Fourier transform of Hermite polynomial times a Gaussian

Question

What is the Fourier transform of an (n-th order Hermite polynomial multiplied by a Gaussian)?

Answer 1

I presume the function you're considering is the solution to the Schrodinger equation for the harmonic oscillator, $$ - \frac 1 2 \frac{d^2}{dx^2} \psi_n(x) + \frac 1 2 x^2 \psi_n (x) = E_n \psi_n (x), \ \ \ \ \ \ \ \psi_n (x) \to 0 {\rm \ \ as \ \ } x \to \pm \infty,$$ where $$ E_n = n + \frac 1 2, \ \ \ \ \ \ n \in \mathbb N. $$ Indeed, the solution to this equation is a hermite polynomial multiplied by a Gaussian, $$ y_n(x) = \exp( - \tfrac {x^2} 2) H_n( x).$$ (See here. Depending on your conventions, you may want to scale $x$ and $\psi_n$ by constants to match the problem you're solving.)

Now take the Fourier transform of both sides of the equation. Remember that, if the Fourier transform of $y(x)$ is $\widetilde y (p)$, then the Fourier transform of $\frac{dy}{dx}(x)$ is $ip \widetilde y(p)$, and the Fourier transform of $x y(x)$ is $i \frac{d \widetilde y}{d p}(p)$. (Depending on your conventions, you may need to include factors of $\sqrt{2\pi}$.)

So we get $$ + \frac 1 2 p^2 \widetilde\psi_n(p) - \frac 1 2 \frac{d^2}{dp^2} \widetilde\psi_n (p) = E_n \widetilde \psi_n(p).$$ But this is really the same equation that we started with!!! Except that we replaced $x$ with $p$ and we replaced $\psi_n(x)$ with $\widetilde \psi_n(p)$. So the Fourier transform $\widetilde \psi_n (p)$ is the same function as the original $\psi_n(x)$!