How particle-hole transformation act on momentum?

1. In Everett You's answer to Particle hole symmetry of single site? he explains how particle-hole transformation acts on a single site. But, we know that the operation of particle-hole transformation $\mathcal{P}$ reverses momentum and spin but leaves position invariant. From equation $\mathcal{P}c^\dagger_{i\sigma}\mathcal{P}^{-1}=c_{i\sigma}$, one can obtain that $$\mathcal{P}c^\dagger_{k\sigma}\mathcal{P}^{-1}=c_{k\sigma}$$ using Fourier transformation. Furthermore, one can obtain $$\mathcal{P}c^\dagger_{k\uparrow}|0\rangle=c^\dagger_{k\downarrow}|0\rangle.$$ Thus, a state with momentum $k$ and spin up $c^\dagger_{k\uparrow}|0\rangle$ transforms into a state $c^\dagger_{k\downarrow}|0\rangle$ with momentum $k$ and spin-down. Under the particle-hole transformation, the momentum of state does not reverse. I am confused by this contradiction. How can I solve this problem?

2. How can I prove the transformation of a Bloch Hamiltonian under the particle-hole reversal $\mathcal{P}$? $$\mathcal{P}h(-k)\mathcal{P}^{-1}=-h(k),$$ where $h(k)$ is a Bloch Hamiltonian.

• Particle-hole transformation is antiunitary, so there is a complex conjugation in it. – FangXie May 9 '18 at 13:21