Equation Icon  &tidle;κtr = 1(κDE )jj, 3 1 jk (&tidle;κe+)jk = -(κDE + κHB ) , 2 jk 1- jk 1-jk ii (24 ) (&tidle;κe− ) = 2(κDE − κHB ) − 3δ (κDE ) , (&tidle;κo+)jk = 1(κDB + κHE )jk, 2 1 jk (&tidle;κo− )jk =-(κDB − κHE ) . 2