# Are perfectly LOCC-indistinguishable states necessarily identical?

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!

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

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 $$\rho = \sigma$$.
• 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. Oct 13 '20 at 14:45
• 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? Oct 13 '20 at 15:15