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


2 Answers 2


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}


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.