1
$\begingroup$

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:

enter image description here

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.)

$\endgroup$
  • $\begingroup$ is the "weakness" of the fields relevant here? The way you write it, this is simply a linear evolution of coherent fields. $\endgroup$ – glS Feb 11 at 15:29
  • $\begingroup$ 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. $\endgroup$ – Steven Sagona Feb 11 at 21:14
  • $\begingroup$ 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). $\endgroup$ – glS Feb 12 at 23:42
  • 1
    $\begingroup$ Consider to spell out acronyms. $\endgroup$ – Qmechanic Feb 13 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.