# According to the Galileo Algebra, space translations commute with time translations. Does this mean that $[\vec P,H]=0$?

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?

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