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?

  • $\begingroup$ 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. $\endgroup$
    – PM 2Ring
    Commented May 22, 2018 at 4:41
  • 1
    $\begingroup$ $E(A,B)=\int A(a,\lambda) B(,b,\lambda) \rho (\lambda) d\lambda$. I don't see to what locality corresponds to the in maths. $\endgroup$
    – reeeeee
    Commented May 22, 2018 at 10:29

1 Answer 1


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 the same with $B$. This means that $A_x$ is the random variable describing the possible measurement outcomes on Alice's side when the measurement choice is $x$, and similarly for $B_y$ describing Bob's outcomes when his measurement choice is $y$. We will label the two possible measurement choices with $x,y\in\{0,1\}$, and the possible measurement outcomes with $\pm1$, meaning $A_x,B_y\in\{-1,1\}$.

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 conditional 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 $0\le \lvert E_\lambda(A_i)\rvert \le 1$, now gives $$\lvert S_\lambda\rvert\le \lvert E_\lambda(B_0)+E_\lambda(B_1)\rvert + \lvert E_\lambda(B_0) - E_\lambda(B_1)\rvert = 2\max(|E_\lambda(B_0)| , |E_\lambda(B_1)|).$$

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.

  • $\begingroup$ related: physics.stackexchange.com/q/114218/58382 $\endgroup$
    – glS
    Commented Jun 27, 2021 at 13:05
  • $\begingroup$ Just a minor correction, I think: OP has $A = \pm 1$, but this answer nominally uses $0 \leq E_{\lambda}(A) \leq 1$. I think what is actually needed is $0 \leq |E_{\lambda}(A)| \leq 1$, which is of course true by hypothesis, if I followed the proof correctly. But please correct me if I missed something $\endgroup$
    – twoform
    Commented Apr 22 at 22:34
  • $\begingroup$ @twoform indeed, fixed, thanks $\endgroup$
    – glS
    Commented Apr 23 at 7:15

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