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Let $\rho,\sigma\in\text{L}(\mathcal{H}_{XAB})$ be given by $$ \rho = \sum_x |x\rangle\langle x|\otimes p_x\rho_x, \quad \sigma = \sum_x |x\rangle\langle x|\otimes q_x\sigma_x, $$ and consider operators $M$ be given by $$ M = \sum_x |x\rangle\langle x|\otimes M_x, \quad\quad M_x\geq 0, \quad \sum_x M_x = id, $$ that is $\{M_x\}$ is a POVM measurement. Now suppose $$ \lVert M(\rho-\sigma)\rVert_1 = \sum_x \left|\operatorname{Tr}M_x(p_x\rho_x - q_x\sigma_x)\right| = 0 $$ for all measurement operators $M$ given as above, where $\{M_x\}$ is implementable by LOCC. Do we necessarily have $\rho = \sigma$?

Using $M_x = id/|X|$ yields $p_x = q_x$, and I have shown it is true for $\rho,\sigma$ pure using Schmidt decomposition, so I do believe it should be possible to prove $\rho_x = \sigma_x$ for all $x$, but I have not managed to do so.

Any help is appreciated!

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  • $\begingroup$ isn't Bennett et al. (1999) (10.1103/PhysRevA.59.1070) a counterexample of this (referring to the question in the title)? $\endgroup$
    – glS
    Jul 11, 2021 at 10:41

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We can take, for example, $ M = |0 \rangle \langle 0| \otimes I $, right? But then:

$$ \Big|\Big| M(\rho - \sigma) \Big|\Big|_1 = \Big|\Big| |0 \rangle \langle 0| \otimes (p_0 \rho_0 - q_0 \sigma_0) \Big|\Big|_1 = \Big|\Big| p_0 \rho_0 - q_0 \sigma_0 \Big|\Big|_1 = 0 \implies p_0 \rho_0 = q_0 \sigma_0 $$

Similary, taking $ M = |x \rangle \langle x| \otimes I $, we get $ p_x \rho_x = q_x \sigma_x, \forall x $. We conclude that $ \rho = \sigma $.

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  • $\begingroup$ Yeah, your are right. However, I just changed my question back to the original, which is slightly different. Sorry about the inconvenience. $\endgroup$
    – user114158
    Oct 13, 2020 at 14:18
  • $\begingroup$ Sorry for that, but could you explain? Which part of the question has changed and the answer is not valid? Also i think you can even take $ M_x = id / |X| $ and conclude since $ ||M(\rho - \sigma)||_1 = \sum_x \text{Tr}\big[ |M_x(p_x \rho_x - q_x \sigma_x)| \big] $, meaning $ | . | $ should be inside the trace. $\endgroup$
    – tsgeorgios
    Oct 13, 2020 at 14:45
  • $\begingroup$ In the sum, I have absolute value of the trace, not trace of the absolute value. I guess, I have used a misleading notation for the norm, but take the assumption as the sum over absolute values of traces being equal to zero. $\endgroup$
    – user114158
    Oct 13, 2020 at 14:50
  • $\begingroup$ Oh, i see now. So just to be clear you are assuming basically that $ \text{Tr} [M_x (\rho_x - \sigma_x) ] = 0 $ for every $ M_x \geq 0 $ and $ \sum_x M_x = I $. right? $\endgroup$
    – tsgeorgios
    Oct 13, 2020 at 15:15
  • $\begingroup$ Yes, that is indeed what I mean! $\endgroup$
    – user114158
    Oct 13, 2020 at 15:21

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