What would be the outcome of conducting the following modified double slit experiment?

Take the normal double slit experiment with a detector on the double-slitted wall, and behind it, add a second wall with three slits in it.

Build the back wall such that a particle passing through one of the front wall slits is likely to continue through either of two of the three back wall slits.

An illustration with gun, detector, walls, and screen: https://sketch.io/mobile/?15344495732574540

______________ <-- Screen

____C_D_E____ <-- Wall with 3 slits (C, D, E)

_____A_B_____ <--- Wall with 2 slits (A, B) and detector

_______|_______ <-- Gun

From A, only C and D should be reachable, and from B, only D and E should be reachable.

The main factor of interest is use of the front wall (with detector) to eliminate one slit in the back wall.

Can you predict and describe the kind of pattern that will appear on the screen, and can you describe/explain the process a particle goes through during the experiment?

It can be with strictly one particle fired at a time, or it can be with an onslaught of fired particles. All fired particles are the same kind.

I don't care for a super precise answer.

Note: I don't know the answer. This is a genuine question I thought of today, not a school assignment (AFAIK, but I have not been taught QM, so what do I know.)

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    $\begingroup$ Of course the answer depends on the size of the slits, the distances between them and the wavelength of the light. How many examples have you worked out? (Also, this has nothing to do with QM; it's just a matter of working out the interference pattern.) $\endgroup$ – WillO Aug 16 '18 at 18:57
  • $\begingroup$ Can you elaborate on how these factors might affect the result? I have not worked out any examples, but I am interested in how this setup affects the tendency towards interference, and the mechanics going on when particles travel through it. $\endgroup$ – Fisk42 Aug 16 '18 at 19:02
  • $\begingroup$ You did notice that I specified there is a detector on the front wall, right? Would this not introduce a soft element of measurement? The idea is to use the front wall to eliminate one slit in the back wall. $\endgroup$ – Fisk42 Aug 16 '18 at 19:05
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    $\begingroup$ Could you draw a picture of your setup? $\endgroup$ – knzhou Aug 16 '18 at 19:33
  • $\begingroup$ Yes. I've tried to sketch it. Sorry, my sketch is a very... sketchy: sketch.io/mobile/?15344489081189360 $\endgroup$ – Fisk42 Aug 16 '18 at 19:51

Since there is a detector at the front wall, you don't get interference between $A$ and $B$. Instead, the patterns simply add. The pattern at the screen from either A and B will be just a double slit, so the pattern here would just be two double slit patterns superposed on top of each other.

As usual, the result of this experiment should not be stated in terms of trajectories of particles, because the particles simply don't have trajectories at all in standard QM.


Like this!

enter image description here

At least, for some particular choices of slit size, separation, and wavelength. Let me explain a little.

There may be many ways of doing this calculation, but by far the most fun is with Feynman path integrals. Actually, the method I used is described very well in his book QED. Essentially, interference is caused by the different lengths of the paths, since a photon traveling a different distance is in a different phase of it's cycle. So, by picking a set of paths for each location on the final screen, I just calculated the phase associated with each path, added them together, and took the absolute value to get the amplitude. Technically, I did:

$$\big|\sum_i \exp(i d_i k)\big |^2,$$ where $d_i$ is the distance of each path and $k$ is the wavenumber.

The arrangement of the slits is as follows:

enter image description here

I have also drawn 4 representative paths, as an illustration. Each point on the screen used 24 paths, as each slit was simply broken down into 2 points, it's edges.

For the actual dimensions, I used arbitrary units, but eventually made choices which roughly correspond to the diffraction of red light. Choosing the wavelength to be 7, the half-slit sizes were each 3500, the center-to-center distances of the slits were all 10000, and the distances between the screens were all 10000000. For the three plots above, I varied the wavelength from 7 (on the left) to 12 to 20 (on the right).

All this is in a spreadsheet, where you can change the various distances to some degree (although you can't improve the resolution, that would require quite a bit more work). You can check out the spreadsheet here. The apparent asymmetry is certainly due to the lack of resolution.

Again, I'm not sure this gives you the answer you want, because everyone who says "it depends on the distances involved" are all correct. But, at least this gives you a flavor of what such a pattern might look like.

