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I am trying to understand Bell's famous paper from 1964 on the EPR paradox.

I stumbled over equation $(9)$ where it says

$$\left.\begin{align}A(\vec a,\vec\lambda)&=\operatorname{sign}\vec a\cdot\vec\lambda\\B(a,b)&=-\operatorname{sign}\vec b\cdot\vec\lambda \end{align}\right\}\tag9$$

Apart from the missing vector arrows I don't understand why it's $B(a, b)$ there. Could Bell actually have meant $B(b, \lambda)$ instead? I wondered about this because I also don't understand how he concludes $(10)$ from this, but this might be another question.

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    $\begingroup$ It's somehow comforting to see even very famous papers have mistakes :) $\endgroup$
    – Andrew
    Aug 11, 2021 at 19:39

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Yes, this is a typo on the left-hand side of the second equation; it is correct that $$B(\vec{b},\vec{\lambda})=-\mathrm{sign}\,\vec{b}\cdot\vec{\lambda}.$$

To see how he concludes (10), let's rewrite and put the normalization in explicitly for this uniform distribution: $$P(\vec{a},\vec{b})=\frac{\int d\lambda A(\vec{a},\vec{\lambda})B(\vec{b},\vec{\lambda})}{\int d\lambda}.$$ The integral is over the surface of a sphere with unit radius, so the denominator will be $4\pi$. The product $A(\vec{a},\vec{\lambda})B(\vec{b},\vec{\lambda})=-\mathrm{sign}\,(\vec{a}\cdot\vec{\lambda})\mathrm{sign}\,(\vec{b}\cdot\vec{\lambda})$ will either be $+1$ or $-1$, and we need to determine for what fraction of the values of $\vec{\lambda}$ each of the $\pm1$ values occur.

The rest is geometry. Set the coordinate system, without loss of generality, to have $\vec{a}$ point to the top of the sphere and $\vec{b}$ be at polar angle $\theta$ and azimuthal angle $0$. This is always allowed because the integral over $\lambda$ is spherically symmetric. The region of values of $\vec{\lambda}$ for which $\mathrm{sign}\vec{a}\cdot\vec{\lambda}$ is $+1$ is the top half of the sphere; the region for which $\mathrm{sign}\vec{b}\cdot\vec{\lambda}$ is $+1$ is the hemisphere whose pole points in the direction of $\vec{b}$. The overlap of the two $+1$ regions forms a spherical wedge with angular width $\pi-\theta$. By symmetry, the spherical wedge on the opposite side of the sphere also has angular width $\pi-\theta$, and there both values are $-1$. (You can check the limits: when $\theta=0$, they fully overlap, and when $\theta=\pi$ they do not.)

From this picture, we see that $\mathrm{sign}\,(\vec{a}\cdot\vec{\lambda})\mathrm{sign}\,(\vec{b}\cdot\vec{\lambda})=1$ for $2\pi-2\theta$ out of the possible $2\pi$ angular width and $\mathrm{sign}\,(\vec{a}\cdot\vec{\lambda})\mathrm{sign}\,(\vec{b}\cdot\vec{\lambda})=-1$ for the other $2\theta$. The total probability is thus (remembering the extra $-$ sign) $$P(\vec{a},\vec{b})=-\frac{2\pi-2\theta}{2\pi}+\frac{2\theta}{2\pi}=-1+\frac{2\theta}{\pi}, $$ as per Bell.

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  • $\begingroup$ Thanks a lot for this detailed explanation! I could follow this, except for the point of the normalization. Where is the $4\pi$ used? $\endgroup$ Aug 11, 2021 at 21:26
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    $\begingroup$ Yeah, I didn't really use it directly. Instead, I concluded that we only have to care about one angular coordinate, so that was normalized by $2\pi$. A more complete mathematical proof would do the integral over all of the surface area and show there to be symmetry about one axis $\endgroup$ Aug 11, 2021 at 21:30

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