I'm studying the paper Quantum dynamical semigroups generated by noncommutative unbounded elliptic operators by Changsoo Bahn and Chul Ki Ko and Yong Moon Park(https://arxiv.org/abs/math-ph/0505026v1).
I have a question on the proof of theorem 3.1 in page 15 of this paper, in which a part of deduction goes as
\begin{align} &[W^2-\sum_{l=1}^d(W_l)_l,W_k\partial_k+\partial_kW_k]\\=&[W^2,W_k\partial_k+\partial_kW_k]-[\sum_{l=1}^d(W_l)_l,W_k\partial_k+\partial_kW_k]\\=&-2W_k((W^2)_k-\sum_{l=1}^d(W_l)_{lk}) \end{align}
where $W=(W_1,\cdots,W_d)$ is a function $:\mathbb{R}^d \to \mathbb{R}^d$, $(W_i)_k$ is the $k^{th}$ derivative of $W_i$ and $W^2=\sum_{i=1}^d (W_i)^2$.
I am stuck in the last equal sign, for which I think we need for $A,B \in \mathcal{D}$:
$$[A,B\partial_k+\partial_k B]=-2B(A)_k.$$
But my calculation for this goes as:
\begin{align} &[A,B\partial_k+\partial_k B]+2B(A)_k\\=&AB\partial_k+A\partial_kB-B\partial_kA-\partial_kBA+2B\partial_kA-2BA\partial_k\\=B(A)_k\\\not = &0 \end{align}
Am I missing something or is the paper wrong?
Using the relation $\partial_k a = a \partial_k + (a)_k$
$$[A, B\partial_k + \partial_k B] = A B\partial_k + A \partial_k B - B \partial_k A -\partial_k BA = AB \partial_k + AB \partial_k + A(B)_k - AB \partial_k -B(A)_k - (AB)_k - AB \partial_k $$ cancelling anything out and using product rule $(ab)_k = (a)_kb + (b)_ka$ $$[A, B\partial_k + \partial_k B] = A(B)_k - B(A)_k - (A)_kB- (B)_kA = -2 (A)_k B$$