Let us first concentrate on one-dimensional systems.

Ψ(x,t) = (2π)

Ψ(x,0) = (2π)

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(ik_{0}x)exp(-ax^{2}).

(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) N^{2}∫_{-∞}^{+∞}exp(-2ax^{2}) dx = N^{2}√(π/2a) = 1

ψ(x,0) = (2a/π)^{1/4}exp(ik_{0}x)exp(-ax^{2}).

(b) |ψ(x,0)|^{2}= (2a/π)^{1/2 }exp(-2ax^{2}).

(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/(2π^{2}))^{1/4}∫_{-∞}^{+∞}exp(-ax^{2}) exp(-i(k-k_{0})x) dx.

ax^{2}+ i(k-k_{0})x = ax^{2}+ i(k-k_{0})x - (k-k_{0})^{2}/(4a) + (k-k_{0})^{2}/(4a)

= (a^{1/2}x + i(k-k_{0})/(2a^{½2}+ (k-k_{0})^{2}/(4a). (completing the square)

Φ(k,0) = (a/(2π^{2}))^{1/4}exp(-(k-k_{0})^{2}/(4a))∫_{-∞}^{+∞}exp(-(ax^{2}+ i(k-k_{0})x - (k-k_{0})^{2}/(4a))dx

= (a/(2π^{2}))^{1/4}exp(-(k-k_{0})^{2}/(4a))∫_{-∞}^{+∞}exp(-(a^{1/2}x + i(k-k_{0})/(2a^{½2}) dx

= (a/(2π^{2}))^{1/4}exp(-(k-k_{0})^{2}/(4a))∫_{-∞}^{+∞}exp(-x'^{2}) dx'

= (a2π)^{-1/4 }exp(-(k-k_{0})^{2}/(4a)).

|Φ(k,0)|^{2}= (a2π)^{-1/2 }exp(-(k-k_{0})^{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 k_{0}
where it reaches a maximum and has a width Δk.

Assume Δk << k_{0}. 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 = k_{0} we may expand α(k) in a Taylor
series expansion,

α(k) = α(k^{0}) + dα(k)/dk|_{k0} (k - k_{0}) + ... .

We are only interested in α(k) in a small range of k
about k_{0}. Keeping only zeroth and first order terms we may write

Ψ(x,0) = (2π)^{-1/2}∫|g(k)| exp(iα(k_{0}))
exp(i(k - k_{0}) dα(k)/dk|_{k0}) exp(ikx)
dk,

and setting x_{0} = -dα(k)/dk|_{k0} this becomes

Ψ(x,0) = (2π)^{-1/2}exp(i(k_{0}x
+ α(k_{0})) ∫|g(k)|
exp(i(k - k_{0})(x - x_{0}))
dk.

Here |g(k)| is nonzero only in a small interval about k_{0}.
Therefore we need to integrate only over this small interval of width ~Δk. In the integral |k - k_{0}|
≤ Δk.

If we write exp(i(k - k_{0})(x - x_{0})) = exp(i2π(k - k_{0})/τ)
with τ = 2π/(x - x_{0}) we find that for
|τ| << Δk or |x - x_{0}| >> Δk^{-1}
the function exp(i(k - k_{0})(x - x_{0})) oscillates very rapidly in the
integration interval and the integral yields approximately zero.

For x - x_{0 }--> 0 |τ| --> ∞
and the function exp(i2π(k-k_{0})/τ)
--> 1 in the integration interval.
Then the
integral yields its maximum value. We therefore have

Ψ(x,0) ≈ 0 for |x - x_{0}|
>> Δk^{-1}, Ψ(x,0)
≈ Ψ_{max} for x = x_{0}.

x_{0} = -dα(k)/dk|_{k0} is the position of the center of the wave packet.

As |x - x_{0}| 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 t_{0}.

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 t_{0} with small width
Δt, then
Ψ(ω) has width Δω
such that Δt Δω ≥
1.