Focussing on classical mechanics of a point particle, WLOG since it captures the relevant information for field theory and generalises to the quantum case, how do we show -- in general -- that conserved charges generate the associated transformations of the variables?
Start with infinitesimal transformations $x(t) \rightarrow x(t) + \epsilon \xi(x)$ which, at least for a potential $V(x)$ would seem to imply $p(t) \rightarrow p + \epsilon m \dot{\xi}(x) = p + \epsilon p \cdot \nabla \xi(x) + \epsilon m \partial_{t} \xi$. Then in one-dimension (for simplicity) the conserved charge associated with this symmetry would be $$ \epsilon Q = \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x - \epsilon f $$ where $f$ is the function with which $\mathcal{L} \rightarrow \mathcal{L} + \epsilon\frac{d}{dt}f$ changes by a total derivative.
Question 1: f will in general depend upon $x$ and $\dot{x}$ but am I supposed to rewrite this in terms of $x$ and $p$?
Continuing, let's write $ Q= p\xi(x) - f$ and consider whether this generates the transformation through Poisson brackets. We'd like $$\delta x = \epsilon \{x, Q\} ~~~\textrm{and}~~~ \delta p = \epsilon \{p, Q\} $$ Using the form of Q we get $$\{x, Q\} = \xi - \frac{\partial f}{\partial p}~~~\textrm{and}~~~ \{p, Q\} = -p \frac{\partial \xi}{\partial x} + \frac{\partial f}{\partial x}.$$ Question 2: To reproduce the transformations of the variables this would seem to require two conditions $$\frac{\partial f}{\partial p} = 0 ~~~\textrm{and}~~~ \frac{\partial f}{\partial x} = m\frac{\partial \xi}{\partial t} + 2p \frac{\partial \xi}{\partial x}.$$ Are these conditions automatically satisfied for some reason? Must they be imposed for consistency? What happens if they're not?
