The [Baker-Campbell-Hausdorff lemma](Baker-Campbell-Hausdorff%20lemma.md) in its derivative form may be shown by [Taylor expanding](Analysis%20(index).md#Taylor%20Series) the expression on the left hand side after parmeterizing the [matrix exponential](Matrix%20exponentials.md) with $t \in \mathbb{R}$. Set $f(t)=e^{tX}Ye^{-tX}$, so that $f(0)=Y,$$f'(0)= (Xe^{tX}Ye^{-tX}-e^{tX}YXe^{-tX})\bigg|_{t=0} =[X,Y],$$f''(0)= \frac{d}{dt}(Xe^{tX}Ye^{-tX}-e^{tX}YXe^{-tX})\bigg|_{t=0}$$=(e^{tX}X^2Ye^{-tX}-e^{tX}XYXe^{-tX}-e^{tX}XYXe^{-tX}+e^{tX}YX^2e^{-tX})\bigg|_{t=0} = [X,[X,Y]]$ therefore, $f(t)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}t^n = Y + t[X,Y]+\frac{t}{2!}[X,[X,Y]]+...+\frac{t^n}{n!}[X,[X,[X,...[X,Y]]]...]+...$ ^163fd4 Plugging in $t=1$, gives us the lemma, $e^XYe^{-X}=Y+[X,Y]+\frac{1}{2!}[X,[X,Y]]+...+\frac{1}{n!}[X,[X,[X,...[X,Y]]]...]+...$ and plugging in $t=i$ gives us the lemma for the [complex matrix exponential.](Baker-Campbell-Hausdorff%20lemma.md#Complex%20BCH%20lemma) $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ ^78258a ### Proof of the integral form of the BCH lemma We can show that $Y(t)=e^{tX}Ye^{-tX}$ solves the 1st order [integral form of the BCH lemma,](Baker-Campbell-Hausdorff%20lemma.md#Integral%20form%20of%20the%20BCH%20lemma) [$Y(t) = Y + \int_0^t d\tau [X,Y(\tau)]$](Baker-Campbell-Hausdorff%20lemma#^5f9ea7) by evaluating the right hand side and setting it equal to $e^{tX}Ye^{-tX}$: $Y+\int_0^t d\tau [X,e^{tX}Ye^{-tX}] = Y + \int_0^t d\tau e^{\tau X}XYe^{-\tau X}- \int_0^t d\tau e^{\tau X}YXe^{-\tau X}$ ^de1943 And we solve each integral individually by integrating by parts: for $\int_0^t d\tau e^{\tau X}XYe^{-\tau X}$ set $dv=e^{\tau X}XY$ and $u=e^{-\tau X}$ obtaining, $\int_0^t d\tau e^{\tau X}XYe^{-\tau X}= e^{\tau X}Ye^{-\tau X}\bigg|_0^t - \int_0^t d\tau e^{\tau X}Y(-X)e^{-\tau X}=e^{t X}Ye^{-tX}-Y+\int_0^t d\tau e^{\tau X}YXe^{-\tau X}$ ^717d80 For $\int_0^t d\tau e^{\tau X}YXe^{-\tau X}$ we make sure to set $dv=YXe^{-\tau X}$ and $u=e^{\tau X}$ obtaining, $\int_0^t d\tau e^{\tau X}YXe^{-\tau X}= -e^{\tau X}Ye^{-\tau X}\bigg|_0^t - \int_0^t d\tau e^{\tau X}XY(-e^{-\tau X})=e^{tX}Ye^{-tX}-Y-\int_0^t d\tau e^{\tau X}XYe^{-\tau X}$ ^cffb6d Combining terms as well as the integrals without evaluating all of them, we obtain $Y(t) = Y -2Y + 2e^{tX}Ye^{-tX}-\int_0^t d\tau [X,e^{tX}Ye^{-tX}]$ By then setting $Y(t)=e^{tX}Ye^{-tX}$ we obtain: $e^{tX}Ye^{-tX}=Y+\int_0^t d\tau [X,Y(\tau)]$ Thus $e^{tX}Ye^{-tX}$ solves the integral equation. ^997f0e We can find what the integral equation would look like up to $n$ orders by repeatedly substituting for $Y(\tau)$ in order to obtain $Y(t) = Y + \int_0^{t} d\tau_1 [X,Y(\tau_1)]+ \frac{t}{2!}\int_0^{t}\int_0^{t} d\tau_1d\tau_2[X,[X,Y(\tau_2)]]+...$ $+\frac{t^{n-1}}{n!}\int_0^{t}...\int_0^{t} d\tau_1...d\tau_n [X,[X,[X,...[X,Y(\tau_n)]]]...]$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ ^c7fef6 #MathematicalFoundations/Algebra/AbstractAlgebra/GroupTheory/Lie/LieGroups #MathematicalFoundations/Algebra/AbstractAlgebra/GroupTheory/Lie/LieGroups/Algebras/LieAlgebras #MathematicalFoundations/Algebra/AbstractAlgebra/LinearAlgebra/Operators/Matrices #Proofs