Degenerate Perturbation: Calculating Eigenvalues

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In the discussion on calculating eigenvalues under degenerate perturbation theory, a model Hamiltonian with unperturbed eigenvalues E1 and E2 = E3 is analyzed. The perturbation V has specific non-zero matrix elements, leading to confusion about diagonalizing the matrix due to many zero elements. It is clarified that the first-order correction is indeed zero, necessitating the use of higher-order corrections to lift the degeneracy. A method is suggested to rotate within the degenerate subspace to simplify the perturbation, allowing for the application of non-degenerate perturbation theory to the remaining states. This approach ultimately helps in determining the corrected eigenvalues effectively.
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Say a model hamiltonian with unperturbed eigenvalues E1 and E2 = E3 is subjected to a perturbation V such that V12 = V21 = x and V13 = V31 = x2, with all other elements zero. I'm having trouble calculating the corrected eigenvalues. In the degenerate subspace spanned by |2> and |3> I need to diagonalise V, but all of these matrix elements are zero?
 
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It sounds like the first-order correction is zero, and you'll need to go to higher orders to lift the degeneracy.
 
If I understand you correctly, you have a perturbed matrix of the form $$H=\begin{pmatrix}
E_1 & V_{12} & V_{13} \\
V_{12}& E_2 & 0 \\
V_{13} & 0 & E_2
\end{pmatrix}.$$Why can you not diagonalize the usual way? Just say$$\det\begin{bmatrix}
E_1-\lambda & V_{12} & V_{13} \\
V_{12}& E_2-\lambda & 0 \\
V_{13} & 0 & E_2-\lambda
\end{bmatrix}=0$$ and solve the characteristic equation. That is easy to do because it factors into ##(E_2-\lambda)## times a quadratic in ##\lambda.## You get three non-degenerate eigenvalues.
 
First of all, there is no reason not to make a rotation in the degenerate subspace such that ##V_{13} = 0##. After that rotation it should be clear that ##E_2## is still an eigenvalue for one state. You can then apply non-degenerate perturbation theory to the remaining 2-dimensional subspace.
 
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