# 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:

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

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

3. $\Psi(x,t+\Delta t) \approx \exp(-\frac{i \hat{V}(x) \Delta t}{2 \hbar}) \mathscr{F}^{-1} (\xi(k))$

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.

• I support Cosmos's answer and I can attest that this is exactly the kind of behavior one sees in the SSFFT. If you're unsure, you can hammer together a version of SSFFT in a few lines in Mathematica or the like. This might give you confidence that all is OK. But, if your FFT is working, the likelihood of coding mistake is low: remember, operator splitting is a very simple, neat algorithm. I'm not altogether sure about the $2\pi$ factor in step 2; I make it to be $\exp\left(-\frac{i \,k^ 2 \,\hbar \Delta t}{2m}\right)$. – Selene Routley Jan 5 '18 at 5:39
• Check this, but a mistake along these lines will simply alter your energy scale, not the fundamental behavior (i.e. like changing units for $V$). – Selene Routley Jan 5 '18 at 5:40
• Rod Vance, thanks for your comment. I was really not confident about the results, it is very good to know that it is something normal and has some meaning also. Well, about the $2 \pi$ that's one thing I got in a mathematics book, never checked for being honest, but I will, thanks again. The ssfft, thanks for mentioning, I'll give it a try, seems to be a way more fast. – Thiago Melo Jan 5 '18 at 10:29

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 $\Delta x \sim \hbar t/m\sqrt{2a}$.