Sunday 25 September 2016

quantum mechanics - 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?



Answer



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.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...