Since I’ve started learning about Bell inequalities, it always bothered me that they are usually present as just some linear combinations of conditional probabilities, without any physical interpretation besides the fundamental fact that the maximal value this linear combination can attain when the probabilities come from local hidden variable models is smaller than when they come from quantum mechanics1.
For example, the most famous Bell inequality, the CHSH inequality, is usually written as
\[ \langle A_0 B_0 \rangle + \langle A_0 B_1 \rangle + \langle A_1 B_0 \rangle – \langle A_1 B_1 \rangle \le 2,\] where
\[\langle A_xB_y\rangle := p(a{=}b|xy)-p(a{\neq}b|xy).\] It looks like just an arbitrary sum of conditional probabilities; it is justified by noting that this is a facet of the local polytope2, but still this doesn’t give an interpretation of what this sum measures.
To make things worse, there’s a lot of arbitrariness in how we write a Bell inequality: we can simply divide both sides by a positive constant3, which clearly doesn’t make a difference in whether the inequality is violated or not, add a combination of probabilities that sum to 1 to the left hand side and 1 to the right hand side, which also clearly doesn’t make a difference, or add a no-signalling constraint, which I’ll talk about later.
As a consequence, the local bound of the CHSH inequality, 2, is just an arbitrary number, and likewise the Tsirelson bound $2\sqrt{2}$, as these transformations change them willy-nilly. Valerio Scarani famously has a football t-shirt with the player number $2\sqrt{2}$. But he might as well have put $\sqrt{2}$ on it, as people sometimes write the CHSH inequality divided by 2, or pretty much any real number – as long as somebody wrote a paper with the corresponding version of the inequality ;)
One can notice, though, that the CHSH expression is maximized when the correlations are such that $a=b$ when the inputs are $(0,0)$, $(0,1)$, or $(1,0)$, and $a\neq b$ when the inputs are $(1,1)$, and reformulate it as the probability that this happens. The result is the inequality
\[ \frac14 \left( p(a{=}b|00) + p(a{=}b|01) + p(a{=}b|10) + p(a{\neq}b|11) \right ) \le \frac34,\]which now has a nice interpretation as a nonlocal game: it measures the probability that Alice and Bob win the game of outputting $(a,b)$ for input $(x,y)$ such that $a\oplus b = xy$, when the inputs are given with uniform probability. The local bound $\frac34$ is no longer arbitrary, but it is the maximal probability of winning the game using local hidden variables, and likewise the Tsirelson bound $\frac{2+\sqrt{2}}{4}$ is the maximal probability of victory with quantum mechanics. As far as I know the first person that formulated the CHSH inequality like this was Boris Tsirelson, which makes it specially appropriate to call $\frac{2+\sqrt{2}}{4}$ a Tsirelson bound.
It turns out that every Bell inequality can be rewritten as a nonlocal game – provided we allow the winning condition of the nonlocal game be probabilistic – so all of them, together with their local and Tsirelson bounds, can be made meaningful in this way. End of story? No, because there is still arbitrariness in the nonlocal game formulation.
For example, one can add the equation
\[ \alpha\left(p(00|00) + p(01|00) + p(10|00) + p(11|00) \right) = \alpha\] to the nonlocal game form of the CHSH inequality (after multiplying it by 4, for convenience). We then divide it by $4+\alpha$, so that it is still a valid nonlocal game, albeit a weird one: when given input $(0,0)$, which happens with probability $\frac{1+\alpha}{4+\alpha}$, Alice and Bob win the game with certainty if their outputs $(a,b)$ are equal, and win the game with probability $\frac{\alpha}{1+\alpha}$ if they are different. That is, the referee will decided probabilistically whether to accept their answers. For the other inputs the winning condition doesn’t change, but they are now distributed with probability $\frac{1}{4+\alpha}$ each.
