# A geometric interpretation of the fractional Fourier transform

I was reading Joe Polchinsky’s autobiography which contains the following anecdote from his time at Caltech (page 18):

Once a week, Feynman led Physics X, where freshman and sophomores could ask their questions about physics, or if we ran out of questions he would talk about some of his ideas. One example of this was, how do you take the square root of the Fourier transformation, so that acting on a function twice with the operation would be the same as the Fourier transform

He goes on to answer in a footnote:

In phase space, the Fourier transformation x → p → −x is a 90◦ rotation. So rotate by 45◦

I’m not a physicist, and not used to thinking of the Fourier transform and time-frequency domains quite so concretely. The picture seems very compelling: it for instance makes intuitive why $$\mathcal{F}^2 = -id$$, or why Fourier and fractional Fourier transforms should preserve $$L^2$$ norms.

After some searching I found that the more general family of linear canonical transforms apparently induces an action of $$SL(2, \mathbb{R})$$ on the time-frequency domain. This begins to explain the analogy, but I don’t feel I fully understood this remark. Some additional questions:

1. How would one find the correct integral transform a priori? In other words is there a more geometric way to arrive at the formula for the fractional Fourier transform (or LCT)? It seems one first finds a symplectic structure on the time-frequency domain, then checks for operators preserving it but I don’t know how to carry that out, and all the sources I can find just define the fractional Fourier transform as a black box. How can we know there is a symplectic structure in the first place, without invoking mechanics?
2. My initial thought for the square root of the Fourier transform was to scale how the Fourier transform acts on a basis of eigenfunctions. I tried deriving the integral transform that way and got stuck again. How could that be done? Is there a way to see that this is compatible with Polchinski’s picture?
• Re 2: the integral transform is given by the Mehler kernel $K(x,y,t)$ --- $t=\pi/2$ gives the Fourier transform and $t=\pi/4$ the square root of it. Commented Jan 23, 2023 at 9:27
• I had gotten a version of Mehler’s formula, but didn’t know the name and could not prove it. I can now find enough resources online, so thank you!
– Luke
Commented Jan 23, 2023 at 11:05

For the record, to answer Q2, let me write down the Mehler kernel for the square root of the Fourier transform, $$K(x,y)=(2\pi)^{-1/2}\sqrt{1-i}\exp\left[{\tfrac{1}{2} i \left(x^2+y^2-2 \sqrt{2} x y\right)}\right].$$ It satisfies $$\int_{-\infty}^\infty K(x,z)K(z,y)\,dz=(2\pi)^{-1/2}e^{-ixy},$$ so application of the transform $$K$$ twice is equivalent to the Fourier transform, defined as $${\cal F}(x)=(2\pi)^{-1/2}\int_{-\infty}^\infty e^{-ixy}f(y)\,dy.$$