# How to understand the failure of Leibniz rule in Lindblad type Heisenberg equation?

Dual to the well-known Lindblad master equation for density matrices, the equation for operators (in the sense of Heisenberg equation) is written as $$\frac{d}{dt}\hat{A}=i[H,\; \hat{A}]+\sum_i \gamma_i\bigg(M_i^\dagger\,\hat{A}\, M_i-\frac{1}{2}\{ M_i^\dagger M_i,\;\hat{A}\}\bigg),$$ where $H$ is the Hamiltonian, $\gamma_i$ and $M_i$ are coefficients and operators that characterise the Lindblad master equation.

However, due to the presence of Lindblad operators, the above equation does not follow the Leibniz rule, that is $$\frac{d}{dt}(\hat{A}\hat{B})\neq \frac{d}{dt}(\hat{A})\,\hat{B}+\hat{A}\frac{d}{dt}(\hat{B}).$$ Actually, it is straightforward to verify that (if my calculation is correct) the left hand side minuses the right hand side is $$\frac{d}{dt}(\hat{A}\hat{B})-\frac{d}{dt}(\hat{A})\,\hat{B}-\hat{A}\frac{d}{dt}(\hat{B})=\sum_i\gamma_i\,[M_i^\dagger,\, \hat{A}]\,[\hat{B},\, M_i].$$

My question is that, what's the consequence of such breakdown of Leibniz rule? I figured out two examples:

1. Since it doesn't define a derivative, how can we integrate it to obtain $\hat{A}(t)$ from the known $\hat{A}(0)$? In other words, is the Heisenberg picture well-defined here? If not, how can we define correlations in the form of $\langle \hat{A}(t_1)\,\hat{B}(t_2) \rangle$ for an open system?

2. If we ignore the failure of defining a derivative and formally integrate it, we can get a $\hat{A}(t)$, but also meet the situation that $$AB(t)\neq A(t)\,B(t).$$ How can we understand that? For example, suppose $H=0$, $\hat{A}$ and $\hat{B}$ both anti-commute with $M$ and $M^\dagger$ (we assume there is only one Lindblad operator), we also assume that $M^\dagger M=1$. Then we have $\frac{d}{dt}\hat{A}=-2\gamma\hat{A}$ and $\frac{d}{dt}\hat{B}=-2\gamma\hat{B}$. So both operators decay exponentially. But at the same time we have $\frac{d}{dt}(\hat{A}\hat{B})=0$, i.e., their product is constant.