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Matter + Antimatter Bound State Mathematics

  1. Apr 11, 2009 #1
    Well, it has been ~ four years ago now I request help with this question in another thread, long dead, so I thought I would bring it to forum again in updated form:

    So, my question is:

    Does anyone know the mathematics that would explain the quantum dynamics of how a matter helium-3 isotope with neutron picture [PNP] and 9-quark bag picture [[uud-ddu-uud] could form a "stable bound state" (WITH NO ANNIHILATION--that is, all 15 quarks remain in quantum superposition) with antimatter antideuteron with pictures, where I use ^=antimatter, [N^P^] and 6-quark bag [d^d^u^-u^u^d^] ?

    So, at quark level this interaction:

    [uud-ddu-uud] + [d^d^u^-u^u^d^]

    My prediction is that quantum mathematics solution will yield what is known as the proton, with up & down valance quarks, and up & down matter and antimatter quarks coexisting within the proton sea. So, I look for my hypothesis to be mathematically falsified. Perhaps good homework project for advanced quantum theory class if any professors here.

    Please, do not reply that the matter and antimatter quarks will annihilate--that is the point of the problem--we assume there is no annihilation and work out the mathematics under this first assumption to see mathematical solution. Perhaps there is no quantum mechanics solution ?, that itself would I think be of interest. Perhaps I just ask goofy question--let me know that also.

    Any comments appreciated.
     
  2. jcsd
  3. Apr 11, 2009 #2
    The annihilation of the quarks is mathematically inevitable. In other words, there is no stable bound state containing those 15 quarks. That said, your prediction:

    "My prediction is that quantum mathematics solution will yield what is known as the proton, with up & down valance quarks, and up & down matter and antimatter quarks coexisting within the proton sea."

    Is basically correct, it's just that the 'proton sea' (the interior of the proton) actually consists of an enormous number of quarks and antiquarks --- when we say that a proton contains three quarks, we mean that (# of quarks) - (# of anti-quarks) = 3 at all times.
     
  4. Apr 11, 2009 #3
    Thanks Isabelle,

    Your reply raises a few questions:

    So, would it perhaps be more accurate to say....as predicted by current mathematical formalism of quantum theory, there is no stable bound state containing those 15 quarks...? In others words, perhaps some other mathematical theory could--we just have no idea what it might be ?

    Do you see any way to modify the current equations of quantum theory--perhaps change a (+) sign to a (-) somewhere, that would allow for a stable bound state of those 15 quarks ?
     
  5. Apr 11, 2009 #4

    Vanadium 50

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    I'm afraid that isn't how science works. You start with a theory - i.e. a mathematical model for how things behave - and see what the model predicts. If you can't do the calculation, you don't have a prediction.
     
  6. Apr 11, 2009 #5
    Excuse me--but "Science does not start with a Theory". Science ends with a Theory.

    A Theory is defined as "a well substantiated explanation of some aspects of the natural world that can incorporate facts, laws, inferences, and tested hypotheses" [National Academy of Sciences, 1998, Teaching about evolution and the nature of science, Washigton, DC, National Academy Press]

    First come the facts, laws, inferences, hypothesis, informed by mathematics--then is formed the theory.

    So, if you not able to derive a mathematical formalism to the OP hypothesis, fine, no problem.

    Since this forum allow cute comments to end posts, such as 2 + 2...here is mine:

    ___________________
    "Science is the belief in the ignorance of experts."
    (Richard Feynman, The Pleasure of Finding Things Out, 1999)
     
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