Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
Replace $h\rightarrow \hbar$ in your formula. – Will Jul 6 '13 at 15:02
up vote 5 down vote accepted

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:

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.