# Looking to compute $e^{\phi}\Delta^2 f e^{-\phi}$ in a coordinate-free way

Let $$f, \phi:\mathbb{R}^n\to\mathbb{R}$$ with $$\phi(x)$$ looking like $$\phi(x)=(x+a)^2$$ for some $$a\in\mathbb{R}^n$$. I am interested in computing $$L_\phi(f)=e^{\phi}\Delta^2 \left(e^{-\phi}\,f\right)$$ in a coordinate-free way. What I mean by that is that the result should be entirely in terms of terms like $$\nabla \phi,\nabla f, \Delta^2 f$$ and the like, with no explicit reference to coordinates. Another way of saying this is that all derivatives must be explicitly acting on either $$\phi$$ or $$f$$ but not some combination thereof. It should also be fully simplified, in other words, it cannot contain terms like $$\Delta (e^{-\phi})$$. For example, you can write $$e^{\phi}\Delta \left(e^{-\phi}\,f\right)=\Delta f+(|\nabla\phi|^2-\Delta\phi)f-2\nabla \phi \cdot \nabla f.$$ The right-hand side is the type of expression I am looking for. As you can probably see, the calculations get quite messy in the case of $$\Delta^2$$, which is why I am hoping someone here can help me compute this with Mathematica.

Eventually, I would like to split $$L(f)$$ into a symmetric and an anti-symmetric (under $$\phi\to -\phi$$) part, $$L_\phi(f)=S_\phi(f)+A_\phi(f),$$ where $$S_\phi(f)=\frac{1}{2}(L_\phi(f)+L_{-\phi}(f))$$ and $$A_\phi(f)=\frac{1}{2}(L_\phi(f)-L_{-\phi}(f))$$, and compute the commutator $$[S,A](f)=S(A(f))-A(S(f)),$$ to give you an idea of what I want to be able to do with the result.