Do Hermite polynomials imply a weight for quantum harmonic oscillator wavefunctions? I know that solutions of quantum harmonic oscillator can be expressed in the form of Hermite polynomials.
But I recently came to know that Hermite polynomials are actually orthogonal polynomials having $e^{-x^2}$ weight function.
So, here is my question (out of curiosity):
Is there a connection between the exponential weight function and the harmonic oscillator solutions? Perhaps the oscillator problem can somehow be transformed into a different space having $e^{-x^2}$ weight function with simpler solutions.
Thanks for any help.
Sorry for missing out some technical details as I am a beginner in Physics.
 A: Absolutely. The Hermite polynomials 
$$
H_n(x) = (-1)^n e^{x^2} \partial_x^n  e^{-x^2} = \left(2x -  \partial_x\right)^n \cdot 1
$$ 
are orthogonalized by
$$
\int_{-\infty}^\infty H_m(x) H_n(x)\, e^{-x^2} \,dx = \sqrt{\pi}\, 2^n n! ~ \delta_{nm} ~,
$$
whereas the (nonpolynomial) Hermite functions
$$
\psi_n(x) = \left (2^n n! \sqrt{\pi} \right )^{-\frac12} e^{-\frac{x^2}{2}} H_n(x) = (-1)^n \left (2^n n! \sqrt{\pi} \right)^{-\frac12} e^{\frac{x^2}{2}}  \partial_x ^n e^{-x^2}
$$
are orthogonalized trivially, 
 $$\int_{-\infty}^\infty \!\psi_n(x) ~ \psi_m(x) \,dx = \delta_{nm} ~,
$$
so they obey the quantum oscillator eigenvalue equation,
$$
\partial_x^2 ~ \psi_n(x) + \left(2n + 1 - x^2\right) \psi_n(x) = 0.$$
It is these functions that are the eigenfunctions of the Fourier transform, 
as the oscillator Hamiltonian is invariant under Fourier transformation.
The Hermite polynomials happened to have been invented first and be more mathematically meaningful by virtue of being an Appel sequence,
$$
\partial_x H_n(x) = 2nH_{n-1}(x),
$$
since 
$$
 H_n(x) = 2^n ~e^{-\partial_x^2 /4} ~ x^n ~.
$$
