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 said:A discrete click may involve a million impacts[the reception of a million impacts--impacts of the distribution function rho]
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 said: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.
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 said: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.
By the mechanism provided in the postings 27 and 42 one will obtain:JesseM said: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 [tex]P(a,b) - P(a,c) = \int d\lambda \rho(\lambda) A(a,\lambda)A(b,\lambda)[4A(b,\lambda)A(c,\lambda) - 1][/tex], that implies [tex]\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[/tex]. Since [tex]\rho(\lambda)[/tex] is always positive and A(a,λ)*A(b,λ)=±(1/4) that becomes [tex]\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[/tex], 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 [tex]\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) (1/4) [1 - 4A(b,\lambda)A(c,\lambda)][/tex] or [tex]\mid P(a,b) - P(a,c) \mid \le \int d\lambda \rho(\lambda) [(1/4) - A(b,\lambda)A(c,\lambda)][/tex]. Bell then notes that [tex]\int d\lambda \rho(\lambda) [-A(b,\lambda)A(c,\lambda)] = P(b,c)[/tex] and since [tex]\int d\lambda \rho(\lambda) = 1[/tex] 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.
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.Anamitra said: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)|
Anamitra said:But one may consider "hidden variables" to be responsible for the association between the two spin values.
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.Anamitra said:Response to Posting 45
I have not meant the expectation values of a single trial.
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.Anamitra said:In the revised formula I have meant the cumulative expectation values over the time intervals concerned.
JesseM said: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.
No, you don't get any increased flexibility, because changing the value of n also changes the values of P(a,b) and P(b,c) and P(a,c) (when they are defined as you define them, as expectation values for the sum of each result over n trials), in such a way as to make the inequality equally impossible to violate under local realism. As I said before if we denote your cumulative expectation values as PA(a,b), PA(b,c) and PA(a,c), while denoting Bell's single-trial expectation values as PB(a,b), PB(b,c) and PB(a,c), then we have:Anamitra said:With the consideration of the cumulative effect of measurement time we have
n+P(a,b)>=Mod[P(a,b)-P(a,c)]
P(a,b) ,P(a,c) and P(b,c) should have the interpretation as I have expounded in postings 47 and 50 . "n" is some positive number greater than one.
Now the violation shown in the example Bell's paper does not exist.Rather it can be avoided.[one may assign the value 2 or 3 to n]. We have got a certain amount of flexibility in the application of the inequality.
If the absolute value of each of them is smaller than 1 while n ≥ 3, then the inequality will be satisfied. But the point is that the inequality cannot be violated in local realism, whereas QM does violate this inequality (i.e. it violates your altered inequality above just like it violates Bell's original inequality). To disprove Bell's theorem you would have to find a local realistic model that allows for the inequality to be violated (under the test conditions Bell outlined), but Bell proved this is impossible.Anamitra said:Let us have a look at the following result:
n + PA(b,c) ≥ |PA(a,b) - PA(a,c)|
If each value like n*P(a,b)<1 [n*P(a,c) and n*P(b,c)<1]even if "'n" is a suitable positive number larger than one[it may be 2,3, 5.9 etc], the testing of Bell's inequalities with Quantum Mechanical results like P(a,b)=-a.b will not produce any contradiction. You may consider the example in Bell's paper.
Anamitra said:You may consider the example in Bell's paper.
The inequality
n + PA(b,c) ≥ |PA(a,b) - PA(a,c)|
is not violated by Quantum Mechanics by the values considered[a.c=0,a.b=b.c=1/[sqrt(2)],that is,P(a.c)=0;P(a.b)=P(b.c)= -1/[sqrt(2)] ]
This applies to local or non-local conditions.
But the inequality
n + PA(b,c) ≥ |PA(a,b) - PA(a,c)|
gets violated by the choice of values considered above.
Incidentally P(a.b)=-a.b is a quantum mechanical result.
The effect of the hidden variables may be operative in the local or non local situation
The action of the hidden variable may be of a complicated nature. It may not depend on distance. It may depend on the nature of the signals received/emitted by the particles and not on the strength of the signals. Even if there is a dependence on the strength of the signals it may be argued that each particle has to take the full share of the signals emitted by the other. We are dealing with a closed system[isolated one] containing only two particles.
As SpectraCat said, the values in Bell's paper are the single-trial expectation values, i.e. there you have PB(a.b) = PB(b.c) = -1/[sqrt(2)]. These are different from the sum-over-cumulative-trials expectation values, which in this case would be PA(a.b) = PA(b.c) = -n/[sqrt(2)]Anamitra said:You may consider the example in Bell's paper.
