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When a photon of light hits a mirror does the exact same photon of light bounce back or is it absorbed then one with the same properties emitted? If the same one is bounced back does it's velocity take all values on $[-c,c]$ or does it just jump from $c$ to $-c$ when it hits the mirror?
Or, is the phenomenon of a mirror better explained using a wave analogy? If so, what is this explanation?
How do mirrors work? is closely related to your question, if not a precise duplicate.
We normally think of photon scattering as absorbing the original photon and emitting a new one with a different momentum, so in your example of the mirror the incoming photon interacts with the free electrons in the metal and is absorbed. The oscillations of the free electrons then emit a new photon headed out from the mirror. Unlike e.g. electrons, photon number isn't conserved and photons can be created and destroyed whenever they interact.
If you think of this in terms of quantum field theory, which is really required to give meaning to the photon, then all you are able to say is that the photon can take any of all possible paths from where it is emitted to where it is absorbed. These paths will contain paths where the photon momentarily splits into an electron positron pair, where the interactions with the electrons in the mirror involve all sorts of virtual particles, where the photon travels in directions which are far from the classical trajectory etc. The total amplitude is given by the sum of all these possibilities and they can all occur. In the classical limit this sum over all paths gets dominated by the contributions closest to the classical straight line path of the photon with velocity $c$, so classically we see light travel in a straight line at velocity $c$, and obey the laws of optics. However if you really wanted to follow the path of an individual photon you would see that it could do any of a spectacular number of things (and unfortunately our attempts to observe the photon would interfere with its path). If you want to understand this better, I highly recommend Feynman's description of it all in his lectures here or in his book taken from the lectures: "QED, the strange theory of light and matter".
I think it can probably be misleading to think of the matter as "knowing" which way to emit the reflected photon. In order to fully describe this process it seems necessary to combine the mechanism of the interaction of light with matter, which allows for the possibility of absorption and radiation by electrons within the lattice of a material, with the Feynman path integral formulation as mentioned already in order to sum the amplitudes for an event to occur. The observed fact of equal angles of incidence and reflection is due to it being the route with the greatest coherence of phases. Reflection at a different points on the mirror will tend to cancel out rapidly as you depart from the point of equal angles. (this observed path is also the shortest path by Fermat). All that is then left to do is to explain how it is exactly that photons induce movement in charges, in a way which will clearly depend on the detailed structure of the material.
Revisited after some negative votes.
The photon is an elementary particle in the standard model of particle physics. This means it is a quantum mechanical "particle" described by a wavefunctions which will give for any interaction the probability of a specific photon to interact. In the case of a mirror, ray optics describe the most probable path of a photon before and after an interaction.
As a particle, when hitting matter in solid state it may scatter elastically with the collective electric field of the medium it hits, To have a mirror all photons must scatter elastically from the solid state lattice that is the mirror.
Elastically means that the photon leaving an interaction only changes direction in the center of mass. The center of mass of a photon and a mirror is effectively the laboratory frame as the mirror is of order ~10^23 molecules in mass. Thus the elastically scattered photon does not lose energy, and the colors of the images it helps to build up do not change. How classical states emerge from the underlying quantum field theory state is described here.
A photon will be absorbed if its energy, given by $E=h\nu$, fits some energy level of the atoms, (molecules, system) it hits and then a re-emitted photon may change both direction and energy with respect to the originating one, i.e. if the reflected one changes frequency because of the re-emission,and loses the phase it cannot contribute to a faithful image. The photon of course goes with velocity $c$ (as all photons) whatever its direction (elastic scattering means only change of direction and not energy).
The diagrams describing photon scattering are similar in first order to the ones below,
where the electrons are virtual, interacting with the mirror lattice and the outgoing photons have the same frequency/energy.
In elementary particles "same" can only have the meaning on specific variables in specific interactions. In elastic scattering the photon entering the interaction and the photon leaving have the same frequency (energy) and each photon has a probability to be scattered at an angle. The classical wave built up by the zillions of photons in superposition of their wavefunctions have to keep the phases so that the macroscopic images can keep their color and dimension, i.e. be "mirrored".
When considering the question "does the exact same photon of light bounce back or is it absorbed then one with the same properties emitted" it's important to remember that photons are indistinguishable bosons. The question suggests that such photons would be distinguishable, when they are not. In the end, It's just "a photon" with X momentum.
