What is the spreading for rectangular wave packets?

Question

I know the spatial width for rectangular wave packets, but what is the spreading for a rectangular wave packet?

Answer 1

In a word (well, two): not pretty. A rectangular wavepacket has a sharp discontinuity, which means that its momentum-space representation (i.e. its Fourier transform) has a significant support at very large momentum, and as soon as you give the time-dependent Schrödinger equation any time to act, those short-wavelength, high-momentum components will whizz away fast.

The wavepacket itself is exactly solvable: you start with the free-particle propagator, which is expliclty known as a Fresnel oscillating exponential, $$ K(x,x';t) ={\frac {1}{2\pi }}\int _{-\infty }^{+\infty }e^{ik(x-x')}e^{-{\frac {i}{2}\hbar k^{2}t}}\mathrm dk =\sqrt{\frac {1}{2\pi it}}e^{-{\frac {(x-x')^{2}}{2i t}}} $$ and then you just integrate to get the wavepacket at later times, \begin{align} \psi(x,t) & = \int_{-\infty}^\infty K(x,x';t)\psi(x',0)\mathrm dt \\ & = \sqrt{\frac {1}{2\pi it}}\int_{-L/2}^{L/2} e^{-{\frac {(x-x')^{2}}{2i t}}}\mathrm dx' \\ & = \frac{i}{2} \left[ \text{erfi}\left(\sqrt{\frac{i}{2t}}\left(x+\frac{L}{2}\right)\right) -\text{erfi}\left(\sqrt{\frac{i}{2t}}\left(x-\frac{L}{2}\right)\right) \right], \end{align} in terms of imaginary error functions evaluated along the complex-plane diagonal at which $|\mathrm{erfi}(\sqrt{i}x)|$ is oscillatory around unit modulus.

At short times, this wavepacket has some extreme ringing caused by the discontinuity, but at longer times this gives way to a behaviour that becomes very similar to that of the propagator itself (i.e. to a wavepacket that starts off as a delta function). More graphically:

^{Mathematica source: Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["https://i.stack.imgur.com/cnnL8.png"]}

(Here the dark and light blue lines are the real and imaginary parts of $\psi(x,t)$, phase-shifted so $\psi(0,t)$ is always real and positive for visual clarity. Mind the logarithmic timescale, though: it makes it easier to see the details of the dynamics, but the animation doesn't accurately represent how the evolution would look like on linear time.)