Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 305 / 2

Take ${\tau }^a={\sigma }^a/2$, $Y_{\varphi }=1/2$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and a Higgs potential with ${\mu }_h^2>0$, $\lambda >0$. The gauge-scalar electroweak interaction isHere $B_{\mu \nu }={\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }$. With the printed plus-sign convention for $D_{\mu }$, define $[D_{\mu },D_{\nu }]=ig{W}_{\mu \nu }^a{\tau }^a+i{g}^{\prime }{Y}_{\varphi }{B}_{\mu \nu }$; thus $W_{\mu \nu }^a={\partial }_{\mu }{W}_{\nu }^a-{\partial }_{\nu }{W}_{\mu }^a-g{ϵ}^{abc}{W}_{\mu }^b{W}_{\nu }^c$. The minus sign in the nonlinear gauge field strength follows from this convention.

The minima have ${\varphi }^{†}\varphi =v^2/2$, with $v^2={\mu }_h^2/\lambda$. A gauge transformation rotates the vacuum to $(0,v{)}^T/\sqrt{2}$. In unitary gauge, the three angular Goldstone bosons are removed, leaving $\varphi =(0,v+H{)}^T/\sqrt{2}$ and one real Higgs boson. The generator $Q=T^3+Y$ annihilates the vacuum, because its lower component has $T^3=-1/2$ and hypercharge $1/2$. This identifies the unbroken electromagnetic gauge group.

The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. It has eigenvalues zero and $(g^2+g^{\prime 2})v^2/4$, with the zero eigenvector giving $A_{\mu }$. Expanding the Higgs potential about its minimum gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso $m_W=m_Z\cos {\theta }_W$ and $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.

Replacing $v^2$ by $(v+H{)}^2$ in the neutral mass term gives all the Higgs boson couplings to Z bosons in unitary gauge:There are exactly the cubic $HZZ$ and quartic $HHZZ$ elementary tree-level Feynman diagrams. Differentiating with respect to the identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$, respectively. There is no elementary $ZHH$ vertex for the real neutral radial field.

The authoritative PDF has $m_Z^2/v$, not the dimensionally incorrect $m_Z^2/v^2$ in the TeX conversion. The two identical Z bosons give the vertex factor two, so the scattering amplitude for the decay isup to an irrelevant overall phase. Apply the massive vector polarization sum, identifying the hint's $M_Z$ with $m_Z$:Since $k_1\cdot k_2=(m_H^2-2m_Z^2)/2$, putting $x=m_Z^2/m_H^2$ givesNo initial-spin average is needed for a Higgs boson. In its rest frame the two-body momentum is $|\mathbf{k}|=(m_H/2)\sqrt{1-4x}$. Integrating the energy and momentum delta functions in the Lorentz-invariant phase-space measure gives $d{\Phi }_2=\sqrt{1-4x}d\Omega /(32{\pi }^2)$, hence $\int d{\Phi }_2=\sqrt{1-4x}/(8\pi )$.

The printed generic width formula treats the daughters as labelled. Here the identical final-state symmetry factor is $1/2!$, without which the same physical configuration is counted twice. Thus the Higgs decay to two Z bosons hasEquivalently, $1/v^2=\sqrt{2}{G}_F$ gives the prefactor $G_F{m}_H^3/(16\sqrt{2}\pi )$. The threshold limit is zero and the large-mass limit is $m_H^3/(32\pi v^2)$. The additional factor for identical daughters is a required specialization of the PDF's generic phase-space formula, not an extra factor in the vertex.