Beside the local $SU(3)$-Color-symmetrie The QCD Lagrangian also has global symmetries:
$$L_{QCD}=\sum_{f,c}\bar{q_{fc}}(i\gamma^\mu D_\mu - m ) q_{fc} - \frac{1}{4}F^a_{\mu \nu} F^{a \mu \nu} $$
- $SU(2)_V$ Isospin, since $m_u \approx m_d$, the conserved current is a vector current, hence "V"
- $SU(2)_A$ Invariance under $q \rightarrow e^{-i \omega T^a \gamma_5}q$. Conserved current is an axial current, hence "A". Would be a perfect symmetry if $m_u=m_d=0$
- $U(1)$ Baryon conservation
For mathematical reasons the $SU(2)_V \times SU(2)_A \times U(1)$ is usually written as a product of two chiral symmetries $SU(2)_R \times SU(2)_L$.
Since the quarks are not masseless, this global symmetry is broken according to the following pattern: $SU(2)_R \times SU(2)_L $\rightarrow$ SU(2)_V,$ where the remaining symmetry is isospin.
The Goldstone theorem says, that every broken global symmetry implies the existence of a number of Goldstone bosons (one for each generator of the broken symmetry group, i.e. Here 3). They must show up in the Lagrangian. After symmetry breaking we expand the field $\phi$ around its vacuum expectation value and insert this expansion into the Lagrangian: $\phi= (0,v+\rho(x))e^{-i\xi^a(x) T^a}$ We can then rewrite the Lagrangian in terms of the deviation fields $\rho(x)$ and $\xi(x)$. We will get a kinetic term for the goldstone bosons $\partial_\mu \xi^a \partial^\mu \xi^a$. Now I have read, in case of the QCD the Goldstone bosons are the three pions. But I do not see where they show up in the QCD Lagrangian. I would be grateful for help!