# Where is locality used in CHSH/Bell's inequality?

A very similar question is asked here, but I'm still confused :(

From Bell, in a hidden variable model, $A = A(\lambda, a)=\pm 1$ is the observed spin of the first particle around axis $a$, and $B = B(\lambda, b)=\pm 1$ is the same for the 2nd particle around $b$. The CHSH proof is then $E(AB)+E(A'B)+E(AB')-E(A'B')=E(AB+A'B+AB'-A'B')\leq 2$, since $|AB+A'B+AB'-A'B'|=|A(B+B')+A'(B-B')|\leq 2$.

But we could do the same trick if $A$ depends on $b$ too, so where is locality used? The link says that $E(AB)+E(A'B)+E(AB')-E(A'B')=E(AB+A'B+AB'-A'B')$ is unjustified, but aren't expectations always linear?

• The hidden variable is a property of each particle, thus it's local to that particle. Bell's doesn't apply to a non-local hidden variable. May 22, 2018 at 4:41
• $E(A,B)=\int A(a,\lambda) B(,b,\lambda) \rho (\lambda) d\lambda$. I don't see to what locality corresponds to the in maths. May 22, 2018 at 10:29

For better clarity I will here be using the notation $$A_0$$ and $$A_1$$, instead of $$A$$ and $$A'$$, to denote the outcomes for different measurement setups, and same with $$B$$.

The expectation values are defined as $$E(A_xB_y)\equiv\int d\lambda\, q(\lambda)\, E_\lambda(A_xB_y) =\int d\lambda\, q(\lambda)\sum_{ab} ab \,p(ab|xy,\lambda),$$ where $$q(\lambda)$$ is the probability of the hidden variable having the value $$\lambda$$, and $$p(ab|xy,\lambda)$$ is the (joint) probability of getting the outcomes $$a$$ and $$b$$ given the measurement settings $$x$$ and $$y$$ and hidden variable $$\lambda$$.

The locality assumption (plus assumption of independence of measurement choices) is embedded in the following factorization for $$p$$: $$p(ab|xy,\lambda)=p(a|x,\lambda)p(b|y,\lambda).\tag A$$

This relation is needed to have $$E_\lambda(A_x B_y)=E_\lambda(A_x)E_\lambda(B_y)$$, so that \begin{aligned} E(A_0B_0)+E(A_0B_1) &=\int d\lambda \,q(\lambda)[E_\lambda(A_0)E_\lambda(B_0)+E_\lambda(A_0)E_\lambda(B_1)] \\ &= \int d\lambda \,q(\lambda)E_\lambda(A_0)[E_\lambda(B_0)+E_\lambda(B_1)], \end{aligned} where $$E_\lambda(A_x)\equiv\sum_a a \,p(a|x,\lambda),\quad E_\lambda(B_y)\equiv\sum_b b \,p(b|y,\lambda).$$ Without locality assumption, the above would not hold.

An analogous argument leads to $$E_\lambda(A_1B_0)-E_\lambda(A_1B_1) = E_\lambda(A_1)[E_\lambda(B_0)-E_\lambda(B_1)].$$

The conclusion is now straightforward from here. Define $$S_\lambda\equiv E_\lambda(A_0B_0)+E_\lambda(A_0B_1)+E_\lambda(A_1B_0)-E_\lambda(A_1B_1).$$ Then, using the above equalities, we have $$S_\lambda= E_\lambda(A_0)[E_\lambda(B_0)+E_\lambda(B_1)] + E_\lambda(A_1)[E_\lambda(B_0)-E_\lambda(B_1)].$$ The triangle inequality, together with the fact that, by definition of the numbers we are attaching to the possible outputs, we have $$0\le E_\lambda(A_i)\le 1$$, now gives $$\lvert S_\lambda\rvert\le \lvert E_\lambda(B_0)+E_\lambda(B_1)\rvert + \lvert E_\lambda(B_0) - E_\lambda(B_1)\rvert = 2 \max(E_\lambda(B_1),E_\lambda(B_2)).$$ Using again that by definition $$0\le E_\lambda(B_i)\le 1$$, we conclude that $$\lvert S_\lambda\rvert \le 2$$.

The full $$S$$ is now defined by averaging $$S_\lambda$$ over the hidden variable $$\lambda$$, and because a convex mixture of numbers in $$[-2,2]$$ remains in $$[-2,2]$$, we reach the conclusion: $$\lvert S\rvert\le 2.$$

But we could do the same trick if A depends on b too, so where is locality used?

No, you could not.

If the outcome of $$A$$ directly depends on $$B$$, then (A) does not hold, thus the probabilities do not factorize ($$E(AB)\neq E(A)E(B)$$), thus $$E(A_0B_0)+E(A_0B_1)\neq E(A_0(B_0+B_1))$$, thus the CHSH argument cannot be applied.

Aren't expectations always linear?

Indeed they are. The locality assumption is needed not for the linearity but to factorize the probabilities/expectation values, in order to put them into the form $$E_\lambda(A_0)[E_\lambda(B_0)+E_\lambda(B_1)]$$, at which point the CHSH argument applies.

• – glS
Jun 27, 2021 at 13:05