Carving up the Fermi surface

[fluidistic]

I would like to know every bit of information one can retrieve by looking at the Fermi surface of a material.

Here's what I think is correct thus far:

1) The fact that the material has a Fermi surface already tells us a lot. The material could be a metal or something that resembles a metal, but it's not an insulator (and not a semiconductor either?)

2) The more surface area, the more the charge carrier density and hence the higher the conductivity.

3) The local curvature of the Fermi surface tells us how the (quasi)electrons (or holes) would react when a small perturbation is applied, such as a thermal gradient, an electric field or a magnetic field. In other words, one gets the know the renormalized or effective mass of the charge carriers.

4) If the surface is actually blurred, so it's not a surface anymore, then the system is not exactly at absolute zero in temperature.


What I do not know but would like to know:

A) If we take for example lithium's FS. It is anisotropic, i.e. it isn't perfectly spherical and so the FS has a greater wavevector k in some direction compared to others. What exactly does this tell us regarding the resistivity of a Li cubic monocrystal? Does this mean that the resistivity is not uniform, i.e. it cannot be described by a single value but instead must be described by at least 2 values (so it's a tensor with at least 2 different entries)?

B) What does the absolute value of the wavevector on the FS tell us? I realize that a big wavector would mean a higher speed of charge carriers (the so-called Fermi speed), but what impact does this have regarding the material?

 

[Henryk]

The more surface area, the more the charge carrier density and hence the higher the conductivity.

Not necessarily so, Electrical conductivity depend on the number of carriers and their scattering time.

If we take for example lithium's FS. It is anisotropic,

I'm not quite certain about lithium but the Fermi surface of alkali metals is as close to a sphere as you can get (true at least in the case of potassium).

B) What does the absolute value of the wavevector on the FS tell us

One thing, for a spherical Fermi surface, the volume of the surface is a cubic function of the magnitude of the Fermi vector ( $V_{sphere} = \frac 4 3 \pi r^3$). In other words, the number of free electrons is a cubic function of the Fermi vector magnitude.

 

[fluidistic]

Not necessarily so, Electrical conductivity depend on the number of carriers and their scattering time.

As far as I understand, the electrical conductivity depends on the Fermi surface (I'm talking about metals, not semiconductors and others). That surface tells us about the number of electrons that are involved in electric current when a current is applied. So one can express the electrical conductivity in terms of a surface integral over the Fermi surface only. As Ziman writes in his famous Solid State Physics textbook (page 218):

From the point of view of linear response theory, it would be even better to write $\sigma=\frac{1}{3}J_F^2 \tau N(E_F)$, showing that the conductivity depends only on the properties of the electrons at the Fermi level, not on the total number of electrons in the metal.

I'm not quite certain about lithium but the Fermi surface of alkali metals is as close to a sphere as you can get (true at least in the case of potassium).

Li and Cs Fermi surface aren't that spherical. Li's one has around 5% anisotropy in certain directions. It means the surface is in some directions much closer to the Brillouin Zone boundaries than in other directions. Check out https://www.phys.ufl.edu/fermisurface/ for a visual picture.

One thing, for a spherical Fermi surface, the volume of the surface is a cubic function of the magnitude of the Fermi vector ( $V_{sphere} = \frac 4 3 \pi r^3$). In other words, the number of free electrons is a cubic function of the Fermi vector magnitude.

Sure, but again, what does it tell us about the properties of the material? The number of free electrons seems to be irrelevant regarding to the determination of the electrical conductivity and other quantities. I thought that this would mean a higher Fermi velocity and hence a higher electrical conductivity, but I am now reading that the closer to the BZ edges, the slower the electrons (not sure why!), and thus the lower the electrical conductivity. I am completely confused.

So I am still lightyears away from understanding the most basics things one can get out of these Fermi surfaces (which make up the cover of many solid state textbooks!).

 

[Dr_Nate]

As far as I understand, the electrical conductivity depends on the Fermi surface (I'm talking about metals, not semiconductors and others). That surface tells us about the number of electrons that are involved in electric current when a current is applied. So one can express the electrical conductivity in terms of a surface integral over the Fermi surface only. As Ziman writes in his famous Solid State Physics textbook (page 218):...

Sure, but again, what does it tell us about the properties of the material? The number of free electrons seems to be irrelevant regarding to the determination of the electrical conductivity and other quantities. I thought that this would mean a higher Fermi velocity and hence a higher electrical conductivity, but I am now reading that the closer to the BZ edges, the slower the electrons (not sure why!), and thus the lower the electrical conductivity. I am completely confused.

The conductivity depends on the density of electrons at the Fermi surface. It doesn't depend on what a chemist would call the valence electrons which I think is what you mean by free electrons.

The velocity of the electrons are different at different points on a Fermi surface. You can't know what these velocities are by the FS alone. You need the electronic band structure. The velocity of an electron up to a constant factor is determined by the slope of the band structure.

You should be really be looking at Fermi surfaces and electronic band structures together to get a fuller picture.