# $4\times4$ Dirac Hamiltonian in Graphene

When linearizing the Hamiltonian of Graphene in reciprocal space around $\vec{q} = \vec{k}-\vec{K}_\pm = \vec{0}$, where $\vec{K}_\pm$ are two independent Dirac points, one can get two Hamiltonians, one corresponding to the Dirac point $\vec{K}_+$ and the other corresponding to $\vec{K}_-$:

$$\mathcal{H}_{\vec{K}_+} = v_F (\vec{\sigma}\cdot\vec{p}) \qquad \text{and} \qquad \mathcal{H}_{\vec{K}_-}= -v_F(\vec{\sigma}\cdot\vec{p})$$

Here, $v_F$ is the Fermi velocity, $\vec{\sigma}$ is the 2D vector of Pauli matrices and $\vec{p} = \hbar\vec{q}$. One thing I've seen very often so far is that people say that the total $4\times 4$ Hamiltonian

$$\mathcal{H} = \begin{pmatrix} \mathcal{H}_{\vec{K}_-} & 0 \\ 0 & \mathcal{H}_{\vec{K}_+} \end{pmatrix} = v_F \begin{pmatrix} -\vec{\sigma}\cdot\vec{p} & 0 \\ 0 & \vec{\sigma}\cdot\vec{p} \end{pmatrix}$$

corresponds to a massless Dirac Hamiltonian in Weyl representation. My question is why one can merge the Hamiltonians for the two Dirac points into one "big" Hamiltionian although the momenta in each Hamiltonian are different:

$$\hbar(\vec{k}-\vec{K}_+) \neq \hbar(\vec{k}-\vec{K}_-)$$

So, the vector $\vec{p}$ in the two Hamiltonians above is not the same in both cases. How can we justify representing two Hamiltonians with different momenta by one single Hamiltonian?

• I understand that but the momenta $\vec{p}$ in each of the two Hamiltonian are measured relative to the Dirac points, right? So why can I 'concatenate' the two Hamiltonians although the momenta are defined differently? – MeMeansMe Aug 20 '18 at 9:14