Quantum mechanics and QFT use extensively Fourier analysis.

When trying to approximate a periodic function by Fourier series (say a rectangular wave), it is possible to increase the number of terms until the approximation seems good enough for our ends.

But Fourier analysis aims to approximate also functions limited to a finite range. The underlying idea is to use the destructive interference of the harmonic functions to get zero out of that range.

Suppose I know $f(p)$, and want to evaluate numerically, (because the integrand is not analytically solvable), the Fourier transform to get $f(x)$.

I can proceed as in Fourier series, choosing an interval, splitting $f(p)$ in a series of coefficients that multiplies the exponentials, and adding all the parts.

But when I go that way, what I get is always a periodic function, what I know it is not the case.

Is there other way to have progressively better estimates as it happens in Fourier series (or Taylor expansions)?

  • $\begingroup$ Is there anything wrong with saying your function is the Fourier series multiplied by a function that is zero outside your specified range? $\endgroup$ Dec 25 '19 at 3:22
  • 1
    $\begingroup$ @Jahan Claes If I know f(p), the range where it is meanignful is also known. But it does't mean that the same range is valid to f(x). In principle I don't know how f(x) spreads to define a range. $\endgroup$ Dec 25 '19 at 14:22
  • $\begingroup$ Why can't you just extrapolate your function smoothly in both directions to make it periodic? I don't see the point in making it zero outside the original range. $\endgroup$
    – D. Halsey
    Dec 25 '19 at 19:02
  • $\begingroup$ @D. Halsey That is the point of Fourier analysis. It is possible to describe a wave pulse (or package), not only periodic waves. $\endgroup$ Dec 25 '19 at 22:11

Periodic functions $f(t)$ (with a time period $T$) can be approximated by a Fourier series, i.e. by summing harmonic oscillations with the discrete frequencies $0, \frac{2\pi}{T}, 2\frac{2\pi}{T}, 3\frac{2\pi}{T}, \ldots$ . $$f(t)=\sum_{n=-\infty}^{+\infty} F_n e^{in\frac{2\pi}{T}t}$$

But, as you already noticed, with a Fourier series you cannot build aperiodic functions (e.g. functions limited to a finite range). For approximating such functions you need a Fourier integral, i.e. summing harmonic oscillations with all frequencies from $0$ to $\infty$. $$f(t)=\int_{-\infty}^{+\infty}F(\omega)e^{i\omega t}\ d\omega$$

Example 1: Periodic function $\to$ discrete spectrum

As an example for $f(t)$ let's choose the rectangular wave (with period $T$) extending from $t=-\infty$ to $t=+\infty$.
periodic function
$f(t)$ is a perfectly periodic function, and therefore can be decomposed into a Fourier series. Its spectrum (the Fourier coefficients $F_n$) look like this:
discrete spectrum

Eaxample 2: Aperiodic function $\to$ continuous spectrum

Now as a second example for $f(t)$ let's choose the rectangular wave (again with period $T$) restricted to a certain range. Outside this range we define $f(t)=0$.
aperiodic (but nearly periodic) function
Obviously this function is quite similar to the first example. Therefore we expect its spectrum to be somehow similar to the spectrum of the first example.

Because $f(t)$ is aperiodic, it cannot be decomposed into a Fourier series. But it can be decomposed into a Fourier integral. Its Fourier spectrum (the function $F(\omega)$) looks like this:
continuous (but nearly discrete) spectrum
(Function plots created with FooPlot)

This spectrum is continuous, but still quite similar to the discrete spectrum of the first eample. It has some pronounced peaks (at the same frequencies as in the first example). But there is also some spectral intensity outside of these peaks.


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