Degenerate Perturbation: Calculating Eigenvalues

[ergospherical]

Homework Statement

See below

Relevant Equations

N/A

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?

 

[vela]

It sounds like the first-order correction is zero, and you'll need to go to higher orders to lift the degeneracy.

 

[kuruman]

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.

 

[Orodruin]

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.