Suppose $Q$ is a non-self-adjoint operator which commutes with it's adjoint, $Q^\dagger$. That is $$[Q,Q^\dagger]=QQ^\dagger-Q^\dagger Q=0$$What can be said about the relationship between eigenvalues of $Q$ and $Q^\dagger$?
Suppose $Q|\psi\rangle=\lambda|\psi\rangle, Q|\phi\rangle=\mu|\phi\rangle$, with $\lambda\ne\mu$. What can be said about $\langle\phi|\psi\rangle$?
My question is in regards to the second part about finding $\langle\phi|\psi\rangle$ - I suspect it is $0$ but cannot get anywhere with proving it. Here are my attempts so far: $$\langle\phi|Q^\dagger Q|\psi\rangle=\langle\phi|QQ^\dagger|\psi\rangle\\\bar\mu\lambda\langle\phi|\psi\rangle=\langle\phi|QQ^\dagger|\psi\rangle$$But I am unable to simplify the RHS. I was hoping to get something like $$(\bar\mu\lambda-\mu\bar\lambda)\langle\phi|\psi\rangle=0$$But perhaps I am wrong in thinking this is the solution. How can I proceed with this problem?
Note: I have tagged with eigenvalues-eigenvectors since this can easily be reformulated as a linear algebra problem.