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

Solution

Σ = σ I + i π^{a} τ^{a}, Σ \mapsto L Σ R^{†} .

A suitable renormalizable potential is

V (Σ) = - \frac{μ ^{2}}{2} Tr (Σ^{†} Σ) + \frac{λ}{4} [Tr (Σ^{†} Σ)]^{2},

with a small symmetry-breaking source proportional to the quark-mass matrix when desired. A vacuum

⟨ Σ ⟩ = v I

is fixed precisely by

L = R

, so the unbroken group is the diagonal isospin

S U (2)_{V}

. The three broken axial generators produce the pion triplet

π^{\pm}, π^{0}

m_{u}, m_{d}

are restored.

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