Past exam of the particle physics course of the University of Oxford / 2022 / 5

The covariant derivative $D_{\mu }$ differentiates a field while preserving its transformation law under the local electroweak gauge symmetry $SU(2{)}_L\times U(1{)}_Y$. Here ${\partial }_{\mu }$ is the ordinary spacetime derivative and $\mu$ a Lorentz index. The gauge fields $W_{\mu }^j$, $j=1,2,3$, are the three SU(2) fields; $B_{\mu }$ is the U(1) field. The corresponding dimensionless coupling constant is $g$ for SU(2) and $g^{\prime }$ for U(1).

The ${\tau }_j$ are weak isospin generators in the representation of the differentiated field, with ${\tau }_j={\sigma }_j/2$ on a doublet and ${\tau }_j=0$ on an SU(2) singlet. The repeated index $j$ is summed. The generator $Y/2$ supplies the field's weak hypercharge in the convention $Q={\tau }_3+Y/2$, where $Q$ is its fractional electric charge. The factor $-i$ follows the convention of Hermitian generators. The two connection terms compensate the derivatives of spacetime-dependent gauge transformations.

Set $G=\sqrt{g^2+g^{\prime 2}}$ and $s_W=\sin {\theta }_W$. First expand the charged term using the ladder operators:Substituting the neutral gauge fields as well, the covariant derivative becomesMatching $W^3$ to $g{\tau }_3$ requiresMatching $B$ to $g^{\prime }(Q-{\tau }_3)=g^{\prime }Y/2$ requiresFor $g,g^{\prime }>0$, elimination yieldsEquivalently, $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$ and $1/e^2=1/g^2+1/g^{\prime 2}$. The conventional Weinberg angle lies in the first quadrant.

Write $f=(v+h)/\sqrt{2}$. The Pauli matrices give the weak isospin operatorsFor the Higgs doublet $\Phi =(0,f{)}^T$ with weak hypercharge $Y=1$,The Higgs field direction, including its vacuum expectation value, is electrically neutral. Consequently the electromagnetic U(1) leaves the vacuum invariant: electroweak symmetry breaking preserves electromagnetism and the photon remains massless. The result concerns this neutral vacuum direction; it does not say that both components of the full Higgs doublet have zero electric charge.

Using c and omitting ${\partial }_{\mu }\Phi$, the covariant derivative isThe two components are orthogonal, so their squared magnitudes add; $(W_{\mu }^{+}{)}^{†}=W_{\mu }^{-}$. Thus the Higgs field kinetic contribution to the Lagrangian density isThe two $v^2$ terms are gauge boson mass terms. Comparing with $m_W^2{W}^{+}{W}^{-}+\frac{1}{2}{m}_Z^2{Z}^2$ gives $\boxed{m_W=gv/2,m_Z=Gv/2}$ and $m_W/m_Z=\cos {\theta }_W$. The factor $1/2$ for the real Z boson field is essential.

The two $2vh$ terms are the cubic Higgs boson coupling to W bosons and Higgs boson coupling to Z bosons. They produce the three-leg $h{W}^{+}{W}^{-}$ and $hZZ$ Feynman vertices, allowing Higgs decay to two W bosons, Higgs decay to two Z bosons when kinematically available, and the corresponding production mechanisms with virtual gauge bosons. The two $h^2$ terms are quartic $hh{W}^{+}{W}^{-}$ and $hhZZ$ interactions, including vector-boson scattering into a Higgs boson pair. Written in terms of the masses,The corresponding Feynman vertex factors are $2i{m}_V^2{g}_{\mu \nu }/v$ for one $h$ and $2i{m}_V^2{g}_{\mu \nu }/v^2$ for two $h$, for $V=W,Z$, after differentiating with respect to identical fields. These six contributions contain no photon mass or direct $hAA$ interaction because $Q\Phi =0$.