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# 10.2: Fourier Transforms

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The Fourier series applies to periodic functions defined over the interval $$-a/2 \le x < a/2$$. But the concept can be generalized to functions defined over the entire real line, $$x \in \mathbb{R}$$, if we take the limit $$a \rightarrow \infty$$ carefully.

Suppose we have a function $$f$$ defined over the entire real line, $$x \in \mathbb{R}$$, such that $$f(x) \rightarrow 0$$ for $$x \rightarrow \pm\infty$$. Imagine there is a family of periodic functions $$\big\{f_a(x) \,\big|\, a \in\mathbb{R}^+\big\}$$, such that $$f_a(x)$$ has periodicity $$a$$, and approaches $$f(x)$$ in the limit $$a\rightarrow \infty$$. This is illustrated in the figure below: Figure $$\PageIndex{1}$$

In mathematical terms, $f(x) = \lim_{a \rightarrow \infty} f_a(x), \;\;\;\text{where}\;\; f_a(x+a) = f_a(x).$ Since $$f_a$$ is periodic, it can be expanded as a Fourier series: $f_a(x) = \sum_{n=-\infty}^\infty e^{i k_n x}\, f_{an}, \quad\mathrm{where}\;\; k_n = n\Delta k, \;\; \Delta k = \frac{2\pi}{a}.$ Here, $$f_{an}$$ denotes the $$n$$-th complex Fourier coefficient of the function $$f_a(x)$$. Note that each Fourier coefficient depends implicitly on the periodicity $$a$$.

As $$a \rightarrow \infty$$, the wave-number quantum $$\Delta k$$ goes to zero, and the set of discrete $$k_n$$ turns into a continuum. During this process, each individual Fourier coefficient $$f_{an}$$ goes to zero, because there are more and more Fourier components in the vicinity of each $$k$$ value, and each Fourier component contributes less. This implies that we can replace the discrete sum with an integral. To accomplish this, we first multiply the summand by a factor of $$(\Delta k/2\pi) / (\Delta k/2\pi) = 1$$: $f(x) = \lim_{a\rightarrow \infty} \left[\;\,\sum_{n=-\infty}^\infty \frac{\Delta k}{2\pi} \, e^{i k_n x}\, \left(\frac{2\pi \,f_{an}}{\Delta k} \right)\;\,\right].$ (In case you’re wondering, the choice of $$2\pi$$ factors is essentially arbitrary; we are following the usual convention.) Moreover, we define $F(k) \equiv \lim_{a \rightarrow \infty} \left[\frac{2\pi\, f_{an}}{\Delta k}\right]_{k = k_n}.$ In the $$a \rightarrow \infty$$ limit, the $$f_{an}$$ in the numerator and the $$\Delta k$$ in the denominator both go zero, but if their ratio remains finite, we can turn the Fourier sum into the following integral: $f(x) = \int_{-\infty}^{\infty} \frac{dk}{2\pi} \, e^{i k x}\, F(k). \label{Fk}$

## The Fourier relations

The function $$F(k)$$ in Eq. $$\eqref{Fk}$$ is called the Fourier transform of $$f(x)$$. Just as we have expressed $$f(x)$$ in terms of $$F(k)$$, we can also express $$F(k)$$ in terms of $$f(x)$$. To do this, we apply the $$a \rightarrow \infty$$ limit to the inverse relation for the Fourier series in Eq. (10.1.13): \begin{align} F(k_n) &= \lim_{a\rightarrow \infty} \frac{2 \pi\, f_{an}}{\Delta k} \\ &= \lim_{a\rightarrow \infty} \frac{2 \pi}{2\pi/a}\, \left(\frac{1}{a} \int_{-a/2}^{a/2} dx\; e^{-i k_n x}\right) \\ &= \int_{-\infty}^\infty dx\; e^{-i kx}\, f(x).\end{align} Hence, we arrive at a pair of equations called the Fourier relations:

Definition: Fourier relations

\begin{align}\left\{\;\;\begin{aligned}F(k) &= \;\int_{-\infty}^\infty dx\; e^{-ikx}\, f(x) &&\text{(Fourier transform)}\\ f(x) &= \int_{-\infty}^\infty \frac{dk}{2\pi}\; e^{ikx}\, F(k)&&\text{(Inverse Fourier transform).}\end{aligned}\;\;\right. \end{align} \label{fourier-relations}

The first equation is the Fourier transform, and the second equation is called the inverse Fourier transform.

