Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 305 / 2

Let $V_i=\partial V/\partial {\varphi }_i$ and $M_{ij}^2={\partial }_i{\partial }_{j}V{|}_{{\varphi }_0}$ be the scalar mass matrix, which is the Hessian matrix of the scalar potential at the vacuum. Invariance under the infinitesimal Lie group action givesDifferentiate with respect to ${\varphi }_j$ and evaluate at the vacuum expectation value ${\varphi }_0$. Since $V_i({\varphi }_0)=0$ at a minimum,Thus every tangent vector $i{t}^a{\varphi }_0$ generated by a broken Lie algebra generator is a zero eigenvalue eigenvector of the scalar mass matrix. The unbroken generators are precisely those in the stabilizer Lie algebra $\mathfrak{h}$ and give the zero tangent vector; independent broken directions span $\mathfrak{g}/\mathfrak{h}$. Hence Goldstone theorem givesmassless scalar modes at the classical level.

The minima of the scalar potentialsatisfyThe SU(2) group acts transitively on this three-sphere of vacua, and the stabilizer of a nonzero fundamental doublet is trivial. A global transformation may therefore choose ${\varphi }_0=(0,v{)}^T/\sqrt{2}$. Because the symmetry is gauged, unitary gauge removes all three angular Goldstone bosons, leaving only the real radial Higgs mode $h$:Substitution into the gauge-covariant kinetic term giveswithThe interaction terms produce $h^3$, $h^4$, $hBB$, $hhBB$, and the cubic and quartic non-Abelian gauge-boson vertices shown below.

The three broken generators supply the three longitudinal polarizations of the equally massive gauge bosons. No physical massless Goldstone boson remains, and because the unbroken subgroup is trivial there is no massless gauge boson either. The remaining physical spectrum has $3\times 3+1=10$ degrees of freedom, equal to the original $3\times 2+4=10$.