Prove that $$e^{(A+B)\Delta t}=e^{A\Delta t}e^{B\Delta t}e^{-1/2[A,B]\Delta t^2}+O(\Delta t^3)$$
The Baker-Campbell-Hausdorff Formula is $$e^{tX}e^{tY}=e^{t(X+Y)+\frac{t^2}{2}[X,Y]+\frac{t^3}{3!}[[X,[X,Y]]-[Y,[X,Y]]]+\cdots}$$ and when $[X,[X,Y]]=[Y,[X,Y]]=0$ we have the BCH formula in the form $e^{tX}e^{tY}=e^{t(X+Y)+\frac{1}{2}[X,Y]}=e^{t(X+Y)}e^{\frac{1}{2}[X,Y]}$ and I understand the proofs for these.
Please check Notes on Baker-Campbell-Hausdorff (BCH) Formulae and Commutator algebra in exponents for the proofs.
We can also prove that $e^{tX}e^{tY}=e^{t(X+Y)}$ when $[X,Y]=0$
How do we reach the form $e^{(A+B)\Delta t}=e^{A\Delta t}e^{B\Delta t}e^{-1/2[A,B]\Delta t^2}+O(\Delta t^3)$ from the Baker-Campbell-Hausdorff Formula ?
Thanks @Paresseux Nguyen for the hint.
$$ e^{A\Delta t}e^{B\Delta t}e^{\frac{-1}{2}[A,B]\Delta t^2}=\Big(I+A\Delta t+\frac{(A\Delta t)^2}{2}+O(\Delta t^3)\Big)\Big(I+B\Delta t+\frac{(B\Delta t)^2}{2}+O(\Delta t^3)\Big)\\\times\Big(I-\frac{1}{2}[A,B]\Delta t^2+O(\Delta t^4)\Big)\\ =I+(A+B)\Delta t+\Big(AB+\frac{A^2}{2}+\frac{B^2}{2}-\frac{1}{2}[A,B]\Big)\Delta t^2+O(\Delta t^3)\\ =I+(A+B)\Delta t+\frac{(2AB+{A^2}+{B^2}-AB+BA\Big)\Delta t^2}{2}+O(\Delta t^3)\\ =I+(A+B)\Delta t+\frac{({A^2}+{B^2}+AB+BA)\Delta t^2}{2}+O(\Delta t^3)\\ =I+(A+B)\Delta t+\frac{(A+B)^2\Delta t^2}{2}+O(\Delta t^3) $$
$$ e^{(A+B)\Delta t}=I+(A+B)\Delta t+\frac{(A+B)^2\Delta t^2}{2}+\frac{(A+B)^3\Delta t^3}{3!}+\cdots\\ =I+(A+B)\Delta t+\frac{(A+B)^2\Delta t^2}{2}+O(\Delta t^3)-O(\Delta t^3)+O_1(\Delta t^3)\\ =e^{A\Delta t}e^{B\Delta t}e^{\frac{-1}{2}[A,B]\Delta t^2}-O(\Delta t^3)+O_1(\Delta t^3)\\ =e^{A\Delta t}e^{B\Delta t}e^{\frac{-1}{2}[A,B]\Delta t^2}+O_2(\Delta t^3)\\ $$