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:


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}=∫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 p, 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]


F^{-1} δ_k=÷{1}{(2⋅π)³}
δ_k=F (F^{-1} δ_k)=F (÷{1}{(2⋅π)³})

And for Quantum Mechanics:

F (F^{-1} δ_p)=F (÷{1}{(2⋅π⋅ħ)³})