I am reading an article (Nonspreading wave packets, by Michael Berry, Am. J. Phys. 47, 264 (1979), author eprint), and I have some questions about the maths within. First of all they take a wave function packet that at $t = 0$ is

$$\psi(x, 0) = \mathrm{Ai}\mathopen{}\left(\frac{B x}{\hbar ^{2/3}}\right)\tag{1}$$

Where $\mathrm{Ai}$ denotes the Airy function.

$B$ is an arbitrary constant. According to the Schrödinger equation, the wave packet evolves as

$$\psi(x, t) = \mathrm{Ai}\mathopen{}\left[\frac{B}{\hbar^{2/3}}\left(x - \frac{B^3 t^2}{4m^2}\right)\right] \exp\left((i B^3 t/ 2m\hbar)[x - (B^3t^2/6m^2)]\right)\tag{3}$$ and Berry claims that

It's clear that $|\psi|^2$ does propagate without spreading, and accelerates to the right with velocity $B^3t^2/2m^2$.

Question 1: Shouldn't it be $b^3t/2m^2$?

A few lines later, it says something about the centre of gravity of a wave packet: ma question is now: how do I find, mathematically, the centre of gravity of a wave packet?

Also it says that

to explain why the Airy packet is the only one that does not spread, we first note that a necessary condition for ANY probability density to propagate unchanged in form is that its semiclassical representation translates rigidly along $x$ as time elapses.

Question 3 Is this a theorem? I never heard of this "necessary condition", so where does it come from?

• Yes I think the concept of the question is actually important as coherent states are examples of non-spreading wave packets, and Schrodinger actually found an early solution to this problem. It would be much better if your question was self-contained. – ZeroTheHero Aug 30 '18 at 15:57
• A copy of the paper can be found here: fisicafundamental.net/relicario/doc/berry078.pdf – ZeroTheHero Aug 30 '18 at 21:01