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Let $\Sigma =m_e^2+m_t^2$, $Q_e=-1$, $Q_t=2/3$ and $N_c=3$. The single tree-level Feynman diagram is annihilation through a virtual photon. Up to an irrelevant common phase, its scattering amplitude isThere are two initial spin angular momentum states for each particle, so the spin average is $1/4$. The final color charge sum gives ${\sum }_{c,d}|{\delta }_{cd}{|}^2=N_c$. There is no additional photon polarization average for a virtual photon and no separate polarization label beyond the Dirac spinor spin sums.

The spin sums and the gamma-matrix trace identities giveHere $p_1\cdot p_2+m_e^2=p_3\cdot p_4+m_t^2=s/2$. Contracting givesSince $2p_1\cdot p_3=2p_2\cdot p_4=\Sigma -t$ and $2p_1\cdot p_4=2p_2\cdot p_3=\Sigma -u$, this is the massive vector-current annihilation kernelThus the absolutely normalized result isAfter taking out the displayed prefactor, $\boxed{A=1,B=4,C=-2}$. Because the source uses a proportionality sign, only their ratio is fixed until that prefactor is specified. Equivalently, after taking out just $e^4/s^2$, their values are $8/3,32/3,-16/3$. No mass has been neglected. At $m_e=0$ and $s=4m_t^2$, the bracket yields the finite isotropic threshold value of the scattering amplitude; the vanishing threshold scattering cross-section comes from relativistic two-body phase space.

The real scalar field has canonical momentum $\pi =\dot{\varphi }$. Its Legendre transform givesThe first three terms constitute $H_0$, the free massive Hamiltonian operator. The remaining term is $H_I=\lambda \int {\varphi }^4{d}^{3}x/4!=-\int {\mathcal{L}}_I{d}^{3}x$. This sign is a consequence of having no time derivatives in the interaction: the canonical momentum is unchanged, while the interaction enters the Hamiltonian density with the opposite sign to the Lagrangian density.

In the interaction picture, operators evolve with $H_0$ and states evolve with $H_I(t)$ constructed from the free interaction-picture field. Their evolution operator satisfiesIntegrating once gives $U_I=I-i{\int }_{t_0}^t{H}_I(t_1)U_I(t_1,t_0)d{t}_1$. Repeated substitution orders the integrations as $t\ge t_1\ge \cdots \ge t_n\ge t_0$. Replacing the ordered simplex by the full integration cube introduces $1/n!$ and a time-ordered product, which proves the Dyson seriesTaking the asymptotic limits with the usual adiabatic or regulated scattering prescription yieldsThe field in this expression is the free field evolving in the interaction picture, not an independently prescribed classical field. For quantum fields the expansion is a regulated perturbation series; the time-ordering symbol cannot be dropped because the operators at different times need not commute.

Use relativistically normalized two-particle states $|p_1,p_2⟩=a^{†}(p_1)a^{†}(p_2)|0⟩$, with $⟨p|q⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. The connected first-order scattering contribution isHere the connected S-matrix uses a normalized vacuum and physical external one-particle states. At this order, normal ordering the vertex separates the four-external-leg contribution from vacuum and tadpole pieces; those disconnected or external-mass contributions are not the two-to-two scattering amplitude.

Two fields annihilate the incoming particles and two create the outgoing ones. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ choices of the annihilating fields, two assignments of the incoming momenta and two assignments of the outgoing momenta. Hence the total factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=4!$, the external-leg factorial cancellation at a quartic scalar vertex. Each attachment contributes its plane-wave factor, soThe spacetime integral is the four-momentum conservation delta function, givingIn particular, the tree scattering amplitude is independent of angle.

