# Quick way of showing visibility of HOM interference for weak coherent states?

It's shown in this paper how weak coherent states, can at-most have a visibility (using the "special visibility" of HOM interference) of 1/2 (as compared to single photons which have a visibility of 1).

In the paper, the derivation is fairly involved, mostly because it finds this visibility as a function of the polarization of the two fields.

# The problem:

If you assume the coherent states are identical (in their degrees of freedom like polarization), what's a simple way of showing that the visibility is 1/2 or lower?

That is, if I have two identical coherent states (with some phase difference between them), and I interfere them on a beamsplitter like so:

And we consider the function:

$$V_{\text{HOM}} = 1 - \frac{P^{(\text{coin})}}{P^{(c)}P^{(d)}},$$ where $$P(c)$$ is the probability of seeing a click on detector $$1$$ (corresponding to states $$|1\rangle_c, |2\rangle_c, ....$$), $$P(d)$$ is the probability of seeing a click on detector $$2$$ (corresponding to states $$|1\rangle_d, |2\rangle_d, ....$$), and $$P(\text{coin})$$ is the probability of measuring a coincidence in detector $$1$$ and $$2$$ at the same time. How can I show that the value of my the function $$V_{HOM}$$ below is $$1/2$$?

# My attempt:

(I've spent some time trying to work this out, and will outline where I'm stuck. I'll add this in the future when I have some more time.)

• is the "weakness" of the fields relevant here? The way you write it, this is simply a linear evolution of coherent fields. – glS Feb 11 at 15:29
• I'm not sure. In the paper I linked they show that the value of the visibility decreases with larger values of $\alpha$. But I think that could be because of some technical, more advanced complicated model. Right now I'm first looking for a simple answer, so it's fine if it doesn't resolve that at all. – Steven Sagona Feb 11 at 21:14
• there is a general way to express a linear evolution of coherent states. I was writing an answer along these lines, but then, looking at your definition of $V_{\text{HOM}}$, don't you always have $V_{\text{HOM}}=0$? Upon linear evolution the output is a state of the form $|\beta_1\rangle\otimes|\beta_2\rangle$ for some coherent states $|\beta_i\rangle$. But then you always have $P^{(\text{coin})}=P^{(c)}P^{(d)}$: you don't have correlation between detections at different outputs (although this changes when the inputs are not coherent states). – glS Feb 12 at 23:42
• Consider to spell out acronyms. – Qmechanic Feb 13 at 5:06