Suppose I have two matrices $e^{H_1t/n},e^{H_2t/n}$ for $t$ a small integer, and $n$ some positive integer, and $H_1,H_2 \in \mathbb{C}^{N \times N}$. I have given in a paper that by using the second order of the Baker-Campbell-Hausdorff formula it is true that :
$e^{[H_1,H_2]t + O(t^2/n^2)} = (e^{H_1t/n}e^{H_2t/n}e^{-H_1t/n}e^{-H_2t/n})^{n^2/t}$ where $[A,B]$ is the matrix commutator.
The author then states that $(e^{H_1t/n}e^{H_2t/n}e^{-H_1t/n}e^{-H_2t/n})(e^{-H_1t/n}e^{-H_2t/n}e^{H_1t/n}e^{H_2t/n}) = e^{2[H_1,H_2]t^2/n^2 + O((t/n)^4)}$.
I'm not sure how the author arrived at this last equality. Insights appreciated.
Best Answer
I think you are almost certainly misreading the paper to reverse the flow of logic involved.
You simply evaluate the lowest order CBH expansion mindful of the omitted orders, $$e^{H_1t/n}e^{H_2t/n}= e^{(H_1+H_2)(t/n) +\frac{1}{2} [H_1,H_2](t/n)^2 + O((t/n)^3)} \\ e^{-H_1t/n}e^{-H_2t/n}= e^{-(H_1+H_2)(t/n) +\frac{1}{2} [H_1,H_2](t/n)^2 + O((t/n)^3)} \\ f(t)\equiv (e^{H_1t/n}e^{H_2t/n}e^{-H_1t/n}e^{-H_2t/n})= e^{[H_1,H_2](t/n)^2 + O((t/n)^3)}.$$ (Observe the $O(t/n)$ terms mutually commute, being opposite, so they do not contribute to $O((t/n)^2)$.)
It is then evident that $$ f(t) f(-t)= e^{2[H_1,H_2]t^2/n^2 + O((t/n)^4)}, $$ since the $O((t/n)^3$ odd terms cancel each other in the sum, leaving $O((t/n)^4$.
I'm not sure how you concluded your mystery author needed the first equation to derive this.
(Note, instead, that $ f(t)^m= e^{m[H_1,H_2](t/n)^2 + O((t/n)^3)} $, but glibly taking $m=n^2/t$ will net an ill-defined error. I suspect there is essential information in your text you omitted.)