As an editorial note: this was super fun, but I'm sure not it was the easiest way to do this calculation!

  • $\begingroup$ Wow, that's pretty good work. 2 comments: 1) minor point, the author (Fisk42) did put detectors in the first wall, so wave functions were destroyed there. But I like your experiment better! 2) The interpretation of Feyman's paths (I think) are that photon travel/detection is not likely to the places where cancellation has been calculated, photon travel is likely where the phase calculation is high in magnitude. Every photon that begins and completes a journey thru the slits does get detected, no energy is mysteriously lost. $\endgroup$ – PhysicsDave Aug 23 '18 at 18:31
  • $\begingroup$ @PhysicsDave: 1) Yeah you are correct, I assumed it wasn't their actual intention to say that, so this answer if off base if they really meant for there to be additional detectors. 2) What you describe is basically what I did. The amplitude of each path is $exp(i d_i k)$, so for each location on the screen you sum up the amplitudes and square them to get a (relative) probability. So every photon contributes, but photons close in phase lead to constructive interference, and those out of phase lead to more destructive interference. I only counted 24 paths, instead of an infinite number. $\endgroup$ – levitopher Aug 24 '18 at 13:40
  • $\begingroup$ 2) So you pick a point on the screen and calculate 24 paths to it .... I would like to say that when the "possible paths' $\endgroup$ – PhysicsDave Aug 25 '18 at 2:07
  • $\begingroup$ 2) So you pick a point on the screen and calculate 24 paths to it .... I would like to say that when the "possible paths" (not photons) are out of phase the probability is low. When "possible paths" are in phase they are more probable and these paths are the ones photons are more likely to follow. Feyman's work is an evaluation of the possible valid or probable trajectories and that's where the photons go. I don't think the photons interfere at all, if 100 photons go into this experiment you get the same total intensity as if 100 photons were hitting the screen with no slits. $\endgroup$ – PhysicsDave Aug 25 '18 at 2:18
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    $\begingroup$ Ok that sounds good. I always get obsessed by the single photon/electron double slit experiments where the diffraction pattern still emerges. So what your pointing out is the single (but multi-part) photon wave seeing all paths and these interfere to get the final probability at a point on the screen. Thanks. $\endgroup$ – PhysicsDave Aug 26 '18 at 2:32

It would likely look very mixed on the screen. The particles passing on the left 1st wall get diffracted and become multiple sources for the 2nd wall. The particles where the detector is on the right side lose all diffraction properties and it appears as a point source for the 2nd wall. So finally the 3 slits see multiple sources from the left slit and a point source from the right slit, these sources would be diffracted by the 3 slits which would leave a complicated mixed pattern on the screen.

  • $\begingroup$ I had not given any thought to the detector's side. I have no idea about the workings of such a detector in a real experiment, but my intention was for the detector to have identical effect on a particle's journey regardless of which of the first slits it went through. So with that change, the 3 slits would see 2 point sources, I reckon. Would a diffracted source just miss the slits more often, or is there a deeper significance to "multiple sources"? $\endgroup$ – Fisk42 Aug 16 '18 at 21:08
  • $\begingroup$ Yes its impossible to detect the particle or photon without fully interacting with it thereby collapsing its wave function and making it start a new one ( un-diffracted). No real significance to multiple sources, it's just the diffraction pattern caused by the particle's wave interaction with the first slit(s). Many particles will hit the walls. $\endgroup$ – PhysicsDave Aug 16 '18 at 21:51
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    $\begingroup$ I just always assumed that when a detector was mentioned in a double slit experiment, it was automatically implied that both slits were being observed. I meant for that to be implicit here. Otherwise you're not really measuring which slit a particle goes through. I didn't think I needed to mention two separate devices to be clear. $\endgroup$ – Fisk42 Aug 16 '18 at 23:15

Like WILLO said "the answer depends on the size of the slits, the distances between them and the wavelength of the light" Answering your question also depends on the location of the 1st wall detector. If it blocks photons at (A) then you have a single source from (B) hitting a three slit wall and creating a three slit pattern on the final screen. If the detector blocks photons at (B) then (A) becomes a single source and creates the same three slit pattern at the final screen. If the detector blocks both A and B then you have no source, no light and no pattern.


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