$\newcommand{\ad}{\operatorname{ad}}$As my recent post ([here](https://mathoverflow.net/questions/202163/nilpotent-operator-of-the-weyl-algebra)) 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!



Locally nilpotent operators of the Weyl algebra
Dixmier gives a complete characterization of ad-locally nilpotent elements of $A=A\_1$ in his paper ["Sur les algèbres de Weyl"](http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1968__96_/BSMF_1968__96__209_0/BSMF_1968__96__209_0.pdf), Bulletin de la S. M. F., tome 96 (1968), Theorem 9.1: 
An element $q \in A\_1$ is ad-locally nilpotent iff it belongs to the orbit of the constant coefficient differential operators under the automorphism group of $A\_1$.


