Baker–Campbell–Hausdorff formula for [exp(x),exp(y)] Can someone provide a explicit (the first priority with leading orders, then the secondary consider as complete as possible, or) expansion like Baker–Campbell–Hausdorff formula for the commutator:
$$[\exp(x),\exp(y)]=?$$
where the commutator $[x,y]\neq 0$.
I am expecting an answer in terms of $[x,y]$ and its commutator with $x$ and $y$ and go on: $[x,[x,y]]$, $[y,[x,y]]$, etc.... 
Thanks for reply.  
 A: Ok, so $exp(x)exp(y) = exp(x+y+ \frac{1}{2}[x,y] +\cdots )$ and $exp(y)exp(x) = exp(y+x+ \frac{1}{2}[y,x] +\cdots )$. Therefore, using $[y,x]=-[x,y]$,
$$ [exp(x),exp(y)] =  exp(x+y+ \frac{1}{2}[x,y] +\cdots ) - exp(x+y- \frac{1}{2}[x,y] +\cdots )$$
then,
$$ [exp(x),exp(y)] =  [x,y] +\cdots. $$
The answer would seem to fall out of the BCH relation for higher orders if you want higher orders.
An outline of how to continue: use BCH to expand as: (the right equality is meant to indicate $A$ is the terms in the argument of the exponential)
$$ exp(x)exp(y) = exp(x+y+\frac{1}{2}[x,y]+\frac{1}{12}[x,[x,y]]-\frac{1}{12}[y,[y,x]] + \cdots ) =exp(A)$$
Likewise
$$ exp(y)exp(x) = exp(x+y+\frac{1}{2}[y,x]+\frac{1}{12}[y,[y,x]]-\frac{1}{12}[x,[x,y]] + \cdots ) $$
Note, (the right equality is meant to indicate $B$ is the terms in the argument of the exponential)
$$ exp(y)exp(x) = exp(x+y-\frac{1}{2}[x,y]-\frac{1}{12}[x,[x,y]] + \frac{1}{12}[y,[y,x]]\cdots ) = exp(B) $$
We would like to simplify the quantity $exp(A)-exp(B)$ order by order in $x,y$. Recall $exp(A) = 1+A+\frac{1}{2}A^2 + \cdots$ and $exp(B) = 1+B+\frac{1}{2}B^2 + \cdots$. Clearly the constant and first order terms in $x,y$ cancel. I think at second order everything cancels except $[x,y]$. At third order, I'm not sure without a lot more calculation, however, I hope the path is partly clear now, granted it is tedious. There is probably a better way. Perhaps a program of successive differentiation and evaluation.
A: The set of x, y and their nested commutators comprise a Lie algebra. A product of group elements, $\exp x ~ \exp y$ is in the group, ($\exp x ~ \exp y =\exp z $), so its logarithm is in the Lie algebra: $z(x,y)$ is the CBH expansion, so linear combs of x, y and nested commutators. 
Note the important symmetry $z(y,x)=-z(-x,-y)$. 
However, the commutator of two group elements, such as the one you are interested in, is, in general, not in the group. It is in the universal Lie algebra, so not just nested commutators. In fact, expanding expz to cubic order, and discarding the x-y symmetric terms, by virtue of the above identity,  you readily get, to cubic order in x and y ,
$$
\exp z(x,y) -\exp z(y,x) = [x,y] + \tfrac{1}{2}\{ (x+y),[x,y]\}+...
$$
which is not reducible to nested commutators--it is not in the Lie algebra. The { } bracket is the anticommutator; note how cleverly the surviving cross term upgrades the antisymmetry of one of its constituent components to yield the cubic term.
