How does $[H,H] = 0$ imply $dH/dt = 0$? I have a conceptual question about how $[H,H] = 0$ implies $dH/dt = 0$. Does this relation work for both time-dependent and -independent Hamiltonians, or a general observable $H$?
 A: $\newcommand{\bra}[1]{\langle #1 \rvert}$
$\newcommand{\ket}[1]{\lvert #1 \rangle}$

I have a conceptual question about how $[H,H] = 0$ implies $dH/dt = 0$. Does this relation work for both time-dependent and independent Hamiltonian, or general observable $H$?

The Hamiltonian has a special place in Quantum Mechanics because it determines the time evolution of the wave function (here denoted by $\ket{\Psi_S}$):
$$
i\hbar\frac{\partial \ket{\Psi_S}}{\partial t} = H\ket{\Psi_S}\;.\tag{1}
$$
When the Hamiltonian does not depend explicitly on time, we can integrate the above equation and obtain:
$$
|\Psi_S(t)\rangle = e^{-i\hat H t/\hbar}|\Psi_S(0)\rangle
$$
In the Schrodinger picture, operators do not evolve with time (unless there is explicit time dependence). Rather, wavefunctions evolve with time per Eq. (1) above.
In the Heisenberg picture, the time-dependence is moved to the operators and we have:
$$
A_H(t) = e^{iHt/\hbar}A_Se^{-iHt/\hbar}\;,
$$
where $A$ is any operator, and where the $H$ subscript means Heisenberg picture and the $S$ subscript means Schrodinger picture.
Therefore, generally:
$$
\frac{dA_H}{dt} = i[H,A_H]/\hbar +{\left(\frac{\partial A_S}{\partial t}\right)}_H\;,\tag{2}
$$
where the last term accounts for any explicit time dependence.
BTW, we also have:
$$
H_H = H_S
$$
so we don't need to worry about the subscript on $H$.
The general Eq. (2) also holds when the operator is the Hamiltonian:
$$
\frac{dH}{dt} = i[H,H]/\hbar +\frac{\partial H}{\partial t}=0\;,
$$
since $[H,H]=0$, because any operator commutes with itself, and since $\partial H/\partial t = 0$, because we already specified that the Hamiltonian has no explicit time dependence.
