Question on expression in "J.S.Bell : On the Einstein Podolsky Rosen paradox" I have a question on the article
J. S. Bell, On the Einstein Podolsky Rosen paradox, Physics 1, 195, 1964. (link)
My question concerns the expression (3) of the article, at page 196. I don't understand what is the reasoning that leads to this expression of the expectation value... I think I miss something but I don't know what.
This is what I understood from now on : 
$\vec{\sigma_1}$ and $\vec{\sigma_2}$ are the spins of the two particles that move apart and must be exactly opposite according to quantum mechanics when measured in a direction of the component $\vec{a}$.
First, did I understand well and do we really have
$$A(\vec{a},\lambda) = \vec{\sigma_1}.\vec{a} = \pm 1 \\
 B(\vec{b},\lambda) = \vec{\sigma_2}.\vec{b} = \pm 1$$
then ? If not, what does $A(\vec{a},\lambda)$ and $B(\vec{b},\lambda)$ correspond to ? A sort of $sign$ function or something like in the next section?
Secondly, why
$$ <\vec{\sigma_1}.\vec{a}\; \vec{\sigma_2}.\vec{b}> = -\vec{a}.\vec{b}$$
Is it because $\vec{\sigma_1}$ and $\vec{\sigma_2}$ are opposite ?
Thanks !
 A: No, all your $\vec \sigma_1, \vec \sigma_2$ are simply the Pauli matrices, but applying to the first or the second particle, so $\vec \sigma_1. \vec a, \vec \sigma_2. \vec b$ are the measurement operators applying respectively to particles $1$ and $2$.
The outcome of a measurement  can be $1$ or $-1$, but that does not mean that 
$\vec \sigma_1. \vec a = \pm 1$ or $\vec \sigma_2. \vec b = \pm 1$. This is false, this is not a operator equality.
In fact, a better notation for the measurement operator of the 2-particles system, is  $\vec \sigma. \vec a \otimes \vec \sigma. \vec b$.
Here $\vec \sigma$ are also the Pauli matrices
This notation means that the operator $\vec \sigma. \vec a$ is applyed to the first particle, and that the operator $\vec \sigma. \vec b$ is applyed to the second particle.
The mean value is referred to the singlet state : 
$$\psi = \frac{1}{\sqrt{2}} (|+ \rangle |- \rangle - |- \rangle |+ \rangle) \tag{1} $$
So, you have : 
$$M=<\vec \sigma. \vec a \otimes \vec \sigma. \vec b>_{Singlet} = \langle\psi|\vec \sigma. \vec a \otimes \vec \sigma. \vec b|\psi\rangle \tag{2}$$
That is :
$$M = \frac{1}{2}(\langle +| \langle -| - \langle -| \langle +| )|\vec \sigma. \vec a \otimes \vec \sigma. \vec b(|+\rangle|-\rangle - |-\rangle|+\rangle) \tag{3}$$
So, you have four terms for $M$ ($M= M_1 + M_2 +M_3 + M_4$): 
$$M_1 = \frac{1}{2} \langle +|\vec \sigma.\vec a|+\rangle  \langle -|\vec \sigma.\vec b|-\rangle\tag{4}$$ 
$$M_2 = -\frac{1}{2} \langle +|\vec \sigma.\vec a|-\rangle  \langle -|\vec \sigma.\vec b|+\rangle\tag{5}$$
$$M_3 = -\frac{1}{2} \langle -|\vec \sigma.\vec a|+\rangle  \langle +|\vec \sigma.\vec b|-\rangle\tag{6}$$
$$M_4 = \frac{1}{2} \langle -|\vec \sigma.\vec a|-\rangle  \langle +|\vec \sigma.\vec b|+\rangle\tag{7}$$
With the help of the expressions :
$$\langle +|\vec \sigma.\vec c|+\rangle = c_3, \langle +|\vec \sigma.\vec c|-\rangle = c_1 - ic_2,\langle -|\vec \sigma.\vec c|+\rangle = c_1 + ic_2, \\ \langle -|\vec \sigma.\vec c|-\rangle = -c_3\tag{8}$$
you will find easily that : 
$$M = -(a_1b_1 + a_2b_2 + a_3b_3) = - \vec a.\vec b \tag{9}$$
