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.
Best Answer
$\rho_\psi$, the density matrix, is not an observable/operator evolving in the sense of the Heisenberg equation of motion $$ \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t} A = - [H,A]$$ since it is defined, as you correctly write, as a projector on states, hence it is time-dependent in the Schrödinger picture (since there the states it projects on are time-dependent), obeying the von Neumann equation $$ \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t}\rho_\psi = [H,\rho_\psi]$$ as a direct consequence of the Schrödinger equation. The von Neumann equation indeed differs from the Heisenberg equation of motion by a sign because it is not an equation in the Heisenberg picture, but in the Schrödinger picture.
In the Heisenberg picture, the states are time-independent, and the density matrix does consequently not evolve. In particular, it does not obey the Heisenberg equation of motion.
You can also look at this by considering that $\mathrm{Tr}(\rho_\psi A)$ is the expectation value of a operator for a particular state. It takes to inputs - the "state input" $\rho_\psi$ and the "operator input" $A$. In the Schrödinger and Heisenberg pictures, only one of these should be time-dependent, even if they both "look" like operators.