Uncertainty principle

1. Gaussian wave packet

The Gaussian distribution (normal distribution) is written as:

P(x) = 1/σ(2π)^1/2 exp {- (x - μ)²/2σ²}   (1)
Which is already normalized.

σ² is the dispersion or the variance that shows how large data are spread around the location.
Its square root, σ, is called standard deviation. It shows how large each value is distant from the neighbor other value.

The mean value of a variable "x" μ = <x> is called also the average. It is the expectation value E(X) for a random variable X associated to an event E and related to the theoritical probility distribution.

In our case, the gaussian distribution (amplitude) is:
A(k) = 1/σ(2π)^1/2 exp {- (k - k_o)²/2σ²}

The average vulue for all wave numbers is k_o. The good approximation is to take the uncertainty of the variable "k" Δk equal to the dispersion σ². We know that the dispersion of "k" is defined as:
σ_k² = E[(k - μ)²] = <(k - <k>)²> = <k²> - [<k>]²
The wave function of the Gaussian packet is then:
ψ(x,t) = ∫ A(k) exp{i(kx - ωt)} dk = = 1/σ_k(2π)^1/2 ∫ exp {- (k - k_o)²/2σ_k²} exp{i(kx - ωt)} dk
At t = 0, we have:
ψ(x,0) = ψ(x) = N ∫ A(k) exp{i kx} dk = = 1/σ_k(2π)^1/2 ∫ exp {- (k - k_o)²/2σ_k²} exp{ikx} dk
Where N is set to normolize the function ψ(x).

Now: the integrale:
I = ∫ exp {- (k - k_o)²/2σ_k²} exp{ikx} dk
Let k - k_o = κ the dk = dκ
I = exp {ik_ox} ∫ exp {- κ ²/2σ_k²} exp{iκx} dκ = exp {ik_ox J, where :
J = ∫ exp {- κ ²/2σ_k²} exp{iκx} = ∫ exp {[- (κ ² - 2iκσ_k²x)]/2σ_k²} dκ
With :
κ ² - 2iκσ_k²x = [κ - ixσ_k²]² + x²σ_k⁴
We have:
J = exp{- x² σ_k² /2} ∫ exp {[- (κ - ixσ_k²)²]/2σ²} dκ
let: z = [κ - ixσ_k²]/(2)^1/2σ_k , then dz = dκ/(2)^1/2σ_k and thus:
J = ((2)^1/2σ_k) exp{- x² σ_k² /2} ∫ exp {[- z²]} dz
∫ exp {[- z²]} dz = (π)^1/2 Refer to Result of this integrale. Hence:
I = exp {ik_ox (2)^1/2σ_k exp{- x² σ_k²/2} (π)^1/2
ψ(x) = N 1/σ_k(2π)^1/2 exp {ik_ox (2)^1/2σ_k exp{- x² σ_k²/2} (π)^1/2
ψ(x) = N exp{- x² σ_k²/2} exp {ik_ox}

ψ(x) = N exp{- x² σ_k²/2} exp {ik_ox}  (2)


To find the value of N , we normalize the function as follows:
∫ ψ(x) ψ^*(x) = 1. That is:
∫ N² exp{- x² σ_k²} dx = = N² ∫ exp{- x² σ_k²} dx = 1
Let z = x σ_k, then dx = σ_k dz. we have then:
1 = N² σ_k ∫ exp{- z²} dz
∫ exp{- z²} dz = (π)^1/2 → N² σ_k (π)^1/2 = 1 →
N= 1/σ_k^1/2 (π)^1/4

N = 1/σ_k^1/2 (π)^1/4   (3)

2. Uncertainty principle

We obtain an envelope exp{- x² σ_k²/2} and an oscillating factor : exp {ik_ox} for the wave function ψ(x). The envolope has to be a Gaussian distribution and written as:
ψ(x) = 1/σ_x(2π)^1/2 exp {- (x - μ)²/2σ_x²}.
The initial mean μ= <x> is choosen around x = 0 and leads to = 0. Then:
σ_x² = <x²> - 0² = <x²> That is:
exp{- x² σ_k²/2} = exp {- x ²/2σ_x²}
σ_k² = 1/σ_x². That is:
σ_k = 1/σ_x
With: Δk = σ_k ² and Δx = σ_x ². We obtain the uncertainty principle of Heisenberg:
Δx Δk = 1 or according to De Broglie relationship: p = hk/2π, we find:
Δx Δp = h/2π

Δx Δp = h/2π   (4)

The expression of σ_x is given at t = 0. In the ehe dispersive medium, this expression changes and increases with time and the wave packet spreads out.