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?

  • $\begingroup$ Replace $h\rightarrow \hbar$ in your formula. $\endgroup$
    – Will
    Jul 6, 2013 at 15:02

1 Answer 1


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:

$\begin{pmatrix} \psi_+(k) \\ \psi_-(k) \end{pmatrix} \rightarrow \begin{pmatrix} -\psi^*_-(-k) \\ \psi^*_+(-k) \end{pmatrix}$

  • 1
    $\begingroup$ The reference is inaccessible now. $\endgroup$ Jun 21, 2017 at 22:12
  • $\begingroup$ The step going from $e^{−i\frac{\pi}{2}\sigma_y}=−i\sigma_y$ is described here $\endgroup$
    – Adam
    Nov 1, 2019 at 16:09
  • $\begingroup$ This is generally wrong. Time-reversal operator can take all kind of forms and depends on the Hamiltonian $\endgroup$
    – varantir
    Jan 12, 2022 at 15:57

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