Categories: Mathematics, Optics, Physics.

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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