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 Parsevalâ€™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\{ \dvn{n}{f}{x} \Big\} = (- i s k)^n \tilde{f}(k) } \end{aligned}

Derivatives in the frequency domain have an analogous property:

\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 $N$ 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}) \ddn{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}) \ddn{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}

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

\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:

\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 $N > 1$, but the FT is still useful, notably for the Laplacian $\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$. 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}

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

\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 $\idv{f}{x_n}$ are nicely localized. To proceed, we then integrate by parts again:

\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 $f$ is localized, yielding:

\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}

## References

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 O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.