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

For $m_u=m_d=0$ and electromagnetism neglected, the classical flavour symmetry isup to finite quotients. The chiral anomaly breaks $U(1{)}_A$ quantum mechanically, while $U(1{)}_V=U(1{)}_B$ remains. The quark condensatespontaneously breaksSmall quark masses and electromagnetism explicitly break parts of this approximate symmetry.

Introduce the linear sigma model fieldA suitable renormalizable potential iswith a small symmetry-breaking source proportional to the quark-mass matrix when desired. A vacuum $⟨\Sigma ⟩=vI$ 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.

At energies well below ${\Lambda }_{\mathrm{QCD}}$, collect the pion fields intoThe leading chiral perturbation theory Lagrangian iswhere $M=\mathrm{diag}(m_u,m_d)$ is treated as a symmetry-breaking spurion. The derivative and quark-mass expansion organizes the omitted terms.

Exact isospin symmetry would make the proton and neutron degenerate. Their observed mass difference is generated by explicit isospin breaking from $m_d-m_u$ together with electromagnetic effects. The approximate symmetry explains why the splitting is small compared with the nucleon mass, while chiral perturbation theory parametrizes it through symmetry-breaking operators and low-energy constants; symmetry alone does not predict its numerical value.