$e^Ae^B$ when $A$ and $B$ anticommute

Question

I know that if $A,B$ ($n\times n$ real matrices) commute, then $e^{A}e^B = e^{A+B}$. Is there a similar identity when $A,B$ anticommute?

Answer 1

The most general answer is given by the Campbell–Baker–Hausdorff formula:

$$ e^X e^Y = e^{X+Y + \frac{1}{2}[X,Y] + \cdots} $$

Which reduces to $e^X e^Y = e^{X+Y}$ if the commutator $[X,Y] = XY-YX$ is $0$, and to $e^X e^Y = e^{X+Y + \frac{1}{2}[X,Y] }$ if $[X,[X,Y]] = [Y,[Y,X]] = 0$.

However for anti-commutative matrices $XY = -YX$ you will find that in the series expansion infinitely many terms are non-zero, e.g. $[X,[X,\ldots [X,[X,Y]]]] = 2^k X^k Y$