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I am studying time-independent perturbation theory in quantum mechanics and I found myself not understanding how to associate degenerate eigenstates their specific Hamiltonian eigenvalue correction.

Suppose that the full Hamiltonian has the form $$ \widehat{H}= \widehat{H_0} + \lambda \widehat{V} $$ and the unperturbed Hamiltonian $\widehat{H_0}$ has $k$ degenerate eigenstates, $\left| 1 \right\rangle \cdots \left| k \right\rangle $.

The eigenvalues of the degenerate subspace matrix $$ V_{ij}= \left\langle i \right| \widehat{V} \left| j \right\rangle \quad , \quad i,j \in \left[ 1,k \right] $$ represent the first order Hamiltonian eigenvalue corrections in $\lambda$ .

My question is, how is it possible to associate the new, perturbed energy values to the original unperturbed eigenstates?

Ahornach
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1 Answers1

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When you first consider a degenerate subspace of states that all share the same order zero energy, you don't really know how to pick them in such a way that they become the order zero states of your perturbation series. In practice what happens is that the perturbation is the one that selects them and we can read this as "the states you would find if you turn on the perturbation but then you turn it off back again".

In practice, the basis that you use to compute $V_{ij}$ is generically just an orthonormal basis that you found for your degenerate subspace and in principle it does not have anything to do with the order zero states.

But when you compute the first order corrections as the eigenvalues of $V_{ij}$, the corresponding eigenvectors are the actual order zero states! These are the only meaningful vectors and, if the degeneracy is fully lifted, then you can use those states from then on to compute higher order corrections now using non-degenerate perturbation theory. Otherwise, you'll still have degenerate subspaces and you would have to keep using degenerate perturbation theory in higher orders, if you really need to push that far.

secavara
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