Title says it all – recently I encountered a strange homework exercise on de Broglie dual theory with an electron wavelength of few millimeters – which implies the velocity lower than 1 m/s. I personally considered the problem very artificial and unrealistic, but it made me wonder – what is the smallest electron velocity recorded in experiments? What does the theory on the other hand say?

# [Physics] How low can an electron go

electronsquantum mechanicsvelocity

## Best Answer

If you define the speed of your electron in a de Broglie way and just set $v = \frac{h}{m\cdot \lambda}$, you can rewrite that expression to involve the wavenumber $k$, i.e. $v = \frac{\hbar k}{m}$.

$k$ is constrained by the boundary conditions of the electron. In a large solid with periodic boundary conditions, $k$ can be arbitrarily small, so the particular electron would be arbitrarily slow.

However, you don't see these electrons, as, due to the Pauli principle, when you start filling up your solid with electrons, you have to go to ever larger values of $k$. The largest of them is called the Fermi wavenumber $k_F$, which then defines you a Fermi velocity $v_F$ which can be quite large. In experiments, you typically interact with electrons at, or close to, the Fermi level, since these are the ones with the highest energy and hence the ones you can remove the easiest. Therefore, you will not really

seethe slow electrons.If, on the other hand, you don't consider electrons, which are Fermions and thus subject to the Pauli exclusion principle, but

Bosons(such as ${}^{85}Rb$ atoms), you don't have to put them all in different quantum states. You are then free to put themallinto the $k = 0$ state. Indeed, this happens atverylow temperatures. Such a state is called a Bose-Einstein condensate. In time-of-flight experiments, it is then indeed observed that they all (or, at finite temperature, at leastmostof them) have essentially a velocity of $0$.