Inner product of two eigenstates with different eigenvalues

Question

Problem:
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 $\langle\phi|\psi\rangle$?

Attempt:
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?

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Note: I have tagged with eigenvalues-eigenvectors since this can easily be reformulated as a linear algebra problem.

Answer 1

I will answer both parts, since having the first part correct is essential for the second part.

An operator that commutes with its adjoint is called normal. They have various nice properties, most of which stem from the fact that $$ \lVert Q \lvert \psi \rangle \rVert^2 = \langle \psi \rvert Q^{\dagger}Q \lvert \psi \rangle = \langle \psi \rvert QQ^{\dagger} \lvert \psi \rangle = \lVert Q^{\dagger} \lvert \psi \rangle \rVert^2, $$ for any $\lvert \psi \rangle$. Suppose that $\lvert \psi \rangle$ is an eigenvector of $Q$ with eigenvalue $\lambda$. Then $\lvert \psi \rangle$ is also an eigenvector of $Q^{\dagger}$ with eigenvalue $\bar{\lambda}$: $$ 0 = \lVert (Q-\lambda I)\lvert \psi \rangle \rVert^2 = \langle \psi \rvert (Q^{\dagger}-\bar{\lambda}I)(Q-\lambda I)\lvert \psi \rangle \\ = \langle \psi \rvert (Q-\lambda I)(Q^{\dagger}-\bar{\lambda}I)\lvert \psi \rangle = \lVert (Q^{\dagger}-\bar{\lambda}I)\lvert \psi \rangle \rVert^2, $$ where the third equality uses that $Q$ commutes with $Q^{\dagger}$; this argument works in either direction, so any $\lvert \psi \rangle$ that is an eigenvector of one is an eigenvector of the other with conjugate eigenvalue.

Now, if we also have $Q \lvert \phi \rangle = \mu \lvert \phi \rangle $ with $\mu \neq \lambda$, then $$\lambda \langle \phi | \psi \rangle = \langle \phi \rvert Q \lvert \psi \rangle = \overline{\langle \psi \rvert Q^{\dagger} \lvert \phi \rangle} = \overline{\bar{\mu} \langle \psi | \phi \rangle } = \mu\langle \phi | \psi \rangle, $$ and $\lambda \neq \mu$, so $\langle \phi | \psi \rangle=0$; it's exactly the same as the proof about eigenvectors of Hermitian matrices with different eigenvalues being orthogonal.

(Note that a little more is required in addition to the first part to show that Hermitian operators have real eigenvalues: one looks instead at $\langle \psi \rvert Q-\lambda I \lvert \psi \rangle$.)