# What is the condition for local operations on bipartite entangled state?

I have an entangled state between Alice and Bob $|\psi\rangle_{AB}$ ( both Alice and Bob have states in Hiblert space of dimension $n$ ). Alice and Bob can only perform local meaurements. I assumed that POVM for measurements on the combined state will be of the form $E_A \otimes E_B$ where $E_A$ and $E_B$ ( both are operators on $n$ dimensional Hilbert space ) are local POVM's for Alice and Bob respectively . But the paper I am reading currently says the condition for measurement operators to be local is $[E_A,E_B]=0$ and here $E_A$ and $E_B$ are operators on $n^2$ dimensional Hilbert space. I can see my case is a specific case of the latter one but how does one explain this commuting condition for measurement operators to be local? Also can same analogy be extended to local unitary operations ?

• The condition that $E_A$ and $E_B$ commute is a natural generalization of locality, but in general not the same. However, one can indeed do resource theory based on such a commutation relation. What paper are you reading? – Norbert Schuch Jun 23 '15 at 11:41
• @NorbertSchuch arxiv.org/pdf/1303.3081v4.pdf here in the section "Self test" they have used this relation. – sashas Jun 23 '15 at 12:34
• Where exactly? You are not being very specific. – Norbert Schuch Jun 23 '15 at 14:03
• @NorbertSchuch sorry . Under the heading "4.2.3 Self-test of the singlet using the Mayers-Yao statist ics" ( on page 29 ) third line, it says "The locality of the measurement $[M_A,N_B]=0$ is assumed." – sashas Jun 23 '15 at 15:28
• Locality of $M_A$ and $N_B$ certainly implies that $[M_A,N_B]=0$, so the latter is a valid property which can be used to derive properties of local measurements. Does this answer your question? – Norbert Schuch Jun 23 '15 at 15:32

If we consider local operators $M_A$ and $N_B$ acting on Alice's and Bob's part, respectively, then it holds that $[M_A,N_B]=0$, i.e., we can use this property in proofs involving local operations. Note that conversely, however, commutativity need not imply locality (to start with, there need not even be a tensor product structure).
• one last doubt both $M_A$ and $N_B$ act on the hilbert space of the composite system as opposed to the tensor structure ( $m_a \otimes m_B$ ) where $m_a$ and $m_B$ both act on local hilbert spaces of alice and bob respectively. Am I right ? Is it that just $M_A=ma_a \otimes I$ and $M_B=I \otimes ma_b$ ( composite system is AB ) – sashas Jun 23 '15 at 17:19
• @sasha Exactly, $M_A=m_A\otimes I$ and $M_B=I\otimes m_B$. It is however customary (though sloppy) to use $M_A$ both for the operator on $\mathcal H_A$ and $\mathcal H_A\otimes \mathcal H_B$. – Norbert Schuch Jun 23 '15 at 17:37