The following figure shows the 1st Brillouin zone of graphene (shaded area).

enter image description here

At the $K$ and $K'$ points (called Dirac points), the upper (conduction) band touches the lower (valence) band, and therefore at the Dirac points, we have a two-fold degeneracy (valence band and the conduction band are degenerate with zero energy). We know that all three $\mathbf K$ points are equivalent, and also all $\mathbf K'$ points are equivalent. Therefore, we only have two distinct Dirac points, each of them being degenerate with $E=0$.

Now, these two Dirac points, which are located at $\mathbf K = -\mathbf K'$ (as a result of TR-symmetry) both have the same energy (zero) and are degenerate too. Therefore, there are 4 states that have zero energy (conduction and valence bands, at K and K' points), and the zero energy state should be 4-fold degenerate.
But in various resources, I've read that in graphene "there is one pair of Dirac points, and the zero-energy states are doubly degenerate; which is also called a twofold valley degeneracy".

What was wrong with my reasoning? Why are zero-point energy states of graphene two-fold degenerate (instead of four-fold)? In particular, aren't all points of the Brillouin zone with the same energy (and of all bands) counted toward the degeneracy of an energy level?

enter image description here


1 Answer 1


You are definitely on the right track, but the accounting here is a little tricky and easy to mistake.

First of all, and probably most important, there is no difference between the valence and conduction bands at the K points. They are the same; they don't count twice. Of course, like any state, two electrons can fill it with opposite spins.

So then, why are there two K points (and not one or six)? Try the trick for counting atoms in a unit cell, when there are atoms at the edges of the cell (are they in or out? do they count or not?): Imagine shifting the cell boundaries slightly in any direction, keeping the atoms fixed. Then, how many atoms are contained wholly within the cell volume?

The trick works in reciprocal space as well. Do the same for graphene's Brillouin zone, count the K points, and you'll find two, next to each other. By symmetry, these are the same as the ones opposite each other. Alternatively, if you think about it, each K point is really only $1/3$ within the Brillouin zone, and $1/3*6=2$.

So for counting the states for the density of states, for example, you'll find a total degeneracy factor of 4: two for the K points, and two for the spins.

  • $\begingroup$ I'm a little confused. Do you mean, neglecting spin, a Dirac point (or any other point of band touching) per se doesn't have any degeneracy? $\endgroup$
    – Luttinger
    Commented Nov 3, 2017 at 10:38
  • $\begingroup$ @Luttinger Yes, for degeneracy there needs to be some difference in the degenerate states other than energy (Pauli exclusion principle forbids multiple electrons in identical states). Both the eigenvalues and eigenvectors of the conduction and valence bands converge at the K points, meaning that there is no way to distinguish them. Thus they are identical and there is only one state. $\endgroup$
    – Gilbert
    Commented Nov 3, 2017 at 13:16
  • $\begingroup$ My impression was that, since according to the Bloch theorem, we are solving a Schrodinger equation for each $\mathbf k$, the solution (dropping the $e^{i\mathbf{k.r}}$) would be some $\lvert u_{n\mathbf k}\rangle$, and when two bands touch, we have an accidental degeneracy because each band has a different $n$ and therefore is a different state. Can you please explain why am I wrong? Thanks in advance. $\endgroup$
    – Luttinger
    Commented Nov 3, 2017 at 14:36
  • $\begingroup$ I explained my confusion in the above comment in another question. $\endgroup$
    – Luttinger
    Commented Nov 3, 2017 at 22:55
  • $\begingroup$ At the risk of reviving an old question: How would this accounting work out for the M points? To me it seems like there are three distinct M points, M, M' and M'' that I can't get around between using reciprocal lattice vectors. So this leads me to expect a 3-fold degeneracy at the M-points. Is this reasoning correct? I reach a similar conclusion when trying the 'shift the Brillouin zone' trick, as I end up with three M points and two K points in the new BZ. $\endgroup$
    – 1MegaMan1
    Commented Nov 3, 2019 at 18:07

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