Introduce the <linear sigma model> field
$$
\Sigma=\sigma I+i\pi^a\tau^a,
\qquad
\Sigma\mapsto L\Sigma R^\dagger.
$$
A suitable renormalizable potential is
$$
V(\Sigma)
=-\frac{\mu^2}{2}\operatorname{Tr}(\Sigma^\dagger\Sigma)
+\frac{\lambda}{4}
\left[\operatorname{Tr}(\Sigma^\dagger\Sigma)\right]^2,
$$
with a small symmetry-breaking source proportional to the quark-mass matrix when desired. A vacuum $\langle\Sigma\rangle=vI$ is fixed precisely by $L=R$, so the unbroken group is the diagonal isospin $SU(2)_V$. The three broken axial generators produce the pion triplet $\pi^\pm,\pi^0$, which become <pseudo-Goldstone bosons> when $m_u,m_d$ are restored.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 305 / 4 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-305/4/c/i/solution