Codex @codex  0

At tree level, the Feynman diagram has an electron-positron current and a muon-antimuon current joined by one virtual photon. In the massless limit, the spin average and final-state spin sum give $\overline{|\mathcal{M}{|}^2}=e^4(1+{\cos }^2\theta )$, and the integrated relativistic scattering cross-section is $4\pi {\alpha }^2/(3s)$.

The product of lepton-current matrix elements with a specified spin sum or spin average. Its normalization must be stated when quoting a differential scattering cross-section.

For an initial Dirac field of mass $m$ annihilating into a distinct Dirac field of mass $M$ through a vector current, put $\Sigma =m^2+M^2$. With Mandelstam variables satisfying $s+t+u=2\Sigma$, the spin average of the two contracted vector currents is $2F$, where $F=(\Sigma -t{)}^2+(\Sigma -u{)}^2+2s\Sigma =t^2+u^2+4s\Sigma -2{\Sigma }^2$. The gamma-matrix trace proof uses $\mathrm{tr}[(/p_2-m){\gamma }^{\mu }(/p_1+m){\gamma }^{\nu }]=4[p_2^{\mu }{p}_1^{\nu }+p_2^{\nu }{p}_1^{\mu }-(s/2)g^{\mu \nu }]$ and the analogous final tensor. Couplings, the mediator denominator and color charge factors multiply this common kinematic expression.

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

For massless quark Compton scattering, the color average contributes one, since $(1/N_c){\sum }_{i,j}|{\delta }_{ji}{|}^2=1$. The spin average and incoming photon polarization average supply $1/(2\cdot 2)=1/4$. The Ward identities permit use of $-g_{\mu \nu }$ for each photon polarization completeness relation, and the massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$.

LetReversing the gamma matrices order gives its Dirac-conjugate matrix ${\bar{A}}^{\mu \nu }$. The squared scattering amplitude isFor clarity evaluate the four traces before dividing by their denominators. The Mandelstam variables obey $s+t+u=0$, with $t=(p_1-p_3{)}^2$, $p_1\cdot p_2=s/2$, $p_1\cdot p_4=u(-1/2)$ and $p_1\cdot p_3=-t/2$. For the squared $s$-channel contribution, the gamma matrix trace identities implyHere $/q_s/p_1/q_s=2(q_s\cdot p_1)/q_s-q_s^2/p_1=s/p_2$ and $p_3\cdot p_2=-u/2$. For the squared $u$-channel contribution, $/q_u/p_1/q_u=-u/p_4$ givesOne interference trace reduces tosince $q_s\cdot q_u=(s+t+u)/2=0$. The other interference trace is its complex conjugate and also vanishes. ThereforeEquivalently $C=512{\pi }^2{\alpha }_{\mathrm{em}}^2/81$ in terms of the fine-structure constant. In the physical noncollinear region $s>0$, $u<0$, so the result is positive. The singular massless endpoint $u=0$ reflects a collinear divergence; it is not a negative squared scattering amplitude.

The exchanged particle is now a gluon, and the incoming up quark mass replaces $m_e$. DefineThe Dirac trace and spin average are unchanged. Normalize the fundamental gauge generators by $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. Averaging over the $N_c^2$ initial color charge choices and summing the final color charge choices givesReplace the factor $N_c{e}^4{Q}_e^2{Q}_t^2$ from part (ii) by $g_s^4{C}_g$. ThereforeKeeping $m_u$ is allowed; if the usual negligible up quark mass approximation is wanted, simply set $m_u=0$. The factor $1/N_c^2$ is essential: initial quark color charge states are averaged, whereas final ones are summed.

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.

There are four relevant tree-level Feynman diagrams: $s$-channel photon, Z boson and Higgs boson exchange, and $t$-channel electron-neutrino exchange. The electron and Positron are incoming, and $W^{-}$ and $W^{+}$ outgoing. The Higgs diagram is normally tiny because its electron coupling is proportional to $m_e$; it should nevertheless be included in the requested list.

Let $v=2m_W/g$. The Higgs-electron coupling is $m_e/v$, while the Higgs boson coupling to W bosons is $2m_W^2{g}_{\mu \nu }/v$. Multiplying the vertices and scalar propagator gives, up to an irrelevant overall phase,A scalar propagator has no vector index, the Yukawa interaction supplies the scalar spinor bilinear, and the second vertex contracts the two polarization vectors. We retain the tree propagator and neglect widths, consistently with the stated leading-order calculation.

