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When light (photon) is reflected the the original photon is absorbed by an electron and then emitted again. Does this "new" photon have the same wavelength, frequency etc. as the original?

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  • $\begingroup$ What you are describing is not reflection, but photoluminescence. What you want is called Compton scattering. $\endgroup$ – hpekristiansen Jan 19 '14 at 15:59
  • $\begingroup$ @Hans-Peter E. Kristiansen Is reflection a similar process to photoluminescence? $\endgroup$ – threewisemonkeys Jan 19 '14 at 16:32
  • $\begingroup$ Depends on what you mean by similar. In both cases there is an incoming and outgoing photon and an electron(within an atom) involved. You can read about both processes on wikipedia. $\endgroup$ – hpekristiansen Jan 19 '14 at 19:29
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Like any quantum field you can approximate light as either a particle or a wave. However you need to be clear that both are just approximations to the true nature of light and like all approximations they work well in some circumstances and badly in others.

In this case the photon model is a poor way to describe the process of reflection. Reflection doesn't involve photons being absorbed then re-emitted. You're correct that the oscillating light wave of the light interacts with electrons in the reflector, and the resulting oscillating dipoles reradiate light. However, while this is easily described using a wave model it's hard to describe using photons. As a general rule the photon model works well when light is exchanging energy with something so for example it would be a good model if the light was ejecting photoelectrons from the mirror. When we're studying the propagation of light the wave model is a far better approximation.

So I don't think your question can usefully be answered as it's currently phrased. However we can say that the reflected light has approximately the same frequency as the incident light. A say approximately because in principle there are effects that could change the frequency. For example if the mirror is floating in zero-G then the momentum change when the light is reflected will cause the mirror to accelerate by a tiny amount and therefore red shift the reflected light by a tiny amount. In most circumstances we can ignore these small effects.

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  • $\begingroup$ Assuming the mirror was stationary in the first place. And let's not get into choices of reference frame here, shall we? Assume, as I believe the OP does, that the measurements are done in the frame of the emitter. $\endgroup$ – WhatRoughBeast Jun 2 '15 at 17:34
  • $\begingroup$ @WhatRoughBeast: the problem with the original question is that propagating light is poorly described as photons. The reflected wave will be the same as the incident wave except of course that its momentum points in a different direction. $\endgroup$ – John Rennie Jun 3 '15 at 5:09
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There is no absorption and re-emission involved in reflection. None whatsoever.

This can be proved with a laser; any colored laser.

Using a collimated expanded laser beam, incident on the reflecting surface, at an angle, you will get interference between the incident wave and the reflected wave. That is only possible if the two are coherent, which means the incident wave is refracted, and not absorbed. The reflecting surface is not a lasing medium, so there couldn't be any coherent stimulated emission, and moreover, be so for any incident wavelength.

The photons are not absorbed during reflection.

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  • $\begingroup$ If that's true, then there can be no absorption by a laser beam being transmitted through an atmosphere. In that case laser beams must propagate through an atmosphere at c. And they don't. You might want to rethink (or restate) that one. $\endgroup$ – WhatRoughBeast Jun 2 '15 at 21:08
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During reflection does the emitted photon have same properties?

During reflection the color does not change and the phases do not changes otherwise the reflected images would be fuzzy.

I will complete the other answers by examining the mirror material.

Not all materials reflect. Reflecting materials are materials where the microscipic domains in the solid are coherent forming crystal structures. A highly polished metal mirror, for example reflects classically because of the very high reflection coefficient. That is why all mirrors are glass backed be some type of metal in highly reflective mode. (Glass itself, as water too, have a transmission coefficient which reduces the reflection coefficient to a very small number except at certain angles).

Photons are a different framework, a quantum mechanical framework . In quantum mechanics we have solutions of equations with boundary conditions for each given problem. The problem of "a photon hits a highly reflective metal surface and is reflected" has a very specific solution with the given boundary conditions. As you know quantum mechanics has specific energy states that ensure that there is no dissipation of energy at that energy state. Thus, if the photon is reflected and its energy(color) is still the same it means that a total state ( photon/ metal crystal) gave the solution of an elastic scatter where no energy was lost ( within the heisenberg uncertainty principle). Nature is very good at analogue solving of problems :). Good reflectors have many states that can accommodate a solution for reflection,( metals mainly because of the bands of electrons not attached to individual molecules and atoms).

So both classically and quantum mechanically there is consistency: the photon does not lose its energy in interacting with the whole crystal/domain structure of the reflector.

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Your title and question has some in-consistency, for example, photon can also have spin, and the field itself can also carry orbital angular momentum et.al. But from fresnel equation:


r=(n-1+ik)/(n+1+ik)


where n and k are the real part and imaginary part for the materials, you can see the reflection spectra already contain the dispersion of the materials, and not only the amplitude, the reflection phase is also changed, from those information one can detect the properties of the materials: what human eyes see most is those reflected light, right? The photon reflected can also carry information for the spin and orbital angular momentum of the materials et.al.

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