Given two hermitian matrices $A$ and $B$, the Golden - Thompson inequality states: $$tr\left(e^{(A+B)}\right)\le tr\left(e^Ae^B\right)$$ My question is: when the two traces are equal?
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The condition for the identity is $[A,B]=0$.
Jon
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Well, at least a sufficiant condition is $[A,B]=0$. But it's not even necessary for $e^{(A+B)}=e^Ae^B$ – roman Sep 11 '13 at 13:31
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2@roman: $A$ and $B$ are Hermitian, so it is necessary – user8268 Sep 11 '13 at 13:41