Why doesn't a typical beam splitter cause a photon to decohere?

Nobody is answering this question, so I'll take a stab at it.

Consider the mirror. Suppose you started your experiment by (somehow) putting it in a nearly-exact momentum state, meaning there is a large uncertainty in its position. Now, when you send a photon at it, the photon either bounces off or passes through. If the photon bounces off the mirror, it will change the momentum of the mirror. You could theoretically measure the "which-way" information by measuring the momentum of the mirror after you've done the experiment. In this scenario, there wouldn't be any interference.

However, you didn't do that. You started the mirror off in a thermal state at room temperature. This state can be considered as a superposition of different momentum states of the mirror¹, with a phase associated to each one. If you change the momentum by a small amount, the phase associated to this state in the superposition only changes by a small amount. Now, let $p_\gamma$ and $p_m$ be the original momenta of the photon and the mirror, and let $\Delta p_\gamma$ be the change in the momentum when the photon bounces off the mirror. When you send the photon towards the mirror, the original state $p_m$ (photon passes through) will end up in the same configuration as the original state $p'_m = p_m - \Delta p_\gamma$ (photon bounces off). These two states $p_m$ and $p'_m$ had nearly the same phase before you aimed the photon at the mirror, so they will interfere, and if the phase on these two states are really close, the interference will be nearly perfect.

Of course, a change in momentum isn't the only way for the mirror to gain which-way information. However, I think what happens when you consider the other ways is that they behave much like this, only not anywhere near as cleanly, so they're harder to work with.

¹ Technically, it's a mixed state, i.e., a density matrix, and not a pure state. But the basic idea of the above explanation still holds.

Hopefully a sharper restatement of the question is: what's the difference between a mirror and a photocathode? Experimentally, in a Mach-Zender interferometer we can fold light paths with a mirror while maintaining coherent interference, but passing either beam into the photocathode of a photodetector destroys interference effects, even for photons that didn't take that path. In both cases, you have an intimate interaction of a photon with a macroscopic object, so you might naively expect both objects to produce decoherence by irreversibly coupling the photon to the environment.

In either a mirror or a photocathode, incoming photons indeed couple with the free electrons in the material. In a mirror, these electrons immediately re-emit the photons, crucially without being able to absorb any energy. In a photocathode, the post-collision electrons absorb enough energy to leave the material surface and go on to couple with the environment (such as a detector).

In solid-state terms, the difference between these cases is attributed to the photoelectric work function in the material: a mirror reflects light because the photon's energy is insufficient to remove an electron from an atom, so the photon has no choice but to leave unaltered.

In quantum terms, the big difference is again energy: the phase changes in Schrödinger's equation depend completely on the Hamiltonian--the wave function's total energy. A surface electron coupled with an incoming photon has negative energy inside a mirror, so the wavefunction for this coupling dies away exponentially, resulting in net zero impact on the photon. Inside a photocathode, the same electron-photon coupling has positive energy, so it can sustain itself indefinitely, permanently affecting the photon's wavefunction and mixing it with the environment.

If you follow Dr. Shor's answer and imagine situations where the mirror could siphon off some of the photon's energy, for example by recoiling fast enough for the doppler shift to result in a longer-wavelength reflected photon, you have not only built a possible photon detector (by measuring the mirror's recoil), but also clearly have a shorter range for possible interference effects (due to the wavelength shift).

Interestingly, the same object can act as a mirror at long wavelengths and as a photocathode at short wavelengths, so you can get wavelength-dependent decoherence!

In many experiments in quantum mechanics, a single photon is sent to a mirror which it passes through or bounces off with 50% probability, then the same for some more similar mirrors, and at the end we get interference between the various paths. This is fairly easy to observe in the laboratory.

This experimental fact tells me that the photon has 50% probability to scatter back each time and 50% to go through the collective field of the atoms of the mirror.

The elastically scattered photon, by definition keeps its momentum and phases to the universe except for the direction.

Why is it that the mirrors retain no or very little trace of the photon's path, so that very little decoherence occurs?

Your problem then is with the through going photons in a 50% transparent 50% reflective medium.

The elastically scattered ones by definition/solution-of-the-qunatum-mechanical-boundary-condition-problem cannot leave a mark . Classically a ball bouncing off an infinite mass wall does not lose any energy. If there is some diminution in energy of the scattered photon due to momentum conservation, with the whole mirror, it will be very small, because of the size of the mirror and the "size" of the photon.

The ensemble of the incident photons on the mirror finds the medium transparent, because it does not interact with the electrons forming the molecules and structure of the solid . It is as if the solid is not there. Why? Because the energy levels of the molecular, and atomic structure, including vibrational and rotational , do not match the frequency of the wave, ( in this case optical frequency). If they matched it, the medium would not be transparent. Thus the reflected wave and the through going wave keep their original phases and can produce the interference patterns .

Photons pass through the material because they don't have sufficient energy to excite a glass electron to a higher energy level. Physicists sometimes talk about this in terms of band theory, which says energy levels exist together in regions known as energy bands. In between these bands are regions, known as band gaps, where energy levels for electrons don't exist at all. Some materials have larger band gaps than others. Glass is one of those materials, which means its electrons require much more energy before they can skip from one energy band to another and back again. Photons of visible light -- light with wavelengths of 400 to 700 nanometers, corresponding to the colors violet, indigo, blue, green, yellow, orange and red -- simply don't have enough energy to cause this skipping. Consequently, photons of visible light travel through glass instead of being absorbed or reflected, making glass transparent.

In general, how do I look at a physical situation and predict when there will be enough noisy interaction with the environment for a quantum state to decohere?

The phases will be lost if the scattering is inelastic. If inelastic scattering is dominant the beam will decohere, which happens with non reflective surfaces. Reflection and elastic scattering are a different side of the same coin.had

When one reaches the 10^23 or so number of quantum mechanical entities entering collective interactions the loss of coherence will be dependent on the material and the particular boundary conditions of each problem. For light absorption coefficients etc can characterize the decoherence. For other set ups the individual boundary conditions and the collective emergence of phenomena has to be thought about on a case by case basis ( superconductivity comes to mind).