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Degenerate perturbation

  1. Aug 18, 2009 #1
    Hi,

    I know that for degenerate states, we need to apply degenerate perturbation theory by looking at the perturbative hamiltonian in the subspace of the degenerate states.

    What then if the states still degenerate after we cast them the generate subspace. Is there a way to find the zero order 'good' eigenstate?

    Thanks.
     
  2. jcsd
  3. Aug 19, 2009 #2
    I borrow this thread to ask a question that is similar to this one.
    What does it mean that an operator has a degenarate spectrum and how can we see that if the operator is represented by a matrix. Thanx alot and I hope I'm not stepping on anybodys toe by using this thread.
     
  4. Aug 19, 2009 #3

    Avodyne

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    Welcome to the forum, hermitian!

    I believe you can use any basis you like in the subspace if the perturbation does not lift the degeneracy.

    A "degenerate spectrum" means that at least one eigenvalue occurs more than once. This would then also be true of the matrix representing the operator.
     
  5. Aug 19, 2009 #4
    thnx that helped alot. Let me see if I got this right: Eigenvalues are the observables and the eigenvectors are the states of the system. what would degenerate spectrum mean physically.

    thnx
     
  6. Aug 19, 2009 #5

    Avodyne

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    It would mean that knowing the eigenvalue is not sufficient information to determine the state.
     
  7. Aug 19, 2009 #6
    thanks Avodyne,

    i spend sometime to convince myself that I can use any basis I like in the degenerate subspace....
     
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