Rigorous Bell-CHSH?

by AmazedByTheQuantum   Last Updated January 18, 2018 20:19 PM

The usual derivations by physicists of the Bell-CHSH inequalities as a limit imposed on any possible local hidden variables description of quantum systems seems so sloppy, that I'm tempted to ask you statisticians if there is a rigorous mathematical derivation of these results. Also, what would be a proper statistical analysis of the corresponding experiments (such as the one by Aspect et al)?

Tags : probability


Answers 1


This answer is based on the concepts introduced in Gill (2012) (soon to be published in Statistical Science). Of course, Gill's paper has much more to offer. One example is the exponential bound established in his Theorem 1. In what follows we use the new features introduced by Gill to give an elementary proof of a much simpler result.

Let $\Lambda_1,\Lambda_2,\dots$ be a sequence of independent random objects identically distributed with some distribution $\mu$ over a certain measure space $(L,\mathscr{L})$. Denote by $\mathbb{S}^3$ the set of unit vectors of $\mathbb{R}^3$ and let $$ A:\mathbb{S}^3\times L\to\{-1,1\} \quad\text{and}\quad B:\mathbb{S}^3\times L\to\{-1,1\}$$ be two measurable functions. For every $a,b\in\mathbb{S}^3$, the symmetry of the problem imposes that $$ \Pr\{\omega:A(a,\Lambda_i(\omega))=-1\} = \Pr\{\omega:A(a,\Lambda_i(\omega))=1\} = \frac{1}{2} \, , $$ $$ \Pr\{\omega:B(b,\Lambda_i(\omega))=-1\} = \Pr\{\omega:B(b,\Lambda_i(\omega))=1\} = \frac{1}{2} \, , $$ and also $$ \Pr\{\omega:A(a,\Lambda_i(\omega))=-B(a,\Lambda_i(\omega))\} = 1 \, . $$ The $\Lambda_i$'s are the so called hidden variables that would, in principle, completely represent a pair of particles prepared in a singlet state. The functions $A$ and $B$ give the values of the spins of each particle measured by detectors along specified directions. There must be "locality", in the sense that the probability distribution $\mu$ must have no relation with chosen directions.

Defining the process correlation $$ \rho_{ab} = \int A(a,\lambda)\,B(b,\lambda)\,d\mu(\lambda) \, , $$ for $a,b\in\mathbb{S}^3$, the original Bell-CHSH inequality is obtained as follows.

For $a,a',b,b\in\mathbb{S}^3$, we have $$ |\rho_{ab} + \rho_{ab'} + \rho_{a'b} - \rho_{a'b'}| $$ $$ = {\small\left| \int \left( A(a,\lambda)\,B(b,\lambda) + A(a,\lambda)\,B(b',\lambda) + A(a',\lambda)\,B(b,\lambda) - A(a',\lambda)\,B(b',\lambda)\right)d\mu(\lambda)\right|} $$ $$ \leq \int \left| A(a,\lambda)(B(b,\lambda)+B(b',\lambda)) + A(a',\lambda)(B(b,\lambda)-B(b',\lambda)) \right|\,d\mu(\lambda) \, , $$ in which we used the modulus inequality. Outside of a set of zero $\mu$-probability, either $$ B(b,\lambda)=B(b',\lambda) $$ or $$ B(b,\lambda)=-B(b',\lambda) \, . $$ Therefore, by inspection, the expression in the integrand satisfies $$ A(a,\lambda)(B(b,\lambda)+B(b',\lambda)) + A(a',\lambda)(B(b,\lambda)-B(b',\lambda))\in\{-2,2\} \, , $$ and we have the standard Bell-CHSH inequality $$ |\rho_{ab} + \rho_{ab'} + \rho_{a'b} - \rho_{a'b'}| \leq 2 \, . $$

The novelties introduced by Gill (2012) show that this formulation is not enough to describe the state of affairs in Aspect like experiments. We need to formalize the random selection of detectors directions and work explicitly with the sample correlations.

