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One observes that the fact of detecting the location of a particle path (which slit the particle went through) entails a collapse of the particle field (interference fringes disappear on the detection screen).

But, in every condition of the experiment, the detection screen does also give some information about the location of the particle. So how comes the detection screen does not trigger any collapse in the condition where slits are not observed?

Shouldn't the fact of hitting the screen entail some change in the particle field ? How is the state of the field in the condition where slits are not monitored ?

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  • $\begingroup$ The slit widths and distance between them is such that it is not possible, within errors, to tell which slit the photon or electron came through (one particle at a time experiments), so the path cannot be detected $\endgroup$ – anna v Sep 22 '19 at 14:51
  • $\begingroup$ @anna v : I meant : how comes, in those same "one particle at a time experiments", the detection screen does not make interference fringes disappear ? Isn't that screen also a type of measurement itself, just like the "which slit" detector is? Where is the difference coming from. $\endgroup$ – Michelange Baudoux Sep 22 '19 at 15:13
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    $\begingroup$ The detection screen is the apparatus to measures the fringes. It tells you that there are fringes. Not sure what you mean that it can make them disappear. $\endgroup$ – Jan Bos Sep 22 '19 at 16:23
  • $\begingroup$ Interference disappears in experiments where the slit through which the particle passed is detected. The spatial accuracy of the points on the screen do not allow to draw a path to one of the slits, as they also are at a very close to each other. So the experiment cannot tell which slit the particle passed. $\endgroup$ – anna v Sep 22 '19 at 16:39
  • $\begingroup$ @annav, Jan Bos : All this was very well understood (see my answer here : physics.stackexchange.com/a/504008/242752). My question is how comes one measurement at the slit is destructive and the detection screen is not. $\endgroup$ – Michelange Baudoux Sep 23 '19 at 5:38
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Any edge deflects photons into areas with higher and lower density. To measure such a result different measurement instruments can be used. Beside a screen one may use a photodiode which is moved and the outcoming current shows the same intensity distribution like the pattern on a screen.

The detection of photons direct behind the slit destroys the interaction between the photons and the edge(s). Since a photon-photon interaction is very improbable you could use electrons, shooting them parallel to the wall with the slits. It will interfere the fringes. But why?

We know that the electrons we use for detection, interact with the photons after the edges. With the same certainty, we should assume that the photons and the surface electrons of the edges interacting.

Let's imagine an experiment. We shoot three electron beam and set up our photon beam perpendicular to it. We should get a result with an intensity distribution similar to the double slit (not the same, because only a small amount of photons meets the electrons).
Now, if you try to find out, which way the photon was going between the electron beams and you do this with an additional electron beam, clearly you’ll disturb the intensity distribution. Simply, we not have the sensitive enough measurement instrument.

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