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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 $|\psi\rangle$ be an eigenstate of $H_1$, then I think if $|\psi\rangle$ is not an eigenstate of $H_2$ then we can conclude that both 2 and -2 are degenerate (thanks for correcting), since $|\psi\rangle$ and $H_2|\psi\rangle$ 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!!

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    $\begingroup$ The eigenvalue does is not necessarily degenerate, e.g. if $|\psi\rangle$ is also an eigenvector of $H_2$. $\endgroup$
    – NDewolf
    Commented Sep 25, 2020 at 16:06
  • $\begingroup$ @NDewolf Oh that's right. Thanks! $\endgroup$
    – ZR-
    Commented Sep 25, 2020 at 16:08
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    $\begingroup$ Re "I think we can conclude...", try $H_1=H_2=\pmatrix{2&0\cr 0&-2\cr}$. $\endgroup$
    – WillO
    Commented Sep 25, 2020 at 16:09
  • $\begingroup$ @WillO Thanks for the correction:) $\endgroup$
    – ZR-
    Commented Sep 25, 2020 at 16:15
  • $\begingroup$ I don't understand your question: take H1= diag(2,2,-2) and H2=diag(-2,2,2). All linear combinations of the first two eigenvectors will be eigenvectors of H1, but not H2, with one exception. What is it you want to do with them? Resolve them? $\endgroup$ Commented Sep 25, 2020 at 21:31

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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, $H_1|\psi_1\rangle = H_2|\psi_1\rangle = 2|\psi_1\rangle$; $H_1|\psi_2\rangle = -H_2|\psi_2\rangle = 2|\psi_2\rangle$; $-H_1|\psi_3\rangle = H_2|\psi_3\rangle = 2|\psi_3\rangle$; $H_1|\psi_4\rangle = H_2|\psi_4\rangle = -2|\psi_4\rangle$.

In general, all vectors $\alpha |\psi_1\rangle + \beta |\psi_2\rangle $ are eigenvectors of $H_1$ with eigenvalue 2, but not of $H_2$, since it acts markedly differently on them, $$ H_2 (\alpha |\psi_1\rangle + \beta |\psi_2\rangle)=2(\alpha |\psi_1\rangle - \beta |\psi_2\rangle), $$ with one (two) exceptions. The exceptions are for either Ξ± or Ξ² vanishing, in which case you have excluded the freak circumstance of unshared eigenvectors.

Likewise for the $\gamma |\psi_3\rangle + \delta |\psi_4\rangle $ 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.

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  • $\begingroup$ Thank you so much for the answer! Should the superposition at the right-hand-side be $2(𝛼|πœ“1⟩+𝛽|πœ“2⟩)$ ? $\endgroup$
    – ZR-
    Commented Sep 25, 2020 at 23:15
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    $\begingroup$ No, not in general: You posited it not be an eigenvector of $H_2$. It collapses to an eigenvector only when Ξ± or Ξ² vanish. $\endgroup$ Commented Sep 26, 2020 at 11:30
  • $\begingroup$ Thanks!! So the equation $H_2(𝛼|πœ“1⟩+𝛽|πœ“2⟩)=2(𝛼|πœ“1βŸ©βˆ’π›½|πœ“2⟩)$ is not a Schrodinger equation unless either $\alpha$ or $\beta$ equal to 0, right? I'm still a little bit confused about how it comes from. $\endgroup$
    – ZR-
    Commented Sep 26, 2020 at 21:16
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    $\begingroup$ Right. The vector acted upon is not an eigenfunction. It comes about by definition/assumption: $(H_2 -2)\psi_1=0=(H_2 -2)\psi_3=(H_2 +2)\psi_2=(H_2 +2)\psi_4$, of course! $\endgroup$ Commented Sep 26, 2020 at 21:27
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    $\begingroup$ Yes, I edited my answer to stress that is what was assumed. For the basis given, simultaneous diagonalization, all four eigenvectors are shared. Conversely, rotations in degenerate subspaces spoil that. $\endgroup$ Commented Sep 26, 2020 at 21:37

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