Reply to Jakulin:Can quantum probabilities be always reduced to joint
In your last Email you discussed another question which of the
fundamental value for comparasion of CI and QI, or that is equivqlent
CP and QP.
> Thus, the joint probability mass P(A,B) is constant. However,
> is not the same as P(B) if A and B are not independent (ie. are
> My suggestion would be to reconsider the wave function and to
> the \"weight\" corresponding to each Hilbert space base vector as a
> classical probability. But, I might be missing something. If I am,
> is the difference between quantum probabilities and joint probability
> distributions? Can quantum probabilities do something that joint
> probability distributions can\'t?
The main mathematical point in Bells arguments (if we forget for a
moment all intriguing stories about quantum nonlocality) is that if one
assume that the joint probability distribution P(A,B,C) for three
different experimental settings exists, then one will get the Bells
inequality and consequently the contradiction with the experiment
(see my book <<INterpretations of Probability>>). But of course the
joint probability distributions P(A,B), P(A,C), and
P(C,B) are well defined since observables on different particles
belonging to the same pair of entangled particles are compatible
(corresponding operators commute). This was the essence of the
EPR-trick: for one particle they are incompatible and Bohr and
Heisenberg could speak as long as they like about irredicible
disturbances. And of course, by using P(A,B) you can always define
P(B|A=a) with the aid of the Bayes formula (I recall for one fixed
experimental arrangement you can always use the classical probability).
Is the process of quantum
> the same as the operation of taking the conditional distribution?
Yes, it is correct, but you could not assume that there exists one fixed
probability measure for incompatible experimental settings.
> Agreed, this relates to the problem of defining what is an event in
> classical probability. In that respect, the definition of an event
> depends on the time window within which we interpret two detections
> relating to the same event. This is how we have chosen to
> the world.
Yes, it is a good formulation that teh definition of an event (and thus
a Kolmogorov model) depends on time window. Thanks! We shall use this
point of view in future.
> I would be interested in how to express A,B,C and D in terms of
> properties. Thus, if we have detectors A and B, and o(A) and o(B) are
> their orientations, I would be interested in the probabilistic model
> P(A,B|o(A),o(B)) or the joint probability distribution of clicks in A
> and B given the orientations of A and B.
This is the simple question until you are speaking only about two
orientations, here you can write
P(A=+1,B=+1|o(A),o(B))= 1/2 cos^2 (a-b)/2, where a,b are angles
determining orientations A and B; P(A=-1,B=+1|o(A),o(B))=
1/2 sin^2 (a-b)/2, and so on. We tokk this answer from QM. But you would
not be able to find a single probability distribution for A,B,C.
Therefore I speak about contextual probability theory.
With Best Regards,
Director of International Center for Mathematical Modeling in Physics,
Engineering, Economy and Cognitive Sc.,
University of Vaxjo, Sweden
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