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In an Oct 2011 Physics StackExchange comment, anna v (hi Anna!) asserted

... the photon [in a transparent crystal] quantum mechanically interacts with the whole crystal in a complex, many-variable way not solvable analytically except by approximations.

The above assertion appears to diverge somewhat from the more prevalent interpretation that the photon interacts specifically, but in a somehow hidden fashion, with random atoms within a crystal.

My best guess is that the hidden-random-atom interpretation is the prevalent textbook explanation, though I've not checked. On the surface, it sounds like a simple and quite reasonable invocation of Feynman's integral of all possible histories. However, on closer examination, and regardless of how often it is used, the hidden-random-atom interpretation goes against Feynman's adamant insistence that for histories to interfere it's more than a matter of "not knowing" or "not caring" which atom was hit. The event must be genuinely indistinguishable by any conceivable experiment.

While one may certainly assert individual photon absorption-emission events are indistinguishable, relying solely on human assertions of ignorance of which-way information gets problematic quickly. Imagine, for example, applying the same argument to neutron diffraction. It is difficult to imagine the large-scale statistically consistent occurrence of some form of random same-energy re-emission of neutrons to explain neutron diffraction and reflection, not to mention that no such loss-free mechanism exists in nuclear physics. Indeed, Feynman neutrons and neutron diffraction in the next section.

The same problem of the non-existence of same-energy re-emission is true also for photons, of course, since as Compton himself pointed out (p.20), the correct term for photon re-emission from a single atom is fluorescence. Compton was sharply aware that there was a prickly classical-quantum transition problem hiding there, which makes it ironic that his name is so frequently to "solve" perplexing coherence conundrums.

Anna, since I see you are very much active here these days, I would love to hear you (and others) elaborate a bit more on what you meant by that. And yes, I realize you may have been saying nothing more than "abandon hope all ye who enter here, use the model that works and stop asking!" But you sound like you've put some thought into this, so it doesn't hurt to ask.

Terry Bollinger
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1 Answers1

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Chad Orzel wrote:

Carrying this picture over to the quantum regime, you would say that a single photon entering the material will potentially be absorbed and re-emitted by each of the atoms making up the first layer of the material. Since we cannot directly measure which atom did the absorbing, though, we treat the situation mathematically as a superposition of all the possible outcomes,

The way I understand Chad Orzel is as follows:
Our theory of interaction between light and matter is in terms of scattering events; contatenation of scattering events. Also, our theory inherently allows/enforces conceptualizing events taking place in the form of a superposition of allowed possible paths.

An exhaustive computation would be to compute all the possibilities of the allowed superposition.


So, in the case of transmission of light through a transparent solid, what do you do?

Presumably an exhaustive computation would take many times the lifetime of the Universe. I presume the only way to implement computation at all is to implement a Monte Carlo algorithm.

Our models go only so far.
Even if it is the case that light is actually 'interacting with the whole crystal', we don't have a model that does directly that.

But it seems to me that a properly implemented Monte Carlo algorithm for propagation of light in a transparent solid does in effect achieve a the goal of being consistent with the light interacting with the crystal as a whole.


Sideways related:
I'm reminded by an answer in which tracks in a cloud chamber are discussed.

Tracks in a cloud chamber are in that rare class of cases where effects arising from the motion of a single particle are visible with the naked eye.

The probability of emission of an $\alpha$-particle is spherically symmetric, yet the track is linear.

Cleonis
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  • Chad Orzel's comment is simply wrong, for a rather blunt reason: The per-atom event that is being superposed does not exist. Quantum superposition only works for events that, individually, make sense. In this case, the nominal event is the absorption and re-emission of a photon by an atom with no energy loss and no loss of information, although there is a phase inversion. The difficulty is that no real atom does that. What you get in real-world atomic absorption and re-emission events is the effective destruction of the photon followed by re-emission at a lower frequency. Think ruby lasers. – Terry Bollinger Dec 05 '21 at 15:21
  • Fog chamber particles are classical since each new droplet is an interference-destroying observation. Most quantum uncertainty disappears with the first droplet. Also, while it is easy to create a spherical function for a lightweight photon, e.g., via fluorescence by an atom in space, the nuclei that emit alpha particles create which-way recoil phonons that narrow the quantum wave. Alpha and larger particles (up to over 2,000 carbon atoms) require considerable effort to create wave functions that diverge enough to interfere, let alone ones that approach even hemispherical form. – Terry Bollinger Dec 05 '21 at 15:55