Non-commuting matrix exponentials

Question

Reading this book, I came across the following formula

$$e^A e^B = e^{A+B}e^{\frac{1}{2}[A,B]}$$

where $A$ and $B$ are two matrices and $[A,B] := AB-BA$.

I tried to find a proof, without success. It is impossible to use the binomial theorem since $A$ and $B$ do not commute. I've thought about developping the product in a power series, but I'm not sure the Cauchy product is allowed when $[A,B] \ne 0$.

I thought that $[A,B^k]$ could be useful so I searched the general expression

$$[A,B^n] = \sum_{i=0}^{n-1} B^i [A,B] B^{n-i-1}$$

Does anyone know a proof of the formula above? If so, is it possible to provide some hints? Or a full proof?

Answer 1

The author probably means an approximation rather than a true equality, as $(A+B)+\frac12[A,B]$ are only the first two terms in the BCH formula Zassenhaus formula.

Answer 2

The equality is true when $A,B$ are quasi-commutative, that is, when $[A,B]$ commutes with both $A$ and $B$.