Notion of anisotropic Fermi surface

Question

I would like to check something.

I know that the Fermi energy is the maximum energy occupied by a Fermion at $T=0$ (if I have $N$ fermion it will be the energy of the Fermion that has the highest single particle energy).

I would like to check something about the anisotropy of the Fermi surface.

Does the anisotropy of the fermi surface occurs only because of the geometry of the material?

Imagine I have a free electron gas, I know that the wavevectors allowed are of the form:

$$ k=2 \pi (\frac{n_x}{L_x},\frac{n_y}{L_y},\frac{n_z}{L_z}).$$

Thus, if $L_x \neq L_y \neq L_z$, the value of $k_F$ can be reached for vectors that do not lie on a sphere.

Thus, in a general case I would have $$k_F^2=4\pi^2(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2})$$

Which is the equation of an ellipsoïd.

In conclusion : the fermi surface is anisotropic only because of the geometrical structure of the crystal and it is always an ellipsoïd.

Am I right?

Also, can we define a fermi surface for an interacting system? Indeed, to define it we need to talk about particle wavevectors. And it is a good quantum number for free electrons. So how is it defined for an interacting system?

Answer 1

Like you said, Fermi energy corresponds to the highest energy occupied by an electron at $ T = 0 $. For the free electron gas, $E = 1/(2m)\times\left(k_x^2+k_y^2+k_z^2\right)$. If you plot the momenta on the Cartesian axes, the constant energy will be a spherical shell.

Here's a more intuitive way of seeing this. Take a 3D box with $L_x\neq L_y\neq L_z$. The Schrodinger equation that you need to solve is

$$ -\frac{\hbar^2\nabla^2}{2m}\Psi\left(x,y,z\right) = E\Psi\left(x,y,z\right) $$

for $x\in \left[0,L_x\right]$, $y\in \left[0,L_y\right]$, $z\in \left[0,L_z\right]$. The differential equation is separable so we get

$$ -\frac{\hbar^2}{2m}\Psi''_x\left(x\right) = E\Psi_x\left(x\right)\,, \\ -\Psi''_x\left(x\right) = \frac{2mE}{\hbar^2}\Psi_x\left(x\right)\,, \\ -\Psi''_x\left(x\right) = k^2\Psi_x\left(x\right)\,. $$

Note that $k$ does not depend on the direction, only on energy and the mass. In $x$-direction, we have the unnormalized solution

$$ \Psi_x\left(x\right) \propto \sin kx\,. $$

Because the wave vanishes at the boundaries, $k = n_x \pi / L_x$. Similarly, for $y$ and $z$ directions, we have $k = n_y \pi / L_y$ and $k = n_z \pi / L_z$.

If we set $E \rightarrow E_F$, $k\rightarrow k_F$ and

$$ n_x = \frac{k_F L_x}{\pi}\,, $$ and similarly for $y$ and $z$. You see that the $n$'s are not the same in the three directions. $k_F$, on the other hand is the same so that your Fermi surface is a sphere.

Now, imagine if the mass were anisotropic. Then, the energy becomes $$ E = \frac{k_x^2}{2m_x} + \frac{k_y^2}{2m_y} + \frac{k_z^2}{2m_z}\,.$$ In this case, the Fermi energy surface is no longer a sphere, but an ellipsoid.

Answer 2

While we would like for Fermi surfaces to be nice ellipsoidal features, even for non-centrosymmetric crystal systems, the fact remains that electrons don't seem to necessarily follow human aesthetics. Any structure to the Fermi surface should, certainly, be related to the symmetries of the crystal and the unit cell (I assume that is what is meant by the 'shape' of the crystal).

Lets start with some of the more free-electron like metals, the fcc metals copper, silver and gold. While much of the Fermi surface is pretty spherical, each one has a Fermi surface that necks out in the $L$ direction, connecting there. From D.J. Roaf, Phil Trans. Roy. Soc. A 255(1052) 135-152 (1962):

OK, lets look at a bcc metal, iron. Iron is magnetic, and in the ferromagnetic state you get separate non-contiguous spin up and spin down pockets of electrons. I'm not quite sure how to explain what the various surfaces look like, so we will use M. Asdente and M. Delitala, Phys. Rev. 163(2) 497-503 (1967):

And, finally, for a hcp metal, Be. This one is truly bizarre, forming what T.L. Loucks and P.H. Cutler, PHys. Rev. 133(3A) A819-A829 call the 'cigar and coronet' Fermi surface. While one could easily consider the 'cigar' as an ellipsoid, the 'coronet' is pretty hard to imagine as anything free-electron like. The 'shape' of the crystal is displayed quite clearly - the six-fold symmetric 'coronet' in the plane, and the 'cigar' in the stacking direction.

So, overall, in many real metals the Fermi surfaces diverge, sometimes wildly, from our naive free-electron like intuition.