For completeness, derive the relativistic two-body phase space in the centre-of-mass frame. Write $\sqrt{s}=2E_{\ast }$, $p_{\ast }^2=E_{\ast }^2-m^2$. The spatial delta function sets the two final momenta opposite, and the energy delta function is $\delta (\sqrt{s}-2\sqrt{p^2+m^2})$. Its radial Jacobian is $E_{\ast }/(2p_{\ast })$. ThusThe invariant flux factor is $\mathcal{F}=4p_{\ast }\sqrt{s}$ for equal initial masses. Using $d\sigma =|T{|}^{2}d{\Phi }_2/\mathcal{F}$, the labeled angular density is ${\lambda }^2/(64{\pi }^{2}s)$. The outgoing quanta belong to one real scalar field, so they are identical. A full-sphere integration counts each event twice and requires the identical final-state symmetry factor $1/2!$. ConsequentlyEquivalently one may integrate the labeled density over a hemisphere representing each event once. The final-state factor is separate from the vertex factorial: applying it at the vertex instead would give the wrong elastic scattering from a quartic scalar contact interaction cross-section.

Use Minkowski metric signature $+---$ and natural units. With no lepton mixing, choose the weak charged current conventionIts conjugate convention exchanges $J$ and $J^{†}$ without changing the interaction. The relevant four-fermion interaction gives the muon decay amplitude, up to an overall phase,Here $G_F$ is the Fermi constant; the factors $1-{\gamma }_5$ are twice the left chiral projector, so there is no additional factor of two to insert.

For the final-spin sum let ${\rho }_{\mu }=(/p+m)(1+{\gamma }_5/s)/2$, where $m=m_{\mu }$. It already describes the specified initial polarization; there is no further initial-spin average. The two gamma matrix traces areThe electron-mass term vanishes in its chiral trace. In the muon trace, anticommutation with ${\gamma }_5$ givesFor the chiral fermion trace contraction, defineThe supplied trace conventions imply $L^{\alpha \beta }=8K^{\alpha \beta }(k,q)$ and $H_{\alpha \beta }=4K_{\alpha \beta }(q^{\prime },r)$, where $r=p-ms$. The factor four in the second trace follows from the polarization projector. The symmetric parts contract to $2[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$. The Levi-Civita symbol contraction, including $i^2=-1$, contributes $2[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Mixed symmetric-antisymmetric contractions vanish. Thus $K^{\alpha \beta }(a,b)K_{\alpha \beta }(c,d)=4(a\cdot c)(b\cdot d)$, and

Now neglect $m_e$ as well. Put $Q=p-k$ and $c=\hat{\mathbf{k}}\cdot \mathbf{s}$. The given moment integral over two-neutrino relativistic two-body phase space yieldsIn the muon rest frame, $k\cdot Q=mE$, $k\cdot r=mE(1+c)$, $r\cdot Q=m^2-mE(1+c)$ and $Q^2=m^2-2mE$. Substitution givesCombining this with the Lorentz-invariant phase-space measure, whose one-particle denominator is $(2\pi {)}^{3}2E$, and $d^{3}k=E^{2}dEd\Omega$, gives the polarized muon decay distributionConsequently $C=-2$, $D=-2$, $E=1$. Integration over the electron direction removes the polarization term, and ${\int }_0^1{x}^2(3-2x)dx=1/2$, recovering $\Gamma =G_F^2{m}^5/(192{\pi }^3)$.

The spin vector is axial and the momentum direction is polar: under parity, $\mathbf{s}$ stays unchanged while $\hat{\mathbf{k}}$ reverses. Their scalar product therefore changes sign. Its nonzero coefficient demonstrates parity violation in polarized muon decay, arising from the left-chiral weak charged current. At the upper endpoint the negative muon's electron is preferentially emitted opposite the spin, providing a sign check on the result.

For an unpolarized initial fermion, average over its two spin states and sum over the unobserved final spin:The initial scalar has only one spin state. One can evaluate this sum directly from normalized Dirac spinors, or use consistent fermion spin sums to express it as a trace. It is the squared sum of both tree scattering amplitudes, not the sum of their separate squares.