There are notable differences between the two formulas. First, there is a factor of $$1/2\pi$$ appears next to $$dk$$, but no such factor for $$dx$$; this is a matter of convention, tied to our earlier definition of $$F(k)$$. Second, the integral over $$x$$ contains a factor of $$e^{-ikx}$$ but the integral over $$k$$ contains a factor of $$e^{ikx}$$. One way to remember which equation has the positive sign in the exponent is to interpret the inverse Fourier transform equation (which has the form of an integral over $$k$$) as the continuum limit of a sum over complex waves. In this sum, $$F(k)$$ plays the role of the series coefficients, and by convention the complex waves have the form $$\exp(ikx)$$ (see Section 6.3).

As noted in Section 10.1, all the functions we deal with are assumed to be square integrable. This includes the $$f_a$$ functions used to define the Fourier transform. In the $$a \rightarrow \infty$$ limit, this implies that we are dealing with functions such that $\int_{-\infty}^{\infty} dx\; \big|\,f(x)\,\big|^2\;\;\text{exists and is finite}.$

## A simple example

Consider the function $f(x) = \left\{\begin{array}{cl}e^{-\eta x}, & x \ge 0 \\ 0, & x < 0,\end{array}\right. \qquad \eta \in \mathbb{R}^+.$ For $$x < 0$$, this is an exponentially-decaying function, and for $$x < 0$$ it is identically zero. The real parameter $$\eta$$ is called the decay constant; for $$\eta > 0$$, the function $$f(x)$$ vanishes as $$x \rightarrow +\infty$$ and can thus be shown to be square-integrable. Larger values of $$\eta$$ correspond to faster exponential decay.

The Fourier transform can be found by directly calculating the Fourier integral: $F(k) \;=\; \;\int_{0}^\infty dx\; e^{-i kx}\, e^{-\kappa x} \;=\; \frac{-i}{k - i \eta}.$ It is useful to plot the squared magnitude of the Fourier transform, $$|F(k)|^2$$, against $$k$$. This is called the Fourier spectrum of $$f(x)$$. In this case, $\big|\,F(k)\,\big|^2 = \frac{1}{k^2 + \eta^2}.$ Figure $$\PageIndex{2}$$

The Fourier spectrum is shown in the right subplot above. It consists of a peak centered at $$k = 0$$, forming a curve called a Lorentzian. The width of the Lorentzian is dependent on the original function’s decay constant $$\eta$$. For small $$\eta$$, i.e. weakly-decaying $$f(x)$$, the peak is narrow; for large $$\eta$$, i.e. rapidly-decaying $$f(x)$$, the peak is broad.

We can quantify the width of the Lorentzian by defining the full-width at half-maximum (FWHM)—the width of the curve at half the value of its maximum. In this case, the maximum of the Lorentzian curve occurs at $$k=0$$ and has the value of $$1/\eta^2$$. The half-maximum, $$1/2\eta^2$$, occurs when $$\delta k = \pm \eta$$. Hence, the original function’s decay constant, $$\eta$$, is directly proportional to the FWHM of the Fourier spectrum, which is $$2\eta$$.

To wrap up this example, let’s evaluate the inverse Fourier transform: $f(x) \; = \; -i\int_{-\infty}^\infty \frac{dk}{2\pi} \; \frac{e^{i kx}}{k-i\eta}.$ This can be solved by contour integration. The analytic continuation of the integrand has a simple pole at $$k = i\eta$$. For $$x < 0$$, the numerator $$\exp(ikx)$$ vanishes far from the origin in the lower half-plane, so we close the contour below. This encloses no pole, so the integral is zero. For $$x > 0$$, the numerator vanishes far from the origin in the upper half-plane, so we close the contour above, with a counter-clockwise arc, and the residue theorem gives $f(x) = \left(\frac{-i}{2\pi}\right) \, \left(2\pi i\right) \, \mathrm{Res}\left[ \frac{e^{ikx}}{k-i\eta}\right]_{k=i\eta} = e^{-\eta x} \qquad(x > 0),$ as expected.

This page titled 10.2: Fourier Transforms is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.