 Solution

For a Hermitian operator $Q$, the operator exponential $C_{\lambda }=e^{i\lambda Q}$ is a unitary operator, with $C_{\lambda }^{†}=e^{-i\lambda Q}$. Differentiation and $[Q,C_{\lambda }]=0$ giveWhen $[Q,A]=cA$, this is the first-order ordinary differential equation $A^{\prime }(\lambda )=icA(\lambda )$ with $A(0)=A$. Multiplication by $e^{-ic\lambda }$ makes its derivative zero, provingEquivalently, the commutator expansion for exponential conjugation has successive nested commutators $c^{r}A$, whose exponential series sums to this expression.