Notion of anisotropic Fermi surface

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?

• @thermomagneticcondensedboson - Do you limit answers to the question to free electron systems (which no actual crystal truly is), or do you mean for a general crystal (which is not a free electron gas)? Commented Jun 28, 2019 at 17:03
• @JonCuster I do not limit the answers. Feel free to give insights. I do not realize how the answer would change if we're not dealing with a free electron gas. Commented Jun 29, 2019 at 18:32
• @thermomagneticcondensedboson - there are many metals with non-contiguous Fermi surfaces, which in no way resemble an ellipsis. I’ll work on an answer if I have time. Commented Jun 29, 2019 at 19:19
• @JonCuster Ah, I know this! But the question is mainly whether the shape of the material has an effect on the shape of the Fermi surface. In that case, it doesn't matter whether the metal behaves like a free electron gas or not, the answer is always the same, namely that the FS doesn't depend on the shape of the material. Does this sound correct? Commented Jun 29, 2019 at 20:13

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.

• That's the consequence of the boundary conditions, but, as I said, it has nothing to do with the Fermi surface. If you have a macroscopic piece of metal, the Fermi surface does not change when you bend it or cut it in two Commented Jun 23, 2019 at 0:36
• I wouldn't say it's wrong. It's just linking wrong things. For a Fermi surface to be anisotropic, the momentum prefactor in the energy expression has to be direction-dependent. I expanded the answer a bit. Does it make more sense? Commented Jun 23, 2019 at 15:11
• That's the point, the $n_x$, $n_y$, and $n_z$ are not equal. Instead, $k_x$, $k_y$, and $k_z$ are. For the Fermi surface, we obtain the $k_x$, $k_y$ and $k_z$ first and from that we get the corresponding harmonic number in different directions. Because the lengths are generally not equal, the same $k$ gives different $n$ in different directions:) Commented Jun 24, 2019 at 6:29
• I see! But is there any intuitive reason why the $k_i$ are equal and the thus the ratios $n_i/L_i$ rather than the $n_i$'s? Let me think. If I make a side twice as large as another one, say $L_x = 2L_y$, then we must have $n_x = 2n_y$. Does this mean that the state density in $k$-space, is isotropic, i.e. the same for the x and y directions in that case? (assume the effective masses are the same in all directions). Commented Jun 24, 2019 at 16:32
• Thank you for the kind words! Commented Jun 25, 2019 at 16:16

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.

• While I appreciate very much these examples of real Fermi surfaces, the answer does not treat the part of the question that asks whether the shape of the material has an impact on the Fermi surface shape. The other answer showed that the material shape has no impact on the FS at least when the FS is spherical. Commented Jul 1, 2019 at 16:00
• @thermomagneticcondensedboson - thanks for the reminder. Now, I will assume you mean the shape of the unit cell (assuming Wigner-Seitz), and the obvious answer is for beryllium - that 6-fold crown is a direct result of the hexagonal crystal lattice. Commented Jul 1, 2019 at 16:05
• I do not mean the shape of the unit cell. I mean what I wrote. One can have a Fermi surface without any crystal structure by the way. Commented Jul 2, 2019 at 6:30
• @thermomagneticcondensedboson - of course - regardless of whether the material is crystalline, polycrystalline, or amorphous there is always a last-filled electron level. So, I would suggest that the one look at non-continuous materials shapes, starting with a donut. Commented Jul 2, 2019 at 13:02