Fourier transform of squared Gaussian Hermite polynomial

Question

I'm attempting to calculate the first-order perturbation energy shift for the quantum harmonic oscillator with a perturbing potential of $V(x)=A\cos(kx)$. Omitting the relevant physical factors, I've gotten to the point where I need to calculate: \begin{equation} \int_{-\infty}^{\infty} e^{-x^2}(H_n (x))^2 e^{ikx}dx \end{equation}

Now, I know that the Gaussian Hermite polynomial $e^{-x^2 /2}H_n(x)$ is an eigenfunction of the Fourier transform. However, it's not immediately clear to me what the Fourier transform of the square of this function would be, as is the case for my problem. I'm assuming that I may need to use the convolution theorem, but I've yet to come up with anything that seems correct. Looking for a suggested solution method to calculating this integral.

Answer 1

Starting with the generating function $$ e^{2xt-t^2}=\sum_{n\geq 0}H_n(x)\frac{t^n}{n!}\tag{1} $$ then replacing $t$ with $t e^{i\theta}$ we have $$ \exp\left[2xt e^{i\theta}-t^2 e^{2i\theta}\right] = \sum_{n\geq 0}H_n(x) e^{ni\theta}\frac{t^n}{n!}\tag{2} $$ and by Parseval's identity $$ \int_{-\pi}^{\pi}\exp\left[4xt\cos\theta-2t^2\cos(2\theta)\right]\,d\theta=2\pi\sum_{n\geq 0}H_n(x)^2 \frac{t^{2n}}{n!^2}. \tag{3} $$ Now we can multiply both sides of $(3)$ by $e^{-x^2}e^{kix}$ and apply $\int_{\mathbb{R}}(\ldots)\, dx$ to get $$ \sqrt{\pi}e^{-k^2/4}\int_{-\pi}^{\pi}e^{2t^2-2ikt\cos\theta}\,d\theta = 2\pi\sum_{n\geq 0}\left(\int_{\mathbb{R}}H_n(x)^2 e^{-x^2} e^{kix}\,dx\right)\frac{t^{2n}}{n!^2}\tag{4} $$ which simplifies into $$\sqrt{\pi}e^{-k^2/4}e^{2t^2}\,J_0(2kt)= \sum_{n\geq 0}\left(\int_{\mathbb{R}}H_n(x)^2 e^{-x^2} e^{kix}\,dx\right)\frac{t^{2n}}{n!^2}\tag{5} $$ and the wanted integral can be recovered from the Cauchy product between the Taylor series of $e^{2t^2}$ and the Taylor series of $J_0(2kt)$: $$\left(\int_{\mathbb{R}}H_n(x)^2 e^{-x^2} e^{kix}\,dx\right)=\sqrt{\pi}e^{-k^2/4}n!^2\cdot [t^{2n}]\sum_{a,b\geq 0}\frac{2^a (-1)^b k^{2b} t^{2a+2b}}{a!b!^2}\tag{6}$$ such that: $$\boxed{\int_{\mathbb{R}}H_n(x)^2 e^{-x^2} e^{kix}\,dx=\sqrt{\pi}e^{-k^2/4}n!^2\sum_{b=0}^{n}\frac{2^{n-b} (-1)^b k^{2b}}{(n-b)!b!^2}.}\tag{7}$$ ^{Holy cow, it really worked!}

Answer 2

An alternative answer, based off setting n=m in the answer Fourier transform of a product of the Hermite polynomials

There, it is shown that the integral $$ I_{nm}=\int_{-\infty}^{\infty}e^{\pm ikx}H_n(x)H_m(x)e^{-x^2}dx $$ is equal to $$ I_{nm}=\sqrt{\pi}e^{-\frac{1}{4}k^2}m!n!\sum_{l=0}^{\operatorname{min}(n,m)}\frac{2^l\left(\mp i\right)^{n+m-2l}}{l!(m-l)!(n-l)!}k^{n+m-2l}. $$

Setting $n=m$ and relabelling $b=n-l$ leads to your desired result.