Could you give me an example of a measurement which is LOCC (Local Operations Classical Communication) but not separable? Or better, one which is separable but not LOCC?

Given an ensable of states $\rho^{N}$, a separable measurement on it is a POVM $\lbrace N_i \rbrace$ where the effects $N_i$ are all of the form $N_i = A_i^{1} \otimes A_i^{2} \otimes \dots \otimes A_i^{N}$. So they are a separable product of effects acting on each state $\rho$ in $\rho^{N}$.

Is every separable measurement LOCC?

  • $\begingroup$ Cross-posted on QCSE. $\endgroup$ – S.D. Apr 2 '19 at 20:53
  • $\begingroup$ What do you mean by "separable"? $\endgroup$ – Norbert Schuch Apr 2 '19 at 23:37
  • $\begingroup$ Aren't separable POVMs more powerful than LOCC ones? $\endgroup$ – Norbert Schuch Apr 3 '19 at 10:03
  • $\begingroup$ Yes indeed, I made a bad mistake in formulating the question. I want a separable operation that is non LOCC. I corrected the mistake. I just have never seen a counterexample to all separables are LOCC. $\endgroup$ – MrRobot Apr 3 '19 at 10:14
  • 1
    $\begingroup$ MrRobot, edits which change the meaning of a post --- especially edits which invalidate existing answers --- are discouraged. I've tried to preserve both your original question and your intended meaning. $\endgroup$ – rob Apr 3 '19 at 20:13

There isn't: Any LOCC measurement is also a separable measurement. This is easy to see: Alice's first measurement has POVM elements $A_{i_1}\otimes I$. Alice then communicates her outcome $i_1$ to Bob. Bob's subsequent measurement has elements $I\otimes A^{i_1}_{i_2}$, where $i_2$ enumerates Bob's outcomes, and $A^{i_1}$ indicates that Bob's POVM can depend on Alice's outcome. The total POVM of both has then elements $$ N_{i_1,i_2}=A_{i_1}\otimes B^{i_1}_{i_2}\ , $$ which is a separable POVM with double index ${i_1,i_2}$. Clearly, this can be iterated to an arbitrary number of rounds, and generalized to an arbitrary number of parties, and will always have POVM elements of the form $N_i=A_i\otimes B_i\otimes \cdots$.

Conversely, not every separable POVM can be written as a LOCC POVM. A counterexample is given in Bennett et al., Quantum Nonlocality without Entanglement, Phys. Rev. A. 59, 1070 (1999).

| cite | improve this answer | |

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.