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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: $$\sum_{k=0}^{\infty}\frac{1}{k+a}h_k(x)h_k(y)$$

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    $\begingroup$ 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 $1$? $\endgroup$ Commented May 9, 2023 at 20:20
  • $\begingroup$ Yes, but what is the close formula of this $$\int^1_0 w^{a}(1-w^2)^{\frac{-1}{2}} e^{ -\frac{1+w^2}{2(1-w^2)} (x^2+y^2)+ \frac{2w}{1-w^2}xy } dw$$ $\endgroup$ Commented May 10, 2023 at 11:21

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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 $a>0$.

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  • $\begingroup$ Thank you a lot. Yes, but what is the close formula of this $$\int^1_0 w^{a}(1-w^2)^{\frac{-1}{2}} e^{ -\frac{1+w^2}{2(1-w^2)} (x^2+y^2)+ \frac{2w}{1-w^2}xy } dw$$ $\endgroup$ Commented May 10, 2023 at 11:21
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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 $G(x,y,E)=f_l(x,E)f_r(y,E)/Wr[f_l,f_r], x<y$. 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.

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  • $\begingroup$ Thank you a lot $\endgroup$ Commented Mar 12 at 22:38

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