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So to break the axial part of [itex] SU(2)_L \times SU(2)_R [/itex] we use the composite field [itex] \chi_{\alpha i} \xi^{\alpha \bar{j}} [/itex] (since this is Lorentz scalar, colour singlet, and has the right transformation under the flavour symmetries, only breaking the axial part etc). We assume that [itex] \langle 0| \chi_{\alpha i} \xi^{\alpha \bar{j}} |0\rangle=-v^3 \delta^{\bar{j}}_i [/itex]. (I understand roughly the mathematics of this operator does the job of breaking axial but not vector generators, but from my reading of symmetry breaking I don't really understand how we just *choose* a field in some arbitrary way like this; I thought the field had to be field involved in your Lagrangian and then you would find a continuous family of ground states of it etc)

Then my text talks about constructing a low energy effective lagrangian for the three pions (psuedo goldstone bosons) by letting [itex] |U\rangle [/itex] be a low energy state for which the expectation value of [itex] \chi_{\alpha i} \xi^{\alpha \bar{j}} [/itex] varies slowly in flavour space as a function of spacetime:

[tex] \langle U|\chi_{\alpha i} \xi^{\alpha \bar{j}}|U\rangle=-v^3 U^{\bar{j}}_i(x) [/tex]

with U(x) a spacetime dependent unitary matrix that can be written [itex] U(x)=exp\left[2i\pi^a(x)T^a/f_\pi\right] [/itex] ([itex]\pi[/itex] are pion fields, T generators, f is pion decay const)

Then why do we think of U(x) as an effective low energy field? why does specifiying it's Lagrangian to be most gen consistent with symmetries, say anything about the pions? what is really going on here? I don't see why we are forming a Lagrangian from this object that was just on the RHS of an expectation value of this composite field.

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# Chiral sym breaking questions

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