Matrix Equality Can you help me to prove this equality ? 
Let $A,B$ be $n\times n$ matrices. Let $[A,B]$ denote the usual matrix commutator and $e^{A}$ the usual matrix exponential.  By hypothesis, let's say that $[A,[A,B]]=[B,[A,B]]=0$. We want to prove that, $\forall t\in \mathbb{R}$ we have that $e^{tA}e^{tB}=e^{t(A+B)}e^{\frac{t^{2}}{2}[A,B]}$.
EDIT : Thank you for the suggestion of Zassenhaus Formula, but I think, since this is homework question, I should do a little bit more than just say this is a direct consequence of that result. Instead of proving Zassenhaus Formula, I think it's easier if I prove that $e^{-t(A+B)}e^{tA}e^{tB}$ is the solution of the following differential equation : $\frac{d}{dt}x = t[A,B]x$. Can somebody help me doing this ?
Thank you very much :)
 A: The Baker-Campbell-Hausdorff formula says that
$$
e^{tX}e^{tY}=e^Z
$$
where
$$
Z=tX+tY+\frac{t^2}{2}[X,Y]+\frac{t^3}{12}\big([X,[X,Y]]-[Y,[X,Y]]\big)+\dots
$$
and further terms involve higher powers of $t$ and higher order commutators. Since all commutators of order higher than $1$ are $0$, this gives the result. That is, because $[X,Y]$ commutes with both $X$ and $Y$,
$$
\begin{align}
e^{tX}e^{tY}
&=e^{tX+tY+\frac{t^2}{2}[X,Y]}\\
&=e^{tX+tY}e^{\frac{t^2}{2}[X,Y]}
\end{align}
$$

The Baker-Campbell-Hausdorff Formula
Suppose $X$ and $Y$ are elements in a non-commutative algebra. Define the operator $\mathrm{ad}(X)$ by $\mathrm{ad}(X)Y = [X,Y] = XY-YX$.  Then we get the following:
Lemma 1: $\displaystyle X^kY=\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^jYX^{k-j}$
Proof: The case $k=0$ is trivial.  Suppose the statement is true for some $k$, then
$$
\begin{align}
X^{k+1}Y
&=\mathrm{ad}(X)\left(X^kY\right)+\left(X^kY\right)X\\
&=\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^{j+1}YX^{k-j}
+\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^jYX^{k-j+1}\\
&=\sum_{j=0}^{k+1}\binom{k}{j-1}\mathrm{ad}(X)^{j}YX^{k-j+1}
+\sum_{j=0}^{k+1}\binom{k}{j}\mathrm{ad}(X)^jYX^{k-j+1}\\
&=\sum_{j=0}^{k+1}\binom{k+1}{j}\mathrm{ad}(X)^jYX^{k-j+1}
\end{align}
$$
Thus, the statement is true for $k+1$.$\quad\square$
Let $\mathrm{D}$ be the usual derivative: $\mathrm{D}(XY) = X\,\mathrm{D}(Y) +\mathrm{D}(X)\,Y$.  Then
$$
\begin{align}
\mathrm{D}\left(X^n\right)
&=\sum_{k=0}^{n-1}X^k\mathrm{D}(X)X^{n-k-1}\\
&=\sum_{k=0}^{n-1}\sum_{j=0}^k\binom{k}{j}\mathrm{ad}(X)^j\mathrm{D}(X)X^{k-j}X^{n-k-1}\\
&=\sum_{j=0}^{n-1}\sum_{k=j}^{n-1}\binom{k}{j}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\\
&=\sum_{j=0}^{n-1}\binom{n}{j+1}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\tag{1}
\end{align}
$$
Using the power series for $e^X$, we get
$$
\begin{align}
\mathrm{D}\left(e^X\right)
&=\sum_{n=0}^\infty\sum_{j=0}^{n-1}\frac1{n!}\binom{n}{j+1}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\\
&=\sum_{j=0}^\infty\sum_{n=j+1}^\infty\frac1{(j+1)!}\frac1{(n-j-1)!}\mathrm{ad}(X)^j\mathrm{D}(X)X^{n-j-1}\\
&=\sum_{j=0}^\infty\sum_{n=0}^\infty\frac1{(j+1)!}\frac1{n!}\mathrm{ad}(X)^j\mathrm{D}(X)X^n\\
&=\left[\frac{e^{\mathrm{ad}(X)}-1}{\mathrm{ad}(X)}\mathrm{D}(X)\right]e^X\tag{2}
\end{align}
$$
Because $\frac{\mathrm{d}}{\mathrm{d}t}e^{tX}=Xe^{tX}$, we get the following
Lemma 2: $e^{tX}Ye^{-tX}=e^{t\,\mathrm{ad}(X)}Y$
Proof: Note that
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{tX}Ye^{-tX}\right)
&=X\left(e^{tX}Ye^{-tX}\right)-\left(e^{tX}Ye^{-tX}\right)X\\
&=\mathrm{ad}(X)\left(e^{tX}Ye^{-tX}\right)
\end{align}
$$
Thus, $e^{tX}Ye^{-tX}=e^{t\,\mathrm{ad}(X)}Y$.$\quad\square$
From this, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{tX}e^{tY}\right)
&=X\left(e^{tX}e^{tY}\right)+\left(e^{tX}Ye^{tY}\right)\\
&=\left(X+e^{t\,\mathrm{ad}(X)}Y\right)\left(e^{tX}e^{tY}\right)\tag{3}
\end{align}
$$
Using the power series for $\log$, we see that formally, there is a $Z$ so that $e^Z=e^{tX}e^{tY}$.
Combining
$$
\frac{\mathrm{d}}{\mathrm{d}t}\left(e^Z\right)e^{-Z}
=\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{tX}e^{tY}\right)e^{-tY}e^{tX}\tag{4}
$$
with $(2)$ and $(3)$, we get that
$$
\frac{e^{\mathrm{ad}(Z)}-1}{\mathrm{ad}(Z)}\frac{\mathrm{d}}{\mathrm{d}t}Z
=X+e^{t\,\mathrm{ad}(X)}Y\tag{5}
$$
and thus,
$$
\frac{\mathrm{d}}{\mathrm{d}t}Z
=\frac{\mathrm{ad}(Z)}{e^{\mathrm{ad}(Z)}-1}\left(X+e^{t\,\mathrm{ad}(X)}Y\right)\tag{6}
$$
If we iterate $(6)$, the power series in $t$ for $Z$ converges at least one order of $t$, hence one order of commutator, per iteration to yield
$$
Z=tX+tY+\frac{t^2}{2}[X,Y]+\frac{t^3}{12}\big([X,[X,Y]]-[Y,[X,Y]]\big)+\dots\tag{7}
$$
where only terms with commutators of third or higher order have been omitted.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$

