Can 3 photons bunch, Hanbury Brown and Twiss effect?

[Spinnor]

Could we split the beam of light in the Hanbury Brown and Twiss effect setup into 3 equal beams each, with each beam having a detector, and measure three photons arriving at nearly the same time at all three detectors?

Thanks for any help!

 

[Cthugha]

Yes, one can do that and there have been experiments on that. What you find is that the relative coincidence count rate shows a factorial increase with the number of beams for thermal light. The 2-photon coincidence rate shows the well known increase of 2!=2. For three photons, you get 3!=6, for four photons 4!=24, for five you will get 5!=120 and so on.

 

[Spinnor]

Thanks for your reply!

I'm confused though, if we have really weak light, say a photon per hour on average, and split the incoming beam into say three beams won't we most likely only measure one photon at one of the detectors? Is there another factor that takes the intensity of the light into account besides the n! factor?

Thanks for your help!

 

[Cthugha]

The HBT effect is about the relative coincidence rate only. Assume you do measurements every second, but only have one photon per hour. This means in any single beam 1 out of 3600 measurements will give you a photon on average. If the beams are statistically independent, you would expect to detect photons simultaneously on average in 1 out of 12960000 (simply 3600*3600) measurements. That is one pair in 150 days. The HBT effect just increases the average photon pair detection rate to two pairs in 150 days. Quite analogous, you would expect to detect three photons simultaneously in 1 out of 3600*3600*3600 measurements and the HBT effect will increase the rate to 6 out of 3600*3600*3600 measurements.

As this is about relative count rates, the effects in terms of absolute counts will be drastic only for large mean photon numbers.

 

[Spinnor]

That example made it very clear, thank you!