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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?

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    $\begingroup$ The reason we can in many cases neglect interactions in solid state systems is a rather long story eventually resolved by Fermi liquid theory $\endgroup$ Oct 24, 2023 at 10:45

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