#### SpectraCat

We may consider the function[integral]for the expectation value of spin:

$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$

We may broaden the choice of the integral by including weights f(lambda,t) and g(lambda,t) for the spin values A and B. rho is normalized to unity as usual but we make,

Integral [rho*g*f]d lammda =n [which is not unity but some finite number suited to our testings]

[ By the above weights I have tried to take care of the time of measurement. Suppose for each instant we have the same distribution function rho for lambda[Or we may consider distribution functions].The weights are taking care of the cumulative effects of the distribution functions over the small interval of measurement.]
None of that makes any difference, because by introducing new parameters, you change the system, and you must take those changes into account and re-derive the appropriate Bell inequalities. Anyway, why not just work within the idealized framework that Bell provided? Why are you introducing these new parameters?

#### Anamitra

Suppose a lottery is repeated several times and you are allowed to participate in it on the same ticket. Your expectation simply increases ,by a factor n.

Consider the expectation of spin given below:
$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$

We assume that the same normalized function rho[lambda] holds for each moment or instant of the measurement. So the overall expectation covering the time interval of measurement becomes n times . "n" is some weight to take care of the time interval of measurement.Now we have:
$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{n}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$

and $${\int}{\rho}{(}{\lambda}{)}{n}{d}{\lambda}{=}{n}$$

This could give us some advantage!

#### JesseM

Suppose a lottery is repeated several times and you are allowed to participate in it on the same ticket. Your expectation simply increases ,by a factor n.
No it doesn't, because "expectation value" is supposed to be what expect if you conduct a large number of trials, add the results of each trial, then divide by the number of trials. So simply adding more trials would increase your chance of getting a large result on at least one trial, but it would also increase the number you are dividing by at the end, so the expectation value isn't different.

#### Anamitra

What you are saying is the expectation for each trial or one trial

I am thinking in terms of the total impact of a large number of trials in a very short interval of time.

You would definitely consider yourself luckier if you were allowed to participate in the same lottery several times on the same ticket!

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#### JesseM

What you are saying is the expectation for each trial or one trial

I am thinking in terms of the total impact of a large number of trials in a very short interval of time.
But how is that relevant to Bell's theorem? The "expectation values" in Bell inequalities are also expectations for each trial. If you want some cumulative sum of results over many trials then again you'd have to use a different inequality, and it would depend on the precise number N of trials you were thinking of.

#### Anamitra

Any measurement process usually involves a huge number of trials/impacts. It is not possible to know the exact number of such trials..One may use a weight to represent such effects[no of trials] and then try to calculate it [or get some estimate of it]from indirect evidences like conformity with accepted principles/theories.

#### Anamitra

Each moment our measuring system is experiencing the distribution function[or its effect].Each moment it is experiencing the effect of the expected value of an individual trial. In a short interval of time it is giving a report of such cumulative values over the interval

#### Anamitra

If the time of measurement in an experiment is very long or sufficiently then of course we get a steady value of the physical quantity because the past effects [of the initial moments] are lost/dissipated.

[The past effects are likely to be lost if the interval is long enough]

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#### Anamitra

Suppose we are measuring temperature using an ordinary mercury thermometer. We have to provide "sufficient time" so that the temperature may be recorded

#### JesseM

Any measurement process usually involves a huge number of trials/impacts. It is not possible to know the exact number of such trials..
Why do you think it's not possible? Each particle measurement is based on a single discrete "click" at some detector, so you can just count the number of clicks, we aren't talking about measuring a continuously-varying quantity like temperature.

#### Anamitra

A discrete click may involve a million impacts[the reception of a million impacts--impacts of the distribution function rho]---a discrete click and a single impact are not identical ideas.

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#### JesseM

A discrete click may involve a million impacts[the reception of a million impacts--impacts of the distribution function rho]
Impacts of what? A "function" isn't a physical entity that can impact with anything. In any case, each "trial" consists of a single click which is recorded as +1 or -1 (or +1/2 and -1/2 if you prefer), regardless of what is really going on physically in each click.

#### Anamitra

In a single click of your gadget,you may have to consider the same distribution function[the normalized rho] several million times--the theoretical analysis of a click[what it measures]--involves such a consideration.

#### JesseM

In a single click of your gadget,you may have to consider the same distribution function[the normalized rho] several million times--the theoretical analysis of a click[what it measures]--involves such a consideration.
I don't get it, what does it mean to "consider" a function several million times? Do you mean considering several million possible values of lambda which are each assigned probabilities (or probability densities) by the distribution function, or something else?

#### Anamitra

What happens "inside" the click is important.We have "hidden variables" that can exist physically and contribute to the result of the click. The result of a click is not a magical thing that you are possibly inclined to believe in. If it is not magical ,how does it occur[I mean how the result of the measurement takes place]? We need to investigate this and the hidden variables can play an important role, a definite role--in a manner I have indicated.

