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With two slits experiment shooting electrons when we place a detector on one of the slits the electrons act as particles.

What will happen if we use three splits and one detector? Will we have an interference pattern? If we do how will it look like?

Also in general if we have $n$ slits and $m; m < n - 1$ detectors how would the pater look like?

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  • $\begingroup$ When you "place a detector on one of the slits", does it obstruct the slit? $\endgroup$
    – R.W. Bird
    Commented May 24, 2020 at 19:27
  • $\begingroup$ @R.W.Bird it's possible, but then the question is not so interesting. $\endgroup$ Commented May 24, 2020 at 22:13

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If there is a detector at one of the slits, it corresponds to a measurement with the associated eigenvalues, say, $1$ (the particle passes through the given slit) and $0$ (the particle doesn't pass through the given slit). Now if there are two other slits, the eigensubspace corresponding to the eigenvalue $0$ would be two-fold degenerate corresponding to the states of particle passing through slit $2$, slit $3$, and superpositions thereof.

Thus, if the detector measures $1$ then the particle would not be in any of the "slit-superpositions" and wouldn't interfer with itself. It simply go to the screen and produce a diffraction pattern. But if the detector measures $0$ then the degeneracy of the eigensubspace will make sure that the particle would still be in a superposition of going through slit $2$ and slit $3$ and thus, it would interfere with itself and produce a double-slit interference pattern on the screen.

So when you do the experiment, the end result would be that the pattern on the screen would be a simple summation of the double-slit interference pattern corresponding to the slits without detectors and the diffraction pattern corresponding to the slit with a detector.

I'll leave the generalization of this to you which should is easy enough. :)

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  • $\begingroup$ Thank you for your answer, it answers the general case as well. Can you please elaborate more on the summation? Is there a formula to calculate the general case? $\endgroup$ Commented May 24, 2020 at 1:15
  • $\begingroup$ @IlyaGazman Yes, I just added the line that the generalization is left as an exercise to the reader ;) The summation I mention is a non-interfering summation. If the intensity at a spot is calculated to be $I_1$ due to the case of passing through slit $1$ and $I_0$ due to the case of not passing through slit $1$ then the total intensity at the spot would simply be $I_1+I_0$. Notice that $I_0$ itself is an interference generated intensity but there will be no interference between the cases of having passed through slit 1 and not having passed through it, thus the simple summation $I_1+I_0$. $\endgroup$
    – user87745
    Commented May 24, 2020 at 1:19
  • $\begingroup$ In general, the formula would depend on the placement of slits, etc. In the usual planar case, the source is such that the superposition of passing through slit 1 and slit 2 is an equiprobable superposition of the two cases. If you add three slits on a planar screen, this cannot be arranged so it would depend on how the slits are placed w.r.t. the source, etc. I'm not aware of such a formula but it can be easily worked out. $\endgroup$
    – user87745
    Commented May 24, 2020 at 1:22
  • $\begingroup$ It sounds like the answer is what's expected in theory. Can someone post an image gotten from an actual experiment? $\endgroup$
    – Noel Yap
    Commented Aug 21, 2022 at 18:16
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With two slits experiment shooting electrons when we place a detector on one of the slits the electrons act as particles.

That isn’t what happens. Instead of slits the first experiments were done with one wire only. No slits at all. The electrons get bended around it and on the screen they saw an intensity distribution:

enter image description here

What will happen if we use three splits and one detector? Will we have an interference pattern? If we do how will it look like?

Single edges are enough to get an intensity distribution with electron-poor and with electron-rich areas. Some years earlier single edge deflection was observed.

enter image description here Fresnel electron diffraction structures at an edge

Also in general if we have $n$ slits and $m; m < n - 1$ detectors how would the pater look like?

Take it this way: Whenever you prevent - for example by a measurement or the close of a slit - the electrons to interact with edges, the intensity distribution pattern disappear for that edge / edges / slit. The other edges or slits work as awaited, they deflect the electrons and the superposition of all edges deflections are on the screen.

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  • $\begingroup$ Is there a fundamental difference between the interference patterns produced by electrons and those produced by photons? $\endgroup$
    – R.W. Bird
    Commented May 24, 2020 at 19:05
  • $\begingroup$ @R.W.Bird It is. Photons get deflected behind and away from the geometrical shadow. Electrons get deflected only away. Follow the primary sources from the beginning of the 1900th. $\endgroup$ Commented May 24, 2020 at 19:14
  • $\begingroup$ It would appear that the waves associated with electrons do not follow Hygen's pricinple. $\endgroup$
    – R.W. Bird
    Commented May 25, 2020 at 15:01
  • $\begingroup$ The experiment shows, that electrons get deflected and this deflection has a swelling character. Furthermore one will not be able to detect these electrons near the back side of the wall. For water that is natural because the dispersion depends from the elasticity of the medium and this is equal in all directions for water. $\endgroup$ Commented May 25, 2020 at 15:08
  • $\begingroup$ And the wave associated with electrons has nothing to do with the ray. The interaction between each electron and the electrons on the surface of the slits are the interaction which makes the intensity distribution. Would be nice to measure the phonons in the slits material from the going nearby electrons. One by one electrons perhaps induce phonons or on the left or on the right slit. See here physics.stackexchange.com/questions/484160/… $\endgroup$ Commented May 25, 2020 at 15:11

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