Assumption: Screen detector is much closer to the slits than in "standard experiment" and the small angle approximation can't be used to determine the interference fringe maxima, but the interference pattern still occurs. The point of detector screen being closer is to make the difference between path lengths (from slits to the same point on screen) relatively greater so they could not be approximated as parallel, to make this difference more significant in comparison to the distance between the slits. But it still must be small enough to keep the interference.

I want to know as much as I can about the difference between QM probability wave function propagation of photon interfering with itself and CM paths of photon considered as a classical EM wave or a particle. I think that travel time is an important aspect in this respect.

As I reckon, any measurement at the slits would destroy the interference pattern, so I need to measure the time of photon leaving the gun, so I need the first detector right there.

Is it possible to check the following assumption experimentally: Travel time of SINGLE photon from the gun outlet to the fringe maxima on screen has to be equal or greater than the longer path divided by speed (c for photon). If the time was shorter, the interference pattern could not occur.

image description

My reasoning: Photon, as a probability wave, takes both paths to interfere with itself. I assumed, that it can't travel faster than light even as a probability wave. In other words - probability of detecting photon, that moves faster than light, is zero. Interference pattern on screen requires contributions from both slits, so it can't occur until both paths are traversed by the probability wave, hence the longer path in assumption.

I already know, that I can't measure the time with uncertainty less than $Δt=(l_2 - l_1)/c$ to keep the interference, so let's assume, that we can get one the following, statistical results:



$t_0=(a+(l_1 + l_2)/2)/c$
$Δt=(l_2 - l_1)/c$

$T_1$ and $T_2$ ranges overlap in 75% with $T_0$ and in 50% with each other, so their middles are within the other ranges, but they are "statistically" distinct, right?
image desc
I want to check, which of them will be the closest to the actual experiment result, obtained from multiple, single photon shots. For the sake of argument, let's assume that $Δt$ is slightly greater.

If such setup would make the time measurement inaccurate due to the uncertainty principle or because of any other reason, I still want to know is it possible to determine the probability of detection in certain point and time by solving the Schrodinger equation or it's modification, or derive it from path-integral formulation. If so, I could know the most probable detection time for certain points on screen, especially fringe maxima, and verify the assumption.

  • 3
    $\begingroup$ This is an interesting approach to the double slit experiment. Excellent questionA complication is that the Fraunhofer approximation no longer holds but that does not invalidate the experiment. My best guess is that the time of flight will be $c/a+c/{\cal l}_0$, the distance from the center of the slit system. $\endgroup$
    – my2cts
    Commented Mar 29, 2020 at 22:30
  • $\begingroup$ @my2cts Thx :) It was also my first guess, but I came to conclusion, that the interference pattern requires the time >= travel time of longer path. $\endgroup$
    – Marcin
    Commented Mar 29, 2020 at 22:50
  • 1
    $\begingroup$ How did you reach that conclusion? You should edit your question if you did. $\endgroup$
    – my2cts
    Commented Mar 30, 2020 at 5:37
  • $\begingroup$ @my2cts I've added the explanation below the picture. $\endgroup$
    – Marcin
    Commented Mar 30, 2020 at 7:13
  • $\begingroup$ Hi Marcin -- we've noticed you are making a lot of edits to the question. Each edit bumps the question to the front page, so we want to make sure all edits are required. In the future, please try to make changes in as few edits as possible. Thanks, and welcome to our site! If you get a chance, check out our Tour page to see other tips about out site! $\endgroup$
    – tpg2114
    Commented Apr 1, 2020 at 9:23

1 Answer 1


Another way of saying "a photon interferes with itself" is to say "somehow the photon knows which path to take". The photon "interference" is an interpretation, not a fact and it does lead to inconsistencies. For example 100 photons passing the 2 slits does result is 100 photons on the screen, there is no cancelling. (i.e no photons in the dark areas).

Feynman postulated that photons take certain paths that are n multiples of its wavelength, other paths are prohibited. A possible interpretation is that an excited atom/electron is acting virtually with the EM field, until at some point the field (influenced by receiving atoms?) accepts the energy and transmits it at the speed of light to a receiving atom/electron. This chosen path is high probability with at least one consideration that the path is n times the wavelength.

  • $\begingroup$ Thx! I didn't know this interpretation. Would you agree, that it implies the time distribution around $t_0$? I understand, that $l_1=nλ$ and $l_2=mλ$ where $m$ and $n$ are integers. $\endgroup$
    – Marcin
    Commented Apr 1, 2020 at 6:44
  • $\begingroup$ You don't need I1 and I2 because this is based on the "interference" principle, full of nasty issues and inconsistencies. The interference principle is great for high school and 1st year university (or if you live in the 1600s like Huygens did). Every photon is doing its own thing but because of the slits and the high coherence (by coherence you can say high geometric constraints of the source) you get the photons doing similar things, i.e high probability paths (bright) and low probability paths (dark). Upset a laser cavity by misalignment of the mirrors and voila ... no photons! $\endgroup$ Commented Apr 1, 2020 at 14:15
  • $\begingroup$ You can google Feynman double slit to get some interesting articles. It's not really QM, he looked at the results of summing many paths and the resulting path was n times lamda. It seems as though the EM field wants to resonate like a guitar string, depending on lamda certain paths resonate. $\endgroup$ Commented Apr 1, 2020 at 14:22
  • $\begingroup$ Did Feynman propose a possible range of n as a function of other parameter(s)? I am particularly interested in values of n for fringe maxima. Did he give any distribution of n? $\endgroup$
    – Marcin
    Commented Apr 1, 2020 at 15:10
  • $\begingroup$ I see a problem with Feynman interpretation. The distribution on screen is rather continuous, like this: i.sstatic.net/UiLAp.gif not just thin fringes at maxima, corresponding to $nλ$. $\endgroup$
    – Marcin
    Commented Apr 2, 2020 at 14:04

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