I want to show that if $(M, g, \omega)$ is a Kähler manifold, then the following formula holds: $$ (*) \qquad \qquad \big\langle i\partial \bar{\partial}\varphi, \omega \big\rangle =\Delta\varphi=-g^{j, \bar{k}}\frac{\partial^2\varphi}{\partial z^j\partial\bar{z}^k} . $$ Here, $\langle\cdot, \cdot\rangle $ is the induced inner product on forms, $\varphi$ is a real function (let's say, at least $C^2$) and $\Delta \varphi:=(\bar{\partial}\bar{\partial}^*+\bar{\partial}^*\bar{\partial})\varphi$ is the usual $\bar{\partial}$-laplacian. I've encountered this identity in the context of a short seminar on complex Monge–Ampère equations, but they simply use it as a plain known and "trivial" fact. In fact, I already know that $\Delta\varphi=-g^{j, \bar{k}}\frac{\partial^2\varphi}{\partial z^j\partial\bar{z}^k}$, but the lecturer said that the identity $(*)$ gives it a proof on its own (he didn't know a proof neither, he just told me not to be bothered by it because it is just a "geometry thing")
Can you provide me any hints or insights on why this formula holds? It looks like a quite nice identity on its own.
$$ \big\langle i\partial \bar{\partial}\varphi, \omega \big\rangle = \big\langle i\varphi_{i\bar j}\,dz^i\wedge d\bar z^j, i g_{j \bar{k}} d z^{j} \wedge d \bar{z}^{k}\big\rangle=\pm\varphi_{i\bar j}g^{i \bar{j}},$$ since $\big\langle dz^i, d\bar z^j\big\rangle=g^{i \bar{j}}$.