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I've been reading the degenerate perturbation theory section of Griffiths QM. He introduces the idea that, if we can find an operator $\hat A$ which commutes with $\hat H^0$ and $\hat H'$, then simultaneous eigenfunctions of $\hat A$ and $\hat H^0$ (say, $\psi_a$ and $\psi_b$) will be "good" eigenstates to use in the degenerate theory, satisfying $W_{ab} \equiv \langle \psi_a^0 | \hat H' \psi_b^0\rangle = 0$ (which resolves computational issues involving division by zero).

In other words, if the operator $\hat A$ commutes with the new Hamiltonian $\hat H = \hat H^0 + \hat H'$ and is therefore conserved under the perturbation in the sense that $$ \frac{\mathrm d\langle A\rangle}{\mathrm dt} = \frac i \hbar \langle [\hat H, \hat A] \rangle + \left\langle \frac{\partial \hat A}{\partial t}\right\rangle = 0, $$ then the simultaneous eigenstates of $\hat A$ and $\hat H^0$ are "good".

Intuitively, I understand that the underlying reason why certain states are "good" is that they remain stationary after the perturbation is turned on, whereas other states in the degenerate subspace are no longer stationary.

My question is, how can I concretely prove that $\langle A\rangle$ being conserved after the perturbation is turned on leads to its stationary eigenstates remaining stationary under the perturbation?

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Good quantum states are those that happen to diagonalize the perturbation. The effect of the perturbation on the original Hamiltonian is then merely an increase in the diagonal entries, which are the energies. The "good" basis doesn't call for further diagonalization.

This is useful even if said basis does not diagonalize the full perturbed hamiltonian. Sometimes having a subspace be diagonal is already very helpful.

As for the mathematical proof, you already got it: if $\langle A \rangle$ is conserved, $A$ commutes with the full Hamiltonian, which implies that they can be simultaneously diagonalized. The basis that simultaneously diagonalizes these two operators is, by definition, what you called a "good" basis.

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  • $\begingroup$ Thanks for the answer, but I still don't see the connection to stationary states. $\endgroup$
    – Chris Yang
    Jun 5, 2022 at 12:37
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    $\begingroup$ A stationary state is one that solves the time-independent Schrodinger equation. But these are just the eigenstates of the Hamiltonian. Under the assumptions above, these eigenstates are the good states. It seems that you are using many different terms to describe the same physics, maybe try to condense your language a little bit. $\endgroup$ Jun 5, 2022 at 23:36
  • $\begingroup$ Thanks. On further inspection of the concept, I realize that by using the time-independent Schrödinger equation in the first place to develop the degenerate perturbation theory, we are assuming that the states are stationary. It is from this that everything else (constructing and diagonalizing $\mathbf W$) is derived. $\endgroup$
    – Chris Yang
    Jun 10, 2022 at 17:37

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