Fourier Transforms

Consider a function $F(x)$ that is periodic in $x$ with period $L$. In other words,

$$F(x+L) = F(x)$$ (8.1)

for all $x$. Recall, from Section 5.5, that we can represent such a function as a Fourier series; that is,

$$F(x)= \sum_{n=1,\infty} \left[C_n\,\cos(n\,\delta k\,x) + S_n\,\sin(n\,\delta k\,x)\right],$$ (8.2)

where

$$\delta k = \frac{2\pi}{L}.$$ (8.3)

[We have neglected the $n=0$ term in Equation (8.2), for the sake of convenience.] Equation (8.2) automatically satisfies the periodicity constraint (8.1), because $\cos(\theta+n\,2\pi)=\cos\theta$ and $\sin(\theta+n\,2\pi)=\sin\theta$ for all $\theta$ and $n$ (with the proviso that $n$ is an integer). The so-called Fourier coefficients, $C_n$ and $S_n$, appearing in Equation (8.2), can be determined from the function $F(x)$ by means of the following readily demonstrated (see Exercise 1) results:

$$\frac{2}{L}\int_{-L/2}^{L/2}\cos(n\,\delta k\,x)\,\cos(n'\,\delta k \,x)\,dx = \delta_{n,n'},$$ (8.4)

$$\frac{2}{L}\int_{-L/2}^{L/2}\sin(n\,\delta k\,x)\,\sin(n'\,\delta k \,x)\,dx = \delta_{n,n'},$$ (8.5)

$$\frac{2}{L}\int_{-L/2}^{L/2}\cos(n\,\delta k\,x)\,\sin(n'\,\delta k \,x)\,dx =0,$$ (8.6)

where $n$, $n'$ are positive integers. Here, $\delta_{n,n'}$ is a Kronecker delta function. In fact,

$$C_n =\frac{2}{L}\int_{-L/2}^{L/2} F(x)\,\cos(n\,\delta k\,x)\,dx,$$ (8.7)

$$S_n =\frac{2}{L}\int_{-L/2}^{L/2} F(x)\,\sin(n\,\delta k\,x)\,dx.$$ (8.8)

(See Exercise 1.) Incidentally, any periodic function of $x$ can be represented as a Fourier series.

Suppose, however, that we are dealing with a function $F(x)$ that is not periodic in $x$. We can think of such a function as one that is periodic in $x$ with a period $L$ that tends to infinity. Does this mean that we can still represent $F(x)$ as a Fourier series? Consider what happens to the series (8.2) in the limit $L\rightarrow\infty$, or, equivalently, $\delta k\rightarrow 0$. The series is basically a weighted sum of sinusoidal functions whose wavenumbers take the quantized values $k_n=n\,\delta k$. Moreover, as $\delta k\rightarrow 0$, these values become more and more closely spaced. In fact, we can write

$$F(x) = \sum_{n=1,\infty} \frac{C_n}{\delta k}\,\cos(n\,\delta k\,... ...lta k+ \sum_{n=1,\infty} \frac{S_n}{\delta k} \,\sin(n\,\delta k\,x)\,\delta k.$$ (8.9)

In the continuum limit, $\delta k\rightarrow 0$, the summations in the previous expression become integrals, and we obtain

$$F(x)=\int_{-\infty}^\infty C(k)\,\cos(k\,x)\,dk + \int_{-\infty}^\infty S(k)\,\sin(k\,x)\,dk,$$ (8.10)

where $k=n\,\delta k$, $C(k)=C(-k)= C_n/(2\,\delta k)$, and $S(k)=-S(-k)=S_n/(2\,\delta k)$. Thus, for the case of an aperiodic function, the Fourier series (8.2) morphs into the so-called Fourier transform (8.10). This transform can be inverted using the continuum limits (i.e., the limit $\delta k\rightarrow 0$) of Equations (8.7) and (8.8), which are readily shown to be

$$C(k) =\frac{1}{2\pi}\int_{-\infty}^\infty F(x)\,\cos(k\,x)\,dx,$$ (8.11)

$$S(k) =\frac{1}{2\pi}\int_{-\infty}^\infty F(x)\,\sin(k\,x)\,dx,$$ (8.12)

respectively. (See Exercise 5.) The previous equations confirm that $C(-k)=C(k)$ and $S(-k)=-S(k)$. The Fourier-space (i.e., $k$-space) functions $C(k)$ and $S(k)$ are known as the cosine Fourier transform and the sine Fourier transform of the real-space (i.e., $x$-space) function $F(x)$, respectively. Furthermore, because we already know that any periodic function can be represented as a Fourier series, it seems plausible that any aperiodic function can be represented as a Fourier transform. This is indeed the case.

