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Symmetry breaking/ degrees of freedom

  1. May 9, 2012 #1
    Dear PF...

    Please help me with basic question more I think more I get confused...

    In O(n) space there are n(n-1)/2 generators...
    suppose I have symmetric tensor in O(n) space, it will have n(n+1)/2 independent components... and i am building invariant potential from it (quartic polynomial expression), then I will find which components minimize my potential and will have symmetry breaking.

    Broken generators will transform into Goldstone boson.
    So initially i have n(n-1)/2 goldstonic degrees of freedom.

    But before i minimize my potential i want to diagonalize my tensor...
    Does diagonalization consume goldstonic degrees of freedom/consumes generators?

    After diagonalisation shall i have again n(n-1)/2 goldstonic degrees of freedom in my bydget or less ?

    thank you
  2. jcsd
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