Free Gaussian wave packet evolution (${\sigma }_x(t{)}^2={\sigma }_0^2+{\hslash }^2{t}^2/(4m^2{\sigma }_0^2)$)

Let the initial Gaussian wave packet beFourier transforming, multiplying each momentum amplitude by $e^{-i{p}^{2}t/(2m\hslash )}$ and evaluating the Gaussian integral givesIts probability density is a normalized Gaussian with center $x_0+p_{0}t/m$ and width ${\sigma }_x={\sigma }_0\sqrt{1+{\eta }^2}$; the momentum width remains ${\sigma }_p=\hslash /(2{\sigma }_0)$. Its symmetrized position-momentum covariance is $C_{xp}=t{\sigma }_p^2/m$, soThus free spreading preserves equality in the Robertson-Schrödinger uncertainty relation while producing a nonzero quadratic position-dependent phase and a product ${\sigma }_x{\sigma }_p$ larger than $\hslash /2$. More generally, ${\sigma }_x(t{)}^2={\sigma }_x(0{)}^2+2t{C}_{xp}(0)/m+t^2{\sigma }_p(0{)}^2/m^2$, so an initially correlated Gaussian can first contract.