 Solution

Write $D=[A,B]$, so $[A,D]=[B,D]=0$. Repeated use of the product rule for the commutator gives $[A,B^n]=nD{B}^{n-1}$. Summing the exponential series yieldsThese manipulations hold for bounded operators, or as identities on a common invariant domain where the series and derivatives are justified.

The same commutation relation implies $e^{\lambda A}B{e}^{-\lambda A}=B+\lambda D$: its derivative is $e^{\lambda A}[A,B]e^{-\lambda A}=D$ and its initial value is $B$. Put $G(\lambda )=e^{\lambda A}{e}^{\lambda B}$. ThenFor the specified $F(\lambda )=G(\lambda )e^{-\lambda (A+B)}$, the operator $G$ commutes with $A+B$. Indeed conjugation by $e^{\lambda B}$ sends $A+B$ to $A+B-\lambda D$, and subsequent conjugation by $e^{\lambda A}$ restores $A+B$. Thereforeso $F(\lambda )=e^{{\lambda }^{2}D/2}$. At $\lambda =1$ this proves the central-commutator case of the Baker--Campbell--Hausdorff formula:The last equality holds because $D$ commutes with $A+B$.