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Given the Fourier transform defined like $$ \mathcal F[f](\omega) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(t) e^{-iwt} \mathrm{d}t $$ how could one define a square root $\mathcal G$ of the operator $\mathcal F$ so that $$ \mathcal G^2[f] = F[f] $$ My idea is to literally take the square root of $\mathcal F$ like $$ \mathcal G[f](\omega) = \sqrt{\mathcal F[f](\omega)} = \sqrt{\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(t) e^{-iwt} \mathrm{d}t} = \frac{1}{\sqrt{\sqrt{2 \pi}}} \int_{-\infty}^\infty \sqrt{f(t)} e^{-\frac{iwt}{2}} \mathrm{d}t $$ Then the square of $\mathcal G[f]$ would be $$ \mathcal G^2[f](\omega) = \frac{1}{\sqrt{\sqrt{2 \pi}}} \frac{1}{\sqrt{\sqrt{2 \pi}}} \int_{-\infty}^\infty \int_{-\infty}^\infty \sqrt{f(t)} \sqrt{f(t')} e^{-\frac{iwt}{2}} e^{-\frac{iw't'}{2}} dt' dt $$ But this doesn't quite equate to $\mathcal F[f]$ again. So I guess my approach of literally taking the square root is a bit too naive. How then could one define the square root of a Fourier transform?

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    $\begingroup$ how do get the square root inside the integral? $\endgroup$ – tired Oct 22 '17 at 21:26
  • $\begingroup$ if it should be an integral transform, then you want to find $g$ candidate functions for which $f(t)e^{-iwt} = \int f(t) g(w,t) dt$ $\endgroup$ – mathreadler Oct 22 '17 at 21:46
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    $\begingroup$ To see the problem with your approach, consider the much easier problem of finding a 'square root' of the square function, that is a function $f()$ such that $f(f(x))=x^2$ (There are many such functions; $f(x)=x^{\sqrt{2}}$ is probably the easiest.). Your approach here would suggest using $f(x)=\sqrt{x^2}=x$, which clearly doesn't work (because $f(f(x))=f(x)=x$.) In general finding explicit 'compositional square roots' for any function or operator is likely to be a hard problem. $\endgroup$ – Steven Stadnicki Oct 22 '17 at 22:04
  • $\begingroup$ This is the fractional Fourier transform. $\endgroup$ – Henricus V. Oct 23 '17 at 4:15
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The Fourier transform can be diagonalised. Let $h_n$ be the Hermite functions, then we have $$ \mathcal F[h_n] = (-i)^n h_n. $$ So a good definition for a square root of the Fourier transform would be $$ \mathcal G[f] = \sum_{n=0}^\infty \langle f, h_n \rangle \mathcal G[h_n ] = \sum_{n=0}^\infty \langle f, h_n \rangle (-i)^{n/2} h_n. $$ A closed form expression can be found in the other answers.

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You have an operator $\mathcal{F}$ that is unitary and satisfies $\mathcal{F}^4=1$. Let $r=e^{2\pi i/4}=i$. Define polynomials $$ p_k(z) = \frac{\prod_{l=0,l\ne k}^{3}(z-r^l)}{\prod_{l=0,l\ne k}^3(r^k-r^l)}. $$ Then $p_k(r^l)=\delta_{k,l}$ for $0 \le k,l \le 3$. Therefore, $$ p_1+p_2+p_3+p_4 = 1. $$ (This is because this third order polynomial sum is $1$ at the four points $r,r^2,r^3,r^4$.) Likewise, $$ p_1(\mathcal{F})+p_2(\mathcal{F})+p_3(\mathcal{F})+p_4(\mathcal{F})=I, \\ p_k(\mathcal{F})p_l(\mathcal{F})=0,\;\; k\ne l. \\ \mathcal{F}p_k(\mathcal{F})=r^{k-1}p_k(\mathcal{F}). $$ Consequently, $$ \mathcal{F} = p_1(\mathcal{F})+rp_2(\mathcal{F})+r^2p_3(\mathcal{F})+r^3p_3(\mathcal{F}). $$ Therefore, $$ S = p_1(\mathcal{F})+r^{1/2}p_2(\mathcal{F})+rp_3(\mathcal{F})+r^{3/2}p_4(\mathcal{F}) $$ satisfies $$ S^2 = p_1(\mathcal{F})+rp_2(\mathcal{F})+r^2p_3(\mathcal{F})+r^3p_4(\mathcal{F})=\mathcal{F}. $$ So there is a third-order polynomial in $\mathcal{F}$ that is a square root of $\mathcal{F}$.

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  • $\begingroup$ Interesting approach! Would it make sense to try and do something similar with non-scalars as coefficients for polynomials? $\endgroup$ – mathreadler Oct 23 '17 at 7:46
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    $\begingroup$ @mathreadler : This polynomial calculus allows you to construct a large class of functions of $\mathcal{F}$ because it is the spectral resolution of the unitary operator $\mathcal{F}$. For example $e^{\mathcal{F}}$ can be defined in this way, etc.. I'm not sure what has been done with non-scalar coefficients in spectral resolutions. $\endgroup$ – DisintegratingByParts Oct 23 '17 at 14:11
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I think what you are looking for is the Fractional Fourier Transform, which is a generalization of the Fourier transform to the $n^{\textrm{th}}$ power. They are a subset of the Linear Canonical Transforms.

For any value $\alpha=\frac{\pi}{2} \cdot n$, the definition of the Fractional Fourier Transform of a function $f$ is $$ \mathcal{F}_\alpha[f](u) = \sqrt{1-i\cot(\alpha)} \cdot e^{i \pi \cot(\alpha) u^2} \int_{-\infty}^\infty e^{-i2\pi \left(\csc(\alpha) u x - \frac{\cot(\alpha)}{2} x^2\right)} f(x)\, \mathrm{d}x $$

For the square root, filling in $\alpha=\frac{\pi}{4}$ gives: $$ \mathcal{F}_\alpha[f](u) = \sqrt{1-i} \cdot e^{i \pi u^2} \int_{-\infty}^\infty e^{-i2\pi \left(\sqrt{2} u x - \frac{1}{2} x^2\right)} f(x)\, \mathrm{d}x $$

For integer multiples of $\pi$, the cotangent and cosecant functions diverge, so you need to take the limit instead.

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    $\begingroup$ In particular, the results of taking those limits are $F_{2n\pi}[f](u) = f(u)$ and $F_{(2n+1)\pi}[f](u) = f(-u)$. $\endgroup$ – eyeballfrog Oct 23 '17 at 22:29
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@Demophilus shows probably the most straight forward way to do it, but you can also add for example a shuffle (permutation) of the frequency components. If we assume the canonical decomposition by Demophilus is ${\bf SDS}^{-1}$ You can then bake it in the canonical transformation like this:

$${\bf SDPS}^{-1}$$

Where $\bf D$ is the diagonal matrix with the modified eigenvalues, for example like @Demophilus proposed in his answer. Furthermore, will need ${\bf P}^2 = \bf I$. And of course $$({\bf SDPS}^{-1})^2 = \bf F$$ The Fourier transform matrix.

To show that such a $\bf P$ exists, here is one for a 6 point FFT $${\bf P} = \left[\begin{array}{cccccc}0&0&0&1&0&0\\0&0&0&0&0&1\\0&0&1&0&0&0\\1&0&0&0&0&0\\0&0&0&0&1&0\\ 0&1&0&0&0&0\end{array}\right]$$

This one was found by ocular inspection on the eigenvalues. But we know results about the eigenvalues to the Fourier transform in general which pave the way to construct this in a more general setting.

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