The inequality
n + PA(b,c) ≥ |PA(a,b) - PA(a,c)|
is not violated by Quantum Mechanics by the values considered[a.c=0,a.b=b.c=1/[sqrt(2)],that is,P(a.c)=0;P(a.b)=P(b.c)= -1/[sqrt(2)] ]
JesseM said:As SpectraCat said, the values in Bell's paper are the single-trial expectation values, i.e. there you have PB(a.b) = PB(b.c) = -1/[sqrt(2)]. These are different from the sum-over-cumulative-trials expectation values, which in this case would be PA(a.b) = PA(b.c) = -n/[sqrt(2)]
I wasn't measuring a single trial result when I gave PA(a.b) = PA(b.c) = -n/[sqrt(2)], I was talking about the expectation value for a sum of results over n trials, for example if I had n=4 trials my expectation value would be -4/[sqrt(2)], while if the actual trials gave results -1, -1, +1, -1 then my sum in this case would be -2.Anamitra said:This is not a correct assertion. You simply cannot measure a single trial result.
A "trial" is defined as a single recorded outcome, like a single "click" of the detector. It doesn't matter if at some hidden level unknown to us, the detector is really caused to click by a million brief interactions with a cloud of particles, it's still only a single trial if we only have one outcome. If you want the word "trial" to represent something else then this would once again change the meaning of P(a.b) and so forth, and there's no reason to expect the same inequality would be derived.Anamitra said:There may be one million trials in a measurement spanning across a short or an infinitesimally small interval of time.
That doesn't make much sense. What is a "trial of the distribution function" supposed to mean in physical terms? Physically the distribution function just tells you probability the hidden variables will take various values (each value of lambda represents a complete state of hidden variables), it's true these hidden variables could be rapidly changing during the measurement period, but in his more carefully worded paper La nouvelle cuisine he defined lambda to give the values of the hidden variables in at every point in space time in some complete cross-section of the past light cone of the region of spacetime where the measurement happened, like region "3" in the diagram at the top of this page. So in this case lambda isn't even meant to tell you the value of any hidden variables during the measurement period itself.Anamitra said:The Quantum Mechanical Expectation[which may be obtained in a physical experiment] covers such a span of time which may include one million trials of the distribution function rho[lambda]
JesseM said:I wasn't measuring a single trial result when I gave PA(a.b) = PA(b.c) = -n/[sqrt(2)], I was talking about the expectation value for a sum of results over n trials, for example if I had n=4 trials my expectation value would be -4/[sqrt(2)], while if the actual trials gave results -1, -1, +1, -1 then my sum in this case would be -2.
A "trial" is defined as a single recorded outcome, like a single "click" of the detector. It doesn't matter if at some hidden level unknown to us, the detector is really caused to click by a million brief interactions with a cloud of particles, it's still only a single trial if we only have one outcome.
JesseM said:That doesn't make much sense. What is a "trial of the distribution function" supposed to mean in physical terms? Physically the distribution function just tells you probability the hidden variables will take various values (each value of lambda represents a complete state of hidden variables), it's true these hidden variables could be rapidly changing during the measurement period, but in his more carefully worded paper La nouvelle cuisine he defined lambda to give the values of the hidden variables in at every point in space time in some complete cross-section of the past light cone of the region of spacetime where the measurement happened, like region "3" in the diagram at the top of this page. So in this case lambda isn't even meant to tell you the value of any hidden variables during the measurement period itself.
Then you can no longer say that your PA(a.b) is equal to n*PB(b.c), since Bell's PB(b.c) is the result of a single recorded outcome. If you redefine "trial" then you must redefine the meaning of the "expectation value", and there is no reason to expect the same inequality would still apply, since that inequality was derived under the assumption we were talking about a trial as a recorded outcome.Anamitra said:I have not used the word trial with the meaning of a single recorded outcome.
Your language is completely incomprehensible. How can a "distribution function" have multiple "impacts"? Have you ever heard someone say "Ouch! I've just been impacted in the head by a distribution function"? You need to explain your ideas in more physical terms, the distribution function is just mean to give the probability that lambda will take various values, where each value of lambda represents the state of some hidden variables (in Bell's argument it gives the value of these variables in a cross-section of the past light cone of a region of spacetime where a single measurement was performed).Anamitra said:It is simply the impact[or the influence] of the distribution function----and we are considering several such impacts---to get a single recorded outcome.
That doesn't tell me what the distribution function is mean to represent a distribution of, in physical terms! The distribution is assigning probabilities, yes? So what is do you think it is assigning probabilities to, in physical terms?Anamitra said:What is a "trial of the distribution function" supposed to mean in physical terms?
We are considering the same normalized distribution function to be valid for each instant of time in the measuring interval.One may consider different distribution functions[normalized ones] for different instants.
