Time-dependence of density operator in quantum statistical mechanics I'm struggling to understand a couple of textbook explanations relating to the density operator in quantum statistical mechanics.
Firstly, in Huang's book "Statistical Mechanics" it says that "The density operator (rho) contains all the information about an ensemble. It is independent of time if it commutes with the Hamiltonian of the system and if the Hamiltonian is independent of time."
Now I can see why the first part is true, as if rho commutes with the Hamiltonian, then the commutator [H,rho] will be zero, and hence so will i.hbar.(partial)d(rho)/(partial)dt in the quantum Liouville equation of motion for rho. But why do we need the second part? Does the Hamiltonian need to be time-independent in order for it to always commute with rho?
Secondly, in Pathria and Beale's book "Statistical Mechanics" it says that "If the given system is known to be in a state of equilibrium, the corresponding ensemble must be stationary... [the quantum Liouville equation tells us] for this to be the case (i) the density operator rho must be an explicit function of the Hamiltonian operator H (for then the two operators will necessarily commute) and (ii) the Hamiltonian must not depend explicitly on time."
Can anyone explain why rho must be an explicit function of H for it to commute with H, as this is not at all clear to me? And (ii) is the same unexplained statement made in Huang's book.
 A: Density operator in Dirac-Ket formalism
$$\hat\rho = \sum_j w_j |\Psi_j\rangle \langle\Psi_j|$$
So the time dependence will be
$$i\hbar\frac{d\hat\rho}{dt} = \sum_j \left( \hat H w_j|\Psi_j\rangle \langle\Psi_j| - w_j|\Psi_j\rangle \langle\Psi_j| \hat H + i\hbar\frac{dw_j}{dt} \right)$$
or
$$i\hbar\frac{d\hat\rho}{dt} = [\hat H, \hat\rho ]$$
since total weight is conserved $\sum_jw_j=1$.
Therefore, the density operator is independent of time $d\hat\rho/dt=0$ if the Hamiltonian commutes with it $[\hat H, \hat\rho ]=0$. Time-independence of the Hamiltonian is not required.
Equilibrium requires $d\hat\rho/dt=0$ which implies density and Hamiltonian must commute; an explicit function of the Hamiltonian is not needed. In fact 
$$\hat\rho = 1/\Omega \sum_j |\Psi_j\rangle \langle\Psi_j|$$
with $\Omega$ the total number of quantum states is just the expression for the microcanonical density operator.
A: In the Schrödinger picture $\rho (t)$ satisfies
\begin{equation*}
\partial _{t}\rho (t)=-i[H(t),\rho (t)]=-iL(t)\rho (t).
\end{equation*}
In case $H$ and hence $L$ is time-independent the solution is
\begin{equation*}
\rho (t)=\exp [-iHt]\rho (0)\exp [+iHt]=\exp [-iLt]\rho (0).
\end{equation*}
If $H$ depends on $t$, $H=H(t)$, it depends. If $H(t_{1})$ and $H(t_{2})$
commute for $t_{1}\neq t_{2}$ then
\begin{equation*}
\rho (t)=\exp \left[-i\int_{0}^{t}dsL(s)\right]\rho (0),
\end{equation*}
but if $[H(t_{1}),H(t_{2})]\neq 0$ the exponential changes into a
time-ordered one
\begin{equation*}
\rho (t)=\exp \left[-iT\int_{0}^{t}dsL(s)\right]\rho (0)
\end{equation*}
which is far more complicated. Returning to the first case
\begin{equation*}
\lbrack H(t_{1}),H(t_{2})]=0,\;\forall t_{1},t_{2}
\end{equation*}
this means that both have common orthonormal eigenvectors $u_{n}$,
\begin{equation*}
H(t)=\sum_{n}\lambda _{n}(t)|u_{n}\rangle\langle u_{n}|
\end{equation*}
Suppose that $\rho (0)$ commutes with $H(0)$. Then also
\begin{equation*}
\rho (0)=\sum_{n}\rho _{n}(0)|u_{n}\rangle\langle u_{n}|
\end{equation*}
and
\begin{eqnarray*}
\lbrack \rho (0),H(t)] &=&[\sum_{m}\rho
_{m}(0)|u_{m}\rangle\langle u_{m}|,\sum_{n}\lambda _{n}(t)|u_{n} \rangle\langle u_{n}|] \\
&=&\sum_{m}\sum_{n}\rho _{m}(0)\lambda
_{n}(t)\{|u_{m} \rangle\langle u_{m}|u_{n} \rangle\langle u_{n}|-|u_{n} \rangle\langle u_{n}|u_{m} \rangle\langle u_{m}|\} \\
&=&\sum_{m}\sum_{n}\rho _{m}(0)\lambda _{n}(t)\delta
_{mn}\{|u_{m} \rangle\langle u_{n}|-|u_{n} \rangle\langle u_{m}|\}=0
\end{eqnarray*}
so $\rho (0)$ commutes with $H(t)$ for all $t$. But then
\begin{equation*}
\rho (t)=\exp \left[-i\int_{0}^{t}dsL(s)\right]\rho (0)=\rho (0)
\end{equation*}
since $L(s)\rho (0)=0$ for all $s$.
Thus the point (ii) of your book is too strong. Note further that if $H$ has continuous spectrum then $\rho$ cannot commute with $H$ since $\rho$ being a density operator, has discrete spectrum.
