About the C*-algebra of the Schrodinger representation of the Weyl C*-algebra We start with the Weyl C*-algebra $\mathcal{W}$ for a finite dimensional symplectic space and we consider the irreducible Schrodinger representation $\pi:\mathcal{W}\rightarrow \mathcal{B}(\mathcal{H})$ where $\mathcal{H}=L^2(\mathbb{R}^n)$. Since such representation is irreducible, the von Neumann algebra generated by such represenation is $\pi(\mathcal{W})''=\mathcal{B}(\mathcal{H})$. The question now is:
¿Is the C*-algebra generated by such representation $\pi(\mathcal{W})=\mathcal{B}(\mathcal{H})$, or is it strictly smaller, i.e. $\pi(\mathcal{W}) \subsetneq \mathcal{B}(\mathcal{H})$? In such case, ¿Is there any useful characterization for such C*-algebra $\pi(\mathcal{W}) \subsetneq \mathcal{B}(\mathcal{H})$?

$\pi(\mathcal{W})$ is strictly smaller than $B(H)$.
Since the Weyl algebra $\mathcal{W}$ is simple, every nontrivial representation is faithful. Therefore $\pi:\mathcal{W}\rightarrow\pi(\mathcal{W})\subseteq B(H)$ is a bijective *-homomorphism (surjective to its image), hence $\pi(\mathcal{W})\cong\mathcal{W}$ as C${}^\ast$-algebras. I hope this answers your second question.
As for your first, $\pi(\mathcal{W})$ does not contain the compact operators (because it's simple), so $\pi(\mathcal{W})\subsetneq B(H)$.
For more info, see Photons In Fock Space And Beyond by Honegger and Rieckers.
