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Time-reversal
Let us first consider operators
which are either even (invariant)
or odd (antiinvariant) with respect to the time-reversal:
![\begin{displaymath}
\hat{\cal{T}}^\dagger\hat{\cal{O}}\hat{\cal{T}}= \epsilon_T \hat{\cal{O}}, \qquad \epsilon_T=\pm1.
\end{displaymath}](img112.gif) |
(11) |
From Eqs. (9) and (11) one gets
![\begin{displaymath}
\langle \mbox{{\boldmath {$n$ }}}\,\zeta\vert\hat{\cal{O}}\...
...t{\cal{O}}\vert\mbox{{\boldmath {$n$ }}}'\,-\!\zeta'\rangle^*.
\end{displaymath}](img113.gif) |
(12) |
It follows, that the matrix corresponding to
reads
![\begin{displaymath}
{\cal{O}}= \left(\begin{array}{cc}
A & Y \\
-\epsilon_T Y^* & \epsilon_T A^*
\end{array} \right),
\end{displaymath}](img114.gif) |
(13) |
where A is hermitian, and Y is antisymmetric
or symmetric for
=+1 and
=-1, respectively.
No block-diagonal structure appears, nevertheless, only two
instead of four
submatrices, A and Y, complex in general, have to be calculated.
Jacek Dobaczewski
2000-02-05