= Solution
Put $s_W=\sin\theta_W$ and $c_W=\cos\theta_W$. The left-handed doublet has
$$
T_L^3=\begin{pmatrix}\frac12&0\\0&-\frac12\end{pmatrix},\qquad
Q_L=\begin{pmatrix}0&0\\0&-1\end{pmatrix},
$$
while the right-handed charged-<lepton> singlet has $T_R^3=0$ and $Q_R=-1$. There is no weakly coupled right-handed <neutrino> in the stated field content. For a <fermion> $f$, write $g_L^f=T_{3L}^f-Q_fs_W^2$ and $g_R^f=T_{3R}^f-Q_fs_W^2$, with $g_R^\nu=0$ here. Since $\bar f_L\gamma^\mu f_L=\bar f\gamma^\mu P_Lf$ and similarly for $R$,
$$
\bar f\gamma^\mu(g_L^fP_L+g_R^fP_R)f
=\bar f\gamma^\mu(c_V^f-c_A^f\gamma^5)f,
\quad c_V^f=\frac{g_L^f+g_R^f}{2},\quad c_A^f=\frac{g_L^f-g_R^f}{2}.
$$
Consequently the <neutral-current vector and axial couplings> in the specified overall normalization are
$$
\boxed{c_V^\ell=-\frac14+s_W^2,\qquad c_A^\ell=-\frac14,\qquad
c_V^\nu=c_A^\nu=\frac14.}
$$
These coefficients accompany $g/c_W$. A convention with $g/(2c_W)$ uses twice these vector and axial coefficients; mixing the two conventions would make the widths wrong by a factor of four. For the <neutrino>, $c_V-c_A\gamma^5=\frac12P_L$, so the chiral projection is already built into the vertex.
For either massless final pair with momenta $k_1,k_2$, $p=k_1+k_2$, the invariant amplitude is, up to an irrelevant overall phase,
$$
\mathcal M_f=\frac g{c_W}\epsilon_\mu(p)\bar u_f(k_1)\gamma^\mu(c_V^f-c_A^f\gamma^5)v_f(k_2).
$$
The <spin>-summed <fermion> tensor is
$$
T_f^{\mu\nu}=\operatorname{tr}\!\left[\not k_1\gamma^\mu(c_V^f-c_A^f\gamma^5)
\not k_2\gamma^\nu(c_V^f-c_A^f\gamma^5)\right].
$$
The symmetric part, using the four-gamma <trace>, is
$$
(T_f^{\mu\nu})_{\rm sym}=4[(c_V^f)^2+(c_A^f)^2]
\left(k_1^\mu k_2^\nu+k_1^\nu k_2^\mu-\eta^{\mu\nu}k_1\cdot k_2\right).
$$
The vector-axial interference is antisymmetric in $\mu,\nu$, so it drops out of the symmetric $Z$ polarization sum. Also $p_\mu T_f^{\mu\nu}=0$ for massless external <fermions>, by their <Dirac equations>; hence the $p_\mu p_\nu/m_Z^2$ part contributes zero. Using $k_1\cdot k_2=m_Z^2/2$ gives
$$
\boxed{\overline{|\mathcal M_f|^2}
=\frac{g^2}{3c_W^2}(-\eta_{\mu\nu})T_f^{\mu\nu}
=\frac{4g^2m_Z^2}{3c_W^2}[(c_V^f)^2+(c_A^f)^2].}
$$
The <massive-vector spin average> supplies the factor $1/3$ for the three <spin> states of the massive $Z$. The massless <two-body Lorentz-invariant phase space> is $d\Phi_2=d\Omega/(32\pi^2)$, and so $\int d\Phi_2=1/(8\pi)$. The <spin>-averaged amplitude above is angle-independent in the rest frame, giving
$$
\boxed{\Gamma_f=\frac{g^2m_Z}{12\pi c_W^2}[(c_V^f)^2+(c_A^f)^2].}
$$
The <massless leptonic Z decay widths> for one charged-<lepton> flavor and one active <neutrino> flavor respectively are
$$
\boxed{\Gamma(Z\to\ell\bar\ell)=\frac{g^2m_Z}{96\pi c_W^2}
(1-4s_W^2+8s_W^4),\qquad
\Gamma(Z\to\nu\bar\nu)=\frac{g^2m_Z}{96\pi c_W^2}.}
$$
There is no color factor for <leptons> and no extra final-state identical-particle factor for a particle-antiparticle pair. The <neutrino> projectors already exclude sterile helicity states, so no additional factor of two is needed. These are tree-level massless widths; electroweak radiative corrections and charged-<lepton> masses have not been included. If desired, using $G_F=g^2/(4\sqrt2m_W^2)$ and $m_W=c_Wm_Z$ converts the common prefactor to $G_Fm_Z^3/(12\sqrt2\pi)$.
Back to article page
