In the CHSH game where Alice and Bob share one EPR the optimal strategy is to measure the following observables $A_0=Z,A_1=X,B_0=1/\sqrt{2}(X+Z),B_1=1/\sqrt{2}(X-Z)$ depending on their question. We can see that the optimal strategy is projective measurement instead of general measurement. For other nonlocal games does it suffice to consider only projective measurement to get a optimal strategy?

For example in this paper equation (3) only the projective measurement of dimension 4 is considered for CHSH games sharing a general state.

In the QCQI section 2.2.8 it is proved that any general measurement can be changed into a projective measurement where an auxiliary bit is introduced, so the dimension of the operator is enlarged. To get an optimal strategy, do we need to search the projective measurement in the enlarged Hilbert space?


1 Answer 1


As you correctly state, POVMs can be dilated into a projective measurement on a larger Hilbert space. Because we do not restrict the possible dimension of the Hilbert spaces in the game, we can therefore always turn a non-projective measurement strategy into a projective measurement strategy without changing the statistics (and thus the game score). This implies that it is sufficient to only search for projective measurement strategies (although you must remember that a priori they can be in any dimension).

The case of the CHSH game is particularly nice wherein the optimal strategy is achieved by a projective measurement strategy in dimension $2 \times 2$. This is a feature of binary input/output games and follows from a result known in the community as Jordan's Lemma (see also here for a sketch derivation of the reduction for nonlocal games / Bell inequalities).

  • $\begingroup$ Thanks for your wonderful answer. I am reading the related reference arxiv.org/pdf/1303.3081v3 by V. Scarani. On page 33 note 29 he said that there are still some complexity left after using the Jordan's lemma, so is the reduction to single qubit case for general 2-input 2-output nonlocal game still a conjecture? $\endgroup$
    – qmww987
    Commented yesterday
  • $\begingroup$ No, it is not conjecture. I linked a derivation in the answer that works for all binary games. $\endgroup$
    – Rammus
    Commented 19 hours ago

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