3
$\begingroup$

I've recently noticed that the Poisson brackets of the three dimensional angular momentum $$\{L_i,L_j\}$$ in classical mechanics follow the same commutator relations as the standard basis of the Lie algebra $\mathfrak{so}(3)$. This means that these to Lie algebras are isomorphic.

Also $\mathfrak{so}(3)$ is the Lie algebra of the Lie group ${\rm SO}(3)$, which is the group of three dimensional rotations.

This seems very geometric to me. My question is therefore: Is there a geometric way to interpret Poisson brackets (of angular momenta or in general)?

$\endgroup$
0
1
$\begingroup$

Nice observation! This is the geometric picture: in this example you have a Hamiltonian action of the Lie group $SO(3)$ on the phase space $T^\star \mathbb R^3$ associated to a 3D particle. The angular momenta $L_i$ collectively define the moment map for this map. See the wikipedia article on moment maps and in particular the second paragraph under "Examples of momentum maps" which treats precisely this example.

The general case of course is that of a Hamiltonian action of a Lie group on the phase space under consideration.

$\endgroup$
6
$\begingroup$
  1. The $j$th component of the angular momentum $L_j$ is the Noether charge for 3D rotation around the $j$th axis. It satisfies the $so(3)$ Lie algebra $$\begin{align}\{L_j,L_k\}~=~& \epsilon_{jk\ell}L_{\ell},\cr j,k,\ell~\in~&\{1,2,3\}.\end{align}\tag{1}$$ where the curly bracket $\{\cdot,\cdot\}$ denotes the Poisson bracket.

  2. The corresponding Hamiltonian vector field $X_{L_j}\equiv\{L_j,\cdot\}$ is the generator of 3D rotations around the $j$th axis, cf. this Phys.SE post. It satisfies the $so(3)$ Lie algebra $$\begin{align}[X_{L_j},X_{L_k}]~\stackrel{\text{Jac. id.}}{=}&~X_{\{L_j,L_k\}}\cr ~\stackrel{(1)}{=}~~& X_{\epsilon_{jk\ell}L_{\ell}}~\cr \stackrel{\text{linearity}}{=}&~ \epsilon_{jk\ell} X_{L_{\ell}}.\end{align}\tag{2}$$

  3. Under quantization $L_j\to \hat{L}_j$ from functions to operators, the Poisson bracket (1) is replaced with commutators (divided by $i\hbar$): $$[\hat{L}_j,\hat{L}_k]~=~ i\hbar\epsilon_{jk\ell}\hat{L}_{\ell},\tag{3}$$ cf. this Phys.SE post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.