In almost every book about solid state theory, electrons in a periodic potential are introduced and calculations for the dispersion relation $E(k)$ are presented. One obtains the usual pictures in reciprocal space, where the parabolas for the free electron gas are sketched and the impact of a weak periodic potential is displayed by band gaps at the edges of the first Brillouin zone.

Now one also often finds pictures like this one: Fermi surface and Brillouin zones

One can see the Fermi surfaces for the cases where no potential is applied (left) and where a weak periodic potential is added. The fermi surface, which was a sphere before, now bends a little bit at the edges of the BZ. In the two sketches on the right side of the image, the first and second BZ are drawn. However, they are labeled with first and second band.

I also heard someone saying that the Brillouin zones equal the energy bands. As I understood it, the BZs are more a geometric construction and I cannot see how they should equal bands.

So the question is: What is the exact relation between energy bands and the Brillouin zones?


1 Answer 1


Brillioun zone (BZ) generally refers to a domain in the space of $k$-vectors, and in that sense are geometric. Band generally refers to energy levels. The concepts are closely related, but they are different, so the glib statement that a BZ is a band is not quite correct, but most people understand what is meant. Every point in a BZ maps to one or more bands.

A dispersion relation (or "energy curve") maps points in the domain of $k$ space to the range of energies.

In the diagram you present, the shaded areas indicate regions of $k$ space that map to energies that are occupied. The unshaded regions map to energies that are unoccupied. Two bands are shown in the example in two different representation schemes: the two on the left are in the extended zone scheme, and the two on the right are in the reduced zone scheme. Bands are discussed in the context of the reduced zone scheme. In that scheme, the set of all energies in the range mapped to by $k$ vectors in the domain (the BZ, the square) constitutes a band. Note that each point in $k$ space maps to two energies, that is, to states in two bands.

  • $\begingroup$ So it is wrong to say that $n$th Brillouin zone = $n$th energy band? And can you elaborate what you mean with "closely related"? $\endgroup$
    – Merlin1896
    Jan 13, 2016 at 12:00
  • $\begingroup$ It's not quite correct to say that. The Brillouin zones are the domains in $k$ space. The bands are the ranges. $\endgroup$
    – garyp
    Dec 3, 2016 at 3:48
  • $\begingroup$ Answer given by "garyp" helped me clear up my confusion. I also realised that going back to 1D case makes everything much easier to understand. In the 1D case, bands are drawn as lines in an $E(k)$ plot; where as in the 2D case, bans are drawn as colored patches in the $xy$-plane. $\endgroup$ Nov 10, 2022 at 10:59

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