Split-step FFT: Evolving a wave freely makes it spread Is there a "quantum" meaning in the wave spreading away while evolving it in time?
For instance, we use a wave like:
$\Psi(x,0) = \frac{1}{\sqrt[4]{2 \pi {\Delta x}^ 2}} \exp \left( i k_0 x - \frac{(x-x_0)^ 2}{4 {\Delta x}^ 2} \right)$
where $k_0$ as I understand is a proxy for the wave energy (for a free particle):
$E_0 = \frac{k_0^2 \hbar^2}{2m}$
Then we use split-step and FFT for propagating it in time:
$\Psi(x,t+\Delta t) \approx \exp(-\frac{i \hat{V}(x) \Delta t}{2 \hbar}) \exp(-\frac{i \hat{K} \Delta t}{\hbar}) \exp(-\frac{i \hat{V}(x) \Delta t}{2 \hbar}) \Psi(x,t)$
We approach this as follows:


*

*$\eta(x) = \exp(-\frac{i \hat{V}(x) \Delta t}{2 \hbar}) \Psi(x,t)$

*$\xi(k) = \exp(-\frac{i (2 \pi k)^ 2 \hbar \Delta t}{2m}) \mathscr{F} (\eta(x)) $

*$\Psi(x,t+\Delta t) \approx \exp(-\frac{i \hat{V}(x) \Delta t}{2 \hbar}) \mathscr{F}^{-1} (\xi(k))$
The steps 1-3 are repeated many times for small time steps, usually $\Delta t \approx 1\times 10^{-18} s$
What we observe is that the wave spreads away like:



I am sorry for the bad images, I hope it is possible to understand them.
But, the thing is, even for waves propagating without potentials, or waves with very high energies, they always behave like this.
Also, when we try to simulate solid state devices, we use the effective mass approximation, and it makes the wave persist for a little longer.
Sure, it is possible that I am making something wrong.
 A: Yes, free wave packet solutions of the dispersive 
Schrödinger equation almost always (*) spread like this, independently of your group velocity $k_0$ of the wavepacket. If you set that to 0, they'd still spread like this, in place.  It is the heart of quantum mechanics.
The intuitive reason is that the initial width a spreads to 
$$ \sqrt{a^2 + (\hbar t/m)^2 \over a}~,$$
so, eventually (very quickly, in practice) it
grows linearly in time, as   $\hbar t/(m\sqrt{a})$. Why?
This linear growth is a reflection of the (time-invariant) momentum uncertainty: the wave packet is confined at first to a narrow region $Δx\sim \sqrt{a/2}$, and so has a momentum which is uncertain (according to the  uncertainty principle ) by the amount  $\Delta p\sim\hbar/\sqrt{2a}$; thus, a spread in velocity of  $\hbar /m\sqrt{2a}$; and thus in the subsequent position by $\Delta x \sim \hbar t/m\sqrt{2a}$.  
The uncertainty relation is then a strict inequality, very far from saturation. The initial uncertainty  ΔxΔp=ħ/2  has now increased by a factor of  ħt/ma  for large  t. This is seen as a generic property of Fourier analysis, then.



*

*(*) Almost always: sometimes (rarely) the introduction of interaction terms in dispersive equations, such as for the quantum harmonic oscillator potential, may result in the emergence of envelope-non-dispersive, classical-looking solutions (coherent states). They don't spread, sort of like classical objects! Such "minimum uncertainty states" do saturate the uncertainty principle, permanently. Also see the peculiar Airy wave-train.

