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?


1 Answer 1


The idea behind the 4 by 4 matrix is that you want an effective way to describe your system close to an energy level (because of the Fermi level). Let's say for the graphene you want an effective model for the energy E=0. In the tight-binding model of the graphene you will find that there is 2 distinct points where E=0 in the reciprocal space. So your effective Hamiltonian near the energy E=0 need to 'concatenate' the two Hamiltonian at theses points and that's why you have a 4 by 4 and not a 2 by 2

  • $\begingroup$ 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? $\endgroup$
    – MeMeansMe
    Commented Aug 20, 2018 at 9:14
  • 1
    $\begingroup$ Let's take the problem the other way: Force the periodicity of the Hamiltonian to be 2 time larger. You will get 4 band that matches perfectly two by two. By doing so, the two dirac point will be, in the k space, sitting at the same k. And then you can derive the effective Hamiltonian near E=0 $\endgroup$
    – sailx
    Commented Aug 24, 2018 at 14:43

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