Non-commuting matrix exponentials 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?
 A: The equality is true when $A,B$ are quasi-commutative, that is, when $[A,B]$ commutes with both $A$ and $B$.
A: 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.
A: The book you referenced was "Quantum Physics in One Dimension". The operators in quantum mechanics have to obey particular commutation relations. For instance the operators $x$ and $p$ obey the canonical commutation relation, 
$$ \left[ x, p \right] = i I. $$
This allows us to simplify the Zassenhaus formula. The operator $[x,p]$ commutes with every operator because it is proportional to the identity; if you consider the higher order terms in the Zassenhaus formula you will see they involve commutators which must be zero.
Meaning that, 
$$\exp(x+p)= \exp(x)\exp(p)\exp(-[x,p]/2),$$ 
is an exact relationship for the operators $x$ and $p$. 
Knowledge of specific commutation relations can allow you to simplify the Zassenhaus formula significantly. 
