Time Reversal Operator I know that time reversal operator is an antiunitary operator.
How does it work on wavefunctions?
I believe in this way:
$$T \psi (k,+)=e^{i\pi S_y/\hbar} K \psi (k,+) = \psi^*(-k,-),$$ 
but I am not sure. ("+" and "-" represent spin-up or spin-down states)
Does anybody have a good explanation?
 A: Reference (page $13$, formula $17.71$)
The time-reversal operator is $\Theta = Ke^{-i\pi S_y/\hbar}$, where $K$ is the complex conjugation operator.
Taking a spin $1/2$, we have a wavefunction which is a $2$- component spinor $\psi(x) =  \begin{pmatrix}
\psi_+(x) \\
\psi_-(x)
\end{pmatrix}$, 
Note that, for spin $1/2$, $e^{-i\pi S_y/\hbar} = e^{-i \large \frac{\pi}{2} \sigma_y} = -i\sigma_y =\begin{pmatrix}
0&&-1 \\
1&&0
\end{pmatrix}$
So time-reversal gives: $\begin{pmatrix}
\psi_+(x) \\
\psi_-(x)
\end{pmatrix} \rightarrow  K\begin{pmatrix}
0&&-1 \\
1&&0
\end{pmatrix}\begin{pmatrix}
\psi_+(x) \\
\psi_-(x)
\end{pmatrix} =  K\begin{pmatrix}
-\psi_-(x) \\
\psi_+(x)
\end{pmatrix} = \begin{pmatrix}
-\psi^*_-(x) \\
\psi^*_+(x)
\end{pmatrix}$
By using Fourier transform $\psi(k) \sim \int  \psi(x) e^{ -i k.x}$, we may notice that the Fourier transform of $\psi^*(x)$ is $\psi^*(-k)$. So we get the time-reversal operation: 
$\begin{pmatrix}
\psi_+(k) \\
\psi_-(k)
\end{pmatrix} \rightarrow \begin{pmatrix}
-\psi^*_-(-k) \\
\psi^*_+(-k)
\end{pmatrix}$