The relativistic scattering cross-section is obtained by integrating the Lorentz-invariant phase-space measure and dividing by the invariant flux factor:There is no identical-final-particle factor, because the outgoing scalar and fermion are distinct. All energies are positive, with $E_q^2=|\mathbf{q}{|}^2+M^2$ and $E_l^2=|\mathbf{l}{|}^2+m^2$.

Equivalently, in the centre-of-momentum frame set $s=-(p+k{)}^2$. Integrating the energy delta function in the relativistic two-body phase space givesFor this elastic process $k_f=k_i$, so integrate $\overline{|T{|}^2}/(64{\pi }^{2}s)$ over the full solid angle. This supplies the requested prescription without evaluating the angular integral.

Use the momentum-space Feynman rules with the scalar Feynman propagator $i/(p^2-m^2+i0)$. The four-leg vertex of a factorial-normalized scalar interaction has weight $i\lambda$ for the positive interaction sign printed here. There are $4!$ Wick contractions assigning four external legs to its four fields, cancelling the factorial in its coefficient. A negative interaction sign would give $-i\lambda$; its squared tree amplitude is the same.

At each vertex include $(2\pi {)}^4{\delta }^4(\sum p)$ with all incident momenta taken incoming. Assign an internal momentum to each line and integrate each independent loop with $\int d^4\ell /(2\pi {)}^4$. Divide a graph by its Feynman-diagram symmetry factor, sum the graphs at the chosen order, and omit disconnected vacuum graphs from normalized amplitudes. For an S-matrix element, amputate external propagators and put the external momenta on shell as in the LSZ reduction formula; the external one-particle residues are one at tree level.

Define the invariant amplitude by the relativistically normalized matrix elementwith $⟨p|p^{\prime }⟩=2E_{\mathbf{p}}(2\pi {)}^3{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The lowest-order connected four-point graph is one contact vertex:There is no exchange graph at this order because there is no three-field interaction.

For the elastic scattering from a quartic scalar contact interaction, write $\sqrt{s}$ for the total centre-of-mass energy and $E_p=\sqrt{s}/2$ for the energy of each incoming particle. The incoming and outgoing spatial momentum magnitudes both equal $k=\sqrt{E_p^2-m^2}$, with $E_p>m$. The invariant incident flux isThe Lorentz-invariant phase-space measure for two outgoing particles isIn the centre-of-mass frame, the spatial delta function sets ${\mathbf{p}}_4=-{\mathbf{p}}_3$, while the energy delta function has radial derivative $2k/E_p$. Therefore the relativistic two-body phase space satisfiesThe two outgoing real-scalar particles are identical. Integrating over the full solid angle counts each unordered pair twice, so include the identical final-state symmetry factor $1/2!$. This givesThis is isotropic. When $E$ denotes each particle's energy, $s=4E^2$ and the full-sphere event density is ${\lambda }^2/(512{\pi }^2{E}^2)$. If $E$ denotes the total energy of the pair, $s=E^2$ and it is ${\lambda }^2/(128{\pi }^2{E}^2)$. Stating the answer in $s$ removes that energy-label ambiguity.

An equally valid angular convention selects one outgoing particle in a hemisphere, so each event is represented once. In that convention omit $1/2!$ and use $d\sigma /d\Omega ={\lambda }^2/(64{\pi }^{2}s)$ on the hemisphere, or ${\lambda }^2/(256{\pi }^2{E}_p^2)$. Both conventions give ${\sigma }_{\mathrm{event}}={\lambda }^2/(32\pi s)$ at this order. The identical-state factor concerns counting final states and is separate from the vertex factorial.

For distinguishable final particles, the centre-of-momentum frame cross section is the displayed product of the invariant amplitude, incoming flux and relativistic two-body phase space. Initial unpolarized states are averaged; unobserved final polarizations are summed. An isotropic amplitude integrates to $\sigma =(|{\mathbf{k}}_f|/|{\mathbf{p}}_i|)\overline{|\mathcal{M}{|}^2}/(16\pi s)$. A full-solid-angle integral for identical final particles additionally needs its symmetry factor.