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I have been study the minimal (maximal) of a $f-$divergence. Fumio Hiai introduced the $\widehat{S}_f (\rho \| \sigma)$ divergence in his article.

$$\widehat{S}_f (\rho \| \sigma) := \text{Tr} \sigma^{1/2} f(\sigma^{-1/2} \rho \sigma^{-1/2}) \sigma^{1/2}$$

This fomular can be rewrite as follow

$$\widehat{S}_f (\rho \| \sigma) = \text{Tr} \sigma f(\sigma^{-1/2} \rho \sigma^{-1/2})$$

For $\sigma, \rho \in \mathbb{D}_n$ ($\mathbb{D}_n$ is the set of density matrices of order $n$), $U \in \mathfrak{U_n}$ with $\mathfrak{U_n}$ be the set of unitary matrices. In the article, Hiai assume that $f: (0; +\infty) \rightarrow \mathbb{R}$ is a continuous function such that the limits

$$f(0^+) := \displaystyle\lim_{x \searrow 0} f(x) \text{ and } f'(+\infty) := \displaystyle\lim_{x \rightarrow +\infty} \dfrac{f(x)}{x}$$

exist in $\mathbb{R} \cup \{\pm\infty\}$, and they are not both infinity with opposite signs. Then he showed that

$$f(x) = f(0^+) + ax + bx^2 + \displaystyle\int_{(0,+\infty)} \left( \dfrac{x}{1+s} - \dfrac{x}{x+s} \right) d \mu_f (s)$$

I would like to find the minimal and maximal of $\widehat{S}_f (\rho \| U^* \sigma U)$ with the function $f$ above, that is find

$$\displaystyle\min_{U \in \mathfrak{U_n}} \widehat{S}_f (\rho \| U^* \sigma U) \text{ and } \displaystyle\max_{U \in \mathfrak{U_n}} \widehat{S}_f (\rho \| U^* \sigma U)$$.

I think maybe $f(\sigma^{-1/2} \rho \sigma^{-1/2}) = f(\sigma^{-1} \rho)$, but I have no proof. To find the extremely values, we can use some characterizations of operator monotone and operator convex functions in this book and the von Neumann trace inequality to calculate.

Thank for read, if you have some idea, please let me know. Thank all!

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  • $\begingroup$ Thanks for the question, interesting topic. A few comments/questions: (1) Do you know that there's a closed form expression for these optimizations? (2) What constraints do you put on the function f? (3) Have you thought about the classical case? (4) This post is probably more likely to get an answer on the mathematics stackexchange. $\endgroup$
    – Rammus
    Commented Apr 4, 2023 at 9:06
  • $\begingroup$ Sorry for the inconvenience. You indicated my thoughts, it will be great if post this question on the mathematics stackexchange. My problem is mathematical calculation. Thank you! $\endgroup$
    – Minh
    Commented Apr 4, 2023 at 9:20

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