How is homodyne detection a quantum measurement?
In quantum mechanics, the way I'm used to think about obtaining the expectation value of an operator $A$ when the state is $| \psi \rangle $ is as follows. You prepare many copies of the state $| \psi \rangle$(*). Then, you measure $A$ for each of the copies. Every measurement projects $|\psi\rangle$ to some eigenstate of $A$, and the measurement outcome is the corresponding eigenvalue. Then, a numerical estimate of $\langle \psi |A |\psi \rangle$ is obtained by taking the average of all the measurement outcomes.
How does this relate to the homodyne detection of light? In homodyne detection, the measurement outcome is some classical current which is proportional to, e.g. $\langle q \rangle$, with $q$ the `position' quadrature operator. Reading books on quantum optics (**), it seems to me that this expectation value is obtained without making many copies of the state of light that is measured, or projecting any state to an eigenstate of $q$. It somehow seems that $\langle q \rangle$ is obtained without ever measuring $q$. I do not understand how this is to be reconciled with my understanding of measurement as described in the previous paragraph. Also, I'm not really confident with the meaning of the angular brackets in $\langle q \rangle$, which is maybe another source of confusion. I do understand how homodyne detection works classically.
(*) I know about the no-cloning theorem, but the unknown state $| \psi \rangle$ could be the result of some preparation process. If the preparation process repeated is many times with the same settings, the assumption is that the process produces $|\psi\rangle$ each time.
(**) I've tried the books by Scully & Zubairy, Gerry & Knight and Grynberg & Aspect & Fabre, but their explainations are all similar and do not help me.