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?