Gaps between roots of consecutive Hermite polynomials

Question

Let $H_k(x)$ be (probabilists' or physicists', does not matter for this question) Hermite polynomials.

It is well-known that all the gaps between consecutive roots of $H_k(x)$ are at least a multiple of $1 / \sqrt{k}$ (see, e.g., Szego, "Orthogonal Polynomials", page 130).

Is the same known to be true if we consider the roots of $H_k(x)$ and $H_{k+1}(x)$ together? Equivalently, is it true that the zeros of $H_{k+1}(x)$ are roughly half-way between its respective local minima and maxima?

Looking at the numerical values for $k \leq 100$, the desired bound seems to be true.

Answer 1

The answers are yes and yes.

Consider the Hermite Gauss functions : $\psi_n(x)=e^{-x^2/2}H_n(x) $, we have two properties: $$\psi_n''(x)+(2n+1-x^2)\psi_n(x)=0 $$ and $$\psi_{n+1}=\psi_n'(x) +x\psi_n(x) $$ For $n$ large, we take a lenght $L>0$ which is very small (for example $L=1/\log{n}$). On any $[x_0-L/2,x_0+L/2]\subset ]- \sqrt{2n},\sqrt{2n}[$, $\psi_n$ is rountly the solution of $$\psi_n''(x)+(2n+1-x_0^2)\psi_n(x)=0 $$ and therefore $\psi_n(x)=A\sin(\omega_0 x+\alpha)$ on $[x_0-L/2,x_0+L/2]$ with $\omega_0=\sqrt{2n+1-x_0^2}$ and $\psi_{n+1}(x)=-\omega_0 A \cos(\omega_0 x+\alpha)+x_0A\sin(\omega_0 x+\alpha)$. In particular the zeros of $\psi_n$ are placed periodically every $\pi/\omega_0$ and the zeros of $\psi_{n+1}$ are just shifted by $\omega_0 (\pi/2 + \arctan(x_0/\omega_0))$

For $x_0=\pm \sqrt{2n}$, the $\sin$ has to be replaced by the Airy function with a scale $1/\sqrt{n}$. And the distribution of its zeros and the zeros of its derivative are the same (with the scale $1/\sqrt{n}$) https://en.wikipedia.org/wiki/Airy_function