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

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 $$\hat Q$$?

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 $$H$$).