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Symmetry breaking and the ground state of a quantum field theory.

  1. Jan 5, 2012 #1
    when a continuous symmetry is broken, we say that the ground state is just one of the possible ground states, and there is no energy cost in moving from one to the other..
    why doesn't the state keep changing with the slightest perturbation (production of goldstone boson).
    why don't we have a different ground state every time we do an experiment.
    and what if we had different possible ground states that are not symmetric (but equal in energy),, then every time we have a phase transition we could get a field with a different mass.

    sorry for the question being unclear, I barely understand this.
  2. jcsd
  3. Jan 5, 2012 #2


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  4. Jan 10, 2012 #3


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    To begin with I think one should differentiate whether the symmetry broken is a "true" symmetry or merely a gauge symmetry.
    I the latter case, there is only but one ground state.
    In the first case (e.g. broken rotational symmetry in a magnet or broken translational symmetry in a crystal) there are different ground states which owe their stability to the ground state being broken on a macroscopic scale, while all (or most ) of the experiments we are doing only act locally.
  5. Jan 12, 2012 #4
    Different vacuum expectation values (vevs) may disrupt the vacuum stability. One can not choose arbitrary or multiple vevs. that's the reason why some theories (ex:Technicolor) have been discredited. Furthermore, it may effect the naturalness and triviality of your theory.
    Last edited: Jan 12, 2012
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