JesseM said:Then you can no longer say that your PA(a.b) is equal to n*PB(b.c), since Bell's PB(b.c) is the result of a single recorded outcome. If you redefine "trial" then you must redefine the meaning of the "expectation value", and there is no reason to expect the same inequality would still apply, since that inequality was derived under the assumption we were talking about a trial as a recorded outcome.
JesseM said:Your language is completely incomprehensible. How can a "distribution function" have multiple "impacts"? Have you ever heard someone say "Ouch! I've just been impacted in the head by a distribution function"? You need to explain your ideas in more physical terms, the distribution function is just mean to give the probability that lambda will take various values, where each value of lambda represents the state of some hidden variables (in Bell's argument it gives the value of these variables in a cross-section of the past light cone of a region of spacetime where a single measurement was performed).
Expectation value of what physical quantity, if not an observed "click"?Anamitra said:The expectation value for each moment/instant may be denoted by P(a,b).
What is the physical meaning of "the variable lambda" in your mind, if it doesn't have the same meaning that Bell assigns to itAnamitra said:The manner in which the variable lambda
Sure, but so what? In Bell's terminology lambda does not represent the value of any hidden variables at a single time coordinate in the measurement region. Rather a single value of lambda tells you the value of all hidden variables at all points in spacetime in another region that's in the past light cone of the measurement, like region 3 in the diagram. Region 3 is not the measurement region, that's region 1 in the diagram. Of course region 3 lasts an extended period of time in Bell's diagram too (though he could have made it just a single instantaneous spacelike cross-section of the past light cone), but that doesn't mean lambda is changing because lambda does not represent the value of the hidden variables at a single instant of time, rather a single value of lambda represents the values of the hidden variables at every point in region 3.Anamitra said:You have talked of a region of spacetime where a single measurement is made. Such a region can have millions of time coordinates.
Anamitra said:Bell's treatment/formulation is of a general type, intended to cover all possible situations concerning the hidden variable.
It is quite interesting to observe the valiant attempt of the Scientific Adviser to contradict this basic general nature of the paper.I have reasons to thank him--and he would find it very difficult to understand this.
Not true, if you define lambda in arbitrary ways then you may not have a basis for claiming that the result A at one detector can be deduced in a deterministic way from only the detector setting a and lambda, in other words you may need to use a probabilistic function P(A|a,lambda) rather than a deterministic function A(a,lambda), and in fact Bell does use a probabilistic function in most of his later papers. But even if you use a probabilistic function, to derive a Bell inequality you still need a step like the on on p. 15 of this paper where you say P(A,B|a,b,lambda)=P(A|a,lambda)*P(B|b,lambda) which depends on the assumption that lambda "screens off" any statistical correlation between the result/setting A/a and the result/setting B/b due to influences from the region where the past light cones of both measurements overlap (because of the possibility of such influences, you could not say that that P(A,B|a,b)=P(A|a)*P(B|b), for example). If you don't make some assumption like treating lambda as telling you all hidden variables in a cross-section of the past light cone of the measurements this step may not be justifiable. And of course here we are defining "A" and "B" as the observable measurement outcomes, whereas you seem to be defining them differently yet you refuse to actually explain what physical quantity you are calculating an "expectation value" for if not the observable measurement outcome. In this case there is obviously no justification for either the claim that this "expectation value" is equal to an integral involving deterministic functions (A|a,lambda) and (B|b,lambda), or the probabilistic claim that P(A,B|a,b,lambda)=P(A|a,lambda)*P(B|b,lambda). Neither of these steps is justified on the basis of pure probability theory, they both depend on physical assumptions about the physical meaning of expectation values P, so if you change the meaning you can't justify these steps unless you provide a clear definition of what physical quantity you are computing an expectation value for.Anamitra said:Bell's treatment/formulation is of a general type, intended to cover all possible situations concerning the hidden variable.
Certainly the paper is fairly general, but you only show your lack of comprehension if you think it's so general that you don't have to worry at all about the physical meaning of various terms like lambda and the expectation values P(a,b) etc. The paper involves multiple steps that can't be justified on the basis of abstract mathematics alone, you can't arbitrarily change the physical meaning of the symbols and expect it to still work.Anamitra said:It is quite interesting to observe the valiant attempt of the Scientific Adviser to contradict this basic general nature of the paper.
JesseM said:Not true, if you define lambda in arbitrary ways then you may not have a basis for claiming that the result A at one detector can be deduced in a deterministic way from only the detector setting a and lambda, in other words you may need to use a probabilistic function P(A|a,lambda) rather than a deterministic function A(a,lambda), and in fact Bell does use a probabilistic function in most of his later papers.