maximum of the probability distribution $T_{\rm max}$. To make the expressions as simple as possible, let us assume that $T_{\rm max}(t)=t$, i.e. the peak of the probability distribution is simply at $t$. More general dependences can of course be considered, altering the formulas minimally (for a more complete treatment see [@njp]). We will also assume that we can approximate $f$ reasonably well by a Dirac delta, namely, $$f(T-t)=\delta(T-t)+ a(T)\delta'(T-t)+b(T)\delta''(T-t)+\ldots,$$ where the first term has a unit coefficient so the integral of the probability is unit and we assume $b(T)>0$ so it represents extra width with respect to the Dirac delta.

We now consider the evolution of the density matrix, $$\rho(T)=\int_{-\infty}^\infty dt\,\rho_{\rm sys}(t) {\cal P}_t(T)=
\int_{-\infty}^\infty dt\,\rho_{\rm sys}(t) f(T-t)$$ and associating a Hamiltonian with the evolution operator $U(t)=\exp(iHt)$, we get, $$\rho(T)=\rho_{\rm sys}(T)+a(T) [H,\rho_{\rm sys}(T)]-b(T) [H,[H,\rho_{\rm sys}(t)]],$$ and we notice that there would be terms involving further commutators if we had kept further terms in the expansion of $f(T-t)$ in terms of the Dirac deltas.

We can now consider the time derivative of this expression, and get, $${\partial \rho(T)\over \partial T} =i\left(-1+{\partial a(T)\over \partial T}\right) 
[H,\rho(T)] +\left(a(T)-{\partial b(T)\over \partial T}\right) [H,[H,\rho(T)]].$$ If we had considered a symmetric distribution (it is natural to 