A question on Liouville's theorem and time-dependence of the Hamiltonian

The condition of equilibrium in statistical mechanics is $$\frac{\partial \rho}{\partial t}=0$$ where $$\rho$$ is the phase space density. By virtue of Liouville's theorem, this is equivalent to the statement $$\{\rho,H\}=0$$. Mathematically, does this mean that $$H$$ must also be explicitly time-independent like $$\rho$$ is?

• Also be careful with explicit vs. total time dependence... – Qmechanic Feb 14 '20 at 17:15
• @ValterMoretti Thanks. So it does not mean that H have to be time-independent. – mithusengupta123 Feb 14 '20 at 17:17
• Yes, see my explicit answer. – Valter Moretti Feb 14 '20 at 17:24

If $$H$$ is time independent and satisfies $$\{\rho, H\}=0$$, also $$H(t)=f(t)H+ g(t)$$ does for every pairs of smooth maps $$f,g$$. Hence the answer is negative. $$H$$ can be time dependent.
• Does the derivation of Liouville's theorem hold for a general $H$ with explicit $t$- dependence? – mithusengupta123 Feb 14 '20 at 17:28
• Yes, in the proof of the Liouville theorem it is not required that $H$ does not explicitly depend on time. – Valter Moretti Feb 14 '20 at 17:31