Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 305 / 1 / ii / Solution

Let $x_a\ge 0$ be the eigenvalues of the positive semidefinite matrix $M^{†}M$. The potential isFor $k<0$ and $\lambda >0$, each summand is minimized atso $M_0^{†}{M}_0=v^{2}1$. A symmetry transformation preserves the representative $M_0=v1$ precisely when $L=R$, givingThe number of broken generators is $2N^2-N^2=N^2$, so the Goldstone theorem predicts $N^2$ modes, each a Goldstone boson.