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The Weyl equations in 3+1D are: $$ i(\partial_t\pm\sigma\cdot\nabla)\psi_{L/R}=0 $$ and low energy Hamiltonian in graphene are (2 branches): $$ -iv_F\sigma\cdot\nabla \,\,\,\,\,\,\,\,\, and \,\,\,\,\,\,\,\,\, -iv_F\sigma^*\cdot\nabla $$ which are almost the 2+1D version of the Weyl fermion excitation.

Can we say (some of) the low energy excitation in graphene are Weyl fermions? Show me the reason, thank you!

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  • $\begingroup$ The difference is in the detail. Weyl equation is not the same as massless Dirac equation. Graphene follows the latter, so no Weyl. $\endgroup$ – Mauricio Nov 23 '17 at 23:23
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The answer is no!

Indeed we have emergent massless Dirac fermion $\partial\!\!\!/ \psi=0$ in (2+1)D graphene, however there is no proper Weyl spinor (chiral basis) decomposition, while in (3+1)D spacetime it is sufficient to do so with a Lorentz invariant $\gamma^5$ matrix: $\psi_L=\frac{1-\gamma^5}{2}\psi$,$\psi_R=\frac{1+\gamma^5}{2}\psi$.

Generally speaking, Weyl representation is a diagonal representation of $\gamma^{D+1}$, and only in even dimensional spacetime can one define a $\gamma^{D+1}$ well in corresponding Clifford algebra. Hence they are essentially different with respect to the representation of group.

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