# Showing that $\operatorname{div}\Psi\nabla\Psi^*-\operatorname{div}\Psi^*\nabla\Psi=\operatorname{div}[\Psi\nabla\Psi^*-\Psi^*\nabla\Psi]$

Can anyone show me how to prove that:$$\operatorname{div}\Psi \nabla\Psi^*-\operatorname{div}\Psi^* \nabla\Psi=\operatorname{div}\left[\Psi\nabla\Psi^*-\Psi^*\nabla\Psi\right]$$

I have this in my notes but don't know how to obtain it. Thanks.

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Divergence is a linear operator no? –  anon Mar 11 '13 at 3:59

Let $F=(F_1(x_1,x_2,x_3),F_2(x_1,x_2,x_3),F_3(x_1,x_2,x_3))$ be a $\mathcal C^1$ vector field in $\mathbb R^3$. We have
$$\operatorname{div}(F):=\sum_{i=1}^3 \frac{\partial F_i}{\partial x_i}.$$
As the partial derivative operator $\frac{\partial }{\partial x_i}$ is linear by its very definition, then $\operatorname{div}(\cdot)$ is linear as well. In fact, $\operatorname{div}$ is a finite sum of linear operators. In other words
$$\operatorname{div}(F+G)= \operatorname{div}(F)+\operatorname{div}(G),$$
for all vector fields $F$ and $G$ as above.