Hi Hosein,

We have a collection of tools for working with continuum models in Kwant 1.3. If we give it a Dirac equation, can we get the honeycomb lattice?

Discretizer doesn't know anything about honeycomb lattices; it just uses finite differences on a square lattice to construct the tight binding model.

It seems that we can only get a tight-binding model of square lattice, so it will not work for Dirac equation Hamiltonian?

You can of course discretize the Dirac Hamiltonian on a square lattice, although you will run into the Fermion doubling problem at the edge of the Brillouin zone. The discretizer does not know anything about Fermion doubling, so you would need to "get rid" of the unphysical states by other means (e.g. adding extra Hamiltonian terms to "gap out" the unphysical states). Happy Kwanting, Joe