Common eigenfunctions of commuting operators: case of degeneracy 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:

*

*$\phi=c\psi_{A_i}$, with $c$ a constant. Hence, commuting operators have simultaneous eigenstates.

*$\phi\neq c\psi_{A_i}$. 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?
 A: 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.
A: 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 $\hat{p}$.
A: 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.
