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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?

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I think the original title was clearer... –  David Z Apr 27 '11 at 20:39
    
Right, but it was so inviting....:=) –  Georg Apr 27 '11 at 21:28
    
This would allow us to actually test quantum gravity effects in the from of zwitterbewegung, the 21st century equivalent of Millikan's oil drop experiment ... (insert "perhaps", "but" and "maybe" wherever required ;) –  user346 Apr 28 '11 at 6:14
    
@Deepak: Nitpick, it's Zitterbewegung. –  Raskolnikov Apr 28 '11 at 11:52
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Actually, Zwitterbewegung means "hermaphrodites' movement" :-) –  jdm Jan 18 '12 at 15:19
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6 Answers

up vote 4 down vote accepted

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 see the 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 them all into the $k = 0$ state. Indeed, this happens at very low 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 least most of them) have essentially a velocity of $0$.

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I hope you mean Rb! (I'm no nuclear physicist, but I imagine $^{85}$Ru is not an easy thing to work with...) –  wsc Apr 28 '11 at 3:33
    
Yep, I meant Rb. Sorry for the mix-up –  Lagerbaer Apr 28 '11 at 4:21
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Smallest recorded in experiments? The experiment that comes to my mind is the work done in the Gabrielse lab at Harvard, where they've trapped a single in a Penning trap, and cooled it down to less than a tenth of a Kelvin (which pretty much puts the electron in the ground state of the trap).

The electron has fast cyclotron motion, but I'd estimate that in one direction it should have a velocity around 1000 m/s.

To get a bare electron with a velocity of 1 m/s would require cooling the thing down below a microkelvin (if I'm doing my arithmetic right). As far as I know, that's never been done experimentally.

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If this surprises you it means you need to understand the role of Special Relativity!

The wavelength of a free electron is infinite in its rest frame. Furthermore, the direction of the wavefront of an electron depends on the reference frame. Whatever direction you choose, there is a reference frame in which the wavefront of the electron has that particular direction.

Believe me, it all makes sense.

1) The wavefront is about the phase, not the magnitude. So the density of the electron is the same at all places of the wavefront.

2) Think of the phase as a local clock. In the rest frame all clocks show the same time. If you choose a different reference frame then some clocks in a certain direction show a later time while the clocks in the opposite direction show an earlier time. This direction is the same direction as that of the wavefront.

There are images in my book which illustrate this.

http://physics-quest.org/Book_Chapter_Klein_Gordon.pdf

Look for instance at figure 9.2 and 9.3 to start with. For the rotation of the wavefront you can look at 9.4 and 9.5.

Regards, Hans

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I turns out that when you find the expectation value of free electrons speed that tend to always get c, when interpreted by the dirac equation!

This phenomenon is known as zitterbewegung, The idea being that you have an electron moving at c, but around a path with a low drift velocity.

That probably didn't help but I think it is nice.

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There is nothing nice in it but a confusion. –  Vladimir Kalitvianski Apr 28 '11 at 16:06
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Electrons in a white dwarf star should be static as atoms in a condensed matter.

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The minimum electron velocity depends on the external potential and can be as low as zero m/s.

Consider, for example an electron bouncing on a horizontal mirror in the gravitational field. Such a system has a minimal energy $E_0$ and therefore a minimal non-zero velocity.

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protected by Qmechanic Jan 30 at 14:23

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