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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?

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$E$ 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.

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  • $\begingroup$ 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

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