 Solution

Strictly, the normalized functions form the unit sphere rather than a vector space; let $V=L^2(\mathbb{R})$ and restrict to normalized states when interpreting wavefunctions. The group law givesTranslation preserves Lebesgue measure and both exponential factors have unit modulus, so $D(g)$ preserves the inner product and is a unitary representation. This is the Schrödinger representation of the Heisenberg group: $q$ translates position, $r$ translates momentum, and $s$ contributes the physically irrelevant overall phase.