# What exactly does $S$ represent in the CHSH inequality $-2\leq S\leq 2$?

What exactly does $S$ represent in the CHSH inequality

$$-2~~\leq ~S~\leq ~2?$$

Sorry I've been reading for a couple days and I can't figure out what exactly $S$ is and the math is a bit over my head.

Any help is much appreciated.

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The $S$ in this inequality is defined as $$S = E(a b) − E(a b') + E(a' b) + E(a' b')$$ where $E(M)$ is the "expectation value of $M$" which means the average value calculated from many repetitions of the same experiment (empirically) or from the probability distributions (theoretically).

All the expectation values are taken from the products of two quantities in the list $\{a,a',b,b'\}$. Each of these four quantities – which represent the spin measurement of the first particle with respect to two axes ($a,a'$) or the second particle with respect to two axes ($b,b'$) – is equal to either $+1$ or $-1$ so all the four products above are either $+1$ or $-1$, too. The positive value means that the two factors in the product yield the same sign; the negative value means that they have the opposite sign.

For example, $E(ab)$ may be interpreted as $1-2P_{a\neq b}$ where $P_{a \neq b}$ is the probability that $a,b$ will be measured with the opposite sign if you decide to measure the unprimed spin for the $a$ particle and unprimed spin for the $b$ particle, too. The other two $E(\dots)$ terms are similarly linked to the analogous probabilities $P_{i,j}$. So $$S = 2 - P_{a \neq b} + P_{a\neq b'} - P_{a'\neq b} - P_{a'\neq b'}$$ You saw that $S$ is composed of four terms and each of them is in between $-1$ and $+1$. So a priori, $S$ could be anything between $-4$ and $+4$. However, the assumptions of realism and locality are enough to prove that in local realist theories, $S$ is always between $-2$ and $+2$; quantum mechanics allows $S$ to be as high as $2\sqrt{2}\sim 2.8$ which exceeds the interval allowed by local realist theories and experiments confirm that this $2.8$ is realized in the appropriate spin experiments, thus falsifying local realist theories.

An intuitive sketch why $S$ can't be greater than two in local realist theories is this one: to maximize $S$, you want the terms with the plus signs to be close to $1$ and the second term to be close to $-1$. However, it's not possible. If the first, third, and fourth terms are very close to one, it means that $a$ is highly correlated with $b$, $a'$ is also correlated with $b$, and $a'$ is correlated with $b'$. Transitive laws are enough to see that the first two conditions imply that $a$ is highly correlated with $a'$ and in combination with the last correlation, $a'$ with $b'$, it means that $a$ is correlated with $b'$, too. So the second term $-E(ab')$ is large negative, i.e. close to $-1$, and it gets mostly subtracted. Therefore, we get a number close to $S=2$ and not $S=4$.

Maths is needed to show that $2$ is the upper bound in local realist theories and $2.8$ is the prediction in quantum mechanics (confirmed experimentally). In quantum mechanics, it is impossible to deduce that $a$ is highly correlated with $a'$ out of the first two conditions. For a perpendicular pair of axes, they may be totally uncorrelated even though the previous two correlations may be very high.

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Does "upper bound" means that if i will calculate S, based on for example 4 (or more) experiments, it can be less than 2 in quantum mechanics, but some next calculated S will be more than 2? In other words, does violate CHSH equation means measure |S| > 2 at least once or always? – dk14 Oct 1 '14 at 0:47
Dear dk, local realist theories imply that $|S|\leq 2$ is true, i.e. it always hold. The negation of this statement is that $|S|\gt 2$ holds at least sometimes. So it's enough to find one example where $|S|\gt 2$ and all local realist theories are ruled out. – Luboš Motl Oct 1 '14 at 7:20
Thanks. I realized that it's just a result of Boole–Fréchet inequality so it's like i will receive head 1000000 times when fliping a quantum coin. physics.stackexchange.com/questions/138080/… – dk14 Oct 1 '14 at 10:00
Just one more question. What if i have an hypothetical experiment where i measured states of particles 1000 times and calculated S = 1.8, then (maybe in same experiment) i measured states of particles 1000 times and calculated S = 2.21, does it means i violated CHSH (with such dynamic mathematical expectation)? – dk14 Oct 2 '14 at 0:19
dear @dk14, if you want to violate (rule out) an inequality by an experiment, you have to keep track of the inevitable error margin of the experiment. With the distribution of the "true" value of $S$ indicated by your measured mean value and error margin, you must calculate the probability that you got such a high value just by chance. And only if this probability is tiny, like smaller than $10^{-6}$ (the usual 5-sigma criterion), you may say that the experiment has established something. 1,000 measurements yielding $S=2.21\pm \epsilon$ is probably more than enough to falsify the $S\lt 2$ law. – Luboš Motl Oct 2 '14 at 6:37