Are perfectly LOCC-indistinguishable states necessarily identical?

Question

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!

Answer 1

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 $.