Let $h_0(x)=e^{-x^2/2}$ and $h_k=B^kh_0$, where $B=-\dfrac{d}{dx}+x$. Show that the $\dfrac{h_k}{\|h_k\|_2}$'s form a complete orthogonal system.
(Hint: We have $\langle Af,g\rangle=\langle f,Bg\rangle$, where $A=\dfrac{d}{dx}+x$. Consider $\langle B^kh_0,B^lh_0\rangle$, and use the commutator formula $[A,B^n]=ncB^{n-1}$.)
We can show that $h_k(x)=H_k(x)e^{-x^2/2}$, where $H_k(x)$ is a polynomial of degree $k$ defined by $H_k(x)=2xH_{k-1}(x)-H'_{k-1}(x)$.
We have $\langle B^kh_0,B^lh_0\rangle=\langle A^lB^kh_0,h_0\rangle$. I don't see why this should be equal to $0$. Also, what does the commutator formula mean?
Using the suggested hint, we have, using the commutator formula $$\langle B^kh_0,B^lh_0\rangle =\langle AB^kh_0,B^{l-1}h_0\rangle\\ =\langle B^kAh_0,B^{l-1}h_0\rangle+nc\langle B^{k-1}h_0,B^{l-1}h_0\rangle.$$ The interest of the commutator formula is that $Ah_0=0$, hence using it the $A$ is on the good side. We thus have, for each $k,l\geqslant 1$, $$\langle B^kh_0,B^lh_0\rangle =nc\langle B^{k-1}h_0,B^{l-1}h_0\rangle.$$ This reduces the proof to the verification of orthogonality in the case $l=0$ and $k\geqslant 1$, which follows from the already used facts that $\langle Af,g\rangle=\langle f,Bg\rangle$ and $Ah_0=0$.