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In an answer to a another question, the poster states without sources the following:

From a quantum mechanical perspective, all light scattering is a form of absorption and re-emission of light energy. Photons don't bounce off a surface.

If this is true, what is the evidence for it, or is it a theoretical postulate under quantum mechanics?

The reason I ask this is that the statement would seem to be contradicted by the phenomena of specular reflection. Since specular reflections always faithfully reproduce the spectrum of the incident light, that would suggest that no light is be absorbed and that, indeed, photons are "bouncing" off of the surface.

Ambrose Swasey
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  • Possibly how you always see yourself AND the object behind a pane of glass when light shines on it. If light simply bounced off a surface, then you would see your flashlight itself when it shines on, say, concrete, or even asphalt. – Curious Fish Oct 31 '18 at 16:07
  • Your edit does contradict my previous statement, but if you look at the object/surface from a different angle, you can still see right through the specular surface. Even if you look at it from the same angle, careful observation may indicate subtle translucency. – Curious Fish Oct 31 '18 at 16:17
  • https://physics.stackexchange.com/a/13504/174203 this may be an answer to your question. – Curious Fish Oct 31 '18 at 16:26

2 Answers2

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The fundamental theory that described the interaction between light and matter is quantum electrodynamics (QED). It has been exhaustively tested and found to be in excellent agreement with experimental observation.

According to QED the fundamntal interaction is given by a vertex with three legs: two for the charged fermion (electron) and one for the photon. It means that the photon is either absorbed or emitted - not reflected.

But here is the catch: a single vertex cannot describe a physical process, because by itself it cannot satisfy momentum conservation. Therefore, a physical process will always involve an even number of vertices, the simplest being a case with two vertices, connected by a (virtual) particle that is "off-shell" (it does not obey the dispersion relations). So, the combination of two vertices can represent a "reflection." Moreover, because the combination of the two vertices must be computed together, the coherence of the photon can be maintian during the whole process.

flippiefanus
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Reflection is just some type of coherent scattering. Scattering happens because of electric dipoles oscillating and radiating their energy to all directions. But how do they oscillate? They absorb the energy of the electric wave, meaning the electric wave scattered is the result of energy being absorbed and re-emitted from dipoles.

Ofek Gillon
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  • Please give references for this absorption reemission idea. I come across this often in thus site, and always without any reference. – my2cts Nov 02 '18 at 15:26
  • I think any advanced book in EM fields will do (But especially books that are focused on scattering). I think the derivations of Rayleigh scattering in Griffith or Jackson could help – Ofek Gillon Nov 02 '18 at 15:35
  • I don't think this. Also, Rayleigh scattering is not connected to specular reflection. @Ofek Gillon – my2cts Nov 02 '18 at 22:30
  • Specular reflection is the result of a superposition from infinite dense emitters due to Rayleigh scattering – Ofek Gillon Nov 03 '18 at 06:27
  • Again, no reference. I'd like to read how specular reflection can be framed like this. – my2cts Nov 03 '18 at 10:14