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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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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.

spinachflakes
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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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