Given $e^{A} = \sum_{i=0}^{\infty}\frac {A^{i}}{i!}$ where: $A^{0} = I$ (identity matrix).
Prove that: $e^{A}e^{B} = e^{A+B+\frac{1}{2}([A,B])}$ if: $[A,[A,B]] = [B,[A,B]] = 0$ ; where: $[A,B] = AB-BA$ .
If there is any method other than arranging the term properly? I am trying to apply brute force but cannot get $\frac12(AB-BA) $ in the desired expression.
We use the Zassenhaus formula, a corollary of the BCH formula - cf:
https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#The_Zassenhaus_formula
When $A,B$ commute with $[A,B]$ and $t$ is a small complex, we obtain the following closed form:
$\exp(t(A+B))=\exp(tA)\exp(tB)\exp(-t^2/2[A,B])$; since $A+B$ and $[A,B]$ commute, $\exp(t(A+B)+t^2/2[A,B])=\exp(tA)\exp(tB)$.
Now, our functions are holomorphic over $\mathbb{C}$; according to the "extension of equalities" theorem, the previous equality holds for $t=1$.