can be estimated by calculating the so-called *standard deviation* $\Delta Q_t$ of the position observable ${\bf Q}$ (associated with the center of mass), which by definition is given by the square root: $\Delta Q_t =\sqrt{\langle {\bf Q}^2\rangle_t- \langle {\bf Q}\rangle_t^2}$, where $\langle\cdots\rangle_t$ denotes the quantum average relative to the state of the center of mass entity. Using *Ehrenfest theorem*, with which one can calculate the average values of quantum observables, one can show (following a little long but not difficult calculation) that by judiciously choosing the origin of the time axis the spreading $\Delta Q_t$ of the center of mass wave packet, at time $t$, is given by: $$\Delta Q_t = \sqrt{{\Delta P_0^2\over M^2} \, t^2 + \Delta Q_0^2},
\label{eq1}$$ where $\Delta Q_0$ is the spatial spreading at time $t=0$, and $\Delta P_0$ the spreading with respect to the momentum observable ${\bf P}$, at time $t=0$.

For the initial width $\Delta Q_0$ of the packet, we can choose the typical value of the *Bohr radius* (which in the semiclassical model of the Danish physicist corresponds to the radius of the innermost electron), i.e., about $5.3 \times 10^{-11}$ m ($0.53$ angstroms). For the value of $\Delta P_0$, we can instead consider 