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I've come up with a thought experiment to help myself understand the classic double slit test, and specifically to understand what happens to the particles that do not get detected on the other side of the slits due to destructive interference. I couldn't find anything quite like this in popular literature, but if it has been done already even better.

In my setup there are two detection surfaces placed one behind the other, A and B. They are both sensitive to electrons and can record the X position of the hit. In surface A there are double slits. The electrons are pointed directly at the center of A. The assumption is that the electrons we do not detect at B will be detected at A.

In the first setup - W for wave - there is no path detection, so we would expect the interference pattern at B. In the second setup - P for particle - we have lights placed right above the slits at the exit point so that we can see the paths of each electron; in this case we do not expect the interference pattern at B. In both setups we fire off the exact same number of electrons, one at a time.

Here are illustrations:

Setup W (wave)

Setup P (particle)

And here are the questions:

  1. Do I have the correct rough detection patterns at both A and B in both the W and P setups?
  2. Will the total number of detections N in each setup W and P be the same? My understanding is yes.
  3. Will Nb (the total number of detections at B) in the W setup be less than Nb in the P setup? My understanding is that it will.
  4. Assuming I am correct on (2) and (3), then it must follow that Na will be greater in the W setup than the P setup. Is that correct? Otherwise how do we account for the "missing" electrons that do not get detected at B in the W setup?
Peter Moore
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  • There is a body of literature on phase rigidity and phase rigidity breaking in which path interferometers (which are similar to two-slit experiments). In these systems losses of electrons have to be artificially introduced to break the Onsager-Büttiker relations. See, e.g., this answer and reference in it – Roger V. Nov 23 '22 at 08:43
  • @RogerVadim could you elaborate on how that relates specifically to my questions? Are you suggesting the detection counts at Nb will differ between the W and P setups? – Peter Moore Nov 23 '22 at 18:07
  • The comment "to understand what happens to the particles that do not get detected on the other side of the slits due to destructive interference" ... is incorrect. All particles that pass thru the slits get detected .... the term "interference" is historical and misleading ( but still taught today). Nothing is destroyed. – PhysicsDave Nov 24 '22 at 18:38
  • As I mentioned in the comment to the accepted answer, I was misinformed as to the detection count changing between the W vs. P rig. Given that the detection count does not change, dwelling on this further seems as though it would only confuse the issue. – Peter Moore Nov 24 '22 at 20:02
  • Moreover, I never suggested that the particles annihilated or anything similar. My - incorrect - hypothesis was that the presence or absence of detectors at the slits would impact the detection rate at both A and B, but I never imagined the total detection count would differ between the two setups. Of course that would break all sorts of very good laws. But again, since the detection rates at both respective planes do not change between W vs. P, then there's really nothing else to say I don't think. – Peter Moore Nov 24 '22 at 20:08

2 Answers2

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  1. Do I have the correct rough detection patterns at both A and B in both the W and P setups?

Yes (though it would of course be strictly zero where the slits are)

  1. Will the total number of detections N in each setup W and P be the same? My understanding is yes.

Yes

  1. Will Nb (the total number of detections at B) in the W setup be less than Nb in the P setup? My understanding is that it will.

No it will not. The type of interference pattern does not change the number of photons hitting the back screen. It just causes them to be arranged with a different distribution, one which mirrors two waves which can interfere destructively.

  1. Assuming I am correct on (2) and (3), then it must follow that Na will be greater in the W setup than the P setup. Is that correct? Otherwise how do we account for the "missing" electrons that do not get detected at B in the W setup?

Since 3) was not correct, 4) does not follow.

doublefelix
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  • Thank you. In another question I asked (3) more directly, and someone with a very high reputation commented and was seeming to say that the detection rate WOULD be different. https://physics.stackexchange.com/questions/737522/does-double-slit-detection-rate-change-when-path-is-known-vs-unknown?noredirect=1#comment1650976_737522. They also pointed me to a source that seemed to say the same thing. Will link to that in a separate comment. Any way to reconcile all this? – Peter Moore Nov 22 '22 at 21:12
  • This was the source they cited. https://www.sps.ch/artikel/progresses/wave-particle-duality-of-light-for-the-classroom-13/ It says "when each photon is forced to follow a specific path (right), its probability to be detected is higher than when it is left with a choice of two possible paths (left)". – Peter Moore Nov 22 '22 at 21:13
  • I read the source. It is also correct, but we are speaking about different things. The source is talking about the number of photons hitting the back screen at a particular point (or small area) corresponding to where they placed the photomultiplier (=photon detector). In my answer I spoke of the total number of photons hitting a back screen as in the question you wrote. In the sps.ch paragraph where they say that the detection rate is dramatically increased, they have placed the photomultiplier at a minimum in the interference pattern, which is why the detection rate is so low. – doublefelix Nov 22 '22 at 21:25
  • So, in that paragraph, they are referring to a setup without a screen but where they sample small areas of the screen one at a time until they build up the interference pattern of the whole screen. It's not exactly the same setup as in your question but you can still infer the same results by making a "collage" of the PM (photomultiplier) click rates – doublefelix Nov 22 '22 at 21:28
  • Ah ok. It seems that I was led astray by the rather cryptic comments being given to my other question and my attempts to get clarification from the author were not successful. I appreciate the clarification. If you wouldn't mind answering that question as well I'll be happy to accept these both. Thank you again. – Peter Moore Nov 22 '22 at 21:30
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The root of your question is that you are thinking because there is "interference" particles are cancelling and therefore there should be less of them ....?

Electrons are particles and they do not annihilate themselves (that would be a violation of conservation of mass).... Photons are packets of energy in the EM field and they also do not annihilate themselves (that would be a violation of conservation of energy) .... all photons are eventually absorbed/observed.

In your experiments (with electrons) the excited electrons in the electrode are already interacting with the EM field due to EM forces, one can say the forces are shown to be strongest at the bright spots (and these can be calculated per Feynman). When the electron is emitted these forces influence its path .... hence the "interference" pattern. Note that the pattern is caused by forces .... i.e. virtual. Note that electrons always have wave properties within the EM field.

In the second experiment the electron will interact with a photon after the slit ..... the virtual forces are working with the photon field after the slit .... hence the the slit has no effect and no pattern.

PhysicsDave
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