Why does having two hermitian operators commute imply that there is no degeneracy?

Question

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?

1. In a new basis of simultaneous eigenbasis of $\hat Q$ and $\hat H$, two degenerate energy eigenstates are not degenerate anymore, or
2. They are still degenerate, but can be distinguished by $\hat Q$?

Answer 1

The second one is the correct viewpoint. Given a Hamiltonian $H$ and another hermitian operator $S$, which commute you can find a simultaneous eigenbasis for both operators. Let's call them $|n,s>$. Acting on these states with $H$ we have $H|n,s>=E_n|n,s>$ for some value(s) of $s$, which means that the eigenspace for the eigenvalue $E_n$ might be more than one-dimensional: $\dim(V_{E_n})=\# \text{Allowed Values of s}$. If this is the case, this means degeneracy regarding the H-Op.

Given any hamiltonian and hilbertspace there is no way by finding other operators to reduce the dimensionality of the eigenspace of $H$. The dimensionality is an algebraic property of $H$ itself.

However, if there is a degeneracy, it is very useful to find an operator, which commutes with $H$, to classify the degeneracy i.e. to learn about it and also to get control about the degenerate eigenspaces i.e. to find a basis in the eigenspaces labelled by a conserved quantum number (eigenvalue of an operator, which commutes with $H$).