Orthogonal eigenvectors and Green functions

  • Thread starter paolorossi
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Hi you all. I have to diagonalize a hermitian operator (hamiltonian), that has both discrete and continuous spectrum. If ψ is an eigenvector with eigenvalue in the continuous spectrum, and χ is an eigenvector with eigenvalue in the discrete spectrum, is correct to say that ψ and χ are always mutually orthogonal? I think the answer is yes. But if I numerically calculate the inner product between ψ and χ, then I find that this is far from zero.

PS
I work in this way.
I calculate the eigenvalues and eigenvectors from the Green operator.
Specifically, I have the hamiltonian operator

H = H0 + V

H0 has only continuous spectrum. Using a perturbative expansion, I find the Green operator

G(z) = 1/(z-H)

in terms of V and of the Green operator of H0

G0(z) = 1/(z-H0)

So I find that G(z) has a branch cut and one simple pole. This is consistent with various works and books. Then I calculate the eigenvalues and the corrispondent eigenvectors. So I express ψ and χ in terms of a common basis, and using the Fourier coefficients I can calculate the inner product.
 

Answers and Replies

  • #2
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ok I solve it. I have calculated analytically the inner product and I see that it is zero. In fact, I had made ​​a mistake in the numerical calculation.
 

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