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Are photons avoiding the angles of diffraction where destructive interference would occur if they weren't alone in a one photon at a time double slit experiment? Is the number of photons detected on the wall equal to the number of photons went through the slits in a single photon at a time experiment?

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The answer to your first question is, in a sense, yes, the photon avoids the angles that would produce destructive interference. But probably not in the way you are thinking.

Photons are not little balls of something. They are not particles as we understand the word in our classical macroscopic world. The word "particle" has a different meaning in the quantum microscopic world. A particle is an excitation of a field, in this case, the EM field. When the field gives up a quantum of excitation during an interaction with another object, it does so 1.) at a point 2.) gives up it's energy to the other object and 2.) gives up its momentum. That's just what happens in a classical collision. But unlike the classical collision, the "particle" vanishes. Clearly the classical particle picture has a problem.

In the quantum mechanical picture, we start with the EM field. The field fills all of the space between the source and the screen (even if there is no source, the zero-point field exists), and it exists in whatever slits are available. If there are two slits there are nodes (points of zero field) on the screen due to what you might call interference, but you might also call it diffraction, the way the field arranges itself around objects. The field, and thus the excitation (photon) fills all of space, but the field, and thus the excitation, has zero amplitude at a node. No interaction will occur at the points on the screen where a node exists. As if those locations are avoided, even though there is only one photon. (By the way, this picture explains the statement "A photon interferes only with itself.")

I don't know quite how to interpret your second question. I'll say this: In the quantum mechanical picture, the field exists in both slits, so the excitation exists in both slits. If you call the excitation a photon, then the photon goes through both slits. But this is a dangerous statement because it mixes up classical, quantum, and everyday meanings of words.

By the way, this is not the only way to "explain" these things. There are other mental pictures that are consistent with the theoretical physics. There's no one correct picture. I think (my opinion) that this fact speaks to our brain's limitations, it's inability to understand the "true" nature of what's going on. Our limited brains create metaphors based on things that we can understand or visualize. But this is just my personal point of view.

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  • $\begingroup$ This is enlightening. My second question simply if there are two counters at the source and screen c1 and c2 is c1= c2 at the end of a single photon at a time experiment (assuming all emitted photons will make it through the slits regardless of which slit or both at the same time) $\endgroup$ – Amr Berag Mar 27 at 2:42
  • $\begingroup$ This version of the question really requires a long answer. The presence of two detectors complicates the picture. If the detector near the source "detects a photon" (records an interaction), that excitation (photon) is gone. If the detector near the screen records an interaction, it will be a different excitation of the same field. Is this a different photon? Uhhh... the language is getting in the way. A time-resolved experiment will not detect that second photon a time $d/c$ later. But there will be a tendency for detections to occur near each other. We're off the rails now! $\endgroup$ – garyp Mar 27 at 13:21
  • $\begingroup$ Excuse my ignorance but I meant counting them before or when leaving the source and after or when hitting the screen i.e., c1 is connected to the mechanism used to produce a single photon and c2 is connected to the screen it self. $\endgroup$ – Amr Berag Mar 27 at 14:26
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Photons like to travel distances that are multiples of their wavelength, just like a note on a guitar string, it has a certain length for a certain frequency. Thus the EM field is max in some locations and null in others. And to your second question all photons that pass thru the slits will create light, i.e if 1000 photons pass thru than 1000 photons will be observed, energy must be conserved, photons do not animate each other. The concept of interference is an historical one and it explained much of the observed behaviour and was convenient to teach (it also created a fair amount of discussion and controversy back then), but there was no denying the wave nature of light so the interference term persisted. Consider 2 water waves, or 2 tsunamis, one from Tokyo and one from LA but out of phase (one up wave from Tokyo and one down LA), when the waves meet at Hawaii they cancel and everyone is saved but a minute later the 2 waves are observable and real again and the waves hit Tokyo (down) and LA (up).

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  • $\begingroup$ Hi Dave. I found this video describing what happens when there are way more than two holes. m.youtube.com/watch?v=F6dZjuw1KUo from 9:00 $\endgroup$ – Amr Berag Mar 27 at 14:42
  • $\begingroup$ Yes it is very cool but the video teaches the traditional "interference" method of explanation, which works out to predict peak positions but does not really answer the net total intensity question which you posed, which is important. What's interesting is if you fire one photon at a time, say one per hour or one per even day you will eventually still get the same pattern as when the beam is full on. The photons or waves will prefer certain positions on the screen not because of "interference" but because they are looking to travel an integer multiple of their wavelength (Feynman principle). $\endgroup$ – PhysicsDave Mar 28 at 2:46
  • $\begingroup$ There's a lot of math, when there are more holes they need to be identically spaced apart, somehow this makes the small spots, but the central spot is still brightest, if fact in can be ~ 10 times brighter that the 2 hole example. Diffraction gratings are used in practice to concentrate a broader light beam, into a spike in many spectrometers so it can be analyzed more effectively by the photodetector chip. (better S:N) $\endgroup$ – PhysicsDave Mar 28 at 2:46
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You can interpret the classical interference pattern as the average distribution of photons on the "wall". At low intensity you will see random deviations from this average because of the finite number of photons. This phenomenon is known as photon shot noise. In this sense you can say that photons avoid the regions of destructive interference.

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