# Time evolution of mixed state?

Suppose I have a quantum statistical mechanics system in the grand-canonical ensemble. It is given by some Hamiltonian $$H = H_{0} + V$$, where $$H_{0}$$ is the free part and $$V$$ an interaction. The state of the system is described by a mixed state (i.e. density matrix) $$\rho_{\beta,\mu} := e^{-\beta H - \mu N}$$, with $$N$$ the number operator. The average of an observable (self-adjoint operator) $$A$$ in this state is given by: $$\langle A\rangle_{\beta,\mu} := \frac{\operatorname{Tr}(\rho_{\beta,\mu}A)}{\operatorname{Tr}(\rho_{\beta,\mu})}$$

I am confused about the time evolution of this system. In the Heisenberg picture, observables evolve in time and states do not. In this case, I would consider $$A(t) = e^{-itH}Ae^{itH}$$. What about $$\rho_{\beta,\mu}$$? It is an operator, but also a state. Should the time average now be: $$\langle A(t)\rangle_{\beta,\mu} = \frac{\operatorname{Tr}(\rho_{\beta,\mu}A(t))}{\operatorname{Tr}(\rho_{\beta,\mu})}$$ with $$\rho_{\beta,\mu})$$ time-independent or: $$\langle A(t)\rangle_{\beta,\mu} = \frac{\operatorname{Tr}(\rho_{\beta,\mu}(t)A)}{\operatorname{Tr}(\rho_{\beta,\mu}(t))}$$ with time-dependent $$\rho_{\beta,\mu})(t) = e^{-itH}\rho_{\beta,\mu}e^{itH}$$ instead?

• Hint: von Neumann equation. Commented Jan 28 at 12:41

The first option is the correct one in the Heisenberg picture. While $$\rho$$ can be regarded as an operator, it is a state, and states do not evolve in the Heisenberg picture. This may seem a bit weird when you are dealing with mixed states, but the important thing to keep in mind is that the density matrix is not a regular operator, since it has a very different physical meaning.