What happens at the QM level with a photon hitting a mirror? I have read these questions:


*

*Reflection of light at a microscopic level

*About photons and mirrors

*Photons and perfect mirror

*Do photons bounce at mirrors?

*How do mirrors work?

*What happens when a photon hits a mirror?
And specifically Anna V's and John Rennie's answers. 
John Rennie says when a photon hits a mirror, it is absorbed and re-emitted.
Anna V says it is not absorption and re-emission but it is elastic scattering.
She says in her comment to John Rennie's answer: 

John, please look at this again, as it is a chosen answer. is not correct for a mirror. Absorption and re-emission would change the phases (the reemitting source would have random direction) and no images would be transmitted to the eye , to call it a mirror. It has to be elastic scattering for a mirror

She says that with absorption and re-emission the problem is, that the re-emitted photon can change direction and energy too, and so the phase of the photon will change too, and that is not a mirror image. Elastic scattering is the one that will keep the energy level of the photon and its phase too.

re-emitted photon can change both direction and energy with respect to the originating one, and the originating one loses energy, i.e. changes frequency. If it is reflected, it of course goes with velocity c (as all photons) whatever its direction (elastic scattering means only change of direction and not energy).

So there can be two cases here:


*

*When a photon hits a mirror, the photon gets absorbed and re-emitted.

*There is no absorption and re-emission, but only elastic scattering (Rayleigh).
None of these answers give an explanation to what happens when a photon hits a mirror.
Question:


*

*Which one of the two is it when a photon hits a mirror?

 A: The problem with any attempt to try and explain some macroscopic phenomenon in terms of the detailed interactions at the lowest level is that one encounters lots of complexity. If one looks at the interactions of photons with the electrons at the fundamental level, one finds that there are no vertices for elastic scattering; photons are always absorbed and re-emitted. Yet, it is true that a mirror must be doing some form of elastic scattering. How does that work?
Well, it is the result of an infinite number of such absorption and re-emission processes that happen in a quantum superposition. These different events interfere with each other in such a way that constructive interference builds up the picture of an elastic scattering process.
A: Here  are the compton scattering  lowest order diagrams

**There is nothing that forbids elastic scattering **, i.e. energies unchanged in the center of mass and only angles different.
Elastic scattering is logically necessary because if photons were absorbed and re-emitted, the phases between the photons from which the macroscopic classical electromagnetic light wave emerges would be lost, image coherence would be lost in mirrors. This is what happens with non-mirrors, mainly the scattered photons become point sources incoherently reflected and absorbed. Also, it is necessary for mirror reflection to keep the colors. Absorption and reemission would change the energy, if not elastic, and the frequency of the photons would change and thus the colors built up macroscopically.
To answer the title:
The mirror is macroscopic and its lattice has an effective electric field from the surface atoms. The photon hits the field by exchanging a virtual electron, where also the electron on the left is virtual to represent the field because it is tied up in a lattice. The photon scatters elastically in the center of mass, keeping the phases of the emergent beam  and thus gives a faithful image back. The center of mass is almost identical to the laboratory center because the mirror is macroscopic and effectively it is the lattices mass that enters the kinematics (similar to a ball hitting a wall classically).
The question remaining is how the phases that define the classical beam as a superposition of photons are kept. My opinion is that one should go to the emergent E and B fields from the individual wave functions of the zillions of photons which also collectively interact with the lattice fields, making most probable scattering direction the classical ray direction of reflection, but I have no link or proof for this. Motl's link above justs deals with the emergence of the classical electromagnetic wave from the quantum mechanical fields and particles.
The type of surface decides whether coherent elastic scattering will dominate or diffuse scattering because of surface anomalies, or absorption and re-emission. 
What is more interesting is when the lattice is transparent, where the photon has to interact with the whole lattice and keep color and phases. Here it is more evident that emissions and re-absorptions on individual charge centers cannot explain real transparent media that transfer colors and images. These enter in distortions and color changes.
There is this publication in Russian which calculates elastic scattering from bound electrons too.
A: Most metal, when polished, appear grey and reflect all visible light frequencies equally well.
The reason is the excellent electrical conductivity which presents an impedance of a few milliohms to a wave that usually, in the vacuum, encounter about 340 ohms.
The large mismatch force the wave to reflect back toward the source in the case of wires or cylinders.
For the flat 2D surfaces such as the mirror, the reflection happen on the z-direction as demonstrated eloquently by Physics Girl and Veritassium.
I notice that many answers are stating absorption/emission of virtual photons which is different from fluorescence.
