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Another Bell's Theorem Question

I am trying to follow the simple model of Bell's Theorem outlined in this paper: https://people.eecs.berkeley.edu/~vazirani/s07quantum/notes/lecture1.pdf. Please read section 5 for details, but basically he outlines a communication protocol where two people $A$ and $B$ receive a bit each ($X_a$ and $X_b$). They than each have to independently produce new bits ($a$ and $b$). $A$ and $B$ are trying to cooperatively maximize the probability that $$a\ XOR\ b = X_a\ AND\ X_b$$

There is a trivial strategy that wins 75 percent of the time. Always produce $a$ and $b$ of 0. The last section describes the strategy that wins more than 75% of the time in quantum mechanics and disproves 'local hidden variables'. From my understanding a local hidden variable theory would be one where the two particles before they were separated planned out every possible result of any possible experiment that could be performed once they are separated. This way there was no FTL communication.

My questions are as follows:

If in the quantum version of the game they always have the EPR pair, than why can't they win 100 percent of the time? If A measures 1 he produces 1 and B always produces 0. I don't get what this has to do with quantum mechanics since this strategy could be devised even if hidden variable theory was 'true' and the two particles 'agreed' on there configurations before they were separated in space.

In the protocol at the end he says if $X_a$ = 0 than do a certain measurement. What is the output produced by the strategy? It is very unclear. Is the output the result of the measurement?

Also don't you have to perform a measurement to know that $X_a$ = 0? Wouldn't this already affect the state EPR pair before you made the second measurement?

Am I fundamentally missing something?

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  • $\begingroup$ You link to lecture notes of Vazirani and say it’s outlined by Aaronson... By the way, I think this version of the CHSH inequality as a XOR game is neither due to Aaronson nor Vazirani. I think changing your title to something like “Bell inequality as a game: Why is it impossible to always win?” would be clearer $\endgroup$ – Frédéric Grosshans Nov 19 '18 at 13:50
  • $\begingroup$ @FrédéricGrosshans Agreed and updated. $\endgroup$ – Jacob Schneider Nov 20 '18 at 16:21
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According to the source: Alice and Bob are each handed a single random classical bit $X_a$ and $X_b$ and they also share a pair or maximally entangled qubits in the state $|\psi\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$.

Now based on this $X_a ∧ X_b = 0$ unless both $X_a$ and $X_b=1$ (definition of the logical "and" operator, $∧$). Because $X_a$ and $X_b$ are random and independent, this means that $X_a ∧ X_b$ is random and $=0$ 75% of the time and =1 only 25% of the time.

Now you are correct that Alice and Bob can always guarentee that the outcome of their measurements is the same if they both measure in the same basis or always the opposite (e.g. one measure in the $|0\rangle$ or $|1\rangle$ basis and the other measures in the swapped, $|1\rangle$ or $|0\rangle$ basis). However, taking the xor operation on the output means that Alice and Bob can choose to always have $a⊕b=0$ (by measuring in the same bases) or $a⊕b=1$ (by measuring the swapped bases). Therefore, always choosing a deterministicly correlated/anti-correlated measurement scheme means that you only win either 25% or 75% of the time (because remember $X_a ∧ X_b =0$ 75% of the time and 1 25%).

Because $X_a$ and $X_b$ are random and independent, Alice and Bob gain no information from the actual value of their random bit and so can't use this information alone to do better than 75%. However, since the $X_a ∧ X_b$ outcome is random but biased, they can use the rules of quantum mechanics to create random bits $a$ and $b$ that are biased in such a way as to get higher than 75% by measuring in different bases as given by the example in the reference. However, maximizing over possible measurement schemes never gives you 100% (and the answer in the reference is the maximum probability you can win).

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  • $\begingroup$ Wow, I didn't realize the EPR pair was separate from the two random bits. This makes sense, but I will have to think about it a bit further to understand why a local hidden variable theory could not explain the EPR pairs behavior. $\endgroup$ – Jacob Schneider Nov 15 '18 at 23:33

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