4
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\langle S_{l}^{(GHZ_N)} + S_{m}^{(GHZ_N)} \rangle > 1$.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.