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In the context of degenerate perturbation theory, for a perturbed Hamiltonian $H_0 + \lambda H_1$, I've heard of a very useful tool:

"If $[H_0,\lambda H_1] = 0$ holds (both parts of the hamiltonian commute) then $H_1$'s matrix must be diagonal, this is, the $(\lambda H_1)_{ij} = <n_0^i| H_1 |n_0^j>$ (where $|n_0^i>$ are the unperturbed eigenstates of $H_0$) must be zero outside of the diagonal."

This would allow us to apply non-degenerate perturbation theory to the problem. I'm writting a text about degenerate perturbation theory and I'm not sure how to explicitly demonstrate this tool.

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  • $\begingroup$ "I'm writting a text about degenerat perturbation theory..." Oh really? Who is the publisher and do they have a proofreader? $\endgroup$
    – hft
    Commented Feb 11, 2023 at 17:45
  • $\begingroup$ @htf Mea culpa. I dropped the OP's participle endings myself without finishing the job. What if the proofreader is in Portuguese? In any case, the "I've heard quote" merits more scrutiny... I wish I knew where it came from... $\endgroup$ Commented Feb 11, 2023 at 21:36

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Your quote is an exaggeration: $H_1$ may be diagonalized, which then fixes the ambiguous parts of the basis of the degenerate $H_0$. "Already" is out of place in an ambiguous basis.

If $[H_0,H_1]=0$, then $H_0,~H_1, ~ H=H_0+\lambda H_1$ can be simultaneously diagonal matrices. The zeroth order basis is ambiguous in the degenerate subspace, but you have already diagonalized the hamiltonian, if you have diagonalized $H_1$ which breaks the degeneracy, $\langle n_0^i | H_1|n_0^j \rangle \propto \delta_{ij}$.

You have thus found the full Hamiltonian's full eigenstates, to all orders in λ, hence merely the first: the first order shifted energy is an exact expression, $$ E^j= E^j_0 + \lambda \langle n_0^j | H_1|n_0^j \rangle. $$

  • No perturbative approximation is needed, as you have already diagonalized the full hamiltonian.

The point is $H_1$, assuming it is not degenerate itself, helps you specify the orthogonal eigenstates it dictates by breaking the degeneracy, i.e. rotates the arbitrary zeroth order eigenstates to eigenstates of it as well.

Illustrate by $H_0=$diag(1,1,2), and $$ H_1=\begin{bmatrix} 3/2 &-1/2&0\\ -1/2&3/2&0\\ 0&0&3\end{bmatrix}. $$ subsequently diagonalized to $H_1=$diag(1,2,3), without affecting $H_0$. It is $H_1$ that fixes the above basis dubbed $|n_0^j\rangle$, not $H_0$.

(Freak lark logical possibility: You need both pieces to break each others's degeneracy in a common diagonal outcome, e.g., for $H_0=$diag(1,1,2), $H_1=$diag(1,2,2). One should never get one's students there... They should spend their time on useful things...)

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