Consider a globally hyperbolic manifold $(M,\tilde{g})$, then we know that $M \cong \mathbb{R}\times \Sigma$ and $$\tilde{g}=\beta dt^2-\tilde{h},$$ with $\beta$ and $h$ that doesn't depend on t.$\newcommand{\det}{{\operatorname{det}}}$
Construct the conformal metric $$g:=\beta^{-1}\tilde{g}=dt^2-h,$$ with $h:=\beta^{-1}\tilde{h}$.
I want to prove that the two D'Alembert wave operators $\square_{\tilde{g}}$ and $\square_g$, where $$\square_g:=g^{ab}\partial_a\partial_b=\frac{1}{\sqrt{|\det g|}}\partial_a\left(\sqrt{|\det g|}g^{ab}\partial_b\right)\;?$$
are related by $$\beta^{\frac{2+n}{4}}\square_{\tilde{g}}\beta^{\frac{2-n}{4}}=\square_g+V$$ where $n=\operatorname{dim} M$ and $V$ is the operator defined by $$V=\beta^{-\frac{2-n}{4}}\left(\Delta_h\beta^{\frac{2-n}{4}}\right)$$ and $\Delta_h$ is the Laplace-Beltrami operator on $\Sigma$ associated with $h$.
I present here my calculations: If $\tilde{g}=e^{2\varphi}g$, then we get $$\square_{\tilde{g}}=e^{-2\varphi}\square_g+e^{-2\varphi}(n-2)g^{lm}\partial_m\varphi\frac{\partial}{\partial x^l},$$ then for our notation $e^{2\varphi}=\beta$, hence $\varphi=1/2\ln(\beta)$, then the previous equation becomes $$\square_\tilde{g}=\beta^{-1}\left[\square_g+\frac{n-2}{2\beta}g^{lm}\partial_m\beta\frac{\partial}{\partial x^l}\right]$$ It is easy to show that $\square(fg)=\square(f)g+\square(g)f+2g^{ij}\partial_if\partial_jg$. Now I try to put this expression in the result: let any $f\in\mathcal{C}^\infty(M)$, then $$\beta^{\frac{2+n}{4}}\square_\tilde{g}\left(\beta^\frac{2-n}{4}f\right)= \beta^{\frac{2+n}{4}}\beta^{-1}\left[\square_g\left(\beta^\frac{2-n}{4}f\right)+\frac{n-2}{2\beta}g^{lm}\partial_m\beta\frac{\partial\left(\beta^\frac{2-n}{4}f\right)}{\partial x^l}\right]= \beta^{\frac{-2+n}{4}}\left[\square_g\left(\beta^\frac{2-n}{4}\right)f+\beta^\frac{2-n}{4}\square_g(f)+2g^{lm}\frac{\partial \beta^\frac{2-n}{4}}{\partial x^l}\frac{\partial f}{\partial x^m}+\frac{n-2}{2\beta}g^{lm}\partial_m\beta\frac{\partial \beta^\frac{2-n}{4}}{\partial x^l}f+\frac{n-2}{2\beta}g^{lm}\partial_m\beta\beta^\frac{2-n}{4}\partial_l f\right]= \beta^{\frac{-2+n}{4}}\square_g\left(\beta^\frac{2-n}{4}\right)f+\square_g(f)+\beta^{\frac{-2+n}{4}}\left[2\frac{2-n}{4}g^{lm}\partial_l\beta\beta^{\frac{-2-n}{4}}\partial_m f-\frac{(n-2)^2}{8}g^{lm}\beta^{\frac{-6-n}{4}}\partial_m\beta\partial_l\beta f+\frac{n-2}{2}g^{lm}\partial_m\beta\beta^\frac{-2-n}{4}\partial_l f\right]=\\ \square_g(f)+\beta^{\frac{-2+n}{4}}\square_g\left(\beta^\frac{2-n}{4}\right)f-\frac{(n-2)^2}{8\beta^2}g^{lm}f\partial_m\beta\partial_l\beta$$
As you can see the last term shouldn't be there. Where am I wrong? Thanks in advance.