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Degenerate Perturbation Theory

  1. Feb 15, 2012 #1

    I am reading about Degenerate Perburbation Theory, and I have come across a question. We all know that the good quantum numbers in DPT are basically the eigenstates of the conserved quantity under the perburbation. As Griffiths he says in his book: "... look around for some hermitian operator A that communtes with H0 and H'". Now say I form a linear combination ψ of my (unperturbed) degenerate states. It satisfies

    H0ψ = E0ψ.

    Assuming it is the "correct" linear combination in the sense that it diagonalizes my perturbation H', I can write

    (H0+H')ψ = Eψ,

    where E = E0+ΔE. But this is my question: If the relation (H0+H')ψ = Eψ holds, then what about any first order correction to the wavefunction due to the perturbation? An example of this is the Helium-atom. Here we find that the good linear combinations are symmetric and anti-symmetric, but they are only correct to zeroth order, right?

    Last edited: Feb 15, 2012
  2. jcsd
  3. Feb 15, 2012 #2
    I may be missing your point here, but: If we could really find simultaneous eigenstates of H0 and H' we wouldn't need to use perturbation theory, because these states would be the eigenstates of the full Hamiltonian, which are what we are trying to find.

    When we use degenerate pertubation theory, it is because there is some subspace of states that are degenerate with respect to H0. We then take linear combinations of these states that diagonalize H' *within that subspace*--that is, the matrix elements of H' between these linear combinations should be zero. But the matrix element of H' between one of these linear combinations and some other state that isn't in the degenerate subspace will in general be nonzero. Thus in general these linear combinations we have taken will *not* be eigenstates of H'. We haven't diagonalized H', we've only diagonalized it within the degenerate subspace.
  4. Feb 15, 2012 #3
    Ah, I see. Thanks for clarifying that for me. So these linear combination are still only zero-order solutions to the perturbation?
  5. Feb 15, 2012 #4
    Yes; to get the first-order perturbed eigenstates you still have to do the regular perturbation theory calculation, just now you do it with your carefully chosen basis of H0 eigenstates.
  6. Feb 15, 2012 #5
    Thanks for that! OK, so in the case of e.g. the exicted states of Helium where we find the antisymmetric and symmetric linear combinations, then they are still only correct to zero order. But how does this justify that we just tack on a symmetric/antisymmetric spin wavefunction to these? I mean, if they are only correct to zeroth order, it must be very imprecise.
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