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A State Space and Probability Theory

  1. Jun 15, 2017 #1

    bhobba

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    Hi All

    This is in relation to the folllowing paper:
    https://arxiv.org/pdf/1402.6562.pdf

    See section 3 on examples where standard probability theory is discussed. Is it valid? To me its rather obvious but I had had a retired professor of probability say probability theory doesn't have a state space. This has me totality flummoxed. Is he right and if so what am I missing?

    Thanks
    Bill
     
  2. jcsd
  3. Jun 15, 2017 #2

    Stephen Tashi

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    Mathematical probability theory defines a "probability space" and does not define a "state space".

    What the article calls "Classical probability" is not an exposition of mathematical probability theory. Instead it is an application of mathematical probability theory to "state space" as defined in physics - presumably as defined in "classical" physics.
     
  4. Jun 15, 2017 #3

    bhobba

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    Yes - agreed. But its a perfectly legit space

    Its easy to define. If your outcomes are definite you simply define each as a vector with 1 in the outcome. They are by definition the pure states. Then you create all the convex sums of them. That forms a space. The sums are by definition called mixed states and of course the 'weight' in that sum can easily be interpreted as the probability of that pure state. Or have I made a mistake?

    Its purpose is to bring out the difference to QM. There states are positive operators of trace 1. Pure states are defined to be of the form |u><u|. You can also form mixed states from them with the same interpretation. The difference is the pure states are not definite - there is no way to tell what a quantum pure state is by observation.

    This is all part of a class of theories based on probability but generalize it in all sorts of ways

    Thanks
    Bill
     
    Last edited: Jun 15, 2017
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