Equivalent definitions of Hermite polynomials

It's not easy to derive one definition from the other, but here's one possible way: the Hermite polynomials can also be expressed in terms of their generating function, that is a function $g(t,x)$ such that $$ g(t,x) = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}. $$ It follows that $$ H_n(x) = \left.\frac{\partial^n}{\partial t^n}g(t,x)\right|_{t=0} $$ Now, if we take the first definition $$ H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}, $$ we see that it looks quite similar. We need to find a way to change the derivatives to $x$ into derivatives to $t$, and we can do that by noting that $$ \frac{\partial}{\partial t}e^{-(t-x)^2} = -\frac{\partial}{\partial x}e^{-(t-x)^2} $$ Thanks to this trick, the function $g(t,x)$ follows immediately: $$ g(t,x) = e^{x^2}e^{-(t-x)^2} = e^{2xt-t^2}. $$ This allows us to derive yet another form for the Hermite polynomials, by equating the terms in $$ e^{2xt-t^2} = \sum_{n=0}^\infty\frac{(2xt-t^2)^n}{n!} = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}. $$ Let us examine the $(n-k)$th term in the first series: $$ (2x-t)^{n-k}\frac{t^{n-k}}{(n-k)!} = \sum_{l=0}^{n-k}\binom{n-k}{l}(2x)^{n-k-l}(-1)^l\frac{t^{n-k+l}}{(n-k)!}. $$ The $l=k$ term in this sum contains $t^n$: $$ (-1)^k(2x)^{n-2k}\frac{t^n}{k!\,(n-2k)!}, $$ provided that $2k\leqslant n$. In other words, all the terms that contribute to the $t^n$ term are $$ H_n(x)\frac{t^n}{n!} = \sum_{k=0}^{[n/2]}(-1)^k(2x)^{n-2k}\frac{t^n}{k!\,(n-2k)!}, $$ so that we've found another expression for $H_n(x)$: $$ H_n(x) = \sum_{k=0}^{[n/2]}(-1)^k(2x)^{n-2k}\frac{n!}{k!\,(n-2k)!}. $$ We can write this as $$ H_n(x) = \sum_{k=0}^{[n/2]}\binom{n}{k}\frac{(n-k)!}{(n-2k)!}(-1)^k(2x)^{n-2k}, $$ which is $$ H_n(x) = \sum_{k=0}^{[n/2]}\binom{n}{k}(-1)^k\frac{d^k}{dx^k}(2x)^{n-k}, $$ and this is a more explicit notation of the other definition $$ \left(2x - \frac{d}{dx}\right)^n (1) $$ In principle, you can follow the steps in the reverse order to derive the first definition from the second, but it's certainly not intuitive. Alternatively, one might use the recurrence relations $$ H_{n+1}(x) = 2xH_n(x) - 2n H_{n-1}(x) $$ $$ H'_n(x) = 2nH_{n-1}(x). $$ It's easy to verify that both definitions satisfy these relations.