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Assume $H=H^0+V$ with $V$ a perturbation and $H^0$ has multiple degeneracies (Think of $H^0$ as diagonal and various diagonal elements are equal, e.g. $H^0_{11}=H^0_{22}\neq H^0_{33}=H^0_{44}\neq H^0_{55} $).

Now in textbooks/internet what I can find are examples of one d-fold degeneracy. One diagonalizes $V$ given in the basis of $H_0$ in the degenerate $d$-dimensional subset. The eigenvalues are the first order energy corrections and linear combinations of the eigenvectors in the $H^0$ basis give the correct zeroth order states (just a basis transformation in the respective subset).

Then to get the first order state corrections I apply the normal perturbation theory, namely

$|n> = |n^{(0)}>+ \sum\limits_{j \neq n} \frac{<j^{(0)}|V|n^{(0)}>}{E_n-E_j} |j^{(0)}>$

but in the summation over $j$ do not include the degenerate subset. However I use this state correction also for the correct zeroth order states ( $|n^{(0)}>$ is the corresponding linear combination then). Correct until now?

Now to my question. Of course I can just do diagonalize each degenerate subset on its own. I get energy corrections and zeroth order states but what does this mean for the state corrections? Do we just sum over all non-degenerate $j$'s or do we actually mix the different zeroth order linear combinations of the different subsets?

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