I cannot give quite a satisfactory answer. However, I do believe an essential ingredient is the fact that the interaction of any photon with any particular atom in the beam splitter is exceedingly small. The reflection arises as a collective, or coherent, effect. For this reason, at least, the large number of atoms does not necessarily lead to decoherence. As for temperature, I would argue that room temperature does not lead to the presence of optical thermal photons, so that for an optical photon things are OK. In other words, thermal lattice vibrations do not interact with the incident photon, because of a considerable energy and frequency mismatch.

Initially I might have said: it is because no energy gets transferred from the photon to the beam splitter. This, however, is not true, since there are plenty of instances in which decoherence arises without energy transfer. 

If the photon in no way becomes entangled with the beam splitter (so that measuring the beam splitter variables would determine which way the photon went), then the input and output states of the photon transform as a unitary operation within the photon's state space. Then there is no decoherence.

Whil 

Oh no. All my previous typing in vain...

While I agree at first sight with Michael's answer, I wonder what "in no way entangled" means, and if the statement is mathematically correct, considering also mixed input states. With regard to the initial question, I would like to offer another view. Every interaction with an environment (by definition not part of the observed system) leads to decoherence, the question is to what degree. Thermally oscillating atoms move by much less than the optical wavelength. This leads to minimal distortion of the classical em wave packet of the puls containing the photon. The overlap of the actually reflected with the ideally reflected pulse will be almost 100%, equivalent to minimal decoherence. But also the recoil of the whole mirror has to be considered here. A light beamsplitter would distort (chirp) the reflected puls as it is accelerated during the reflection. If it is possible to determine the path of the photon by distinghuishing the beamsplitter's state before and after (with high probability), no subsequent interference experiment involving the two paths will lead to a predictable result (with low probability). This leads back to Michael's answer.

Best regards, Thomas.

I agree with T.'s and F.'s analyses. If the beamsplitter is heavy enough then the recoil will be negligible. And the light reflects collectively from all atoms together, so no one atom recoils or gets excited. Therefore no trace of information regarding the photons' interacting with the beamsplitter will be recorded. Thus no entanglement. If no entanglement, no decoherence.

The following is leading away from the initial discussion. But I would like to nitpick and ask Michael, if his statement is really correct. I would agree that for decoherence to happen an entangling operation is required, in the sense that separable input states exist for which the output state will be entangled. But if beamsplitter and incoming light field are in a mixed, separable state, I would expect that further decoherence can occur without resulting in an entangled state. Repeating the same process might eventually end up in an entangled state. Is this correct? 

Regards,

Thomas

Thomas, I think you are right. I was talking about the case of pure states. A typical beam splitter is never in a pure state, although recent experiments with micro-mirrors in the area of opto-mechanics do work with pure state beamsplitters. For a macro beamplistter at room temperature, a classically-induced random motion of the beamplitter can decohere the state o f the photon, as is well known in the case of a noisy interferometer. But in a sense that is not fundamental, because the information needed to control the location of the beamsplitter is classical, so it could be stored up and be used to undue the supposed decoherence. Still, it is real decoherence from a practical point of view. (At least this is how I look at it.)

This is a rather beautiful thread following on from what seems like a simple question, and I must confess to having had to rethink some of my assumptions here.  Michael Raymer's point about the beamsplitter being 'heavy' enough is important in this context, as of course the beamsplitter must not be able to extract information about the change in momentum of the photon.

There appears to be one aspect of the discussion that I believe is missing from the comments above - and that is a fairly obvious one.  The beamsplitting operation itself does not cause interference.  Instead it is the recombination that shows the interference.  So we need to consider energy transfer from both the reflected and transmitted parts (there's an interaction in each case, but presumably the statistical result is no information extraction).

We can also see that interference must occur from the viewpoint of bounding the energy loss and momentum change of the photons.  If there is no significant energy lost to the mirror, then we understand this by comparing the spectrum of the single photons from each pathway.  If the overlap of the spectra is still ~100% we say that the photon pathways are indistinguishable, and it is the indistinguishability that means we see interference.  Momentum also must be balanced and certainly the momenta associated with the pathways are distinguishable, although this is removed with the photon recombination.

I think that an important way to view this problem is by analogy with reflection that includes momentum transfer to the mirror (for example off a cantilever).  Here one would expect that the cantilever gains one phonon on reflection, none on transmission.  This causes the reflected photon mode to have lost energy and hence be reduced in energy.  This should destroy any possible interference if the phonon energy is large compared to the linewidth of the photon.

One would also guess that in the limit of a quantum cantilever, the same energy loss would also mean that if you upshifted the reflected photon by the same energy it would have lost (e.g. by scattering off an acoust-optic modulator), the interference pattern would NOT be recovered, because the quantum cantilever is entangled with the mode and provides which-path information despite the fact that the energy spectra of the two photon modes overlap.