# Shared eigenbasis of commuting Operators

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!!

• The eigenvalue does is not necessarily degenerate, e.g. if $|\psi\rangle$ is also an eigenvector of $H_2$. Commented Sep 25, 2020 at 16:06
• @NDewolf Oh that's right. Thanks!
– ZR-
Commented Sep 25, 2020 at 16:08
• Re "I think we can conclude...", try $H_1=H_2=\pmatrix{2&0\cr 0&-2\cr}$. Commented Sep 25, 2020 at 16:09
• @WillO Thanks for the correction:)
– ZR-
Commented Sep 25, 2020 at 16:15
• 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? Commented Sep 25, 2020 at 21:31

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 α or β 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.

• Thank you so much for the answer! Should the superposition at the right-hand-side be $2(𝛼|𝜓1⟩+𝛽|𝜓2⟩)$ ?
– ZR-
Commented Sep 25, 2020 at 23:15
• No, not in general: You posited it not be an eigenvector of $H_2$. It collapses to an eigenvector only when α or β vanish. Commented Sep 26, 2020 at 11:30
• Thanks!! So the equation $H_2(𝛼|𝜓1⟩+𝛽|𝜓2⟩)=2(𝛼|𝜓1⟩−𝛽|𝜓2⟩)$ 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.
– ZR-
Commented Sep 26, 2020 at 21:16
• 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! Commented Sep 26, 2020 at 21:27
• 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. Commented Sep 26, 2020 at 21:37