/ Home / math / Fourier_Transform /

Fourier Transform

Periodic functions can be expressed in their frequency domain.

First, the function (let's call it f) must be square integrable (at least over the periodic interval):

Then a base can be found and the function be expressed using a (continuous) linear combination:

$f[x⃗]=÷{1}{(2⋅π)³}⋅∫d³k⋅\hat{f}[k⃗]⋅\exp[i⋅k⃗∙x⃗]$

This operation is called the inverse transform $F^{-1}$ .

$f[x⃗]=(F^{-1} \hat{f}[k⃗])[x⃗]:=÷{1}{(2⋅π)³}⋅∫d³k⋅\hat{f}[k⃗]⋅\exp[i⋅k⃗∙x⃗]$

The coefficients $\hat{f}[k⃗]$ can be calculated by:

$\hat{f}[k⃗]=(F f[x⃗])[k⃗]:=∫d³x⋅f[x⃗]⋅\exp[-i⋅k⃗∙x⃗]$

Delta Distribution and Fourier Transform

The Fourier Transform can also be used with the δ Distribution:

First, it holds that:

$\hat{δ}=1$

$\hat{δ}=∫d³x⋅δ[x⃗]⋅\exp[-i⋅k⃗∙x⃗]=\exp[-i⋅k⃗∙0⃗]=1$ Proof.

This also works in the other direction (!!)

$F^{-1} δ_k=÷{1}{(2⋅π)³}$

$F^{-1} δ_k=÷{1}{(2⋅π)³}⋅∫d³k⋅δ_k[k⃗]⋅\exp[i⋅k⃗∙x⃗]=÷{1}{(2⋅π)³}⋅\exp[-i⋅k⃗∙0⃗]=÷{1}{(2⋅π)³}$ Proof.

In Quantum Mechanics it is customary to instead use the momentum , in that case it's slightly different:

$F^{-1} δ_p=÷{1}{(2⋅π)³}⋅∫(÷{d³p}{ħ³})⋅δ_p[p⃗]⋅\exp[÷{i}{ħ}⋅p⃗∙x⃗]=÷{1}{(2⋅π⋅ħ)³}⋅\exp[-÷{i}{ħ}⋅p⃗∙0⃗]=÷{1}{(2⋅π⋅ħ)³}$

Note that δ is not linear.

Triple Triple Integrals

Sometimes multiple integrations can be simplified by observing that:

$F^{-1} (F δ)=δ$

$F^{-1} 1=δ[x⃗]$

$÷{1}{(2⋅π)³}⋅∫d³k⋅\exp[i⋅k⃗∙x⃗]⋅1=δ[x⃗]$

$÷{1}{(2⋅π)³}⋅∫d³k⋅\exp[i⋅k⃗∙(x⃗-x⃗')]⋅1=δ[x⃗-x⃗']$

Likewise:

$F^{-1} δ_k=÷{1}{(2⋅π)³}$

$δ_k=F (F^{-1} δ_k)=F (÷{1}{(2⋅π)³})$

$δ_k[k⃗]=∫d³x⋅(÷{1}{(2⋅π)³})⋅\exp[-i⋅k⃗⋅x⃗]$

$δ_k[k⃗-k⃗']=∫d³x⋅(÷{1}{(2⋅π)³})⋅\exp[-i⋅(k⃗-k⃗')⋅x⃗]$

$δ_k[k⃗-k⃗']=∫d³x⋅(÷{1}{(2⋅π)³})⋅\exp[i⋅(k⃗'-k⃗)⋅x⃗]$

And for Quantum Mechanics:

$F (F^{-1} δ_p)=F (÷{1}{(2⋅π⋅ħ)³})$

$δ_p[p⃗-p⃗']=∫d³x⋅(÷{1}{(2⋅π⋅ħ)³})⋅\exp[÷{i}{ħ}⋅(p⃗'-p⃗)⋅x⃗]$

$(2⋅π⋅ħ)³⋅δ_p[p⃗-p⃗']=∫d³x⋅\exp[÷{i}{ħ}⋅(p⃗'-p⃗)⋅x⃗]$

Author: Danny (remove the ".nospam" to send)

Last modification on: Sat, 04 May 2024 .