# How are electrons distributed on the screen in the double-slit experiment with an observer?

As I understand it, when both slits are open, the electrons spread out in an interference pattern, and if one of the slits is closed then the electrons are spread out according to a single-slit diffraction pattern.

In his conference about the double-slit experiment, Feynman said that when both slits are open and you look at which slit the electrons passed through, the electron distribution is the sum of the distribution if slit (1) is closed and the distribution if slit (2) is closed. The electrons are therefore spread out over the entire screen, with a higher concentration in front of the slits.

However, I saw in several images that, in the presence of a detector, the electrons were split into two bands, each in front of a slit. This makes sense, since measuring the electron's position should force it to behave like a particle, and thus cancel out the diffraction effect.

Am I missing something ? Unfortunately, I haven't found any photos on the Internet that show the results of the experiment with an observer. I would be very grateful if you could help me.

• Sum of slit 1 (with s2 closed) and S2 (with S1 closed) is also 2 bands ... I think that's what Feynman meant. The 2 situations are the same. Commented May 31, 2023 at 21:36
• youtube.com/watch?v=b0EChbwSuuQ Go to 35:00, it doesn't really look like 2 separate bands like in the second picture of my post. If what you say is right, then he drew the electron distribution in a pretty misleading way... So there's no electron diffraction when a single slit is closed ? Commented May 31, 2023 at 21:55
• @PhysicsDave see AXensen's answer below. It should be one band, not two. Commented Jun 1, 2023 at 0:34
• Each of the two separate bands has individually a distribution as when only a single slit is open. It's still the sum of the two, it's just that the individual distributions that are being added together are narrower and more separated - how exactly it's going to look (overlap or no obvious overlap) depends on the details of the experimental setup, and is not important. (Imagine it's bullets - it would depend on how much they ricochet off of slit walls, how far apart are the slits, etc.) The point is that you get a simple sum of individual distributions, and not an interference pattern. Commented Jun 1, 2023 at 2:44
• "it doesn't really look like 2 separate bands like in the second picture of my post. If what you say is right, then he drew the electron distribution in a pretty misleading way" - actually, that second picture in your post is schematic; it's "morally correct" as they say, but don't take it too literally when it comes to where the electrons hit and how they arrive to the screen - all schematic depictions are to a certain extent misleading when it comes to details. Commented Jun 1, 2023 at 3:15

The first picture you posted shows in (b-left) the result of the electrons only going through slit 1. (b-right) is the result of electrons going through slit 2. The result with electrons going through both slits, but with an observer is the sum of the two pictures.

The second picture is wrong - if electrons going through either slit don't have overlap in the positions they might land, there won't be any interference. In the dark interference fringes, electrons could have landed there if they went through either slit, but the dynamic phase they accumulate through either slit is different by $$\pi$$; the two components of the wavefunction cancel out.

the rest kind of goes beyond the original question, but it might be useful to show some back of the envelope math that explains a bit more quantitatively how this works

Making these two distributions overlap while keeping electron travel distances short enough that they aren't likely to collide with a background gas molecule and lose coherence is required to make one of these experiments.

After going through a slit of width $$w$$, the velocity uncertainty in the horizontal direction is roughly $$\hbar/(2mw)$$. So if the slits are separated by a distance $$d$$, they should fly for a time $$t=2dmw/\hbar$$ so the distributions overlap. I frankly don't know what parameters were historically used in such experiments, but $$E=10\text{ eV}$$ electrons ($$t=D/\sqrt{2E/m}$$ where $$D$$ is the distance to the screen) and $$1\text{ }\mu\text{m}=w=d$$ seem plausible (comment below) and give a distance to the screen of $$D=3\text{ cm}$$. It's pretty easy to make the vacuum good enough to have a free path of $$3\text{ cm}$$ for electrons.

• Thanks a lot ! Could you tell me if my interpretation is correct? When we look at which slit the electron passed through, we force it to pass through just one slit with a 50/50 probability, and it was "as if" the other slit was closed. And so the electron continues to propagate like a wave diffracted by a single slit. The result of the experiment with an observer is therefore the sum of the results with a slit closed and with the other slit closed. Is that right ? Commented Jun 1, 2023 at 6:55
• @RoyalSalmon that sounds good to me Commented Jun 1, 2023 at 9:32