I know that if two hermitian operators commute, then there is complete set of simultaneous eigenstates. But, why does this imply that they have no degeneracy in the spectrum ? For example, if the hamiltonian $\hat H$ and a symmetry operator $\hat Q$ commute,(if $\hat Q$ is only a single symmetry operator) are there no degeneracy in the spectrum?
Or, I guess that it is not the case that the degeneracy is lifted, but we can find another operator that distinguishes the degenerate states. Which is right?
- In a new basis of simultaneous eigenbasis of $\hat Q$ and $\hat H$, two degenerate energy eigenstates are not degenerate anymore, or
- They are still degenerate, but can be distinguished by $\hat Q$?