Chiral Symmetry Breaking

1. Feb 18, 2006

Neitrino

Dear PF,

Could anyone help me to understand the following issue:

I dont understand the formation of Goldston bosons in chiral symmmetry breaking. Namely - suppose there is only fermion fields Lagrangian and due to some reasons the fermion field obteins VEV - <0|psi^bar psi|0> is non zero. And left/right chiral symmetry is reduced to its diagonal group.
I dont understant how Goldstone bosons are formed - I mean as folows"
In Scalar field case if we have multiple fields phi1, phi2, phi3.... and when
this scalalar field develops some VEV - v, then for convience we can put only phi1 equals to this VEV -v, rest fields we can put zero...we will say this VEV of field phi1 is v, VEVs of rest field are zero: <0|phi1|0>=v,
<0|phi2|0>=0 etc. We say we have phi1 massive scalar field and rest Goldsone Bosons from Broken Symmetry.

So I dont understand what happens when suppose fermion fields develop VEV.

Thanks
George

2. Feb 18, 2006

Physics Monkey

The story goes as follows, the vacuum expectation value of certain composite fields break a continuous symmetry of the Lagrangian and Goldstone bosons emerge for each of the unbroken generators. I suspect this is what you have in mind anyway, so let me talk specifically about chiral QCD with the light quarks (u, d, s). The Lagrangian for the quark part of the theory is $$\mathcal{L} = \bar{Q}_R i \gamma^\mu D_\mu Q_R + \bar{Q}_L i \gamma^\mu D_\mu Q_L$$. Clearly the whole thing is invariant under the chiral transformation $$SU(3)_L \times SU(3)_R$$ because we've left out the quark mass terms. Now, when you look at composite operators like $$\bar{Q}^a_R Q^b_L$$ (a,b are flavor indices) you find that these operators acquire a vacuum expectation. In this case the vev originates from the non-perturbative part of QCD. The full chiral symmetry is broken down to the subgroup $$SU(3)_V$$ just as you said, and the vacuum expectation value can be written like $$\langle \bar{Q}^a_R Q^b_L \rangle = v \delta^{a b}$$. In order to describe fluctuations of the operator \bar{Q}^a_R Q^b_L one introduces another field $$\Phi$$ which is an $$SU(3)$$ matrix defined via $$\bar{Q}_R Q_L \rightarrow v \Phi$$. Because it is a member of $$SU(3)$$ it can be written in terms the exponential of (i times) a traceless hermitian matrix. Such a hermitian matrix has 8 independent components, one for each of the broken generators, and each one corresponds to one of the massless Goldstone bosons. These fields are your pions, your kaons, and the eta, and they represent the low energy excitations of the system. That's how the Goldstone modes appear in the theory. They can be given mass by including the effects of quark masses to leading order, and then they are referred to as pseudo-Goldstone bosons because the chiral symmetry which would be spontaneuously broken by the vev is explicitly broken by the quark masses.

Hope this helps.

Last edited: Feb 18, 2006

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