A picture in my text book shows a three dimensional wave packet dispersing, "resulting from the fact that the phase velocity of the individual waves making up the packet depends on the wavelength of the waves." Does this mean a particle moving through space has a gradually diminishing probability of being in it's location? Also, why does the wavelength of the the wave change the speed for a probability wave? I thought the dispersion was characteristic of the medium, and I thought for things like vacuums&light, air&sound, and also probability waves and space, that they wouldn't disperse.
Answer
The dispersion of a wave is a result of the relationship between its frequency and its wavelength, which is appropriately known as the dispersion relation for the wave. For classical waves this depends on the medium: light, for example, will be dispersionless in vacuum and will have dispersion inside material media because the medium affects the dispersion relation. Quantum mechanical waves, on the other hand, have dispersion fundamentally built in.
Let's have an equation look at how light behaves. The dispersion relation for light is $$\omega=\frac c n k,$$ where $k=2\pi/\lambda$ is the wavenumber, $c$ is the speed of light in vacuum, and $n=\sqrt{\varepsilon_r\mu_r}$ is the medium's refractive index. In vacuum, $n\equiv1$ and there is no dispersion: the phase and group velocities, $\frac \omega k$ and $\frac{d\omega}{dk}$ are equal, constant, and independent of $k$, which are the mathematical conditions for dispersionless waves. In material media, though, $n$ will depend on the wavelength - it has to depend on the wavelength - and there will be dispersion.
Matter waves, on the other hand, are quite different. What are their frequency and wavelength, anyway? Well, the first is given by Planck's postulate that $E=h\nu$, and the second by de Broglie's relation $p=h/\lambda$; both should really be phrased as $$E=\hbar \omega\text{ and }p=\hbar k.$$ How are $\omega$ and $k$ related? The same way that $E$ and $p$ are: for nonrelativistic mechanics, as $E=\frac{p^2}{2m}$. Thus the dispersion relation for matter waves in free space reads $$\hbar\omega=\frac{\hbar^2k^2}{2m},\text{ or }\omega=\frac \hbar{2m}k^2.$$ Note how different this is to the one above! The phase velocity is now $v_\phi=\frac\omega k=\frac{\hbar k}{2m}$, and it is different for different wavelengths.
Now, why exactly does that imply dispersion?
Let's first look at the phase velocity, which is the velocity of the wavefronts, which are the planes that have constant phase. Since the phase goes as $e^{i(kx-\omega t)}$, the phase velocity is $\omega/k$. This, of course, is for a single plane wave, and doesn't apply to a general wavepacket for which phase is not as well defined, and for which the different wavefronts might be doing different speeds.
How then do we deal with wavepackets? The approach that works best with the formalism above is to think of a wavepacket $\psi(x,t)$ as a superposition of different plane waves $e^{i(kx-\omega t)}$, each with its own weight $\tilde\psi(k)$: $$\psi(x,0)=\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{ikx}.\tag{1}$$ Now, if all the different plane-wave components $\tilde \psi(k)e^{ikx}$ moved at the same speed then their sum would just move at that speed and would not change shape.
(More mathematically: if $v_\phi=\omega/k$ is constant, then $$\psi(x,t)=\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{i(kx-\omega t)} =\int_{-\infty}^\infty \frac{dk}{\sqrt{2\pi}} \tilde \psi(k)e^{ik(x-v_\phi t)} =\psi(x-v_\phi t,0),$$ so the functional form is preserved.)
For a matter wave in free space, however, the different phase speeds are not the same, and the different plane-wave components move at different speed. It is the interference of all these different components that makes them sum to $\psi(x,0)$ in equation (1), and if you mess with the relative phases you will get a different sum. Thus, with longer waves moving slower and shorter ones going faster, wavepackets with lots of detail encoded in long high-$k$ tails of their Fourier transform will change shape very fast.
In general it is hard to predict what the evolution of a wavepacket will do to it in detail. However, it is very clear that all wavepackets will (eventually) spread, since some components are going faster than others. Since the total probability is conserved, this must mean that the probability density will in general decrease.
If I put a particle with zero net momentum localized in some interval, then the probability of it remaining there will decrease. Note, though, that this is no surprise! The Uncertainty Principle demands that there be uncertainty in the particle's momentum. There is then some chance that the particle was moving to the left or to the right, so who's surprised to eventually find it out of the original interval?
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