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Maps on quantum states preserving Bregman and Jensen divergences. (English) Zbl 1387.47022
Let $$\mathcal{H}$$ be a finite dimensional complex Hilbert space and $$\mathcal{B}(\mathcal{H})$$ the set of all bounded linear operators on $$\mathcal{H}$$. Denote by $$\mathcal{S}(\mathcal{H})$$ the state space of $$\mathcal{H}$$ (i.e., the set of all positive semi-definite operators in $$\mathcal{B}(\mathcal{H})$$ having unit trace) and by $$\mathcal{P}_{1}(\mathcal{H})$$ the set of all rank-one projections (rank-one self-adjoined idempotents) in $$\mathcal{B}(\mathcal{H})$$. Let $$\operatorname{Tr}$$ denote the usual trace functional on $$\mathcal{B}(\mathcal{H})$$.
Any unitary or antiunitary conjugation leaves the Bergman divergences invariant. The first result of the paper states that the converse is also true, i.e., the preservers of Bergman divergences are necessarily unitary or antiunitary conjugations.
Let $$f \in C^{1}((0 ,\infty )) \cap C^{0}([0 ,\infty ))$$ be a strictly convex function. Let $$\Phi :$$ $$\mathcal{S}(\mathcal{H}) \rightarrow \mathcal{S}(\mathcal{H})$$ be a bijective map which preserves the Bergman $$f$$-divergence $$H_{f}$$, that is, $H_{f}(\Phi (A) ,\Phi (B)) =H_{f}(A ,B)\tag{1}$ for every $$A ,B \in \mathcal{S}(\mathcal{H})$$. Then there exists a unitary or antiunitary operator $$U \in \mathcal{B}(\mathcal{H})$$ such that $\Phi (A) =UAU^{ \ast } ,\quad A \in \mathcal{S}(\mathcal{H}).\tag{2}$ The main tool in the proof of this result is Wigner’s theorem which states that any bijective map $$\xi :\mathcal{P}_{1}(\mathcal{H}) \rightarrow \mathcal{P}_{1}(\mathcal{H})$$ which preserves the transition probability, i.e., $$\operatorname{Tr}(\xi (P)\xi (Q)) =\operatorname{Tr}(PQ)$$, holds for any $$P ,Q \in \mathcal{P}_{1}(\mathcal{H})$$, is implemented by a unitary or antiunitary operator.

##### MSC:
 47B49 Transformers, preservers (linear operators on spaces of linear operators) 46L30 States of selfadjoint operator algebras
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