1.Have systems involving say spin 1/2 particles even been implemented which verify or would exhibit the contextual characteristics of the three box paradox and or spectors parable of the seer, and other odd (very/oddly) strongly contextual, systems.
2.Although I am not sure here, can speckers scenario be realized or is this an example of what quantum theory disallows?
Is it supposed to be characteristic of a spin 1 particle with three outcome and dependent on which path you place the detector; or rather characteristic of three identically prepared spin 1/2 particles which are prepared and measured so that the probability of spin up in any direction for any individual particle is 1/3. I presume that its not a case of standard contextuality, even the bases of the three boxes or particles are different.
I presume that the total state is the same (not outcome state) but pre-measurement state, of the combined system regardless of which box you look in. So in the case of particles, for them to remain entangled in that basis it would still have to be that all three are premeasured and become entanged along that same basis, regardless of which box you look in; so that what does looking in one box rather then another correspond to; inspecting the outcome detection for any two of the particles versus inspecting the outcome detection of some other two of the particles;
I presume the other box/particle is still premeasured or becomes entangled along the same basis, its just that one does go to look and see what outcome it exhibit; or is that the point of the whole thing that the third particle is not premeasured at all, and so it cannot be said to be in any definite three way entangled state, along some definite three basis although all three are nontheless maximally entangled. is it as if it converts the two particles which should have 1/3 probability for spin up into a singlet state of just those two particles with probability one for spin up; as if as it were, spin up has to show up somewhere, and only that which is measured exists, and as that (the third) box is not measured, it transmits its probability mass onto the other two particles that are measured, thus ensuring that outcome still occurs, else it would not occur at all (as it would occur in an unmeasured experiment) or a case where the more probable event has to occur (probability of two disjuncts) are greater then the probability of the presumably equi-probable non measured dis-junct,
Would this be a case of frequency steering without violating the probabilities; because if you repeated say the spekker scenario over and over again, by measuring always the same two boxes, the probability of the disjunction of finding the outcome in one of the two boxes so measured would be 2/3 not one; and or the probability of any given box in that set of two boxes is now half or relative frequency so obtained is necessarily 1/2 not 1 or do these oscillate in such a way that they are not lebesque measurable or rather do no show individual frequencies (which would satisfy the weak law of large numbers but not the strong law)and as there is no quantum state for the di-sjunction of the two as such, its not much a failure of the law of additivity as such but that its not a borel field; or rather for the frequentist, the probabilities for each disjunct are not well defined so perhaps the disjunctions probability is not well defined (or rather is, is as its just a well defined frequency but no failure of additivity, as the disjuncts are not defined), the kolmogorovian they might be able to rely on the weak law of large numbersand then say that its not a field, so that the disjunction's probability is not defined but non measurable, and this in some sense, there is no counter exampley to additivty (if non measurable sets are compatible with necessary converegence)
The the dis-junction does not have well defined probability value; which is compatible with it 'necessarily' having a relative frequency of 1 or likewise if we can simply change the scenario so that long run frequency of it being in either of two boxes is 'necessarily' 3/4 instead this is not compatible with any probability value including the scenario where the disjunction should sum to 3/4 or 2/3).
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his is because unless one is a frequentist, for no standard probability measure is possible, for IID variables; that is no relative frequency occurs as matter of necessity; including one that matches the probability value, and no probability value is consistent with it being impossible for frequencies to diverge, or failing to 'converge'(in fact a requirement, if its probability would have summed up to 2/3 other than 1, which requires that an infinite limiting relative frequency of 2/3 should be at least possible). The only difference between this alternative and the standard case, is that the with a probability of absolute 1, (not just almost sure) but absolute certainty, generally requires that assignment to be the probability of the corresponding proposition; that is its not only compatible but requires it, in cases of absolute necessary convergence to relative frequency 1 .
I presume that lebesque non-measurable events cant say this about frequencies but I presume they are compatible with anything as you cannot say anything about them. Its a bit out of my depth, however, although its unclear whether they are compatible with modal propositions such as with certainty relative frequency is 1.
Although one is not inferring this from the non-measurable set, but the other way around. I can see them being compatible with 'necessary' convergence to any other relative frequency value; that is, other then 1, as no other relative frequency occurs with necessity as a result of any probability value, and thus cannot be a probability or a probability function at all. when IID;
Thus, in those cases, it must be non-measurable;
but necessary modal frequency values are compatible with said t aforementioned probability value of absolute 1, and thus are compatible with a probability value which expresses necessary convergence. But does it necessitate that there exists a measurable probability value, and thus one which takes value 1; are non-measurable sets compatible with this.
Otherwise if necessary convergence does require it to be have a well defined probability value of 1, one so that might be able to say that additivity does fail, That is, in that one cannot say its a mere case of it not being a field and that the disjunction probability value is not defined, and take the back door option of saying there does not exist is well defined probability value of the disjunction wherein additivity fails or where value of the disjunction is something other then probability value of the disjuncts' as there not exist any value and thus no there exists no value which exhbits this difference. If that is how you define additivity at least; as there exist no probability value of the disjunction at all and thus there does not exist any value such that it takes a value something which violates additivity.
Thus, any absolute convergence to any value, from the non-measurable set (as opposed to the relative frequency just being one) is compatible with in some sense with the probability axioms if they are expressed as only holding when events are lebesgue measurable ( i note that pitowski gave a case of non-measurable sets, but I do not know if the probability axioms themselves rule these out, as opposed to the extra conditions in the model theory, unless it due to these necessary convergence cases which requires it to take a measurable value of one, yet its a non measurable set). Maybe non standard analysis would help here? ie; defining away all necessary convergence and defining all certainties as 1- (very small less then the inverse of uncountable) infinitesimal number or by saying that the disjunction does not necessarily converge, its has by saying it almost surely has a relative frequency of absolute one; ie that almost all possible ensembles do not have a single failure, but in this case convergence is necessary thus it must be non-measurable; and then try to escape the issue that necessary events must have a well defined probability insofar that it the relative frequency of the disjunction is not necessary but just 'very likel'y as is compatible perhaps with non-measurable sets (without contradicting oneself; one would have to say perhaps in the first horn that disjunction likewas not necessary and so the reason it must be non measurable is because no no hyperfinite probability function allows for convergence to occur with that degree of certainty, depending on the type of infinitesimal used) or that is has a probability of absolute one, but without depending on frequentism one can say that the disjuncts dont have defined probability values, as some nonstandard models i think suggest that convergence is necessary so that any possibility of divergence is incompatible with the disjuncts being a probability function (instead of using the weak law of large numbers to say that the probability values could be defined)
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