= Solution
The minima of the <scalar potential>
$$
V(\phi)=m^2\phi^\dagger\phi+\frac\lambda2(\phi^\dagger\phi)^2
$$
satisfy
$$
\phi^\dagger\phi=-\frac{m^2}{\lambda}=\frac{v^2}{2},
\qquad
v^2=-\frac{2m^2}{\lambda}.
$$
The <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 $\phi_0=(0,v)^T/\sqrt2$. Because the symmetry is gauged, <unitary gauge> removes all three angular <Goldstone bosons>, leaving only the real radial <Higgs mode> $h$:
$$
\phi(x)=\frac1{\sqrt2}\binom0{v+h(x)}.
$$
Substitution into the <gauge-covariant kinetic term> gives
$$
\boxed{
\mathcal L=-\frac14F^a_{\mu\nu}F^{a\mu\nu}
+\frac12(\partial h)^2+\frac{g^2}{8}(v+h)^2B^a_\mu B^{a\mu}
-\frac12m_h^2h^2-\frac{\lambda v}{2}h^3-\frac\lambda8h^4+\text{constant}
}
$$
with
$$
\boxed{m_h^2=\lambda v^2=-2m^2,\qquad m_B^2=\frac{g^2v^2}{4}.}
$$
The interaction terms produce $h^3$, $h^4$, $hBB$, $hhBB$, and the cubic and quartic non-Abelian gauge-boson vertices shown below.
\Image[/media/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-305-higgs-interactions.png]
{height=650}
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 $3\times3+1=10$ degrees of freedom, equal to the original $3\times2+4=10$.
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