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Why is goldstone's theorem incorrect in gauge theories?

  1. Feb 7, 2014 #1
    Hello, I am currently studying spontaneous symmetry breaking in qft. Several textbooks I've read prove Goldstone's theorem under supposing that
    1) There exists a continuous global symmetry under which the Lagrangian is invariant.
    2) The vacuum state is not annihilated by the conserved charge(or, alternatively, a field has a non-zero vacuum expectation).

    Later it is said that theories with a gauge symmetry do not satisfy these hypothesis and so the goldstone theorem is invalid. In fact, a massive boson appears and not a massless one.
    My question is how does a gauge symmetry violate the two hypothesis. Since it is a local symmetry, it also contains the global symmetry(the transformation is independent of spacetime) and so it should have the same conserved currents and charges.
    I am guessing this is why Higgs won the nobel prize xD
    Thank you
  2. jcsd
  3. Feb 7, 2014 #2


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    These are the necessary conditions for spontaneous semmetry breaking.

    Which book says that? This is incorrect. Without those conditions the symmetry does not get hidden and the massless gauge (vector) bosons stay massless. However, local gauge invariance allows us the freedom to gauge away "the would be massless Goldstone (scalar) bosons" by simply redefining the fields in the theory by making a clever gauge transformations.
    Have you not heard Sidney Coleman famous saying :The gauge fields have "eaten up" the masselss Goldstone's bosons and become massive. The scalar degrees of freedom become the longitudinal polarization of the vector gauge bosons.
    They don't.
    And they do have the same currents and charges.

  4. Feb 7, 2014 #3


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    I think in general there are many more conditions for Goldstones theorem. Physically the most important one is that of the hamiltonian being sufficiently local. For example the BCS model of superconductivity does not contain Goldstone bosons because the reduced hamiltonian considered by BCS is too non-local. That was quite a lucky coincidence as a true superconductor also has no Goldstone bosen. However in the latter situation this is due to the Anderson Higgs mechanism.
    An interesting read on that topic is the book "Symmetry breaking" by Franco Strocchi.
  5. Feb 8, 2014 #4
    Ok, I think I understand. Thank you!
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