Hamiltonian system and exponential map -- backward

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Consider a linear ODE $$ \dot x(t) = A\,x(t). $$ A solution is $x(t)=\exp(t\,A)$ where $\exp$ is defined by $$ \exp(A) = \sum_{n=1}^n\frac{A^n}{n!}. $$

Consider a Hamiltonian system $$ \dot x(t) = \frac{\partial H}{\partial p}(x(t),p(t)), \quad \dot p(t) = -\frac{\partial H}{\partial x}(x(t),p(t)). $$ I found that solution is often denoted as $z(t) = (x(t),p(t)) = \exp(t\vec{H}(z(t))$ where $\vec{H}$ is the Hamiltonian vector field.

Do you have any reference why in this case, we still use the exponential notation? Is there a definition of such exponential like in the linear case or it's just a notation?


Motivation: I want to write the solution of this Hamiltonian system which was integrated backwards between $T$ and $0$ where $T>0$. This exponential notation suggests that this solution can be written as $z(t)=\exp(-t\vec{H}(z(t))$.

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To every one-parameter group of diffeomorphism (the solution of Hamilton equations) you can associate a unitary operator $U_t$ that advance time. This point of view is sometime called Koopmanism.

If $g^t$ is your dynamical flow

$$ U_t f = f \circ g^t $$

For a function $f$. This unitary is again a one parameter group and by Stone's theorem you can associate to it a generator $G$ such that $$ U_t = e^{i t G} $$

In this way $G$ is a self-adjoint operator. I believe in your paper $\vec{H}=G$.