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

Question

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.

Answer 1

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.

Answer 2

Your intuition is correct:

The picture you have shown appears in many a pop-sci YouTube video, but it is just wrong. When the particle goes through one slit, it behaves like a wave going through one slit: it diffracts. It always propagates like a wave between emission and detection.

Note that diffraction is the coherent sum of propagation through all infinitesimal slices of the slit. In fact, a slit is the same as two half-slits in the limit of zero separation. It is simply not possible for it suddenly "be a particle" (which is a wave following geometric optics), unless the slits are too wide, in which case there won't be an interference pattern anyway, because there is no final-state overlap in propagation through each slit.

Answer 3

Despite the prevailing consensus, I argue that the observed patterns are attributable to photons, as if the double-slit experiment were conducted with photons. When electrons are accelerated through the experiment, they emit billions of photons that produce familiar patterns. As the electrons move toward the detection screen, they are guided or corralled by reflecting photons to specific locations on the screen, aligning with the interference pattern of photons. If observer means blocking one of the slits then you get what you would expect from a normal double slit experiment with photons.