# 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....

• sure, $[exp(x),exp(y)]=exp(x)exp(y)-exp(y)exp(x)$. But, I'm guessing you want an expansion in $x,y$. Commented Mar 2, 2014 at 0:56
• I am expecting an answer in terms of $[x,y]$ and its commutator with $x$ and $y$ and go on: $[x,[x,y]]$, $[y,[x,y]]$. Thanks for reply. Commented Mar 2, 2014 at 1:02
• Hi again, I forgot about this question when I came upon math.stackexchange.com/q/711308/36530 but, it seems to me, there may be a way of using the technique there to help out here, or at a minimum it's worth looking at. Commented Apr 6, 2014 at 20:08
• You did not say why you expect an answer in the Lie algebra (nested commutators). The commutator you are considering need not be in the group. The answer will only be in the universal Lie Algebra, by the looks of it. Commented Apr 6, 2017 at 19:12
• When considering small perturbations, check out "Lyndon basis".
– mdot
Commented Apr 7, 2017 at 15:13

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.

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.

• I liked to know higher order terms to the third order. Commented Mar 2, 2014 at 1:05
• And whether there is an analytic expression for the coefficients for n-th order. Commented Mar 2, 2014 at 1:06
• well, I'm still thinking about my answer as it stands, have I missed other second order terms from $(x+y+\frac{1}{2}[x,y])^2$ or do they all cancel against the $(x+y-\frac{1}{2}[x,y])^2$ as I implicitly claim? Commented Mar 2, 2014 at 1:07
• I give you credit first +1, but please fill in a more satisfactory answer ASAP, which hopefully reveals more structure than I already knew... Commented Mar 2, 2014 at 1:15
• en.wikipedia.org/wiki/… gives the next terms with $1/12$. However, I need to sit and calculate a bit before I'm sure it is my knee-jerk conjecture of $\frac{1}{6}([x,[x,y]]-[y,[x,y]])$. Commented Mar 2, 2014 at 1:18