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


  • 1
    $\begingroup$ Can you please explain explain the action of $\hat{\mathcal C}$ in more details? What is $U$? I also cannot follow the passages. $\endgroup$
    – fra_pero
    May 8, 2020 at 10:03
  • $\begingroup$ If you have $N$ independent fermions in your system, your Hamiltonian should be of dimension $2^N$, and so should $\mathcal{C}$, no? $\endgroup$
    – Ronan
    May 8, 2020 at 13:11

1 Answer 1


I'll do it the other way round and show that, given the one-particle Hamiltonan condion, the many-particle Hamiltonian is left invariant.

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 $$ {\mathcal C}H_{\alpha\beta}{\mathcal C}^{-1}= H_{\alpha\beta}. $$

  • $\begingroup$ and going from the fourth to the fifth line you used the fermionic anti-commutation relationship as well as the fact that the trace of the single-particle Hamiltonian vanishes, right? $\endgroup$
    – Milarepa
    May 8, 2020 at 18:00
  • $\begingroup$ Yes. I used fermionic property, but omitted the trace $c$-number contribution as unimportant. $\endgroup$
    – mike stone
    May 8, 2020 at 20:55

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