Solution of Lindblad equation when the dissipative term commutes with Hamiltonian parts Let us define Liouvillian superoperator that has time-dependnet Hamiltonian parts as $\mathcal{L} = -\text{ad}_{H(t)} + L_D$. If we assume $[{L}_D, -i\text{ad}_{H(t)}] = 0$, then the total time propagator becomes
$\mathcal{T}\exp\Big(\int_0^T dt (L_D + -i\text{ad}_{H(t)})\Big) = e^{L_DT} \mathcal{T}\exp(\int_0^T dt (-i\text{ad}_{H(t)}))$.
The solution of Lindblad equation $(\dot{\rho} = \mathcal{L}\rho)$ is therefore $\rho(t) = e^{L_DT} \mathcal{T}\exp(\int_0^T dt (-i\text{ad}_{H(t)}))\rho = e^{L_DT} (U\rho U^\dagger)$ for some unitary $U$ that describes the time-dependent Hamiltonian part.
My question: Is $e^{L_DT} (U\rho U^\dagger) = U e^{L_DT}(\rho) U^\dagger$?
Update: Here is my guess: According to this post, we have $e^{-i[H,\cdot]t}= e^{-iHt} \cdot e^{iHt}$ so that $\rho(t) = \mathcal{T}\exp(\int_0^T dt (-i\text{ad}_{H(t)})) (e^{L_DT}\rho) = U(e^{L_DT}\rho)U^\dagger$. So I suspect $U(e^{L_DT}\rho)U^\dagger = e^{L_DT}(U\rho U^\dagger)$
 A: Yes. You note that the solution is "$\rho(t) = e^{L_DT} \mathcal{T}\exp(\int_0^T dt (-i\text{ad}_{H(t)}))\rho = e^{L_DT} (U\rho U^\dagger)$ for some unitary $U$ that describes the time-dependent Hamiltonian part." To show that $\rho (t) = U e^{L_DT} \rho U^\dagger$ as well, it's enough to show that the two superoperators $A \equiv e^{L_DT}$ and $B \equiv \mathcal{T}\exp(\int_0^T dt (-i\text{ad}_{H(t)})) = U \cdot U^\dagger$ commute, since then:
$$\rho (t) = A B \rho = B A \rho  =  U e^{L_DT} \rho U^\dagger$$
To show this, we just have to remember the physical meaning of the time-ordered exponential $B \equiv \mathcal{T}\exp(\int_0^T dt (-i\text{ad}_{H(t)}))$. This exponential generates the time evolution of $\rho$ under the Hamiltonian $H(t)$ over the interval $t=0$ to $t=T$. We can think of this time evolution as split into $N$ small (infinitesimal in the limit $N \to \infty$) timesteps of duration $\Delta t = T/N$. If we take $\Delta t$ small enough so that $H(t)$ is essentially constant over each timestep, then the evolution under $H(t)$ over each infinitesimal timestep will be generated by the superoperators $\exp( \Delta t (-i\text{ad}_{H(t_1)}))$, $\exp( \Delta t (-i\text{ad}_{H(t_2)}))$, $\exp( \Delta t (-i\text{ad}_{H(t_3)}))$... $\exp( \Delta t (-i\text{ad}_{H(t_N)}))$ , where $t_1 = 0$, $t_2 = \Delta t$, $t_3 = 2 \Delta t$... $t_N = (N-1) \Delta t$ are the beginnings of each time interval. Therefore, the total time evolution superoperator $B$ can be reexpressed as (in the limit $N \to \infty$):
$$B = \lim_{N \to \infty} \left[ \exp( \Delta t (-i\text{ad}_{H(t_N)}))... \exp( \Delta t (-i\text{ad}_{H(t_2)}))\exp( \Delta t (-i\text{ad}_{H(t_1)}))  \right]$$
But since $A$ commutes with each factor in this product of exponentials, it commutes with $B$ itself. This is what we needed to show.
