# Biorthogonality of vectors

This question is equal parts math and physics, though I chose to ask it here because I am more concerned with the mathematics behind it, rather than physical implications.

Let $\hat{K}$ be a non-Hermitian operator and \begin{align*} \hat{K}|\phi_n\rangle&=\kappa_n|\phi_n\rangle\,,\\ \hat{K}^{\dagger}|\chi_n\rangle&=\nu_n|\chi_n\rangle\,,\\ \end{align*} where $\kappa_n=\bar{\nu}_n$ are complex conjugates and $|\phi_n\rangle$, $|\chi_n\rangle$ are eigenvectors. If $\langle\chi_m|\phi_n\rangle=\delta_{mn}$, then we would call the sets $\{|\phi_n\rangle\},\{|\chi_n\rangle\}$ biorthogonal. The resultant wave vectors can be formed from these sets as

\begin{align*} |\psi\rangle=\sum c_n|\phi_n\rangle\,, \end{align*} and it's dual \begin{align*} \langle\tilde{\psi}|=\sum d_n\langle\chi_n|\,. \end{align*} Looking at Brody and Curtright, it looks as if, when $c_n=d_n$ and $\langle\chi_n|=\langle\phi_n|$, so that $\langle\tilde{\psi}|=\langle\psi|$, regular quantum mechanics is recovered.

My question regards going in the opposite direction. Let's say that I have the expectation value, $\langle\psi|\hat{A}|\psi\rangle$. To make the expectation value biorthogonal, is it valid to just let $\langle\psi|\to\langle\tilde{\psi}|$? From equation (27) in Brody, it looks as if this will give the right answer, but I'm concerned about any subtleties. In going from regular quantum mechanics to biorthogonal quantum mechanics, will I also need to change $\hat{A}$ so it becomes non-Hermitian?

What happens if $\hat{A}$ is a function of position and momentum operators, which themselves are Hermitian?