Celia Lucas

2022-06-14

Why do wave packets spread out over time?
Why do wave functions spread out over time? Where in the math does quantum mechanics state this? As far as I've seen, the waves are not required to spread, and what does this mean if they do?

lilao8x

Suppose you have an infinite plane wave. To find the momentum of this wave you Fourier transform it. Because it's an infinite wave the Fourier transform is a delta function and the wave has a well defined single value for the momentum.
Now take a wave packet i.e. the same infinite plane wave but now multipled by some envelope function. When you Fourier transform this you get the original delta function but now convolved with the Fourier transform of the envelope function. The packet is made up from waves with a range of different frequencies/momenta. The longer the packet the smaller the range of momenta, but for any finite wave packet there will always be a spread of momenta.
For a massive particle the spread of momenta means there is a spread of velocities, and therefore the wave packet broadens away from its average position.

Eden Solomon

An easy way to make this intuitively plausible is by remarking that the Schroedinger equation in the absence of a potential is as follows
$\frac{\mathrm{\partial }}{\mathrm{\partial }t}\mathrm{\Psi }={\mathrm{\nabla }}^{2}\mathrm{\Psi }$
up to constants, which is the heat equation if we ignore the fact that the omitted constants are complex numbers rather than real and of the right sign.
If you consider your wave function to be a heat map, then it is intuitively clear that it should spread out.

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