$\newcommand{\ad}{\operatorname{ad}}$As my recent post (here) did not receive any answers yet, I thought I would ask a similar question in which I'm also interested.

Let $A=$ $^{k \langle x,y\rangle }\Big/_{(yx-xy-1)}$ be the Weyl Algebra over a field $k$ of characteristic $0$. For an element $q(x,y) \in A$, we have the operator $\ad_q(h):=[q,h]=qh-hq$. Also, we say that $\ad_q$ is locally nilpotent if $\forall h \in A$ $\exists n_h \in \mathbb{N}$ such that $\ad_q^{n_h}(h)=0$. My question is simply:

For which $q$ is $\ad_q$ locally nilpotent?

If $q \implies \ad_q$ locally nilpotent, I say that $q$ is *good*. Here's what I have:

- It's simple to show that $q(x,y)=r(x)+s(y)$ is
*good*if $\deg(r)=0,1$ or $\deg(s)=0,1$. - If $q$ is a polynomial in $\theta=xy$, I can show that $q$ is
*good*iff $q$ is constant.

As $\ad_q$ is a derivation (it satisfies the Leibniz rule), it might be useful to notice that $q$ is *good* iff $\exists n_x,n_y$ such $\ad_q^{n_x}(x)=0$ and $\ad_q^{n_y}(y)=0$.

I conjecture that the polynomials of the form $r(x)+s(y)$ with $\deg(r)=0,1$ or $\deg(s)=0,1$ are in fact the only *good* polynomials, but currently I'm not seeing a direct way to prove this. Any ideas/hints? Thank you!