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Could you please rigorously define Bayesian probability in mathematical formula? What is the main differences between Bayesian probability and quantum probability theories?

I have seen in literature like https://arxiv.org/pdf/1601.02480.pdf that quantum probability theory does not assume probabilistic communitivity and distributivity:

$$A\cap B \neq B\cap A$$

High GPA
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  • The first question may require several book chapters (e.g. Chow and Teicher's Probability Theory for rigorous measure theoretic definitions). – Björn Jun 17 '17 at 07:15
  • Related https://stats.stackexchange.com/q/173056/35989 – Tim Jun 17 '17 at 07:28
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    The title and the question don't match and so it is unclear what the question is. Basic principles of Bayesian statistics and/or Bayesian interpretation of probability are (Bayesian-)probably found in many existing question/answers. If "rigorous" here means measure-theoretic foundations (such as issues regarding conditioning on zero-measure sets etc.), this would be a question on its own. But then this question actually seems to be after differences between quantum probability and Bayesian probability. Thus I voted to close as unclear (could be "too broad" also). – Juho Kokkala Jun 17 '17 at 12:52

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I'll try to explain it rather in simple terms. Understanding things for real needs to get into the basics of QM rather seriously. Count a few years :-)

There is a loose connection (and only loose) between QM and probabilities. For example, the momentum $p$ (or speed if you prefer) of a particle is not assumed to have a precise value. It is assumed to have a probability distribution. Thus it could be seen as a random variable. Same for the position $x$. But if QM obeyed classical probability theory then the pair $(p,x)$ would be a random variable as well and would have a joint distribution. It doesn't. This is the Heisenberg principle. See https://en.wikipedia.org/wiki/Wigner_quasiprobability_distribution

In a QM experiment, classical probabilities do not apply until the decoherence happens. Only events post decoherence can be seen with normal probability theory. A framework made it formally clearer : https://en.wikipedia.org/wiki/Consistent_histories

In Bell's experiment(s), one of the key moments when QM was understood to not be just classical probabilities for sure, the Bayesian question was natural. Can the result of the experiment be seen as an illusion caused by a Bayesian effect: probabilities of measurements on Bob side would be impacted by the measurements on Alice side only apparently : would it be just a conditioning ? In a way it's what Einstein believed (he formalized it as "hidden variable"). Bell proved this was impossible no matter how you model the microscopic reality. See https://en.wikipedia.org/wiki/Bell%27s_theorem

Unless a genius comes along and says "you fools haven't seen that...", QM is strongly disbelieved to be just probabilities. There is a way to use probabilities a bit with it, but fundamentally, the good formalization is Hilbert spaces with observables being self-adjoint operators and not probability spaces with observables being random variables. These two formalizations are incompatible and the Hilbert one has been verified experimentally... just as much as you can imagine. Because everyone (including theoretical geniuses, brilliant experimentalists and curious amateurs like me), is puzzled by the paradoxes of QM and everyone has doubts this can be real : just an illusion ?

In a word QM is not (really) probabilities. Bayes is just a small theorem in probabilities and of course it does not change the problem. It's just that we... naively believe this could be it sometimes.

Benoit Sanchez
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    Some parts in this answer seem to be trying to say that QM is just a different kind of probability theory, not classical. For exmaple _"if QM obeyed classical probability theory then the pair $(p,x)$ would be a random variable as well and would have a joint distribution"_. But the Heisenberg uncertainty principle is orthogonal to whether QM "obeys classical probability theory" or not. It's just that QM says that the pair $(p,x)$ _is not a thing that can have a value_. – JiK Jun 17 '17 at 13:05
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    And this answers seems to be completely irrelevant to any question that was asked. This answer doesn't talk about formalism of either Bayesian probability or "quantum probability". The latter is a mathematical formalism, but is not required to understand or study things like Heisenberg uncertainty princpile or decoherence or Bell's theorem, and the question didn't ask about it's applications. – JiK Jun 17 '17 at 13:11