TITLE: Baker-Hausdorff Lemma from Sakurai's book
QUESTION [15 upvotes]: I'd like to show that, given to hermitian operators $A,G$ on a Hilbert space $\mathscr{H}$, the following identity holds:
$$
e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]+\ldots
$$
where $\lambda$ denotes a real parameter and $[\ \!,]$ indicates the commutator.
This is a proof left to the reader by Sakurai in his books on Modern Quantum Mechanics.

REPLY [3 votes]: Let
$ f(i\lambda) = e^{iG\lambda}Ae^{-iG\lambda} $
We Taylor expand
$f(i\lambda) = f(0) + i\lambda f'(0) + \frac{(i\lambda)^2}{2}  f''(0) + ... + \frac{(i\lambda)^n}{n!}f^{n}(0) + ...$
We see that
$ f'(i\lambda) = \frac{d}{di\lambda}(e^{iG\lambda})A e^{-iG\lambda} + e^{iG\lambda} A \frac{d}{di\lambda} (e^{-iG\lambda}) $
             $ = e^{iG\lambda}GAe^{-iG\lambda} - e^{iG\lambda}AGe^{-iG\lambda} $
             $ = e^{iG\lambda}[G,A]e^{-iG\lambda} $
$f'(0) = [G,A]$
In general, if
$f^{n}(i\lambda) = e^{iG\lambda}\underbrace{[G,...,[G}_{\text{$n$ times}},A]...]e^{-iG\lambda} $
Then
$f^{n+1}(i\lambda) = \frac{d}{di\lambda}(e^{iG\lambda})\underbrace{[G,...,[G}_{\text{$n$ times}},A]...]e^{-iG\lambda} + e^{iG\lambda}\underbrace{[G,...,[G}_{\text{$n$ times}},A]...]\frac{d}{di\lambda}(e^{-iG\lambda}) $
             $ = e^{iG\lambda}G\underbrace{[G,...,[G}_{\text{$n$ times}},A]...]e^{-iG\lambda} - e^{iG\lambda}\underbrace{[G,...,[G}_{\text{$n$ times}},A]...]Ge^{-iG\lambda} $
             $ = e^{iG\lambda}\underbrace{[G,...,[G}_{\text{$n+1$ times}},A]...]e^{-iG\lambda} $
$ f^{n+1}(0) = \underbrace{[G,...,[G}_{\text{$n+1$ times}},A]...] $
Putting this all together we get
$ e^{iG\lambda} A e^{-iG\lambda} = A + i\lambda[G,A] - \frac{(i\lambda)^2}{2}[G,[G,A]] + ... \frac{(i\lambda)^n}{n!} \underbrace{[G,...,[G}_{\text{$n$ times}},A]...] + ... $