Entanglement Witnesses close to GHZ states

[Danny Boy]

TL;DR Summary

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.

 

[azntoon]

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.