# Is this two forms of Hubbard model equivalent?

I have seen two form of Hubbard model, one is: $$H=-t\sum_{<ij>s}c_{is}^\dagger c_{js}+h.c.+U\sum_i(n_{i\uparrow}-1/2)(n_{i\downarrow}-1/2)-\mu\sum_{is}n_{is}$$

The other is a more familiar one, which writes:

$$H=-t\sum_{<ij>s}c_{is}^\dagger c_{js}+h.c.+U\sum_i n_{i\uparrow}n_{i\downarrow}-\mu\sum_{is}n_{is}$$

I want to know if these two forms are equivalent. If not, on what conditions they are equivalent?(lattice type, filling,type of ground-state,attractive U of repulsive U etc.)

• Are you sure that the third term in the first equation is proportional to (n_up-1/2)(n_down-1/2), but not to n_up(n_down-1/2)? Mar 28 '14 at 9:13
• @freude Yeah, I am quite sure I have seen a lot of Hubbard model in this form. For example: here Mar 28 '14 at 9:55
• I have a feeling that the first case is related to the representation of electrons and holes, since, depending on the population probability $n_i$, the third term can take positive (repulsion) as well as negative (attraction) values. The second equation is written in the purely electron representation. See Haug H., Koch S. W., "Quantum Theory of the Optical and Electronic Properties of Semiconductors," p. 80-81 Mar 28 '14 at 10:09

One can notice that: $$(n_{i\uparrow}-1/2)(n_{i\downarrow}-1/2) = n_{i\uparrow}n_{i\downarrow} -\frac{1}{2}(n_{i\uparrow}+n_{i\downarrow}) +\frac{1}{4}$$ To show the equivalence you can absorb the $(n_{i\uparrow}+n_{i\downarrow})$ term in the chemical potential.
We don't care about the kinetic term, and have: $$U\sum_i(n_{i\uparrow}-1/2)(n_{i\downarrow}-1/2) - \mu\sum_{is} n_{is} ={ U\sum_i n_{i\uparrow}n_{i\downarrow} - (\mu+U/2)\sum_{is}n_{is} +UN/2}$$ where $N$ is the total number of sites $N=\sum_i 1$. Since the last term is only a constant you can get rid of it in the hamiltonian and change the chemical potential to have your equivalence.
These two models are exactly the same. The first model has the additional advantage that particle-hole symmetry takes place at $\mu = 0$. The second model takes place at $\mu = U/2$.