Dispersion of Probability Wave Packets 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.
 A: 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?
A: Actually there is an instructive take (at least I thought so) on Emilio Pisanty's answer that yields further intuition for the property of "mass" of a lone, first quantized particle. As well as inertia, mass in this case can be interpreted as a "tendency to stay put" in this sense: if a massive particle is in a state that is well localized in space (i.e. behaving in a way we might think of as a classical particle), then the more massive it is, the more it will "stay put" and the longer it takes to delocalize or "spread out": i.e. the longer it takes to take on the unclassical property of delocalisation. This is all to do with how the mass property comes into the dispersion relationships. 
Let's look at the difference between the electron (or other massive fermion) and the photon. The curl Maxwell equations in freespace can be written:
$\left(c^{-1}\partial_t + \sigma_1 \partial_x + \sigma_2 \partial_y + \sigma_3 \partial_z\right) \,\Psi_+ = {\bf 0}$
$\left(c^{-1}\partial_t - \sigma_1 \partial_x - \sigma_2 \partial_y - \sigma_3 \partial_z\right) \,\Psi_- = {\bf 0}$
where $\sigma_j$ are the Pauli spin matrices and the electromagnetic field components are:
$\begin{array}{lcl}\Psi_\pm &=& \left(\begin{array}{cc}E_z & E_x - i E_y\\E_x + i E_y & -E_z\end{array}\right) \pm i \,c\,\left(\begin{array}{cc}B_z & B_x - i B_y\\B_x + i B_y & -B_z\end{array}\right)\\
& =& E_x \sigma_1 + E_y \sigma_2+E_z\sigma_3 + i\,c\,\left(B_x \sigma_1 + B_y \sigma_2+B_z\sigma_3\right)\end{array}$
and represent the left and right-hand polarized light field whereas the Dirac equation for a massive Fermion can be written:
$\left(c^{-1}\partial_t + \sigma_1 \partial_x + \sigma_2 \partial_y + \sigma_3 \partial_z\right) \, \Psi_+ = \frac{m\,c}{i\,\hbar} \Psi_-$
$\left(c^{-1}\partial_t - \sigma_1 \partial_x - \sigma_2 \partial_y - \sigma_3 \partial_z\right) \, \Psi_- = \frac{m\,c}{i\,\hbar} \Psi_+$
where now the $\Psi_\pm$ are $2\times1$ complex column vectors rather than $2\times2$ matrices (and also they are not quite the same as the so called spinors wontedly used to write the Dirac equation: they are sums and differences of the more usual spinors, but this need not worry us here). The main point of these fancy equations, even if you don't fully understand all their meaning is that they are exactly the same but for one thing: the Maxwell equations are two uncoupled equations for the left and right polarized light fields: the Dirac equations are the same, but now there is a mass term on the right that couples the otherwise uncoupled equations. One can think of an electron as two massless particles (Roger Penrose quaintly calls these "Zig" and "Zag") that are bound to one another such that the electron's state oscillates back and forth between the two (indeed I rather believe Penrose with his cute names might be playing on Schrödinger's word for this "jitteriness" of the electron: the Zitterbewegung).
This coupling also ties in with your knowledge that the medium disperses light, but your slight quizzing of how the same thing can happen for massive particles in vacuum. Guess what: the medium (other than vacuum) enters the Maxwell equations in the above form in almost exactly the same way that mass enters the Dirac equation, namely, it couples the two uncoupled components of the light. 
This mutual coupling begets the mass of the twin-particle system. When you confine massless particles in some way, this gives the confined system an inertia $E/c^2$ - just as a photon caught in between two mirrors can be shown to have this inertia - see https://physics.stackexchange.com/a/70948/26076 or comments on method 2 of https://physics.stackexchange.com/a/72688/26076. Lets now look at mass from the point of view of dispersion. By doing some further manipulation on the Maxwell equations, you can come up with:
$c^{-2}\partial_t^2 \psi - \nabla^2 \psi = 0$
which, when you Fourier transform this yields, as in the other answer, the dispersion relationship $\omega = c\,k$. This is a dispersionless entity travelling always at the speed of light: in one dimension you get the general solution of $\psi = f_1(x - c\,t) + f(x +c\,t)$: waves which exactly keep their shape and always run at speed $c$.
In contrast, if we do the same to the Dirac equation, we get the Klein Gordon equation
$c^{-2}\partial_t^2 \psi - \nabla^2 \psi + \frac{m^2\,c^2}{\hbar^2}\psi = 0$
which, when Fourier transformed gives us the all-important dispersion relationship:
$-\frac{\omega^2}{c^2} + k^2 + \frac{m^2\,c^2}{\hbar^2} = 0$
OR: 
$\omega = c \sqrt{k^2 + \frac{m^2 c^2}{\hbar^2}}$
Let's look at the intuition behind this one:
Firstly, as $m\rightarrow0$ we get our original, dispersionless, constant phase speed $c = \omega/k$
Secondly: if $m$ is appreciable, the group velocity $v_g = \frac{d\,\omega}{d\,k}$ is almost zero, even for particles with quite high values of $k$. The group velocity plays a similar but more general role than the phase velocity in Emilio's Fourier integral: if a wave is modulated with an envelope, the envelope's shape propagates with speed $v_g$. Now, how do we make a localized particle? We need a wide spectrum of spatial Fourier components (i.e. a spread of $k$ values up to roughly $\frac{2\pi}{L}$, where $L$ is the width of the region the particle is localized within) to do this. A big value of $m$ allows us to do this and still have the particle frequency $\omega \approx \frac{m c^2}{\hbar}$ almost unchanged and $v_g = \frac{d\,\omega}{d\,k} \approx 0$. (If $k_{max} \ll \frac{m c}{\hbar}$ then this behavior follows).
The particle will stay put, localized for long periods of time for large values of $m$. Thus the mass $m$ shows itself to be a measure of a particle's "unwillingness" to spread (disperse) as well as its "unwillingness" to shift (inertia).
