free particle wave packet dispersion

First look at a single plane wave. Re(ψ) = cos(kx − ωt) is a ripple that goes on without end. Change k to change how tight the ripple is, and send the crests forward in time. Here k = 2π/λ is the wavenumber. The speed of the crests reads as the phase speed v_p = ℏk/2m, and raising k makes the crests run faster.

Add several cosines of nearby k values and make one packet. Let time run and the two speeds split. A marked crest goes slowly at the phase speed, and the peak of the whole packet leads at the group speed v_g, twice as fast. You can see the peak overtaking the crest. For a matter wave it is always v_g = 2v_p, and the speed at which the particle really travels is the group speed.

Both speeds live in the dispersion curve ω(k) = ℏk²/2m. This ω(k) is the dispersion frequency that ties frequency to wavenumber, not the oscillator ω and not a beat frequency. Move a point on the curve and the slope of the tangent is the group speed v_g, and the slope from the origin to that point is the phase speed. Because it is a parabola the tangent is twice as steep as the chord, and v_g = 2v_p is confirmed.

A free particle spreads as time passes even if it sits still. |ψ(x,t)|² grows wider and flatter as you push time. It happens because the dispersion curve is curved, that is because d²ω/dk² = ℏ/m is not 0. The narrower you make the packet at the start, the faster it spreads. Narrowing the position widens the momentum, which ties to the uncertainty.

Set the free particle next to the box. A free particle allows any k, so E(k) = ℏ²k²/2m is an unbroken continuous curve. A trapped box allows only a separated ladder E ∝ n². Switch the two and the difference is plain: free is a smooth continuum, the box is discrete steps. Trap it and it quantizes. Release it and it is continuous.