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.
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$\newcommand{\dag}{\dagger}$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.