Graphene and Hubbard model

Question

I am trying to understand the graphene Hamiltonian. As I see, the Hubbard model can be presented as follows

$$ \mathcal{H} = \sum_{i,j,\sigma} - t_{ij} \hat{c}^\dagger_{j,\sigma}\hat{c}_{i,\sigma} + h.c. + U \sum_i \hat{n}_{i\uparrow}\hat{n}_{i\downarrow}, $$ where we have kinetic and potential terms. I saw the graphene is well approximated only by the first term. I saw on Wikipedia, that $U=0$ means that we consider the Fermi gas. So, do we assume in graphene that we consider Fermi gas or are there any other reasons that we consider only kinetic term?

Answer 1

The Hubbard model is a relatively simple model to describe properties of materials. $U$ is used to model the interaction strength between the electrons. To model different materials, we should use different values of $U$. E.g. for cuprates, a family of materials containing copper, usually an intermediate to high interaction strength $U \approx 8$ is used to describe their behavior like superconductivity, pseudogap or anti-ferromagnetism. For graphene however, the "free" model, i.e. the model without $U$ already works pretty well. Additionally, in contrast to the classic 2D Hubbard model of the square lattice, graphene has 2 atoms per unit cell, so it has 2 bands in its Brillouin zone. It also exhibits Dirac points at the high symmetry points of the BZ, making already the "free", i.e. non-interacting model interesting and rich in physics.