How would one prove $[f,[\nabla^2,f]]=-2(\nabla f)^2$? How would one prove this equation:
$$[f,[\nabla^2,f]]=-2(\nabla f)^2 $$

And I'm confused that $\nabla f\nabla f$ equals $(\nabla f)^2$ or $\nabla(f\nabla f)$.
 A: This is pretty straightforward if you assume $\nabla$ fulfills the Leibnitz rule (so the relationship works for any covariant as well as co-ordinate derivative - this is really the definition of $\nabla$) and also take heed that $−2(\nabla f)^2$  is to be interpreted as a multiplication operator. $\nabla f \nabla f$  is, strictly speaking, ambiguous without brackets: it could be either of your expressions.
A: Step 1
$[\nabla^2,f]g=\nabla \cdot \nabla(fg)-f\nabla^2g=\nabla \cdot ((\nabla f)g+f(\nabla g))-f\nabla^2 g=(\nabla^2 f)g+2(\nabla f)\cdot(\nabla g)+f\nabla^2 g-f\nabla^2g=(\nabla^2f)g+2(\nabla f)\cdot(\nabla g)$
$\Rightarrow [\nabla^2,f]=(\nabla^2 f)+2(\nabla f)\cdot \nabla$
Step 2
\begin{eqnarray*}
[f,[\nabla^2,f]]g & = & [f,(\nabla^2 f)+2(\nabla f) \cdot \nabla]g \\
& = & [f,2(\nabla f) \cdot \nabla]g \\
& = & 2(\nabla f) \cdot [f,\nabla]g \\
& = & 2(\nabla f) \cdot (f\nabla g-\nabla(fg)) \\
& = & 2(\nabla f) \cdot (-(\nabla f)g)
\end{eqnarray*}
$\Rightarrow$
$[f,[\nabla^2,f]]=-2(\nabla f) \cdot (\nabla f)=-2(\nabla f)^2$
Pls correct me if I'm wrong anywhere.
