I came across the following problem:
Suppose that a certain material consists of $N$ atoms that are ordered in a 2D lattice with a lattice constant $a$, and that each atom donates two conduction electrons at $s$-level. Determine whether the material is a conductor or an insulator, using NFE and the tight-binding models.
And the following solution which I don't fully understand:
Using the tight-binding model: there is one energy band of the form
$$ E(k)=E_0 - \beta - 2 \gamma \left ( \cos(k_x a) + \cos(k_y a) \right ) $$
The first Brillouin zone contains $N$ momentum states. Accounting for two possible spin orientations we deduce that the band has $2N$ electron states. But since there are $N$ atoms and each atom donates exactly two electrons we have $2N$ electrons which means that the band would be completely filled. Therefore we get an insulator. (Notice that the exact shape of the band doesn't matter in this case).
Using the nearly free electron model: the starting point is the assumption that the electrons are free and therefore the occupancy of energy levels is determined by the Fermi surface. In order to calculate the radius of the Fermi surface we note that there are $2N$ electrons in the crystal. Therefore:
$$ 2N = \underbrace{2}_{\mathrm{spin}}\frac{\pi k_F^2}{\frac{(2\pi)^2}{A}} $$
Thus
$$ k_F = \sqrt{4\pi n} = \sqrt{\frac{4\pi}{v_p}} = \sqrt{\frac{4\pi}{a^2}} = \frac{2}{\sqrt{\pi}} \frac{\pi}{a} > \frac{\pi}{a} $$
So the Fermi surface is slightly bigger than the Brillouin zone (with small deformations at the edges). In other words, BZ1 will be almost completely filled and BZ2 will be almost completely empty --> Two partially-filled bands --> the material is a conductor.
A couple of questions:
Why can't we use the concept of Fermi surface in the tight-binding approximation? How the assumption that the electrons are (almost) free (in the NFE model) implies that the occupancy is determined by the Fermi surface? Isn't it always determined by the Fermi surface, regardless of which model we use?
If I recall correctly the number of $k$-states in a single Brillouin zone is equal to the number of unit cells in the entire system ($N$ in this case), i.e. it has nothing to do with the model we use to describe electrons. So why this fact is only used in the tight-binding model but not in the NFE model?
Shouldn't the two models agree on whether the material is a conductor or an insulator? And if not, which one gives the correct result?