Let $X,Y$ be two matrices, and we define $$ e^X:=\sum_{k=0}^{\infty}\frac{1}{k!}X^k $$ In a problem about Lie algebras, I need to show if $[X,Y]=\alpha Y,\alpha\neq 2\pi ik$, then $$ e^XY=\frac{\alpha}{1-e^{\alpha}}Ye^X $$
I want to expand $e^X$ and see $$ (I+X+\frac12X^2+\dots)Y=*Y(I+X+\frac12X^2+\dots) $$ But things seem to be difficult. Is there any way easy to solve this? Thanks in advance!
Use the standard identity $$ e^{X}Y e^{-X} = e^{\operatorname{ad} _X} Y =Y+\left[X,Y\right]+\frac{1}{2!}[X,[X,Y]]+\frac{1}{3!}[X,[X,[X,Y]]]+\cdots \\ =Y+\alpha Y +\frac{1}{2!}\alpha^2 Y +\frac{1}{3!} \alpha^3Y+\cdots = e^\alpha ~Y~~~\leadsto \\ e^{X}Y= e^\alpha Y e^{X}, \tag{*} $$ which is not what you are seeking, but your target expression is off. Convince yourself of that, perturbatively in α!
There is the more important identity in that article, $$ e^X e^Y= e^{X+{\alpha \over 1-e^{-\alpha}}Y}, $$ but I can't tell what you might be up to...
NB The above identity (*) is more trivial than it looks. For $[\partial_z , e^{\alpha z} ]= \alpha e^{\alpha z}$, you obtain a straightforward translation by 1, $$ e^{\partial} e^{\alpha z}=e^{\alpha ( z+1)} e^{\partial} =e^{\alpha}e^{\alpha z } e^{\partial}, $$ used routinely in physics...