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)
  = (a^1/2x + i(k-k₀)/(2a^½2 + (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(-(a^1/2x + i(k-k₀)/(2a^½2) 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) ≈ Ψ_max 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.