Fermionic Operator under time-reversal symmetry 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.
 A: 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:
https://arxiv.org/abs/0912.2157
https://arxiv.org/abs/1512.08882
https://doi.org/10.1103/RevModPhys.88.035005
