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.

  • $\begingroup$ "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? $\endgroup$ – Bart Wisialowski Aug 23 '19 at 5:53

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)


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)! $\endgroup$ – Dilip Sarwate Jul 25 '15 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. Jul 25 '15 at 17:05
  • $\begingroup$ Unitary factor should be $\frac{1}{\sqrt {2 \pi}}$, mistyped in the last paragraph. $\endgroup$ – mbaitoff Jul 25 '15 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). $\endgroup$ – Dilip Sarwate Jul 25 '15 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. $\endgroup$ – robert bristow-johnson Sep 29 '19 at 17:42

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