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I have a very normal yet a curious question about the double slit experiment. No one takes much interest in my college to explain the stuff. Instead they are all just busy to complete the syllabus so no point in asking them. At first I thought that this question is stupid but now am asking. How does the particle coming out of the particle accelerator doesn't hit the space in between the two slits? Does it diverge in a curved path before going through any one of the slit? And if it does diverge, then why?

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There was a question of mine about Can the intensity distribution behind edges and slits be explaint by the interaction with the surface electrons of the edges? – HolgerFiedler Jan 11 at 20:54
Some historical facts about electron diffraction experiments In the double slit experiment what, exactly, is a slit? – HolgerFiedler Jan 11 at 21:01
up vote 7 down vote accepted

Most particles do hit that space, and are blocked. The source of particles must be such that it emits in all directions. Often, something is done to simulate a point source. An ideal point source emits in all directions. For example, in the case of light, a single slit is put in place before the double slit. The single slit simulates a point source and radiates in all forward directions.

Update after @Jaywalker's answer

@Jaywalker brings up an important aspect that I glossed over: How to reconcile the wave picture and the particle picture? How can a single "particle" (say, an electron) be launched toward both slits? The answer depends on what we think about when we say "particle". If we mean a tiny little object that moves through space like a projectile, then the answer is "it can't". This is primary evidence that that picture of "particle" is not a good description of what is going on. In short, all "particles", including electrons and photons, are quantized excitations of a field. Energy and momentum are transferred at particular locations, giving the impression that a object hit something. The field exists everywhere in space except where space is occupied by an obstruction. The excitation of the field (the "particle") exists everywhere, but the interactions occur at particular locations. The field obeys some wave equation, and thus exhibits interference.

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I am sorry, but the electron IS an excitation of the field, and it does not exist everywhere, but where the quantum mechanical probability of its existence/excitation is high. . It is the electron field that exists everywhere with an expectation value of zero unless the probability of an electron passingt that (x,y,z) at time t is high. – anna v Jan 11 at 15:58
Quantum mechanics tells us only where the particle might be found. It might be found anywhere. The expectation value of it's location is large only after it's been observed to be there. There is no additional information available as to where or what path the particle takes. – garyp Jan 11 at 16:18
@annav I'm no expert in the interpretation of QM, but I don't see anything in the theory that suggests that it's valid to say that "the electron exists at some location, but we just don't know where". To me, that sounds like a hidden variable. – garyp Jan 11 at 16:45
@garyp It cannot be found anywhere - it must still obey all the fundamental rules. One such limit is the speed of light - the particle simply cannot "move" faster than light, which limits the volume where (and when) it might be found. And of course, even within this volume (of space and of time), there are places where it's much more likely to be. The precise meaning of "where" is one of the places where the competing QM theories differ quite a bit, so you'll get a different answer from a casual multi-worlder than from a string-theorist. The difference isn't simple to observe... – Luaan Jan 11 at 17:39
@radiantshaw Aren't you glad you asked such a simple to answer question? ;-) – Cort Ammon Jan 11 at 19:59

Really when dealing with this problem, the particle can not be considered as a particle but a wave. The wave is emitted in all directions and a small portion of it goes through the slits. Another part reflects off the wall inbetween the slits. If a source of single particles like an electron gun was pointed at the slits, indeed a portion of the total number of particles have a chance of reflecting off the barrier between the slits.

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Yes. I've updated my answer to address this important point. – garyp Jan 11 at 15:32
I am sorry, but electrons go through the double slits "whole" no "partial electrons" hit the screen. You are making the common mistake to think that the wave nature is in the mass/energy distribution. It is a PROBABILITY distribution wave. Each time one sends an electron towers the double slits, there is a probability for it to go either through one or the other or hit the space in between or outside the slits. This is given by the square of the wavefunction , the solution of the appropriate quantum mechanical equation with boundary conditions "two slits". – anna v Jan 11 at 15:56
I dont see how you come to the conclusion that My explanation has anything to do with "partial particles"! I am fully aware that the wave is not a mass distribution – Jaywalker Jan 11 at 16:13
I agree with Jaywalker but I would drop the wave description and change the statement to say: The accelerated electron emits photons in all directions and a small portion of them go through the slits. And some reflects off the wall in between the slits. – Bill Alsept Jan 11 at 17:36
because of " small portion of it goes through the slits". pies have portions, elementary particles are mathematically described either as point particles or probability waves . (They interact as point particles in the Feynman diagrams). – anna v Jan 11 at 18:00

The problem here is to distinguish between theory, facts and interpretations. The facts related to the two-slit experiment is that one particle always arrives as a point. If you have enough particles in sequence, a diffraction pattern becomes discernible that is consistent with the proposition that a wave diffraction pattern gives the probability that a given particle will arrive at a given point. We also believe the experiment complies with the laws of conservation of energy and of momentum. We also know if we shine strong light on the particles as they exit the slits, the diffraction pattern disappears, and we get the pattern, more or less, of one electron having gone through one known slit.

Either there is a wave and a particle, or there is not. In the Copenhagen Interpretation, there appears to be not a wave as such, and the effect happens merely to comply with an equation, and the whole issue is left afloat. The premise that there is a wave was followed separately by de Broglie and Bohm, and this is the interpretation I follow, although I have made some alterations in that I add the requirement that the phase term follows Euler and becomes real at the antinode (and I attach physical significance to that at times) and that the phase velocity must equal the particle velocity to affect the wave. That requires the wave to transmit energy, and I assume it guides the wave through an energy field. Where the energy is is admittedly a problem, and it effectively requires another dimension, which some will regard as ugly. I call these guidance waves, to slightly differentiate them from the pilot wave.

The important point is that weak measurements (Kocsis, S. and 6 others. 2011. Observing the Average Trajectories of Single Photons in a Two-Slit Interferometer Science 332: 1170 – 1173.) indicate that when emerging from the slit the photon follows a trajectory in accord with that predicted by Bohm, which, in my opinion, is strong evidence in support of the wave plus particle concept, and does not sit at all well with the distributed quantum field, or the probability distribution until observed concept.

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I fail to see how this answers the questions asked by OP: "How does the particle coming out of the particle accelerator doesn't hit the space in between the two slits? Does it diverge in a curved path before going through any one of the slit? And if it does diverge, then why?" – AccidentalFourierTransform Jan 11 at 21:30
Who says it does not strike the solid? I am unaware of any experiment that counts precisely the particles emitted, and the particles counted. I have always assumed that some will hit the solid, but they are then irrelevant. If I am wrong, I would've to see the relevant reference. – Ian Miller Jan 12 at 23:05
Also the interference pattern appears when single electrons are fired! Im sorry but your answer is wrong – Jaywalker Jan 14 at 15:14

protected by Qmechanic Apr 17 at 22:35

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