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Flavour issue in the printed effective interaction. The displayed charged-current factor uses an electron field even for the muon-neutrino term. Taken literally, this introduces an extra ${\nu }_{\mu }e$ charged-current interaction, contradicting the Standard Model diagrams in part (a). In the Standard Model that factor uses the charged lepton of the same flavour; for electron scattering the charged-current contribution is therefore present only for ${\nu }_e$. The weak flavour selection in neutrino-electron scattering fixes the physically consistent assignment used below. The literal printed consequence is given afterwards, so the discrepancy is explicit.

At leading order in the electron mass and in the four-fermion regime, define ${\Gamma }_L^{\alpha }={\gamma }^{\alpha }(1-{\gamma }^5)$. The weak neutral current yields the scattering amplitudeAn overall amplitude phase has no observable effect. The Fermi interaction is valid when the exchanged invariants are small compared with the squared weak-boson masses.

Average over the two initial electron spin states, and sum over final spins. There is no averaging over a sterile right-handed incoming neutrino. Using the fermion spin sum, the averaged square isTo display the trace contraction, defineThe supplied gamma matrix trace identities give the two tensors $8(S(k^{\prime },k)+iE(k^{\prime },k))$ and $4[(c_V^2+c_A^2)S(p^{\prime },p)+2i{c}_V{c}_{A}E(p^{\prime },p)]$. Symmetric-antisymmetric mixed contractions vanish. Set $X=(k^{\prime }\cdot p^{\prime })(k\cdot p)$ and $Y=(k^{\prime }\cdot p)(k\cdot p^{\prime })$. The remaining contractions are $S(k^{\prime },k)S(p^{\prime },p)=2(X+Y)$ and $E(k^{\prime },k)E(p^{\prime },p)=-2(X-Y)$, using the stated negative epsilon-contraction sign. Their sum givesThe massless Mandelstam variables satisfy $s+t+u=0$, and the two products are $s^2/4$ and $u^2/4$, respectively. Therefore

In the centre-of-momentum frame, momentum conservation gives $E=\sqrt{s}/2$. If $z=\cos \theta$ is the angle cosine between $p$ and $p^{\prime }$, then $u=-s(1+z)/2$. The massless two-body Lorentz-invariant phase space is $d{\rho }_2=d\Omega /(32{\pi }^2)$, while the flux is $2s$. ThusIntegrate ${\int }_{-1}^{1}dz=2$ and ${\int }_{-1}^1(1+z{)}^{2}dz=8/3$:Consequently the intended coefficients areThis is the massless neutrino-electron contact cross-section.

For completeness, the literal printed muon-neutrino charged-current term would instead shift both couplings by one, exactly as in part (c):For arbitrary real $c_V,c_A$, its linear terms cannot be represented by the requested quadratic form with coupling-independent constants. The Standard Model result and the literal printed interaction are not the same problem.

Evaluate the two-body Lorentz-invariant phase space in the decaying particle's rest frame, $p=(m,0)$. The spatial delta function sets ${\mathbf{q}}_2=-{\mathbf{q}}_1$. Both massless particles have energy $q=|{\mathbf{q}}_1|$, leavingHere the factor $1/2$ in the second equality is the Jacobian of the energy delta function. Integrating the solid angle gives $\int d{\Pi }_2=1/(8\pi )$. If the decay is isotropic, the squared amplitude is angle independent, soThis uses distinguishable final particles or an amplitude/phase-space convention with any required identical-particle symmetry factor already included. Two genuinely identical final bosons require an additional $1/2!$ if the full angular domain is integrated without such a factor. The complex particle and antiparticle in the following parts are distinct, so no such division is required there.

Evaluate the two-body Lorentz-invariant phase space in the parent's rest frame, $p=(m,0)$. The spatial Dirac delta function sets ${\mathbf{q}}_2=-{\mathbf{q}}_1$. For massless daughters $E_1=E_2=q$, soThe factor $1/2$ in the last step is the derivative of the delta-function argument. Substituting into the decay width gives, for labelled or distinct daughters,The angular domain is one sphere: $0\le \phi <2\pi$ and $-1\le \cos \theta \le 1$, equivalently $0\le \theta \le \pi$. The printed negative-to-positive polar-angle interval is not a second independent coverage of the sphere.

For identical daughters, integration over the full labelled two-body phase space counts each physical final state twice. One must therefore include an additional factor $1/2!$, or equivalently integrate over an unordered half-domain. This distinction is necessary for the two identical photons in solution.

For a resting W boson with real spatial polarization vector along $z$, the fermion spin sum gives $\sum |\mathcal{M}{|}^2=g^2{M}_W^2{\sin }^2\theta /2$ for massless charged lepton and antineutrino momenta. The two-body Lorentz-invariant phase space gives the displayed angular decay rate. Since $\int {\sin }^2\theta d\Omega =8\pi /3$, the integrated width is $g^2{M}_W/(48\pi )$, equal to the unpolarized width. The massive-vector spin average changes the angular distribution, not the total width. No initial spin average is taken for a specified pure polarization.

Here $x=m_W^2/m_t^2$, the bottom quark mass is neglected, and the W boson is on shell. The weak charged current gives the spin-averaged squared scattering amplitude $g_W^2|V_{tb}{|}^2(m_t^2-m_W^2)(m_t^2+2m_W^2)/(4m_W^2)$. The two-body Lorentz-invariant phase space has $p^{\ast }=(m_t^2-m_W^2)/(2m_t)$; multiplying by $p^{\ast }/(8\pi m_t^2)$ gives the displayed decay width. Its longitudinal-vector contribution explains the large-$m_t$ cubic growth, and its threshold factor makes the width vanish as $m_t\downarrow m_W$. The formula requires $m_t>m_W$ and receives finite-mass and radiative corrections.