Shared eigenbasis of commuting Operators

Question

Suppose I have two Hamiltonian pieces $H_1$ and $H_2$ such that $[H_1,H_2]=0$. Then we know that the two pieces have shared eigenbasis. Assume both $H_1$ and $H_2$ have eigenvalues 2 and -2. Let $|psirangle$ be an eigenstate of $H_1$, then I think if $|psirangle$ is not an eigenstate of $H_2$ then we can conclude that both 2 and -2 are degenerate (thanks for correcting), since $|psirangle$ and $H_2|psirangle$ have the same eigenvalue. However, I'm still a bit confused about how can I find the shared eigenbasis between the two Hamiltonians? Do I need to consider the superposition (linear combination) of the degenerate states? Thanks!!

Cosmas Zachos · Accepted Answer

I don't fully understand what you are after, but from the comment I understand that you assume something like
$$ H_1=operatorname{diag} (2,2,-2,-2), qquad H_2=operatorname{diag} (2,-2,2,-2),
$$
in a space parameterized by eigenvectors $psi_{1,2,3,4}$,
that is, $H_1|psi_1rangle = H_2|psi_1rangle = 2|psi_1rangle$;  $H_1|psi_2rangle = -H_2|psi_2rangle = 2|psi_2rangle$;  $-H_1|psi_3rangle = H_2|psi_3rangle = 2|psi_3rangle$;  $H_1|psi_4rangle = H_2|psi_4rangle = -2|psi_4rangle$.
In general, all vectors $alpha |psi_1rangle + beta  |psi_2rangle $ are eigenvectors of $H_1$ with eigenvalue 2, but not of $H_2$, since it acts markedly differently on them,
$$
H_2 (alpha |psi_1rangle + beta  |psi_2rangle)=2(alpha |psi_1rangle - beta  |psi_2rangle), 
$$
with one (two) exceptions.
The exceptions are for either α or β vanishing, in which case you have excluded the freak circumstance of unshared eigenvectors.
Likewise for the $gamma |psi_3rangle + delta  |psi_4rangle $
subspace.
So, you parameterize the eigenvectors of $H_1$ corresponding to eigenvalues 2 and -2, respectively, and then run through each set finding the special two representatives which are also, exceptionally, eigenvectors of $H_2$ as well.

Shared eigenbasis of commuting Operators

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