I know that tight-binding graphene with $\text{p}_\text{z}$-, $\text{d}_\text{xz}$- and $\text{d}_\text{yz}$-orbitals and spin-orbit coupling is a $\mathbb{Z}_2$-topological insulator. But I want to categorise it according to the periodic table of topological invariants. My problem now is that I don't know which of the ten Altland-Zirnbauer classes is the correct one. Based on my logic, it has to be AII:

  • Graphene satisfies time-reversal symmetry, and $T^2=-1$. So it has to be one of the symmetry classes AII, DIII and CII.
  • Graphene is a two-dimensional $\mathbb{Z}_2$-topological insulator. So, it must be either AII or DIII.
  • If I'm not mistaken, the d-orbitals break particle-hole symmetry as the band structure is no longer symmetric around the Fermi energy. So it has to be AII.

I've also computed $\sigma_z H(-\vec{k})\sigma_z \neq -H(\vec{k})$ (with $\sigma_z$ acting on the sublattices). As far as I know, equality here would imply particle-hole symmetry in graphene. The two terms $\sigma_z H(-\vec{k})\sigma_z$ and $-H(\vec{k})$ differ by twice the d-orbital energy, which I consider reassuring.

My question can now be summarised as: Is graphene, as I described it, in the symmetry class AII? Furthermore, did I make any mistakes with my arguments above?


1 Answer 1


Yes, that's correct. Graphene with spin-orbit coupling falls in the symplectic symmetry class (AII).

Reference: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.98.256801

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    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Miyase
    Jul 4, 2023 at 20:14

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