Time-reversal

Let us first consider operators $\hat{\cal{O}}$ which are either even (invariant) or odd (antiinvariant) with respect to the time-reversal:

$$\hat{\cal{T}}^\dagger\hat{\cal{O}}\hat{\cal{T}}= \epsilon_T \hat{\cal{O}}, \qquad \epsilon_T=\pm1.$$ (11)

From Eqs. (9) and (11) one gets

$$\langle \mbox{{\boldmath {$n$ }}}\,\zeta\vert\hat{\cal{O}}\... ...t{\cal{O}}\vert\mbox{{\boldmath {$n$ }}}'\,-\!\zeta'\rangle^*.$$ (12)

It follows, that the matrix corresponding to $\hat{\cal{O}}$ reads

$${\cal{O}}= \left(\begin{array}{cc} A & Y \\ -\epsilon_T Y^* & \epsilon_T A^* \end{array} \right),$$ (13)

where A is hermitian, and Y is antisymmetric or symmetric for $\epsilon_T$=+1 and $\epsilon_T$=-1, respectively. No block-diagonal structure appears, nevertheless, only two instead of four submatrices, A and Y, complex in general, have to be calculated.