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The Zassenhaus formula is

$$e^{t(X+Y)}= e^{tX}~ e^{tY} ~e^{-\frac{t^2}{2} [X,Y]} ~ e^{\frac{t^3}{6}(2[Y,[X,Y]]+ [X,[X,Y]] )} ~ e^{\frac{-t^4}{24}([[[X,Y],X],X] + 3[[[X,Y],X],Y] + 3[[[X,Y],Y],Y]) } \cdots$$

from this Wikipedia page.

$X$ and $Y$ are linear operators, and $[X,Y]$ is their commutator.

I mostly want to prove it for the case where the commutator of $X$ and $Y$ is a constant, or simply the general proof.

What I'm looking for is either a reference to a place where it's proven or at least some shove in the right direction.

So far I've tried to just expand the exponential to see if I see anything but have no ideas so far.

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up vote 2 down vote accepted

If the case $[X,Y]=c$ ($c$ a constant) is sufficient for you: notice that $$\frac{d}{dt}e^{tX}e^{tY}e^{-t^2c/2}=Xe^{tX}e^{tY}e^{-t^2c/2}+e^{tX}Ye^{tY}e^{-t^2c/2} -tc\,e^{tX}e^{tY}e^{-t^2c/2} =(*).$$ As $e^{tX}Ye^{-tX}=Y+ct$ (to see it: the derivative of both sides is $c$ and the value for $t=0$ is $Y$), we have $$(*)=(X+Y)e^{tX}e^{tY}e^{-t^2c/2}.$$ If $F(t)=e^{tX}e^{tY}e^{-t^2c/2}$, we have $F(0)=1$ and $dF/dt=(X+Y)F(t)$, which are the defining relations for $e^{tX+tY}$.

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thank you I understand it, do you have any idea of a place here the full proof is shown? Do you know how this topic of mathematics is called, cause I got to it from quantum mechanics and I'm not sure what the name of the branch is – fiftyeight Jul 3 '12 at 21:36
How is there a asymmetry in the original formula? The fourth factor as 2 in front of $[Y,[X,Y]]$, but 1 infront of $[X,[X,Y]]$. If I swap X and Y shoudn't the formula remain the same, as $X+Y=Y+X$ – user23238 Dec 21 '12 at 8:39
@ramanujan_dirac: surely $X+Y=Y+X$, but $e^{tX}e^{tY}\neq e^{tY}e^{tX}$ – user8268 Dec 21 '12 at 21:53
oops, sorry. My mistake. – user23238 Dec 22 '12 at 0:41

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