In the double-slit experiment, do electrons need a minimum velocity to produce an interference pattern?

Question

For electrons to produce interference patterns, do they need a minimum velocity?

Answer 1

The interference pattern formed by an electron in a double-slit experiment is actually easier to observe when the electrons are moving slower. This is because the angular distance between successive interference peaks is given by $$ \Delta \theta \approx \frac{\lambda}{d} = \frac{h}{p d} $$ where $d$ is the distance between the slits. (This approximation holds when $\theta \ll 1$.) We see that for smaller momentum $p$, the angular spread will be greater; so all other things being equal, we want the electrons to be moving as slow as possible.

This said, there can be experimental limits that make it impractical to reduce the electron speed too much. The slower the electrons are traveling, the more time they will take to travel from the slits to wherever we are detecting them. (It's also easier to spatially separate the interference peaks if the detector is farther from the screen, which also advantages a longer time of flight.)

But the longer the electrons are in flight, the more likely it is that the electrons' states will be influenced by outside noise. These outside influences effectively "measure" the electron while it's in flight, changing its state and destroying its ability to interfere between the two paths. So for a given experimental setup with a given level of external noise, one may find that the interference pattern disappears below a certain electron velocity due to these "decoherence" effects.

Answer 2

Yes, but it is very small.

Let's consider a double slit experiment. What matters there is the ratio of wavelength and slit distance. If you start making your wavelength longer (by slowing down the electrons) the minima in the diffraction pattern will be moving further apart. At some point you will not get any diffraction minima any more, just one broad maximum. You could say that the diffraction "disappears". There will still be some intensity modulation due to diffraction effects but the characteristic pattern of minima and maxima will be gone.

The position of the diffraction minima is given by: $$ d \sin\theta = \left(n+\frac{1}{2}\right) \lambda,\ \ \ n = 0, 1, 2, 3, ... $$

If we put $d = \lambda/2$ then the first minimum will be at $\theta = \pi/2$ and we will be at the edge of diffraction disappearance (in the above sense). So, for diffraction to disappear we need:

$$ \lambda > 2d $$

The wavelength is linked to the velocity by:

$$ \lambda = \frac{h}{p} = \frac{h}{mv} $$

So we have: $$ v < \frac{h}{2md} $$

For slit spacing $d = 0.1$ mm (a small but still macroscopic double slit) you end up with $v<3.6 \ \mathrm{m}/\mathrm{s}$. That looks reasonable until you realise that an electron with 1 eV kinetic energy travels at almost 600 km/s. Getting electrons to travel so slow that you could actually observe the diffraction on such a double slit disappearing would be quite a challenge.

On the other hand, if you were to look at diffraction on a crystal lattice then your $d$ would be on the order of 0.1-1 nm. The speeds would become reasonable - you would essentially be doing low energy electron diffraction and could try to make it disappear by reducing the electron energy.

Answer 3

The de Broglie wavelegth of the matter particle has to be of the order of slit spacing to observe wave effects like interference, diffraction etc. which then sets the energy scale, which then sets the velocity (as @Karel points out), which in turn leads to practical considerations (as @MichaelSiefert notes). But I don't think that's what you really meant by your question (as @JonCuster commented on your question, it is a bit open-ended)

Instead, let me assume that you meant to ask if there is a fundamental minimal velocity requirement to observe interference with matter particles (using an appropriate slit). There isn't $^1$. Since wave nature of matter particles is a purely quantum phenomenon and notion of velocity and rest is ill-defined and replaced with those of energy and momentum and localization in quantum mechanics, the best one can say is that as long as the particle has any non-zero energy, in other words, as long as it can go through and therefore be localized by, a howsoever large yet finite slit(s), it'll show diffraction(interference) after an equally long observation time.

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$^1$ apart from the ground state energy of the electron-slit system which is irrelevant since in an idealization, we consider the electrons to be free.