Particle-hole symmetry in 2nd quantization

Question

In second quantization one the Particle-hole transformation is defined as \begin{align} \hat{\mathcal{C}} \hat{\psi}_A \hat{\mathcal{C}}^{-1} &= \sum_B U^{*\dagger}_{A,B} \hat{\psi}^{\dagger}_B \\ \hat{\mathcal{C}} \hat{\psi}_A^{\dagger} \hat{\mathcal{C}}^{-1} &= \sum_B \hat{\psi}_B U^{*}_{B,A} \\ \hat{\mathcal{C}} i \hat{\mathcal{C}}^{-1} &= +i \end{align} And if in a 2nd quantized Fermionic Hamiltonian ($\hat{\mathcal{H}} $) Particle Hole symmetry is present then $$ \hat{\mathcal{C}} \hat{\mathcal{H}} \hat{\mathcal{C}}^{-1} = \hat{\mathcal{H}} $$ I want to see what this equation means in single particle basis. In single particle basis I can write the 2nd quantized Hamiltonian ($\hat{\mathcal{H}}$) as $$ \hat{\mathcal{H}}=\sum_{A,B}\hat{\psi}^\dagger_{A}H_{A,B}\hat{\psi}_B $$ Here the matrix $H$ is the Hamiltonian in single particle basis. Now, with the transformation rules on should get $$ U H^{*} U^{\dagger} = - H $$ In the single-particle basis. But what I am getting using the transformation rules is $$ U^* H U^{*\dagger} = -H $$ Now I have started to think whether the transformation rules given here are right or not. I wanted to know if the transformation rule or my calculation is wrong.
Source: Topological phases: Classification of topological insulators and superconductors of non-interacting fermions, and beyond Equation 17

Answer 1

The transformation rules given are correct. We have

\begin{align} \hat{\mathcal{C}} \hat{\mathcal{H}} \hat{\mathcal{C}}^{-1} & = \sum_{AB} \hat{\mathcal{C}} \hat{\psi}^{\dagger}_A \hat{\mathcal{C}}^{-1} \hat{\mathcal{C}} H_{AB} \hat{\mathcal{C}}^{-1} \hat{\mathcal{C}} \hat{\psi}_B \hat{\mathcal{C}}^{-1} = \sum_{AB} \, (\sum_{C}\hat{\psi}_C U^{*}_{CA}) \, H_{AB} \, (\sum_{D}U^{*\dagger}_{BD}\hat{\psi}^{\dagger}_D) \nonumber\\ & = \sum_{ABCD} \hat{\psi}_C \, U^{*}_{CA} \, H_{AB} \, U^{*\dagger}_{BD} \, \hat{\psi}^{\dagger}_D = \sum_{CD} \hat{\psi}_C \, (U^* H U^{*\dagger})_{CD} \, \hat{\psi}^{\dagger}_D \nonumber\\ & = -\sum_{CD} \hat{\psi}^{\dagger}_D \, (U^* H U^{*\dagger})_{CD} \, \hat{\psi}_C = -\sum_{CD} \hat{\psi}^{\dagger}_D \, (U^* H U^{*\dagger})^T_{DC} \, \hat{\psi}_C = -\hat{\psi}^{\dagger} (U^* H U^{*\dagger})^T \hat{\psi} \nonumber\\ & = -\hat{\psi}^{\dagger} (U H^T U^{\dagger}) \hat{\psi} = -\hat{\psi}^{\dagger} (U H^* U^{\dagger}) \hat{\psi} \,, \end{align}

where in the 1st line 2nd equality, I've used your transformation rules. Note that everything is in component form, so $H_{AB}$ is just a complex number, and with $\hat{\mathcal{C}} i \hat{\mathcal{C}}^{-1} = i$, we have $\hat{\mathcal{C}} H_{AB} \hat{\mathcal{C}}^{-1} = H_{AB}$. In the 3rd line 1st equality I have used the fermion anticommutation rules. In the same line last equality I have written everything from component form into matrix multiplication form. In the last line last equality I have used the Hermiticity of $H$.

