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A pion decays to a singlet electron/positron state. We will measure the component of the electron's spin in the $\vec a$ direction and positron's spin in the $\vec b$ direction. If there exists a local hidden variable(s) $\lambda$, then outcomes of the measurements must be given by functions $A(\vec a,\lambda)$ and $B(\vec b,\lambda)$ respectively. these functions can only output the values $\pm1$ (in units of $\hbar/2$.) Due to entanglement and the case of $\vec a=\vec b$, we may write $B(\vec b,\lambda)=-A(\vec b,\lambda)$. We also assume a non-negative probability density $\rho(\lambda)$ which integrates to unity over all of space. Then Bell's inequality follows in the usual way.

This is my question: What is it in the above definitions for $\lambda$ that makes it "local"?

Griffiths writes the following:

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Is the locality condition only that $A$ is strictly not a function of $\vec b$ (and $B$ not of $\vec a$), which could not be ruled out for FTL signals?

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  • $\begingroup$ A great resource for understanding Bell's theorem and the notion of locality it assumes is Travis Noreen's articles on the subject (you can search for them on arxiv); this one in particular is relevant: arxiv.org/abs/0707.0401 $\endgroup$
    – Matt Dickau
    Commented Jul 1, 2022 at 13:54

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In addition to $A$ strictly not being a function of $b$ (and vice versa), the locality condition is ensures that $\vec{a}$, $\vec{b}$, and $\lambda$ are completely independent of each other. In particular, $\vec{a}$, $\vec{b}$ are not functions of $\lambda$ (or correlated with $\lambda$ in any way). Any such correlation would spoil the proof of Bell's theorem.

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What makes the functions $A(a,\lambda)$ and $B(b,\lambda)$ local are two factors

  1. As you said before the fact that $A(a,\lambda)$ does not depend on b.
  2. The $\lambda$ variables are fixed at the source at the moment the particles are created. So their behavior is "locally" explained or predetermined by these variables.

Notice that "predetermined" in a general sense does not imply determinism because the model may also be stochastic.

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  • $\begingroup$ For a given direction in $A(a,\lambda)$ we must put $a=\theta$ and for the opposite direction $a=\theta+\pi$. So the function is well defined. $\endgroup$
    – facenian
    Commented Jul 20, 2022 at 20:05
  • $\begingroup$ Suppose we consider variable 1 in a and 2 in b, if we look at thd setup in a spacetime diagram, then after the meadurement 1 can "know" the direction a. So looking at it through time the lambda1 can depend on a after ghe measurement and be transmitted to the averager, like in $cov(a,b,l1,l2)=1/(l1*l2)\int_0^{l1}Cos(A-S)ds\int_0^{l2}cos(b-t)dt$ $\endgroup$
    – QuantumPotatoïd
    Commented Aug 7, 2023 at 10:36

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