Why particle hole symmetry and chiral symmetry are called symmetries?

Question

$PHP^{-1}=-H$ (particle-hole symmetry) and

$\Gamma H \Gamma^{-1}=-H$ (chiral symmetry)

I understand why we get the negative signs but im just a bit confused as to why such equalities mean $H$ is particle hole and chiral symmetric.

When we say $H$ is symmetric under some operation, like time reversal, don't we usually mean that $H$ is invariant under the said transformation? $THT^{-1}=H$ if $H$ is time reversal symmetric.

On the contrary, we say $H$ is symmetric under particle hole and chiral operations when $H$ picks up a negative sign. This seems to contradict our usual convention of what we mean by $H$ being symmetric(invariant).

Answer 1

They are called symmetries because (when the symmetry exists) they commute with the second quantized Hamiltonian:

$$\hat{H} = \sum_{AB}\hat{\psi}^{\dagger}_A H_{AB} \hat{\psi}_B,$$

where $H_{AB}$ are the matrix elements of the single particle Hamiltonian:

Time reversal:

$$\hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} = \hat{H}$$

Particle hole:

$$\hat{\mathcal{C}}\hat{H}\hat{\mathcal{C}}^{-1} = \hat{H}$$

and chiral

$$\hat{\mathcal{S}} = \hat{\mathcal{C}} \hat{\mathcal{T}}$$

The only extra data needed to obtain their action on the single particle Hamiltonian, as written in the question, is how they are implemented on the creation and annihilation operators:

$$\hat{\mathcal{T}}\hat{\psi}_A\hat{\mathcal{T}}^{-1} = \sum_B (U_T)_{AB} \hat{\psi}_B$$

$$\hat{\mathcal{C}}\hat{\psi}_A\hat{\mathcal{C}}^{-1} = \sum_B (U_C^*)_{AB} \hat{\psi}^{\dagger}_B$$

and in addition whether they are anti-unitary

$$\hat{\mathcal{T}}i\hat{\mathcal{T}}^{-1} = -i$$

($U_T$ and $U_C$ are unitary matrices)

Please see the review by Ludwig (sections 1-2), and Ryu, Schnyder, Akira and Ludwig (The big footnote after equation (5)), where the above conditions are elaborated into the required action on the single particle Hamiltonian, and the further elaboration of the discrete symmetries properties.

Elaboration

The time reversal case

Acting by the time reversal operator on the second quantized Hamiltonian, we obtain $$ \begin{align} \hat{\mathcal{T}}\hat{H} \hat{\mathcal{T}}^{-1} &= \hat{\mathcal{T}}\hat{\psi}_A^{\dagger} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}}H_{AB} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}}\hat{\psi}_B\hat{\mathcal{T}}^{-1} \\ &= (U_T^*)_{AC} \hat{\psi}_C^{\dagger} H_{AB}^{*} (U_T)_{BD} \hat{\psi}_D = \hat{H} = \hat{\psi}^{\dagger}_C H_{CD} \hat{\psi}_D \end{align} $$ (Please notice that when $\hat{\mathcal{T}}$ acts on the numerical parameters $H_{AB}$, it reverses the sign of $i$ and produces the complex conjugate. Therefore, we obtain:

$$(U_T^*)_{AC} H_{AB}^{*} (U_T)_{BD} = H_{CD}$$

which are the components of the matrix equation:

$$U_T^{\dagger} H^{*} U_T = H$$

The particle hole (charge conjugation) case,

Here:

$$ \begin{align} \hat{\mathcal{C}} \hat{H} \hat{\mathcal{C}}^{-1} &= \hat{\mathcal{C}} \hat{\psi}_A^{\dagger} \hat{\mathcal{C}}^{-1} \hat{\mathcal{C}} H_{AB} \hat{\mathcal{C}}^{-1} \hat{\mathcal{C}} \hat{\psi}_B \hat{\mathcal{C}}^{-1} \\ &= (U_C)_{AD} \hat{\psi}_D H_{AB} (U_C^*)_{BC} \hat{\psi}_C^{\dagger} \\ &=-\hat{\psi}_C^{\dagger} (U_C^t)_{DA}H_{AB} (U_C^*)_{BC} \hat{\psi}_D = \hat{H} = \hat{\psi}^{\dagger}_C H_{CD} \hat{\psi}_D \end{align} $$

(Here the action of $\hat{\mathcal{C}}$ on the numerical parameters $H_{AB}$, is trivial because charge conjugation is a unitary operator). The minus sign is obtained from reversing the order of $\psi$ and $\psi^{\dagger}$ which are Grassmann variables. The last equality is equivalent to the matrix equation:

$$-U_C^{t} H (U_C)^*= H^t$$

Taking the complex conjugate of both sides, we obtain:

$$U_C^{\dagger} H^* {U_C}= -H^{\dagger} = -H$$