# Closed formula for Hermite polynomials

Hermite polynomials $$H_k(x), x \in \mathbb{R}, k \in \mathbb{N}$$ are defined by the formula $$H_k(x)=(-1)^k e^{x^2} \frac{d^k}{d x^k}\left(e^{-x^2}\right) .$$ Each $$H_k(x)$$ is a polynomial of exact degree $$k$$. The Hermite polynomials are also given by the generating function $$e^{2 x w-w^2}=\sum_{k=0}^{\infty} \frac{H_k(x)}{k !} w^k$$

Define the Hermite functions $$\tilde{h}_k(x)$$ by $$\tilde{h}_k(x)=H_k(x) e^{-\frac{1}{2} x^2} .$$

We have the Mehler's formula, for the Hermite functions $$\tilde{h}_k(x)$$.

Proposition For $$w \in \mathbb{C},|w|<1$$ and $$x, y \in \mathbb{R}$$, $$\sum_{k=0}^{\infty} \frac{\tilde{h}_k(x) \tilde{h}_k(y)}{2^k k !} w^k=\left(1-w^2\right)^{-\frac{1}{2}} e^{-\frac{1}{2} \frac{1+w^2}{1-w^2}\left(x^2+y^2\right)+\frac{2 w}{1-w^2} x y}$$

One has $$\int_\Bbb R\left(\tilde{h}_k(x)\right)^2 d x=2^k k ! \sqrt{\pi} .$$

Thus we can normalise $$\tilde{h}_k(x)$$ by defining $$h_k(x)=\left(2^k k ! \sqrt{\pi}\right)^{-\frac{1}{2}} \tilde{h}_k(x)$$

This family $$\{h_k(x): k \in \mathbb{N}\}$$ is an orthonormal system in $$L^2(\mathbb{R})$$. But we can say more.

Theorem The system $$\{h_k(x): k \in \mathbb{N}\}$$ is an orthonormal basis for $$L^2(\mathbb{R})$$. Consequently, every $$f \in L^2(\mathbb{R})$$ has an expansion $$f(x)=\sum_{k=0}^{\infty}\left(f, h_k\right) h_k(x)$$ where the series converges to $$f$$ in the $$L^2$$ norm.

My question is there a close formula for this sum: $$\sum_{k=0}^{\infty}\frac{1}{k+a}h_k(x)h_k(y)$$

• Wouldn't you just expect this to be whatever you get by multiplying Mehler's formula by $\pi w^{a - 1}$ and integrating with respect to $w$ from $0$ to $1$? Commented May 9, 2023 at 20:20
• Yes, but what is the close formula of this $$\int^1_0 w^{a}(1-w^2)^{\frac{-1}{2}} e^{ -\frac{1+w^2}{2(1-w^2)} (x^2+y^2)+ \frac{2w}{1-w^2}xy } dw$$ Commented May 10, 2023 at 11:21

Up to some normalization, the harmonic oscillator $$H$$ is self-adjoint such that $$\langle Hu, u\rangle=\sum_{k\ge 0}(\frac12+k) \vert u_k\vert^2,$$ and thus defining a self-adjoint $$A$$ by the equality $$\langle Au, u\rangle=\sum_{k\ge 0}(a+k) \vert u_k\vert^2, \quad\text{implying}\ A=H+a-\frac12.$$ As a result your sum is the kernel of the operator $$(H+a-\frac12)^{-1},$$ which makes sense for $$a>0$$.
• Thank you a lot. Yes, but what is the close formula of this $$\int^1_0 w^{a}(1-w^2)^{\frac{-1}{2}} e^{ -\frac{1+w^2}{2(1-w^2)} (x^2+y^2)+ \frac{2w}{1-w^2}xy } dw$$ Commented May 10, 2023 at 11:21
It was already written above that the required sum is the kernel of the operator $$(H+a−1/2)^{−1}$$
For your problem, this will be the Green's function for a second-order linear ordinary differential equation (stationary Schrödinger equation on the entire axis with a harmonic oscillator potential) $$\left(-\frac{d^2}{dx^2}+x^2-1/2-E\right) G(x,y,E)=\delta(x-y).$$ In this case, the Green's function $$G(x,y,E)=f_l(x,E)f_r(y,E)/Wr[f_l,f_r], x. where $$f_l(x,E)$$,$$f_r(x,E)$$ are (left/right) solutions of the underlying ODE, with the following asymptotics: left - $$f_l(x\to -\infty,E)\to 0$$ and right $$f_r(x \to\infty,E)\to 0$$. For the harmonic oscillator, the corresponding solutions are parabolic cylinder functions https://en.wikipedia.org/wiki/Parabolic_cylinder_function. Note that $$a=-E$$, and some rescaling and a constant shift of the potential x^2 may be needed.