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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 experiment, 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 $t = (\text{path length})/c$ that we get the measurement/click at the detector?

I also understand that the wavefunction cannot be given a physical meaning but I've seen some sources mention 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!

Vincent Thacker
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Weezy
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  • Highly related 165020/66086. – Cosmas Zachos Aug 01 '17 at 19:06
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    I don't think there is a single correct way to look at anything quantum mechanical. But I do favor your second approach. The field is set up when the apparatus is built. Spatial pattern of the field can be found by solving the Helmholtz equation, and the field pattern will have fringes where you "expect" them to be. Even with no excitation, the field exists and is "occupied" by zero point fluctuations. Exciting the field by one photon excites the entire field, whose spatial pattern already exists. The detector fires where the field exists, and doesn't at the nodes. – garyp Jul 25 '20 at 20:27
  • @garyp That sounds a lot like the Bohmian approach. – Weezy Jul 26 '20 at 08:05
  • I wouldn’t try to visualize a wave function either. Light is made of billions of individual photons. A light wave is made of billions of individual coherent photons. Single coherent photons projected through a slit or multiple slits One at a time can impact A detection screen and eventually build up a Fringe pattern. – Bill Alsept Sep 07 '21 at 06:31

4 Answers4

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The confusion usually arises from trying to treat the double slit experiment from a particle point of view, where photons/electrons are injected on one side of the screen and detected on the other side. If we consider photons/electrons as waves, this problem does not arise: we solve the wave equation (Maxwell or Schrödinger) in all the space, and its eigenmodes already contain the interference, due to the boundary conditions on the screen. We can then think of a photon/electron injected at one point as a wave packet, which we expand in this eigenmodes. The narrower this packet is, the more modes it includes - thus, in this extreme particle view, the photon/electron is present on the other side of the screen from the very beginning, but we have to wait till its amplitude on the other side becomes detectable (i.e., results in the finite probability of detection).

Finally, one could explicitly construct a wave packet localized on one side of the screen. In this case we will have a wave front propagating in the direction of the detector with the speed less or equal the speed of light, but interference is already present in it.

Roger V.
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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.

S. McGrew
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I believe you are placing too much emphasis on stating that in any interference experiment the slits define two different distances to the screen/detector. The wave properties associated with a single photon experiment are based on the wavefunction reaching both slits at exactly the same time. Once the photon has passed through the slits (pictured as having passed through both slits) there may indeed be a small difference in the distance traveled from the two slits to the detector/screen (depending on where on the screen the detection occurs). However, this difference in distance is much less than the wavelength of the photon. Therefore, it is not accurate to say that the one traveling the shorter distance got there first because you must take into account the uncertainty within the experiment. The slits are always set to be only a tiny fraction of the wavelength of the incident particle. The exact time when the detection occurs at the screen is uncertain within the wavelength of the particle. Therefore, you cannot say with certainty when the particle will arrive at the detector for either of the two possible path. If fact, you can argue that the photon traveling either path will arrive at the detection screen at the same time (within the uncertainly of the experiment).

JRL
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  • Would this mean that there is some small chance that the photon could arrive at the target at a location where we expect maximum interference? If it arrived at the earliest possible time at the target, would this correspond to a part of the wave that is too early for the more distant slit to interfere? (In other words, is the very front of the wave less affected by interference?) [Edit] This appears to be answered by this other answer in the affirmative. https://physics.stackexchange.com/a/561739/369277 – cesoid Sep 25 '23 at 22:53
  • Didn't mean to create a link for editing the answer, but now I can't edit the word "Edit" because it's been too long. – cesoid Sep 25 '23 at 23:01
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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.

Jan Bos
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  • 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). – PhysicsDave Jun 26 '20 at 17:50
  • 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. – Jan Bos Jun 30 '20 at 12:19
  • The EM field is always present everywhere, we could say a photon is a disturbance that adds energy to the field .... in the dark areas of the pattern nothing has happened to the EM field ... there was never any energy directed there. The classical equations with "interference" imply non conservation of energy in the DSE which is out dated. – PhysicsDave Jul 07 '20 at 16:21
  • The field is clearly not present (or reduced) in areas with dark fringes in an interference experiment. If one puts charges in areas with dark fringes the probability of creation or annihilation of photon quanta when interacting with the field is greatly reduced or practically zero. How so do classical equations imply non conservation of energy in the DSE? – Jan Bos Jul 07 '20 at 23:56
  • The problem is the opposite of what you say. The photon is a Particle and it has been proven over and over again. You can derive or explain any light phenomena with a particle theory. Wave theory can only explain some. Wave function is a good mathematical tool but does not explain anything physically. – Bill Alsept Aug 06 '21 at 20:33
  • @Bill Alsept Okay describe the physics of the double slit experiment with the photon as a particle – Jan Bos Aug 09 '21 at 12:27
  • @JanBos There are several ways to physically describe the double slit experiment. One way is the way I describe it in the link at the top of my page at billalsept.com – Bill Alsept Aug 09 '21 at 14:53