# Quantum non-locality with commuting measurements?

We consider the Bell scenario, in which Alice and Bob share an entangled pure quantum state $\mid \Psi \rangle_{AB}$. Alice gets an input in the set $\{1,2\ldots X\}$ and Bob gets an input in the set $\{1,2\ldots Y\}$. Based on her input $x \in \{1,2\ldots X\}$, Alice performs a projective measurement $\{E_x^a\}_{a=1}^A$ with $A$ number of outcomes. Based on his input $y \in \{1,2\ldots Y\}$, Bob performs a projective measurement $\{F_y^b\}_{b=1}^B$ with $B$ number of outcomes. Probability of obtaining outcome $(a,b)$ given inputs $(x,y)$ is represented as $p(a,b|x,y) = \langle \Psi \mid E_x^a \otimes F^b_y \mid\Psi\rangle$.

In the famous case of CHSH-inequality (http://en.wikipedia.org/wiki/CHSH_inequality), when $\mid \Psi \rangle$ is EPR state and $A=B=X=Y=2$, Alice measures either in pauli-$X$ basis or pauli-$Z$ basis and Bob measures in bases $\frac{X+Z}{\sqrt{2}}$ or $\frac{X-Z}{\sqrt{2}}$ to obtain a probability distribution $p(a,b|x,y)$ which cannot be described by local hidden variable model. Such a probability distribution (which cannot be described by local hidden variable model) is called non-local.

I am interested in the case where the measurements of Alice or Bob mutually commute. In fact there are two cases:

1) For all $a,a',x,x'$ we have $[E^a_x,E^{a'}_{x'}]=0$. (But this is not required on Bob's side). Then does there exist an entangled state $\mid \Psi\rangle$ such that the probability distribution $p(a,b|x,y)$ is non-local?

2) For all $a,a',x,x'$ we have $[E^a_x,E^{a'}_{x'}]=0$. Furthermore, for all $b,b',y,y'$ we have $[F^b_y,F^{b'}_{y'}]=0$. Then does there exist an entangled state $\mid \Psi\rangle$ such that the probability distribution $p(a,b|x,y)$ is non-local?

Note that in CHSH case, none of the above cases are true.

I also have a similar question in mind when measurements are POVMs and shared quantum state can be mixed, so I ll also appreciate if something can be said about this case.

In all these discussions about entanglement, all the measured observables of Alice always commute with those of Bob. Their degrees of freedom describe two factors ${\mathcal H}_A$ and ${\mathcal H}_B$ of the overall Hilbert space of possibilities which is the tensor product $${\mathcal H} = {\mathcal H}_A\otimes {\mathcal H}_B$$ This factorization of the Hilbert space has to be the case by definition, otherwise the notion of the entanglement cannot be defined at all.
I think that you got confused by some forgotten indices. The observables may be the Pauli matrices but they still need a label that indicates whether it's Alice's or Bob's Pauli matrices. So we have three matrices $\sigma_i^A$ belonging to Alice, and three $\sigma_i^B$ to Bob. In the whole Hilbert space, these six operators may be written as the tensor products $$\sigma_i \otimes 1, \quad 1 \otimes \sigma_i$$ The first three operators obey the usual Pauli matrix algebra among themselves, and so do the last three. But each of the first three commutes with each of the last three.
Quantum field theory strictly bans any non-locality; non-relativistic quantum mechanical theories may allow them but these non-localities play absolutely no role in the explanation of the entanglement and the resulting correlations. Indeed, the locality of these theories is tightly linked to the fact that all Alice's observables commute with all Bob's observables. In quantum field theory, locality is mathematically expressed by the fact that the quantum fields exactly commute at spacelike separation: $$\forall x,y\in M^4,\,\,(x-y)^2\lt 0: \quad [\phi(x^\alpha), \phi (x^{\prime \beta})]=0$$ And the Wikipedia page you linked to doesn't ever use the adjective "non-local". It's something you added (unfortunately, popular books and pop-science articles do so all the time) and it is not correct.
Otherwise for a given choice of "type of the experiment" that you called $x$ or $y$, the projection operators always commute with each other. For two different values of $x$ but the same experimenter (e.g. both Alice), the projection operators generally don't commute, at least some of them don't. Otherwise they wouldn't be different. If all of the projection operators of Alice's for the values of $x_1$ and $x_2$ commuted with all others, one could unify these measurements into one and remove e.g. $x_2$ from the list of choices for $x$.