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For a time reversal symmetry operator $\hat{T}$, we have $$\hat{T}^2=(-1)^{N_f}$$ for a fermionic (electronic) system. How do we understand $\hat{T}^2=(-1)^{N_f}$ in terms of operator matrices on the Hilbert space? $N_f$ is the number of fermions in the system.

How could we write down the expression of $\hat{T}$ in the complex fermion or Majorana fermion basese?

Naively, we have $$\hat{T}= i \sigma_y K$$ for acting on a single 2-component spin-(1/2) system, but how can the complex conjugation $K$ be realized in terms of $2 \times 2$ matrix (since a spin-(1/2) object takes a 2-dimensional Hilbert space)? Or do we need to enlarge the Hilbert space to $4 \times 4$ matrix in order to realize the complex conjugation $K$?

For a many-body electron system, how do we write down $\hat{T}$ and $\hat{T}^2=(-1)^{N_f}$ in terms of operator matrices on the $N$-dimensional Hilbert space?

p.s. Please do not give Refs. Please you should explain the answer by explicit results. Thanks in advance,

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    $\begingroup$ You can’t write $K$ as a matrix because it’s not a linear operator; the one-dimensional analogue of this is asking for a complex number $w$ so that $wz = \bar{z}$, which is impossible. $\endgroup$ – knzhou Oct 24 '17 at 18:58
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This does not exactly answer your question but I think you want to justify $T^2=(-1)^N$

You best understand it by the action of Time-reversal operators on second quantized fermionic field operators. They act like

\begin{align} \mathcal{T}\Psi_{\alpha}(\mathbf{k})\mathcal{T}^{-1} & =\sum_{\alpha'}(U^{\dagger}_{\operatorname{T}})_{\alpha,\alpha'}\Psi_{\alpha'}(-\mathbf{k}),\label{eq:-29}\\ \mathcal{T}\Psi_{\alpha}^{\dagger}(\mathbf{k})\mathcal{T}^{-1} & =\sum_{\alpha'}\Psi^{\dagger}_{\alpha'}(-\mathbf{k})(U_{\operatorname{T}})_{\alpha',\alpha},\label{eq: timereversal} \end{align}

where $\alpha$ is some degree of freedom and $U_T$ is a unitary matrix. For example if I have a many-body state with $N$ electrons it can be given by $$|N\rangle=\prod_{i}^{N}\Psi^{\dagger}_{\alpha}(k_i)|\mathrm{vac}\rangle,$$ Now it is clear that $$\mathcal{T}^2|N\rangle=(U_T^2)^N|N\rangle, $$ finally if your system consists of pin one half particles $U_T$ squares to minus one.

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