3
$\begingroup$

In Quantum Mechanics (QM), angular momentum turn out to be the generator of rotational symmetry. This is trivial to see because in QM, angular momenta are defined by the commutation relations $$[J_j,J_k]=i\hbar\epsilon_{jkl}J_l.$$ One immediately recognises these as the generators of the rotation group. But in classical mechanics, angular momenta $L_i$ are numbers, or at best, functions of $x_j$ and $p_k$ as $L_i=\epsilon_{ijk}x_j p_k$. Can they be called the generators of some symmetry group because generators are usually differential operators or matrices?

$\endgroup$
4
$\begingroup$

As you mention, they're not "just numbers" - they are functions of the coordinates and momenta. And, as such, they can indeed be used as generators of some symmetry group via the usual tool for the job in hamiltonian mechanics: the Poisson bracket.

Here, you won't be much surprised to learn that the mutual relationships between the angular momentum components are $$ \{L_i,L_j\}=\epsilon_{ijk}L_k, $$ where the Poisson bracket is defined as $$ \{f,g\} = \sum_i \left[ \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial x_i} \right], $$ and the group action generated by the angular momentum on a function $f$ works as $$ f \mapsto f+\{L_i,f\} $$ in its infinitesimal version, and as $$ f \mapsto\exp\left(\theta\{L_i,\cdot\}\right) f $$ for a finite angle $\theta$.

And, of course, the group that they generate is simply $\rm SO(3)$, acting on the space of real functions on phase space.

$\endgroup$
  • $\begingroup$ Thanks! I didn't know that such group actions also occur in classical mechanics. What does the notation $\{L_i,\cdot\}$ stand for? @EmilioPisanty $\endgroup$ – SRS Feb 10 at 14:15
  • $\begingroup$ @SRS Generally, the center-dot notation implies a function, with the variable going in where the dot is: $$\{L_i,\cdot\}: f\mapsto \{L_i, f\}.$$ Here this notation is necessary to denote that we're not taking the exponential of the number-valued function $\{L_i, f\}$, but of the operator $\{L_i,\cdot\}$, so \begin{align}\exp(\theta\{L_i,\cdot\}) & = \sum_{n=0^\infty} \frac{1}{n!} \{L_i,\cdot\}^n \\ & = 1+ \theta \{L_i,\cdot\} + \frac{1}{2!}\theta^2\{L_i,\{L_i,\cdot\}\} + \frac{1}{3!} \theta^3 \{L_i,\{L_i,\{L_i,\cdot\}\}\} + \cdots .\end{align} $\endgroup$ – Emilio Pisanty Feb 10 at 14:49

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.