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I understand the double slit for waves but for electrons do we have a beam for each slit so each beam is responsible le to shoot electrons through its own slit.

Or do we have just one beam? Which slit do we place the beam to? If it’s in the middle, wouldn’t all the electrons hit the wall between the two slits?

Sorry Ian very confused as how to carry out the experiment for electrons

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  • $\begingroup$ one beam of electrons covers both elite, the same as with light. $\endgroup$ – trula Feb 19 at 22:16
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There is only one source or one beam of electrons. They are directed toward the slits, with some hitting the middle and some making contact with the edges of the openings.

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  • $\begingroup$ But the experiment is not really done with slits. It is either electron diffraction from a surface with atom rows, or with a wire at negative voltage acting as a biprism. $\endgroup$ – Pieter Feb 19 at 23:35
  • $\begingroup$ @Pieter You still need slits to be a double slit experiment or holes to be a double hole experiment. $\endgroup$ – Bill Alsept Feb 19 at 23:54
  • $\begingroup$ But how would you make such a thing? $\endgroup$ – Pieter Feb 20 at 0:55
  • $\begingroup$ @Pieter I’m not sure what you’re asking but creating two slits in any mass should do it. Two slits that are small enough and close enough to each other $\endgroup$ – Bill Alsept Feb 20 at 0:59
  • $\begingroup$ @Pieter look at my answer $\endgroup$ – anna v Feb 20 at 5:32
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The question is right to ask about how to do this in practice. Yes, this is done using two beams, but those beams come from the same source: enter image description here

The electron beam (in a modified electron microscope) is sent through electron optics where it is split in two by a thin wire at negative voltage. After that the beams focused again at the same spot where it produces interference fringes. The electron optics works like a biprism.

In a way it is the same as a demonstration experiment that I show every year with a laser beam that is split in two by reflection from the front and the back of a thick glass plate. Those beams are then two centimeters apart. But when those beams are diverging (because of a lens in front of the laser) they will overlap at a large enough distance, where an interference pattern is then produced. It is is the pattern of interference from two point sources. The data from the electron experiment (shown in Anna V's answer) also look like that. A single-slit envelope pattern is not apparent.

Edit: I just came across this article with 600 eV electrons (50 pm wavelength) and real milled slits:"The individual slits are 62 nm wide and separated by 272 nm." Bach, Pope, Liou and Batelaan (2013)

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  • $\begingroup$ You still have one source, one beam and the wire splits the opening into two slits. That’s why it’s called a double slit experiment. $\endgroup$ – Bill Alsept Feb 20 at 15:25
  • $\begingroup$ Not at all. Your second sentence says “this is done using two beams”. That is wrong or at the very least confusing because there is only one source. Also the wire splits the opening into two slits. $\endgroup$ – Bill Alsept Feb 20 at 15:38
  • $\begingroup$ I wrote:"this is done using two beams, but those beams come from the same source". This is illustrated in a nice schematic and any normal person can understand that. It also addresses directly OP's question. $\endgroup$ – Pieter Feb 20 at 15:42
  • $\begingroup$ Up top you were claiming that there are not two slits. You also use the wording two beams, which is also confusing. If in the long run you were trying to say the same thing as my answer above then of course I agree. $\endgroup$ – Bill Alsept Feb 20 at 15:50
  • $\begingroup$ @BillAlsept There is no slit width, there are no slits. $\endgroup$ – Pieter Feb 20 at 15:51
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This is not a direct answer to the question, because the question is built on some misconceptions:

  1. There's nothing special about "double slit" interferometry. All that's required to produce interference fringes is for two waves from the same source to be superimposed at an angle. This occurs in Young's double-slit interferometer, and it occurs in practically any other interferometer unless the two beams (from the same source) are perfectly superimposed (though interference still occurs in that case, it's more difficult to detect the interference).

  2. Two slits or two pinholes a small distance apart allow a single beam to be divided into two beams which spread out so that they overlap. Something analogous happens with a single wire. A diffraction grating can do the same thing. The important thing is that two physically separated portions of the incoming beam be diverted in such a way that they overlap, with the path length differences between the two paths being less than the coherence length of the source light.

In more direct response to the question, there are many ways to prove electron interference. Fabricating a sufficiently tiny version of Young's double slit, and finding a way to get a sufficiently coherent source of electrons is one way. Another way would be to use diffraction in a crystal to generate two beam from one, then another crystal to combine the beams at a very small angle ( again, by diffraction).

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The distance between slits and the slits' width for the experiment are calculated to be smaller than the de Broglie wavelength of the electrons. (similar to the double slit for photons)

This allows to have a double slit experiment with one electron at a time.

enter image description here

The interference effects show up in the probability distribution accumulated in the last slide.

So it is one beam with the appropriate DB wavelength for the geometry of the experiment to show the interference effects.

Edit after discussions in the comments:

this more recent experiment , found by Pieter, managed to create the double slit experiment as described by Feynman himself, electrons impinging on real slits.

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  • $\begingroup$ I just came across this article with 600 eV electrons (50 pm wavelength) and real milled slits:"The individual slits are 62 nm wide and separated by 272 nm." iopscience.iop.org/article/10.1088/1367-2630/15/3/033018 $\endgroup$ – Pieter Feb 22 at 21:21
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    $\begingroup$ @Pieter thanks, it is good to know that a real double slit for electrons exists. The experiment referred in wikipedia brings up a number of questions that are really irrelevant to the particle wave discussion, as all can be resolved with the mathematics describing the system. $\endgroup$ – anna v Feb 23 at 5:16
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In the Copenhagen interpretation, the principle is that there is a wave and that wave difracts through the holes like any other wave. The electron as a particle then appears at the screen with a probability that depends on the square magnitude of the wave. In this context, it is not suggested that the electron definitely went through one or other slit. Rather - if you adjust the experiment to determine which one it went through then you adjust the wave and so change the pattern on the screen.

The interference of "particles" in quantum mechanics is, mathematically, the interference of some kind of waves.

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  • $\begingroup$ This does not answer the question about how this is done. $\endgroup$ – Pieter Feb 20 at 8:44
  • $\begingroup$ I was mainly responding to this part of the question -- "If it’s in the middle, wouldn’t all the electrons hit the wall between the two slits?" Which appears to show the OP misunderstands what the experiment is all about. $\endgroup$ – Ponder Stibbons Feb 20 at 23:51
  • $\begingroup$ @PonderStibbons I do not think the OP misunderstands. Electrons are real and physical. In a double slit experiment an electron will pass through one of the slits or it will hit something like the OP suggests. When the OP says “if it’s in the middle” He’s not talking about the detection screen. The OP is talking about that little piece of mass between the two slits. And again the OP specifically said double slit experiment. $\endgroup$ – Bill Alsept Feb 23 at 16:42

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