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Degenerate time indep. perturbation theory

  1. Jan 23, 2009 #1
    why in time independent degenerate perturbation we diagonalize the matrix of the perturbation part of the hamilitonian and not the original hamiltonian?
     
  2. jcsd
  3. Jan 24, 2009 #2
    Because the unperturbed part is just proportional to the unit matrix since all states in the degenerate subspace have the same energy.
     
  4. Jan 24, 2009 #3
    what? the bold part is true and i agree with.

    the italicized part isn't true, firstly since i just did a problem where the matrix representation of the unperturbed part had off diagonals, secondly because an unperturbed part which was proportional to the identity would imply that it's already diagonalized and hence no degeneracy and hence no need for degenerate theory.

    i'm thinking it's because the the operator/matrix that diagonalizes the perturbed part also diagonalizes the unperturbed part. but i'm probably wrong.
     
  5. Jan 24, 2009 #4
    Of course it depends on what basis you are using if the matrix representation is diagonal or not. I assumed that you are working in an eigen-basis of the unperturbed Hamiltonian. In this basis the matrix representation of the unperturbed Hamiltonian is diagonal with entries corresponding to the eigen-energies, right? As far as the degenerate subspace is concerned the hamiltonian is diagonal with the same eigen-energies, hence it is proportional to the unit matrix.

    The point of degenerate perturbation theory is that can choose a basis of the degenerate subspace arbitrarily and obviously it is most convenient to choose this basis such that it diagonalizes the perturbation. Then once we have made this choice of basis we can perform regular (non-degenerate) perturbation.

    The reason why we do not care about diagonalizing the unperturbed part is because any linear combination of states within the degenerate subspace will also be an eigenstate of the unperturbed part with the degenerate energy. This is just another way of saying that the unperturbed Hamiltonian in this subspace is represented by a matrix proportional to the unit matrix.
     
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