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First, the context for this question can be found in Wilde's Quantum Shannon Theory text on the arXiv, specifically the section starting at thebottom of page 98, entitled "". My particular question relates toEntanglement in the CHSH GameExercise 3.5.11 on page 100of the same document... I am working through this book independently and have moved well beyond this point, but frustratingly still can't solve the seemingly simple exercise 3.5.11, and I think it's because I'm missing something basic about how to calculate probabilities of the different outcomes!

My understanding:Alice and Bob share a maximally entangled Bell state, |φ^{+}> = 1/sqrt(2) [|00> + |11>]. We are given Alice and Bob's measurement strategies for each possible scenario and asked to show that their winning probability is cos^{2}(pi/8), the well-known result of the chsh experiment... If Alice gets x=0 from the ref, she will measure the Z operator (phase flip in the computational basis) and return the result as "a", if she gets x=1 she will measure the X operator (a NOT operation in the computational basis) and return the result as "a". For Bob, upon receiving y=0 he will measure 1/sqrt(2) [ X+Z ] and for y=1 he will measure 1/sqrt(2) [ Z - X ], returning the result as "b" in each case.

Alice and Bob "win" if aXORb = xANDy, where x and y are 0 or 1, uniformly chosen by the ref. It's straightforward to see that this only holds when Alice and Bob each return a 0 or each return a 1 for three of the four cases. For the final case, corresponding to 1AND1 we need Alice and Bob to return opposite pairs, (0,1) or (1,0).

My approach (definitely wrong) has been to take each of the four cases, form the tensor product in terms of computational basis operators (not matrices), and separate the resulting operator into measurement operators that result in the different possible outcomes.For example, if Alice and Bob both receive x=y=0, the global measurement operator is

Z⊗(1/√2)(X+Z) = (1/√2)(|0><0| - |1><1|)⊗(|1><0| + |0><1| + |0><0| - |1><1|)

= (1/√2) [ (|00><00| + |00><01|) + (|11><11| - |11><10|) + (|01><00| - |01><01|) - (|10><10| + |10><11|)]

= (1/√2) [ M_{1}+ M_{2}+ M_{3}- M_{4}]

It's easy to check that these measurement operators, while not projectors, "resolve" the Bell basis:

ΣM^{†}_{i}M_{i}= I_{Bell}

So I thought that, to find the probability that Alice and Bob collectively return 00 or 11 would just be the sum of the probabilities for M1 and M2 above, namely <φ^{+}|M_{1}|φ^{+}>, and for M2, but this gives the mixed result of 1/2...

Should I instead be performing local measurements only, as in en. 4.123 of pg 127 of the document? This approach seems to get me closer, but still gives trouble.

Thanks for taking the time!

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# Calculating measurement probabilities for the CHSH game

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