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  States diagonal in the tensor product of Bell states.

+ 6 like - 0 dislike

Bell-diagonal states are 2-qubit states that are diagonal in the Bell basis. Since those states lie in $\mathbb{C}^{2} \otimes \mathbb{C}^{2}$, the Peres-Horodecki criterion is a sufficient condition to show separability and it's also pretty easy to check: $\rho = \sum_{i \in [0,3]}\lambda_i|\psi_{i}\rangle\langle\psi_{i}| $ is PPT (or separable) if and only if $Tr(\rho) \geq 2\lambda_i \geq 0$ for every $i$. (Here {$|\psi_{i}\rangle$} are the Bell states)

In my research I am dealing with a generalization of those states. In particular, my question is about states in $\mathbb{C}^{2^d} \otimes \mathbb{C}^{2^d}$ that are diagonal in the basis given by the $d$-fold tensor product of Bell states.

For example, for $d=2$, the states I am considering are diagonal in the basis: $$ |\psi_{0}\rangle\otimes|\psi_{0}\rangle,|\psi_{0}\rangle\otimes|\psi_{1}\rangle, \ldots,|\psi_{3}\rangle\otimes|\psi_{3}\rangle. $$

I am wondering the following:

are there some nice criteria already known to check when these states are PPT or separable?

Notice that those states are in general different from the states diagonal in what is called the generalized Bell basis in literature.

This post has been migrated from (A51.SE)
asked Mar 30, 2012 in Theoretical Physics by Alessandro Cosentino (30 points) [ no revision ]
retagged Mar 18, 2014 by dimension10

1 Answer

+ 4 like - 0 dislike

Try Theorem 12 in (the arXiv version) of http://arxiv.org/abs/quant-ph/0411098. What I call "lattice states" there should exactly be the class of states you are interested in.

This post has been migrated from (A51.SE)
answered Apr 2, 2012 by Marco (260 points) [ no revision ]
thanks Marco! this is what I was looking for.

This post has been migrated from (A51.SE)

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