With the invariant expansion method, we can use the symmetry to find the low effective Hamiltonian easily. But first we should choose several orbitals which belong to an irreducible representation. My question is what are the orbitals we should use in graphene to get the Dirac equation near K(or K') point finally. And Preferably, gives the subsequent steps.


A good reference on this topic is the article

Invariant expansion for the trigonal band structure of graphene
R. Winkler and U. Zülicke, Phys. Rev. B 82, 245313

which is also available on the arxiv.

To answer your question, the relevant basis functions transform according to the irreducible representation $\Gamma_5$ of the point group $D_{\text{3h}}$ at $K$ (see the part III.A of the mentioned article and references therein if you are interested in why), so the corresponding effective Hamiltonian is constructed from the IR in $$\Gamma_5 \times \Gamma_5^* = \Gamma_1 + \Gamma_2 + \Gamma_6.$$ Basis matrices can be chosen as $\mathbb{1}$ for $\Gamma_1$, $\sigma_z$ for $\Gamma_2$ and $(\sigma_x,\sigma_y)$ for $\Gamma_6$ on the one hand. On the other hand, invariant tensors at first order in $k$ are $(k_x,k_y)$ in $\Gamma_6$ (and $k_z$ in $\Gamma_4$ but this one is not relevant as graphene is planar). Combining invariant tensors with basis matrices transfoming according to the same irreducible representation produces the low-energy effective Hamiltonian (at lowest order in $k$, near the $K$ point where $k=0$)

$$H(k) = \mu \, \mathbb{1} + \hbar v_{\text{F}} \, (k_x \sigma_x + k_y \sigma_y) + \mathcal{O}(k^2)$$

where $\mu$ and $v_{\text{F}}$ are free parameters.

The article of Winkler and Zülicke gives a lot of other terms (e.g. taking into account spin-orbit effects).

  • $\begingroup$ can you futher explain why the basis transform according to the representation /Gamma_5? $\endgroup$ – Vivian Feb 17 '17 at 14:38

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