#### JesseM

What happens "inside" the click is important.We have "hidden variables" that can exist physically and contribute to the result of the click. The result of a click is not a magical thing that you are possibly inclined to believe in. If it is not magical ,how does it occur[I mean how the result of the measurement takes place]? We need to investigate this and the hidden variables can play an important role, a definite role--in a manner I have indicated.
Huh? This is a totally vague answer, you didn't address my specific questions above about the meaning of "considering" the distribution function several million times. And of course in a hidden variables theory the variables might causally influence the result in a complicated way, but that doesn't change the fact that the result itself is only recorded as one of two possible outcomes, a click at one detector or a click at another. It is the probabilities or expectation values for these recorded outcomes that Bell inequalities are dealing with, an inequality involving hidden variables would be useless since we wouldn't have a way to measure these variables so there'd be no way to test whether the inequality was respected or violated.

#### Anamitra

We have the two detectors measuring the spin values of the two particles.These values are not independent.They are related to each other. Some may believe that the connection is a spooky one in its very existence or in relation to the fact that such a relation might lead to the violation of Special Relativity concept of finite speed of signal transmission.
But one may consider "hidden variables" to be responsible for the association between the two spin values.
The two measurements are not instantaneous in the mathematical sense. They involve a huge number of instants[we may think of a short interval broken up into a number of even shorter ones]---and the hidden variables may be operating on the spin states through each these instants.We make a theoretical estimate of the expected spin product[A*B] for each instant and them multiply this expectation by a weight factor to take care of the interval of measurement.
We have the formula:

$${P}_{h}{(}{a}{,}{b}{)}{=}{\int}{n}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$

Where,
$${\int}{\rho}{(}{\lambda}{)}{n}{d}{\lambda}{=}{n}$$

#### Anamitra

Of course it does, the inequality would just be modified if you use different numbers to label "spin-up" and "spin-down" measurements. Look at the step at the top of p. 406 (p. 4 of the pdf) of Bell's derivation, he invokes equation (1) which says that A(b,λ) = ±1 in order to justify equating the integrand [A(a,λ)*A(c,λ) - A(a,λ)*A(b,λ)] to the integrand A(a,λ)*A(b,λ)*[A(b,λ)*A(c,λ) - 1]...in other words, he's using the fact that A(b,λ)*A(b,λ)=+1. If you instead labeled the results with +(1/2) and -(1/2), then A(b,λ)*A(b,λ)=+(1/4), so that step wouldn't work. Instead the second integrand would have to be A(a,λ)*A(b,λ)*[4*A(b,λ)*A(c,λ) - 1].

Continuing on with steps analogous to the ones in Bell's paper, we can then apply the triangle inequality (which works as well for integrals as it does for discrete sums) to show that if $$P(a,b) - P(a,c) = \int d\lambda \rho(\lambda) A(a,\lambda)A(b,\lambda)[4A(b,\lambda)A(c,\lambda) - 1]$$, that implies $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \mid \rho(\lambda) A(a,\lambda)A(b,\lambda)[4A(b,\lambda)A(c,\lambda) - 1] \mid$$. Since $$\rho(\lambda)$$ is always positive and A(a,λ)*A(b,λ)=±(1/4) that becomes $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) (1/4) \mid [4A(b,\lambda)A(c,\lambda) - 1] \mid$$, then since 4*A(b,λ)*A(c,λ) - 1 is always equal to -2 or 0, its absolute value is always equal to 1 - 4*A(b,λ)*A(c,λ) so we now have $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) (1/4) [1 - 4A(b,\lambda)A(c,\lambda)]$$ or $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) [(1/4) - A(b,\lambda)A(c,\lambda)]$$. Bell then notes that $$\int d\lambda \rho(\lambda) [-A(b,\lambda)A(c,\lambda)] = P(b,c)$$ and since $$\int d\lambda \rho(\lambda) = 1$$ this gives an inequality equivalent to the one Bell derives:

(1/4) + P(b,c) ≥ |P(a,b) - P(a,c)|

With the measurement results labeled +(1/2) or -(1/2), there's no way this inequality can be violated in a realist theory that respects relativity.

By the mechanism provided in the postings 27 and 42 one will obtain:
(1/4)n + P(b,c) ≥ |P(a,b) - P(a,c)|
(1/4) + P(b,c) ≥ |P(a,b) - P(a,c)|
The first result will not show any conflict between classical intuition and quantum mechanics
at least in relation to the current discussion.[EPR paradox and Bell's inequalities]

#### Anamitra

In the previous posting the values of A and B have been assumed to be +1/2 or -1/2for each instead of +1 or -1[for each]

For A and B equal to +1 or -1 [for each] and this is conventional, we have
n + P(b,c) ≥ |P(a,b) - P(a,c)| instead of

1 + P(b,c) ≥ |P(a,b) - P(a,c)|

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#### JesseM

By the mechanism provided in the postings 27 and 42 one will obtain:
(1/4)n + P(b,c) ≥ |P(a,b) - P(a,c)|
Are you sure? Terms like P(b,c) and P(a,b) represent expectation values for a single trial, if you want them to represent an expectation value for a sum of results over many trials that would probably change aspects of the derivation, I'm not sure you would actually end up with the equation above. You need to show your work, give the steps in deriving the final inequality like I did.

