Particle-Hole symmetry of the Hubbard Hamiltonian In Condensed Matter Field Theory (by Altland and Simons), section 2.2, the authors state that the Hamiltonian of the Hubbard model
$$\hat{\mathcal{H}} = -t\sum_{<ij>,\sigma} a_{i\sigma}^\dagger a_{j\sigma} + U\sum_i \hat{n}_{i\downarrow} \hat{n}_{i\uparrow}$$
is symmetric under the exchange of particles and holes. The proof of this is left as an exercise. Now, to verify this, I want to check what happens under the action of the charge conjugation operation 
$\hat{\mathcal{C}}$ that exchanges creation operators with annihilation operators.
Let's express the kinetic Hamiltonian as $$\hat{\mathcal{H}_t} = -t\sum_{<ij>,\sigma} a_{i\sigma}^\dagger a_{j\sigma}$$ and the interaction Hamiltonian as $$ \hat{\mathcal{H}_U} =  U\sum_i \hat{n}_{i\downarrow} \hat{n}_{i\uparrow}$$ 
Then $\hat{\mathcal{H}_t} $ changes under charge conjugation as
\begin{align} 
\hat{\mathcal{H}_t} \to \hat{\mathcal{H}_t}' &= \hat{\mathcal{C}}\hat{\mathcal{H}_t}\hat{\mathcal{C}}^{-1}\\
&= -t\sum_{<ij>,\sigma} a_{i\sigma} a_{j\sigma}^\dagger\\
&= -t\sum_{<ij>,\sigma} (-a_{j\sigma}^\dagger a_{i\sigma}+\{a_{i\sigma},a_{j\sigma}^\dagger\})\\
&= -\hat{\mathcal{H}_t}-2tzN_{\text{sites}}
\end{align}
where $z$ is the coordination number of the lattice and $N_{\text{sites}}$ is the number of sites in the lattice. Factor of 2 in the second term is due to the two spins.
Similarly, $\hat{\mathcal{H}_U}$ changes as
\begin{align}
\hat{\mathcal{H}_U} \to \hat{\mathcal{H}_U}' &= \hat{\mathcal{C}}\hat{\mathcal{H}_U}\hat{\mathcal{C}}^{-1}\\
&= U\sum_i (1-\hat{n}_{i\downarrow})(1- \hat{n}_{i\uparrow})\\
&= \hat{\mathcal{H}_U} - U\sum_i (\hat{n}_{i\downarrow}+\hat{n}_{i\uparrow}) + U N_{\text{sites}}
\end{align}
Thus the full Hamiltonian changes as
$$ \hat{\mathcal{H}} \to \hat{\mathcal{H}}' \equiv \hat{\mathcal{C}}\hat{\mathcal{H}}\hat{\mathcal{C}}^{-1} = -\hat{\mathcal{H}_t} + \hat{\mathcal{H}_U} - U\sum_i (\hat{n}_{i\downarrow}+\hat{n}_{i\uparrow}) + (\text{constant terms})$$
Having reached this point, I can't see how the particle-hole symmetry condition (which, as far as I understand, is expressed as $\hat{\mathcal{C}}\hat{\mathcal{H}}\hat{\mathcal{C}}^{-1}=\hat{\mathcal{H}}$) would be satisfied. How should I correctly interpret the statement that the authors made?
 A: The particle-hole transformation you are looking for is a little more complicated than exchanging creation and annihilation operators - as you found this has the undesirable effect of changing the sign of the hopping. Instead, let's introduce the operators:
$$d_{j \sigma}^\dagger = (-1)^j c_{j \sigma} \ .$$
On a 1D chain, clearly the factor $(-1)^j$ takes values of $\pm 1$ on odd/even sites. A square lattice is bipartite, and this factor takes a value of +1 on one sublattice, and -1 on the other.
Clearly $d_{j \sigma}^\dagger d_{j \sigma} = 1 - c_{j \sigma}^\dagger c_{j \sigma}$, so this indeed describes a particle-hole transformation. If we now look at the hopping term:
$$c_{i \sigma}^\dagger c_{j \sigma} = (-1)^{i+j} d_{i \sigma}^\dagger d_{j \sigma} = d_{i \sigma}^\dagger d_{j \sigma} \ ,$$ since the minus sign arising from the anticommutator (which you obtained) combines with the minus sign arising from the transformation.
The interaction term transforms in the way you found, giving the extra term $U (n_{i \uparrow} + n_{i \downarrow})$. This can, however, simply be absorbed into a chemical potential term. Indeed, the chemical potential is not invariant in general under the particle-hole transformation, except at the special case of half-filling when the number of electrons and holes is equal.
