Looking for the functions $\alpha_n(x)$ that satisfy $\sum_{n=0}^{\infty} e^{i \alpha_n(x)}\frac{H_n(x)}{\sqrt{2^n n!}}=0$

Question

I am looking for the functions $\alpha_n(x)$ that satisfy $\sum_{n=0}^{\infty} e^{i \alpha_n(x)}\phi_n(x)=0 \text{ or } \delta(x)$ where $\phi_n(x)=\frac{H_n(x)}{\sqrt{2^n n!}}$ and $H_n(x)$ is the $n^{\small\mbox{th}}$ Hermite Polynomial (physicist's convention). Do these functions exist or is there a method to find them numerically?

Answer 1

You mentioned that you would be interested in seeing complex $\alpha_{n}(x)$ solutions; I believe I have the answer in the $\delta(x)$ case. We will use the fact that:

$$ \int_{-\infty}^{\infty}H_{n}(x)H_{m}(x)e^{-x^2}\,\text{d}x = \sqrt{\pi}2^n n! \delta_{n m} \tag{1} $$

We wish to find $\alpha_{n}(x)$ such that:

$$ \sum_{n=0}^{\infty}\frac{e^{i\alpha_{n}(x)}}{\sqrt{2^n n!}}H_{n}(x)=\delta(x) \tag{2} $$

Begin by multiplying both sides by $H_{m}(x)$ and integrating over the real line:

$$ \sum_{n=0}^{\infty}\frac{1}{\sqrt{2^n n!}}\int_{-\infty}^{\infty}H_{n}(x)H_{m}(x)e^{ i\alpha_{n}(x)}\,\text{d}x =\int_{-\infty}^{\infty}H_{m}(x)\delta(x)\,\text{d}x $$

Motivated by $(1)$, let us assume that the $\alpha_{n}(x) = i(x^2 - C_{n})$ for some constant $C_{n}$. Then, $e^{i\alpha_{n}(x)}=e^{-x^2 + C_{n}}=B_{n}e^{-x^2}$, so:

$$ \sum_{n=0}^{\infty}\frac{B_{n}}{\sqrt{2^n n!}}\int_{-\infty}^{\infty}H_{n}(x)H_{m}(x)e^{-x^2}\,\text{d}x =\int_{-\infty}^{\infty}H_{m}(x)\delta(x)\,\text{d}x $$

Now use $(1)$ and the fact that $\int f(x) \delta(x)\,\text{d} x = f(0)$:

$$ \sum_{n=0}^{\infty}\frac{B_{n}}{\sqrt{2^n n!}}\cdot \sqrt{\pi}2^n n! \delta_{n m} = B_{m} \sqrt{\pi 2^m m!} =H_{m}(0)\implies B_{m} = \frac{H_{m}(0)}{\sqrt{\pi 2^m m!}} $$

Therefore, we have $\alpha_{n}(x) = i\left(x^2 - \ln(\frac{H_{n}(0)}{\sqrt{\pi 2^n n!}})\right)$ And therefore:

$$ \sum_{n=0}^{\infty}\frac{H_{n}(0)}{\sqrt{\pi} 2^n n!}H_{n}(x)e^{-x^2}=\delta(x) \\ \sum_{n=0}^{\infty}\frac{H_{2n}(0)}{\sqrt{\pi} 4^n (2n)!}H_{2n}(x)e^{-x^2}=\delta(x) \\ $$

Since $H_{2n+1}(0)=0$. Note that this trick won't work for the $0$ case since the $B_{n}$ would all be $0$.