 By Koch V.

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Extra info for Aspects of chiral symmetry

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E. that chiral symmetry is explicitly broken. g. ref. ). 4 fm, and we do not expect domains of large size to be created. (a) (b) Figure 10: System in restored phase (a) and in quench scenario (b). Domains of considerable size, however, can be formed by a dynamical process. Assume chiral symmetry is restored initially. Thus, the expectation values of the σ as well as pion fields are zero and the effective Hamiltonian has the shape depicted in figure 10 (a) (for a 20 This will only work, however, if the pions from the DCC do not reinteract with the thermal pions surrounding them.

In this approach, the problem is shifted to the determination of an unknown off-shell parameter appearing in the Delta-propagator. 30 the sigma-direction (see figure below ). This confines the dynamics to the circle, defined by the minimum of the potential. σ 2 + π 2 = fπ2 V(σ, π=0) (139) V(σ, π=0) fπ σ fπ σ This additional condition removes one degree of freedom, which close to the ground state, where < σ >= fπ , is the sigma field, and we are left with pionic excitations only. Because of the above constraint (139), the dynamics is now restricted to rotation on the so called chiral circle (actually it is a sphere).

There are several answers on the market. Georgi  argues, based on renormalization arguments, that the scale should be 4πfπ ∼ 1 GeV, whereas others argue [19, 17], that the mass of the lowest lying resonance should give the scale, since this is the energy, where the entire game seizes to work. This seems to be a reasonable argument and, assuming that there is no σ-meson of mass ∼ 500 MeV, the mass of the ρ-meson should provide a reasonable benchmark. e. assuming that the pion mass vanishes. The explicit breaking of chiral symmetry is introduced by terms of the form ∼ tr(U + + U ) and and the simplest symmetry breaking is δLXχSB = fπ2 m2π tr(U + + U ) 4 1 4 − m2π Φ2 + O(Φ4 ) 2 (163) which to leading order in the pion-fields corresponds to a pion mass-term (the constant term does not contribute to the dynamics).