Symbol W^dagger

The passage around this formula

The dual of conjugation by W is conjugation by $W^\dagger$ : from the defining pairing, $\mathrm{Tr}$[W$\rho$ $W^\dagger$ $O_c$] = $\mathrm{Tr}$[$\rho$ $W^\dagger$ $O_c$ W] holds for every $\rho$ , so $\Lambda^\dagger(O_c)$ = $W^\dagger$ $O_c$ W . Substituting, $\Lambda^\dagger(H_c) = W^\dagger\big(W H_f W^\dagger\big) W = (W^\dagger W) H_f (W^\dagger W) = H_f$. using only $W^\dagger$ W = I . So $D_\Lambda$ = $H_f$ - $H_f$ = 0 , identically, as an operator, for every unitary W and every $H_f$ . This is not a limit, an approximation, or a property of some special state; it is an algebraic identity that holds before any state is chosen at all.

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