# 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 :)

• Shouldn't this follow directly from Baker-Campbell-Hausdorff formula ? In special it is called Zassenhausformula Oct 22, 2013 at 9:03
• Hi Dominic. I have this as homework exercise and I never heard about Baker-Campbell-Hausdorff formula. Anyway, thank you for the reference :)
– KLMN
Oct 22, 2013 at 9:07

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.

• what an amazing answer ... thank you ! (i recovered my old account, i'm the same as the KLMN user) Oct 22, 2013 at 15:36
• @thetruth: if you wish, you can merge the KLMN account into this account. Just look here for instructions.
– robjohn
Oct 22, 2013 at 16:30


$\ds{\large% \expo{tA}\expo{tB} = \expo{t\pars{A + B}}\expo{\bracks{A,B}t^{2}/2}:\ {\Large ?}}$. 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 $\bracks{A,B} = -\bracks{B,A}$:

$$U\pars{t} = \expo{Bt} \expo{-\bracks{A,B}t^{2}/2}$$

• Thank you for the answer Felix. I'm sorry i cannot upvote, but i'm not registered. Have a nice day :)
– KLMN
Oct 22, 2013 at 11:49
• @KLMN You are registered now, so you can upvote... Oct 22, 2013 at 13:31
• How do you infer that $U(t) = e^{Bt + [B,A]t^2/2}$ from $\partiald{U(t)}{t}=\braces{B + [B,A]t}U(t)$? I can see that the former implies the latter. Are you using some sort of uniqueness theorem? Oct 22, 2013 at 15:55
• @user1551 The general solution involves a time ordered exponential. However, $\large \bracks{B + \bracks{B,A}t,B + \bracks{B,A}t'} = 0\,,\ \forall\ t,t'.$ In this case, the ordered product is not necessary any more and it's like any 'ordinary equation'. Oct 22, 2013 at 21:55

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.

• Inspired by a similar question on this site, I think it's easier if I prove that $e^{-t(A+B)}e^{tA}e^{tB}$ is the solution of $\frac{d}{dt}x = t[A,B]x$. Still, i'm having trouble doing this. If you have time and patience could you help me ? I will edit my question with this suggestion as well. Thanks so much for the attention :)
– KLMN
Oct 22, 2013 at 10:53