= Commutator expansion for exponential conjugation
{title2=$e^ABe^{-A}=\sum_{n\ge0}\operatorname{ad}_A^n(B)/n!$}
For bounded <linear operators>, or formally in an associative algebra,
$$
e^{\lambda A}Be^{-\lambda A}=\sum_{n\ge0}\frac{\lambda^n}{n!}\operatorname{ad}_A^n(B),\qquad \operatorname{ad}_A(B)=[A,B].
$$
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 $e^{\lambda A}\operatorname{ad}_A^n(B)e^{-\lambda A}$, and the <Taylor series> at zero proves the formula. For bounded operators convergence follows from $\|\operatorname{ad}_A^n(B)\|\le(2\|A\|)^n\|B\|$. 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.
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