# Factorization of $\exp(A+B)$ for non-commuting operators $A$ and $B$

I am trying to understand the following expression

$$\exp(A+B) = \exp(A/2) \exp(B) \exp(A/2) + O((A,B)^3)$$

where $$A$$ and $$B$$ are in general non-commuting operators. The formula appears in the following link as a variant of the Baker-Campbell-Hausdorff formula:

I would like to have a proof of the expression and to know how to figure out if the approximation is a good one given $$A$$ and $$B$$, so that I can understand the method in the link. I have no clue how to prove this. The only clue I have is the Baker-Campbell-Hausdorff formula, but I cant see the relation between them (saying it is a variant may not even mean that you can get the later from the BCH, i don't know).

• Hello Rodrigo. I don't have any clue on another reference. I found the formula in the site provided and the author makes no comment on the validity of it. I think it may be implicit in the $O(A,B)^3$ error estimation, but I don't even understand what (A, B) is. It seems powers of A and B (or both). I completely clueless. Before posting the question I tried to find a derivation in the internet. No lucky.
– Blue
Commented Nov 17, 2022 at 12:48
• I assumed this may be standard notation. Also, I assumed this would be something simple to prove. So I thought the problem was me, and my lack of knowledge on the topic. something on that way. But I think his explanation on the method is the clearest I found, apart this formula of course.
– Blue
Commented Nov 17, 2022 at 13:08

Dual to the Baker-Campbell-Hausdorff formula is the Zassenhaus formula $$e^{A + B} = e^A e^B e^{-\frac{1}{2}[A, B]} \dots.$$ Using this on $$A + B = A/2 + (B + A/2)$$ twice, we get \begin{align*} e^{A + B} &= e^{A/2} e^{B + A/2} e^{-1/2[A/2, B+A/2]} \dots \\ &= e^{A/2} e^B e^{A/2} e^{-\frac{1}{2} [B,A/2]} e^{-1/2[A/2, B+A/2]} \dots \end{align*} Looking at the terms with commutators at the end, we have $$-\frac{1}{2} \Bigl( \frac{B A}{2} - \frac{A B}{2} \Bigr) - \frac{1}{2} \Bigl( \frac{A B}{2} + \frac{A^2}{4} - \frac{B A}{2} - \frac{A^2}{4} \Bigr) = 0.$$ Hence there are no "second order" terms when we use this symmetric arrangement of $$A + B = A/2 + B + A/2$$. Of course there are still third order terms I haven't bothered writing out, involving things like $$[A/2, [B+A/2, A/2]]$$ and similar. Presumably the author of the text uses $$(A, B)^3$$ as some kind of abbreviation for terms of that shape.