# Free electron gas model and electrons in real metals

I sometimes come across the statement that the free electron Fermi gas model best describes the electronic properties (e.g. electron heat capacity and electron thermal conductivity) of metals that are 'simple'. What does a 'simple' metal mean? And why is the free electron gas model more suitable for describing the electronic properties of these metals?

• The noble metals (Cu, Ag, Au) are considered simple metals, primarily because they are well described by using the free electron model. The free electron model just means that the band structure of the conduction electrons has energy vs momentum being roughly what one would expect of truly free electrons. This makes all the math on electronic properties easy, or, well, 'simple' to calculate. See Ashcroft and Mermin, Solid State Physics. – Jon Custer Aug 2 '17 at 13:46
• @JonCuster, compared to the noble metals you mentioned, the electron density of states (DOS) of Al is much more similar to the DOS predicted by the free electron gas (FEG) model. So FEG predicts the electronic properties of Al better than those of the noble metals. Ref. : Z. Lin, L. V. Zhigilei and V. Celli, Phys. Rev. B 77, 075133 (2008) – apadana Aug 2 '17 at 19:53
• very nice paper - many thanks. I've incorporated some info using it below, with a shout out to you. – Jon Custer Aug 2 '17 at 20:48

To expand a bit on my (potentially ephemeral) comment, I'll add a bit more background. One good blanket reference is Ashcroft and Mermin's Solid State Physics book, but lets just focus on what 'simple' means for a metal.

For this purpose, there are two classic papers to consider. The first is J.C. Slater's Reviews of Modern Physics 1934, and the second is J.C. Slater's Physical Review 1934. The first is a long and complete discussion, while the second is focused on calculating the band structure of sodium.

Lets step back a bit first. What would a 'free electron gas' (FEG) behave like? Every electron would have $E = p^{2}/2m$ ($E = \frac 1 2 mv^{2}$), so the energy would be parabolic with electron momentum. Further, no direction in the lattice would be favored more than any other, since it is 'free' of the constraints of atomic positions. Well, in the Phys. Rev. paper, Slater calculates the band structure of sodium following along from Wigner and Seitz's theory to get Bloch functions (essentially).

So, what are the results? Quoting from the Phys. Rev paper:

If we look at the figure in a broader way, however, we see that the lines of constant energy are roughly circles. If the electrons were free, the lines would be exactly circular, the energy depending only on the magnitude of the momentum, not on its direction. From the resemblance of the curves to circles, we see that the free electron picture is not entirely incorrect.

So, a rigorous wave function calculation of the band structure results in a free-electron looking band structure. Further,

The comparison is better shown in Fig. 3, in which we plot energy as a function of the magnitude of k, for the 110 direction, or the 45 degree direction of Fig. 2. The free electron distribution would correspond to a parabolic curve, the kinetic energy being proportional to the square of the momentum. As we see, the actual curve agrees rather closely with the free electron parabola, which is drawn with the correct constants. In fact, for the lower half of the bottom zone, which alone is filled with electrons in the normal state of the metal, the agreement is practically perfect. This is an unexpected and significant result of the present calculations. It has been expected that the true curve would be represented by a parabola in its lower part, but it was generally supposed that the curvature of the parabola would be less than for free electrons. As a rnatter of fact, if we draw similar curves for a distance of separation decidedly greater than the normal distance, say twice as great, we do find a decidedly smaller curvature, the gaps becoming much larger in proportion, and the curves for the occupied regions much flatter, so that these regions are narrower than for the case of the free electron distribution, as pointed out by signer and Seitz. But at, or even in the general neighborhood of, the equilibrium distance, the free electron energy is a good approximation, except in the immediate neighborhood of the gaps.

In other words, there are some slight issues near the Brillouin zone boundaries, but it looks awfully free-electron-like.

Turning to Ashcroft and Mermin, they discuss the band structures of metals in Chapter 15. A brief quote:

...their Fermi surfaces are closely related to the free electron sphere; however, in the <111> directions contact is actually made with the zone faces...

Except for those contacts, the rest of the surfaces for Cu, Ag, and Au are shown to be nearly circular (free-electron like) in Figure 15.5 in A&M. As for aluminum (more below), they say:

The Fermi surface of aluminum is very close to the free electron surface for a face centered cubic monatomic Bravais lattice with three conduction electrons per atom... Once can verify ... that the free electron Fermi surface is entirely contained in the second, third, and fourth zones.

(Note that the three electrons per atom does lead to other weirdness, including a positive high-field Hall coefficient, but that is a discussion for another day.)

So, in some metals (alkali metals and the noble metals, all having single electrons to consider for conduction), the behavior of the electrons in the crystal is similar to that of 'free' electrons. For other metals, this stops being the case, the $E$ vs $p$ band structure gets weirder, you get decidedly non-spherical Fermi surfaces (and even disconnected Fermi pockets), and life becomes harder. Or at least not very 'simple'.

As an added bonus, a comment from @Arham points to a (slightly!) more recent paper, Zhibin Lin et al., 2008 which uses DFT modeling to look at the electronic heat capacity which is, of course, also closely related to the 'free-electron'-ness and the Fermi surface. Now, the paper's major focus is on far-from-equilibrium processes, so when the show how well Al matches a FEG while the noble metals perhaps not so much, that is in regions far from the Fermi surface at room temperatures. Still, the fact that Al is essentially a FEG 5eV or more out from the Fermi surface is quite remarkable.

(As a side note, the figures in Lin et al. on the density of states are very nice in that they show the positions of the lower-lying d-levels for Cu, Ag, and Au, showing how they are responsible for the colors of Cu and Au, and, if we could see just a bit further into the UV, Ag as well. Those who argue that the color of Au has something to do with relativistic electrons should spend time looking over these figures.)

In other comments, the question was further expanded to ask why a metal might not be well described by a free electron gas. Ultimately, this comes down to the band structure ($E$ vs $p$ in all directions), which comes out of the atomic electronic configuration and the crystal structure of the solid. One good example (used by me in an answer over on Chemistry SE) would be iron. It crystallizes in a bcc crystal structure and has d-electrons and an atomic magnetic moment. The band structure is discussed in J. Callaway and C.S. Wang, Physical Review B 1977, and is fairly ugly - multiple isolated Fermi surfaces exist, and they exist separately for spin-up vs spin-down. One cannot begin to describe the Fermi surface as free-electron like - it just does not work. Most bcc crystalline metals have similarly ugly Fermi surfaces, whether they are magnetic or not.

• Thank you for this extended answer. So we can say that FEG explains the behavior of electrons in some metals better than in other metals. But why? What is common among those metals that are described better by FEG? Probably there is something common to the electronic structure of the constituent atoms of these metals. – apadana Aug 3 '17 at 11:20