$PHP^{-1}=-H$ (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).

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    $\begingroup$ The former is a dynamical symmetry in proper sense, as it is an anti unitary operator, thus a Wigner symmetry, and anticommutes with the Hamiltonian, thus preserving time evolution. The latter is a Wigner symmetry it being unitary, but it is not a dynamical symmetry as it does not preserve time evolution (it reverses it). So the notion of "symmetry" is used here in an extended version. There are many terminologies also incompatible, and every statement about symmetries should be always written into a clear form (see the vague statements about CPT symmetry for instance). $\endgroup$ Commented Jan 21, 2018 at 11:02

1 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.


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,


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

(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:

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

  • $\begingroup$ For particle hole symmetry, shouldnt you have a negative sign on the right hand side? $$\hat{\mathcal{C}}\hat{H}\hat{\mathcal{C}}^{-1} = -\hat{H}$$ $\endgroup$
    – Blackwidow
    Commented Jan 21, 2018 at 14:43
  • $\begingroup$ For the second quantized theory, the particle hole is a symmetry, thus no minus sign exists, however, when the action on the single particle Hamiltonian is elaborated, we get the minus sign. This is given in the attached references. $\endgroup$ Commented Jan 21, 2018 at 14:49
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    $\begingroup$ I have added the elaboration for the time reversal and particle hole symmetries. $\endgroup$ Commented Jan 22, 2018 at 11:53
  • $\begingroup$ Could you elaborate why the equation (5) in the paper holds, i.e. the way that the second quantized operator $\Psi_A$ is transformed. This is essential for the following discussion in your answer. How can I understand that formula? $\endgroup$ Commented May 4, 2021 at 12:00
  • $\begingroup$ These equations describe group actions. You can prove this fact by applying two consecutive transformations. $\endgroup$ Commented Jul 1, 2021 at 8:49

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