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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).

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  • $\begingroup$ 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. $\endgroup$
    – Blue
    Commented Nov 17, 2022 at 12:48
  • $\begingroup$ 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. $\endgroup$
    – Blue
    Commented Nov 17, 2022 at 13:08

1 Answer 1

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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.

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  • $\begingroup$ Your comment on the abbreviation makes sense to me now! But given those terms are weird commutators, how can I know, given A and B, the approximation is good? In the site the author gives the two operators. But how can I know the approximation is valid? $\endgroup$
    – Blue
    Commented Nov 17, 2022 at 13:42

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