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:

$$ 2N = \underbrace{2}_{\mathrm{spin}}\frac{\pi k_F^2}{\frac{(2\pi)^2}{A}} $$


$$ k_F = \sqrt{4\pi n} = \sqrt{\frac{4\pi}{v_p}} = \sqrt{\frac{4\pi}{a^2}} = \frac{2}{\sqrt{\pi}} \frac{\pi}{a} > \frac{\pi}{a} $$

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:

  1. 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?

  2. 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?

  3. 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?

  • 1
    $\begingroup$ A divalent material always has enough electrons to fill one band. But its possible the bands overlap such that the lowest energy filling of 2 bands is two have two partially occupied bands, not a single full one. Otoh I don't really follow what is going on in that NFE argument right now. $\endgroup$
    – jacob1729
    Commented Mar 22, 2021 at 10:49
  • $\begingroup$ @jacob1729 - In this particular problem there is a single orbital per atom, which implies a single energy band (in the TB model at least), if I'm not mistaken. $\endgroup$
    – grjj3
    Commented Mar 22, 2021 at 10:56

1 Answer 1


Why not use the Fermi surface in the tight-binding model?

By convention, the Fermi surface separates filled from empty states, where these are infinitesimally close – that is, where the boundary falls in the middle of a band. If the boundary (the Fermi level) instead falls in a band gap, we say that a material has no Fermi surface. Thus one definition of a metallic conductor is “a solid with a Fermi surface”.

In the NFE model, there are only very small band gaps, which are treated as perturbations to the free electron gas, and there will always be a Fermi surface. On the other hand, in the tight-binding approximation, there are large gaps between bands, and if (as in this case) we have a filled band, there will be no Fermi surface.

Why not use the number of $k$-states in a Brillouin zone in the NFE model?

You are correct that the number of $k$-states is the same in both cases, but it's only important to determine the conductivity in the TB case.

In both, the occupied region in reciprocal space has the same area as the first Brillouin zone, because each atom contributes two electrons.

In the NFE case, there is (almost) no band gap and therefore (almost) no penalty for leaving the first Brillouin zone. Thus the Fermi surface has (almost) its natural shape of a circle (and we would predict metallic behaviour no matter how many electrons each atom contributed).

In the TB case, there is a large band gap and therefore a large penalty for leaving the first BZ. Thus the occupied region has the same square shape as the first BZ and we have an insulator in this case. (But if each atom only contributed one electron, we'd have a half-full band and a metal; that's why we need to consider the number of $k$-states in this case alone.)

Why don’t they give the same answer?

Ultimately these two models are approaching the same place from opposite directions: the NFE model says that solids are basically metals, with some small perturbations that could make them insulators, and the TB model says that solids are basically insulators, with some small perturbations that could make them metals. You could think of them like the Taylor series for the same function expanded about different points. They’re both approximations to the same thing, but that doesn’t mean they have to agree with one another – each is better than the other in different regions.

In order to decide which model is more appropriate for a real system, we’d need to get more quantitative. In the NFE model, this would mean specifying the weak periodic potential (or, more specifically, its Fourier components). In the TB model, it would mean introducing more orbitals and being precise about their overlap.

Given quantitative values for each model, it is indeed possible for them to predict the same bands and therefore the same behaviour. But with as little information as you were given in this problem, we can’t go any further than the worked solution that you gave – including the fact that the models give different predictions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.