TITLE: Solving $\left(x-c_1\frac{d}{dx}\right)^nf(x)=0$ for $f(x)$
QUESTION [11 upvotes]: I'm given that
$$\left(x-c_1\frac{d}{dx}\right)^nf(x) =  0$$
I have to solve for $f(x)$ in terms of $n$.


For $n=0$:
$$f(x)=0 \tag{0}$$

For $n= 1$:


$$\begin{align}
xf(x) - c_1f'(x) &= 0 \\
\quad\implies\quad f(x) &= c_2\exp\left(\frac{x^2}{2c_1}\right) \tag{1}
\end{align}$$

For $n=2$:
$$\begin{align}
\left(x-c_1\frac{d}{dx}\right)^1(xf(x) - c_1f'(x)) &= 0 \\[4pt]
\quad\implies\quad x^2f(x) -xc_1f'(x) -c_1(f(x)+xf'(x)) +c_1^2f''(x) &=0 \\[4pt]
\quad\implies\quad f(x) = k_1\exp\left(\frac{-x^2}{2c_1}\right)
+ k_2x\exp\left(\frac{-x^2}{2c_1}\right) & \tag{2}
\end{align}$$

The case for $n= 3$ gets so complicated that I haven't put up the solution.


The solution is based on hermitian polynomials.

REPLY [3 votes]: Here is a fast track. Writing $D=\partial/\partial x$, $c=c_1$ we have the commutation relation:
$$ ( x - c D) \; e^{\frac{x^2}{2c}} = e^{\frac{x^2}{2c}} (-cD) $$
Thus, $$ 0 = ( x - c D)^n f_n(x) =  ( x - c D)^n  e^{\frac{x^2}{2c}}  e^{-\frac{x^2}{2c}} f_n(x) = e^{\frac{x^2}{2c}}  (- c D)^n  e^{-\frac{x^2}{2c}} f_n(x) \Leftrightarrow $$
$$  (- c D)^n  e^{-\frac{x^2}{2c}} f_n(x) = 0\Leftrightarrow$$
$$ e^{-\frac{x^2}{2c}} f_n(x) = P_n(x),  \ \ P_n\in {\Bbb R}_n[x]\Leftrightarrow$$
$$  f = P_n(x)e^{\frac{x^2}{2c}},  \ \ P_n\in {\Bbb R}_n[x].$$
A fairly general formula (less well-known but with the same proof) is obtained by considering $q\in C^\infty({\Bbb R})$. Then
$$ (D - q'(x))^n f_n(x)=0 \ \ \Leftrightarrow  \ \
  f_n = P_n(x) e^{q(x)}, \ \ P_n\in {\Bbb R}_n[x].$$