 Schrödinger representation of the Heisenberg group

Use the Heisenberg group law $(q,r,s)(q^{\prime },r^{\prime },s^{\prime })=(q+q^{\prime },r+r^{\prime },s+s^{\prime }+q{r}^{\prime })$. For a fixed real $\lambda \ne 0$, its Schrödinger representation on the Hilbert space $L^2(\mathbb{R})$ isTranslations preserve the Lebesgue measure and the phase has modulus one, so these are unitary operators. Composing two such operators introduces the phase $\lambda q{r}^{\prime }$, exactly the cross term in the Heisenberg group law. The center of a group therefore acts by ${\pi }_{\lambda }(0,0,s)=e^{i\lambda s}I$.

On smooth rapidly decreasing functions, the differentiated Lie algebra representation sends the coordinate generators to $d/dt$, multiplication by $i\lambda t$, and $i\lambda I$. Their commutator is $[d/dt,i\lambda t]=i\lambda I$. This is the representation behind the canonical commutation relation between position operator and momentum operator; their normalization fixes the central parameter.