There is a meaningful distinction to be made about the reflection process though - There's no absorption and re-emission process going on here, as Anna V points out this would imply very different behavior. Photons and their evolution in time (up until, debatably, their measurement) is properly described as a wave, not as a particle which "bounces".
My classical picture is that when the incident electric field hits the metal (usually silver) surface of a mirror, the electrons, feeling the electric field, oscillate. As the electric field is oscillating very fast, the electrons tend oscillate out of phase with the field like a spring that's being wiggled too fast. The out of phase oscillation of the electrons produces a new wave which interferes deconstructively with the incident wave and cancels the field amplitude going into the metal, but produces a new field going away from the metal. Considering a large beam hitting a large area of a mirror, and therefore many electrons, all the electrons oscillate together and produce more collective interference effects, but the same picture holds. Tracking this down precisely to the quantum field theory level should be possible, although I personally would probably struggle to gain much intuition from those explanations.
The mirror has to be sufficiently cool enough to be contained in a mirror box. You might not think this answers your question, but in a way it does. Consider the energy put in to keep those mirrors nice cold, it will absorb less energy. But if the box heats up, energy will tend to escape and absorb into the mirror.
We should also include effects of moving mirrors. While the photons bounce off a moving mirror, they can lose or even gain energy (!), depending on the motion of the mirror.
Like in our original case, when the mirror is not moving, then the energy will generally stay the same, but the number of reflected photons will usually be lower, than the number of incident photons, so the total energy of reflected light will be slightly lower.
The question is excellent, and is a variation on the question, "What is a photon?", which is a variation on the question, "What is a quantum?".
A step towards an answer is to consider the implications of the "no-hiding theorem". One implication is that if a photon is absorbed, the packet of uncertainty that it represents simply takes a different form; it does not vanish. It is not unreasonable to consider that the "quantum" is the "packet of uncertainty", and that it can be passed along from one particle to another or from one particle to an assembly of particles.
The answer I give below is based on that point of view.
The answer depends on whether the term "photon" refers 1) to a particular packet of electromagnetic energy, or 2) to the packet of uncertainty (that is, the "quantum") associated with a particular packet of electromagnetic energy.
In the first case, we need to decide whether or not the packet of electromagnetic energy loses its identity in the process of being reflected from a mirror. I'd say it does not lose its identity, just as the "packet of energy" in a swinging pendulum is always the same energy even though at one moment it's in the form of gravitational potential energy and at another moment it's in the form of kinetic energy. In other words, in the first case I'd say the answer is "yes, it's the same photon before and after reflection".
In the second case, the answer is "yes", because essentially all of the quantum uncertainty of the incident photon survives in the reflected photon.
The answer here was correctly given by John Rennie and Anna V. But some parts were not covered, and I will try to explain it in more everyday understandable form. So the individual photons are absorbed, and the a 'new' photon is re-emitted, with the same properties, except for the direction (which is in a normal mirror opposite, or can be something little bit bent depending on the type of the mirror). So the answer to your question is:
And then before the re-emission happens something very important happens that the above answers on the page did not address, and this is the key to your question as to why the EM wave itself slows down in material, but the individual photons still move at speed c. And this thing that happens is the excited state of the electron/atom. See the absorbed photon transmitted it's energy into the electron, so the electron (and the atom) is in excited state. In order to do a re-emission, the electron needs to go back to ground state (or it's previous state). But what was not addressed, is for how long is the electron/atom in the excited state? See the absorption/ re-emission itself is instantaneous in QM. But the excited state itself is not always. It has an average lifetime. If you do enough experiments, the average lifetime of the excited state of the atom will be around 10^-8sec.
So why is this 10^-8sec so important? Because the specific photon (the one absorbed/re-emitted) does not move for this period. On average, it will slow down that photon ONLY if we would consider the absorbed/re-emitted photon the same thing. But we don't say that. We say that the photons are slowed down in a herd (called the EM wave). Because the absorbed/re-emitted photon is still not considered moving through the excited state. It is in form of energy in the electron for average 10^-8sec. And we do not calculate that time into the speed=distance/time calculation. Why? Because seemingly for calculation reasons either #1that photon re-emitted is still not the same identity as the on absorbed before or #2 we don't consider the photon existing as a photon (only for speed calculation reasons) throughout the excited state while it is in form of the electron's excess energy. We usually say that it is the same photon. So we must say the #2. But the experimental reason is just that the EM wave's measured speed in the glass for example is just slower. And now you see what is the reason for it, the excited state's average lifetime.
