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Degenerate Perturbation Theory Wavefunction Correction

  1. Aug 30, 2011 #1
    Hi,

    If we have a non degenerate solution to a Hamiltonian and we perturb it with a perturbation V, we get the new solution by

    [tex] |\psi_{n}^{(1)}> = \sum \frac{<\psi_{m}^{(0)}|V|\psi_{n}^{(0)}>}{E_n^{(0)} - E_m^{(0)}}\psi_m^{(0)}[/tex]

    where we sum over all [itex]m[/itex] such that [itex]m\neq n[/itex].

    When we do the same for a degenerate perturbation theory, we simply exclude all the [itex]\psi_m[/itex] that belongs to the degenerate subspace. That is

    [tex] |\psi_{n}^{(1)}> = \sum \frac{<\psi_{m}^{(0)}|V|\psi_{n}^{(0)}>}{E_n^{(0)} - E_m^{(0)}}\psi_m^{(0)}[/tex]

    only this time the summation exculdes m if [itex]\psi_{m}[/itex] is in the degenerate subspace. Am I right so far?

    If so, my question is this. Assuming we are taking the "good" states that Griffiths mentions (those which are the eigenstates of V in the degenerate subspace), then each one has a different energy correction. But the wavefunction correction for all of them is exactly the same according to the expression given above. That seems a little fishy. The first order corrections to each of the "good" states is exactly the same but the first order energy correction is different?

    Thank you for your help.
     
  2. jcsd
  3. Aug 30, 2011 #2

    DrDu

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    The matrix elements [tex]\langle \psi_m^{(0)}|V|\psi_n^{0)}\rangle[/tex] will be different for different "good states" n from the degenerate subspace, no?
     
  4. Aug 30, 2011 #3
    Oh dear, I don't know what I was thinking! Sorry about that. It is obviously different for different n. Thank you DrDu.
     
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