Mori–Zwanzig formalism

Remarks

The decomposition follows from the Dyson identity $e^{tL} = e^{tQL} + \int_0^t e^{(t - s)L} P L e^{sQL}\, ds$ applied to $QLA$, after splitting $\tfrac{d}{dt}A(t) = e^{tL}PLA + e^{tL}QLA$. The noise satisfies $P F(t) = 0$ for all $t$: $QLA$ lies in the range of $Q$, and $QL$ maps $V$ into that range, so $e^{tQL}$ preserves it. Example: let $V = \R^2$ with basis $e_1, e_2$, let $L = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, let $P$ be the projection onto $\R e_1$ along $e_2$, and let $A = e_1$. Then $LA = -e_2$, so $PLA = 0$ and the Markovian term vanishes. Next $QLe_2 = 0$, so $F(t) = -e_2$ for all $t$, and $PLF(s) = -e_1$, so the memory term is $-\int_0^t A(u)\, du$. The formalism gives $\tfrac{d}{dt}A(t) = -\int_0^t A(u)\, du - e_2$, and indeed $A(t) = (\cos t, -\sin t)$ satisfies it. Non-example: take the non-idempotent map $P = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$ in the same example. The same formula can be written, but $Q = I - P$ is not a projection. Here $PF(t) = 2t e_1 + O(t^2) \ne 0$ for small $t > 0$, so the noise is no longer orthogonal to the relevant variables. In the usual applications $V$ is a Hilbert space of phase-space functions and $L$ is the Liouville operator of a Hamiltonian system. Then $P$ is the orthogonal projection onto the chosen variables in the equilibrium inner product (Mori), or a conditional expectation (Zwanzig).