Ladder Operations and Rodriguez formula

Question

During my attempt to prove the Rodriguez Formula for Hermite Polynomials by using the Ladder Operators, $H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$, I arrived to the formula $H_n(x) = e^{x^2/2}(-\frac{d}{dx}+x)^ne^{-x^2/2}$.Where $(-\frac{d}{dx}+x)^n$ represents the operator being applied n times. Now I have no idea on how to prove these two to be equivalent. Any hints or solutions would be appreciated

Answer 1

$\textbf{Hint}:$ Notice that $e^{x^2/2} = e^{x^2}e^{-x^2/2}$ and

$$\left[e^{-x^2/2},\left(-\frac{d}{dx}+x\right)\right] = -xe^{-x^2/2}$$

Answer 2

This is completely equivalent to @Ninad Munshi 's fine operatorial calculus identity, so then $$ e^{x^2} \partial_x e^{-x^2}=e^{x^2/2} (\partial_x -x) e^{-x^2/2}~~~\implies \\ (e^{x^2} \partial_x e^{-x^2})^n =(e^{x^2/2} (\partial_x -x) e^{-x^2/2})^n. $$ Multiplying on the right by 1 to terminate further action of the derivatives, you just recover $(-)^n H_n(x)$. I suspect that conversion was at the root of your puzzlement. In fact, WP has the shorter reduced form, $$ H_n(x) = \left(2x - \partial_x \right)^n \cdot 1 $$