Is density operator an observable?

Question

Studying the time evolution of the density operator I've encountered the following equation:
$$partial_t rho = -frac{i}{hbar}[H,rho].$$
The text defined it as the time evolution of $rho$ in the Schrödinger scheme, which I find confusing since $rho$ is an operator hence it should be time independent in the Schrödinger scheme. As far as I've understood, this is because the representation of $rho$ is proportional to the sum of the projectors on the system's states, which do evolve over time in the Schrödinger scheme.
The second thing I find confusing is the fact that $rho$ is an hermitian operator and thus should be an observable (?) (its eigenvalues are the occupation frequencies of the eigenstatets), so it should satisfy the Heisenberg equation for its time evolution which differs from the previous one by a minus sign. On any text I've consulted, the minus sign is said to be an important difference but no one explains its meaning.
Have I misunderstood something or am I missing a link between the two equations? Is $rho$ an observable, or should it satisfies other conditions aside being hermitian?

DerHutmacher · Accepted Answer

I will try a conceptual answer with the hope it clarifies a little more, maybe you will find it too dumb at some points but who knows.
To every System we assign a Hilbert Space and an Hamilton Operator on this space. Now Imagine just the Hilbert space without any physics. Then you have a bunch of states and a bunch of operators which come with the definition of the Hilbert space. Now to capture the time evolution of your system one imposes the time evolution of a system (or the state the system is described by) via the Schrödinger equation on the state of the system. This imposin is what is meant by "using the Schrödinger picture". By now you have only imposed things which are not essential to the Hilbert space. The Hilbert space and all operators exist mathematically without any physics.
Now we say: okay, but how do we obtain physical information about our system with the description we chose so far. Well we say that to every observable A there corresponds a specific operator $hat{A} = sum_{m,n}a_{m,n}|m><n|$ on our hilbert space. With $|m>$ being some basis of the Hilbert space. Now we have to every observable an corresponding operator which is time independent. But what is an observable and what is its operator is imposed from you by the way you define your theory or quantize a classical theory. You see that all the time dependence is imposed on the quantities we use to describe our system. So if we say that observables are time independent it is because we define them to be. Its not that operators are intrinsically time dependent or independent but we say that the operators describing observables are time independent in our description(Schrödinger Picture). The Heisenberg picture is just saying: okay instead of imposing the Schrödinger equation on States, we impose the Heisenberg Equation on operators corresponding to observables and all predictions we get are the same, so both point of views are equivalent.
The density matrix on the other hand is defined by the state. If you describe physics in the Schrödinger picture this matrix will be time dependent by definition. It just turn out that this time dependents is given by the von-Neumann equation. If you switch to the Heisenberg picture, the density matrix will be time independent, because your state is. The fact that the density matrix is no observable in this description is then given by the fact that it is time dependent in the Schrödinger picture.

Michael Riberdy · Answer

The density operator represents the state of the system. In the Schrodinger picture, which is the picture in which your equation (The Liouville von-Neumann equation) is valid, the state evolves with time. Therefore, in the Schrodinger picture, the density operator evolves with time.
The Heisenberg picture equation you reference does have a relative minus sign, but it lives in a different picture, in which operators evolve and states remain time-independent. Given that the density operator represents the state of the system, it does not evolve in the Heisenberg picture. Therefore, the relative sign creates no contradiction.

Is density operator an observable?

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