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Suppose you have some material which has a low reflectance and is opaque to an incomming photon when the angle of incidence is small.

Now take that material and make a narrow (20:1 width:length ratio or narrower) wedge with a fine edge - 1/10th the wavelength of said photon or less, the very edge itself should be transparent - and shoot said photon directly down the edge of the wedge like so.

How does the photon react? Does the photon treat the edge as slicing a probability wave, resolving to having been deflected randomly to one side of the wedge or the other when observed... or does it treat the edge as if it was no finer than the wavelength and so act as a perpendicular strike on a blunt edge and be absorbed? Is there other quantum weridness involved, or other information required to answer fully?

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    $\begingroup$ I think this is answerable, but just to make it a bit more clear, do you mind making a diagram of your setup? $\endgroup$
    – knzhou
    Commented Feb 21, 2018 at 10:32
  • $\begingroup$ I believe if you fix the reflectance coefficients then you don't need to consider any quantum weirdness for this specific problem, Maxwell's equations alone will do the job (or more likely the Helmholtz equation) $\endgroup$
    – KF Gauss
    Commented Feb 21, 2018 at 14:39
  • $\begingroup$ Maxwell is just about enough because treating it as a single photon or a light wave, quantum electrodynamics will give you the same result (at least to the fourth order) to that of classical and the interpretation will be almost the same. A diagram to specify your question will really be helpful... $\endgroup$
    – Yuzuriha Inori
    Commented Feb 21, 2018 at 15:44
  • $\begingroup$ Diagram included. The wavelength is much greater than the tip's width, but not the base width, so it should interact at some point, but how? $\endgroup$
    – Steven Armstrong
    Commented Feb 22, 2018 at 22:01

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This situation is really little different from most beamsplitter "which way" thought experiments or the explanation behind almost anything you can cook up on an optics table.

There's no particular quantum "weirdness" here. I sense that what may be bothering you is that you may be thinking of the wedge's interaction with the photon as some kind of quantum measurement. This interaction is not a quantum measurement and excited matter states in the wedge's glass simply exist in quantum superposition with the pure, one photon light states. Quantum measurement only happens when the photon is detected, say at a screen behind the wedge.

Unless we're dealing with entangled light states, one simply conceives of the theoretically foretold experimental outcome in the standard way:

  1. Describe and analyze the problem with a standard, linear Maxwellian description;

  2. At our detection plane, calculate the intensity field $\vec{E}\times\vec{H}$ from the Maxwellian solution above. Or, if you have a resonant situation and you want to ask where in the volume you'll detect the photon, you calculate the energy density field $\vec{E}\cdot\vec{D}+\vec{H}\cdot\vec{B}$.

  3. The quantities calculated in 2. above are proportional to the probability density to destructively detect a photon at each point in space when we do the experiment as a repeated, one-photon-present-at-a-time experiment.

This is because there is a one to one correspondence between one photon states and classical solutions to Maxwell's equations, as I discuss further here and here.

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  • $\begingroup$ I'm very close to marking this as accepted and awarding the bounty. Could you perhaps clarify one further point? I'm not adept enough to analyze the problem with the Maxwellian equations, so telling me that I can, unfortunately, doesn't actually give me an answer. If it's something that depends on the properties of the material, that's also a satisfactory answer, though if there's a more probable outcome that would be good to know. Basically, is it possible to say in general if this is more like a beam splitter, or more like a perpendicular impact? Could we design it to behave like either? $\endgroup$
    – Steven Armstrong
    Commented Feb 24, 2018 at 4:28
  • $\begingroup$ @StevenArmstrong Another way to state my answer is as follows. Shine a beam, say a laser, onto the edge and then look at the pattern behind the wedge on a screen orthogonal to the input beam. You'll see an intensity distribution with two equally bright regions on either side of the wedge, dark in the middle, and probably with some interference fringes in the transition regions. This intensity distribution is the same as the probability density to detect a photon when you turn the intensity down so much as to have only one photon fields in the apparatus at a time. So you'll see one photon .... $\endgroup$
    – Selene Routley
    Commented Feb 24, 2018 at 8:51
  • $\begingroup$ ... experiments trigger detectors on either side of the wedge with roughly equal probability; half the time you'd interpret a split to the left, half the time to the right. $\endgroup$
    – Selene Routley
    Commented Feb 24, 2018 at 8:53
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For the reasons explained by @WetSavannaAnimal aka Rod Vance you're just asking a classical optics question, and I'll supplement that by trying to answer the classical optics question.

A small fraction of light will get scattered or absorbed at the sharp tip, but as the tip gets sharper and sharper, that fraction gets lower and lower. The rest of the light will get absorbed along the two sides of the triangle. (If it was a triangle of glass, the light would reflect off the sides of the triangle, but you said in the question that the triangle is made of a very black material with negligible reflectance.)

What is the exact probability distribution for where along the edges the photon is likeliest to be absorbed? It depends on the detailed wavefunction of the incoming photon (or in classical-speak, the incident light wave's phase and intensity profile).

See also knife edge prisms.

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  • $\begingroup$ Ah, perhaps I didn't clarify part of it well after all, the material is supposed to highly absorb at low angles of incidence and highly reflect at high angles of incidence, and then be a question of what the tip behaves like - high angle or low angle. $\endgroup$
    – Steven Armstrong
    Commented Feb 24, 2018 at 4:34
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I'm glad to see this question has gotten some good answers! The other answers have explained that in this particular case, we can understand the result with classical electromagnetism.

I want to explain a bit more why quantum mechanics isn't necessary. The key is that interaction with the edge is not a quantum mechanical measurement. Why not? The general principle is that the interaction of $A$ with $B$ counts as a measurement of $A$ if we can figure out the state of $A$, in principle, from the state of $B$.

First, consider a proton in a superposition of momenta. If the proton crashes into an air molecule, then it no longer behaves as a superposition, because we can infer the momentum from the way the air molecule recoils. This is true even if we don't bother actually looking at the air molecule, or even if we can't; it is the air molecule itself that does the measuring.

Next, consider turning on an electric field. This accelerates the proton, but it does not collapse the superposition, even if we model the interaction as the absorption of a photon. That's because a classical field is made of an enormous number of photons, and it is basically identical after taking a photon out; you can't tell the difference even in principle.

Given a typical wavelength for light and a typical material, the interaction with your edge is much more like the second case. The photon smoothly interacts with many atoms at once, so that the full interaction is not a measurement.

Now you might ask if we've succeeded in 'slicing a photon in half' even though they're supposed to be indivisible. But in this case the quantum nature only comes out when we measure, because individual photons behave classically. If you do choose to try to detect the photon, you will always detect it on one side or the other, never both.

Some extra caveats for the electric field case:

  • Formally, if we model the electric field as a coherent state it's literally completely impossible to tell the difference; the field minus one photon is exactly the same as the original field!
  • You might also think we can detect the recoil on whatever's making the field. The issue there is that if a macroscopic object is responsible for the field, its momentum uncertainty is large enough that the recoil does not have a detectable effect, even in principle; see here.
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