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The QCD lagrangian with two massless flavours of quarks takes the form $$\mathcal L = \sum_{i=u,d} i \bar q_i \gamma_{\mu} D^{\mu} q_i.$$ Defining operators to project out the left and right handed components of the quark fields and rewriting the lagrangian in terms of these components we see that there is no coupling between them. Because of the anomalous $U(1)_A$, the symmetry of massless QCD is then said to be $SU(2)_L \otimes SU(2)_R$.

In Matthew Schwartz's book, P.619 he writes that 'Strong dynamics of QCD induces condensates $\langle \bar \psi_u \psi_u \rangle \approx \langle \bar \psi_d \psi_d \rangle \neq 0$ which spontaneosly break $SU(2)_L \otimes SU(2)_R$ down to $SU(2)_{\text{isospin}}$. Thus, in the low energy theory, particles only form multiplets of $SU(2)_{\text{isospin}}$.'

I am trying to understand what is the relevance of the non vanishing VEV of the quark operators $\bar q q$ to imply SSB of the chiral symmetry group? In the chiral lagrangian above, there is no operator of the form $\bar q q$ since the mass term has been set to zero.

My understanding is that SSB of a symmetry means that the lagrangian of the theory exhibits the symmetry but the ground state of the theory does not. In this case our theory is chiral QCD described by the lagrangian at the top of this post . Are we then adding the term $\bar q q m$ onto the chiral lagrangian which can be seen as the analogous term of the Higgs potential like in SSB of free scalar field theories? The VEV of all terms are then taken and $\langle \bar q q \rangle$ is seen to be non vanishing? The ground state VEV is thus non zero and writing the $\bar q q$ in terms of left and right handed components we see that this term couples them thus the symmetry is no longer the direct product of two $SU(2)$'s.

Is that way of thinking about it anywhere near correct? The more I think about it the more it seems to me that it describes more of an explicit breaking (chiral symmetry softly broken by masses) than spontaneous breaking.

Thanks!

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  • $\begingroup$ This kind of symmetry breaking is usually called dynamical symmetry breaking. It is not a form of explicit symmetry breaking. Not sure what the physical manifestation of the VEV is. One way to investigate this kind of situation is to derive an effective action for the theory, where you introduce a source for $\bar{q}q$. From that one can then often derive a gap equation. $\endgroup$ – flippiefanus Aug 8 '16 at 10:45
  • $\begingroup$ Do you know of a reference where I can read more about how the dynamical formation of the condensate gives rise to SSB of the chiral symmetry? I see that a dynamical symmetry breaking is a form of SSB but I have not encountered the former concept before, only SSB in the context of scalar and gauge field theories. Thanks! $\endgroup$ – CAF Aug 9 '16 at 10:03
  • $\begingroup$ Related to 315061. $\endgroup$ – Cosmas Zachos Feb 27 '17 at 22:38
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The kind of symmetry breaking you are considering is an example of spontaneously broken "global symmetry" which is different from spontaneously broken "gauge symmetry" with which you are familiar with. If the quark masses are non zero then a term like $m\bar u u$ and $m\bar d d$ obviously break separate left-handed and right-handed symmetries and only vector isospin survives. It is in this sense you should understand it.

The quark condensate has direct relation to quark masses and coupling and can be calculated using lattice QCD however it is better to obtain it from experiments because lattice calculations are not very precise.

Separate chapters on spontaneously broken global symmetries and gauge symmetries are treated in Weinberg vol.2.

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