If $[A,B]=B$. Calculate $e^{iA}Be^{-iA}$

Question

If $[A,B]=B$. Calculate \begin{equation*} e^{iA}Be^{-iA} \end{equation*}

From this, We know that \begin{equation*} \begin{gathered} e^{iA}Be^{-iA} = (\mathbb{I} + iA -\frac{1}{2}A^2+...)B(\mathbb{I}-iA+\frac{1}{2}A^2+...)) \\ = (B +iAB -\frac{1}{2}A^2B+...)(\mathbb{I}-iA+\frac{1}{2}A^2+...)\\ = B -iBA + \frac{BA^2}{2} +iAB + ABA + \frac{iABA^2}{2}-\frac{A^2B}{2}+\frac{iA^2BA}{2}-\frac{A^2BA^2}{4}+... \\ = B +iB + \frac{BA^2}{2} + ABA + \frac{iABA^2}{2}-\frac{A^2B}{2}+\frac{iA^2BA}{2}-\frac{A^2BA^2}{4}+... \end{gathered} \end{equation*}

From here, I'm not sure I'm going the right way.

Answer 1

I assume that these are matrices with $[A,B] = AB - BA$.

Hint: Using the fact that $\frac d{dt} e^{tA} = Ae^{tA} = e^{tA}A$, verify that $$ \frac d{dt} [e^{tA}Be^{-tA}] = e^{tA}[A,B]e^{-tA} $$