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I was glancing through "The Fourier Transform & Its Applications" by Ronald N. Bracewell, which is a good intro book on Fourier Transforms. In it, he says that if you take the Fourier transform of a function 4 times, you get back the original function, i.e. $$\mathcal F\Bigg\{ \mathcal F\bigg\{ \mathcal F\big\{ \mathcal F\left\{ g(x) \right\} \big\} \bigg\} \Bigg\} = g(x)\,. $$

Could someone kindly show me how this is possible? I'm assuming the above statement is for complex $x$, and this has something to do with $i^0=1$, $i^1=i$, $i^2=-1$, $i^3 = -i$, $i^4=1$?

Thank you for your enlightenment.

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  • $\begingroup$ "Equivalent to time inversion" -- this got me thinking. If you have the Fourier transform of a particle's wave function, would the inverse Fourier transform give you the wave function of the anti-particle? $\endgroup$ Commented Aug 23, 2019 at 5:53

2 Answers 2

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I'll use the non-unitary Fourier transform (but this is not important, it's just a preference):

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-i\omega t}dt\tag{1}$$

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{i\omega t}d\omega\tag{2}$$

where (1) is the Fourier transform, and (2) is the inverse Fourier transform.

Now if you formally take the Fourier transform of $X(\omega)$ you get

$$\mathcal{F}\{X(\omega)\}=\mathcal{F}^2\{x(t)\}=\int_{-\infty}^{\infty}X(\omega)e^{-i\omega t}d\omega\tag{3}$$

Comparing (3) with (2) we have

$$\mathcal{F}^2\{x(t)\}=2\pi x(-t)\tag{4}$$

So the Fourier transform equals an inverse Fourier transform with a sign change of the independent variable (apart from a scale factor due to the use of the non-unitary Fourier transform).

Since the Fourier transform of $x(-t)$ equals $X(-\omega)$, the Fourier transform of (4) is

$$\mathcal{F}^3\{x(t)\}=2\pi X(-\omega)\tag{5}$$

And, by an argument similar to the one used in (3) and (4), the Fourier transform of $X(-\omega)$ equals $2\pi x(t)$. So we obtain for the Fourier transform of (5)

$$\mathcal{F}^4\{x(t)\}=2\pi\mathcal{F}\{X(-\omega)\}=(2\pi)^2x(t)\tag{6}$$

which is the desired result. Note that the factor $(2\pi)^2$ in (6) is a consequence of using the non-unitary Fourier transform. If you use the unitary Fourier transform (where both the transform and its inverse get a factor $1/\sqrt{2\pi}$) this factor would disappear.

In sum, apart from irrelevant constant factors, you get

$$\bbox[#f8f1ea, 0.6em, border: 0.15em solid #fd8105]{x(t)\overset{\mathcal{F}}{\Longrightarrow} X(\omega)\overset{\mathcal{F}}{\Longrightarrow} x(-t)\overset{\mathcal{F}}{\Longrightarrow} X(-\omega)\overset{\mathcal{F}}{\Longrightarrow} x(t)}$$

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    $\begingroup$ In fact, $(6)$ suggests a fantastical idea about how one can design an amplifier that converts computation into amplitude gain: just take the non-unitary Fourier transform of $x(t)$ 4 times to amplify the signal by a factor of 39 or so (or 31 dB gain)! $\endgroup$ Commented Jul 25, 2015 at 16:44
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    $\begingroup$ @DilipSarwate: How can I have missed that! I'd contact a patent attorney before someone here steals this brilliant idea! $\endgroup$
    – Matt L.
    Commented Jul 25, 2015 at 17:05
  • $\begingroup$ Unitary factor should be $\frac{1}{\sqrt {2 \pi}}$, mistyped in the last paragraph. $\endgroup$
    – mbaitoff
    Commented Jul 25, 2015 at 18:56
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    $\begingroup$ Too late! A patent has already been granted for an even better method (using FFT to reduce total computation from $4N^2$ to $4N\log N$ instead of plain vanilla Fourier transform). $\endgroup$ Commented Jul 25, 2015 at 20:25
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    $\begingroup$ i haven't seen this question or answer before. i would sau that "constant factors" are not "irrelevant". because of that i would recommend the unity Fourier transform. $\endgroup$ Commented Sep 29, 2019 at 17:42
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In addition to the above answer, the original question also asked whether there is any relationship with $i^0=1, i^1=i, i^2=-1, i^3=-i$, and the answer is: yes, these four numbers are the four possible eigenvalues of the unitary Fourier transform, and the corresponding eigenfunctions are the Hermite functions. The deeper reason for this lies in the fact that unitary transforms are generated by a Hermitian operator via exponentiation and the Hermitian operator for the Fourier transform is the number operator N of the quantum harmonic oscillator. Correspondingly, each Fourier transform corresponds to a transform resulting from the Mehler kernel of the number operator for a time span of $t=\pi/2$ such that after four transforms the original function is obtained again. Fractional Fourier transforms correspond to other time spans. For a detailed account of these facts cf. i.a. the text leading up to section 7.5.10 of the book Integral Transforms in Science and Engineering by Kurt B. Wolf (also available here), or this article on Hermite Functions and Fourier Series.

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