# 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:

1. 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?
2. Would this calculated rate be the actual rate at which a beam of particles (all of mass $m$) disperses?
• "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?" note that the HUP states that $\Delta x$ times $\Delta p$ is grater or equal to $\hbar/2$. This means that $\Delta x$ can increase over time, while keeping $\Delta p$ constant, for example. Therefore, we cannot conclude that the momentum distribution gets narrower. You can try to calculate $\Delta p$ as a function of time, and try to check this claim yourself (which might or might not be true) Jan 21 '16 at 19:41

$$|\psi\rangle = \sum_k c_{x,p}|p\rangle$$ 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.
• [note that you can use \langle for $\langle$ and \rangle for $\rangle$, which look much better that <>; also, the symbol for a dagger is \dagger] In your first equation, what is $x$ and what is $k$? Jan 21 '16 at 20:10