Are electrons on a conductor, in electrostatic equilibrium, stationary?

Electrostatics basically means dealing with time independent electric fields (which was produced by stationary charges)

Now consider a neutral conductor. We know that putting a net negative charge on the conductor, the charge will very quickly spread over the surface till electrostatic equilibrium is reached.

What does this exactly mean? does it mean that electrons on the surface are not "moving" anymore and they become stationary? So in this situation we have 2 kinds of electrons: 1-electrons inside the meat of the conductor whose dynamics is described by quantum mechanics 2-electron on the surface which are not moving in the classical sense?

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The electrons in a conductor are free, i.e. not bound in atoms, and have nonzero energy, so they move around. But we do not measure any macroscopic current because they move chaotically so that the contributions to the current from individual electrons cancels.

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Are you saying that stationary electrons means they are moving but randomly? – Revo Oct 21 '12 at 21:13
@Revo Of course. Even at the temperature of absolute zero they still have some momentum uncertainty, so they still move. Even electrons in atoms move, the only difference being the bound electrons cannot move freely. – Ondřej Černotík Oct 22 '12 at 7:12

All the electrons in the metal obey quantum mechanics at the microscopic level, regardless of whether it is the bulk or the surface. However, depending on the system you are studying, and the conditions you have created, you can safely employ a semiclassical description of electron motion. For most typical metals of macroscopic scale, the single-particle semiclassical description works well (Drude model is the simplest). My explanation below relies on this assumption:

Metals have half filled bands; or in other words, the chemical potential lies in between a band. As a result, the electrons near the chemical potential can move around the metal, while the electrons that are much deeper, i.e. way below the chemical potential, stay bound to the positive ion cores. If you introduce an excess negative charge, the extra electrons will go into the half filled band, and consequently the chemical potential will rise slightly; the band will still be half filled. You can safely ignore filling up of this band by the excess electrons. I am saying this because even if you introduce one extra electron per atom, and you have an Avogadro number of atoms, you will get some ridiculous values of net excess charge $6\times10^{23}\times1.6\times10^{-19}=9.6\times10^4 C$. In the end you have a system which is still like a metal except there will be a net charge. Since the extra electrons lie in a half filled band, and lie close to the chemical potential, these electrons will also be delocalized. So yes, like Ondřej Černotík was implying, the electrons will move around randomly such that the net vector sum of their velocities will add up to zero.

To sum up I would like to say that there are indeed two classes of electrons: (1) moving (delocalized) (2) bound. The delocalized electrons will exist on the surface as well as in the bulk. The excess electrons, however, will only exist on the surface. I think that can be explained classically by minimization of capacitive energy $E = Q^2/2C$. The capacitance of a cylindrical shell is proportional to the radius, i.e. $C \propto R$. Therefore it is energetically favorable for the electrons to be as far away from the center of the object.

By the way, all this band description in the second paragraph is simply to make the point that the excess electrons will be delocalized, i.e. they fall in class (1) of electrons. That explanation simply supports my claim that electrons are moving around randomly.

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