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:
- Y. D. Chong, The split-step Fourier method, April 30, 2021.
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).
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.