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Do we know which slit the photon went through in Afshar's experiment? I've been reading up on this experiment and it seems to be that even measuring the photon to only go through one of the slits, it still creates the interference pattern. Do we really know which slit the photon passed through?

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    $\begingroup$ Photons don't "go" places, at all. You are trying to describe a non-classical system in semi-classical terms. That will simply fail. $\endgroup$ Oct 7, 2022 at 22:10

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From this link's overview :

Afshar's experiment uses a variant of Thomas Young's classic double-slit experiment to create interference patterns to investigate complementarity. Such interferometer experiments typically have two "arms" or paths a photon may take.1 One of Afshar's assertions is that, in his experiment, it is possible to check for interference fringes of a photon stream (a measurement of the wave nature of the photons) while at the same time observing each photon's path (a measurement of the particle nature of the photons) .

From this :

The experiment uses a setup similar to that for the double-slit experiment. In Afshar's variant, light generated by a laser passes through two closely spaced circular pinholes (not slits). After the dual pinholes, a lens refocuses the light so that the image of each pinhole falls on separate photon-detectors (Fig. 1). A photon that goes through pinhole number one impinges only on detector number one, and similarly, if it goes through pinhole two it impinges only on detector number two, which is why we see the pinholes separately in the image plane close to the mirrors before the photon-detectors.

So yes, if you performed the experiment one photon at a time, you would know which slit the photon went through.

What generates a conceptual confusion is this "stream" business. Quantum mechanics is one particle at a time and its probability of manifesting at a point (x,y,z). To get a probability distribution one needs a "stream" of photons, as probability distributions are properties of an ensemble of whatever is under observation. One instance does not make a probability distribution, and thus a single photon cannot display wave properties, which are a characteristic of probability distributions in quantum mechanics, the only measurement tool available at that framework.

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It is difficult to give a completely definite answer to this question, since calculations are needed here.

But for sure at the qualitative level there is no reason why we should certainly know through which hole the photon passed. Of course, when one of the holes is closed, all photons hit a certain screen. But the fact is that the phenomenon of interference in quantum mechanics is that when two holes are opened, the whole process proceeds in a completely different way — even if at first glance it seems that there is no interference and, hence, no difference.

So, at the qualitative level, only an unreasonable transfer of reasoning from the case when one hole is open to the case when both holes open pushes us to the conclusion that we know the path of the photon.

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Afshar's conclusion is that, when both pinholes are open, the light exhibits wave-like behavior when going past the wires, since the light goes through the spaces between the wires but avoids the wires themselves, but also exhibits particle-like behavior after going through the lens, with photons going to a correlated photo-detector. Afshar argues that this behavior contradicts the principle of complementarity to the extent that it shows both wave and particle characteristics in the same experiment for the same photons.

It is not a measurement, we must see (detect) the interferences, not assume them to exist. The principle of complimentarity is about measurements, not assumptions and deductions.

If we want to do the opposite of the experiment, instead of wires, bands of width $I= \frac{\lambda a}{D}$ where the interference((the shiny fringes) will appear, the detectors will not pick up anything, what can we deduce from this? (two different experiments to show different behaviors)

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