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Consider two states $\sigma_0,\sigma_1\in\text{L}(\mathcal{H}_{AB})$, and suppose $\sigma_0,\sigma_1$ are separable and orthogonal. Is it possible to distinguish between $\sigma_0,\sigma_1$ through LOCC?

My approach so far har been to write out $$ \sigma_0 = \sum_{i=1}^n p_i |a_ib_i\rangle\langle a_ib_i| \quad\text{and}\quad \sigma_1 = \sum_{j=1}^n q_j |a_jb_j\rangle\langle a_jb_j|, \quad \text{where } p_i,q_j\geq 0, $$ and since $$ 0 = \text{Tr}(\sigma_0^\dagger\sigma_1) = \sum_{i,j}^n p_iq_j \lvert\langle a_ib_i | a_jb_j\rangle\rvert^2, $$ it follows that all terms in the decomposition of $\sigma_0$ are orthogonal to all terms in the decomposition of $\sigma_1$. My idea was now to measure using projections onto the two subspaces spanned by the terms in each decomposition, and these are separable projections. I am stuck at implementing this as an LOCC protocol, so any help with this or giving an alternative approach is appreciated!

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  • $\begingroup$ One of the problems that you have is that while the vectors $|a_i\rangle|b_i\rangle$ may be orthonormal and separable, that does not guarantee that the sets $\{|a_i\rangle\}$ and $\{|b_i\rangle\}$ form orthonormal bases, so you may not be able to set them up as local measurements. It doesn't mean that there isn't an LOCC protocol that distinguishes them, but it does suggest that a general protocol is, at best, problematic. $\endgroup$ – DaftWullie Jun 25 at 8:49
  • $\begingroup$ Does "distinguish" mean "with certainty", or "with some chance"? That's a major difference. $\endgroup$ – Norbert Schuch Jun 25 at 9:59
  • $\begingroup$ @NorbertSchuch Good point, I mean with certainty. $\endgroup$ – user114158 Jun 25 at 17:27
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Consider these two states $$ \sigma_0 = \frac{1}{2}(|11\rangle\langle 11| + |++\rangle\langle ++|) $$ $$ \sigma_1 = \frac{1}{2}(|0-\rangle\langle 0-| + |-0\rangle\langle -0|) $$

I believe they are indistinguishable (with certainty), though to be sure it's better to find the exact proof.

Also check this paper https://arxiv.org/abs/quant-ph/9804053.
It's impossible to distinguish a set of product states in general, though this is not directly answers your question.

Update
As John Watrous explained in the comments, $\sigma_0$,$\sigma_1$ are indeed indistinguishable. They have orthogonal images that span the whole space. So, the only way to distinguish them with certainty is to use the two-outcome projective measurement where projections correspond to the images. But these projections are not separable, we can use PPT criterion to check this. For example, the projection on $\text{Im}(\sigma_0)$ is $$ P_0 = \frac{1}{3}\begin{pmatrix} 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 3 \\ \end{pmatrix} $$ and you can check that the partial transpose $$ P_0^{T_2} = \frac{1}{3}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 3 \\ \end{pmatrix} $$ has a negative eigenvalue.

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    $\begingroup$ Regarding the quoted paper: What does it say? (And if it is about a larger set, then even is a strange word: Is is clearly harder to distinguish more states.) $\endgroup$ – Norbert Schuch Jun 25 at 9:58
  • $\begingroup$ Yes, it's not exactly comparable. On the one hand product state is more restrictive condition than separable, but on the other hand it is harder to distinguish more states. This is why I presented a different example. $\endgroup$ – Danylo Y Jun 25 at 10:21
  • $\begingroup$ I have to admit that "I don't see a way to distinguish them" is not a very compelling argument. (To quote Reinhard Werner: Lack of imagination is not an argument.) For sure, you can probabilistically distinguish them. Also, note that LOCC protocols can be very complicated (multi-round protocols etc.!), so I'm not sure whether "some calculations" are sufficient. $\endgroup$ – Norbert Schuch Jun 25 at 10:33
  • $\begingroup$ I'm not very familiar with such kind of proofs. But the question is asking about some help. Maybe I'll find the exact proof later. $\endgroup$ – Danylo Y Jun 25 at 10:48
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    $\begingroup$ This is s great guess. Indeed these two states cannot be perfectly discriminated by an LOCC measurement, or even by a separable measurement, which is surprising given that the states themselves are separable. The reason is that the states have orthogonal images that together span the entire two-qubit space, so the only measurement that can perfectly discriminate them is the two-outcome projective measurement corresponding to the projections onto their images. These projections, however, are not separable, as the PPT test reveals. $\endgroup$ – John Watrous Jun 30 at 1:28

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