quantum mechanics - Why does the density matrix $rho$ obey a wrong-signed Heisenberg equation of motion?

The density matrix is defined as $$\rho_\psi ~:=~ \frac{\lvert\psi(t)\rangle \langle \psi(t)\vert}{ \langle \psi(t) |\psi(t)\rangle }$$ in the Schrödinger picture. $\rho_\psi$ is obviously a time dependent projector, and the equation of motion on these projectors become:
$$i\hbar\frac{d}{dt} \rho_\psi ~=~ [H,\rho_\psi] \tag{S}$$ but my book also reports that the Heisenberg equation of motion on the operators/observables is: $$\mathrm{i}\hbar\frac{d}{dt} A ~=~ [A,H] . \tag{H}$$
Why are the signs in eqs. (S) and (H) opposite?

Isn't $A$ an operator like $\rho_\psi$, although time independent? They belong to the same operator space, so I don't think I can apply duality, but I know that $A$ operate on the states to give us the expectation value through the relation $$\mathrm{Tr}( \rho_\psi A)$$ so it should be in the dual space of the observables.

Answer

Actually, you can use duality:

the normal states of quantum mechanics are objects of the (unique) predual of the von Neumann algebra of quantum observables.

Using a concrete example: if the algebra of observables are the bounded operators on a Hilbert space, the predual are the trace class operators. Of them, the normal states are the ones positive, self-adjoint and of trace norm one.

It is then clear that by mutual duality the evolution on observables/states induces the evolution of states/observables; and that takes into account of the "minus sign" in the generator that is different between the two.