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Can one "interpret" Snell's law in terms of QED and the photon picture? How would one justifiy this interpretation with some degree of mathematical rigour? At the end I would like to have a direct path from QED to Snell's law as an approximation which is mathematically exact to some degree and gives a deeper physical insight (i.e. from a microscopic = qft perspective) to Snell's law.

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  • $\begingroup$ Maybe a more interesting question would be: Can one "interpret" Snell's law in terms of QED and the photon picture? I.e. can we describe what a single photon does in processes like refraction? $\endgroup$
    – Lagerbaer
    Mar 6, 2011 at 15:47
  • $\begingroup$ Thanks, I have incorporated your suggestion and modified the question. Regarding your second sentence notice the problematic nature of the term "single photon".. $\endgroup$
    – student
    Mar 6, 2011 at 16:06
  • $\begingroup$ Aww, that renders my answer quite stupid... I don't think there is much to interpret then, it's not like a single photon has a well-defined trajectory that suddenly changes direction. (also @Lagerbaer) $\endgroup$ Mar 6, 2011 at 16:09
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    $\begingroup$ Related: physics.stackexchange.com/q/2041/2451 $\endgroup$
    – Qmechanic
    Oct 7, 2011 at 8:50
  • $\begingroup$ There are really two questions here. "Can we understand the index of refraction in terms of QED?" and "Does QED imply ray optics in the appropriate limit once the index of refraction is known". Feynman give a nice answer the second is his pop-sci book on QED. $\endgroup$ Mar 10, 2012 at 16:45

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Sure. Start with QED, obtain Maxwell's equations, do the paraxial approximation and finally use Fermat's principle.

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    $\begingroup$ Isn't using Fermat 'cheating' in a way? Using the Fermat/Lagrangian-approach is the textbook way of proving Snell's law. I'm guessing 'student' is rather looking for a more direct path integral-approach from QFT. $\endgroup$
    – Gerben
    Mar 6, 2011 at 15:46
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    $\begingroup$ @Gerben well, you could use Maxwell's equations and from the continuity derive the Fresnel equations which include Snell's law as a by-product which is less cheating. A more direct path integral-approach doesn't come to my mind so quickly, but it's sunday, maybe tomorrow... $\endgroup$ Mar 6, 2011 at 15:57
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    $\begingroup$ Isn't Fermat's principle pretty much the same thing as the path-integral approach? The E&M Lagrangian density is proportional to $E^{2} + B^{2}$, which works out to a constant proportional to the square of the wave amplitude in the case of the wave. In the classical limit, you expect the action to be minimized, and in the case of a well-defined beam to which Snell's Law would apply, you'd expect the path to be a curve in space. Clearly, the only way to get a stationary phase for this action is to look for the shortest path. Ergo, Fermat's principle. $\endgroup$ Mar 6, 2011 at 16:08
  • $\begingroup$ Though I would say that QED is obtained from quantizing the Maxwell Lagrangian, not the other way around. $\endgroup$ Mar 6, 2011 at 22:12
  • $\begingroup$ @Jerry not necessarily, you can also start with the free particle path integral and use the Yang-Mills theory for U(1), thus starting with local symmetry and obtaining the Maxwell Lagrangian as a result. $\endgroup$ Mar 7, 2011 at 6:50
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This appears to be explained in detail in Feynman's "QED the strange theory of light and matter" in Chapter 2, page 39 to 45, of the 2006 edition, in more or less plain English.

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In regard to the single photon aspect of the question, I speculate that the explanation is similar to the Mossbauer effect, ie the photon is absorbed and re-emitted by the entire mirror/crystal rather than a single atom. If you insist on thinking of a single photon being absorbed and re-emitted by a single atom, you are going to have to invoke the regularity of the mirror or crystal and use constructive and destructive interference, as mentioned in other answers.

You can see the necessity of a coherent many body solution from the fact that a single atom does not refract or reflect, it scatters, and similarly for a rough mirror.

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