Compute the Fourier transform of a function

Question

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?

Answer 1

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 $\sigma(\mathscr{F})=\{1,i,-1,-i\}$.