What's the reason for particle-hole symmetry operator to be anti-unitary? I have been looking at some literature on Topological Superconductor, where the BdG Hamiltonian is frequently used, the $H_{BdG}$ has the so-called particle-hole symmetry, which is commonly defined through $C=\sigma_x \mathcal{K}$, where $C^{-1}HC=-H^*$.
As a beginner, I'm really curious about the basic definition of this particle-hole "transformation".Why should it be defined like this? Hope someone can answer this question.
 A: Particle-hole symmetry is only antilinear in the one-particle space. It is linear and unitary when acting on the many-particle Fock space. See footnote after eq 4 in  S.Ryu, A.Schnyder, A.Furusaki, A.Ludwig, Topological insulators and superconductors: ten-fold way and dimensional hierarchy  New J. Phys. 12, 065010 (2010). ArXiv:0912.2157.
Supose that the one particle Hamiltonian has the property that
$$
C H^* C^{-1}= -H
$$
for some unitary matrix $C$. Then
$$
Hu_n=\lambda_n u_n\quad\Rightarrow \quad HCu_n^* = -\lambda_n Cu^*_n,
$$
so, when $\lambda$ is non zero, the single-particle eigenfunctions come in opposite-eigenvalue  pairs.     In the absence of zero energy states the ground state $|0\rangle$ has all negative-energy states occupied and   is non-degenerate.
We define the action of a  unitary particle-hole operator ${\mathsf C}$ on the many-body Fock space by
$$
{\mathsf C}\Psi_\beta {\mathsf C}^{-1}= \Psi_\alpha^\dagger C_{\alpha\beta}, \quad {\mathsf C}\Psi^\dagger_\beta {\mathsf C}^{-1}= C^\dagger_{\beta\alpha}\Psi_\alpha.
$$
When
${\mathsf C}$ acts on the Hamiltonian  we have
$$ 
{\mathsf C}\hat H {\mathsf C}^{-1}=
{\mathsf C}\Psi^\dagger_\alpha H_{\alpha\beta} \Psi_\beta {\mathsf C}^{-1}\\
={\mathsf C}\Psi^\dagger_\alpha {\mathsf C}^{-1}{\mathsf C}H_{\alpha\beta}{\mathsf C}^{-1}{\mathsf C}  \Psi_\beta {\mathsf C}^{-1}\\
= {\mathsf C}\Psi^\dagger_\alpha {\mathsf C}^{-1}H_{\alpha\beta}{\mathsf C} \Psi_\beta {\mathsf C}^{-1}\\
= C^\dagger_{\alpha\rho}\Psi_{\rho}  H_{\alpha\beta} \Psi^\dagger_\sigma C_{\sigma\beta}\\
=- \Psi^\dagger_\sigma C_{\sigma\beta} H_{\alpha\beta}C^\dagger_{\alpha\rho}\Psi_{\rho}\\
=- \Psi^\dagger_\sigma C_{\sigma\beta} H^T_{\beta \alpha}C^\dagger_{\alpha\rho}\Psi_{\rho}\nonumber\\
=- \Psi^\dagger_\sigma C_{\sigma\beta} H^*_{\beta \alpha}C^\dagger_{\alpha\rho}\Psi_{\rho}\nonumber\\
=+ \Psi^\dagger_\sigma H_{\sigma\rho} \Psi_{\rho}.\\
= \hat H.
$$
So the one-particle transformation on $H$ leaves the many-particle hamiltonian invariant.
Note: In expressions like $ \Psi^\dagger_\sigma C_{\sigma\beta} H_{\alpha\beta}C^\dagger_{\alpha\rho}\Psi_{\rho}$, the quantities $\Psi^\dagger_\sigma$ and $\Psi_{\rho}$ are second quantized operators, whereas $C_{\sigma\beta}$ or $H_{\alpha\beta}$ are regular complex numbers (they are the components of the relevant matrices).
Note that we have used $C^{*\dagger} = C^T$ and the tracelessness (line 5 $\to$ 6) and hermiticity of $H$ in the above manipulations.
