Does physics account for interactions between light and matter ever being "not completed" or backed out of?

Here's what led me to the question. In learning about interference in light, I ended up concluding that destructive interference must occur in the realm of the light's interaction with matter. In other words, that the light is not interfering while traveling in space, but always in conjunction with landing on some surface. So that, in a sense, destructive interference is as much a property of matter as it of light.

But according to this picture that I formed, the absorption of a light-wave or photon by an atom (maybe not in all interactions) can be reversed, that is, before being completed. For, when we look at how the field of optics deals with destructive interference in, say, the double-slit experiment, we see that it's always a function of the difference in distance, and hence time traveled by the waves to a single point, that gives the interference. So that one wave has already made contact with the point before the other arrives...yet they're both rejected/ejected, resulting in the effect of destructive interference (no delivery of energy) at that point and moment.

That's how I arrived at my question: Is there any such concept in physics of interactions between light and matter, especially light-absorption, being backed out of, or "reversed," before being completed?

  • 1
    $\begingroup$ Interference does not necessarily occur when the light falls on a matter. It can occur at any point where the two waves superimpose. $\endgroup$ Apr 25, 2016 at 19:43
  • $\begingroup$ In total internal reflection there is an evanescent field which extends outside of the boundary at which TIR occurs. If another dense medium is placed close enough, light can cross the gap, but unless this happens the evanescent wave does not leak any energy out of the 1st dense medium. It is similar to tunnelling in quantum mechanics. en.wikipedia.org/wiki/Evanescent_field $\endgroup$ Apr 25, 2016 at 19:52
  • 1
    $\begingroup$ While I am sure that one could construct a theory using that concept, the better question is whether it would be of increased utility or not. I see no reason why your viewpoint is more useful. 'Not interacting' seems to always be an option, without 'backing out of' being needed. $\endgroup$
    – Jon Custer
    Apr 25, 2016 at 20:03
  • 1
    $\begingroup$ Yes, one can reverse interactions, as long as no measurement has occurred, but there is absolutely no physical content in this. A measurement is, by definition, irreversible and the free dynamics is, by observation, in principle reversible. What you are talking about here are essentially just system boundary questions. $\endgroup$
    – CuriousOne
    Apr 25, 2016 at 20:20
  • $\begingroup$ @Pranav Rastogi, I guess you could say that I'm referring specifically to light/matter interactions and, in the case of interference, to destructive interference that is observable, such as that dealt with in the field of optics. $\endgroup$ Apr 25, 2016 at 21:23

1 Answer 1


Maybe there IS an effect similar to what you describe. In photoelectric absorption, the permissible departing electron states in a material determine the probability of absorption (the rate). This is easily seen in X-ray absorption near a photoelectric excitation threshold (like a K-edge), and is the basis of XAFS (X-ray absorption fine structure) phenomena.

The kicker is, the ejection of an inner electron into the bulk material is not something that happens AFTER the X-ray is absorbed, but is simultaneous with that absorption. The outgoing electron wave, far from the inner K shell of the absorbing atom, can reflect back from nearby atoms and prevent the X-ray photon from ever being absorbed. If the outgoing electron DOES reflect back, and does so in response to atomic surroundings far outside the K shell ... did that inner electron eject and return, or did it never eject because the X-ray photon was not absorbed?

Well, as in a diffraction, it BOTH ejected and didn't. The X-ray photon both was absorbed, and wasn't. We cannot separate these as a sequence of events in time, but it DOES appear like the long-range outgoing electron left the K shell, then probed the surroundings, then returned.

  • $\begingroup$ I'm flabbergasted...first of all by the clarity and helpfulness of your answer, but second of all by the fact that what I imagined to be the case appears indeed to be regarded as true, or at least plausible. I'd be much obliged if you'd follow up with 3 questions that I have: 1) What do you mean when you say, "as in a diffraction..."? 2) Why can't we separate the process as a sequence of events in time? Because we simply can't discern it yet, or some more fundamental reason? 3) What about the unabsorbed X-ray photon? Is it now back in play? And if so, does it have a new direction of travel? $\endgroup$ Apr 26, 2016 at 16:57
  • $\begingroup$ Also, I found a paper introducing XAFS but it was a bit beyond me. The idea that you develop throughout your answer, can it be read in other sources too (and if so, perhaps you could point me to one or two), or is that your own logic and it hasn't really been elucidated like that before? $\endgroup$ Apr 26, 2016 at 17:15
  • $\begingroup$ I mentioned 'as in diffraction' because the outgoing wave is cancelled in some directions, but enhanced in others, in both situations. The dissimilarity, of course, is that the photon/slit interaction is local (only the interaction with the sides of the slit cause the wave propogation fluctuations), whereas the XAFS case cannot be made to merely be local to the inner electron shell (where the electron is initially located), but must be extended to the neighboring atoms around the periphery. Unoccupied electron orbitals which DO hit the neighboring atoms are just as important as the electron $\endgroup$
    – Whit3rd
    May 11, 2016 at 0:46
  • $\begingroup$ ...as the electron ground state. The X-ray photon does not generally scatter (change direction or energy) when not absorbed, because there is no way to conserve both energy and momentum that is as important (fast, likely) as the photoelectric effect, in the vicinity of the 'edge' energy. So, the absorption never happened, and THAT is why there can be no multiplicity of time-distinct events: photoelectric absorption near an edge is dominated by the very simple two-particle (photon in, electron out) reaction. $\endgroup$
    – Whit3rd
    May 11, 2016 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.