I do not understand the transformation rule of spinful fermionic creation/annihilation operators under time-reversal symmetry. According to the notes written by Prof. Grushin, equation(41) and (42) state the following: \begin{equation} \hat{T} c_{i \uparrow} \hat{T}^{-1} = c_{i \downarrow} ~~,~~ \hat{T} c_{i \uparrow} \hat{T}^{-1} = -c_{i \uparrow} \end{equation} where $\hat{T}$ is the anti-unitary time reversal operator $\hat{T} = U_{T} \mathcal{K}$, $\mathcal{K}$ is the complex conjugation. I do not understand how to derive these two equations.

$\textbf{Attempt}$: I view $c_{i \uparrow}$ and $c_{i \downarrow}$ are the component of a doublet(e.g ($c_{i \uparrow} ~~ c_{i \downarrow} )^{T}$, where $T$ in here means transpose. ). Then, for spin-1/2 fermion, we usually write the time-reversal symmetry as following: \begin{equation} \hat{T} = e^{i\pi \sigma_{y}/2} \mathcal{K} \end{equation} Since we interpret that $c_{i \uparrow}$ and $c_{i \downarrow}$ are the components of a doublet, we can think the time-reversal symmetry is kind of a rotation in the spin-basis of $c_{i \uparrow}$ and $c_{i \downarrow}$. Therefore, we want to compute the following: \begin{equation} \hat{T} \begin{pmatrix} c_{i \uparrow} \\ c_{i \downarrow} \end{pmatrix} \hat{T}^{-1} = e^{i \pi \sigma_{y}/2 } \mathcal{K} \begin{pmatrix} c_{i \uparrow} \\ c_{i \downarrow} \end{pmatrix} \mathcal{K} e^{- i \pi \sigma_{y}/2} \end{equation}

Then, I can rewrite $\sigma_{y}$ in terms of second quantization(e.g. $ \sigma^{y} = \sum_{i \sigma \sigma'} c^{i \sigma} \sigma^{y}_{\sigma, \sigma'}c_{i\sigma'}$). Using Baker-Campbell-Hausdorff formula, I may evaluate the above formula. However, with the complex conjugation, I do not understand how to evaluate the complex conjugate of spinful fermion operators $\mathcal{K} \begin{pmatrix} c_{i \uparrow} \\ c_{i \downarrow} \end{pmatrix} \mathcal{K}$. Could anyone teach me how to evaluate this complex conjugation? I appreciate any comment.

  • $\begingroup$ What is the Hamiltonian? without the Hamiltonian the question is meaningless, time-reversal (or any other transformation) can act in any way you want. $\endgroup$ Commented Sep 17, 2021 at 14:56
  • $\begingroup$ The complex conjugation is defined to act as the identity on creation/annihilation operators. $\endgroup$ Commented Sep 17, 2021 at 14:58
  • $\begingroup$ @AccidentalFourierTransform. Thank you for your comment. According to Prof. Grushin notes, I think that he did not mention the exact Hamiltonian that he considered. He only considered a general free fermionic Hamiltonian like eq(35) in his notes. $\endgroup$
    – Ricky Pang
    Commented Sep 17, 2021 at 15:40
  • $\begingroup$ Thank you for your comment, @JahanClaes. May I ask that why the complex conjugation acts like identity? Is there any intuitive explanation that can help us to understand more? Actually, I agree that the complex conjugation acts like identity. However, if $c$ is a complex fermion(e.g, linear combination of two Majorana modes), I do not think that complex conjugation acts like identity. $\endgroup$
    – Ricky Pang
    Commented Sep 17, 2021 at 15:45
  • 2
    $\begingroup$ You have to define complex conjugation with respect to some basis of operators (it has to consider some basis to be canonically "real"). If you define it in the fermionic basis, it acts at the identity on $c$ but not on Majoranas. If you define it in the Majorana basis, it acts as the identity on Majoranas but not on $c$. If you define it in the Majorana basis, it's easy to check that it would take $c\rightarrow c^\dagger$, which is not what time reversal does. So we want to define it in the $c$ basis, so it takes $c\rightarrow c$, because this is what time-reversal does. $\endgroup$ Commented Sep 17, 2021 at 15:58

1 Answer 1


This is essentially an issue of first quantization vs. second quantization. In second quantization we define the action of the time-reversal operator on the fermionic creation and annihilation operators to be

\begin{equation} \begin{split} \hat{\mathcal{T}}\hat{c}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}_{i, \downarrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}^{\dagger}_{i, \downarrow} \\ \hat{\mathcal{T}}\hat{c}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}_{i, \uparrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}^{\dagger}_{i, \uparrow}. \end{split} \end{equation}

where $\hat{c}^{\dagger}_{i, \sigma}/\hat{c}_{i, \sigma}$ are the fermionic creation and annihilation operators acting on site $i$ and spin state $\sigma=\uparrow/\downarrow$.

