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Symmetry of eigenstates

  1. Jun 28, 2009 #1
    It is common knowledge in Physics that eigenstates share the symmetries of the Hamiltonian.

    And it is trivial to show that this is true for the eigenspaces. Let g be an element of a symmetry group of Hamiltonian H, [tex]M_g[/tex] its representation, [tex]\left| \phi \right>[/tex] an eigenvector and [tex]\lambda[/tex] the corresponding eigenvalue. If g is an element from a symmetry group of H it is given that:

    [tex]M_gHM_{g^{-1}}\left|\phi\right>=\lambda\left|\phi\right>[/tex]
    Thus
    [tex]HM_{g^{-1}}\left|\phi\right> = \lambda M_{g^{-1}}\left|\phi\right>[/tex]

    So we see that the eigenspace for [tex]\lambda[/tex] is closed under the operations of [tex]M_g[/tex]
    Is there some theorem, that I can decompose this eigenspace into eigenstates of M or how do I proceed from here? This is important for Bloch functions and crystallography. I read something about Schur's lemma being involved.

    So to phrase a proper question:
    Is every Eigenfunction of a Hamiltonian,invariant up to a scale factor of unity magnitude under the operation of the Hamiltonian's symmetry groups? And if this is so how do I show this?
     
    Last edited: Jun 28, 2009
  2. jcsd
  3. Jun 28, 2009 #2
    Ok I think I have answered my question using spherical harmonics. It is really only the eigenspaces that are invariant. But then I don't know how to argue that on a lattice the eigenfunctions must be Bloch functions.
     
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