4
$\begingroup$

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!

$\endgroup$
1

1 Answer 1

5
$\begingroup$

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.

A systematic treatment of the low-energy limit of QCD is provided by "Chiral Perturbation Theory", the low-energy effective quantum field theory of the strong interactions. For an introduction into this field of research (pioneered by S. Weinberg, J. Gasser, H. Leutwyler et al.) see e.g. the lecture by G. Ecker, "Chiral Perturbation Theory", hep-ph/9608226 in the arXiv.

$\endgroup$
4
  • $\begingroup$ 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 $\partial_\mu\xi \partial^\mu \xi$. $\endgroup$
    – taxus1
    Commented Jan 13, 2023 at 15:33
  • 1
    $\begingroup$ @taxus1 Because of colour confinement, the fields in the QCD Lagrangian (quark and gluon fields) do not correspond to asymptotic fields (corresponding to particles the experimentalist "see" in their detectors). It can be shown, however, that the low-energy limit of QCD is mathematically equivalent (shown by Weinberg and Leutwyler) to an effective quantum field theory written in terms of the asymptotic fields (pions, in the case of chiral SU(2)), being just what is called "Chiral Perturbation Theory". See also the references to the "classical papers" quoted in the lecture I mentioned above. $\endgroup$
    – Hyperon
    Commented Jan 13, 2023 at 15:48
  • $\begingroup$ Ok, thank you very much! I understand that I need to look for a different theory when I want to find the pions. However I still do not understand how the symmetry breaking procedure can be performed and the Goldstonebosons do not show up in the QCD Lagrangian. If we expand the Quarkfields around their Vacuum expectation value we necessarily have to introduce two deviation fields which thus have to show up somewhere. $\endgroup$
    – taxus1
    Commented Jan 13, 2023 at 20:37
  • 1
    $\begingroup$ @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. $\endgroup$
    – Hyperon
    Commented Jan 13, 2023 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.