$\ds{\large%
\expo{tA}\expo{tB}
=
\expo{t\pars{A + B}}\expo{\bracks{A,B}t^{2}/2}:\ {\Large ?}}$.
Let's define $U\pars{t} = \expo{-tA}\expo{t\pars{A + B}}$. Then, we have to prove that $\ds{\large U\pars{t} = \expo{tB}\expo{-\bracks{A,B}t^{2}/2}}$.

\begin{align}
\partiald{U\pars{t}}{t}
&=
-\expo{-tA}A\expo{t\pars{A + B}} + \expo{-tA}\pars{A + B}\expo{t\pars{A + B}}
=
\expo{-tA}B\expo{tA}\expo{-tA}\expo{t\pars{A + B}}
=
B\pars{t}U\pars{t}
\end{align}
where $B\pars{t} \equiv \expo{-At}B\expo{At}$.

Notice that
$\partiald{B\pars{t}}{t}
=
-A\expo{-At}B\expo{At} + \expo{-At}B\expo{At}A = \bracks{B\pars{t},A}$
$$
B\pars{0} = B\,,
\quad
\left.\partiald{B\pars{t}}{t}\right\vert_{t = 0}
=
\bracks{B,A}\,,
\quad
\left.\partiald[2]{B\pars{t}}{t}\right\vert_{t = 0}
=
\bracks{\bracks{B,A},A} = 0
$$
$$
\left.\partiald[3]{B\pars{t}}{t}\right\vert_{t = 0}
=
\bracks{\bracks{\bracks{B,A},A},A} = 0\,,
\quad\ldots\quad
\left.\partiald[n]{B\pars{t}}{t}\right\vert_{t = 0}
=
0
$$
Then
$$
B\pars{t} = B + \bracks{B,A}t
$$
\begin{align}
\partiald{U\pars{t}}{t}
=
\braces{B + \bracks{B,A}t}U\pars{t}
\quad\imp\quad
U\pars{t} = \expo{Bt + \bracks{B,A}t^{2}/2}
\end{align}
Since $\bracks{B,\bracks{A,B}} = 0$ and $\bracks{A,B} = -\bracks{B,A}$:
$$
U\pars{t} = \expo{Bt} \expo{-\bracks{A,B}t^{2}/2}
$$
A: Hint: 
\begin{align*}
\exp(t(A+B))&= \sum_{i=0}^\infty \frac{t^i (A+B)^i}{i!}
\end{align*}
Now start multiplying out $(A+B)^i$ and think what terms are zero and write the left hand side. 
