# Chiral perturbation theory: what is the Quark Condensate? why expand in $U$ rather than Goldstone fields?

I'm studying Chiral Perturbation Theory ($\chi PT$) from Scherer's Introduction to Chiral Perturbation Theory.

What I am currently having some trouble understanding are two things:

1. The quark condensate. What is this and why is it a sufficient condition for spontaneous breaking of chiral symmetry? What I do not really understand is where the operator $S_a=\bar{q} \lambda_a q$ comes from ($\lambda_a$ are the Gell-Mann matrices) and why the expectation value of this (which I gather is zero) gives us this thing called the quark condensate.
2. The formulation of the effective Lagrangian. There is some stuff in Scherer about the coset $G/H$ where in this case G is the full chiral group and $H$ is the vector subgroup which is left after spontaneous symmetry breaking, but I do not really follow how this discussion explains why the Lagrangian is given in terms of the SU(3) matrix $U=\exp{\frac{i}{F_0} \Phi} = \exp{\frac{i}{F_0} \phi_a\lambda_a}$ for (individual) Goldstone fields $\phi_a$? Why can't we write down the effective Lagrangian in terms of the actual degrees of freedom in the theory, i.e. the Goldstone fields? I've read something about them not transforming non-linearly (and the $U$ transforming linearly) but could not really follow so if someone could elaborate on this I would be very glad.

A big thanks in advance for all help given!

And another thing - if anyone has another tip for an introductory reference to $\chi PT$, I would be very grateful. Scherer works decently but it's always good to read about things from a different viewpoint.

## 1 Answer

1. The operator $S=\bar{q}\lambda_a q$ is the so-called scalar quark density, and together with its pseudoscalar counterpart, it enters the expressions for the divergence of the vector and axial-vector currents (see section 2.3.6).
Spontaneous symmetry breaking occurs if $n$ generators of a symmetry transformation do not annihilate the ground state, resulting in the existence of $n$ massless Goldstone bosons. As derived in section 4.1.2, the action of the generator on the ground state, $Q_a^A\mid0\rangle$, is related to the scalar quark condensate $\langle\bar{q}q\rangle$. This relation indicates that a non-vanishing condensate is a sufficient condition for spontaneous symmetry breaking.
2. The main point here is that the Lagrangian formulated in terms of U is invariant (and therefore transforms) under global $SU(3)_L\times SU(3)_R\times U(1)_V$ (corresponding to $G$) while the fields $\phi$ only transform as an octet under the subgroup $SU(3)_V$ (corresponding to $G/H$). In terms of U, it can also be easily shown (section 4.2.2) that the ground state is invariant under vector but not under axial transformations.
• Greetings; if I may ask a question: is it correct to say that we assume a non-vanishing singlet scalar condensate, that is $Q_{\alpha}^A |0> \neq 0$ or do we show it somehow? I think that we must assume it based on the phenomenology and the fact that although we know that $[Q_V ^{\alpha} , S_0(y)] = 0$ we do not know theoretically the action of the just-mentioned commutator on the vacuum(we cannot express $S_0$ by $Q_V ^{\alpha}$). The same for the $Q_{\alpha} ^A$. That is, we show that if $Q_{\alpha} ^A |0> \neq 0$ then $<\bar q q> \neq 0$ or the other way around; correct? . Thanks. – Constantine Black Mar 26 '17 at 16:59
• @ConstantineBlack Yes, you are correct. – Arturo don Juan May 19 '19 at 1:58