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I'm having trouble following Bell's derivation of equation 22 in his original paper. Particularly, how to go from

$$ | \overline{P} (\vec{a}, \vec{b}) - \overline{P}(\vec{a}, \vec{c}) | \leq 1 + \overline{P}(\vec{b}, \vec{c}) + \varepsilon + \delta $$

to

$$ | \vec{a} \cdot \vec{c} - \vec{a} \cdot \vec{b} | - 2 (\varepsilon + \delta) \leq 1 - \vec{b} \cdot \vec{c} + 2(\varepsilon + \delta) $$

using

\begin{equation} | \overline{P}(\vec{a}, \vec{b}) + \vec{a} \cdot \vec{b} | \leq \varepsilon + \delta. \qquad(1) \end{equation}

My attempt was

$$ \begin{equation} \begin{split} &\overline{P}(\vec{b}, \vec{c}) + \vec{b} \cdot \vec{c} \leq | \overline{P}(\vec{b}, \vec{c}) + \vec{b} \cdot \vec{c} | \leq \varepsilon + \delta\\ \implies & \overline{P}(\vec{b}, \vec{c}) \leq (\varepsilon + \delta) - \vec{b} \cdot \vec{c}\\ \implies & 1 + (\varepsilon + \delta) + \overline{P}(\vec{b}, \vec{c}) \leq 1 - \vec{b} \cdot \vec{c} + 2(\varepsilon + \delta)\\ \implies& | \overline{P} (\vec{a}, \vec{b}) - \overline{P}(\vec{a}, \vec{c})| \leq 1 - \vec{b} \cdot \vec{c} + 2(\varepsilon + \delta) \end{split} \end{equation} $$

How do I deal with the left hand side, i.e., the absolute value of difference of probabilities using $(1)$?

Edited to add more info

We are dealing with a pair of spin one-half particles in a singlet state moving freely in opposite directions.

$$ P (\vec{a}, \vec{b}) \equiv \int d \lambda \rho(\lambda)A(\vec{a}, \lambda) B(\vec{b}, \lambda) $$

where $\lambda$ is a continuous parameter, $\rho(\lambda)$ is its probability distributions and $A(\vec{a}, \lambda) = \pm 1$, $B(\vec{a}, \lambda) \pm 1$ are the possibles results of measuring spin components $\vec{\sigma}_{1}, \vec{\sigma}_{2}$ along $\vec{a}, \vec{b}$ respectively.

The probabilities $\overline{P}$s are nonnegative, the vectors $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and I'm assuning $\varepsilon, \delta > 0$.

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    $\begingroup$ You will need to add details. What is P, how is it defined? You can't expect people to go and read the paper for you. $\endgroup$ Commented Jan 15, 2023 at 18:06
  • $\begingroup$ But, looking at it, isn't it just some triangle inequalities? $\endgroup$ Commented Jan 15, 2023 at 18:08
  • $\begingroup$ I wasn't expecting people to go and read the paper, I was hoping that someone who read the paper could help. Will add more information, but I guess it's not the right place for this kind of specific questions about a specific paper. $\endgroup$ Commented Jan 15, 2023 at 18:17
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    $\begingroup$ I remember having and answering this question when I read this paper, but I no longer have my notes from 2005. I'll try and write an answer later. $\endgroup$
    – rob
    Commented Jan 15, 2023 at 18:33
  • $\begingroup$ @rob As far as I can tell you just insert (1) into the first line and get the second (in fact, with one factor of 2 only being 1). As far as I can tell this is homework-type. $\endgroup$ Commented Jan 15, 2023 at 18:50

1 Answer 1

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What Equation (1) says is that if you replace $P(\vec x,\vec y)$ by $-\vec x\cdot \vec y$, you incur an error of at most $\pm (\epsilon+\delta)$.

Now do this both on the rhs and the lhs of your first equation:

  1. On the lhs of the "$\le$", in the worst case the error will make the value larger, which means you need to put a minus sign to still satisfy the inequality. You do two replacements, thus $-2(\epsilon+\delta)$.

  2. On the rhs, in the worst case the error will make the value smaller. Thus, you get an extra $+(\epsilon+\delta)$.

If you want to formalize this, this can be done by the triangle inequality, plus the following variant thereof: If $|a-b|\le \eta$, then $b-\eta \le a \le b+\eta$.

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