Equilibrium in Quantum Thermodynamics

Suppose I have a isolated system. Will it not be described by a time independent Hamiltonian? For an quantum system the time evolution of the density of states will be given by the standard equation.
\[ \frac{d \rho}{dt} = [H,\rho]\]
If we take \(\rho\) at time \(t=0\) to be some function of \(H\), it will be a constant of motion. To me this appears to be an equilibrium
situation. If yes, what is meant by the temperature of such state?

What I have written above applies to, for example, a single particle
state system in an energy eigen state. And also to an assembly of
{\bf non-interacting} particles, each particle being in some energy eigen state.

Note that  both the cases correspond to a pure state with definite energy eigenvalue.

Even if a system is not in pure state, the density matrix can be written as a sum \(\sum_n  p_n|{E_n}\langle{E_n}|\). This is not an equilibrium state, because the density matrix will keep evolving according to Schrodinger equation. It is not at all clear the quantum evolution of such a system can give rise to density matrix of the form that will correspond to a definite temperature.
\begin{equation} \rho = \frac{e^{-\beta H}}{Z}\end{equation}

Let us look at the whole issue from a different angle. Suppose I have
quantum system of several particles. Let us assume that the total Hamiltonian is independent of time. Let us assume that the probability of system being in energy state \(E_n\) has some initial value \(p_n\). Then under quantum evolution, the probability of energy being \(E_n\) will remain
constant. These probabilities will never evolve to that given by Boltzmann distribution.

Thus to me it seems that for a statistical system of large number of particles, it is not a correct model the system as an ensemble of particles and to assume that the Hamiltonian of the system is independent of time.

It thus seems that there are no stationary states for a real system with a large
number, \(10^{23}\), of particles.