Proof for the Laplacian of scalar fields using index notation For scalar fields Φ and Ψ, the Laplacian is defined by
$$∇^2(ΦΨ)=(∇^2Φ)Ψ+2∇Φ\cdot∇Ψ+Φ∇^2Ψ$$
where $∇$ is the usual del operator and $∇^2$ is the Laplacian.
How can I prove this relation? I tried the brute-force method and it became rather messy by getting an expression for the RHS and LHS. I believe that using Einstien notation would be a better approach.
 A: I think the easiest/best approach is completely coordinate free, using well-known properties of $\nabla$ which need not be expressed in terms of coordinates:
Recall the definition of the Laplacian $\nabla^2$ in terms of $\nabla$ itself:
$\nabla^2 \Theta = \nabla \cdot \nabla \Theta; \tag 1$
the Leibniz rule when applied to $\nabla(\Psi \Upsilon)$ yields the formula
$\nabla(\Psi \Upsilon) = \Upsilon \nabla \Psi + \Psi \nabla \Upsilon; \tag 2$
for any scalar function $f$ and vector field X we have
$\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X, \tag 3$
see https://en.wikipedia.org/wiki/Vector_calculus_identities; we apply this to (2):
$\nabla \cdot (\Upsilon \nabla \Psi) = \nabla \Upsilon \cdot \nabla \Psi + \Upsilon \nabla \cdot \nabla \Psi = \nabla \Upsilon \cdot \nabla \Psi + \Upsilon \nabla^2 \Psi, \tag 4$
$\nabla \cdot (\Psi\nabla \Upsilon) = \nabla \Psi \cdot \nabla \Upsilon + \Psi \nabla \cdot \nabla \Upsilon = \nabla \Psi \cdot \nabla \Upsilon + \Psi \nabla^2\Upsilon; \tag 5$
thus,
$\nabla^2(\Psi \Upsilon) = \nabla \cdot \nabla(\Psi \Upsilon) = \nabla \cdot (\Upsilon \nabla \Psi + \Psi \nabla \Upsilon) = \nabla \cdot (\Upsilon \nabla \Psi) + \nabla \cdot (\Psi \nabla \Upsilon)$
$= \nabla \Upsilon \cdot \nabla \Psi + \Upsilon \nabla^2 \Psi + \nabla \Psi \cdot \nabla \Upsilon + \Psi \nabla^2\Upsilon = \Psi \nabla^2\Upsilon + 2\nabla \Psi \cdot \nabla \Upsilon + \Upsilon\nabla^2 \Psi, \tag 6$
as per request.  QED!!!
A: Doing this with index notation is like killing a fly with a hammer, but here we go.
First task: Show that $\Delta$ is a scalar, i.e, it does not transform under coordinate transformations. In other words, it is the same regardless what basis you use. I'll leave this as an exercise.
Now, choose a (normalized) basis such that all the Christoffel symbols vanish (i.e, $\mathbf g=\operatorname{diag(1,\dots,1)}$) and such that the covariant derivative is just
$$\nabla_i=\partial_i$$
Now,
$$\Delta=\nabla_i\nabla^i$$
So
$$\Delta(\lambda\kappa)\\ =\partial_i(\partial^i(\lambda\kappa))\\ =\partial_i(\lambda\partial^i\kappa+\kappa\partial^i\lambda)\\ =\partial_i(\lambda\partial^i\kappa)+\partial_i(\kappa\partial^i\lambda)\\ =(\partial_i\lambda)(\partial^i\kappa)+\lambda\partial_i\partial^i\kappa+(\partial_i\kappa)(\partial^i\lambda)+\kappa\partial_i\partial^i\lambda\\=\lambda\Delta\kappa+2\langle\nabla\lambda,\nabla\kappa\rangle+\kappa\Delta\lambda$$
Done.
