No Solid State Triangular Lattice Hubbard Model?

This is a record of what Ashvin taught me during cookie time one day: why is it so hard to get triangular lattice models in solid state systems.

Assume that there is already a two-dimensional atomic triangular lattice, particularly with \(C_3\) symmetry. Note that Hubbard models usually arise out of \(d-\)wave orbitals: \(s\)-orbitals are too close to the nucleus that the hopping \(t\) is too small (not sure what happens with \(p\) orbitals). To get a clean Hubbard model one can only have one orbital per atom. This means that we should study the singlet representations of \(C_3\) consisting of \(d\)-orbitals. Now:

• \(d_{xz}\) and \(d_{yz}\) form a two-dimensional irrep of \(C_3\).

• \(d_{x^2-y^2}\) and \(d_{xy}\) form a two-dimensional irrep of \(C_3\).

• \(d_{2z^2-x^2-y^2}\) is a singlet under \(C_3\); however, its orbital shape is such that it does not have a large in-plane hopping \(t\). This means that we also cannot construct a Hubbard model with these orbitals.

Hence we are not able to construct triangular lattice Hubbard models atomically. \(C_4\) evades this problem: \(d_{x^2-y^2}\) and \(d_{xy}\) are singlets under \(C_4\).