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Degenerate basis, incomplete set of operators

  1. Apr 8, 2009 #1


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    I may have heard this or understood this incorrectly, so if i am asking the wrong question, feel free to correct me. As I understand it, if you have a degenerate set of simultaneous eigenvectors, you haven't specified a complete set of operators. For example, the hydrogen atom. You typically have the Hamiltonian, angular momentum, spin in elementary discussions. However, as I understand it, the eigenfunctions aren't complete and a final operator is needed (I forget the name). The thing I don't understand is a) how you would determine that final operator and b) how exactly adding operators would even help to prevent a set of eigenvectors from being degenerate.
  2. jcsd
  3. Apr 8, 2009 #2


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    I'm guessing you mean something involving the Laplace-Lenz-Runge vector?

    Part (b) is just the standard spectral theorem from linear algebra. A single self-adjoint
    operator's eigenspaces span the original space. Then, given another self-adjoint operator
    that commutes with the first, you can re-apply the theorem to each of these eigenspaces.
    For the finite-dimensional case, this process must eventually terminate, given enough
    mutually-commuting self-adjoint operators. (The inf-dim case is trickier of course.)

    As for part (a), that boils down to finding the dynamical group containing the Hamiltonian.
    Alternatively, there are procedures for decomposing the Hamiltonian in terms of a "chain"
    of subalgebras and the Casimir operators thereof. I.e., figure out the symmetry group
    of the degenerate eigenspaces, construct projection operators from these eigenspaces,
    and try to recast the original Hamiltonian in terms of these. Typically, you'll find there's
    something leftover, and that gives the necessary hint.
  4. Apr 10, 2009 #3


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    I'm not sure what you're getting at, but the solutions to the Hydrogen atom do not form a complete set, not due to degeneracy but due to the continuum of states for E > 0, where the thing is no longer square-integrable.
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