A sigle wave
Superposition of waves
© The scientific sentence. 2010

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}/2σ^{2}} (1)
Which is already normalized.
σ^{2} 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}/2σ^{2}}
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 σ^{2}. We know that the dispersion of "k" is defined as:
σ_{k}^{2} = E[(k  μ)^{2}] = <(k  <k>)^{2}>
= <k^{2}>  [<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_{k}^{2}} 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_{k}^{2}} exp{ikx} dk
Where N is set to normolize the function ψ(x).
Now: the integrale:
I = ∫ exp { (k  k_{o})^{2}/2σ_{k}^{2}} exp{ikx} dk
Let k  k_{o} = κ the dk = dκ
I = exp {ik_{o}x} ∫ exp { κ ^{2}/2σ_{k}^{2}} exp{iκx} dκ
= exp {ik_{o}x J, where :
J = ∫ exp { κ ^{2}/2σ_{k}^{2}} exp{iκx} =
∫ exp {[ (κ ^{2}  2iκσ_{k}^{2}x)]/2σ_{k}^{2}} dκ
With :
κ ^{2}  2iκσ_{k}^{2}x = [κ  ixσ_{k}^{2}]^{2} +
x^{2}σ_{k}^{4}
We have:
J = exp{ x^{2} σ_{k}^{2} /2} ∫ exp {[ (κ  ixσ_{k}^{2})^{2}]/2σ^{2}} dκ
let: z = [κ  ixσ_{k}^{2}]/(2)^{1/2}σ_{k} , then dz = dκ/(2)^{1/2}σ_{k} and thus:
J = ((2)^{1/2}σ_{k}) exp{ x^{2} σ_{k}^{2} /2}
∫ exp {[ z^{2}]} dz
∫ exp {[ z^{2}]} dz = (π)^{1/2}
Refer to Result of
this integrale.
Hence:
I = exp {ik_{o}x (2)^{1/2}σ_{k} exp{ x^{2} σ_{k}^{2}/2}
(π)^{1/2}
ψ(x) = N 1/σ_{k}(2π)^{1/2} exp {ik_{o}x
(2)^{1/2}σ_{k} exp{ x^{2} σ_{k}^{2}/2}
(π)^{1/2}
ψ(x) = N exp{ x^{2} σ_{k}^{2}/2} exp {ik_{o}x}
ψ(x) = N exp{ x^{2} σ_{k}^{2}/2} exp {ik_{o}x} (2)
To find the value of N , we normalize the function as follows:
∫ ψ(x) ψ^{*}(x) = 1. That is:
∫ N^{2} exp{ x^{2} σ_{k}^{2}} dx =
= N^{2} ∫ exp{ x^{2} σ_{k}^{2}} dx = 1
Let z = x σ_{k}, then dx = σ_{k} dz. we have then:
1 = N^{2} σ_{k} ∫ exp{ z^{2}} dz
∫ exp{ z^{2}} dz = (π)^{1/2} →
N^{2} 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{ x^{2} σ_{k}^{2}/2} and
an oscillating factor : exp {ik_{o}x} 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_{x}^{2}}.
The initial mean μ= <x> is choosen around x = 0 and leads to = 0. Then:
σ_{x}^{2} = <x^{2}>  0^{2} = <x^{2}>
That is:
exp{ x^{2} σ_{k}^{2}/2} =
exp { x ^{2}/2σ_{x}^{2}}
σ_{k}^{2} = 1/σ_{x}^{2}.
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.
©: The scientificsentence.net. 2007.


