# Typo in Bell's original paper on the EPR paradox?

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.

• It's somehow comforting to see even very famous papers have mistakes :) Aug 11, 2021 at 19:39

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.
• Thanks a lot for this detailed explanation! I could follow this, except for the point of the normalization. Where is the $4\pi$ used? Aug 11, 2021 at 21:26
• 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 Aug 11, 2021 at 21:30