More importantly, and   despite the appearance of ``$*$" in the action  on $H$, the many-body operator  ${\mathsf C}$ must act  on the Fock space  linearly:
$$
{\mathsf C}(\lambda |\psi_1\rangle+\mu |\psi_2\rangle)= \lambda {\mathsf C}|\psi_1\rangle+\mu {\mathsf C}|\psi_2\rangle.
$$
The linearity is  required  in the step
$$
{\mathsf C}H_{\alpha\beta}{\mathsf C}^{-1}= H_{\alpha\beta}.
$$
A: I find it conceptually more simple to think of particle-hole symmetry as defined in second quantization notation. Indeed: the very meaning of a particle-hole transformation should mean that it should interchanging particles and holes, i.e. we want $\mathcal C c^\dagger \mathcal C = c$ (where $\mathcal C^2 = 1$). The anti-unitarity then follows from wanting $\mathcal C$ to preserve the $U(1)$ symmetry of fermions: if $c \to e^{i\alpha} c$, then $\mathcal C (e^{-i\alpha} c^\dagger )\mathcal C = e^{i\alpha} c$. I.e. we want that $\mathcal C e^{-i\alpha} \mathcal C = e^{i\alpha}$. This then naturally and completely defines the particle-hole transformation $\mathcal C$ !
Then how to define being invariant under this symmetry? Naively we would say $\mathcal C H \mathcal C = H$. However this is not the right notion. To see this, take the simple case of $H = \sum t_{ij} c_i^\dagger c_j + \mu c_i^\dagger c_i$. Intuitively we see that this should be particle-hole symmetric if $\mu = 0$. (To convince yourself, consider the case of nearest-neighbour hopping, in which case we know the spectrum is just a cosine, which is clearly particle-hole symmetric at half-filling, i.e. $\mu = 0$.) Using our above definition, $\mathcal C H\mathcal C = \sum t_{ij}^* c_i c_j^\dagger + \mu c_i c_i^\dagger$, which by the fermionic commutation rules is the same as $- \sum \left( t_{ji}^* c_i^\dagger c_j + \mu c_i^\dagger c_i \right) + \mu N_\textrm{sites}$. Then again by the fact that $H$ must be Hermitian, we know that $t_{ji}^* = t_{ij}$, so we see that
$$ \mathcal C H\mathcal C = -H + \mu N_\textrm{sites}$$ 
I.e. in the particle-hole symmetric case, we have that $ \mathcal C H\mathcal C = -H$. It is then natural to take this as our definition of particle-hole symmetry!
A: A more fundamental answer to why the particle-hole operator is taken that way is to look towards QFT. There we have particles and antiparticles and we can assign a charge of $+q$ to a particle and $-q$ to the antiparticle. Thus a operator that interchanges particles and antiparticles $\mathcal{C}$  must satisfy $\mathcal{C}|p\rangle=|\bar{p}\rangle$ so if we have a charge operator $\mathcal{Q}$ it must satisfy: $\mathcal{Q}|p\rangle=q|p\rangle$ and $\mathcal{Q}|\bar{p}\rangle=-q|\bar{p}\rangle$. Now we apply the $\mathcal{C}$ operator:
$$\mathcal{C} \mathcal{Q}|p\rangle= \mathcal{C} q|p\rangle= q|\bar{p}\rangle$$
While simultaneously:
$$\mathcal{Q}\mathcal{C}|p\rangle= \mathcal{Q}|\bar{p}\rangle= -q|\bar{p}\rangle$$
Thus we must have
$$\mathcal{Q}\mathcal{C}=-\mathcal{C} \mathcal{Q}$$
In this sense if an operator is consistent with how charges change under  $\mathcal{C}$  then:
$$\mathcal{C}^{-1}\mathcal{A}\mathcal{C}=-\mathcal{A}$$
Now this we can apply to the Hamiltonian , note that we can't have $\mathcal{H}\mathcal{C}=\mathcal{C} \mathcal{H}$ since this would imply simultaneous eigenstates of the Hamiltonian and $\mathcal{C}$, but the only eigenstates of $\mathcal{C}$ are particles whose antiparticle is themselves since $\mathcal{C}^2=1$. So to establish consistency of particles and antiparticles the Hamiltonian must satisfy:
$$\mathcal{C}\mathcal{H}\mathcal{C}=-\mathcal{H}$$
This actually links to the above answer since maintaining $U(1)$ symmetry implies charge conservation so $\mathcal{C}$ must be anti-unitary. For the case of Condensed Matter we take, conveniently, holes to be the antiparticle of electrons since clearly they have opposite charge. Then all of the above follows.
