Quantum Computing Asked by draks ... on August 20, 2021

In "SEPARABILITY CRITERION FOR MULTIPARTITE QUANTUM STATES BASED ON THE BLOCH REPRESENTATION OF DENSITY MATRICES" by Hassan and Joag, I found this remarkable thing about entanglement of mixed GHZ states, saying:

We consider [a] $N$-qubit state

$$rho_{noisy}^{N} = frac{1-p}{2^N}I + p|psiranglelanglepsi|, 0le ple1 $$

where $|psirangle$ is a $N$-qubit … GHZ state.

**Does anyone know how these value would evolve when $N$ grows?**

Their entanglement/separability criterion is given as Theorem 1.

If a $N$-partite quantum state of dimension $d_1d_2…d_N$ with Bloch representation $(8)$ is fully separable, then

$$

||mathcal{T}^{(N)}||_{KF}le sqrt{frac1{2^N}prod_{k=1}^N d_k(d_k-1)}

$$

$mathcal{T}^{(N)}$ is given as last term in $(8)$:

$||mathcal T||_{KF} = max{||T_{(n)}^N||_{KF}}, n=1,…,N;$ is a Ky-Fan norm, which is the sum of the $k$ largest singular values of the matrix unfoldings of $mathcal T$.

I’m not interested in $|Wrangle$ states…

I don't know the details of this paper, although there are much better things that you can say about the GHZ case (in general, properties of GHZ states are much easier to analyse than W states). I'll summarise the key result in this context below, but further details are available in my paper, here.

There are some very simple entanglement criteria that one can apply. In particular, pick any bipartition of the system. If that bipartition contains entanglement (which might be found using the partial transpose criterion), then the state is certainly not fully separable, because it is not separable across that bipartition. For the GHZ state, this threshold occurs at $$ p=frac{1}{2^{N-1}+1}. $$ These values are lower than those stated in the table, and so give a stronger claim. It actually turns out that one can prove this is the threshold for the state becoming fully separable, so it's an exact result.

Correct answer by DaftWullie on August 20, 2021

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