TITLE: $\exp(A+B)$ and Baker-Campbell-Hausdorff
QUESTION [10 upvotes]: A few years ago, I did research in quantum mechanics, specifically dealing with generalized displacement operators. In such musings, BCH lights (or gets in, depending on your viewpoint) the way. A question that struck me today was: does $\exp(A+B) = \exp(A)\exp(B)$ hold if and only if $A$ and $B$ commute? Clearly, if they do commute this is true but I have not seen anything detailing the opposite direction. Given the complexity of BCH, I would be inclined to think that it's not simple to prove if it is true. I've thought about it on my own but haven't been able to come to any sort of conclusion one way or the other.

REPLY [2 votes]: I just read this file.
The fact that $\exp(A+B) = \exp(A)\exp(B)$ is not equivalent to $AB=BA$ is well-known.
More interesting (and difficult) is: when $n\geq 3$, what is the set $\{A,B\in M_n(\mathbb{C})|\exp(A+B) = \exp(A)\exp(B)\}$. Since nobody knows the answer, this question seems to be a good subject of research.