How Can I Prove the Zassenhaus-Like Formula?

[Haorong Wu]

Homework Statement

Suppose $[A,B]=C$, and $C$ does not commute with $A$ and $B$. Prove $$e^{A+B}=e^Ae^B \exp (-\frac 1 2 C -\frac 1 6 [C,A] - \frac 1 3 [C,B] + \dots) .$$

Relevant Equations

NaN

I find in Wikipedia that the equation is very like the Zassenhaus formula $$e^{t(X+Y)}=e^{tX}e^{tY}e^{-\frac {t^2} 2 [X,Y]} e^{\frac {t^3} 6 (2[Y,[X,Y]]+[X,[X,Y]])}\dots .$$ But I still have no clues.

I try to prove it as the prove of Glauber formula. I start with letting $f(\lambda)=e^{\lambda A}e^{\lambda B}$ and $g(\lambda)=e^{\lambda A} B e^{-\lambda A}$. Then $f'(\lambda)=Ae^{\lambda A}e^{\lambda B} +e^{\lambda A}e^{\lambda B} B$. and $g'(\lambda)=e^{\lambda A} C e^{-\lambda A}$.

Next, I try to switch $A$ and $C$. First, I derive that $$ [C,e^{-\lambda A}]=-\lambda [C,A] e^{-\lambda A}.$$ So $g'(\lambda)=C-\lambda [C,A] e^{-\lambda A}$.

Then integrating $g'(\lambda)$ does not give me a clear result. I do not know how to proceed.

 

[BigJDubs]

The proof of Glauber formula is based on the Baker-Campbell-Hausdorff formula which is not applicable here. So I think there should be another approach to prove this equation. Can someone tell me what is the approach?The Zassenhaus formula (also known as the Magnus formula) can be used to prove the desired equation. It states that $$e^{t(X+Y)}=e^{tX}e^{tY}e^{-\frac {t^2} 2 [X,Y]} e^{\frac {t^3} 6 (2[Y,[X,Y]]+[X,[X,Y]])}\dots .$$ By setting $X=A$ and $Y=B$, we have $$e^{t(A+B)}=e^{tA}e^{tB}e^{-\frac {t^2} 2 [A,B]} e^{\frac {t^3} 6 (2[B,[A,B]]+[A,[A,B]])}\dots ,$$ which is exactly the desired equation.