Compute the Fourier transform of a function How do you go about computing the Fourier transform of the following function: 
$(\frac{d}{dx}-x)^ke^{-x^2/2}$? 
Do we go about it using the binomial formula? 
 A: There is no problem with transforming this function, which is a polynomial times $e^{-x^2/2}$. For a smooth and rapidly decreasing function $f$,
\begin{align}
    &\mathscr{F}\left((\frac{d}{dx}-x)f\right) \\
   &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-is x}(\frac{d}{dx}-x)f(x)dx \\
   &= is\hat{f}(s)-i\frac{d}{ds}\hat{f}(s) \\
   &= -i \left(\frac{d}{ds}-s\right)\mathscr{F}f.
\end{align}
So,
$$
           \mathscr{F}\left((\frac{d}{dx}-x)^ke^{-x^2/2}\right)=(-i)^k(\frac{d}{ds}-s)^k\mathscr{F}e^{-x^2/2}.
$$
Because $\mathscr{F}e^{-x^2/2}=e^{-s^2/2}$, it follows that the Fourier transform of
$$
                      h_k(x)=\left(\frac{d}{dx}-x\right)^k e^{-x^2/2}
$$
is $(-i)^k h_k(s)$.
Note: These functions are eigenfunctions of the unitary Fourier transform on $L^2(\mathbb{R})$:
$$
                      \mathscr{F} h_k = (-i)^k h_k.
$$
These functions $\{ h_k \}_{k=0}^{\infty}$, when normalized in $L^2$ length, form an orthonormal basis of $L^2(\mathbb{R})$ that diagonalizes the unitary Fourier transform. The unitary Fourier transform has spectrum $\sigma(\mathscr{F})=\{1,i,-1,-i\}$.
