Calculating the product of the exponential application using the Baker-Campbell formula

Question

If $[X,Y]=sY$ with $s\neq 2\pi i k$ for all $k\in\mathbb{Z}$, then $\mathrm{e}^X\mathrm{e}^Y=\mathrm{e}^{X+\frac{s}{1-e^{-s}}Y}$ (https://es.wikipedia.org/wiki/Fórmula_de_Baker-Campbell-Hausdorff)

There is something similar when $[X,Y]=sX$ for $\mathrm{e}^X\mathrm{e}^Y?$

Answer 1

Assume that $\mathrm{ad}_XY = [X,Y] = sX$. Then one has $$ e^XYe^{-X} = e^{\mathrm{ad}_X}Y = \sum_{n\ge0}\frac{\mathrm{ad}_X^n}{n!}Y = Y + sX, $$ hence $e^Xe^Ye^{-X} = \exp\left(e^XYe^{-X}\right) = e^{sX+Y}$ and $$ e^Xe^Y = e^Xe^Ye^{-X}e^X = e^{sX+Y}e^X = \exp\left((sX+Y) - \frac{s}{1-e^s}X\right) = \exp\left(\frac{s}{1-e^{-s}}X+Y\right), $$ since $Z := sX+Y$ satisfies $[Z,X] = -sX$, which permits to apply the first formula you mentioned.

Answer 2

Let $[X,Y]=sX$. Then $e^Xe^Y=e^Z$ for some $Z$?

$$[-Y,-X]=[Y,X]=-[X,Y]=-sY=s(-Y)\Rightarrow e^{-Y}e^{-X}=e^{-Y+\frac{s}{1-e^{-s}}(-X)}$$

Now $$1=e^{-Y+\frac{s}{1-e^{-s}}(-X)} e^Z$$

Therefore

$$e^{Y+\frac{s}{1-e^{-s}}X}=e^Z$$