Commutator expansion for exponential conjugation ($e^{A}B{e}^{-A}={\sum }_{n\ge 0}{\mathrm{ad}}_A^n(B)/n!$)

For bounded linear operators, or formally in an associative algebra,Differentiate the two exponentials using the product rule: the result is $e^{\lambda A}[A,B]e^{-\lambda A}$. Induction gives the $n$th derivative $e^{\lambda A}{\mathrm{ad}}_A^n(B)e^{-\lambda A}$, and the Taylor series at zero proves the formula. For bounded operators convergence follows from $\Vert {\mathrm{ad}}_A^n(B)\Vert \le (2\Vert A\Vert {)}^n\Vert B\Vert$. For unbounded quantum operators, use a common invariant domain and justify the series there, or use the finite-dimensional system of commutators when their span closes.

For finite-dimensional complex linear operators with $C=[A,B]$ commuting with both $A$ and $B$, differentiating $e^{-vB}A{e}^{vB}$ gives the constant derivative $C$. Hence $e^{-B}A{e}^B=A+C$, and conjugating the matrix exponential gives $e^{-B}{e}^{-A}{e}^B=e^{-A-C}$. Since $[A,C]=0$, multiplication by $e^A$ proves the identity. The commutator order determines the minus sign.