Equivalency of Hermite Polynomial Definitions

Hermite polynomials are a set of polynomials which arise frequently in probability calculations and in quantum physics when finding the wave functions of harmonic oscillators. Often, Hermite polynomials are defined by two seemingly different mathematical definitions. Here, it is proved that these definitions in fact yield equivalent results. Without further ado, the two common definitions of Hermite polynomials, $H_n(x)$, are as follows:

\begin{align}
H_n(x) &= (-1)^n e^{x^2} \frac{d^n}{{dx}^n} e^{-x^2} \\
H_n(x) &= e^{x^2/2} \left(x - \frac{d}{dx}\right)^n e^{-x^2/2}
\end{align}

To prove the equivalency of the above two definitions, define operators $A$ and $B$ as:

\begin{align}
A &= - e^{x^2} \frac{d}{dx} e^{-x^2} \\
B &= e^{x^2/2} \left(x - \frac{d}{dx}\right) e^{-x^2/2}
\end{align}

In this way, we can write:

\begin{align}
A^n &= (-1)^n e^{x^2} \frac{d^n}{{dx}^n} e^{-x^2} \\
B^n &= e^{x^2/2} \left(x - \frac{d}{dx}\right)^n e^{-x^2/2}
\end{align}

Thus, if we prove $A=B$, then definitions $(1)$ and $(2)$ are proved to be equivalent. Let $f$ be an arbitrary function.

\begin{align}
Af &= - e^{x^2} \frac{d}{dx} \left(e^{-x^2}f\right) \\
Bf &= e^{x^2/2} \left(x - \frac{d}{dx}\right) \left(e^{-x^2/2}f\right)
\end{align}
\begin{align}
Af &= 2xf - \frac{d}{dx}f \\
Bf &= xf + xf - \frac{d}{dx} f
\end{align}
\begin{equation}
A=B
\end{equation}

Therefore, the two definitions, $(1)$ and $(2)$, of Hermite polynomials are proved to be equivalent.