Role of chiral symmetry in QCD vacuum

Let us consider the low-energy limit of QCD with 2 massless quarks (u,d). In terms of the left-handed and the right-handed quark fields, the lagrangian can be written as:

With m=0, the following symmetry transformation holds:

where V_i &isin U(2); i=L,R since the gluon interactions do not change the helicity of quarks. Then, according to Noether's theorem, each continuous symmetry parameter has a corresponding conserved current, there will be 8 conserved currents. Consider six of these currents:

(The other two are U(1)_A and U(1)_B), out of which the former is anomalous. When m_q=0, Q^a_V,A will commute with H_QCD:

This means that for a QCD eigenstate, H_QCD|&Psi&rang = E |&Psi&rang, the states Q_V^a |&Psi&rang and Q_A^a |&Psi&rang would be degenerate. But such parity doubling does not occur in Nature! The symmetry of the Hamiltonian is not respected in nature and one of the flavour SU(2)is spontaneously broken. As evidence for that, there are Goldstone bosons, three in number: &pi⁰; &pi^∓
(Here is the Goldstone theorem: For a generic continuous symmetry that is spontaneously broken, (i.e. the currents are conserved) the action of the charge on the ground state does not leave it invariant. New massless particles appear in the spectrum for each symmetry generator that is broken.)
The vacuum condenses by producing a non-zero chiral condensate. It turns out that a low energy effective theory (Chiral perturbation theory) can be formulated to study the low energy physics.

Casher's argument: There is an argument due to Casher, which states that in the confining regime, chiral symmetry is necessarily broken. First, lets recall the definition of helicity: it is the projection of the spin onto the direction of momentum. With an equation:

Consider a quark with three momentum p moving along the z-axis. Its massless, so helicity is the same as chirality. For simplicity, choose p ∥ s. Confinement implies that the quark must turn back. Now, if chiral symmetry is un-broken, then the spin must also be flipped, so: &Delta s _z = -1. The angular momentum needs to be compensated somehow! But QCD vacuum cannot support this angular momentum conservation; it has no L_z.
Therefore, we can naively conclude that if the chiral symmetry is unbroken, the quark cannot turn back and there is no confinement. On the other hand, if the quark does turn back that at the point of turning back, the helicity (which is the same as chirality in the massless limit) must flip and hence a chiral symmetry breaking term must appear in the quark Green function. But quarks are massless, so this term must arise dynamically. This is called the dynamic chiral symmetry breaking of the QCD vacuum.