we put forth the following identity, whose proof is a simple application of the chain rule. Here {} denotes the Poisson Brackets.

$\{- \Delta s H, x\}  = -\Delta s \frac{\partial H}{\partial x}\frac{\partial x}{\partial p_x} + \Delta s \frac{\partial H}{\partial p_x}\frac{\partial x}{\partial x} =  \Delta s \frac{dp_x}{ds}\frac{\partial x}{\partial p_x} + \Delta s \frac{dx}{ds}\frac{\partial x}{\partial x} = \Delta s \frac{dx}{ds}$.

A simple mathematical induction yields

$\{- \Delta s H, x\}^n = \{-\Delta s H, \{ -\Delta s H ,\{..., \{-\Delta s H,x\}...\}\}\} =  \Delta s^n \frac{d^nx}{ds^n}$,

and thus our Taylor expansion may be written

$x(\Delta s) = \displaystyle\sum_{n=0}^{\infty} \frac{\Delta s^n}{n!} \frac{d^nx}{ds^n}\Bigr|_{\substack{s = 0}} = \displaystyle\sum_{n=0}^{\infty} \frac{\{-\Delta s H,x\}^n}{n!}$.

We now define the Lie exponential operator: $\exp(:f:)g \equiv \sum_{n=0}^{\infty} \frac{\{f,g\}^n}{n!}$, so that\

$x(\Delta s) = \exp(-\Delta s:H:)x(0)$.

Therefore, since $x(\Delta s) = \mathcal{M}x(0)$, we have that

$$\mathcal{M} = \exp(- \Delta s:H:).$$

Though we do not prove it, we note that Lie exponential operators and their compositions are symplectic. The rest of the derivation involves manipulating Lie exponential operators. In particular, suppose that we have a Hamiltonian which may be written $H = H_1 + H_2$. Invoking the Campbell-Baker-Hausdorff (CBH) theorem we may write

$$\begin{gathered}
\mathcal{M} = \exp(-\Delta s:H_1 + H_2:) = \exp(-\frac{\Delta 