How do we prove Hausdorff's formula in the following form by pulling the operator B in $e^A B$ to the left side in the following form ?
$e^ABe^{-A} = B + [A,B]+\frac{1}{2!}[A,[A,B]]+...$
I made some progress as far as to acquire the B factor on the right hand side of the equation but I am unable to get the rest. Any ideas ?
Claim: $$\frac{\mathrm{d}}{\mathrm{d}\lambda} \left( e^{\lambda A} B e^{-\lambda A}\right) = e^{\lambda A} [A,B] e^{-\lambda A} $$
This claim is easily proven by differentiation. Denoting the function above by $f(\lambda)$, we note further that $$f''(\lambda) = e^{\lambda A} [A,[A,B]] e^{-\lambda A}$$ and so on. We can write $f(\lambda)$ as a Taylor Series around $\lambda = 0$: $$f(\lambda) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}\lambda^n$$ and the result follows by letting $\lambda = 1$.