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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?

Thanks.

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1 Answer 1

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The condition for the identity is $[A,B]=0$.

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  • $\begingroup$ Well, at least a sufficiant condition is $[A,B]=0$. But it's not even necessary for $e^{(A+B)}=e^Ae^B$ $\endgroup$
    – roman
    Sep 11, 2013 at 13:31
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    $\begingroup$ @roman: $A$ and $B$ are Hermitian, so it is necessary $\endgroup$
    – user8268
    Sep 11, 2013 at 13:41

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