Let $Df=x^2f-f''$ and $Af=xf-f'$ on $L^2(\Bbb R)$, and define $e_0(x)=\pi^{-1/4}e^{-x^2/2}$ and $e_n=\dfrac1{\sqrt{2n}}A(e_{n-1})$. I would like to show that $e_n$ are eigenvectors (eigenfunctions) of $D$. If I compute $De_0(x)$ and $De_1(x)$ by definition above, I get
$D(e_0)(x)=\pi^{-1/4}e^{-x^2/2}=e_0(x)$, and since
$e_1(x)=\dfrac1{\sqrt{2}}x\pi^{-1/4}e^{-x^2/2}+\dfrac1{\sqrt{2}}x\pi^{-1/4}e^{-x^2/2}=\sqrt2x\pi^{-1/4}e^{-x^2/2}$, we have
$De_1(x)=x^2e_1(x)-e_1''(x)=x^3\sqrt2\pi^{-1/4}e^{-x^2/2}-\sqrt2\pi^{-1/4}e^{-x^2/2}(x^3-2x)=2e_1(x)$
But the commutator $[D,A]=(-\dfrac{d^2}{dx^2}+x^2)(x-\dfrac{d}{dx})-(x-\dfrac{d}{dx})(-\dfrac{d^2}{dx^2}+x^2)=2A$. Hence, $D(e_n)=D(\dfrac1{\sqrt{2n}}A(e_{n-1}))=\dfrac1{\sqrt{2n}}(AD+2A)(e_{n-1})=\dfrac1{\sqrt{2n}}(\lambda+2) A(e_{n-1})=(\lambda+2)e_{n-1}$, where $\lambda$ is the eigenvalue corresponding to $e_{n-1}$, provided $e_{n-1}$ is an eigenfunction of course. But then for $n=1$, that is, $\lambda=1$, I get $D(e_1)=3e_1\ne 2e_1$.
Where have I made a mistake?
Consider two operators $$A=-D+x, \quad B=D+x.$$ One has $$AB=(-D+x)(D+x)=-D^{2}+x^{2}-1.$$ Moreover $$AB-BA=[A,B]=-2.$$ So your original operator is $$L=AB+1.$$ Consider a vector $$e_{n}=\frac{1}{\sqrt{2n}}Ae_{n-1}=\frac{1}{\sqrt{2^{n}n!}}A^{n}e_{0}.$$ We have $$Le_{n}=ABe_{n}+e_{n}=\frac{1}{\sqrt{2^{n}n!}}ABA^{n}e_{0}+e_{n}.$$ Note that $$ABA^{n}=A^{2}BA^{n-1}+2A^{n}=A^{3}BA^{n-2}+4A^{n}$$ $$=A^{4}BA^{n-3}+6A^{n}=...=A^{n}AB+2nA^{n}.$$ Further we noted $Be_{0}=0$ and thus $$Le_{n}=\frac{2n}{\sqrt{2^{n}n!}}A^{n}e_{0}+e_{n}=(2n+1)e_{n}.$$
This solves your problem!