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In any interference experiment, whether it be an unequal arm length interferometer or the classical double slit set up, we have two unequal distances from source (or slits) to screen/detector. Single photon interference in the regime of quantum mechanics is explained by saying that every photon interferes with itself. This generally gives rise to the following misconception that this interference seems impossible since the photon going through the shorter path always gets absorbed at the detector before the one at the longer path makes it.

I have gone through this question which seems to be asking a similar question to this but allow me to explain.

In the double slit exp., is it correct to say that just after the photon is ejected from our source, the wavefunction, which may have a speed greater than the speed of light ** or have no speed at all and is present at all points in space, from the moment the photon was released, has already interfered at the detector before the photon would make it to the detector at velocity $c$ and it's only after time $t = \frac{(Path Length)}{c}$ that we get the measurement/click at the detector?

** I also understand that the wavefunction cannot be attributed a physical meaning but I've seen some sources mentioning about the wavefunction propagating as a spherical wavefront. I'm not sure what to make of this. I find it easier to not even try to visualise the wavefunction at all!

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I guess this paper could be interesting for you:Nature Physics 11, 539–542 (2015) doi:10.1038/nphys3343. There is a physics world article available as well and which you might want to read first.

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The wavefunction of a particle is a representation of the relative likelihood of detecting the particle at any point in space and time.

If a particle is emitted from a source at a given instant, and a portion of the leading edge wavefunction arrives at a detector through the right slit earlier than through the left slit because of path length distances, then the two portions of the wavefunction do not overlap at the detector until the leading edge of the wavefunction also arrives at the detector. Because there is no overlap during that brief interval, we know the wavefunction arrived from the right slit and we also know there is no interference during the interval.

At the trailing edge of the wavefunction, the wavefunction will arrive at the detector only via the left slit for a brief interval, so again there will be no interference during that interval and we will know that the wavefunction arrived only via the left slit.

However, during the time portions of the wavefunction arrive via both slits, there is interference at the detector and the particle has an indeterminate trajectory: it goes through both slits.

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The problem is with describing the photon as a particle which it isn't. The right description would be the presence of electromagnetic fields interacting with charges. In quantum field theory with Fermi's golden rules there is a probably a photon quanta is removed or created but it has nothing to do with particle interfering with itself. The field obeys rules like the superposition principle which leads to interference. The concept of a photon interfering with itself does not make sense and does not need to be worried about.

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  • $\begingroup$ I agree the "photon interfering with itself" is an out dated expression at least. But your answer seems to ignore the single photon DS experiments where the "pattern" (note I don't use the word interference) is also observed. What is superimposing in your answer when it comes to single photons? I am a believer in the Feynman approach, photons want to travel a path length n times there wavelength (n=integer). $\endgroup$ – PhysicsDave Jun 26 at 17:50
  • $\begingroup$ You mean a single photon detected in a DS experiment? These are the results of the interaction of the EM field with charges in the detector. By Fermi golden rule there is a probability a photon quanta is removed from the field. One can not predict where it will happen but it only happens in the regions the EM field is present which again can be calculated using the well known equations. Again interference results in well known interference patterns seen. $\endgroup$ – Jan Bos Jun 30 at 12:19

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