I was seeing this question in which 2 definitions of the Hermite polynomials are given:
$$H_n(x) = (-1)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$
and $$ H_n(x) = \left( 2x - \frac{d}{dx} \right)^n(1) $$
Now, I found another definition:
$$ H_n(x) = e^{x^2/2}\left(x-\frac{d}{dx}\right)^n e^{-x^2/2} $$
There is plenty of literature on the first two definitions, but I have found none on the third one. I would like to know if there is a way of deducing the first or second definitions from the third one. I tried expanding in a binomial expansion but got nowhere. I appreciate your help. Or is there any way of showing that the last expression matches the coefficients of the generating function expansion?
The generating function for Hermite polynomials is
$$ e^{2xt -t^2} = \sum_{n = 0}^{+\infty} H_n(x)\frac{t^n}{n!} $$
where the "coefficients" $H_n(x)$ are obtained via a Taylor expansion of the l.h.s
$$ H_n(x) = \left[\left(\frac{\partial}{\partial t}\right)^n e^{2xt - t^2} \right]_{t= 0} = \left[e^{x^2}\left(\frac{\partial}{\partial t}\right)^n e^{-(x-t)^2} \right]_{t= 0} $$
note that the exponential is a function of $x-t$, so we can use the fact that $\partial f(x -t)/\partial t = -\partial f(x-t)/\partial x$, therefore
$$ H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2} \tag{1} $$
To see this is equivalent to your last expression, define the operator $P$ as
$$ \hat{P} = -e^{x^2}\frac{d}{dx}e^{-x^2} $$
and
$$ \hat{Q} = e^{x^2/2}\left(x - \frac{d}{dx} \right)e^{-x^2/2} $$
and calculate
$$ \hat{P}f = -e^{x^2}\frac{d}{dx}\left(fe^{-x^2}\right) = 2xf - \frac{df}{dx} $$
and
$$ \hat{Q}f = e^{x^2/2}\left(x - \frac{d}{dx}\right)e^{-x^2/2} = 2xf - \frac{df}{dx} = \hat{P}f $$
this means that $\hat{P} = \hat{Q}$. Replacing this in Eq. (1) you get
\begin{eqnarray} H_n(x) &=& (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2} \\ &=& e^{x^2/2}\left(x - \frac{d}{dx}\right)^ne^{-x^2/2} \tag{2} \end{eqnarray}
And I believe that should answer your question. Your second equation follows from making a transformation of the form $x\to \sqrt{2}x$