Consistency of Lindblad-type operator evolution equations One frequently comes across Lindblad-type operator evolution equations in the Heisenberg picture of the form
$$
\frac{\mathrm{d}}{\mathrm{d}t} A =\mathcal{L}^{\dagger}(A),
$$
where the adjoint Liouvillian reads
$$
\mathcal{L}^{\dagger}(A)= \frac{i}{\hbar}[H,A]+\sum_k \left( L_k^{\dagger} A L_k - \frac{1}{2}\{L_k^{\dagger}L_k,  A \}\right).
$$
The technique for the transition from the Schrödinger picture Lindblad equation (for the reduced density operator) to the Heisenberg picture (for an operator $A$) is described in several books (e.g., Breuer & Petruccione) or for example here:
https://physics.stackexchange.com/a/280406/321332
I emphasize again that the above equation is an operator-valued evolution equation (as opposed to an equation for the mean $\langle A \rangle$).
My question is the following:
Is this (dissipative) Heisenberg equation valid at all from a quantum mechanical point of view? The consistency requirements for operator-valued evolution equations are quite strict, since in particular canonical commutator relations must be preserved. Typically, this requires the introduction of a noise operator that guarantees all these things via a fluctuation-dissipation relation (see Scully & Zubairy, Zoller & Gardiner etc.). The above equation, however, appears to apply only to expectation values and should not be stated as an operator evolution equation ... right?
Or have I missed something? I appreciate any comments ...
 A: The adjoint Liouvillian generates a perfectly acceptable operator evolution from a quantum-mechanical point of view. However, the Leibniz rule no longer  applies with respect to the operator product, i.e. it is not generally true that
$$\mathcal{L}^\dagger(\hat{A}\hat{B}) = \mathcal{L}^\dagger(\hat{A})\hat{B}+\hat{A}\mathcal{L}^\dagger(\hat{B}),$$
unless the Liouvillian is of the purely Hamiltonian form $\mathcal{L}^\dagger = ({\rm i}/\hbar)[\hat{H},\bullet]$. Therefore, one cannot assume that a product initial condition $\hat{C}(0) = \hat{A}\hat{B}$ will preserve its product form over time, i.e.
$$\hat{C}(t) = {\rm e}^{\mathcal{L}^\dagger t}\left( \hat{A} \hat{B}\right) \neq \hat{A}(t) \hat{B}(t),$$
in contrast to the case of unitary evolution. With this understood, there is no conflict with the canonical commutation relations $[\hat{x},\hat{p}] = {\rm i}\hbar\mathbb{1}$. Indeed,
$${\rm e}^{\mathcal{L}^\dagger t}\left([\hat{x},\hat{p}]\right) = {\rm e}^{\mathcal{L}^\dagger t}\left( {\rm i}\hbar\mathbb{1}\right) = {\rm i}\hbar\mathbb{1},$$
since ${\rm e}^{\mathcal{L}^\dagger t}$ is a unital map ($\mathcal{L}^\dagger (\mathbb{1})=0$) whenever ${\rm e}^{\mathcal{L} t}$ is trace-preserving.
While violation of the standard Leibniz rule may appear somewhat distasteful, it is a common feature of Markovian descriptions of both classical (e.g. Ito calculus) and quantum dissipative systems, as described at length in textbooks by Gardiner, Zoller and others.