In addition $\hat{\mathcal{T}}$ is an antiunitary operator, therefore $\hat{\mathcal{T}}i\hat{\mathcal{T}}^{-1} = -i$ (we can show this by considering the Heisenberg uncertainty relationship and using the relations $\hat{\mathcal{T}}\hat{x}\hat{\mathcal{T}}^{-1} = \hat{x}$ and $\hat{\mathcal{T}}\hat{p}\hat{\mathcal{T}}^{-1} = -\hat{p}$).

We can summarise the actions of $\hat{\mathcal{T}}$ on the Fock space by first converting $\hat{c}_{i, \uparrow}, \hat{c}_{i, \downarrow}, \hat{c}^{\dagger}_{i, \uparrow}, \hat{c}^{\dagger}_{i, \downarrow}$, ... to $\hat{\psi}_{1}, \hat{\psi}_{2}, \hat{\psi}_{3}, \hat{\psi}_{4}$... The action of $\hat{\mathcal{T}}$ is then surmised as $\hat{\mathcal{T}}\hat{\psi}_{A}\hat{\mathcal{T}}^{-1} = \sum_{B} U_{A, B}\hat{\psi}_{B}$ where $A, B$ are indices labelling the site and other relevant quantum numbers such as spin and $U$ is some unitary matrix. See Kitaev's paper for more details https://arxiv.org/abs/0901.2686. It should be emphasised that, so far, we have defined the action of time-reversal on the Fock space in the formalism of second quantization.

To move to the first quantized picture, we write the second quantized Hamiltonian, $\hat{H}$ in terms of the single-particle Hamiltonian as

\begin{equation} \hat{H} = \sum_{A, B} \hat{\psi}_{A}^{\dagger}H_{A, B}\hat{\psi}_{B} \end{equation}

where the operators $\hat{\psi}^{\dagger}_{A}/\hat{\psi}_{A}$ satisfy the usual anticommutation relations and $H$ is the first quantized Hamiltonian (basically just an $N\times N$ matrix). If the second quantized Hamiltonian possesses time reversal symmetry then $\hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} = \hat{H}$. Using $\hat{\mathcal{T}}\hat{\psi}_{A}\hat{\mathcal{T}}^{-1} = \sum_{B} U_{A, B}\hat{\psi}_{B}$ we find

\begin{equation} \begin{split} \hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} &= \sum_{A, B} \hat{\mathcal{T}} \hat{\psi}^{\dagger}_{A} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}} H_{A, B} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}} \hat{\psi}_{B} \hat{\mathcal{T}}^{-1} \\ &= \sum_{A, B}\sum_{C, D}U^{*}_{A, C} \hat{\psi}^{\dagger}_{C} H^{*}_{A, B} U_{B, D} \hat{\psi}_{D} \\ &= \sum_{C, D} \hat{\psi}^{\dagger}_{C} H_{C, D} \hat{\psi}_{D} = \hat{H} \end{split} \end{equation}

where $H_{C, D} = U^{*}_{A, C} H^{*}_{A, B} U_{B, D}$, i.e. $H = U^{\dagger}H^{*}U$. Here, the single particle Hamiltonian is complex conjugated since $\hat{\mathcal{T}}$ acts on the numerical parameters and reverses the sign of i. Therefore, we may define a first quantized version of $\hat{\mathcal{T}}$ which acts on the single particle space

\begin{equation} T = \hat{\mathcal{T}}_{first quantized} \end{equation}

We may then rewrite the action of time reversal on the first quantized Hamiltonian as

\begin{equation} THT^{-1} = H \quad \text{where} \quad T = UK \end{equation}

In summary, complex conjugation acts on numbers rather than operators.

For more information on the time reversal operator and its use in classifying topological phases of matter see these excellent review papers:




  • $\begingroup$ Thanks for this very helpful answer! Would you mind taking a look at this related question? $\endgroup$
    – xiaohuamao
    Commented Jan 9, 2023 at 6:17

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