Graphene has a honeycomb lattice which can be described as a triangular lattice but with two atoms per unit cell. Therefore, when solving for the band structure of the graphene, we expand the wavefunction like this: (full description in the lecture notes)


where $\psi_k^A$ and $\psi_k^B$ are Bloch wavefunctions. Expanding $\psi_k^A$ and $\psi_k^B$ based on one of the atomic orbitals (say $p_z$ orbitals, which lead to Dirac cones), we finally can write the Schrodinger equation using a $2\times 2$ Hamiltonian (because of the form of the wavefunction and the two $a$ and $b$ coefficients in $(1)$, which is caused by the existence of two atoms per unit cell). Finding the eigenvalues of this $2\times 2$ Hamiltonian, we arrive at the two $\pi$ energy bands:

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We can even write the above wavefunction in the form of a spinor (see the lecture notes, equation $(2.29)$: $$\Psi_k=\begin{pmatrix} a_k \\ b_k\end{pmatrix}$$ and therefore define a pseudospin.

While the math seems pretty clear, I don't know how to think of the two resulting bands, since they're obtained for two electrons instead of one.

Are the two resulting bands really two different bands, just like two different bands of a simple one-atom-per-lattice crystal like silicon? If not, what are the differences?
For example, when two different bands touch at some $k$ for silicon, we have a 2-fold degeneracy. Is it the same here at a Dirac cone where the two $\pi$ bands touch?

  • $\begingroup$ I think the two $\pi$ bands can be thought of as descendants of bonding and antobonding $\pi$ orbitals (see, e.g., isite.lps.org/sputnam/LHS_IB/IBChemistry/UNIT4ChemBonding/…). Or, in other words, you can imagine that the elementary cell with two carbon atoms and two electrons is one composite "atom" with two composite "orbitals", which are the symmetric and antisymmetric superpositions of $\pi$ orbitals on two carbon atoms (and have different energies!). Bringing these composite "atoms" together results in formation of two energy bands from these composite "orbitals". $\endgroup$ Commented Nov 4, 2017 at 1:11
  • $\begingroup$ @AlexeySokolik Seems like a plausible clue, as the notation for the bands is $\pi$ and $\pi^*$ too. Thanks. $\endgroup$
    – Luttinger
    Commented Nov 4, 2017 at 11:16

1 Answer 1


The two bands don't mean two electrons. In fact, we are solving the band structure for a single electron. The single electron states are discrete points on surface of your plot with spacing $\frac{2\pi}{L}$ where $L$ is your system size. The key assumption we made here is:

The interaction between electrons is weak and can be ignored.

Therefore, we are able to fill electrons to these states from lower energy to higher ones.

To find the Fermi level $E_F$, we can count the number of states (Below Fermi level, all single electron states are filled). Assume your graphene has $N$ cells and each cell contains 2 atoms and contributes 2 electrons. Totally, there are $2N$ electrons. We know number of k points in the $1$st Brillouin zone is equal to $N$ (cf. solid state physics textbook such as Kittel). Therefore, the number of states in the lower band is $2N$ (two comes from two spins), which happens to be the number of electrons.

The many-body picture of the graphene is your plot with all states in the lower band filled ( Fermi level $E_F=0$).

We said that graphene has two bands touching and form Dirac cones. There is an ambiguity here since two bands usually mean a nonzero gap between two energy spectrum. In graphene, at the two Dirac points the gap is closed. Because they only touch at two points and don't really overlap, we still call them two bands. This is just a terminology issue and doesn't affect its physics.

For the case of silicon you mentioned, different bands may cross other and have 2-fold degeneracy. This kind of degeneracy can be removed by small perturbation. However, the Dirac cones in graphene is robust in the sense that no small perturbation can open the gap. This is related to graphene's $C_3$ rotation symmetry and inversion symmetry. The detailed argument can be found in Bernevig's topological insulators and topological superconductors Chapter 7.

  • $\begingroup$ Thanks for the answer. Another confusion: What is the degeneracy of the energy $E=0$ in this case for graphene? Is it two-fold degenerate (as stated in my notes), or four-fold, as we have two distinct cones each of them being two-fold degenerate at $E=0$ (because $\pi$ and $\pi^*$ touch at $E=0$)? $\endgroup$
    – Luttinger
    Commented Nov 5, 2017 at 12:48
  • $\begingroup$ At one K point, there is 4-fold degeneracy due to pseudospins and spins. One may write down low energy effective Hamiltonian for grahpene-like system. It's 8x8 matrix labelled by spin, points $K$, $K^\prime$, and sublattice $A$,$B$. In this case, they have 8 zero energy states (if we don't break the symmetry of grahpene) since it includes two $K$ points. $\endgroup$
    – Yu-An Chen
    Commented Nov 5, 2017 at 18:59

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