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In its wave-form a particle should pass through every time because it propagates in all directions. So there shouldn't be any losses of particles landing in between the slits, right?

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Absolutely, but those do not register at the detection plane.

In some sense we are talking only about the data that comes out of the experiment and particles lost this way do not contribute to the data.

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Just to mess things up a little, would a detector in between the slits alter the result observed at the detector plane behind them? – Jaime Dec 24 '12 at 4:02
Depends on how the detector acts. If it lets particles through after detecting them, then yes, it would affect the result in some way, but if it blocks/reflects particles, it would just act the same as a wall. – David Z Dec 24 '12 at 6:25
Thanks for the answers. It makes me wonder why those particles' wave function collapses before passing through the holes. Nobody is observing them at that point and they should technically behave as waves i.e. pass every time through whatever openings available. – Lisbeth Dec 24 '12 at 22:04
@Lisbeth The issue is what is meant by "observe". The detector work by some physical mechanismin which the particle interacts in it's entirety in one particular place. There is nothing that prevents that same physi cs from happening at the screen. There is no complete theory of when and why that should be, but "decoherence" is the word that get batted around a lot. – dmckee Dec 24 '12 at 22:31
I always wondered why they don't decohere when they bounce off the mirrors further down the path during the experiment. – Lisbeth Dec 25 '12 at 23:24

Maybe I failed to understand your question or dmckee's answer, but it is my understanding that, while some particles pass through the screen, there are a lot of particles reflected from the screen. In the limit of very narrow slits, I guess, all particles are reflected, unless I miss something simple.

EDIT(12/25/2012): In the following article: Zeilinger e.a., Rev. Mod. Phys., v.60, #4, p. 1067 (1988) ( ), in a double-slit experiment with neutrons, some neutrons are definitely absorbed in a boron wire between the two slits.

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The weird thing is, why are some of them reflected while others pass through BOTH openings (seemingly). If the wave is spread out enough to pass through both slits, then no particles should get reflected at all. – Lisbeth Dec 24 '12 at 22:12
@Lisbeth: Before I try to answer, could you please explain something I don't quite understand: are you saying that, according to quantum theory, there will be no reflected particles (and I disagree with that - I believe there will be reflected particles, according to quantum theory), or that, while quantum theory predicts there will be reflected particles, this prediction is weird? – akhmeteli Dec 24 '12 at 22:29
I really don't know what QT predicts in that situation. I was just wondering what happens in actual experiments, and how are these spontaneous collapses explained – Lisbeth Dec 25 '12 at 9:38
@Lisbeth: It is my understanding that QT predicts both transmission and reflection, otherwise you cannot satisfy the boundary conditions on the screen for the wave equation (see also the edit to my answer). As for what happens... There are different opinions. You may wish to look at this article: Phys. Rev. Lett. 97, 154101 (2006), where, in particular, two-slit diffraction is successfully modeled by classical objects. I guess the following recent article of the same authors is publicly available:… – akhmeteli Dec 26 '12 at 2:36
@Lisbeth The wave is not a matter wave in the way of acoustics etc. It is a probability wave , and the interference is an interference in the probability amplitude. It is wrong to think of a particle/wave as having its mass/energy spread all over the space of the wave function. The probability of finding it there is what the square of the wave function predicts. Collapse is a misleading term. Nothing collapses. A probable outcome is realized. See the experiment shown in… . – anna v Dec 26 '12 at 5:51

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