When I was learning the interaction picture of the Hamiltonian, the usual way to convert it from the Schrodinger picture is using the Baker–Campbell–Hausdorff formula, \begin{equation} e^ABe^{-A}=B+[A,B]+\frac{1}{2!}[A,[A,B]]+\ldots. \end{equation} I am wondering if $A$ and $B$ can be separated to several parts, say $A=A_1+A_2$ and $B=B_1+B_2+B_3$, how the Baker–Campbell–Hausdorff formula looks like. I guess it should be like \begin{equation} e^A(B_1+B_2+B_3)e^{-A}=e^{A_2}e^{A_1}B_1e^{-A_1}e^{-A_2}+e^{A_2}e^{A_1}B_2e^{-A_1}e^{-A_2}+e^{A_2}e^{A_1}B_3e^{-A_1}e^{-A_2}. \end{equation} I am a physicist. May I ask whether this is mathematically true?
Thank you!