When a time local quantum master equation is completely positive? I know that a time-local quantum master equation can be put in a "Lindblad-like" form, which is a Lindblad equation whose coefficients can be negative and time-dependent. If all the coefficients are nonnegative, the evolution is completely positive (since we recover the Lindblad form). But what about the converse? What is the most general condition on these coefficients in order to guarantee complete positivity? 
 A: In all generality, this is an open question.
The "Lindblad-like" Master equation you are referring to is
$$ \dot \rho_t = \hat L_t\, \rho_t = -\mathrm i\, [H_t, \rho_t] + \sum_\mu \gamma^\mu_t \left( L^\mu_t \rho_t L^{\mu\dagger}_t - \frac 1 2 \left\{ L^{\mu\dagger}_t L^\mu_t, \rho_t \right\} \right) $$
with a system Hamiltonian $H$, Lindblad operators $L^\mu$ and (not necessarily positive) rates $\gamma^\mu$.
All of these quantities can a priori be time-dependent.
The Master equation gives then rise to dynamical maps
$$ \hat \Phi_{t,s} = \mathcal T \exp\left( \int_s^t \hat L_\tau \, \mathrm d\tau \right) $$
(where $\mathcal T$ denotes time ordering) such that $\rho_t = \hat \Phi_{t,s}\, \rho_s$.
I am aware of the following results (see Breuer, HP et al: Non-Markovian dynamics in open quantum systems, Rev. Mod. Phys. 88, 021002 (2016), especially subsection II.A.2):


*

*The dynamical maps form a semigroup if and only if $H$, $L^\mu$ and $\gamma^\mu$ are time-independent.
Let $\hat \Phi_t = \hat \Phi_{t,0}$, semigroup means that
$$ \hat \Phi_t \hat \Phi_s = \hat \Phi_{t+s} . $$

*If the dynamical maps form a semigroup, they are completely positive if and only if the time-independent rates are all positive,
$$ \gamma^\mu \geq 0 . $$

*If the generator $\hat L_t$ is time-dependent, we know that the dynamical maps are CP-divisible if and only if
$$ \gamma^\mu_t \geq 0 \qquad \text{for all times } t . \tag{1} $$
CP-divisible means that all dynamical maps $\hat \Phi_{t,s}$ are completely positive.

*The open question is under which conditions $\hat \Phi_t$ is completely positive for all $t$. Obviously, (1) is sufficient, but a necessary condition is not known.

