What's the microscopic and macroscopic effect of wavefunction dispersion? In Quantum Mechanics (Merzbacher 2nd ed.), problem 2.1 asks us to derive the time evolution of a one-dimensional Gaussian wavefunction (formula given for $t=0$), assuming the velocity is in the $+x$ direction. It then asks us to apply the result to calculate the effects of wavefunction spreading to  microscopic and macroscopic experiments. 
I did the calculation for a single particle of mass $m$ and, WLOG setting the mean wavenumber $\bar{k}=0$, came up with the result that the spreading of $\|\psi^2\|$ happened at a rate of about $\hbar/2m$, which is extremely small on macroscopic scales even for many elementary particles and molecules.
My questions are:


*

*Since the width of the wavefunction is inversely related to that of its Fourier transform (HUP), does this mean that over time the particle's momentum becomes better defined?

*Would this calculated rate be the actual rate at which a beam of particles (all of mass $m$) disperses?
 A: So to your first question, when you calculate the time-evolution of a wavepacket, you first transform your wavepacket representation from a positional basis to a momentum basis.
$$
|\psi\rangle = \sum_k c_{x,p}|p\rangle
$$
This moment basis is useful as they are eigenstates to your hamiltonian and satisfy
$$
H |p\rangle = E_p |p\rangle
$$
This in turn allows you apply the time-evolution operator to your initial wavepacket giving
$$
|\psi(t)\rangle = \sum_p c_{x,p}e^{i H t/\hbar}|p\rangle = \sum_p c_{x,p}e^{i E_p t/\hbar}|p\rangle
$$
using this basis, you can easily calculate the expectation value of your momentum p.
$$
\langle\psi(t)|\hat{p}|\psi(t)\rangle = \sum_{p,p'} c_{x,p}c^{\dagger}_{x,p'}e^{-i E_{p'} t/\hbar}e^{i E_p t/\hbar}\langle p'|\hat{p}|p\rangle
$$
So now notice that the last term is simply the momentum times a delta function!  This mean that p = p' and one of the sums disappear.  Also, $E_p = E_{p'}$ so the term that has time is also killed! Thus you are left with
$$
\langle\psi(t)|\hat{p}|\psi(t)\rangle = \sum_{p} p c_{x,p}c^{\dagger}_{x,p}
$$
which does not change with time.  You can apply the same argument to a p^2 operator as well to find that you do not gain or lose information on the momentum across time.
