Prove a Hermite polynomial property by linear algebra

Question

Let $$ X=\begin{pmatrix} 0 & & & &\\ 1 & 0 & & &\\ & 1 & 0 & &\\ & & \ddots & \ddots &\\ & & & 1 & 0 \end{pmatrix}, D=\begin{pmatrix} 0 & 1 & & &\\ & 0 & 2 & &\\ & & \ddots & \ddots &\\ & & & 0 & n\\ & & & & 0 & \end{pmatrix}\in\mathbb{R}^{(n+1)\times(n+1)} $$ and $$ \mathbf{e}_0=(1,0,\cdots,0)^\mathrm{T},\mathbf{e}_n=(0,\cdots,0,1)^\mathrm{T}\in\mathbb{R}^{n+1} $$ Prove that $$ (X-D)^n\mathbf{e}_0=\exp\left(-\frac{D^2}{2}\right)\mathbf{e}_n $$

Background: it directly comes from two representations of (probabilists') Hermite polynomial: $H_n(x)=(x-D)^n\cdot 1$ and $H_n(x)=e^{-D^2/2}\cdot x^n$, where $D$ is the differential operator. The equivalence of the two representation is not difficult but it seems techniques like generating functions are necessary at least for me. I wonder whether the identity can be proved by linear algebra only without explicitly going into Hermite polynomials.

Answer 1

We first see that $X{\bf e}_k = {\bf e}_{k+1}$ and $D{\bf e}_k = k {\bf e}_{k-1}$ so $X$ acts as multiplication by $x$ and $D \sim \frac{d}{dx}$. The commutator of $[X,D] = \text{diag}(-1,-1,-1,\ldots,-1,n)$ so $[X,D] = -1$ (and higher order correlators vanish) when acting on ${\bf e}_k$ for $k<n$. By Baker–Campbell–Hausdorff we therefore have

$$e^{t(X-D)}{\bf e}_0 = e^{tX}e^{-tD}e^{\frac{t^2}{2}[X,D]}{\bf e}_0$$

Expanding both sides in a power-series and comparing the coefficient of the $t^n$ term gives us

$$\frac{(X-D)^n}{n!}{\bf e}_0 = \sum_{k+\ell+2m = n} \frac{X^k}{k!}\frac{(-D)^\ell}{\ell!}\frac{(-1)^m}{2^mm!}{\bf e}_0$$

Since $D^\ell {\bf e}_0 = 0$ unless $\ell = 0$ the sum above is over all $m$ satisfying $0\leq 2m \leq n$ with $k = n - 2m$ which gives us

$$(X-D)^n{\bf e}_0 = \sum_{0\leq 2m \leq n} \frac{n!}{(n-2m)!}\frac{(-1)^m}{2^mm!}{\bf e}_{n-2m}$$

where we have used $X^{n-2m}{\bf e}_0 = {\bf e}_{n-2m}$ to simplify. Since the right hand side of the identity we want to prove is a function of $D$ acting on ${\bf e}_n$ we try to rewrite the expression above as a power-series in $D$ acting on ${\bf e}_n$. This can be done by using $D^{2m}{\bf e}_n = \frac{n!}{(n-2m)!}{\bf e}_{n-2m}$ to get

$$(X-D)^n{\bf e}_0 = \sum_{0\leq 2m \leq n} \frac{1}{m!}\left(-\frac{D^2}{2}\right)^m {\bf e}_n$$

Finally since $D^{2m}{\bf e}_n = 0$ for $2m>n$ we can extend the upper limit of the sum above from $n$ to infinity and use the power-series definition for $e^{-\frac{D^2}{2}}$, namely $e^{-\frac{D^2}{2}} \equiv \sum_{m=0}^\infty \frac{1}{m!}\left(-\frac{D^2}{2}\right)^m$, to get the desired result

$$(X-D)^n{\bf e}_0 = e^{-\frac{D^2}{2}}{\bf e}_n$$