Quantum fun: how identical must “identical photons” be?

The basic idea of the experiment is simple. Take two objects that emit single photons that are identical to each other. Detect these photons after they have been mixed together using a beam splitter. If they are identical, both photons will appear on the same output port of the beam splitter and set off the same detector. If they are different, they can appear on different output ports and set off different detectors.

Something the article really doesn’t delve into is a discussion of what is interfering and why you get his articular correlation. This isn’t the familiar case of two waves that are 180º out of phase giving you destructive interference, which is called second-order interference (by some, anyway, it’s the same authors using the terminology across multiple papers, so I don’t know if it’s standard or them hoping it will catch on). One reason we know that this isn’t interference of this type is that this has been tested with independent sources, such that the beamsplitter where they combine is far enough away that the travel time exceeds the coherence time of the light. There can not a be a well-defined phase in that case, so you will not see an interference pattern emerge from this effect.

This is fourth-order interference, involving Fock states* (photon-occupation states) and this is QM with no classical analogue, so I don’t really have a good ball-and-stick description which I can share to explain why this happens. (That is to say I don’t have one, not that I have one and am precluded from sharing it for some reason of national security. As far as you know.) You have photon occupation states leaving the beamsplitter, for travel in the two directions, A and B. You can have |2,0> or |0,2>, for two photons going to one detector or the other, and there’s a |1,1> state, for a photon going to each detector. But really there are two |1,1> states, that look the same when we can’t tell the difference between the photons. And since you add amplitudes before “squaring” them (multiplying by the complex conjugate), the |1,1> terms cancel if the amplitudes are the same, i.e. if you have a 50/50 beamsplitter, leaving you with the photon pairs as the only possibility, as long the photons are identical. Which then raises the question which the paper is trying to address.

* Yes, really. Innuendo galore. Anything you can think of has probably been done before.

WIkipedia article on the HOM effect, as this is called.