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Symmetry Breaking

  1. Mar 6, 2007 #1
    I have done classical symmetry breaking and now want to understand the quantum one. I have seen the statement that the symmetry is broken if and only if Q|0> not 0. Where |0> is the vacuum and Q is the associated charge of the broken symmetry. Why does this imply symmetry breaking? The way I know it a symmetry group G is broken to a subgroup H if the theory is invariant under G whereas the vacuum is invariant only under H, meaning h|0>=|0> for any h element H and g|0> not equal |0> for any g not in H. Is this the right definition? And if so does this imply Q|0> not 0 in the case of a broken symmetry?
  2. jcsd
  3. Mar 6, 2007 #2
    I think I figured it out myself:
    Any symmetry transformation can be represented by either a unitary operator U or an antiunitary operator A acting on the space of states, i.e. U|phi> is the transformed state(for the case of a unitary operator) now if I am not mistaken we can write U = exp(i*x*Q) (x being a parameter here and not spacetime position) where Q is the associated charge of the symmetry. Then for the symmetry to be mainfest we require U|0>=|0> which is equivalent to saying Q|0>=0 when expanding the exponential. A question I have about this is: Does this also hold for symetries being represented by antihermitian operators, i.e. if the states transform under the symmetry as A|phi> can I still write A=exp(i*x*Q) where Q is the conserved charge? In this case the charge would then be antihermitian instead of hermitian, I guess.
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