# $\exp(A+B)$ and Baker-Campbell-Hausdorff

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.

• What are $A,B$ here? Matrices? Bounded operators? Unbounded operators? Elements of an arbitrary Lie algebra? Commented Jun 18, 2014 at 22:17

Take $A=\begin{pmatrix}i\pi & 0 \\ 0 & - i\pi\end{pmatrix}$ and $B=\begin{pmatrix}i\pi & 1 \\ 0 & - i\pi\end{pmatrix}$. Then $A$ and $B$ do not commute and $\exp(A)\,\exp(B)=\exp(A+B)$.

• This does answer the question. +1
– robjohn
Commented Jun 19, 2014 at 2:05

The proof is actually not that hard to follow. Granted, there are proofs which require a great deal of Mathematics to follow the proof. However, there is an excellent proof given by Eichler and is explained in Stillwell's excellent book, Naive Lie Theory, which can be found here:

Naive Lie Theory by Stillwell

What you are looking for begins at page 152. However, I recommend the whole book as a read. He is an excellent author and the book is a nice undergraduate level approach to Lie Groups/Algebras.

• Thanks for the reference. I will definitely be using this in the future. I guess I can see why I haven't come across a proof though since it seems like it's actually not that difficult, but I'm not sure why I haven't seen this explicitly stated before. Commented Jan 14, 2014 at 22:04
• Maybe I'm missing something subtle but it seems to me that it simply doesn't hold so it can hardly be proved.. What is true is that $e^{tA} e^{tB}=e^{t(A+B)}$ for all $t$ iff $A$ and $B$ commute. Commented Jun 18, 2014 at 22:17
• @PeterFranek It was a general question asking about both directions. Certainly the opposite direction is not true, as your example shows. However, in the spirit of being informative, the Campbell-Baker-Hausdorff Theorem offers the form for the opposite direction which also shows it is true when the elements commute. The Theorem also furnishes you with the missing terms which 'fix' this so that the opposite direction follows as well. This is what the OP brought up themselves- the B.C.H. - so I assumed they knew the opposite direction doesn't quite work. Commented Jun 19, 2014 at 0:33

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.