Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
  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.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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