One-dimensional wave packets representing free particles
Consider a free particle which is described at t = 0 by the normalized
Gaussian wave function ψ(x,0) = Nexp(ik0x)exp(-ax2).
(a) Normalize the wave function.
(b) Find the probability density |ψ(x,0)|2 of the particle.
(c) Find its Fourier transform Φ(k,0) of the wave function and the probability density
|Φ(k,0)|2 in k-space.
Now consider a wave packet Ψ(x,0)
= (2π)-1/2∫g(k) exp(ikx)
dk, and assume |g(k)| is centered at k0
where it reaches a maximum and has a width Δk.
Assume Δk << k0. Assume g(k) = |g(k)|exp(iα(k)). Such a wave packet represents a particle with fairly well defined momentum p (Δp << p).
In the neighborhood of k = k0 we may expand α(k) in a Taylor series expansion,
α(k) = α(k0) + dα(k)/dk|k0 (k - k0) + ... .
We are only interested in α(k) in a small range of k about k0. Keeping only zeroth and first order terms we may write
Ψ(x,0) = (2π)-1/2∫|g(k)| exp(iα(k0)) exp(i(k - k0) dα(k)/dk|k0) exp(ikx) dk,
and setting x0 = -dα(k)/dk|k0 this becomes
Ψ(x,0) = (2π)-1/2exp(i(k0x + α(k0)) ∫|g(k)| exp(i(k - k0)(x - x0)) dk.
Here |g(k)| is nonzero only in a small interval about k0. Therefore we need to integrate only over this small interval of width ~Δk. In the integral |k - k0| ≤ Δk.
If we write exp(i(k - k0)(x - x0)) = exp(i2π(k - k0)/τ) with τ = 2π/(x - x0) we find that for |τ| << Δk or |x - x0| >> Δk-1 the function exp(i(k - k0)(x - x0)) oscillates very rapidly in the integration interval and the integral yields approximately zero.
For x - x0 --> 0 |τ| --> ∞ and the function exp(i2π(k-k0)/τ) --> 1 in the integration interval. Then the integral yields its maximum value. We therefore have
Ψ(x,0) ≈ 0 for |x - x0| >> Δk-1, Ψ(x,0) ≈ Ψmax for x = x0.
x0 = -dα(k)/dk|k0 is the position of the center of the wave packet.
As |x - x0| increases |Ψ(x,0)| decreases. We expect the width of the wave packet Δx in coordinate space to be ~Δx = Δk-1. Then Δx Δk ~ 1. This is the classical relation for two functions which are Fourier transforms of each other. The product Δx Δk has a lower bound. The exact value of this bound depends on the precise definition of Δx and Δk.
Now consider the wave function at some fixed position, say x = 0, as a
function of time. Assume we want to build a wave packet representing a particle which has a high
probability of being observed at position r = 0 in a small time interval Δt about t0.
Ψ(0,t) = (2π)-1/2∫Ψ(ω) exp(iωt) dω and Ψ(ω) = (2π)-1/2∫Ψ(0,t) exp(-iωt) dt.
Here Ψ(ω) and Ψ(0,t) are Fourier transforms of each other. If Ψ(0,t) is centered at t0 with small width Δt, then Ψ(ω) has width Δω such that Δt Δω ≥ 1.