Let $M$ be a linear operator on a finite-dimensional complex Hilbert space (thus, $M$ is just a matrix). Assume that $\text{Tr}M = 1$, $M$ is self-adjoint and $M \ge 0$ (positive semi-definite).
In quantum mechanics, one postulates that every such $M$ might depend on $t$, in which case it satisfies the following ODE: $$\frac{dM(t)}{dt} = -i[H, M] \tag{1} \label{1}$$ where $H$ is a fixed self-adjoint operator (the Hamiltonian operator) and $[\cdot, \cdot]$ is the commutator $[A,B] := AB-BA$.
I have two questions about this equation.
First Question: I am considering $M$ as a bounded linear operator on an underlying Hilbert space with norm operator. A natural way to define $dM(t)/dt$ is as follows. Given $\varepsilon > 0$, there exists $\delta > 0$ and an operator $L = L(t)$ such that if $0 < |h| < \delta$ then $$\bigg{\|} \frac{M(t+h)-M(t)}{h}-L(t)\bigg{\|} = \bigg{\|} \frac{M(h)}{h}-L(t)\bigg{\|} < \varepsilon$$ in which case $L(t) := dM(t)/dt$.
On the other hand, we could interpret (\ref{1}) as a collection of ODEs given by the entries of $M$, that is, if $M_{ij}$ is the $ij$-th entry of $M$ then (\ref{1}) means: $$\frac{dM_{ij}(t)}{dt} = -i(H_{ij}M_{ij}-M_{ij}H_{ij}) \tag{2}\label{2}$$ Are these two interpretations equivalent?
Second Question: If $M$ is fixed and we define $M(t) := e^{-iHt}Me^{iHt}$, we see that $M(t)$ satisfies (\ref{1}) with initial condition $M(0) = M$. Is it a unique solution? If we interpret (\ref{1}) as being (\ref{2}), then the answer seems to be a yes. However, I don't know any uniqueness results if (\ref{1}) is treated as an ODE for operators (or elements of a Banach space).