Let $A$ and $B$ be Hermitian matrices.
- If $AB=BA$, we know that $e^{A+B} = e^A e^B$.
- In this paper, the author showed that $\text{Tr } e^{A+B} = \text{Tr } e^A e^B$ iff. $AB=BA$.
As such, $e^{A+B} = e^A e^B$ is equivalent to $\text{Tr } e^{A+B} = \text{Tr } e^A e^B$ in the context of Hermitian matrices.
My question is how we can derive the commutation relation between $A$ and $B$ directly from $e^{A+B}=e^A e^B$ without bringing in the Golden-Thompson inequality (as in the paper I linked). Since the condition $e^{A+B} = e^A e^B$ has a simpler form than that involving the trace, I think there should be some way.
Edit: rephrase the question
Best Answer
The idea here is $A+B$ is Hermitian and the exponential map preserves Hermicity. Taking the conjugate transpose of each side, we have
$e^Ae^B = e^{A+B} = \big(e^{A+B}\big)^*=\big(e^B\big)^*\big(e^A\big)^*=e^Be^A$
so $e^A$ and $e^B$ commute.
Now call on a lemma twice:
for Hermitian $X,Y$
$e^XY= Ye^X$
iff $XY=YX$
proof sketch: the same unitary matrix $U$ that simultaneously diaogonalizes $e^X$ and $Y$ must diagonalize $X$ as well since all are Hermitian. And the same argument also runs backwards. (Underlying idea: the exponential map is injective on reals and Hermitian matrices are diagonalizable with real spectrum. So $e^X \mathbf v = \sigma \cdot \mathbf v\implies X\mathbf v = \log(\sigma)\cdot \mathbf v$ and of course $X \mathbf v = \lambda \cdot \mathbf v\implies e^X\mathbf v = e^{\lambda}\cdot \mathbf v$)
after applying the lemma once, with $Y:=e^B$, $X=A$, we know $Ae^B = e^BA$
and a 2nd application of the lemma, with $Y:=A$ and $X:= B$, tells us $AB = BA$