Introduce the random direction selectors $\{S_i\}_{i\geq 1}$ and $\{T_i\}_{i\geq 1}$ with distributions $$ \mathrm{Pr}\{S_i=a\}=p\, , \quad \mathrm{Pr}\{S_i=a'\}= 1-p\, , \quad \mathrm{Pr}\{T_i=b\}=q \, , \quad \mathrm{Pr}\{T_i=b'\} = 1-q \, , $$ in which $a,a',b,b\in\mathbb{S}^3$, and $0<p,q<1$. Note that in what follows there is no need to assume "fair" selectors with $p=q=1/2$.

We will make the assumption that all the $S_i$'s, $T_i$'s and $\Lambda_i$'s are independent.

For $a,b\in\mathbb{S}^3$, define the sample correlation $$ \hat{\rho}_{\!ab} = \frac{1}{N_{ab}}\sum_{i=1}^n\,A(a,\Lambda_i)\,B(b,\Lambda_i)\,I_{\{(a,b)\}}(S_i,T_i) \, , $$ in which $$ N_{ab} = \sum_{i=1}^n I_{\{(a,b)\}}(S_i,T_i) \, . $$

Lemma. $ |\hat{\rho}_{\!ab} + \hat{\rho}_{\!ab'} + \hat{\rho}_{\!a'b} - \hat{\rho}_{\!a'b'}| \to |\rho_{ab} + \rho_{ab'} + \rho_{a'b} - \rho_{a'b'}|$ almost surely when $n\to\infty$.

Proof. Since the random variables $\{I_{\{(a,b)\}}(S_i,T_i)\}_{i\geq 1}$ are independent and identically distributed, the Strong Law of Large Numbers says that $$ \frac{N_{ab}}{n} = \frac{1}{n}\sum_{i=1}^n I_{\{(a,b)\}}(S_i,T_i)\to \mathrm{E}\left[I_{\{(a,b)\}}(S_1,T_1)\right] = p\cdot q $$ almost surely when $n\to\infty$. Analogously, we find that $$ \frac{1}{n}\sum_{i=1}^n\,A(a,\Lambda_i)\,B(b,\Lambda_i)\,I_{\{(a,b)\}}(S_i,T_i) \to \mathrm{E}\left[ A(a,\Lambda_1)\,B(b,\Lambda_1)\,I_{\{(a,b)\}}(S_1,T_1) \right] $$ $$ = \mathrm{E}\left[ A(a,\Lambda_1)\,B(b,\Lambda_1)\right]\,\mathrm{E}\left[\,I_{\{(a,b)\}}(S_1,T_1) \right] $$ $$ = (p\cdot q) \int A(a,\lambda)\,B(b,\lambda)\,d\mu(\lambda) = (p\cdot q)\;\rho_{ab} $$ almost surely when $n\to\infty$. Hence, rewriting the sample correlation, we have $$ \hat{\rho}_{\!ab} = \frac{n}{N_{ab}} \times \frac{1}{n}\sum_{i=1}^n\,A(a,\Lambda_i)\,B(b,\Lambda_i)\,I_{\{(a,b)\}}(S_i,T_i) \to \frac{1}{p\cdot q}\times(p\cdot q)\;\rho_{ab} = \rho_{ab} $$ almost surely when $n\to\infty$. The same reasoning, applied to the sample correlations computed in the other directions, and the continuity of the operations involved yield that $$ |\hat{\rho}_{\!ab} + \hat{\rho}_{\!ab'} + \hat{\rho}_{\!a'b} - \hat{\rho}_{\!a'b'}| \to |\rho_{ab} + \rho_{ab'} + \rho_{a'b} - \rho_{a'b'}| $$ almost surely when $n\to\infty$. $\;\Box$

One implication of this Lemma is that someone who believes that, for very large $n$, a Monte Carlo simulation -- with any desired local $(\mu,L,\mathscr{L})$ and functions $A$ and $B$ -- can produce sample correlations that violate the Bell-CHSH bound substantially should never fly to Vegas.

Zen
Zen
May 05, 2014 14:30 PM

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