Compatible observables and dealing with degenerate states I'm just trying to self-study QM a little bit, and I need to confirm my understanding about the following: "commuting observables can be used to generate additional quantum numbers to label degenerate states"
So suppose we have two compatible observables $A$ and $B$ such that $[A,B]=0$. If $\vert a_{i}\rangle \, \textrm{for} \, i=1,2,\ldots  $ are eigenkets of $A$, then
$$ A\vert a_{i}\rangle = a^{(i)}\vert a_{i}\rangle $$
Suppose there is a degeneracy, there exists $j,k$ such that
$$ A\vert a_{j}\rangle = a\vert a_{j}\rangle $$
$$ A\vert a_{k}\rangle = a\vert a_{k}\rangle $$
where $a^{(j)} = a^{(k)} = a$. Then since $A$ and $B$ are compatible, $\vert a_{i}\rangle \, \textrm{for} \, i=1,2,\ldots  $ are also eigenkets of B. If there are no degeneracies in the case of $B$, then we can use $b_j, b_k$ as additional quantum numbers to label $\vert a_{j}\rangle, \vert a_{k}\rangle $ lifting the degeneracy.
$$ A\vert a_{j}\rangle = a\vert a_{j}b_j\rangle $$
$$ A\vert a_{k}\rangle = a\vert a_{k}b_k\rangle $$
Is this correct?
 A: Almost. $|a_j \rangle$ and $|a_k \rangle$ don't need to be eigenkets of $B$. They can be taken to be eigenkets of $B$. For example, suppose the observables
$$A = \begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix} \quad\text{and}\quad B = \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix}.$$
Consider the kets
$$|\psi\rangle = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} \quad\text{and}\quad |\phi\rangle = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}.$$
Both of them are eigenkets of $A$ with eigenvalue $2$, but neither of them is an eigenket of $B$. However, we could pick a basis of eigenkets of $A$ such that it also is a basis of eigenkets of $B$. For example,
$$\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, \frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix} \quad\text{and}\quad \frac{1}{\sqrt{2}}\begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}.$$
Now all of these are eigenkets of both $A$ and $B$ at the same time. We can label them by using the eigenvalues of $A$ and $B$ to distinguish between them, lifting the degeneracy that was present with only $A$.
In short, you are nearly correct, but beware: commuting observables can be taken to have a common basis, but this doesn't mean that every basis of eigenkets of $A$ is a basis of eigenkets of $B$. It means only that there is some basis of eigenkets of $A$ that is a basis of eigenkets of $B$.
