# Do photons bounce at mirrors? [duplicate]

I am referring to the simple mirror. How can a photon be reflected and what does that mean? Does it instantaneously reverse direction (in the case of 180 degree bounce), does it loose energy? What effect does this have on the photon wavelength? Heat may be generated (energy lost by the photon, why are mirrors so go at "reflection" Or does it not "reflect", but is absorbed by a mirror and re-emitted?

It's properly reflected, at least in QED (which is a step behind the present best model, which is called QFT).

In QED you do quantum mechanics by trying to assess the probability that a detector which Alice is holding is triggered, given that an emissions source which Bob is holding emits a particle of light. In order to do this, you assign every possible path of photons between Alice and Bob a complex number called an "amplitude." The rules for amplitudes are simple: first, if you can divide a path into several parts, by saying "first it has to go from X to Y, then it has to go from Y to Z," then you can multiply those two amplitudes to get the path from X to Z. Second, you can collect several paths into one amplitude to reach a point (or whatever) by summing their amplitudes. Third, by default an amplitude for a photon is this: the scale factor for the complex number goes like $1/d$ where $d$ is the distance traveled; second for simple paths, the phase of the complex number goes like $2\pi~f~t$ where $f$ is the frequency of the photon and $t$ is the time it travels for.

So when we want to include a contribution from reflecting off of a mirror, we sum up over all of the places on the mirror which a photon could hit, which is the entire mirror. But something very important happens: $f$ is a very high number, so high that even the wavelength $c/f$ is in the hundreds of nanometers for visible light, it's tiny. So it turns out most of these complex numbers totally cancel, in the final sum of the amplitude, from paths that hit the mirror within a few microns to either side, which have the exact opposite phase.

However there is also a narrow region where these amplitudes are not changing very much, and it's because the first derivative of this phase with respect to the position along the mirror is 0, so the phase isn't changing much as you move along the mirror. We see that the phase comes from the travel time, more or less, and it turns out that this path minimizes the total travel time. This general principle governs all of a photon's behaviors in QED, in the end.

This principle has also been known to be the case for reflections since perhaps 60 CE or so, and for refractions since 1021 CE, and finally was hypothesized to hold for all light behavior in 1662 CE by Fermat, so it is now known as Fermat's principle. So you will find that this "least time" principle perfectly encapsulates the "equal angles" principle of reflection. The fact that the other places on the mirror cancel out due to adding complex numbers with opposite phases, means that you could erase almost all of the mirror except for this part and get an identical amplitude -- hence the photon is "coming from" that part of the mirror.

In normal reflection, the photon's appearance at the detector depends on a phase which does not have contributions from any of these other paths, though all of them collaborate to get the photon where they're going. But the real test is if you start saying "well I'll erase the mirror with some periodicity so that I eliminate some of these places where the amplitude's phase is exactly opposite within a micron's distance, now it should scatter at a completely new angle, right?!" -- and it does; and this turns out to be the source of the rainbows you see in the bottom of CDs and DVDs, as the regular spiral groove etched into the mirror surface reflects different wavelengths at different angles, therefore coming into your eye from different places on the DVD...

The photon effectively does every possible thing to get to the detector, but its wavy nature causes most of those paths to interfere; only a couple of places where the phase doesn't interfere allow for valid reflections. But in any case, in normal reflection the photon perfectly reflects without losing energy or getting absorbed, it is the "same photon" for the entire calculation. If the photon gets absorbed and re-emitted then you will not see a very shiny-looking surface as there will be other detection events possible from other paths; this is a natural consequence of "a photon stays over in this state for a time, then the state decays and you see those photons."

Reflection is a macroscopic effect , as far as images go, and is quite well described by classical electrodynamics.

The electromagnetic wave, which light is also, is riding on zillions of photons in a complicated way.

An individual photon can scatter elastically from an atom, a molecule, and also from a lattice. If the elastic scatter is reversing the photon's direction of motion, it can be called a reflection of that individual photon.

The way the photons combine so that one can have a coherent reflected image depends on the reflecting surface. The individual photons hitting a mirror must not only scatter elastically but also keep the phases with all the other photons so that the image remains coherent and not dispersed. (That is why not all surfaces are reflecting images but special conditions have to hold on the lattice).

Absorption and re-emission loses the phases and cannot carry a reflection of a macroscopic object.

There is nothing instantaneous in physics as special relativity holds and nothing transmitting the interactions can go faster than the velocity of light .

• The nontrivial part of the answer is in your 4th paragraph, but you don't explain: how do photons get absorbed and re-emitted in a coherent way if this process will of course destroy the phase information? – Dwagg Jan 17 '20 at 3:41
• @Dwagg The photns contained in images are not absorbed and reemitted, is what I am saying. They have to scatter elastically, so as not to change color, and keep the phases that make up the image. This will depend on the surface of scatter, that is why all surfaces are not mirrors. Absorption and reemission destroys images., because even if the energy of the photon is kept, the phases between the wavefunctions of the photons that make up the image are lost.. – anna v Jan 17 '20 at 5:38
• When you say elastic scattering of a photon, should I picture a Feynman diagram with virtual electron exchange? If the electron is treated as stationary and backreaction is neglected, then I see that the photon momentum is preserved; is it easy to see in QED that phase is preserved? – Dwagg Jan 17 '20 at 14:38
• Just to clarify, are you saying for photons that mirrors = elastic scattering but general surfaces e.g. apple = absorb & re-emit – Dwagg Jan 17 '20 at 14:41
• @KaranBhatia To have a reflected image , means that all photons carrying the information are coherent, have a fixed phase difference, so that the image is coded in the superposition of photons that make light. It depends on the surface, a regular lattice is needed , so the elastic scattering of photons is "photon+lattice" and the phases between photons can be kept, and thus the image information. – anna v Mar 31 '20 at 8:34