An identity on Kähler manifold I am reading  The Seiberg–Witten equations and applications to the topology of smooth four manifolds by John Morgan. In the calculation for a Kähler surface in page 116, he uses an identity: for a smooth real valued function $\lambda$ on a Kähler surface $\bar\partial\partial\lambda\wedge\omega=\frac{2}{i}\Delta(\lambda)dvol$. However my own calculation doesn't match up with the coefficint, can someone please point out where is the mistake I am making. Here's my calculation. I wonder if there's an easier way to see this.
On an appropriate local chart $\omega=dx_1\wedge dy_1+dx_2\wedge dy_2$.
\begin{align}
&\partial\bar\partial\lambda\wedge\omega\\
&=\bigg(\sum\limits_{k=1}^2\frac{\partial^2\lambda}{\partial z_k\partial\bar{z}_k}  dz_k\wedge d\bar{z}_k\bigg)\wedge\omega\\
&=\bigg(\sum\limits_{k=1}^2\frac{1}{2}\big(\frac{\partial}{\partial x_k}-i\frac{\partial}{\partial y_k}\big)\frac{1}{2} \big(\frac{\partial}{\partial x_k}+i\frac{\partial}{\partial y_k}\big)(\lambda) dz_k\wedge d\bar{z}_k\bigg)\wedge\omega\\
&=\frac{1}{4}\times -2i \sum\limits_{k=1}^2\big(\frac{\partial^2}{\partial x_k^2}+\frac{\partial^2}{\partial y_k^2}\big)(\lambda) dx_k\wedge dy_k\wedge\omega\\
&=\frac{i}{2}\Delta(\lambda)dvol
\end{align}
 A: On any Kahler manifold, the Kahler identities say that $[\Lambda, \bar\partial] = -i \partial^\dagger$, where $\Lambda$ is the adjoint of the Lefchetz operator and $\partial^\dagger$ is the adjoint of $\partial$. For a smooth function $f$, we know that $\partial^\dagger f = 0$ for degree reasons. Thus
$$
\Delta_\partial f
= (\partial \partial^\dagger + \partial^\dagger \partial)f
= \partial^\dagger \partial f
= i(\Lambda \bar\partial - \bar\partial \Lambda) \partial f
= i \Lambda \bar\partial\partial f
$$
as $\Lambda \partial f = 0$ for degree reasons. As $\Lambda$ is complex linear and $\Delta_d = 2\Delta_\partial$, we conclude that
$$
\Delta_d f
= 2i \Lambda \bar\partial\partial f
= \frac 2i \Lambda \partial \bar\partial f
$$
Taking the product with the volume form $\omega^n/n!$ we get
$$
\Delta_d f \, \omega^n/n!
= \Lambda \frac 2i \partial\bar\partial f \, \omega^n/n!
= \frac 2i \, \partial\bar\partial f \wedge \omega^{n-1}/(n-1)!.
$$
On a surface, where $n = 2$, this agrees with what your calculations gave.
