Eigenvectors of commuting hermitian operators?

Question

Didn't know if this belonged here or on the maths StackExchange, let me know if I should switch over.

Currently going through a quantum mechanics class and I'm reading the following theorem (translating from french, apologies for any mistakes):

if $A$ and $B$ are two commuting hermitian operators, there exists an orthonormal basis consisting of eigenvectors shared between the two operators.

This part I get. The problem is with a particular case of this theorem:

Let $a_i$ and $b_i$ be eigenvalues of $A$ and $B$ respectively. For 3D matrices, if $a_1=a_2\equiv a_{12}$ and $b_2=b_3\equiv b_{23}$. $b_1$ is not degenerate, thus $|b_1\rangle$ is determined. $|b_1\rangle$ is also an eigenvector of $A$ with eigenvalue $a_{12}$ such that $|a_{12}^{(1)}\rangle\leftrightarrow|b_1\rangle$

I get the example in 3D, but I'm having a hard time generalizing the theorem for more dimensions. Right now, for instance, I have two 4D operators with $a_1=a_2\neq a_3\neq a_4$ and $b_1\neq b_2 \neq b_3 = b_4$.

How does this theorem work exactly for 4+ dimensions? Is it that since neither $b_1$ or $b_2$ are degenerated, both correspond to a different eigenvector of $A$? Do they both have an eigenvalue of $a_{12}$ as well?

Answer 1

In the proof for :

If A and B are two commuting hermitian operators, there exists an orthonormal basis consisting of eigenvectors shared between the two operators.

the following lemma is often used :

Let $A,B$ be two commuting hermitian operators and $E_\lambda$ be an eigenspace of $A$. Then $E_\lambda$ is stable by $B$.

The eigenspace $E_\lambda$ is the set of kets $|\psi\rangle$ such that $A|\psi\rangle = \lambda|\psi\rangle$. That $E_\lambda$ is stable by $B$ means that if $|\psi\rangle\in E_\lambda$, then $B|\psi\rangle\in E_\lambda$.

If $E_\lambda$ has dimension $\geq 2$, we can use the spectral theorem on $B$ (restricted to $E_\lambda$) and find common eigenvectors of both $A$ and $B$.

If $E_\lambda$ has dimension $1$, then we take a non-zero ket $|\psi\rangle \in E_\lambda$ and $B|\psi\rangle \in E_\lambda$ means that $B|\psi\rangle = \mu|\psi\rangle$ for some $\mu\in\mathbb C$.

So the general result you are looking for is :

Let $A,B$ be two commuting hermitian operators and $|\psi\rangle$ be an eigenvector of $A$ for a non-degenerate eigenvalue. Then $|\psi\rangle$ is also an eigenvector of $B$.

However, we have no information on the other eigenvalue. It could be degenerate or not, anything really.