For $[A,B]=0$, if an eigenfunction of $A$ not an eigenfunction of $B$, does that imply degeneracy of one operator?

Question

When two operators $A$ and $B$ commute, there can be functions which are eigenfunctions of $A$ but not that of $B$.

For example, in case of the one-dimensional harmonic oscillator, any linear combination of the ground state and the second excited state is an eigenfunction of parity operator (with eigenvalue +1) but not that of the hamiltonian even though they commute. In this example, this happens because the parity operator has degenerate eigenfunctions with eigenvalue +1.

Is this true in general? I mean, for this to happen, do we always need one of the operators to have degenerate eigenfunctions? Can we prove this?

Answer 1

Yes, it is true. Let $u$ be the said eigenvector of $A$ with eigenvalue $a$ that is not an eigenvector of $B$. Then, $ABu=BAu=aBu$. But $Bu\neq bu$ for any number $b$. Therefore, $Bu$ is an eigenvector of $A$ with the same eigenvalue $a$ that is linearly independent of $u$. The spectrum of $A$ is therefore degenerate.

Answer 2

Yes it is true in general. To prove it let $\vert\lambda_A\rangle$ be an eigenstate of $\hat A$ but not of $\hat B$. Then \begin{align} \hat A\hat B\vert\lambda_A\rangle = \hat B\hat A\vert\lambda_A\rangle =\lambda_A \hat B\vert\lambda_A\rangle \end{align} showing that $\hat B\vert\lambda_A\rangle$ has the same eigenvalue for $\hat A$ as $\vert\lambda_A\rangle$ since by assumption $\hat B\vert\lambda_A\rangle $ is not a multiple of $\vert\lambda_A\rangle$ else it would also be an eigenvector of $\hat B$.

Answer 3

yes the mathematics put above is correct, and in general way, you can say, whenever two operators commute, they must represent some degeneracy. You can put it in this way.