Let $L$ be a finite-dimentional complex Lie algebra. $\forall x \in L$, one defines the adjoint action of $x$ on $L$ as the map $\mathrm{ad}_x : L \to L, \text{ with } \mathrm{ad}_x(y) = [x,y]$ for any $y$ in $L$. Let $tr (\mathrm{ad}_x)$ be the trace of the adjoint action of $x$ on $L$ for any $x \in L$. I'm interested in the following question.
Is there a non-zero element $e$ in the universal enveloping algebra $\mathrm{U}(L)$ such that $$xe-ex = tr (\mathrm{ad}_x)e$$ in $\mathrm{U}(L)$, for any $x \in L$.
I only know that it is true in the following special cases.
(1) The trace of the adjoint action is zero, eg. $L$ is semisimple or nilpotent.
(2) $\exists$ $n$, such that $L^n$(the $n$th term of lower central series) $\subseteq$ the center of $L^1 = [L,L]$.
(3) $\exists$ $n$ and $\{e_1,\dots,e_m\}$ is a basis of $L^n$, such that $$e = \sum_{\sigma \in S_m} (-1)^{|\sigma|} e_{\sigma1} \cdots e_{\sigma m}$$ is a non-zero element in $\mathrm{U}(L)$.
