Homodyne detection quantum state tomography

In summary, balanced homodyne detection allows for state tomography. However, there is confusion regarding the measurement of discrete photon counts while the spectrum of the operator ##q_{\phi}## is continuous. This is explained by the use of detectors that are not single-photon sensitive, but instead measure photocurrents that are proportional to the photon number. This is due to the high number of photons present in the typical local oscillators used in pulsed operation.
  • #1
Jamister
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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.

1651239602039.png

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?
 
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  • #2
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.
 
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