Now suppose there is particle-hole symmetry so we also have the above equals $\hat{\mathcal{H}} = \hat{\psi}^{\dagger} H \hat{\psi}$, so we need to have

\begin{equation} U H^* U^{\dagger} = -H. \end{equation}

Answer 2

This is a good question that I have also had, so I'm answering this even though it is two years old! It looks like your definitions are all fine, but I would use

\begin{equation} \hat{\mathcal{P}}\hat{\psi}^{}_{A}\hat{\mathcal{P}}^{-1} = \sum_{B} \hat{\psi}_{B}^{\dagger}(U_{P})_{B, A} \end{equation}

as it simplifies notation. I'll also refer to the particle-hole operator as $\hat{\mathcal{P}}$ to keep my notation clear, but that's just a personal preference.

Using the above definition we implement the transformation of $\hat{\mathcal{P}}$ on $\hat{H}$ as

\begin{equation} \begin{split} \hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} &= \sum_{A, B} \hat{\mathcal{P}} \hat{\psi}_{A}^{\dagger} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} H^{}_{A, B} \hat{\mathcal{P}}^{-1} \hat{\mathcal{P}} \hat{\psi}^{}_{B} \hat{\mathcal{P}}^{-1}\\ & =\sum_{A, B} \sum_{C, D} (U_{P}^{})^{\dagger}_{A, C} \hat{\psi}_{C} H_{A, B} \hat{\psi}_{D}^{\dagger} (U_{P}^{})^{}_{D, B} \\ &= \sum_{A, B}\sum_{C, D} \delta^{}_{C, D}(U_{P})^{\dagger}_{A, C}H^{}_{A, B}(U_{P})^{}_{D, B} - \hat{\psi}^{\dagger}_{D}(U_{P})^{}_{D, B}H^{}_{A, B}(U_{P})^{\dagger}_{A, C}\hat{\psi}^{}_{C} \\ & = \sum_{A, B}\sum_{C, D} (U_{P})^{\dagger}_{A, C}H^{}_{A, B}(U_{P})^{}_{C, B} - \hat{\psi}^{\dagger}_{D}(U_{P})^{}_{D, B}H^{T}_{B, A}(U_{P})^{\dagger}_{A, C}\hat{\psi}^{}_{C} \\ & = \sum_{A, B}\left( \delta^{}_{A, B}H_{A, B} - \sum_{C, D}\hat{\psi}^{\dagger}_{D}(U_{P})^{}_{D, B}H^{T}_{B, A}(U_{P})^{\dagger}_{A, C}\hat{\psi}^{}_{C} \right) \\ & = \text{Tr}(H) - \sum_{C, D} \hat{\psi}^{\dagger}_{D}H^{}_{D, C}\hat{\psi}^{}_{C}, \end{split} \end{equation}

where we have used the anticommutation relation $\{\hat{\psi}^{}_{A}, \hat{\psi}^{\dagger}_{B}\} = \delta^{}_{A, B}$ between the second and third lines and the unitarity of $U_{P}$ between the fourth and fifth lines. We require $\hat{\mathcal{P}}\hat{H}\hat{\mathcal{P}}^{-1} = \hat{H}$ if our Hamiltonian respects PHS, demanding the conditions that

\begin{equation} \begin{split} \text{Tr}(H) &= 0 \\ H = -U_{P}&H^{*}U_{P}^{\dagger} \end{split} \end{equation}

i.e. the Hamiltonian is traceless. We have also used the Hermicity of the Hamiltonian in the second equality. Thankfully, these two conditions are complementary, since $H = -U_{P}H^{*}U_{P}^{\dagger}$ implies that our Hamiltonian has a symmetric spectrum.

The interesting thing is that the particle hole operator is unitary and linear in the second quantised picture, but antiunitary and antilinear in the first quantised picture, i.e. $P = U_{P}K$ where $P$ is the first quantised version of the particle-hole operator and $K$ is the operation of complex conjugation.