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I am reading Thomas Vidick, Quantum multiplayer games, testing and rigidity. On top of p.4,

$$\text{E}[a\cdot b] = \sum_{i,j\in \{0,1\}}(-1)^{i+j}\text{Pr}\big((a,b)=(i,j)\big)$$

I do not understand what the notation $a\cdot b$ means and where does the right hand side expression comes from, particularly the reason for the sign $(-1)^{i+j}$.

Could someone please shed light on this question?

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In other words, this says that Alice and Bob each perform a measurement of their own observable ($X$ for Alice and $Y$ for Bob) which has $0$ and $1$ as outcomes. If they get a $0$ they select $-1$ and if they get a $1$ they select $+1$. So the product of their selections $a$ and $b$ will be equal to $1$ if either they both got $0$ or they both got $1$ in their measurements while it will be equal to $-1$ if they got different outcomes. So by the definition of the expected value, you get $$\mathbb{E}[a\cdot b] = (+1) \cdot Pr(\text{Alice's outcome}=\text{Bob's outcome}) + (-1)\cdot Pr(\text{Alice's outcome}\neq\text{Bob's outcome})$$ and if you expand it a little more you'll arrive at what the author states. What's confusing here is that the author chooses to denote the selection and the measurement outcomes with the same variable names (in the LHS $a$ and $b$ are $\pm 1$ while in the RHS they are $0$ or $1$).

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  • $\begingroup$ Ah, so it is simply the expectation of $ab$ for $a,b\in\{-1,1\}$ instead of $a,b\in\{0,1\}$. Had he used that, he could have just written $ab$ instead of $a\cdot b$. Why would you want to sow that confusion, when all the eigenvalues of concern are in $\{-1,1\}$? $\endgroup$
    – Hans
    Commented Sep 25, 2022 at 5:59
  • $\begingroup$ Vidick does write that this is the expectation value of $X \otimes Y$, with $X$ and $Y$ taking values in $\{-1,1\}$. But the context makes it more clear what is going on: he is using this to introduce the concept of observables and eigenvalues to and audience that is used to only probabilities, the computer scientists. $\endgroup$ Commented Sep 25, 2022 at 8:28
  • $\begingroup$ @MateusAraújo: Regarding the eigenvalues, that is precisely my question since all the eigenvalues are in $\{-1,1\}$. On top of that, the $a$'s and $b$'s are taking values from $\{-1,1\}$ which is consistent with the eigenvalue domain in the whole document except in this line. It must be a typo. By the way, I see you changed my original $(-1)^{j-i}$ to $(-1)^{i+j}$. They are equal, right? Are you changing for some aesthetic reason? $\endgroup$
    – Hans
    Commented Sep 25, 2022 at 21:07
  • $\begingroup$ Yes, sorry, they are equal, I changed because everywhere it's written as $(-1)^{i+j}$, when I saw the different formula I had to stop and think about whether it meant something. Since I was fixing another mistake anyway I thought it would be worth it changing this as well $\endgroup$ Commented Sep 26, 2022 at 16:17
  • $\begingroup$ But no, it's not a typo, $a,b$ take values in $\{0,1\}$, not in $\{-1,1\}$. From them you can build the observables $X,Y$ via this formula, but the eigenvalues do not correspond to the values you measure. The underlying reasoning is that the measurement values are just irrelevant labels and should play no role in the Bell inequality. Ditto for the observables, they are just defined so that we have a convenient formula for the bias of the nonlocal game, but they don't correspond to anything physical, and the eigenvalues are similarly irrelevant. $\endgroup$ Commented Sep 26, 2022 at 16:22

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