Entanglement Witnesses close to GHZ states

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SUMMARY

The discussion focuses on the construction of entanglement witnesses for detecting entanglement close to Greenberger-Horne-Zeilinger (GHZ) states, as outlined in Toth's 2005 paper "Entanglement detection in the stabilizer formalism." The entanglement witness is defined as $$\mathcal{W} := c_0 I - \tilde{S}_{k}^{(GHZ_N)} - \tilde{S}_{l}^{(GHZ_N)}$$ where $$c_0$$ is derived from the maximum expectation value of the correlation operators for product states. The key conclusion is that if the correlation operators $$\tilde{S}_{k}^{(GHZ_N)}$$ and $$\tilde{S}_{l}^{(GHZ_N)}$$ commute locally, the maximum expectation values for separable and entangled states coincide.

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Danny Boy
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TL;DR
Theoretical query regarding a statement made in a paper on entanglement detection using entanglement witnesses.
Consider page 2 of Toth's paper Entanglement detection in the stabilizer formalism (2005) . To detect entanglement close to GHZ states, they construct entanglement witnesses of the form $$\mathcal{W} := c_0 I - \tilde{S}_{k}^{(GHZ_N)} - \tilde{S}_{l}^{(GHZ_N)},$$
where ##\tilde{S}_{k/l}^{(GHZ_N)}## are elements of the stabilizer group and $$c_0 := \text{max}_{\rho \in \mathcal{P}}\big( \big\langle \tilde{S}_{k}^{(GHZ_N)} + \tilde{S}_{l}^{(GHZ_N)} \big\rangle_{\rho} \big),$$ where ##\mathcal{P}## denotes the set of product states.

Definition: Two correlation operators of the form ##K = K^{(1)} \otimes K^{(2)} \otimes \cdot \cdot \cdot \otimes K^{(N)}~\text{and}~L = L^{(1)} \otimes L^{(2)} \otimes \cdot \cdot \cdot \otimes L^{(N)}## commute locally if for every ##n \in \{1,...,N\}## it follows ##K^{(n)}L^{(n)} = L^{(n)}K^{(n)}##.

Question: In the paper (page 2), an observation which follows states:
Hence it follows that if ##\tilde{S}_{k}^{(GHZ_N)}## and ##\tilde{S}_{l}^{(GHZ_N)}## commute locally then the maximum of ##\big\langle \tilde{S}_{k}^{(GHZ_N)} + \tilde{S}_{l}^{(GHZ_N)} \big\rangle## for separable and entangled states coincide.

Is it clear why this statement holds true? Thanks for any assistance.

 
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Yes, it is clear why this statement holds true. By definition, if ##\tilde{S}_{k}^{(GHZ_N)}## and ##\tilde{S}_{l}^{(GHZ_N)}## commute locally, then for any state, it can be written as a product of its local states, i.e., $$\rho = \bigotimes_{n=1}^N \rho_n.$$ Since the correlation operators commute, it follows that $$\big\langle \tilde{S}_{k}^{(GHZ_N)} + \tilde{S}_{l}^{(GHZ_N)} \big\rangle_\rho = \prod_{n=1}^N \big\langle \tilde{S}_{k}^{(GHZ_N)} + \tilde{S}_{l}^{(GHZ_N)} \big\rangle_{\rho_n}.$$ Therefore, since the local states are always separable, the maximum of ##\big\langle \tilde{S}_{k}^{(GHZ_N)} + \tilde{S}_{l}^{(GHZ_N)} \big\rangle## for separable and entangled states coincide.
 

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