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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?

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Related to – Waffle's Crazy Peanut Aug 30 '12 at 1:27
up vote 3 down vote accepted

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".

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Sorry, but I think this answer does not answer the question. It just tries to explain classical theory in light of QED... A proper explanation would tell why the peak of the probability is highest for specular reflection... xD – J. C. Leitão May 21 '13 at 10:19

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.

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But how does the emitter know the direction in which to emit the photons of an incoming beam, so that the reflection angle is correct? – Arnold Neumaier Aug 30 '12 at 10:11
The reflected photon interacts with the total fields it sees particularly at optical frequencies which are so low in energy. When absorption and re emission happens the phase of the oritinal photon is lost – anna v Aug 30 '12 at 13:10

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.

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I think a photon is a packet of EM radiation, which affects electron clouds as if they were compasses. – Cees Timmerman Jul 21 at 16:11

Ray optics describe the best the path of a photon. As a particle, when hitting matter solid state it behaves like a billiard ball scattering elastically with the collective electric field of the medium it hits, when it is reflected.

It 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 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).

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Your first paragraph seems to answer in the affirmative that the photon's velocity can take on all values on [-c,c] as it is scattering elastically. This is misleading at best. A photon is not a classical object with "primitive this-ness", it is a vibration in a field. It makes no sense to talk about a photon as though it is slowing down and changing direction. The group velocity is what changes. – user1247 Aug 30 '12 at 8:47
@user1247 !!! elastic scatter means change only in direction not value of momentum, in classical physics also. Elastic scattering crossections exist for all scatterings of elementary particles including photons. When there is a slow down, of course in particles with mass, it is called inelastic. – anna v Aug 30 '12 at 13:06
you gave the billiard ball as an example. Lets consider it scattering in one dimension. Its velocity changes continuously because the acceleration is not infinite. Infinite acceleration is not only non physical, it is also wrong and misleading in the case of a photon. The group velocity can do such things, but a single-photon description breaks down and does not correspond to any physical reality. – user1247 Aug 31 '12 at 16:20
@user1247 what do you mean "scattering in one dimension. The photon is not one dimensional, it is four dimensional. – anna v Aug 31 '12 at 17:48
You gave the billiard ball as an example. When a billiard ball collides with another billiard ball, and scatters elastically, its velocity changes during the collision. This is a simple fact. The number of dimensions doesn't matter, but of course it is simplest to consider a 1d collision. – user1247 Aug 31 '12 at 20:38

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