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

Question

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 $b$).

Answer 1

Your Hubbard portion is still just the on-site energy. So, since now you have the bipartite lattice, you get

$$ H_{Hubbard} = U\sum_{j}\left[a^\dagger_{j\uparrow}a^\dagger_{j\downarrow}a_{j\downarrow}a_{j\uparrow} + b^\dagger_{j\uparrow}b^\dagger_{j\downarrow}b_{j\downarrow}b_{j\uparrow}\right] $$