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, 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.


1 Answer 1


A partial explanation is motivated by the proof in Theorem 1 (bottom of page 2).

Assuming two locally non-commuting stabilizing operators, using the Cauchy-Schwarz inequality, for pure product states it is shown that $$\langle S_{l}^{(GHZ_N)} + S_{m}^{(GHZ_N)} \rangle \leq 1.$$ But since we also assume these are stabilizer operators (GHZ eigenstate with eigenvalue of $1$), we know the GHZ state gives an entangled state example where $\langle S_{l}^{(GHZ_N)} + S_{m}^{(GHZ_N)} \rangle > 1$.

$\lnot$ locally commuting stabilizers $S_l^{(GHZ_N)}$ and $S_k^{(GHZ_N)}$ $\Leftrightarrow$ no pure common product eigenstates $\Rightarrow$ $\text{max}_{\rho_{separable}} \langle S_l^{(GHZ_N)} + S_k^{(GHZ_N)} \rangle < 1 \leq \text{max}_{\rho_{entangled}} \langle S_l^{(GHZ_N)} + S_k^{(GHZ_N)} \rangle \Rightarrow \text{valid entanglement witness}$.


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