One-dimensional wave packets representing free particles
Let us first concentrate on one-dimensional systems.
Ψ(x,t) = (2π)-1/2∫g(k) exp(i(kx - ωkt)) dk is a superposition of plane waves traveling in the ±x direction.  At some instance of time, say t = 0,
Ψ(x,0) = (2π)-1/2∫g(k) exp(ikx) dk and g(k) = (2π)-1/2∫Ψ(x,0) exp(-ikx) dx.
g(k) is the Fourier transform of Ψ(x,0).

Problem:

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.

Solution:

• Concepts:
The Fourier transform
• Reasoning:
We are asked to find the Fourier transform of a wave packet.
• Details of the calculation:
(a)  N2-∞+∞ exp(-2ax2) dx = N2 √(π/2a)   = 1
ψ(x,0) = (2a/π)1/4exp(ik0x)exp(-ax2).
(b)  |ψ(x,0)|2 = (2a/π)1/2 exp(-2ax2).
(c)  ψ(x,0) = (2π)-1/2-∞+∞ Φ(k,0) exp(ikx) dk.
Φ(k,0) = (2π)-1/2-∞+∞ ψ(x,0) exp(-ikx) dx
= (a/(2π2))1/4-∞+∞ exp(-ax2) exp(-i(k-k0)x) dx.
ax2 + i(k-k0)x = ax2 + i(k-k0)x - (k-k0)2/(4a) + (k-k0)2/(4a)
= (a1/2x + i(k-k0)/(2a½2 + (k-k0)2/(4a).  (completing the square)
Φ(k,0) = (a/(2π2))1/4exp(-(k-k0)2/(4a))∫-∞+∞ exp(-(ax2 + i(k-k0)x  - (k-k0)2/(4a))dx
= (a/(2π2))1/4exp(-(k-k0)2/(4a))∫-∞+∞ exp(-(a1/2x + i(k-k0)/(2a½2) dx
= (a/(2π2))1/4exp(-(k-k0)2/(4a))∫-∞+∞ exp(-x'2) dx'
= (a2π)-1/4 exp(-(k-k0)2/(4a)).
|Φ(k,0)|2 = (a2π)-1/2 exp(-(k-k0)2/(2a)) is the probability density 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.