##### Contents

A sigle wave

Superposition of waves

# 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)

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 - ko)2/2σ2}

The average vulue for all wave numbers is ko. 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:
σk2 = 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/σk(2π)1/2 ∫ exp {- (k - ko)2/2σk2} exp{i(kx - ωt)} dk
At t = 0, we have:
ψ(x,0) = ψ(x) = N ∫ A(k) exp{i kx} dk = = 1/σk(2π)1/2 ∫ exp {- (k - ko)2/2σk2} exp{ikx} dk
Where N is set to normolize the function ψ(x).

Now: the integrale:
I = ∫ exp {- (k - ko)2/2σk2} exp{ikx} dk
Let k - ko = κ the dk = dκ
I = exp {ikox} ∫ exp {- κ 2/2σk2} exp{iκx} dκ = exp {ikox J, where :
J = ∫ exp {- κ 2/2σk2} exp{iκx} = ∫ exp {[- (κ 2 - 2iκσk2x)]/2σk2} dκ
With :
κ 2 - 2iκσk2x = [κ - ixσk2]2 + x2σk4
We have:
J = exp{- x2 σk2 /2} ∫ exp {[- (κ - ixσk2)2]/2σ2} dκ
let: z = [κ - ixσk2]/(2)1/2σk , then dz = dκ/(2)1/2σk and thus:
J = ((2)1/2σk) exp{- x2 σk2 /2} ∫ exp {[- z2]} dz
∫ exp {[- z2]} dz = (π)1/2 Refer to Result of this integrale. Hence:
I = exp {ikox (2)1/2σk exp{- x2 σk2/2} (π)1/2
ψ(x) = N 1/σk(2π)1/2 exp {ikox (2)1/2σk exp{- x2 σk2/2} (π)1/2
ψ(x) = N exp{- x2 σk2/2} exp {ikox}

ψ(x) = N exp{- x2 σk2/2} exp {ikox}  (2)

To find the value of N , we normalize the function as follows:
∫ ψ(x) ψ*(x) = 1. That is:
∫ N2 exp{- x2 σk2} dx = = N2 ∫ exp{- x2 σk2} dx = 1
Let z = x σk, then dx = σk dz. we have then:
1 = N2 σk ∫ exp{- z2} dz
∫ exp{- z2} dz = (π)1/2 → N2 σk (π)1/2 = 1 →
N= 1/σk1/2 (π)1/4

N = 1/σk1/2 (π)1/4   (3)

### 2. Uncertainty principle

We obtain an envelope exp{- x2 σk2/2} and an oscillating factor : exp {ikox} for the wave function ψ(x). The envolope has to be a Gaussian distribution and written as:
ψ(x) = 1/σx(2π)1/2 exp {- (x - μ)2/2σx2}.
The initial mean μ= <x> is choosen around x = 0 and leads to = 0. Then:
σx2 = <x2> - 02 = <x2> That is:
exp{- x2 σk2/2} = exp {- x 2/2σx2}
σk2 = 1/σx2. That is:
σk = 1/σx
With: Δk = σk 2 and Δx = σx 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 σx is given at t = 0. In the ehe dispersive medium, this expression changes and increases with time and the wave packet spreads out.