Proving Normalized Hermite Polynomials are the eigenfunctions of Fourier Transform by alternative definition without generating function

Question

I am well aware of the existence of proofs like this and that showing that the normalized Hermite polynomials are the eigenfunctions of Fourier Transform, by manipulating the generating function. However, I am reading Stone and Goldbart Exercise 2.5, which wants the reader to prove this without generating function but instead just using the following definition \begin{align*} H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} \end{align*} such that \begin{align*} \varphi_n(x) = \frac{1}{\sqrt{2^n n!\sqrt{\pi}}} H_n(x) e^{-x^2/2} \end{align*} and hence \begin{align*} \mathcal{F}(\varphi_n) = \imath^n\varphi_n(x) \end{align*} where the Fourier transform is defined as \begin{align*} \mathcal{F}(f) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{\imath xs} f(s) ds \end{align*} I tried to use integrating by parts but I can't proceed. Any help would be appreciated!

Answer 1

An alternative demonstration may use a proof by induction based on the recurrence relation $H_n(x) = xH_{n-1}(x) - H_{n-1}'(x)$, which itself implies $\varphi_n(x) = \frac{x\varphi_{n-1}(x) - \varphi_{n-1}'(x)}{\sqrt{2n}}$. However, we must be careful with the notation, because it can lead to an actual muddle in the present case.

For recall, we would like to prove that $\mathfrak{F}[\varphi_n(x)](k) = (\color{red}{-i})^n\varphi_n(k)$. The base case $n = 0$ is easily checked, since $\varphi_0$ is a gaussian, which is mapped to another gaussian in the Fourier space. Then we assume the wanted relation for $n-1$, so that the induction step gives : $$ \begin{array}{rcl} \mathfrak{F}[\varphi_n(x)](k) &=& \displaystyle \mathfrak{F}\left[\frac{x\varphi_{n-1}(x) - \varphi_{n-1}'(x)}{\sqrt{2n}}\right](k) \\ &=& \displaystyle \frac{1}{\sqrt{2n}}\left(i\frac{\mathrm{d}}{\mathrm{d}k}\mathfrak{F}[\varphi_{n-1}(x)](k) - ik\mathfrak{F}[\varphi_{n-1}(x)](k)\right) \\ &=& \displaystyle \frac{-i}{\sqrt{2n}}\left(k(-i)^{n-1}\varphi_{n-1}(k) - \frac{\mathrm{d}}{\mathrm{d}k}(-i)^{n-1}\varphi_{n-1}(k)\right) \\ &=& \displaystyle (-i)^n\frac{k\varphi_{n-1}(k) - \varphi_{n-1}'(k)}{\sqrt{2n}} \\ &=& \displaystyle (-i)^n\varphi_n(k) \end{array} $$