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:


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 $b$).


1 Answer 1


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

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

  • $\begingroup$ Shouldn't there be a different U for each operator? such as U_a and U_b? $\endgroup$
    – Caterina
    Commented Aug 25, 2019 at 21:12
  • $\begingroup$ I mean, since a only acts on site A and b on site B, I would expect to have different on-site Us? Or am I wrong? $\endgroup$
    – Caterina
    Commented Aug 25, 2019 at 21:13
  • $\begingroup$ 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 $\endgroup$
    – IcyOtter
    Commented Aug 26, 2019 at 2:40
  • $\begingroup$ Thanks a lot! Makes complete sense. I just have another question. Do you know if there is an operator that describes 'on-site' repulsion in double bonds rather than on one atom. Because the Hubbard contribution is not as realistic as considering that the 2 electrons are more likely to be in between the 2 atoms rather than located at either A or B. $\endgroup$
    – Caterina
    Commented Aug 26, 2019 at 9:34
  • $\begingroup$ Well, the idea is that you are using the tight-binding Hamiltonian. This means the electrons do, in fact, spend most of their time around the atom. Otherwise, we can't really write the hopping Hamiltonian since the bonding would substantially modify the onsite energy states $\endgroup$
    – IcyOtter
    Commented Aug 27, 2019 at 13:36

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