Let $d\in\mathbb N$, $y\in\mathbb R^d$, $$g(x):=e^{{\rm i}\langle x,\:y\rangle}\;\;\;\text{for }x\in\mathbb R^d$$ and $C\in\mathbb R^{d\times d}$ be symmetric.
How can we compute $\operatorname{div}(C\nabla g)$?
We've clearly got $$\mathcal Dg(x)h={\rm i}g(x)\langle h,y\rangle\tag1$$ and $$\mathcal D^2g(x)(h_1,h_2)=-g(x)\langle h_1,y\rangle\langle h_2,y\rangle\tag2$$ for all $x,h,h_1,h_2\in\mathbb R^d$. In particular, $$\nabla g={\rm i}gy\tag3.$$ Since $C$ is a constant, we should be able to pull it somehow out of the divergence ...
$$\nabla \cdot (C\nabla g)=\partial_i (C_{ij}\partial_j g) = C_{ij} \partial_i\partial_j g= C:\nabla^2 g.$$ Here, Einstein summation is used, and $A:B=A_{ij}B_{ij}$ is the double contraction. If e.g. $C=I$ i.e. $C_{ij}=\delta_{ij}$ then this is the Laplacian as expected. For the given $g$, $\partial_i\partial_j g=-y_iy_j g$, so the answer is $$ \nabla \cdot (C\nabla g)= -C_{ij}y_iy_j g = -C:(y\otimes y)g. $$ If e.g. $C=I$ then we see the familliar Fourier multiplier definition for the Laplacian $\Delta e^{ixy} = -|y|^2 e^{ixy}$.