When sinusoidal waves of different amplitudes, phases, and wavelengths are superposed, they interfere with one another. In some regions of space, the interference is constructive, and the resulting wave amplitude is comparatively large. In other regions, the interference is destructive, and the resulting wave amplitude is comparatively small, or even zero. Equations (8.10)–(8.12) essentially allow us to construct an interference pattern that mimics any given function of position (in one dimension). Alternatively, they allow us to decompose any given function of position into sinusoidal waves that, when superposed, reconstruct the function. Let us consider some examples.

Figure 8.1: Fourier transform of a top-hat function.

Consider the “top-hat” function,

$$F(x)=\left\{ \begin{array}{ccc} 1 &\mbox{\hspace{1cm}}& \vert x\vert\leq l/2\\ 0 && \vert x\vert>l/2 \end{array}\right..$$ (8.13)

See Figure 8.1. Given that $\cos(-k\,x)=\cos(k\,x)$ and $\sin(-k\,x)=-\sin(k\,x)$, it follows from Equations (8.11) and (8.12) that if $F(x)$ is even in $x$, so that $F(-x)=F(x)$, then $S(k)=0$, and if $F(x)$ is odd in $x$, so that $F(-x)=-F(x)$, then $C(k)=0$. Hence, because the top-hat function (8.13) is even in $x$, its sine Fourier transform is automatically zero. On the other hand, its cosine Fourier transform takes the form

$$C(k) = \frac{1}{2\pi}\int_{-l/2}^{l/2}\,\cos(k\, x)\,dx = \frac{l}{2\pi}\,\frac{\sin(k\,l/2)}{k\,l/2}.$$ (8.14)

Figure 8.1 shows the function $F(x)$, together with its associated cosine transform, $C(k)$.

Figure 8.2: A Gaussian function.

As a second example, consider the so-called Gaussian function,

$$F(x) =\exp\left(-\frac{x^{\,2}}{2\,\sigma_x^{\,2}}\right).$$ (8.15)

As illustrated in Figure 8.2, this is a smoothly-varying even function of $x$ that attains its peak value $1$ at $x=0$, and becomes completely negligible when $\vert x\vert\gtrsim 3\,\sigma_x$. Thus, $\sigma_x$ is a measure of the “width” of the function in real (as opposed to Fourier) space. By symmetry, the sine Fourier transform of the preceding function is zero. On the other hand, the cosine Fourier transform is readily shown to be

$$C(k) = \frac{1}{(2\pi\,\sigma_k^{\,2})^{1/2}}\,\exp\left(-\frac{k^{\,2}}{2\,\sigma_k^{\,2}}\right),$$ (8.16)

where

$$\sigma_k=\frac{1}{\sigma_x}.$$ (8.17)

(See Exercise 2.) This function is a Gaussian in Fourier space of characteristic width $\sigma_k=1/\sigma_x$. The original function $F(x)$ can be reconstructed from its Fourier transform using

$$F(x)=\int_{-\infty}^\infty C(k)\,\cos(k\,x)\,dk.$$ (8.18)

This reconstruction is simply a linear superposition of cosine waves of differing wavenumbers. Moreover, $C(k)$ can be interpreted as the amplitude of waves of wavenumber $k$ within this superposition. The fact that $C(k)$ is a Gaussian of characteristic width $\sigma_k=1/\sigma_x$ [which means that $C(k)$ is negligible for $\vert k\vert\gtrsim 3\,\sigma_k$] implies that in order to reconstruct a real-space function whose width in real space is approximately $\sigma_x$ it is necessary to combine sinusoidal functions with a range of different wavenumbers that is approximately $\sigma_k=1/\sigma_x$ in extent. To be slightly more exact, the real-space Gaussian function $F(x)$ falls to half of its peak value when $\vert x\vert\simeq \sqrt{\pi/2}\,\sigma_x$. Hence, the full width at half maximum of the function is ${\mit\Delta} x \simeq 2\sqrt{\pi/2}\,\sigma_x=\sqrt{2\pi}\,\sigma_x$. Likewise, the full width at half maximum of the Fourier-space Gaussian function $C(k)$ is ${\mit\Delta} k \simeq \sqrt{2\pi}\,\sigma_k$. Thus,

$${\mit\Delta} x\,{\mit\Delta} k \simeq 2\pi,$$ (8.19)

because $\sigma_k\,\sigma_x=1$. We conclude that a function that is highly localized in real space has a transform that is highly delocalized in Fourier space, and vice versa. Finally,

$$\int_{-\infty}^\infty \frac{1}{(2\pi\,\sigma_k^{\,2})^{1/2}}\,\exp\left(-\frac{k^{\,2}}{2\,\sigma_k^{\,2}}\right) dk = 1.$$ (8.20)

(See Exercise 3.) In other words, a Gaussian function in real space, of unit height and characteristic width $\sigma_x$, has a cosine Fourier transform that is a Gaussian in Fourier space, of characteristic width $\sigma_k=1/\sigma_x$, and whose integral over all $k$-space is unity.