The local bound of the nonlocal game becomes $\frac{3+\alpha}{4+\alpha}$, and the Tsirelson bound $\frac{2+\sqrt{2}+\alpha}{4+\alpha}$. This might be surprising, since adding such a normalisation condition shouldn’t change anything! Well, it doesn’t change which correlations violate the inequality or not, but it certainly changes something: since both local hidden variables and quantum mechanics will by definition produce normalized probabilities, the normalisation condition cannot discriminate between them. Hence the larger the $\alpha$, the worse the nonlocal game will be at this task. Indeed both the local and Tsirelson bounds converge to 1 for large $\alpha$, as in this limit the game reduces to only asking the players to produce normalized probabilities, which they answer with “Yes, sure, no problem. But why?”.
Now, this is obviously silly, but we have to be able to deal with this silliness in a systematic fashion in order to remove this arbitrariness from the nonlocal game formulation. The way I propose to do it, in my new paper with Flavien Hirsch and Marco Túlio, is to optimise over the form of the nonlocal game to make the gap between the local and Tsirelson bounds as large as possible. We show that the expected $p$-value you get after playing $n$ rounds of a nonlocal game with gap $\chi$ is upperbounded by $(1-\chi^2)^n$, so there is a precise sense in which maximizing the gap between the local and Tsirelson bounds increases the power of the game for rejecting local hidden variables.
It turns out that optimizing over such normalisation conditions is rather easy: we just need to use them to make the smallest coefficient of each setting $(x,y)$ equal to zero, and rescale the game such that the largest coefficient of each setting is equal to one.
What is more difficult is dealing with the no-signalling constraints: we can also add to the CHSH game an equation like
\[ \gamma\left(p(00|00) + p(01|00) – p(00|01) – p(01|01) \right) = 0,\]which simply says that the marginal probability that Alice gets result 0 doesn’t depend on whether Bob’s input $y$ is 0 or 1. Now both local hidden variables and quantum mechanics will by definition respect this equation, so again it doesn’t change which correlations violate the inequality or not, but again it does change the power of the game. After a bit of work to remove the negative coefficients and renormalise the game, we see that the local bound becomes $\frac{3+\gamma}{4+2\gamma}$ and the Tsirelson bound $\frac{2+\sqrt2+\gamma}{4+2\gamma}$. For large $\gamma$ both bounds converge to $1/2$, which makes sense: in this limit the game reduces to asking Bob to signal to Alice, but since they cannot do it, irrespectively of using local hidden variables or quantum mechanics, they simply guess randomly, with probability of success $1/2$.
In this case it is trivial to do the optimisation, we just set $\gamma = 0$, and we have the maximal possible discriminating power, but in general this is quite hard: the best we could do is formulate the optimisation as a linear program, so at least it is easy to do it numerically. We also managed to find analytical solutions for two particular cases: when the nonlocal game can be turned into a unique game4 (as is the case for CHSH), doing that makes it optimal, and when the nonlocal game can be turned into a CGLMP game5, doing that makes it optimal.
In these two cases the solution is unique, so maximizing the gap $\chi$ gives you the optimal local bound, the optimal Tsirelson bound, and the optimal form of the Bell inequality6. But in general life is not so easy: fixing the gap $\chi$ to be the maximal one is not enough to determine the local and Tsirelson bounds. We can deal with that by maximising the Tsirelson bound as well, to make it easier to win the nonlocal game with quantum mechanics.
This is enough to get you optimal, unique, and physically meaningful local and Tsirelson bounds. Pure bliss, no? A bit awkward is the fact that this is still not enough to fix the form of the nonlocal game itself. Perhaps it doesn’t matter, and the differences between these form are as inconsequential as relabellings. Or perhaps they do matter, and somebody else will figure out how to pick up the best among them.
I’ll show the optimal forms of some Bell inequalities in a separate post, as this one is getting too long already. But there’s one more thing I’d like to say: Valerio, if you’re reading this, put $\frac{2+\sqrt{2}}{4}$ on your t-shirt.