Energy loss of photons crossing a dielectric medium How much energy do photons lose when crossing an "isotropic" dielectric medium? What would be possible mechanisms.
I assume the straight forward answer would be that the photon loses zero energy and experiences no red-shift.
However, here are some ideas for energy losses:


*

*Would it "lose" energy through expansion of the universe while in the medium?

*Is there any other tiny energy transfer (eg. radiation pressure forces against self-induced refractive index contrasts)?
What would be additional mechanisms and what would be the largest contribution to the energy loss?
(please disregard that the photon exiting the material needs to be the same photon entering the material)
 A: Energy loss by photons occurs in a dielectric medium when the photons interact with the atoms of which the dielectric is made. This interaction does not normally result in a red-shift of the photons, but rather an attenuation of a beam of photons due to photons being absorbed.  
Depending on the nature of the material and the energy of the photons, the energy of the photon might be fully absorbed because its energy is equal to the difference in two atomic energy levels; of course the energy is not really "lost" but converted to a different form. 
Microwave energy photons may excite vibrational states of molecules (e.g., water in a microwave) and the energy is ultimately converted to heat.
Very high energy photons might interact with the nucleus and be absorbed, resulting in an excited nuclear state. 
Metals can be described by a dielectric function, and in this case "losses" can be associated with currents produced as the electron gas responds to the changing electric field. And currents flowing result in "ohmic loss" in the classical view, so the energy can be converted to heat. Energy can also be transferred to surface electromagnetic waves (surface plasmons) when metal surfaces are not flat.
