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Bell’s inequality is $$S = P(a,b)-P(a,d)+P(c,b)+P(c,d) \leq 2,$$ which is calculated as $$S = ab – ad + cb + cd \leq 2.$$

The CHSH version is: $$E = \frac{N_{11} + N_{00} - N_{10} -N_{01}} {N_{11} + N_{00} + N_{10} + N_{01}}$$ $$S = {E_1 - E_2 + E_3 + E_4}$$ $N_{11}$ is the number of correlations, etc. $E$ is the correlation coefficient.

There appears to be a significant difference between Bell’s and CHSH’s inequalities. For comparison, converting the correlation coefficient to Probability makes it possible to enter the values directly into Bell’s Inequality. The conversion to probability is the correlation coefficient divided by $2$ then adding $.5$. Using the much mentioned values for $E$ of $-.707, +.707, +.707$ and $+.707$ as $a, b, c$ and $d$ respectively. Bell’s Inequality is calculated as $$S = (-.8535)(.8535) – (-.8535)(.8535) + (.8535)(.8535) + (.8535)(.8535)$$ $$= -.728 + .728 + .728 + .728 = 1.456.$$

So using the real Bell’s Inequality, there is no violation. Shouldn't the CHSH Inequality be calculated as $$S=E_1E_2-E_1E_4+E_3E_2+E_3E_4?$$

**Update 1:**

The essence of the problem with the CHSH experiment is that $E$ is calculated as a correlation coefficient, similar to probability, then inserted into the CHSH inequality as $E(a,b)$, which is a product of 2 independent values. That is a huge mistake. Either this is wrong or the fundamental “proof” of quantum theory is seriously flawed.

**Update 2:**

The $a$ and $b$ in the calculation of $S$ are the results of 2 runs of the experiment from run $a$ and run $b$. The $a$ and $b$ in the calculation of $E$ are the results of 1 run of the experiment from detector a and detector b The two $a$’s are completely different.

**Update 3:**

Please excuse my post that was too nebulous to point out an error. This is a solid description of my problem with the CHSH experiments.

The CHSH inequality is

$$S = E(a,b) - E(a,d) + E(c,b) + E(c,d) \leq 2.$$

It, like Bell’s inequality, is valid for all values from $–1$ to $+1$.

To avoid having 2 $a$’s, separate letters are assigned to each item in the experiment, lower case for values and upper case for the rest.

The letters $a$, $b$, $c$ and $d$ are reserved for the CHSH Inequality.

X and Y are the 2 detectors.

$g$, $h$, $i$ and $j$ are the 4 angles of the detectors.

K, L, M, and N are the 4 runs.

$p$, $q$, $r$ and $t$ are results of K, L, M and N for detector X.

$u$, $v$, $w$ and $z$ are results of K, L, M and N for detector Y.

$E(p,u)$ is inserted into $S$ as $E(a,b)$ by CHSH. $E(q,v)$ is inserted into $S$ as $E(a,d)$. $E(r,w)$ is inserted into $S$ as $E(c,b)$. $E(t,z)$ is inserted into $S$ as $E(c,d)$.

That makes the calculation of $S$,

$$S = E(p,u) - E(q,v) + E(r,w) + E(t,z).$$

So CHSH puts 8 independent values into the inequality. Both Bell’s and CHSH’s inequalities require exactly 4 values. This makes the inequality invalid. What is the error in this description of the CHSH experiment?

**Update 4:**

The last comment said:

“We can also calculate the correlations so, if both choose to measure setting 1, we'd have the value $ab$ being the product of the two answers. If we run this many times, we build up expectation values for each of these entities, eventually giving us things like $E(a,b)$. Then we can evaluate $S=E(a,b)−E(a,d)+E(c,b)+E(c,d)$”.

That has isolated the problem.

The $E(a,b)$ from the runs is not the same as the $E(a,b)$ in the CHSH Inequality. In the runs $a$ is the setting of the "a" polarizer in degrees, i.e. 22.5°. Both Bell’s and CHSH’s inequalities are valid for all numbers from –1 to +1. They are not valid for 22.5. If you disagree, what is the value of $a$ in the $E(a,b)$ of the runs?

**Update 5:**

The reason the CHSH experiment has been able to get away with a fatal error for 50 years is because of clever but erroneous replies to objections like this typical one.

“In a single run of an experiment, there are two boxes. On each box, you can choose one of two settings, and you get an answer $±1$ If Alice (first box) chose setting 1, she would record her answer in the random variable $a$. If she chose setting 2, she records her answer in random variable $c$. Similarly, Bob uses random variables $b$ and $d$. We can also calculate the correlations so, if both choose to measure setting 1, we'd have the value $ab$”

Look what that says, set $a$ and $b$ to the same value and correlate them. The result of the correlation of $a$ and $b$ has to be 1. That is definitely not a justification for the CHSH calculations.

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Hi, Bob. Welcome to Quantum Computing! Please use MathJax for properly typesetting mathematical expressions and use the appropriate tags. Review How to write a good question?. I've [edit]ed it on your behalf this time.

– Sanchayan Dutta – 2019-02-18T16:50:08.9101

what are the $E_i$ in your second definition of $S$? And where does your definition of $E$ come from? To be clear, the typical definition of the CHSH $S$ operator is similar to your first two formulas, see e.g. eq. (4) of Brunner et al.

– glS – 2019-02-21T15:19:48.683E1, E2, E3 and E4 are the Corelation Coeficients as the results of 4 runs of the experiment at different angles. – Bob – 2019-02-21T16:33:16.240

See Wikipedia, CHSH Inequality, Stack Exchange, Detection Angles in Bell-type Experiments or look at some of the descriptions on the internet. – Bob – 2019-02-21T16:36:18.177

The use of (a,b) in the calculation of E does not justify skipping the the need for it in the calculation of S. – Bob – 2019-02-21T16:40:37.873

1if you don't mention me in a comment (with @username) I'll probably not see it anytime soon. Anyway, that was not my point. I'm saying that what you call "Bell's inequality" in the first equation, is actually the typical form of a CHSH inequality. I don't see wikipedia using that formula for Bell's inequalities, so where did you get it? The wikipedia article (as of now) mentions the original Bell's inequality in the form $C_h(a,c)-C_h(b,a)-C_h(b,c)\le1$. More generally, one should keep in mind that Bell's and CHSH inequalities refer to different physical scenarios – glS – 2019-02-25T10:31:50.733

@glS. The version of Bell’s Inequality, S=ab–ad+cb+cd≤2, is valid for all probability values from 0 to 1. You can verify this for yourself by having your computer calculate the result from a range of values from 0 to 1, as I did.

The following is from Bell’s Test Experiment, Wikipedia: “(1) E = (N++ + N−− − N+− − N−+)/(N++ + N−− + N+− + N−+). Once all four E’s have been estimated, an experimental estimate of the test statistic (2) S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) “ E(a,b) is a Correlation Coefficient, which is convertible to probability, not a times b. – Bob – 2019-02-26T23:43:15.960

Where is there any connection between E and E(a,b)? Nowhere are any a and b multiplied. – Bob – 2019-02-26T23:45:23.317

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this is still unclear for me. I don't know what you mean when you say that "

– glS – 2019-02-28T17:21:49.940with the CHSH experiment $E$ is calculated as a correlation coefficient" as opposed to using probabilities. See e.g. this answer for a derivation that only uses probabilities. In the CHSH, $E(a,b)$ is not the product of two independent values, but rather obtained by averaging (over the LHV) a series of products of two independent variables. I also don't get your form of Bell's ineq. Can you add the derivation or cite a source? I've only ever seen that $S$ for CHSH