Is there a relation between the factorisation of the joint conditional probability distribution and Bell inequality?

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[I'm sorry, I've already posted the same question in the physics community, but I haven't received an answer yet.]


I'm approaching the study of Bell's inequalities and I understood the reasoning under the Bell theorem (ON THE EINSTEIN PODOLSKY ROSEN PARADOX (PDF)) and how the postulate of locality was assumed at the start of the demonstration.

However, I find problematic to arrive at the equivalence $$ E(\vec{a},\vec{b}) = \int_{\Lambda}d\lambda \rho(\lambda)A(\vec{a},\lambda)B(\vec{b},\lambda),$$
starting from the point of view expressed by the Clauser and Horne definition of locality.

CH claimed that a system is local if there is a parameter $\lambda$ and a joint conditional probabilities that can be written as follows: $$p(a,b|x,y,\lambda) = p(a|x,\lambda)p(b|y,\lambda),$$ and $$p(a,b|x,y) = \int_\Lambda d\lambda \rho(\lambda) p(a|x,\lambda)p(b|y,\lambda)$$ which make sense since it affirms that the probability of obtaining the value $a$ depends only on the measument $x = \vec{\sigma}\cdot\vec{x} $ and the value of $\lambda$.

However, if I use this expression to write down the expectation value of the products of the two components $\vec{\sigma}\cdot\vec{a}$ and $\vec{\sigma}\cdot\vec{b}$, I obtain as follows:

$$ E (\vec{a},\vec{b}) = \sum_{i,j}a_ib_jp(a,b|x,y) = \\ = \sum_{ij}a_ib_j \int_\Lambda d\lambda \rho(\lambda) p(a|x,\lambda)p(b|y,\lambda) \\ = \int_\Lambda d\lambda \rho(\lambda) (\sum_{i}a_ip(a|x,\lambda))(\sum_{i}b_ip(b|y,\lambda)) $$ where in the last equivalence I've used the fact that if the measument are independent their covariance must be equal to $0$.

At this point, the terms in the RHS in the brackets are equal to: $$ (\sum_{i}a_ip(a|x,\lambda)) = E(a,\lambda) =? = A(\vec{a},\lambda)\quad \quad (\sum_{i}b_ip(b|y,\lambda)) = E(b,\lambda) =?= B(\vec{b},\lambda).$$

That is not the equivalence that I want to find.

In fact in the RHS of the first equation $A(\vec{a},\lambda)$ is, according to Bell original article, the result of measure $\vec{\sigma}\cdot\vec{a}$, and fixing both $\vec{a}$ and $\lambda$ it can assume only the values of $\pm1$. (The same is applied for $B(\vec{b},\lambda)$.)

Some of you knows, where I fail? How can I obtain the original equivalence (that then is proved to be violate in the case of an entangled system) starting from the CH definition of reality?

Edit #1:

I've noted that I obtain the wanted equivalence only if I assume that $p(ab|xy\lambda) = E(\vec{a}\vec{b})$, but is it possible? How can a conditional probability be linked to the mean value of the product of two components?

Edit #2:

Surfing the internet I found an article (https://arxiv.org/abs/1709.04260, page 2, right on the top) which reports the same CH's local condition (to be accurate, the article presents the discrete version) and then affirm that:

Blockquote "The central realization of Bell’s theorem is the fact that there are quantum correlations obtained by local measurements ($M_a^x$ and $M_b^y$) on distant parts of a joint entangled state $\varrho$, that according to quantum theory are described as: $$p_{Q}(a,b,|x,y) = \text{Tr}(\varrho(M_a^x\otimes M_b^y) $$ and cannot be decomposed in the LHV form (i.e. The CH condition for locality)"

So why $p_Q(a,b|x,y)$ is seen as a measure of quantum correlation (that for definition is the mean of the product of the possible output)? It isn't a joint probability distribution (as stating while obtaining the LHV form)? Is there a link between the classical correlation ($E(\vec{a},\vec{b})$) and the joint probability distribution $p(a,b|x,y,\lambda)$?

LadyOfShalott

Posted 2018-09-08T12:21:21.077

Reputation: 71

I don't fully get what you are asking. Why isn't the result you get compatible with the one you wanted to get? If the question is about the difference between $A(a,\lambda)$ and $E(a,\lambda)$, I would have said they are the same thing. How do you define them? – glS – 2018-09-11T10:26:03.673

$E(a,\lambda)$ is expected value to obtain the value $a$ with $\lambda$ (and the measurement choice, say $x$) fixed, so it can have a value $\in [-1,1]$, while $A(a,\lambda)$ is the outcome and so it could be only $\pm 1$. – LadyOfShalott – 2018-09-11T13:04:05.190

ah, I see what you mean now. I don't really see any contradiction though. In the first equation you are denoting with $E(a,b)$ the probability of observing the outcomes $a,b$, while later you denote with $E(a,b)$ the expectation value of those outcomes. It is only naturaly that you get different results. If you think this might be the confusion I can explain it better in an answer – glS – 2018-09-12T10:54:54.860

If it doesn't bother you, it'll help me. (Sorry I thought I answered earlier). – LadyOfShalott – 2018-09-18T00:05:50.603

well I've seen your answer to my comment just now =). I don't get a notification if you don't tag me in the comment – glS – 2019-02-19T10:37:29.120

Answers

2

First of all, you inverted $a,b$ with $x,y$ when trying to draw the analogy.

In Bell's original paper, $\vec a,\vec b$ are used to denote the measurement directions, so the underlying probability distribution should be written as $$p(x,y|\vec a,\vec b,\lambda)=p(x|\vec a,\lambda)p(y|\vec b,\lambda).$$ The expectation values $A(\vec a,\lambda)$ and $B(\vec b,\lambda)$ used in Bell's paper would then be given by $$A(\vec a,\lambda)=\sum_x x p(x|\vec a,\lambda)$$ and similarly for $B$. The sum is here extended over the possible values that can correspond to the measurement choice $\vec a$. In the case of Bell's paper, this amounts to $x=\pm 1$.

To get the expectation value $E(\vec a,\vec b)$ you now simply need to take the average over the possible values of the hidden variable $\lambda$.


This answer was also posted on the duplicate question on physics.SE.

glS

Posted 2018-09-08T12:21:21.077

Reputation: 12 247