Commuting elements in Weyl algebras of characteristic zero? Suppose $k$ is a field of characteristic zero and $P$ and $Q$ are commuting elements of the first Weyl algebra. Is it true that $P$ and $Q$ are polynomials in a some third element $H$?
I asked a similar question before but without the characteristic zero condition. 
 A: No. This counterexample is due to Dixmier, Proposition 5.5 of:
Dixmier, J. Sur les algèbres de Weyl Bull. Soc. Math. France 96 (1968) p. 209–242. 
I learned about it from the early pages of 
Makar-Limanov, L. Centralizers in the quantum plane algebra Studies in Lie Theory,
Progress in Mathematics Volume 243, (2006), p. 411-416.
Let 
$$U = (\partial^3 + x^2 - 1)^2 + 2 \partial$$
$$V = (\partial^3 + x^2 - 1)^3 + \frac{3}{2} \left(  (\partial^3 + x^2 - 1)^2  \partial +  \partial (\partial^3 + x^2 - 1)^2 \right).$$ 
Then 
$$UV=VU \quad \mbox{and} \quad V^2 = U^3 + 1.$$
Disclaimer: I have not checked these identities; I just copy-typed them.
Suppose, for the sake of contradiction, that $U=u(T)$ and $V=v(T)$ for some polynomials $U$ and $V$ and some differential operator $T$. Clearly $T$ is nonconstant. Then $T$ does not obey any polynomial relation within the Weyl algebra (exercise). So we must have
$$v(t)^2 = u(t)^3+1$$
as an equality of polynomials. We claim there are no nonconstant polynomials $u$ and $v$ obeying $v(t)^2 = u(t)^3+1$ (in characteristics not $2$ or $3$).
Conceptual proof: $v^2=u^3+1$ is a genus one curve, so it can't be algebraically parametrized by a genus zero curve. 
Direct proof: Since $0^2 \neq 0^3+1$, $u$ and $v$ have no common zeros. Taking derivatives of both sides, we have
$$2 v v' = 3 u^2 u'.$$
So every zero of $v$ is also a zero of $u'$ (with multiplicity), and we deduce that $\deg v \leq \deg u -1$. But also, every zero of $u$ is a zero of $v'$ (twice, in fact) so $\deg u \leq \deg v-1$. Contradiction.
A: However, the $(1,1)$-leading term of each of your given $P$ and $Q$ is a power of the some $(1,1)$-homogeneous element $R$:
$l_{1,1}(P)=\lambda R^n$ and $l_{1,1}(Q)=\mu R^m$,
$\lambda,\mu \in k^*$; just apply Theorem 1.22 (2)
Notice that you can add to your original question "each of $P$ and $Q$ is homogeneous of some degree" and get a positive answer to your question.
