Representations of Lie group symmetries on Hilbert space

Question

I have some troubles understanding Hilbert space representations for (eg.) the standard free quantum particle.

On the one hand, we can represent the Heisenberg algebra $[X_i,P_j] = \mathrm{i} \delta_{ij}$ on the space of square integrable functions on, say, $\mathbb{R}^3$, with the $X$ operator represented as multiplication and $P$ operator as $\mathrm{i}$ 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 the Heisenberg Lie algebra and the Galilean Lie algebra (in particular because $P$ is both the momentum in $[X,P] = \mathrm{i}$, 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 comes 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 $[X,P]= \mathrm{i} \:$?

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 $X = -\frac{1}{m}G + \frac{t}{m} P$; and we get $[X,P] = -\frac{1}{m}[G,P]$, which is zero unless we add a central charge to the galilean group (the mass), in which case we somehow derive $[X,P]=\mathrm{i}$.

The other way around, we have some “god given” $X$ and $P$ with $[X,P]=\mathrm{i}$, we choose $X$ to represent space, and then construct the generators $P$ for translations, $L = X \wedge P$, $G = t P - m X$, $\,H=\frac{1}{2m} P^2$ 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.

Answer 1

The unitary representation of Galileian group already includes a representation of Weyl-Heisenberg group. The boost generator $K$ and the generator of translations $P$ satisfy $[K,P]= imI$ where $m$ is the mass of the system. Therefore, the subgroup whose generators are $m^{-1}K, P, I$ is the wanted unitary representation of Weyl-Heisenberg group. As a matter of fact $X:= m^{-1}K$ is the position observable (the position of the center of mass for a composite system).

If you deal with QM in a finite interval, there is no (strongly continuous) representation of Weyl-Heisenberg group. This is because, due to Stone-von Neumann theorem, these representations can only be constructed along the whole real line. In practice, looking at the representation of Lie algebra, the formal operator $P$ on a segment is not (essentially) self-adjoint or, it is but on a domain where $[X,P]= iI$ does not hold.