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.


2 Answers 2


I think your confusion relies on a misunderstanding: a homodyne detection does not directly measures $\langle q \rangle$, but a value of the operator $\hat q$, by projecting the state on one of its eigenstates, an infinitely squeezed state.

This is clear when you use a time resolved homodyne detection, as needed for a quantum communication protocol like continuous-variable quantum key distribution. The first step in using such detection is to calibrate it on vacuum: you send nothing — that is the vacuum state $\lvert0\rangle$ — many times into your homodyne detection and make statistics. If everything works correctly, your result will be distributed with a Gaussian distribution and its variance (omitting electronic noise) is the variance of the vacuum $\Delta Q=1$, ans it average should be $\langle q \rangle=0$.

In many cases, quantum optics experimentalist only care about the variance, e.g. to show they’re below unity to prove squeezing and, for practical reasons use frequency resolved homodyne detections. In these case, computation of the average and the variance is often done analogically by a spectrum analyzer and this leads to abuse of language and notation, saying $\langle q\rangle$ instead of $\hat q$, because when you can measure $\hat q$, you can often measure $\langle q \rangle$, the repetition rate being often high enough that the statistics are effectively collected “instantaneously”. I suspect this kind of shortcut to be the source of your confusion.


Nice question. To be sincere, I never thought about the fact that homodyne measurement is, in some sense, an "indirect" measurement of a quadrature. Even if I 'm relatively new, I'll try to answer.

First of all, I think you should reverse your idea about quantum measurement. Let suppose we have a quantum system in the state $|\psi\rangle$. We can interpret the (statistical) mean value of an observable $A$, i.e., $<A>=\langle\psi|A|\psi\rangle$ as the mean value of the results of a measurement of $A$ on a set of independently prepared states $|\psi\rangle$. In other terms, you should think of $<A>$ as a statistical parameter. Your vision is an operational interpretation of $<A>$, which is a mathematical quantity.

Let us now recap the principle of homodyne detection which is composed of a beam splitter (with transmission and reflection coefficients $t$ and $r$) and two photon detectors. Assume now a simple non balanced homodyne. The state at the input ports is given by: $|\phi\rangle=|\psi\rangle_1|\alpha_0\rangle_2$ where the state at the second port is a coherent state generated by a strong laser beam. In the Heisenberg picture, the number operator at the first output port is given by: $${n_1}'=|t|^2{n_1}+|r|^2{n_2}+t^*ra_1^{\dagger}a_2+r^*ta_2^{\dagger}a_1$$ where $n_1,n_2,a_1,a_2$ are the number operators and annihilation operators of, respectively, the first "port" and second "port". You should know this result. Now, given the joint state $|\phi\rangle$, you can compute $<n_1'>$, and using the strong laser beam assumption (i.e., $|\alpha_0|^2\gg\langle\psi|n_1|\psi\rangle) $, you find out that: $$<n_1'>=\langle\phi|n_1'|\phi\rangle = |r|^2|\alpha_0|^2 + \kappa\langle\psi|q_1|\psi\rangle$$ where $\kappa$ is some constant and $q_1$ is the position quadrature of the state at the first port of the BS (more precisely, it is a rotation of that quadrature, but this does not matter now). If you now use a photodiode, the output current $i_D\propto<n_1'>.$ Now, you are correct in saying that we are not measuring $q_1$. Indeed, the effective $<q>$ is embedded in the variation of $<n_1'>$. However, you should interpret this in a statistical view. The result we've found could be rephrased as follows

The mean value of a (simple) homodyne measurement,performed on a set of independently prepared states $|\psi\rangle$, is given by $<n_1'>$.

To conclude, we can synthetize as follows. $n_1'$ is an observable and its eigenvectors are Fock vectors. The beam splitter acts on the input state in order to produce an output state whose probabilities (in the Fock basis) are somewhat 'proportional' to the quadrature values of the input state. More precisely, its mean value is exactly proportional to the mean value of the quadrature $a$. This is by no means intuitive, but it is true as it has been proved mathematically ("Shut up and calculate" vision of QM).


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