Given a linear operator $A$ , we define: \begin{equation} \exp{A} := \sum_{k=0}^{\infty} \frac{1}{k!}A^k. \end{equation}It follows that \begin{equation} \frac{\partial}{\partial \tau} e^{\tau A} = A e^{\tau A}= e^{\tau A} A. \end{equation}Consider another linear operator $B$ . Let $B(\tau)=e^{\tau A} B e^{-\tau A}$ . Then one can prove the following series representation for $B(\tau)$ : \begin{equation}\label{eq:genbch} B(\tau)= \sum_{m=0}^{\infty} \frac{{\tau}^m}{m!}B_m, \end{equation}where $B_m =[A,B]_m := [A,[A,B]_{m-1}]$ and $B_0:=B$ . A very important special case of eq. () is known as the Baker-Campbell-Hausdorff (BCH) formula. Namely, for $\tau =1$ we get: \begin{equation} e^A \; B e^{-A} = \sum_{m=0}^{\infty} \frac{1}{m!} B_m. \end{equation}Alternatively, this expression may be rewritten as \begin{equation} [B,e^{-A}] = e^{-A} \left( [A,B]+\frac{1}{2} [A,[A,B]] + \cdots \right), \end{equation}or \begin{equation} [e^A,B] = \left( [A,B]+\frac{1}{2} [A,[A,B]] + \cdots \right) e^A. \end{equation}There is a descendent of the BCH formula, which often is also referred to as BCH formula. It provides us with the multiplication law for two exponentials of linear operators: Suppose $[A,[A,B]] = [B,[B,A]] = 0$ . Then, \begin{equation} e^A e^B = e^{A+B} e^{\frac{1}{2}[A,B]}. \end{equation}Thus, if we want to commute two exponentials, we get an extra factor \begin{equation} e^A e^B = e^B e^A e^{[A,B]}. \end{equation}