As proved in the answer to this post, if the operators $\hat A$ and $\hat B$ commute, then they have the same eigenstates.

Let $$\hat A\psi_{A_i}=A_i\psi_{A_i}\qquad \Rightarrow\qquad \hat B\hat A\psi_{A_i}=\hat B(A_i\psi_{A_i})=A_i\hat B\psi_{A_i}\equiv > A_i\phi .$$ Now, due to the vanishing of the commutator we have that $$\hat B\hat A\psi_{A_i}=\hat A\hat B\psi_{A_i}=\hat A\phi$$ From the RHS of the last equations, we have that $$\hat A\phi=A_i\phi,$$ meaning that $\phi$ is also an eigenstate of $\hat A$ with eigenvalue $A_i$. This could happen for the following reasons:

  1. $\phi=c\psi_{A_i}$, with $c$ a constant. Hence, commuting operators have simultaneous eigenstates.
  2. $\phi\neq c\psi_{A_i}$. In this case the operator $\hat A$ must have degenerate eigenstates, namely $\phi$ and $\psi_{A_i}$. Even at this case, the non-degenerate eigenstates of $\hat A$ are simultaneously eigenstates of $\hat B$.

However, what happens in the second case, where the eigenvalue $A_i$ is degenerate? Can we say that the eigenfunctions corresponding to the degenerate eigenvalue $A_i$ are not eigenfunctions of $\hat B$? Do we know something else about them?


3 Answers 3


If $[\hat A,\hat B]=0$ and they are both non-degenerate, then every eigenstate of $\hat A$ is an eigenstate of $\hat B$ and vice versa.

If $[\hat A,\hat B]=0$ and $\hat A$ has a degenerate spectrum, then you are guaranteed the existence of one common eigenbasis. However, you are not guaranteed that every eigenstate of $\hat A$ will be an eigenstate of $\hat B$.

As a simple counterexample to illustrate that last statement, take the operators $$ \hat A = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix} \quad\text{and}\quad \hat B = \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}, $$ for which $(1,0,0)^T$ is an eigenstate of $\hat{A}$ but not $\hat B$ even though $\hat A\hat B=\hat B\hat A=\hat B$.

If the information you have is that $[\hat A,\hat B]=0$, $\hat A$ has a degenerate spectrum and $v$ is an eigenstate of $\hat A$ in a space with degenerate eigenvalue, then you cannot make any inferences about its relationship to $\hat B$ $-$ it might be an eigenstate, or it might not.

  • $\begingroup$ Is this true even if the operators aren't Hermitian? $\endgroup$
    – Philip
    Aug 28, 2020 at 13:21
  • 3
    $\begingroup$ @Philip In principle, yes, but you need additional external guarantees that they both have eigenstates to begin with. That pulls you into a game of provide-the-minimal-assumptions which is for mathematicians and people with too much time on their hands to play. $\endgroup$ Aug 28, 2020 at 13:25
  • $\begingroup$ That's what I thought, I wasn't completely sure :) $\endgroup$
    – Philip
    Aug 28, 2020 at 13:26

In your case you seem to have defined $\phi_i = \hat{B}\psi_i$, where $i=1,2,3,\dots N$ is the degree of degeneracy.

It should be clear to you that the states $\phi_i$ are still eigenstates of $\hat{A}$. However, there is no reason for them to, a priori, be eigenstates of $\hat{B}$. In fact, since every $\phi_i$ is an eigenstate of $\hat{A}$, you can write it as a linear combination of the "degenerate" eigenstates of $\hat{A}$, $\psi_i$. The action of $\hat{B}$ could then be for example to take one eigenstate to a different one. (You could have, say, $\hat{B}\psi_1 = \psi_2$, for example.)

Therefore in general I don't think there is anything special that can be said in this case without any further information. However, if the operators $\hat{A}$ and $\hat{B}$ are Hermitian, then we are guaranteed that we can diagonalise $\hat{B}$ within this subspace spanned by $\psi_i$, and therefore there exists at least $N$ linear combinations of the $\psi_i$s that are also eigenstates of $\hat{B}$.

In other words, in the case of Hermitian operators, at least one simultaneous eigenbasis can be found.

Example: Consider the Hamiltonian for a free particle: $$\hat{H} = \frac{\hat{p}^2}{2m}.$$

Clearly, $\hat{H}$ and $\hat{p}$ commute, but not all states of definite energy are states of definite momentum. For example, a state $|E_1\rangle \propto |p\rangle + |-p\rangle$ would have the same energy as the state $|E_2\rangle \propto |p\rangle - |-p\rangle$ and so on. However, clearly there is one basis (the basis of $|p_i \rangle$) which is a simultaneous eigenbasis of both $\hat{H}$ and $\hat{p}$.


When one of the two commuting operators has degenerate eigenfunctions, one can always construct their linear combinations which will be the eigenfunctions of the other operator.


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