Why do electrons in graphene behave as Dirac fermions near the Dirac points?

Question

I've been learning about graphene, and I recently calculated the band structure for it using a nearest-neighbor tight-binding model for the $\pi$ electrons:

$$\varepsilon(\vec k)=\pm t\sqrt{3+2 \cos \left(\frac{\sqrt{3} k_x}{2}-\frac{3 k_y}{2}\right)+2 \cos \left(\frac{\sqrt{3} k_x}{2}+\frac{3 k_y}{2}\right)+2 \cos \left(\sqrt{3} k_x\right)}$$

I was told that the dispersion of graphene around the Dirac points (points in $k$-space where $\varepsilon(\vec k)=0$ -- these turn out to be the vertices of the Brillouin zone) is linear, and this linearity leads to the particle behaving like a Dirac fermion. I have partially confirmed the linearity, but I don't know where to start to verify the second part.

Wikipedia mentions the following, too:

It was realized as early as 1947 by P. R. Wallace that the E–k relation is linear for low energies near the six corners of the two-dimensional hexagonal Brillouin zone, leading to zero effective mass for electrons and holes. Due to this linear (or “conical") dispersion relation at low energies, electrons and holes near these six points, two of which are inequivalent, behave like relativistic particles described by the Dirac equation for spin-1/2 particles. Hence, the electrons and holes are called Dirac fermions also called graphinos, and the six corners of the Brillouin zone are called the Dirac points.

Could someone explain how a locally conical $\epsilon(\vec k)$ graph leads to a Dirac fermion? I have not yet worked with the Dirac equation myself.

Answer 1

In calculating the electron dispersion you probably obtained the diagonalized Hamiltonian in the momentum space

$$ H=\sum_\mathbf{k}\left[c^{\dagger}_{\mathbf{k}A},c^{\dagger}_{\mathbf{k}B}\right]\left[\begin{array}{cc}0 & \Delta(\mathbf{k})\\ \Delta^{\dagger}(\mathbf{k}) &0\end{array}\right]\left[\begin{array}{c}c_{\mathbf{k}A} \\ c_{\mathbf{k}B}\end{array}\right]. $$

If you you chose your $x$ axis along the zigzag direction (arXiv:1004.3396), the two nonequivalent Dirac valleys are $\mathbf{K}_\kappa=\left(\kappa\frac{4\pi}{3\sqrt{3}a},0\right)$, $\kappa=\pm1$ and $\mathbf{K}_{-1}=\mathbf{K}^{\prime}$, where $a$ is the C-C distance. Then $\Delta(\mathbf{k})=-t\left(1+e^{-i\mathbf{k}\cdot\mathbf{a}_1}+e^{-i\mathbf{k}\cdot\mathbf{a}_2}\right)$, where $t$ is the hopping term, and $\mathbf{a}_1=\left(\sqrt{3}a/2,3a/2\right)$ and $\mathbf{a}_2=\left(-\sqrt{3}a/2,3a/2\right)$ are the lattice vectors.

Taylor expanding $\Delta(\mathbf{k})$ up to linear terms around those two points you obtain

$$ \Delta(\mathbf{k})=\kappa\frac{3ta}{2}q_x-i\frac{3ta}{2}q_y $$

where $\mathbf{q}$ is the displacement momenta from the $\mathbf{K}_\kappa$ point. Promoting these displacement momenta to operators you obtain the Hamiltonian

$$ H=\hbar v_F\left[\begin{array}{cccc}0 & q_x-iq_y & 0 & 0\\q_x+iq_y & 0 & 0 & 0\\0 & 0 & 0 & -q_x-iq_y\\0 & 0 & -q_x+iq_y & 0\end{array}\right] $$

where $v_F=\frac{3ta}{2\hbar}$ is the Fermi velocity. This is in $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{A\mathbf{K}^{\prime}},\Psi_{B\mathbf{K}^{\prime}}\right]^T$ basis, if you rearrange your basis as $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{B\mathbf{K}^{\prime}},\Psi_{A\mathbf{K}^{\prime}}\right]^T$ you get the compact form

$$ H=\hbar v_F\tau_z\otimes\boldsymbol{\sigma}\cdot\mathbf{k} $$

where $\tau_z$ acts in the valley space. This is similar to the Dirac-Weyl equation for relativistic massless particles, where instead of $v_F$ you get the speed of light

$$ H=\pm\hbar c\boldsymbol{\sigma}\cdot\mathbf{k} $$

where $+$ denotes right-handed antineutrions, and $-$ denotes left-handed neutrions. The differences are that $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y\right)$ for graphene acts in pseudospin space and $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y,\sigma_z\right)$ for neutrinos acts in real spin space.