e^{ik\phi}## and ##e^{-ik\phi}## are both energy eigenstates of the unperturbed system.

    Two terms; see above.
    Last edited: Jul 11, 2014
  9. Jul 11, 2014 #8
    But since those two distinct eigenfunctions have the same energy, I would need to use the degenerate version of 2nd order perturbation?
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