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I understand that photons do not change their polarization when they pass through glass because they are not absorbed in the glass due to the band gap.

I also now understand that reflection does not involve the absorption of re-emission of a photon, although I am somewhat unclear at the quantum level what is happening in reflection. It is clear to me that the object reflecting the photon (like a solar sail) slightly increases in momentum, and since momentum is conserved, the photon must have a slight decrease in momentum, which means that the photon must decrease in frequency somewhat upon reflection.

So my question is why does polarization not change on reflection. Is this because it's the same photon (whatever that means), so there is no opportunity for its quantum state to change, even though it's momentum vector has changed?

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  • $\begingroup$ "the photon must have a slight decrease in momentum" Incorrect. As the photon is reflected its momentum is reversed, which is a change of -200%. $\endgroup$ – my2cts Nov 10 '19 at 19:01
  • $\begingroup$ @my2cts, if the momentum changes by -200%, and if the thing the photon bounces off starts moving, where does the extra energy come from? $\endgroup$ – vy32 Nov 10 '19 at 19:29
  • $\begingroup$ Aha , you mean that apart from the direction reversal there is a small change in momentum. Sure. $\endgroup$ – my2cts Nov 10 '19 at 20:04
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    $\begingroup$ Correct. There has to be some small decrease in the photon's energy (hf), because some energy is transferred to the object from which it bounces. $\endgroup$ – vy32 Nov 10 '19 at 23:30
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At the quantum level the photon is indeed absorbed and re-emitted ("scattered") all the time in the materials including the mirror. The resulting macroscopic propagation is the result after all the quantum amplitudes of all the scatterings and propagations interferences are summed up.. For practical purposes the classical optics wave description of light reflection works fine, but I guess you ask about the underlying action :)

In either case, polarization might not depend on the linear momentum change but angular momentum can be transferred to the reflecting object, thus changing polarization of the reflecting photon (see "optical torque").

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  • $\begingroup$ I disagree with this absorption reemission picture. It is elastic scattering, not absorption. I have made this point many times on this site. $\endgroup$ – my2cts Nov 10 '19 at 18:56
  • $\begingroup$ Wait, is the photon being absorbed and re-emitted as it moves through glass? $\endgroup$ – vy32 Nov 10 '19 at 19:27
  • $\begingroup$ @my2cts I guess it boils down to the context. From a field-theoretic point of view, scattering is absorption + emission.. though I agree you could make it a difference, but the OP would have to specify in what context the question is asked then. I assumed from a QED point of view. It's certainly true that I didn't mean the photon is absorbed/re-emitted in a higher level kind of way (like in stimulated or spontaneous emission). $\endgroup$ – BjornW Nov 11 '19 at 15:39
  • $\begingroup$ @BjornW The elementary excitations involved are not single atomic but collective excitations. The QED creation and annihilation operators refer to the latter excitations. The context here is condensed matter, not atomic physics. $\endgroup$ – my2cts Nov 11 '19 at 17:02
  • $\begingroup$ @my2cts I assume the context is decided by the OP, and not by us :) $\endgroup$ – BjornW Nov 12 '19 at 18:17
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The answer is that the polarization does change upon reflection unless the medium it reflects off is a perfect metal. Only for perpendicular incidence the polarization remains the same, although effectively right and left turning circular polarization transform into each other as the propagation reverses. For all other angles the reflectivity and transitivity depend on polarization. At the Brewster angle of incidence only TE polarization is reflected. Polaroid shades use this effect to suppress reflected light.

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  • $\begingroup$ can you explain a little more about Polaroid blocking reflected light preferentially. Is this true only of reflected sunlight, and does this has something to do with Raleigh scattering? $\endgroup$ – vy32 Nov 11 '19 at 3:11
  • $\begingroup$ This in principle is true for all EM radiation even at energies as high as 100 $eV/c^2 $. You should perhaps read up on the Fresnel equations. $\endgroup$ – my2cts Nov 11 '19 at 6:41

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