Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 305 / 2

For $q=(u,d{)}^T$, massless two-flavour Quantum chromodynamics hasClassically its continuous global symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V\times U(1{)}_A$, up to finite central quotients. The Adler-Bell-Jackiw anomaly destroys the continuous axial $U(1{)}_A$ in the quantum theory, leaving $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$, together with an anomaly-preserved discrete axial subgroup.

The condensate transforms as $⟨{\bar{q}}_L{q}_R⟩\mapsto L⟨{\bar{q}}_L{q}_R⟩R^{†}$, so its value proportional to the identity produces chiral symmetry breakingThe vacuum manifold is the coset $(SU(2{)}_L\times SU(2{)}_R)/SU(2{)}_V\cong SU(2)$ and has three Goldstone modes. A field $U(x)\in SU(2)$ transforms as $U\mapsto LU{R}^{†}$. Lorentz invariance and chiral symmetry leave, at two-derivative order, one invariant metric on this coset, soSince the Lagrangian density has dimension four and $U$ is dimensionless, the pion decay constant $f_{\pi }$ has mass dimension one.

For $U=e^{2i\pi /f_{\pi }}$, the Maurer-Cartan form expands asSubstituting, using cyclicity of the trace, and collecting even powers givesOdd terms cancel because the sigma-model metric is invariant under $U\leftrightarrow U^{†}$.

Electromagnetism gauges the vector flavour transformation $U\mapsto e^{ie\alpha Q}U{e}^{-ie\alpha Q}$, so covariance requires $D_{\mu }U={\partial }_{\mu }U-ie{A}_{\mu }[Q,U]$. The identity part of $Q=I/6+{\sigma }_3/2$ drops out of the commutator. Writinggives $[Q,\pi {]}_{12}={\pi }_{12}$ and $[Q,\pi {]}_{21}=-{\pi }_{21}$. Hence ${\pi }^{+}$ has electric charge $+e$, ${\pi }^{-}$ has charge $-e$, and ${\pi }^0$ is neutral.