Time evolution of a free wave packet
Let Ψ(x,t) = (2π)-1/2∫g(k) exp(i(kx - ωkt)) dk
and assume that |g(k)| is centered at k0 and has width Δk.
Let g(k) = |g(k)|exp(iα(k)). Then
Ψ(x,t) = (2π)-1/2∫|g(k)|exp(i(α(k) - ωkt)) exp(ikx) dk = (2π)-1/2∫|g(k)|exp(iβ(k)) exp(ikx) dk.
Here β(k) = α(k) - ωkt.
If we again use the use a Taylor series expansion,
β(k) = β(k0) + dβ(k)/dk|k0 (k - k0) + ... .
then we already know that Ψ(x,t) peaks at
-dβ(k)/dk|k0 = -dα(k)/dk|k0 + dω(k)/dk|k0t = x0 + dω(k)/dk|k0t = x0 + vgt.
The peak of the wave packet moves with the group velocity, vg = dω/dk|k=ko.
The velocity of a plane wave exp(i(kx - ωt) is vp = ω/k and is called the phase velocity. We have
ħω = ħ2k2/(2m), ω/k = ħk/(2m), dω/dk = ħk/m, vg = ħk0/m = 2vp(k0).
A wave for which Δx << x and Δp << p can represent a classical particle moving with velocity v = vg.