In the last chapter we saw how it is possible to obtain a quantum Hall state by coupling one-dimensional systems. At the end, our recipe was to first obtain a Dirac cone, add a mass term to it and finally to make this mass change sign. Following this recipe we were able to obtain chiral edge states without applying an external magnetic field.

There is a real (and a very important) two-dimensional system which has Dirac cones: graphene. So in this chapter we will take graphene and make it into a topological system with chiral edge states.

Graphene is a single layer of carbon atoms arranged in a honeycomb lattice. It is a triangular lattice with two atoms per unit cell, type \(A\) and type \(B\), represented by red and blue sites in the figure:

Hence, the wave function in a unit cell can be written as a vector \((\Psi_A, \Psi_B)^T\) of amplitudes on the two sites \(A\) and \(B\). Taking a simple tight-binding model where electrons can hop between neighboring sites with hopping strength \(t\), one obtains the Bloch Hamiltonian:

\[ H_0(\mathbf{k})= \begin{pmatrix} 0 & h(\mathbf{k}) \\ h^\dagger(\mathbf{k}) & 0 \end{pmatrix}\,, \]

with \(\mathbf{k}=(k_x, k_y)\) and

\[h(\mathbf{k}) = t_1\,\sum_i\,\exp\,\left(i\,\mathbf{k}\cdot\mathbf{a}_i\right)\,.\]

Here \(\mathbf{a}_i\) are the three vectors in the figure, connecting nearest neighbors of the lattice [we set the lattice spacing to one, so that for instance \(\mathbf{a}_1=(1,0)\)]. Introducing a set of Pauli matrices \(\sigma\) which act on the sublattice degree of freedom, we can write the Hamiltonian in a compact form as

\[H_0(\mathbf{k}) = t_1\,\sum_i\,\left[\sigma_x\,\cos(\mathbf{k}\cdot\mathbf{a}_i)-\sigma_y \,\sin(\mathbf{k}\cdot\mathbf{a}_i)\right]\,.\]

The energy spectrum \(E(\mathbf{k}) = \pm \,\left|h(\mathbf{k})\right|\) gives rise to the famous band structure of graphene, with the two bands touching at the six corners of the Brillouin zone:

Only two of these six Dirac cones are really distinct, the ones at \(\mathbf{K}=(2\pi/3, 2\pi/3\sqrt{3})\) and \(\mathbf{K}'=(2\pi/3, -2\pi/3\sqrt{3})\). All the others can be obtained by adding some reciprocal lattice vector to \(\mathbf{K}\) and \(\mathbf{K}'\).