Hermite polynomials form complete system 
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.
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 $H_k(x)=2xH_{k-1}(x)-H'_{k-1}(x)$.

(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 derive that $\langle B^kh_0,B^lh_0\rangle=\langle A^lB^kh_0,h_0\rangle$. However,  I don't see why this should be equal to $0$. Also, what does the commutator formula mean?
 A: I don't think using the commutator formula is the best way. In my opinion it is easier to just note how $A$ works on functions of the form $f\cdot h_0$:
$$\begin{align}
A\left(f(x)e^{-x^2/2}\right) &= \frac{d}{dx}\left(f(x)e^{-x^2/2}\right) + xf(x)e^{-x^2/2}\\
&= f'(x)e^{-x^2/2} + f(x)\left(-xe^{-x^2/2}\right) + xf(x)e^{-x^2/2}\\
&= f'(x)e^{-x^2/2}.
\end{align}$$
Then, for $k < m$, you have
$$\langle h_k, h_m\rangle = \langle A^mh_k,h_0\rangle = \langle A^m(H_kh_0),h_0\rangle$$
and the above finishes it in no time.
If you use the commutator formula, you get - as a first step -
$$\langle B^kh_0,B^mh_0\rangle = \langle AB^k h_0,B^{m-1}h_0\rangle = \langle (kcB^{k-1} + B^kA)h_0,B^{m-1}h_0\rangle,$$
and you still need to use $Ah_0 = 0$ to simplify it.
To see the completeness of the system, we show that the orthogonal complement of its span is $\{0\}$.
Let $f \in L^2(\mathbb{R})$. Consider the function $F \colon \mathbb{C}\to\mathbb{C}$ defined by
$$F(z) = \int_\mathbb{R} f(x) e^{zx - x^2/2}\,dx.$$
Since $\left\lvert e^{zx-x^2/2}\right\rvert \leqslant e^{\lvert z\rvert\cdot\lvert x\rvert-x^2/2}$, the exponential factor is in $L^2(\mathbb{R})$ for all $z\in\mathbb{C}$, uniformly dominated for $z$ in any compact subset of $\mathbb{C}$, hence continuous, and by your favourite method (Morera's theorem, dominated convergence theorem, …), you can show that $F$ is holomorphic.
Now suppose $\langle h_k,f\rangle = 0$ for all $k\in \mathbb{N}$. Since each monomial $x^m$ is a linear combination of $H_0,\dotsc,H_m$, we then have
$$\begin{align}
F^{(m)}(0) &= \int_\mathbb{R} f(x) \left(\frac{d}{dz}\right)^m\Biggl\lvert_{z=0}e^{zx-x^2/2}\,dx\\
&= \int_\mathbb{R} f(x)\cdot x^me^{-x^2/2}\,dx\\
&= \langle x^m\cdot h_0,f\rangle\\
&= 0.
\end{align}$$
Thus we have $F \equiv 0$. But
$$\begin{align}
F(-it) &= \int_\mathbb{R} f(x) e^{-itx-x^2/2}\,dx\\
&= \int_\mathbb{R} \left(f(x)e^{-x^2/2}\right)e^{-itx}\,dx\\
&= N\cdot \mathcal{F}\left[f\cdot h_0\right](t),
\end{align}$$
is, modulo a normalisation factor, the Fourier transform of $f\cdot h_0$. So $\mathcal{F}\left[ f\cdot h_0\right] = 0$, whence $f\cdot h_0 = 0$ almost everywhere. Since $h_0 > 0$ everywhere, that means $f = 0$ almost everywhere, and therefore
$$\operatorname{span} \left\lbrace h_k : k \in \mathbb{N}\right\rbrace^\perp = \{0\}.$$
