= Free Gaussian wave packet evolution
{title2=$\sigma_x(t)^2=\sigma_0^2+\hbar^2t^2/(4m^2\sigma_0^2)$}
Let the initial <Gaussian wave packet> be
$$
\psi(x,0)=(2\pi\sigma_0^2)^{-1/4}\exp\left[-\frac{(x-x_0)^2}{4\sigma_0^2}+\frac{ip_0(x-x_0)}\hbar\right].
$$
Fourier transforming, multiplying each momentum amplitude by $e^{-ip^2t/(2m\hbar)}$ and evaluating the <Gaussian integral> gives
$$
\psi(x,t)=\frac{(2\pi\sigma_0^2)^{-1/4}}{\sqrt{1+i\eta}}\exp\left[-\frac{(x-x_0-p_0t/m)^2}{4\sigma_0^2(1+i\eta)}+\frac{ip_0(x-x_0)}\hbar-\frac{ip_0^2t}{2m\hbar}\right],\qquad \eta=\frac{\hbar t}{2m\sigma_0^2}.
$$
Its <probability density> is a normalized Gaussian with center $x_0+p_0t/m$ and width $\sigma_x=\sigma_0\sqrt{1+\eta^2}$; the momentum width remains $\sigma_p=\hbar/(2\sigma_0)$. Its symmetrized position-momentum <covariance> is $C_{xp}=t\sigma_p^2/m$, so
$$
\sigma_x\sigma_p=\frac\hbar2\sqrt{1+\eta^2},\qquad \sigma_x^2\sigma_p^2-C_{xp}^2=\frac{\hbar^2}4.
$$
Thus 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 $\hbar/2$. More generally, $\sigma_x(t)^2=\sigma_x(0)^2+2tC_{xp}(0)/m+t^2\sigma_p(0)^2/m^2$, so an initially correlated Gaussian can first contract.
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