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Chiral sym breaking questions

  1. Jan 16, 2012 #1

    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.
  2. jcsd
  3. Jan 25, 2012 #2
    I'm studying this right now as well and had the same question. I think the answer has to do with the concept of "universality class" from statistical mechanics. Basically, I think the idea is that two Lagrangians that are invariant under the same symmetries will flow to the same fixed point in the IR under the renormalization group. In this case, the IR is pion physics.
  4. Jan 26, 2012 #3


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    U is a field that - when expanded in the pion fields - contains the free kinetic energy terms + higher interaction terms which are - due to power counting - non-renormalizable. You can see that by looking at power counting of [itex]\varphi^n[/itex] theories which are 'generated' by expanding U.
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