Categories: Mathematics, Optics, Physics.

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) \equiv \hat{\mathcal{F}}\{f(x)\} \equiv 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) \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} \equiv 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.

Derivatives

The FT of a derivative has a very useful 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} \dv[n]{\tilde{f}}{k} &= A \dv[n]{k} \int_{-\infty}^\infty f(x) \exp\!(i s k x) \dd{x} \\ &= 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}

Multiple dimensions

The Fourier transform is straightforward to generalize to $$N$$ dimensions. Given a scalar field $$f(\vb{x})$$ with $$\vb{x} = (x_1, ..., x_N)$$, its FT $$\tilde{f}(\vb{k})$$ is defined as follows:

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

Where the wavevector $$\vb{k} = (k_1, ..., k_N)$$. Likewise, the inverse FT is given by:

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

In practice, in $$N$$D, there is not as much disagreement about the constants $$A$$, $$B$$ and $$s$$ as in 1D: typically $$A = 1$$ and $$B = 1 / (2 \pi)^N$$, with $$s = \pm 1$$. Any choice will do, as long as:

\begin{aligned} \boxed{ A B = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N} } \end{aligned}

Differentiation is more complicated for $$N > 1$$, but the FT is still useful, notably for the Laplacian $$\nabla^2 \equiv \dv*{x_1} + ... + \dv*{x_N}$$. Let $$|\vb{k}|$$ be the norm of $$\vb{k}$$, then for a localized $$f$$:

\begin{aligned} \boxed{ \hat{\mathcal{F}}\{\nabla^2 f(\vb{x})\} = - s^2 |\vb{k}|^2 \tilde{f}(\vb{k}) } \end{aligned}

1. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.