How is Bell’s Inequality converted to the CHSH inequality?

Question

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.

Answer 1

... and the question changes again. Re Update 4:

I'm not sure what you intended the variable $a$ to be in the question (I think part of the problem is that you've got muddled between the chosen questions and the given answers), but I have used it very carefully, to always be the $\pm1$ answer given. In a quantum setting, $a$ is the answer, a value $\pm 1$ when Alice's qubit is measured in a particular basis (this basis is the 22.5 degrees you're talking about). A projective measurement on a qubit always gives you one of two answers which you can choose to label as $\pm 1$ no matter what the basis of that measurement is.

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The question has entirely changed from previous versions (i.e. should really be a new question). Please find old responses below.

Bell and CHSH inequalities are nothing to do with detectors, measurement angles etc, because those are all things associated with a quantum experiment. These tests are nothing directly to do with a quantum experiment, requiring knowledge of experimental details. Instead, they are more generic than that and, in particular, tell us what properties the answers have to have if the world is classical. This is about proving that the world is not classical as opposed to proving that it is quantum.

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 $\pm 1$. 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 $d$ to record the answers from his box corresponding to his choice of setting. 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 $$ S=E(a,b)−E(a,d)+E(c,b)+E(c,d). $$

If we assume that for each run of the experiment, all 4 of $a$, $b$, $c$ and $d$ have a value (even though we don't see all of them on a single run), then on each run $a(b-d)+c(b+d)$ is a value $\pm 2$, and hence $\langle S\rangle\leq 2$.

Then, we can do a quantum experiment. You don't actually need to know the details of the experiment, they can entirely be hidden away inside these boxes which act based on the questions asked, and give $\pm 1$ outputs. You do this, and find $\langle S\rangle=2\sqrt{2}$, and that proves that the assumption must be invalid. The 4 values are not predetermined, they must be chosen at random at the moment the question is asked.

With that philosophy out of the way, then you can get stuck into doing the actual quantum calculation, choosing an entangled state and measurement bases of particular angles (these do not need to be variables) and verifying that the $2\sqrt{2}$ answer should indeed come out.

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Previous responses:

You've sort of answered your own question. You've stated that $$ S=E_1-E_2+E_3+E_4 $$ and then you've asked whether $$ S=E_1E_2-E_1E_4+E_3E_2+E_3E_4. $$ No, it's what you stated it was.

You then take the values of $E_i$ to be $\pm1/\sqrt{2}$. The only problem you have is that you've put the minus sign on the wrong one: $$ E_1=E_3=E_4=\frac1{\sqrt{2}}\qquad E_2=-\frac1{\sqrt{2}} $$ Subbing these in to your original statement about $S$, you simply get $$ S=\frac4{\sqrt{2}}=2\sqrt{2}, $$ as you'd expect for the quantum calculation.

For concreteness, let's be clear about the conversion: $P(a,b)$ is the probability of getting answers $a$ and $b$ the same, while $E_1$ is the correlation coefficient between answers $a$ and $b$ $$ E_1=P(a,b)-(1-P(a,b))=2(P(a,b)-1). $$ Hence, if we start from $$ P(a,b)-P(a,d)+P(c,b)+P(c,d)\leq2, $$ then we have $$ (E_1/2+1/2)-(E_2/2+1/2)+(E_3/2+1/2)+(E_4/2+1/2)\leq 2, $$ which is the same as $$ \frac12(E_1-E_2+E_3+E_4)\leq1, $$ or $$ E_1-E_2+E_3+E_4\leq2. $$

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As a final note, please observe that you cannot call both $$ S_1=P(a,b)-P(a,d)+P(c,b)+P(c,d) $$ and $$ S_2=E_1-E_2+E_3+E_4 $$ $S$ because they are actually different quantities ($2S_1-2=S_2$). It just happens that both definitions coincide when they have value 2. I think that conventionally you'd call $S_2$ $S$, and $S_1$ is often called $I$.

Answer 2

The original answer did not address the observation that CHSH calculates $a$ from 2 different scenarios, then uses them interchangeably. A technique that allows the generation of impossible results.

Bell’s Inequality is valid for Correlation Coefficients and the CHSH Inequality is computed the same way as Bell’s, that is $S = ab – ad + cb + cd$ where a, b, c and d are the results of 4 runs of the experiment. $E(a,b)$ is the correlation coefficient of 1 run of the experiment. The problem is that $E(a,b)$ should become “$a$” in the S calculation, not $ab$