In Newton's model of light as being composed of particles, it's easy to imagine reflection as being the rebounding of individual corpuscles off a surface. However, since light can also behave like a wave, it poses a challenge in visualizing reflection.

How does a wave reflect off of a surface, whether it be specular reflection or diffuse reflection? Must the wave be first absorbed, and then re-emitted? Or is there a different mechanism?

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    $\begingroup$ This question discusses reflection and refraction by quantum electrodynamics: physics.stackexchange.com/q/2041 $\endgroup$
    – BjornW
    May 23, 2011 at 23:01
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    $\begingroup$ @Bjorn: So, from what I understood, photons are indeed absorbed and re-emitted during reflection. Why is it, then, that the angle of incidence is equal to the angle of reflection? Logically, there must be a finite amount of time that the electron holds on to the energy. When it is radiated away, why is it not in a random direction? $\endgroup$
    – voithos
    May 23, 2011 at 23:38
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    $\begingroup$ They're not radiated at random because of interference. I remember a good Feynman discussion of it (I think in the 2nd of his public QED lectures) - I just looked at the thread @Bjorn linked to and Feynman's QED is the main reference there. $\endgroup$
    – Simon
    May 24, 2011 at 0:00
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    $\begingroup$ @Simon,@voithos: Yes and no, they are indeed scattered at random but in a superposition of all possible directions. The quantum superposition principle then, by interference, selects the reflecting (non-random) direction as the main classical (most probable by far) result. This is described in a VERY good way in the completely essential reading: "Feynman - QED The strange theory of light and matter." $\endgroup$
    – BjornW
    May 24, 2011 at 0:17
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    $\begingroup$ Related: physics.stackexchange.com/q/32483/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 28, 2014 at 16:28

2 Answers 2


I'll just go ahead and write this down even though it's already been covered in the other thread.. but I didn't post there so :)

First, if you think of light as a (scalar) wave (which is really a semi-classical way of thinking but might be enough to answer your question) you can invoke the Huygen-Fresnel principle which in this case boils down to considering every point on the reflecting surface as an origin of a re-emitted spherical wave with a starting phase directly related to the phase the point got from the incident wavefront.

The superposition of these wavefronts, after you let them destructively interfere with each other, will amount to a new combined wavefront which propagates according to Snells law (angle of incidence = angle of reflectance). See this image for the corresponding illustration of refraction (which is very similar, I couldn't quickly find a good image of reflection):

Image of wavefront refraction

Now, light really doesn't behave "sometimes as a particle, sometimes as a wave". It is always detected as quanta (particles) but the probability amplitudes (phases) propagate in a wavelike fashion. One way of expressing the propagation is to say that a photon is sort of split and takes every possible path between A and B (or, in the case of a reflector, from A to any point on the reflector and then from there to point B by any means). Every path gets a phase contribution, and all the indistinguishable paths are summed. Most paths are simply cancelling each other but some constructively interfere, creating a large contribution (in case you don't know QM, the probability amplitude squared is the probability of the described event so a large contribution means this result will most probably occur). There is a VERY good image and description of this process in Feynman - QED The Strange Theory of Light and Matter (as I wrote in the comment above).

In the case of the reflector, the large contribution occurs at the classical angle of reflectance (Snell's law again). Notice the similarity between this formulation (called the path integral approach) and the semi-classical principle outlined above; this is not a coincidence of course.

Also to briefly digress on your implied question on the per-atom non-zero "reflectance" time - saying that an electron-orbit absorbs the photon energy for a while the re-emits it a non-zero time later is of course also a slight simplification. In reality the electron interacts with the a photon, changes its momentum a bit, the re-emits (interacts) with the new photon and changes its momentum again. This scattering process occurs at all allowed momenta and intermediate times, which are then all superpositioned like above and hence I'm not sure it's meaningful to talk about any appreciable time of reflectance. Notice that this scattering is in practice very different from scattering which can excite the electron to another orbit.

  • $\begingroup$ Ah, interesting. I think the last paragraph was probably the most helpful. And, as I keep reading everyone, it seems like looking up Mr. Feynman's talks and publications is a good way to learn more about QM in general. :D $\endgroup$
    – voithos
    May 24, 2011 at 1:59

Actually, I consider reflection of a wave easier than reflection of a particle: say we have a medium in which the wave can propagate easily, that is, its amplitude can vary freely, fullfilling some kind of wave equation. You can imagine it as a sequence of oscillators, where each of them always passes its energy to the next one.

If we now put a brickwall in the wave's way, we essentially just create a region where there are no, or much fewer, oscillators to take over the energy. So what does the wave do? It can't go on in the original direction, there's no way it could get rid of the energy. So the oscillators have no choice but to send the energy back through the medium.


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