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

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?

## 3 Answers

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

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 $$\hat B$$.

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.