# Conserved charges generate transformations

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?

• – Cosmas Zachos Sep 14 '19 at 20:53
• Thank you, Cosmas. I wonder if you have any thoughts on the current question? I mean I know the statement that charges generate transformations is true but how can it be proved in this way? – lux Sep 15 '19 at 16:01
• No thoughts beyond linked and linked therein. Your time-dependent surmise appears obscure and not canonical. Perhaps the answer to this question might be helpful; perhaps not. – Cosmas Zachos Sep 15 '19 at 16:39
• – Cosmas Zachos Sep 15 '19 at 19:31