Construction of the effective Lagrangian in chiral pertubation theory

Question

The QCD Lagrangian containing quarks and gluons has an approximate $SU(3)_\text{L} \times SU(3)_\text{R}$ which is spontaneously broken by a non-vanishing quark condensate $\langle \bar{q} q \rangle \neq 0$ down to $SU(3)_\text{V}$ and at the same time explicitly broken by quark masses. Former leads to eight Goldstones bosons, that acquire small masses due to the explicit symmetry breaking. Now, this Lagrangian is not really useful in the low energy regime, for instance below $\Lambda_\text{QCD}$, since there the hadrons describe the degrees of freedom. Thus at sufficiently low energies one only needs to take the Goldstone bosons $\pi, K, \eta$ into consideration in order to describe dynamics.

What I am not understanding from here on is: In order to construct the effective Lagrangian containing the Goldstone boson fields (and external fields) one wants $\mathcal{L}_\text{eff}$ to be invariant regarding the chiral group $SU(3)_\text{L} \times SU(3)_\text{R}$ instead of the vector subgroup $SU(3)_\text{V}$. This bothers me, since the Goldstone bosons only "exist" once the chiral symmetry group is broken down to $SU(3)_\text{V}$ which would mean that the chiral group is not a symmetry of the theory anymore. So, why isn't it enough to construct an effective Lagrangian that is invariant under $SU(3)_\text{V}$?

Answer 1

There is a funny thing with spontaneous symmetry breaking: goldstone bosons appearing because of it realize non-linear representation of spontaneously broken symmetry. In Your case, although the Goldstone bosons (pseudo-scalar mesons) parametrize the coset $H \simeq SU_{L}(3)\times SU_{R}(3)/SU_{V}(3)$, whose manifold is $\simeq SU(3)$, they in fact realize the non-linear representation of the underlying $SU_{L}(3)\times SU_{R}(3)$ symmetry group.

You can understand this in more or less straightforward way. First, "extract" the goldstones from the quark field, so that $$ q(x) \equiv (u,d,s)^{T} \equiv U(x)\tilde{q}(x), $$ where $$ V(x) \equiv \text{exp}\left( \frac{i\gamma_{5}\epsilon_{a}(x)}{f_{\pi}}\right) $$ is the goldstones matrix (with fields $\epsilon_{a}$ parametrizing the coordinates of the coset $H$ space), and $\tilde{q}$ doesn't contain goldstones field. The linear combination of $\epsilon_{a}$ are pseudo-scalar mesons.

Since You know the transformation rule for $q$ under the global $SU_{L}(3)\times SU_{R}(3)$ transformation, $$ q(x) \to q'(x) = \text{exp}\big(iP_{L}t_{a}c^{L}_{a} + iP_{R}t_{a}c^{R}_{a}\big)q(x), \quad P_{L/R} = \frac{1 \mp \gamma_{5}}{2} $$ You can determine the transformation rule for $\epsilon_{a}(x)$. By using the fact that $V(x)$ represents the coset $SU_{L}(3)\times SU_{R}(3)$, after slightly tedious but straightforward calculation You obtain $$ U(x) \equiv \text{exp}\big( \frac{2it_{a}\epsilon_{a}}{f_{\pi}}\big) \to U'(x) e^{it_{a}c^{R}_{a}}U(x)e^{-it_{a}c^{L}_{a}}, $$ i.e., the goldstones $\epsilon_{a}$ form the representation $(3, \bar{3})$ of the initial symmetry group $SU_{L}(3)\times SU_{R}(3)$. The fact that they form non-linear representation comes from the non-trivial conditions $$ UU^{\dagger} = 1, \quad \text{det}U = 1 $$