Categories: Mathematics, Optics, Physics.

Fourier transform

The Fourier transform (FT) is an integral transform which converts a function f(x)f(x) into its frequency representation f~(k)\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 AA, BB, and ss are unspecified constants (for now):

f~(k)F^{f(x)}Af(x)exp(iskx)dx\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:

f(x)F^1{f~(k)}Bf~(k)exp(iskx)dk\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)f(x) must equal f(x)f(x) again. Let us verify this, by rearranging the integrals to get the Dirac delta function δ(x)\delta(x):

F^1{F^{f(x)}}=ABexp(iskx)f(x)exp(iskx)dxdk=2πABf(x)(12πexp(isk(xx))dk)dx=2πABf(x)δ(s(xx))dx=2πABsf(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 AA, BB, and ss are subject to the following constraint:

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

But that still gives a lot of freedom. The exact choices of AA and BB are generally motivated by the convolution theorem and Parseval’s theorem.

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


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)f(x) is localized, i.e. f(x)0f(x) \to 0 for x±x \to \pm \infty:

F^{f(x)}=Af(x)exp(iskx)dx=A[f(x)exp(iskx)]iskAf(x)exp(iskx)dx=(isk)f~(k)\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)f(x) is localized, the FT eliminates derivatives of the transformed variable, which makes it useful against PDEs:

F^{f(x)}=(isk)f~(k)\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 xx-domain, which is practically guaranteed if f(x)f(x) itself is localized:

F^{dnfdxn}=(isk)nf~(k)\begin{aligned} \boxed{ \hat{\mathcal{F}} \Big\{ \dvn{n}{f}{x} \Big\} = (- i s k)^n \tilde{f}(k) } \end{aligned}

Derivatives in the frequency domain have an analogous property:

dnf~dkn=Adndknf(x)exp(iskx)dx=A(isx)nf(x)exp(iskx)dx=F^{(isx)nf(x)}\begin{aligned} \dvn{n}{\tilde{f}}{k} &= A \dvn{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 NN dimensions. Given a scalar field f(x)f(\vb{x}) with x=(x1,...,xN)\vb{x} = (x_1, ..., x_N), its FT f~(k)\tilde{f}(\vb{k}) is defined as follows:

f~(k)F^{f(x)}Af(x)exp(iskx)dNx\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}) \ddn{N}{\vb{x}} } \end{aligned}

Where the wavevector k=(k1,...,kN)\vb{k} = (k_1, ..., k_N). Likewise, the inverse FT is given by:

f(x)F^1{f~(k)}Bf~(k)exp(iskx)dNk\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}) \ddn{N}{\vb{k}} } \end{aligned}

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

AB=(s2π) ⁣N\begin{aligned} \boxed{ A B = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N} } \end{aligned}

The inverse FT of the forward FT of f(x)f(\vb{x}) must be equal to f(x)f(\vb{x}) again, so:

F^1{F^{f(x)}}=ABexp(iskx)f(x)exp(iskx)dNxdNk=(2π)NABf(x)(1(2π)Nexp(isk(xx))dNk)dNx=(2π)NABf(x)(n=1N12πexp(iskn(xnxn))dkn)dNx\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\} &= A B \int \exp(- i s \vb{k} \cdot \vb{x}) \int f(\vb{x}') \exp(i s \vb{k} \cdot \vb{x}') \ddn{N}{\vb{x}'} \ddn{N}{\vb{k}} \\ &= (2 \pi)^N A B \int f(\vb{x}') \Big( \frac{1}{(2 \pi)^N} \int \exp(i s \vb{k} \cdot (\vb{x}' - \vb{x})) \ddn{N}{\vb{k}} \Big) \ddn{N}{\vb{x}'} \\ &= (2 \pi)^N A B \int f(\vb{x}') \Big( \prod_{n = 1}^N \frac{1}{2 \pi} \int \exp(i s k_n (x_n' - x_n)) \dd{k_n} \Big) \ddn{N}{\vb{x}'} \end{aligned}

Here, we recognize the definition of the Dirac delta function again, leading to:

F^1{F^{f(x)}}=(2π)NABf(x)(n=1Nδ(s(xnxn)))dNx=(2π)NABsNf(x)δ(xx)dNx=(2π)NABsNf(x)\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\} &= (2 \pi)^N A B \int f(\vb{x}') \Big( \prod_{n = 1}^N \delta(s(x_n' - x_n)) \Big) \ddn{N}{\vb{x}'} \\ &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'} = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x}) \end{aligned}

Differentiation is more complicated for N>1N > 1, but the FT is still useful, notably for the Laplacian 2d2/dx12+...+d2/dxN2\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}. Let k|\vb{k}| be the norm of k\vb{k}, then for a localized ff:

F^{2f(x)}=s2k2f~(k)\begin{aligned} \boxed{ \hat{\mathcal{F}}\{\nabla^2 f(\vb{x})\} = - s^2 |\vb{k}|^2 \tilde{f}(\vb{k}) } \end{aligned}

We insert 2f\nabla^2 f into the FT, decompose the exponential and the Laplacian, and then integrate by parts (limits ±\pm \infty omitted):

F^{2f}=A(2f)exp(iskx)dNx=A(n=1N2fxn2)(m=1Nexp(iskmxm))dNx=An=1N[fxnexp(iskx)]An=1Nisknfxnexp(iskx)dNx\begin{aligned} \hat{\mathcal{F}}\{\nabla^2 f\} &= A \int \big( \nabla^2 f \big) \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}} \\ &= A \int \Big( \sum_{n = 1}^N \pdv{ {}^2 f}{x_n^2} \Big) \Big( \prod_{m = 1}^N \exp(i s k_m x_m) \Big) \ddn{N}{\vb{x}} \\ &= A \sum_{n = 1}^N \bigg[ \pdv{f}{x_n} \exp(i s \vb{k} \cdot \vb{x}) \bigg] - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}} \end{aligned}

Just like in 1D, we get rid of the boundary term by assuming that all derivatives df/dxn\idv{f}{x_n} are nicely localized. To proceed, we then integrate by parts again:

F^{2f}=An=1Nisknfxn(m=1Nexp(iskmxm))dNx=An=1Niskn[fexp(iskx)]+An=1N(iskn)2fexp(iskx)dNx\begin{aligned} \hat{\mathcal{F}}\{\nabla^2 f\} &= - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \Big( \prod_{m = 1}^N \exp(i s k_m x_m) \Big) \ddn{N}{\vb{x}} \\ &= - A \sum_{n = 1}^N i s k_n \bigg[ f \exp(i s \vb{k} \cdot \vb{x}) \bigg] + A \sum_{n = 1}^N (i s k_n)^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}} \end{aligned}

Once again, we remove the boundary term by assuming that ff is localized, yielding:

F^{2f}=As2n=1Nkn2fexp(iskx)dNx=s2n=1Nkn2f~\begin{aligned} \hat{\mathcal{F}}\{\nabla^2 f\} &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}} = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f} \end{aligned}


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