# Inner product of two eigenstates with different eigenvalues

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?

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

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 $$\langle \psi \rvert Q-\lambda I \lvert \psi \rangle$$.)