# Charge conjugation symmetry operation on single-particle Hamiltonian

How can I show that given the second-quantized Hamiltonian of a system of non interacting fermions

$$\hat{\mathcal{H}}=\sum_{\alpha, \beta}\hat{\Psi}_{\alpha}^{\dagger}H_{\alpha\beta}\hat{\Psi}_{\beta}$$

which is particle-hole symmetric, i.e. $$\hat{\mathcal{C}}\hat{\mathcal{H}}\hat{\mathcal{C}}^{-1}=\hat {\mathcal{H}}$$, with the action of $$\hat{\mathcal{C}}$$ being defined as

$$\hat{\mathcal{C}}\hat{\Psi}_{\alpha}^{\dagger}\hat{\mathcal{C}}^{-1}=\sum_{\beta}\hat{\Psi}_{\beta}(U^{*})_{\beta\alpha}$$, $$\hat{\mathcal{C}}\hat{\Psi}_{\alpha}\hat{\mathcal{C}}^{-1}=\sum_{\beta}({U^{*}}^{\dagger})_{\alpha\beta}\hat{\Psi}_{\beta}^{\dagger}$$,

the single particle Hamiltonian fulfills

$$U{H}^{*}U^{\dagger}=-H$$?

• Can you please explain explain the action of $\hat{\mathcal C}$ in more details? What is $U$? I also cannot follow the passages. Commented May 8, 2020 at 10:03
• If you have $N$ independent fermions in your system, your Hamiltonian should be of dimension $2^N$, and so should $\mathcal{C}$, no? Commented May 8, 2020 at 13:11

Let us supose that $$U H^* U^{-1}= -H$$ and compute $${\mathcal C}\Psi^\dagger_\alpha H_{\alpha\beta} \Psi_\beta {\mathcal C}^{-1}\\ ={\mathcal C}\Psi^\dagger_\alpha {\mathcal C}^{-1}{\mathcal C}H_{\alpha\beta}{\mathcal C}^{-1}{\mathcal C} \Psi_\beta {\mathcal C}^{-1}\\ = {\mathcal C}\Psi^\dagger_\alpha {\mathcal C}^{-1}H_{\alpha\beta}{\mathcal C} \Psi_\beta {\mathcal C}^{-1}\\ =\Psi_{\rho} U^*_{\rho\alpha} H_{\alpha\beta} U^{*\dagger}_{\beta\sigma} \Psi^\dagger_\sigma\\ =- \Psi^\dagger_\sigma U^*_{\rho\alpha} H_{\alpha\beta} U^{*\dagger}_{\beta\sigma} \Psi_{\rho}\\ =- \Psi^\dagger_\sigma U_{\sigma\beta} H^T_{\beta \alpha}U^\dagger_{\alpha\rho}\Psi_{\rho}\\ - \Psi^\dagger_\sigma U_{\sigma\beta} H^*_{\beta \alpha}U^\dagger_{\alpha\rho}\Psi_{\rho}\\ = \Psi^\dagger_\sigma H_{\sigma\rho} \Psi_{\rho}.$$ We have used $$U^{*\dagger} = U^T$$ and the hemiticity of $$H$$. So the one-particle transformation on $$H$$ makes the many particle hamiltonian invariant.
Note that the many-body map $${\mathcal C}$$ is a linear map: $${\mathcal C}(\lambda |\psi_1\rangle+\mu |\psi_2\rangle)= \lambda {\mathcal C}|\psi_1\rangle+\mu {\mathcal C}|\psi_2\rangle,$$ on the Fock space despite the appearance of "$$*$$" in the action on $$H$$. This shows up in the step $${\mathcal C}H_{\alpha\beta}{\mathcal C}^{-1}= H_{\alpha\beta}.$$
• Yes. I used fermionic property, but omitted the trace $c$-number contribution as unimportant. Commented May 8, 2020 at 20:55