For $A$ a real symmetric $n\times n$ matrix with nonzero determinant and $x\in\mathbb{R}^n$, let $f(x)=e^{-i(Ax\cdot x)/2}$ be a complex-valued function on $\mathbb{R}^n$; notice it is a tempered distribution since $|f|=1$ everywhere. What is its Fourier transform (in a distributional sense, because $f\not\in L^1$)?
I can do some progress as follows: assume $n=1, A=1$. Then $f=e^{-ix^2/2}$ so $f'=-ixf$. Moreover we know that the Fourier transform "exchanges multiplication and differentiation". So taking the Fourier transform of the previous equation, we get $ix(\hat{f}) = \widehat{f'}=\widehat{-ixf} = (\hat{f})'$. Setting $g=\hat{f}$ we get $g' = ixg$. This forces $g(x)=Ce^{ix^2/2}$, and we can compute $C$ plugging in some "easy function", like $\psi=e^{-x^2}$. But I'm not sure about the rigourosity of this argument.
Thanks for your help!
I'd just like to close this question by posting my solution, which I'm quite sure gives the right answer despite the slightly non-rigorous step involving solving the differential equation for $\hat{f}$. This might be well-known theory, but I'm unaware of it.