Fourier transform

The Fourier transform (FT) is an integral transform which converts a function \(f(x)\) into its frequency representation \(\tilde{f}(k)\). Great volumes have already been written about this subject, so let us focus on the aspects that are useful to physicists.

The forward FT is defined as follows, where \(A\), \(B\), and \(s\) are unspecified constants (for now):

\[\begin{aligned} \boxed{ \tilde{f}(k) = \hat{\mathcal{F}}\{f(x)\} = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} } \end{aligned}\]

The inverse Fourier transform (iFT) undoes the forward FT operation:

\[\begin{aligned} \boxed{ f(x) = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k} } \end{aligned}\]

Clearly, the inverse FT of the forward FT of \(f(x)\) must equal \(f(x)\) again. Let us verify this, by rearranging the integrals to get the Dirac delta function \(\delta(x)\):

\[\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\} &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k} \\ &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'} \\ &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'} = \frac{2 \pi A B}{|s|} f(x) \end{aligned}\]

Therefore, the constants \(A\), \(B\), and \(s\) are subject to the following constraint:

\[\begin{aligned} \boxed{\frac{2\pi A B}{|s|} = 1} \end{aligned}\]

But that still gives a lot of freedom. The exact choices of \(A\) and \(B\) are generally motivated by the convolution theorem and Parseval’s theorem.

The choice of \(|s|\) depends on whether the frequency variable \(k\) represents the angular (\(|s| = 1\)) or the physical (\(|s| = 2\pi\)) frequency. The sign of \(s\) is not so important, but is generally based on whether the analysis is for forward (\(s > 0\)) or backward-propagating (\(s < 0\)) waves.


The FT of a derivative has a very interesting property. Below, after integrating by parts, we remove the boundary term by assuming that \(f(x)\) is localized, i.e. \(f(x) \to 0\) for \(x \to \pm \infty\):

\[\begin{aligned} \hat{\mathcal{F}}\{f'(x)\} &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x} \\ &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} \\ &= (- i s k) \tilde{f}(k) \end{aligned}\]

Therefore, as long as \(f(x)\) is localized, the FT eliminates derivatives of the transformed variable, which makes it useful against PDEs:

\[\begin{aligned} \boxed{ \hat{\mathcal{F}}\{f'(x)\} = (- i s k) \tilde{f}(k) } \end{aligned}\]

This generalizes to higher-order derivatives, as long as these derivatives are also localized in the \(x\)-domain, which is practically guaranteed if \(f(x)\) itself is localized:

\[\begin{aligned} \boxed{ \hat{\mathcal{F}} \Big\{ \dv[n]{f}{x} \Big\} = (- i s k)^n \tilde{f}(k) } \end{aligned}\]

Derivatives in the frequency domain have an analogous property:

\[\begin{aligned} \boxed{ \dv[n]{\tilde{f}}{k} = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x} = \hat{\mathcal{F}}\{ (i s x)^n f(x) \} } \end{aligned}\]

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