# Global symmetries QCD goldstone bosons

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!

• Commented Jan 13, 2023 at 14:25

In the so-called chiral limit of vanishing light quark masses ($$m_u=m_d=0$$), the QCD Lagrangian $$\mathcal{L}_{m_{u,d}=0}$$ is invariant under the global flavour group $${\rm SU(2)_L \times \rm SU(2)_R}$$ (chiral symmetry). This chiral symmetry is $$spontaneously$$ broken to the vectorial subgroup $${\rm SU(2)}_V$$, meaning that the symmetry group of the QCD vacuum, $$\rm{SU(2)}_V$$, is smaller than the symmetry group of the QCD Lagrangian in the chiral limit, namely $${\rm SU(2)_L \times \rm SU(2)_R}$$. As a consequence, the Goldstone theorem predicts the presence of three massless pseudoscalar bosons in the spectrum of QCD in the chiral limit.
In addition to the $$spontaneous$$ breaking of the chiral symmetry, $${\rm SU(2)_L \times SU(2)_R}$$ is also $$explicitly$$ broken by the presence of the quark mass terms $$-m_u (\bar{u}_{\rm L} u_{\rm R} +\bar{u}_{\rm R} u_{\rm L}) - m_d (\bar{d}_{\rm L} d_{\rm R} + \bar{d}_{\rm R} d_{\rm L})$$ in $$\mathcal{L}_{m_{u,d} \ne 0}$$. Thus, in the real world (where $$m_{u,d} \ne 0$$), the Goldstone bosons of chiral symmetry breaking (identified with the lightest pseudoscalar mesons $$\pi^\pm$$ and $$\pi^0$$) acquire small masses with $$M_\pi^2 = B (m_u+m_d)+\mathcal{O}(m_q^2)$$, where $$B$$ is related to the quark condensate $$\langle 0 | \bar{u} u |0\rangle = - F^2 B[1+\mathcal{O}(m_q)]$$, the order parameter of chiral symmetry breaking.
• Thanks alot! Your answer contains a lot of clarifications, especially the distinction between spontaneous and explicit breaking of the chiral symmetry. However, it is still not clear to me, where do I find the terms of the (massive) goldstone bosons, i.e. the pions in the QCD lagrangian (after symmetry breaking)? I am looking for terms like $\partial_\mu\xi \partial^\mu \xi$. Commented Jan 13, 2023 at 15:33
• @taxus1 The quark fields have no vacuum expectation values as they are fermion fields. The quark condensate $\langle 0 | \bar{q} q | 0 \rangle$ is the order parameter of the spontaneous breaking of the chiral symmetry. Note that these are non-perturbative properties of QCD! You find all these questions addressed in the relevant literature. There is also a textbook "A Primer for Chiral Perturbation Theory" by Stefan Scherer and Matthias R. Schindler, which might be a good start. Commented Jan 13, 2023 at 21:02