Homodyne detection quantum state tomography

[Jamister]

In balanced homodyne detection, it is claimed that one can do state tomography. I understand most of the derivation except one part. Here is a figure describing homodyne detection.

the operator that is being measured is
$$ R=N_{1}-N_{2}=a^{\dagger} b+b^{\dagger} a $$.
taking the mode b to be coherent state, we obtain:
$$R=r\left(a^{\dagger} e^{i \phi}+a e^{-i \phi}\right)=\sqrt{2} r q_{\phi}$$.
Now what I don't understand is that what is being measured is discrete photon counts. On the other hand, the spectrum of the operator $q_{\phi}$ is continuous. Therefore, how can obtain $\langle q_{\phi} \vert \rho \vert q_{\phi} \rangle$ from the measurements?

 

[Cthugha]

The detectors used for homodyne detection are usually far from being single-photon sensitive. Accordingly, you measure two photocurrents, not individual clicks. These are of course proportional to the photon number present in either detection arm. However, for pulsed operation typical local oscillators consist of 25 million photons per pulse or much more. This is very far from the regime of counting discrete photon detection events.