We study "Free electron theory in metals" in Condensed Matter Physics. While discussing this theory (both classically and quantum mechanically), we assume the electrons moving inside the metal to be a "free electron gas". But in Condensed Matter Physics, we mainly discuss about the "condensed" phases of matter, then why do we consider a gas in this model? (I know this gas is not same as any ordinary gas and it's concentration is much more than that of an ordinary gas, also by assuming this, the calculation gets simplified by many folds, my question is on what physical grounds do we consider the freely moving electrons in the metal to be a gas?)

  • $\begingroup$ Are you familiar with the tight binding model? $\endgroup$ Oct 22 '21 at 5:24
  • $\begingroup$ Yes I am. But in what context TB Model will be considered here? $\endgroup$ Oct 22 '21 at 6:41
  • $\begingroup$ We consider it to be a Fermi gas, i.e. obeying the Fermi distribution. The reason is that it provides a workable and effective model, which is what physics is all about. $\endgroup$ Oct 22 '21 at 6:45
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    $\begingroup$ I understand that it works out that way and it becomes simpler. But to consider a model to simplify the solution of a problem requires some logical reasons. I did not understand why a gas. It is true that FD distribution works out for gas but are there any other reasons? $\endgroup$ Oct 22 '21 at 6:54

That is simplification based on the Wannier theorem, later generalized by Luttinger and Kohn in their 1955 article "Motion of Electrons and Holes in Perturbed Periodic Fields". Basically, if you have energy band depending on the wave vector as $\varepsilon_{0} (\mathbf{k})$, and apply external potential $V(\mathbf{r})$, weak compared to crystal potential and slowly varying compared to unit cell size, then Schrödinger's equation can be written like that:

$$ \left(\varepsilon_{0} \left(- i \mathbf{\nabla}\right) + V(\mathbf{r})\right) \Psi(\mathbf{r}) = \varepsilon \Psi(\mathbf{r}) $$

Here $\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)$. For details and proof go to the 1955 paper mentioned above or to Anselm's "Semiconductor theory", Ch. IV paragraph 3. The latter is simpler.

In many cases, especially in semiconductors, at low values of $k$ energy is parabolic: $\varepsilon_{0}(\mathbf{k}) \approx C * \left| \mathbf{k} \right| ^2$. Renaming $C = \frac{\hbar^2}{2 m^{\ast}}$, where $m^{\ast}$ is called effective mass, we have

$$ \left(- \frac{\hbar^2}{2 m^{\ast}} \nabla^{2} + V(\mathbf{r})\right) \Psi(r) = \varepsilon \Psi(\mathbf{r}), $$

which is very similar to the equation of actual free electron gas.

Actually, the question you asked puzzled scientists for a while, but they asked "why electrons in materials behave like free electron gas".

  • $\begingroup$ If my question is the one that you have mentioned i.e. why electrons in metals behave like a gas, then what should be the answer? $\endgroup$ Oct 22 '21 at 6:43
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    $\begingroup$ That is not an intuitive derivation, but more mathematical one. The answer is "because electrons are described by the same equation". But if you want to imagine some kind of picture then you need to recall that electrons are represented by wavefunctions, and if a wavefunction is very wide (much larger than the size of a unit cell), the electron can hop from one cell to another if pulled by something, for example, electric field. If it is VERY easy to hop, then the electron doesn't even see the unit cells - it simply flows as if there was no crystal. Not sure if helpful, but I did my best :) $\endgroup$
    – Andrei Z.
    Oct 22 '21 at 6:54
  • $\begingroup$ Sure it was helpful. Thanks a lot. $\endgroup$ Oct 22 '21 at 18:13
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    $\begingroup$ @SAGARMODAK Courses in condensed matter (usually advanced ones) will discuss Landau's Fermi Liquid theory, which is a more appropriate description of electron behaviour in metals. This theory is very closely tied to (rather, built on top of) the free electron gas, so many calculations in the two give similar results. Granted, this isn't an explanation for why electrons behave close to gases in a metal - just pointing out that Fermi Liquid theory deals with that closeness in detail. $\endgroup$
    – Crisco
    Nov 16 '21 at 13:30

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