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Are separable, orthogonal states LOCC distinguishable?

Quantum Computing Asked on April 30, 2021

Consider two states $sigma_0,sigma_1intext{L}(mathcal{H}_{AB})$, 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
$$ sigma_0 = sum_{i=1}^n p_i |a_ib_iranglelangle a_ib_i| quadtext{and}quad
sigma_1 = sum_{j=1}^n q_j |a_jb_jranglelangle a_jb_j|, quad
text{where } p_i,q_jgeq 0, $$

and since
$$ 0 = text{Tr}(sigma_0^daggersigma_1)
= sum_{i,j}^n p_iq_j lvertlangle a_ib_i | a_jb_jranglervert^2, $$

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!

One Answer

Consider these two states $$ sigma_0 = frac{1}{2}(|11ranglelangle 11| + |++ranglelangle ++|) $$ $$ sigma_1 = frac{1}{2}(|0-ranglelangle 0-| + |-0ranglelangle -0|) $$

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.

Correct answer by Danylo Y on April 30, 2021

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