Suppose $X$ and $Y$ are $n\times n$ complex matrices, and that $X$ and $Y$ commute with their commutator:
$$ [X,[X,Y]] = [Y,[X,Y]] = 0 $$
According to the Baker-Campbell-Hausdorff (BCH) formula one gets (Hall 2015, Thm. 5.1)
$$ e^X e^Y = e^{X+Y+\frac{1}{2} [X,Y]}. $$
Is there a similar (closed form) expression known if we consider the weaker assumption $ [X,[X,Y]] = [Y,[X,Y]] \neq 0 $ ?
Best Answer
We have the following first terms: \begin{align} Z(X,Y)& =\log(\exp X\exp Y) \\ &{}= X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}\left ([X,[X,Y]] +[Y,[Y,X]]\right ) \\ &{}\quad - \frac {1}{24}[Y,[X,[X,Y]]] \\ &{}\quad - \frac{1}{720}\left([Y,[Y,[Y,[Y,X]]]] + [X,[X,[X,[X,Y]]]] \right) \\ &{}\quad +\frac{1}{360}\left([X,[Y,[Y,[Y,X]]]] + [Y,[X,[X,[X,Y]]]]\right)\\ &{}\quad + \frac{1}{120}\left([Y,[X,[Y,[X,Y]]]] + [X,[Y,[X,[Y,X]]]]\right)\\ &{}\quad + \frac{1}{240}\left([X,[Y,[X,[Y,[X, Y]]]]] \right)\\ &{}\quad + \frac{1}{720}\left([X,[Y,[X,[X,[X, Y]]]]] - [X,[X,[Y,[Y,[X, Y]]]]] \right)\\ &{}\quad + \frac{1}{1440}\left([X,[Y,[Y,[Y,[X, Y]]]]] - [X,[X,[Y,[X,[X, Y]]]]] \right) \\ &{}\quad + \cdots \end{align} Even if $[X,[X,Y]] = [Y,[X,Y]]$, the higher commutators do not vanish completely in general (find an example!). I don't see an easy finite sum.