# How does the Hubbard hamiltonian change when considering a Peierls distortion (bipartite lattice)?

The following is the Hubbard contribution to the hamiltonian in the Hubbard-Tight Binding model.

$$H_{hubbard}=U \sum_i n_{i \uparrow}n_{i\downarrow}$$

where $$n_{i \sigma}=c_{i\sigma}^\dagger c_{i\sigma}$$

And the tight-binding hopping part:

$$H_{TB}=t \sum_{i\sigma} (c_{i\sigma}^\dagger c_{i+1\sigma}+ c_{i+1\sigma}^\dagger c_{i\sigma})$$

The full hamiltonian is then given by:

$$H=H_{TB}+H_{hubbard}$$

If I want to consider a Peierls distortion (which is considering dimerization, breaking the symmetry and now having two different sites A and B within the unit cell) the Tight Binding part is changed to:

$$H_{Peierls}=t_1 \sum_{i\sigma} (a_{i\sigma}^\dagger b_{i\sigma}+ b_{i\sigma}^\dagger a_{i\sigma})+t_2 \sum_{i\sigma} (a_{i+1\sigma}^\dagger b_{i\sigma}+ b_{i\sigma}^\dagger a_{i+1\sigma})$$

In that case, how would the Hubbard hamiltonian change? (To be described in terms of these new operators $$a$$ and $$b$$).

$$H_{Hubbard} = U\sum_{j}\left[a^\dagger_{j\uparrow}a^\dagger_{j\downarrow}a_{j\downarrow}a_{j\uparrow} + b^\dagger_{j\uparrow}b^\dagger_{j\downarrow}b_{j\downarrow}b_{j\uparrow}\right]$$
• Keep in mind that $U$ is just the Coulomb repulsion between the two spins on a given lattice site. Since you are looking at a Peierls-deformed lattice, we are assuming that the atom species of the $A$ sub lattice is the same as the $B$ sub lattice. This means that the way the electrons interact within an individual atom is sub lattice-independent, which means that $U$ is the same for both – IcyOtter Aug 26 at 2:40