Whilst studying Commutators, I stumbled across this post in which the Baker–Campbell–Hausdorff formula is being proven. In one of the answers, it is stated that a certain step can be proven by induction; however, I struggle to do so:
Let $F(\lambda) = e^{\lambda A}Be^{-\lambda A}$
Inductionhypothesis:
$F^{(k)}(0) = \underbrace{[A,[A,\cdots,[A,B]\cdots]]}_{[A,\cdot]\text{ applied $k$ times}} = [A,\cdot]^k B$
Inductionstart:
$F^{(0)}(0) = F(0) = B = [A,\cdot]^0 B$
Inductionstep:
$F^{(k+1)}(0) = \cdots =[A,\cdot]^{k+1} B$
I suppose there is some series expansion for $F^{(k)}(\lambda)$, but this is also tricky since the operators $A$, $B$, $e^A$, $e^{-A}$ do not necessarily commute.
Any help would be appreciated. Thank you!
Best Answer
Let's prove a little more general statement, namely $F^{(k)}(\lambda) = e^{\lambda A} [A,B]_k e^{-\lambda A}$, with $[\,\cdot,\,\cdot\,]_k$ denotes nested commutators, as you defined it. The base case is trivial, since $[A,B]_0 = B$. Now, the induction step goes as follows : $$ \begin{align} F^{(k+1)}(\lambda) &= \frac{\mathrm{d}}{\mathrm{d}\lambda}\left(e^{\lambda A} [A,B]_k e^{-\lambda A}\right) \\ &= e^{\lambda A}\left(A[A,B]_k - [A,B]_kA\right)e^{-\lambda A} \\ &= e^{\lambda A} [A,[A,B]_k] e^{-\lambda A} \\ &= e^{\lambda A} [A,B]_{k+1} e^{-\lambda A} \end{align} $$ In consequence, one has $F^{(k)}(0) = [A,B]_k$.