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 inelastically eachback each time over and 50% to go through the collective field of the atoms of the mirror, and 50% elastically.
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?
The inelastically scattered photons have left their mark onYour problem then is with the mirror (heat probably, or other excitations)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 that manages to be reflected 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, ( 50% in your setupthis case optical frequency)retains. If they matched it, the phasesmedium would not be transparent. Thus the reflected wave and coherence of the through going wave keep their original beamphases and thuscan produce the interference patterns will appear. It is what "elastic" means.
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).