Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 3 / b / iii / Solution

It is useful first to establish the global minimum, rather than only a stationary point. The positive semidefinite matrix $X={\Phi }^{†}\Phi$ has nonnegative eigenvalues $x_1,\dots ,x_N$. For fixed $r={\sum }_i{x}_i$, the Cauchy-Schwarz inequality gives ${\sum }_i{x}_i^2\ge r^2/N$, with equality exactly when all $x_i$ are equal. Since $\alpha >0$,For $m^2<0$ and $\beta >0$, this lower bound has its minimum atThe bound is attained, so every minimum has ${\Phi }^{†}\Phi =C^{2}1$, or $\Phi =CW$ with $W$ a unitary matrix. A global transformation removes $W$ by decomposing it into an overall phase and an $SU(N)$ matrix. Thus ${\Phi }_0=C1$ is a valid representative of the vacuum manifold.

Preservation of this chosen vacuum expectation value requiresInfinitesimally, with traceless Hermitian generators $L,R$, this reads $\theta 1+L-R=0$. Taking the trace forces $\theta =0$, and then $L=R$. The surviving continuous transformations are consequentlyAll other independent continuous directions change the chosen vacuum expectation value. Thus the number of broken generators isGlobally, taking the determinant of the stabilizer equation permits the discrete phases $e^{iN\theta }=1$, paired with $U_L=e^{-i\theta }{U}_R$. They induce the same conjugation action on $\Phi$. The faithful unbroken conjugation group is $SU(N)/{\mathbb{Z}}_N$; for example, at $N=2$, $U=-1$ acts exactly like $U=1$. Thus the question's description by $SU(N)$ is correct as a parametrization and at the Lie algebra level, with this central redundancy understood. It leaves the number of broken generators unchanged.