After(during) the excited state, the 'new re-emitted' photon with same properties is emitted except the direction. Its wavefunction describes it's motion and it's speed is c. It has no mass. The emission is instantaneous, so the new photon's speed is c.
But why is it's speed c instantaneously. Because it and everything massless gets created/exists at speed c in space. The basic misconception you have is
-You think things exist/are created with 0 spacial speed and then need to speed up. That is not true. The universe and the four-vector is set up so that everything massless gets created/exists at speed c. You have to slow it down. How? by gaining mass or by transmitting it's energy into another material with rest mass by getting absorbed.
-you are trying to imagine a particle (photon or anything) with a gas pedal. It does not have a gas pedal. Nor does anything else, we all only have a brake pedal. Everything massless travels(gets created, exists) at speed c, because of the four-vector and the universe's set up that way. You have to SLOW down if you want, and to do it is to gain mass. You are experiencing time as you do because you have mass, and because you have the ability to interact with ordinary things (material/energy) in space as you do.
-there is no agreement on as to what material with rest-mass consists of. Some say everything has rest-mass because massless gluons in it oscillate in some kind of confinement. if that is true, then everything as it is created travels at speed c, and only the sum of it on bigger scale slows down in space, by gaining rest-mass (gluons oscillating, Higgs fiels etc.).
We do not know if the re-emitting electron is the same electron as the absorbing one. It should be because that is the one in the excited state. that has to fall back to ground state.
We do not know how the electron 'knows' that it needs to emit the 'new re-emitted' photon in the exact opposite direction (for mirror), or in almost the same direction (for glass) it just does. Glass does the same thing, absorbs, re-emits, just the re-emission direction is almost the same as the absorption (that's why light passes through glass, without changing the waves too much, just slows them a little bit down cause of the absorption-re-emission and the average lifetime of the excited state). See in the glass the individual photon gets absorbed-reemitted billions of times as it passes through the glass, depending on the thickness of the glass measured in atoms. Each time it gets absorbed-reemitted it needs time of average 10^-8secs (this varies depending on the type of atom, this exact time is for H atom). So if the thickness is a billion atoms, it gets absorbed-reemitted at that magnitude and the slowdown will be 10^8*10^-8=1sec. Thats an example, but it shows that this is an amount of time that is already measurably slowing down the EM wave. The denser the material/medium is, the more emission/absorption, the more it slows down.
Addition to this, on individual electron/photon level we don't know how it knows which way it as to go, but on the EM wave level the emitted EM wave has some phase shift (π/2π/2) relative to that of incoming wave, that causes it to lag behind.IN QM, EM waves are emitted everywhere (not only along the direction of incoming wave). It's just that the other paths taken by light interfere destructively and cancel out one another. The forward radiation goes along with the wave and the stuff that's reflected backward is what you see as the 4% reflected light (from glass). In a mirror the EM wave is shifted to a phase of ππ which causes the forward radiation to interfere destructively and hence, light doesn't pass through metals. Now, the backward radiation passes through glass, gets a few partner waves and that's how you see your face in the mirror.
So on the individual electron/photon level we don't know how it knows which direction to re-emit the 'new' photon, but on the level of the whole herd, the EM wave, this is the explanation for the direction change that it is not a change, it is more of what is left after EM waves in all direction cancel out except the final direction that you see.
OK now I explained why EM waves slow down in dense material/medium. And that the speed of individual photons is always c (measured locally). But I did not explain if that also slows down the EM wave in case of a mirror. It does. The two way speed of EM waves is on average slower then the one-way speed. But the difference if really not so much measurable, and it would need an experiment being repeated many times. Because the excited state's AVERAGE lifetime is 10^-8 secs. The actual time might even be 0. But if you do enough reflections/two way speed tests, it will show on average. Why is this important? Because if you take the shapiro delay, it will show you that on a mirror/reflection test from the Venus, the EM waves (radar signals) will have a 2*10^-4 slowdown due to GR effects. It is on a approx 20min travel. But see the excited state's average lifetime would only on AVERAGE affect it 10^-8secs. So this effect is not so important as it is so small and does not change the results so much. It is only important when passing through thick material when there are many absorption/re-emissions.
