Is there any one lattice tight banding model (2D) hosts Dirac points as graphene does?

graphene has a two sublattice structure and it hosts two Dirac points. I wonder if there is some two dimensional lattice geometry which is simpler than honeycomb lattice, but also host Dirac points.

• Since you want two bands, you won't be able to avoid a two-site unit cell. – Norbert Schuch Dec 5 '15 at 12:38
• @NorbertSchuch You are absolutely correct. – an offer can't refuse Dec 6 '15 at 1:52
• Although staggered hopping in a square lattice is able to produce Dirac physics, ordinary honeycomb lattice on its own does. Interesting. – xiaohuamao Apr 10 '16 at 11:56

Yes, you can realize 2D Dirac fermions on square lattice, known as the stagger fermion or the $\pi$-flux state. The Hamiltonian is given by $$H=-\sum_{\mathbf{r}}(c_{\mathbf{r}+\hat{\mathbf{x}}}^\dagger c_{\mathbf{r}}+(-1)^x c_{\mathbf{r}+\hat{\mathbf{y}}}^\dagger c_{\mathbf{r}}+h.c.),$$ where $\mathbf{r}=(x,y)$ (with $x,y\in\mathbb{Z}$) label the sites on a 2D square lattice, $\hat{\mathbf{x}}=(1,0)$ and $\hat{\mathbf{y}}=(0,1)$ are unit vectors of the lattice. Due to the stagger sign of the $y$-direction hopping, the unit cell also contains two sites, which gives rise to two bands of the Dirac fermion.
• @buzhidao Eq (1) in the paper is exactly the same Hamiltonian as what I have written here. Just set $t_x=t_y=1$ and rename the fermions on A/B sublattice as $a$/$b$ fermion. Or as illustrated in Fig. 1(a) in the paper, it is the $\pi$-flux state with $\pi$ flux through every square. Note that $(-1)^x$ in my Hamiltonian is $1$ if $x$ is even ($a$ fermion), and is $e^{-i\pi}$ if $x$ is odd ($b$ fermion). – Everett You Dec 6 '15 at 5:23