Proving operator equality with exponentials I'm currently trying to prove that for two Operators $A,B$ with the property $[A,[A,B]]=[B,[A,B]]=0$ ($[\cdot,\cdot]$ being the commutator) the following is true
$$e^{A+B}=e^Ae^Be^{-\frac{1}{2}[A,B]}.$$
The Hint was to consider $f(\lambda) = e^{\lambda A}e^{\lambda B}e^{-\lambda(A+B)}$ and show that $f'(\lambda)=\lambda [A,B]f(\lambda)$. The problem is that when I calculate $f'(\lambda)$ I get the following
$$\begin{align}f'(\lambda) &= \left(\frac{\partial}{\partial \lambda}e^{\lambda A}\right)e^{\lambda B}e^{-\lambda(A+B)} +e^{\lambda A}\left(\left[\frac{\partial}{\partial \lambda}e^{\lambda B}\right]e^{-\lambda (A+B)}+e^{\lambda B}\left[\frac{\partial}{\partial \lambda}e^{-\lambda (A+B)}\right]\right)\\ 
&= Ae^{\lambda A}e^{\lambda B}e^{-\lambda(A+B)} +e^{\lambda A}Be^{\lambda B}e^{-\lambda(A+B)}-e^{\lambda A}e^{\lambda B}(A+B)e^{-\lambda(A+B)},\end{align}$$
which certainly isn't what I should get (especially the missing $\lambda$ is giving me some headache, not sure where that should come from).
But even if I assume $f'(\lambda)=\lambda [A,B]f(\lambda)$ to be true I struggle to come to the desired result. This is obviously a differential equation in $f$ with the solution $f(\lambda)=C\exp(-\lambda^2/2~[A,B])$, where $C$ is some constant. If we now set $C=1$ and $\lambda =-1$ and equal this to the given $f(\lambda)$ we get
$$e^{\frac{1}{2}[A,B]}=e^{-A}e^{-B}e^{A+B} \Leftrightarrow e^{A+B}= e^Be^Ae^{\frac{1}{2}[A,B]}.$$
So the sign of one of the exponentials is wrong as well as the order of the exponentials.

Questions: 


*

*What am I doing wrong when calculating the derivative?

*Is it not the right approach to try to solve the differential equation or am I just doing it wrong?

 A: The approach is sound, but you must pursue it to its logical implications.
You note that by the basic Hadamard lemma, given the centrality of the commutator (it commutes with everything),
$$
\bbox[yellow]{e^{\lambda B} A e^{-\lambda B}= A + \lambda [B,A]},
$$
while the remaining terms vanish! So the adjoint action on A is almost trivial.
By the same token,
$$
e^{\lambda A} (e^{\lambda B} A e^{-\lambda B})e^{-\lambda A}  = A + \lambda [B,A] .
$$
So you simply pursue your calculation,
$$ f'(\lambda) = 
 A(e^{\lambda A}e^{\lambda B}e^{-\lambda(A+B)}) +e^{\lambda A}B    e^{-\lambda A}(e^{\lambda A}e^{\lambda B}e^{-\lambda(A+B)})-e^{\lambda A}e^{\lambda B}(A+B)e^{-\lambda B}e^{-\lambda A}(e^{\lambda A}e^{\lambda B}e^{-\lambda(A+B)})\\
=    (A -(A+\lambda [B,A]) ) f (\lambda)= \lambda [A,B] f (\lambda) . $$
The solution to this differential equation is
$$
f(\lambda) = e^{\frac{\lambda^2}{2} [A,B]}  f(0)= e^{\frac{\lambda^2}{2} [A,B]}.
$$
The degenerate limit of the Zassenhaus formula (see WP section) then follows,
$$
e^{\frac{\lambda^2}{2} [A,B]}=  e^{  \lambda A}e^{ \lambda B} e^{ -\lambda (A+B)} ,
$$
which you should be able to check is completely equivalent to your desideratum formula through inversion, $ e^{- \lambda B}e^{ - \lambda A}e^{\frac{\lambda^2}{2} [A,B]}=e^{ -\lambda (A+B)} $, restriction, $\lambda =-1$, and redefinition, $A\leftrightarrow B$.