#### DrChinese

Gold Member
But one may consider "hidden variables" to be responsible for the association between the two spin values.
Not unless you can show us an example of what they are. You keep providing generic examples. How about something specific? Give us a set of 10 runs of what those values might be for measurement settings 0, 120 and 240 degrees.

You should not make speculative statements as if they are established science. Your ideas have been soundly discredited in the past 30 years by numerous experiments, such as:

http://arxiv.org/abs/quant-ph/9810080

"We observe strong violation of Bell's inequality in an Einstein, Podolsky and Rosen type experiment with independent observers. Our experiment definitely implements the ideas behind the well known work by Aspect et al. We for the first time fully enforce the condition of locality, a central assumption in the derivation of Bell's theorem. The necessary space-like separation of the observations is achieved by sufficient physical distance between the measurement stations, by ultra-fast and random setting of the analyzers, and by completely independent data registration. "

#### Anamitra

Response to Posting 45

I have not meant the expectation values of a single trial. In the revised formula I have meant the cumulative expectation values over the time intervals concerned.

P(a,b) in my formula[the revised one] represents n*P(a,b) [ P(a,b) in the term n*P(a,b)represents the expectation value for a single trial]

#### JesseM

Response to Posting 45

I have not meant the expectation values of a single trial.
I understood that, but the point is that this is what P(a,b) and such meant in the original formula, if you want to change the meaning of the terms you need to provide a new derivation of an inequality involving the terms with revised meanings.
In the revised formula I have meant the cumulative expectation values over the time intervals concerned.
Yes, I understood that as well, which is why I said "if you want them to represent an expectation value for a sum of results over many trials that would probably change aspects of the derivation". My point is that you haven't provided any justification for believing that your modified formula is actually correct under your new definitions, you need to show your work and provide a derivation of that formula.

#### SpectraCat

I understood that, but the point is that this is what P(a,b) and such meant in the original formula, if you want to change the meaning of the terms you need to provide a new derivation of an inequality involving the terms with revised meanings.

Yes, I understood that as well, which is why I said "if you want them to represent an expectation value for a sum of results over many trials that would probably change aspects of the derivation". My point is that you haven't provided any justification for believing that your modified formula is actually correct under your new definitions, you need to show your work and provide a derivation of that formula.
I already explained this, but for whatever reason, that part of my post (concerning the necessity of re-derivation of the inequalities) went unaddressed.

#### Anamitra

The derivation
$${P}{(}{a}{,}{b}{)}{=}{\int}{n}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$
$${P}{(}{a}{,}{c}{)}{=}{\int}{n}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{c}{,}{\lambda}{)}{d}{\lambda}$$
$${P}{(}{b}{,}{c}{)}{=}{\int}{n}{\rho}{(}{\lambda}{)}{A}{(}{b}{,}{\lambda}{)}{B}{(}{c}{,}{\lambda}{)}{d}{\lambda}$$
The same weight factor has been taken in each case assuming the same type of measuring technique.
Important to note that,
$${B}{(}{a}{,}{\lambda}{)}{=}{-}{A}{(}{a}{,}{\lambda}{)}$$
[One may consider Bell’s paper [from the first link in posting 23] for the above relation.
Now we proceed:
$${P}{(}{a}{,}{b}{)}{-}{P}{(}{a}{,}{c}{)}{=}{\int}{n}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{A}{(}{c}{,}{\lambda}{)}{d}{\lambda}{-}{\int}{n}{\rho}{(}{\lambda}{)}{A}{(}{a}{,}{\lambda}{)}{B}{(}{b}{,}{\lambda}{)}{d}{\lambda}$$
$${=}{ \int}{d}{\lambda}{n}{\rho}{(}{\lambda}{A}{(}{a}{,}{\lambda}{)}{A}{(}{b}{,}{\lambda}{)}{[}{A}{(}{b}{,}{\lambda}{ A}{(}{c}{,}{\lambda}{)}{-}{1}{]}$$
[Since $${{[}{A}{(}{b}{,}{\lambda}{]}}^{2}{=}{1}$$ ]
$$\mid P(a,b) - P(a,c) \mid \le \int d\lambda {n}\mid \rho(\lambda) A(a,\lambda)A(b,\lambda)[A(b,\lambda)A(c,\lambda) - 1] \mid$$. Since $$\rho(\lambda)$$ is always positive and A(a,λ)*A(b,λ)=±1 we have$$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) (n) \mid [A(b,\lambda)A(c,\lambda) - 1] \mid$$, then since A(b,λ)*A(c,λ) - 1 is always equal to -2 or 0, its absolute value is always equal to 1 - A(b,λ)*A(c,λ) so we now have $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) {n} [1 - A(b,\lambda)A(c,\lambda)]$$ or $$\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) {[}{n} {-}{ n} A(b,\lambda)A(c,\lambda){]}$$. We have, $$\int d\lambda \rho(\lambda) {n}[-A(b,\lambda)A(c,\lambda)] = P(b,c)$$ and since $$\int d\lambda \rho(\lambda) = 1$$ this gives an inequality equivalent to the one Bell derives:

n + P(b,c) ≥ |P(a,b) - P(a,c)|

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