I have some troubles understanding Hilbert representations for (eg) the standard free quantum particle
On the one hand, we can represent Heisenberg algebra [Xi,Pj]= i delta ij on the space of square integrable functions on, say, R^3, with the X operator represented as multiplication and P operator as i times the gradient. On the other hand, we also represent the Galilean Lie group on the same Hilbert space
So my question is : is that obvious that these two representations are "compatible"? Are they any constraints that one may derive by asking that the Hilbert must be the representation of both Lie Heisenberg Lie algebra + Galilean Lie algebra (in particular because P is both the momentum in [X,P] = i, and the generator of translations of the galilean group)?
For instance, I could choose to represent QM on a finite interval; then it would break translation invariance. So which come first? Do we build the Hilbert as a representation of Heisenberg, and then impose some symmetry, or construct the Hilbert space as a representation of symmetry group and then define X via [X,P]= i ?
Thanks for any help!
EDIT : More specifically. At the algebra level, we can define the generators of the Galilean Lie group and their commutators; H for time translation, P for space translation, G for galilean boosts, and L for rotations. Then we might define a position operator X = -G/m + t P/m; and we get [X,P] = -[G,P]/m, which is zero unless we add a central charge to the galilean group (the mass), in which case we somehow derive [X,P]=i
The other way around, we have some "god given" X and P with [X,P]=i, we choose X to represent space, and then construct the generators P for translations, L= X wedge P, G = t P - m X, H=P^2/2m and we get the Lie algebra of (extended) galilean group (with the mass as a central charge).
In both cases, this seems to me insufficent, for the choice of the Hilbert space on which these operators must act has not been specified, and non trivial stuff may appear on finite intervals or semi-infinite intervals