TITLE: What is the Weitzenböck formula for the $\bar\partial$-Laplacian
QUESTION [6 upvotes]: It is well-known that the Weitzenböck formula for the real Laplacian is
$$\frac12 Δ|∇f|2=|Hessf|2+⟨∇f,∇Δf⟩+Ricci(∇f,∇f)$$
where $Hess$ denotes the Hessian tensor of $f$. and $\nabla f$ denotes the gradient vector of $f$, $Ricci$ denotes the Ricci curvature of the manifold $M$.
If $\Delta_{\bar\partial}$ denotes the $\bar\partial$-Laplacian, it is well-known that it is half of the real Laplacian. So I am wondering is there any formula of the Weitzenböck formula in complex coordinates. (Assume the manifold is Kähler). Apparently one can devided the above formula by 2 to the one, but the expression I want should be expressed by $f_{i\bar j}$ and etc. 
ps. The Comparison Geometry of Ricci Curvature, by Shunhui Zhu, 221-262 had a very nice introduction to this formula in real case.
http://library.msri.org/books/Book30/contents.html
However I am not familiar with Kaehler case, for example, I dont know the such a formula can be derived in the same fashion as in Zhu's paper? Any book or paper with detailed calculation would be helpful.

REPLY [2 votes]: Weitzenböck Identities for Kahler manifolds is as follows
Let $\Theta $ be the Chern curvature form of $h$ for the Hermitian line bundle over Kahler manifold $(X,\omega)$
For any smooth section $\xi ∈ Γ(T^{0,1}X \otimes L)$ we have, 
$$(\bar{\partial^*}\bar\partial+\bar\partial\bar{\partial^*})\xi=\bar{\nabla^*}\bar\nabla\xi+(\Theta+Ric(\omega))(\xi,.)$$
and
$$(\bar{\partial^*}\bar\partial+\bar\partial\bar{\partial^*})\xi={\nabla^*}\nabla\xi+(\Theta(\xi,.)-(tr_\omega\Theta)\xi$$
Note that here Ricci make sense as follows
If we take a section $s\in H^0(X,L)$ and $\xi=\alpha_{\bar i}d\bar z^i\otimes s$ then Ricci can be defined as follows
$$Ric(\xi,.)=R_{i\bar j}\alpha_{\bar i}d\bar z^j\otimes s$$
I learn it from paper of Tian