2
$\begingroup$

The Galileo Algebra is discussed in, for example, the wikipedia article Representation theory of the Galilean group. In that article, we can see that, for example, $$ [E,P^i]=0 $$ which means that translations commute, as one would expect. My question is, does the relation above imply $[H,P^i]=0$?

I would say that the answer is "no", as in general $$ \dot P^i\sim [H,P^i]\neq 0 $$ but I'm not sure how to make sense out of $[E,P^i]=0$ if $[H,P^i]\neq 0$. Is $E\neq H$? or maybe $P$ is not the canonical momentum, but some other operator instead?

Improve this question
$\endgroup$
3
$\begingroup$

$E$ is indeed the Hamiltonian (it says so in the Wikipedia page you linked), and $P$ commutes with it. This is because we're considering that space translation is a symmetry of our system, and for this to happen there must be no external forces. In such a situation, $\dot{P} = 0$ is correct.

Improve this answer
$\endgroup$
  • $\begingroup$ to be more explicit: here $P$ is the total (or centre-of-mass) momentum, and is therefore conserved. $\endgroup$ – AccidentalFourierTransform Aug 6 '17 at 9:18

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.