# Proof of a matrix exponential identity

Let $A,B \in M_n(\mathbb{R})$, and suppose $[A,[A,B]] = [B,[A,B]] = 0$, where $[A,B] = BA-AB$. We want to show $$e^{At}e^{Bt}=e^{(A+B)t}e^{[A,B]t^2/2}, \quad \forall t\in \mathbb{R}.$$ I'm given a hint to show that $x(t) =e^{-(A+B)t}e^{Bt}e^{At}x_0$ is a solution to the ODE $x'= t[A,B]x$ for each $x_0\in \mathbb{R}$.

I haven't been able to prove the hint, let alone see why being a solution helps prove the identity. Taking the derivative of $x(t)$ didn't lead to much, and moreover there's no $t$ outside of the exponent, and substituting it into the RHS of the ODE also didn't help.

The RHS of the identity to be shown solves $x' = (A+B+t[A,B])x$, which maybe gives some intuition about the ODE in the hint, but I'm not really sure where to go.

• Have you heard about the Baker–Campbell–Hausdorff formula? Jul 25, 2017 at 7:42
• I have not. Upon looking at the Wikipedia page, I notice that this identity is the Zassenhaus formula (where the exponentials with cubic or higher terms cancel because of my assumption on the commutators). Perhaps I should look up a proof of that, though, this seems like a little overkill for my problem. Jul 25, 2017 at 7:58
• A proof of the Baker-Campbell-Hausdorff formula is given in this answer.
– robjohn
Jul 25, 2017 at 11:44

(Caution: usually we define the commutator as $[A,B]=AB-BA$; in your question, the notation $[A,B]$ is the negative of the usual definition.)
Let $x(t)=e^{-t(A+B)}e^{tB}e^{tA}x_0$. Then \begin{align*} x'(t) &=e^{-t(A+B)}(-A-B)e^{tB}e^{tA}x_0+e^{-t(A+B)}Be^{tB}e^{tA}x_0+e^{-t(A+B)}e^{tB}Ae^{tA}x_0\\ &=e^{-t(A+B)}(-Ae^{tB}+e^{tB}A)e^{tA}x_0.\tag{1} \end{align*} Using mathematical induction and the fact that $B$ commutes with $[A,B]$, one can prove that $-AB^k=kB^{k-1}[A,B]-B^kA$ for each $k\ge1$. Apply this result to the power series expansion of $e^{tB}$, one obtains $$-Ae^{tB}+e^{tB}A=te^{tB}[A,B].\tag{2}$$ As $[A,B]$ commutes with both $A$ and $B$, $(1)$ and $(2)$ together gives $$x'(t)=t[A,B]x.\tag{3}$$ Since $x(t)=e^{[A,B]t^2/2}x_0$ is also a solution satisfying the same differential equation $(3)$ with the same (arbitrary) initial condition $x_0$, we conclude that $$e^{tB}e^{tA}=e^{t(A+B)}e^{[A,B]t^2/2}.\tag{4}$$ Note that your question asks you to prove that $$e^{t\color{red}{A}}e^{t\color{red}{B}}=e^{t(A+B)}e^{[A,B]t^2/2},\tag{5}$$ but this disagrees with the BCH formula. I suppose $(5)$ is a typo and this typo arises probably because the question setter redefined the commutator as $BA-AB$ but forgot to interchange the roles of $A$ and $B$ in $(5)$.