# Infinitesimal transformations and Poisson brackets [duplicate]

I want to understand how bracket operations in general are related to symmetry and infinitesimal transformations (in hindsight of quantumfieldtheory), so I calculated an example with a particle that is moving on a circle with a generic potential. (I used simple polar coordinates in two dimensions)

$H(r,p_{r})= \frac{p^{2}_{r}}{2m}+V(r)$

$H(\phi, p_{\phi})=\frac{p^{2}_{\phi}}{2mr^{2}}+V(\phi)$

Now I know that if you take the Poisson bracket with the Hamiltonian you just get the infinitesimal transformation in time right? So

$\{r,H(r,p_{r})\}= \frac{p_{r}}{m}$

$\{\phi,H(\phi,p_{\phi})\}= \frac{p_{\phi}}{mr^{2}}$

But what if I want to do an infinitesimal transformation of the $r$ or the $\phi$ coordinates? I know that the generator of translations is just the momentum, and that the generator of rotations is angular momentum. How would I do that with the Poisson bracket in this case? And for example when I do an infinitesimal transformation with the radius $r$, what does that mean? Is it that the radius is infinitesimally transformed, or is it more like a global translation where the whole system is somehow translated? Similarly with the angle $\phi$, is it that the angle is locally changed, or is it that the "whole" system is rotated?

• possible duplicate of Understanding Poisson brackets Commented Feb 12, 2015 at 16:20
• My question is completely different. It doesn't mention quantum mechanics nor do I ask something about involution. You just read the title and thought it is similar. Commented Feb 12, 2015 at 23:17
• The other question is not about QM, but about understand what kind of information Poisson brackets encode, just like yours. I actually do not really get what you are asking, but instead of voting to close as unclear what you're asking, I wanted to supply a duplicate that perhaps clears your confusion. Commented Feb 12, 2015 at 23:30

Suppose $g$ is the generator of a certain symmetry (i.e. the generating function of an infinitesimal canonical transformation) and you are interested to know how the observable $f$ changes after the "action" of $g$. In the Hamiltonian formalism the change is found to be $$\delta f \approx \epsilon\{f,g\}$$ which can be related to the time evolution of an observable, where the generator is the Hamiltonian itself, since $$\delta f \approx\{f, H\}\delta t.$$ If you now replace the Hamiltonian $H$ with the momentum $p_r$, the above becomes $$\delta r = \epsilon\{r,p_r\} = \epsilon\cdot1,$$ and this shows that $p_r$ is translating along the radial coordinate, while it is doing nothing on the angular variable, since $$\delta\phi = \epsilon\{\phi,p_r\} = \epsilon\cdot 0.$$

• Hi, thx for your answer. Now what if I want to do it with the angular variable? Do I need angular momentum in polar coordinates? But $\epsilon \{\phi,p_{r}\}=0$ then doesn't mean that momentum is conserved, it just means that the momentum generator has nothing to say about the angular variable? Ok and conserved quantities just give 0 with the Hamiltonian as generator right? Thank you. Commented Feb 12, 2015 at 16:50
• just use $p_\phi$ and you get $\delta r = 0$ and $\delta\phi = \epsilon$ with exactly the same computations as above. Conserved quantities don't change in time, so the check is the Poisson bracket against the Hamiltonian. Commented Feb 12, 2015 at 16:52
• Ok last question. Now what if I insert the generator of rotations, for example angular momentum? That would give $\{r,L\}=0$ and $\{\phi, L\}=1$ ?? Commented Feb 12, 2015 at 22:08