Consider this example:

Picture of the configuration

We have one single electron cannon that is not obstructed in any way. We have a second parallel single cannon electron that fires simultaneously with the previous one. The second cannon can is obstructed by wall with two slits which then lead to two separate chambers. The top chamber is shared with electrons from the first canon. The bottom chamber is not shared. We have a detector in each of the chambers.

My understanding of this example is that interference on the upper detector constitutes a measurement for the lower detector (lower electron went through upper slit) and lack of interference on the upper detector also constitutes a measurement (right electron went through lower slit). Am I then correct that on the lower detector we will see corpuscular pattern with density equivalent to 50% of the firing rate of a single cannon whilst on the upper detector we will see a sum of 50% density corpuscular pattern and 100% density interference pattern?

  • $\begingroup$ I'm no expert, but are we certain that two different electrons will interfere with each other like that? The usual interference pattern is the result of summing many individual electrons interfering with themselves. Two electrons interfering with each other is a totally different scenario. $\endgroup$ Dec 1, 2016 at 13:50
  • $\begingroup$ But just one electron can interfere with itself in classic double slit experiment. Will it be different with two? $\endgroup$
    – user68634
    Dec 1, 2016 at 14:03
  • 3
    $\begingroup$ If there is no phase coherence from one pair of electrons to another (and there likely won't be any if the two cannons are independent), the final result will be no visible interference (their interferences destruct each other). $\endgroup$
    – user130529
    Dec 1, 2016 at 20:03
  • $\begingroup$ @claude chuber: But suppose they are 'phase coherent' (made with some extra device) travelling from different sources, will they interfere as if they were coming from the same source? $\endgroup$
    – fante
    Dec 3, 2016 at 22:41
  • $\begingroup$ @ghurpost: By the way, I think your question is great!!! I never thought about it. $\endgroup$
    – fante
    Dec 3, 2016 at 22:44

2 Answers 2


You have to understand that in the double slit experiment one electron at a time, the same interference pattern appear as with shooting many electrons at a time.

It is the wave function for "one electron scattering over double slit", with the correct boundary conditions in width etc of the slits the interference appears for a single electron at a time. Thus, as far as interference goes, time is irrelevant, it is the energy of the electron and the boundary conditions.

The top gun is the soluiton of " one electron fired from gun next to a wall". What will happen at best is a diffraction pattern at the edge of the wall, and it will be a pattern dependent on the parameters of the top part. The slit below it will again give a diffraction pattern from one slit. If the interference of the edge diffraction disappears it will be the result of the space boundary conditions, not of time.

The third gun will just have a hole diffraction pattern depending on the distances etc.

p.s. your right and left is not clear to me, I just assumed the electrons come in from the left and the screen is on the right


When the right cannon fires, and the electron goes through the left slit, it is in the same chamber as an electron fired from the right cannon, thus you cannot know from which cannon it has been fired, so the normal pattern between two electrons will occur as if they have been shot by the left cannon.

(Ps. I am not one hundred pro cent sure, it is just that I can't comment yet).


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