# Using particle-hole symmetry of the Hubbard model to study the model at different densities

In Condensed Matter Field Theory by Altland and Simons, they state that the Hubbard Hamiltonian $$H = \sum_{\text{nearest neighbors } ij \text{ and spin } \sigma} a^\dagger_{i\sigma} a_{j\sigma} + U \sum_i \hat{n}_{i\uparrow} \hat{n}_{i\downarrow}$$ is invariant under particle-hole symmetry (defined by the transformation $$a^\dagger_i \to (-1)^i a^\dagger_i$$ and similarly for $$a$$) up to a shift by the number operator, which can be absorbed by a change in the chemical potential. They then claim that because of this symmetry, it suffices to study the system at densities $$n = [0, 1]$$ and then, using particle-hole symmetry, we can calculate the dynamics of the system at $$n = (1, 2]$$. My question is, is this a typo? Shouldn't it be that we can study it from $$n = [0, 1/2]$$ and then particle-hole symmetry gives us the dynamics at $$n = (1/2, 1]$$ (since $$n$$ gets mapped to $$1 - n$$ under the symmetry)? Or am I misunderstanding something here? Thanks.

$$\newcommand{\dag}{\dagger}$$I realized it's because there are two spins that must be accounted for. Under the particle-hole transformation, we have \begin{align*} N = \sum_{i, \sigma} a^\dag_{i, \sigma} a_{i, \sigma} &\to \sum_{i, \sigma} a_{i, \sigma} a^\dag_{i, \sigma}, \\ &= \sum_{i, \sigma} 1 - a_{i, \sigma} a^\dag_{i, \sigma},\\ &= 2 V - N, \end{align*} where we have used the fermionic anti-commutation rule and we have defined $$V = \sum_i 1$$ to be the "volume" of the system, e.g., the number of lattice sites for the case of a discrete lattice. Therefore, under the symmetry transformation, $$N \to 2 V - N$$, which implies $$n \equiv \frac{N}{V} \to 2 - n$$, and thus the logic in Altland and Simons makes sense.