The real Hermite polynomials are given by
$$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$
for $x\in \mathbb R$ and $n=0, 1, 2 …$. The Hermite function $H_n$ of order $n$ is an eigenfunction of the harmonic oscillator $\Delta=-\frac{\partial^2}{\partial x^2}+x^2$ corresponding to the eigenvalue $2n+1$, i.e., $$\Delta H_n (x)=(2n+1) H_n(x) .$$
I would like to know, what happens for the complex Hermite polynomials given by
$$H_n(z)=(-1)^n e^{z^2}\frac{d^n}{dz^n}e^{-z^2}$$
for $z\in \mathbb C$ and $n=0, 1, 2 …$, i.e. it exists an operator $\Delta_z$ such that , i.e., $\Delta H_n (z)=\lambda_n \, H_n(z) ?$
Thank you in advance
Best Answer
There is no difference - i.e. the Hermite polynomials are trivially (and uniquely!) extended to the rest of the complex plane via analytic continuation. It's really that simple, though it may be understandably non-intuitive.
https://en.wikipedia.org/wiki/Hermite_polynomials