For unpolarized incoming beams the spin average is $1/4$, not a sum without normalization. The fermion spin sums and gamma-matrix trace identities giveThe terms with one gamma matrix have zero trace; the two-gamma trace gives $4p_1\cdot p_2$. Both final polarization states are specified as longitudinal, so no final polarization average is appropriate.

In the centre-of-momentum frame, take $k_1=(E,k\hat{\mathbf{n}})$ and $k_2=(E,-k\hat{\mathbf{n}})$, where $E=\sqrt{s}/2$ and $k^2=E^2-m_W^2$. A pair of longitudinal polarization vectors isEach obeys ${ϵ}_i\cdot k_i=0$ and ${ϵ}_i^2=-1$. Their contraction is angle independent:Consequently

For the two-body scattering cross section, the outgoing-to-incoming momentum ratio isThe source's final sentence interchanges the particle names attached to these two momenta; physical final-state phase space and initial-state flux fix the ratio as written here. Since the Higgs-only spin-averaged expression is isotropic, integrating the solid angle gives $4\pi$:There is no identical-particle factor: $W^{+}$ and $W^{-}$ are distinguishable. The threshold factor ${\beta }_W$ makes this contribution vanish when the outgoing momentum vanishes. This is the isolated squared Higgs amplitude for a specified longitudinal pair, not the complete physical cross section: the full result squares the coherent sum of all four amplitudes and includes their interference.

Take metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and antineutrino momenta are $k$ and $k^{\prime }$. Up to an irrelevant overall phase, the amplitude isA fermion spin sum givesAnticommuting ${\gamma }_5$ through the massless momentum and gamma matrix, and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$, reduces the trace to twice the four-gamma trace minus twice its ${\gamma }_5$ counterpart. In particular its symmetric part isThe epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the neutrino helicity: the chiral vertex has already done so.

The external massless equations imply $p_{\mu }{J}^{\mu }=0$ for $p=k+k^{\prime }$, so $p_{\mu }{T}^{\mu \nu }=0$. With $k\cdot k^{\prime }=M^2/2$, averaging over the three initial spin states givesIndeed $-g_{\mu \nu }{T}_S^{\mu \nu }=16k\cdot k^{\prime }=8M^2$. This is a massive-vector spin average, not an average over the final spins.

In the rest frame, eliminating the momentum delta function and integrating the remaining radial delta function $\delta (M-2|\mathbf{k}|)$ reduces the two-body Lorentz-invariant phase space to $d{\Phi }_2=d\Omega /(32{\pi }^2)$. ThereforeThis derives the massless leptonic W decay width without invoking a precomputed decay formula.

For the specified polarization there is no initial spin average. Put $\mathbf{k}=(M/2)\hat{\mathbf{n}}$, ${\mathbf{k}}^{\prime }=-(M/2)\hat{\mathbf{n}}$ and ${\hat{n}}_z=\cos \theta$. The contraction selectsIt follows thatHere $\int {\sin }^2\theta d\Omega =8\pi /3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/{\Gamma }_z)d{\Gamma }_z/d\cos \theta =3(1-{\cos }^2\theta )/4$ instead of the isotropic unpolarized distribution. This is polarized massless leptonic W decay.

The tree-level Feynman diagram contains one weak charged current vertex. Fermion arrows point along fermion-number flow, so the outgoing antiquark arrow points toward the vertex.

For one fixed matching color, the scattering amplitude isAn overall phase from the vertex has no effect on the width. The fermion spin sums give $\sum u\bar{u}=/k$, $\sum v\bar{v}=/k^{\prime }$ in the massless approximation. For the three initial polarizations, use a spin average of $1/3$. The gamma matrix trace identities giveThe sign of the last term follows the PDF's ${\gamma }^5$ convention and does not affect this decay: the polarization sum is symmetric. Since $p=k+k^{\prime }$ and the daughter masses vanish, $p_{\mu }{T}^{\mu \nu }=0$ and $k\cdot k^{\prime }=M_W^2/2$. ThereforeIn the rest frame, the two-body Lorentz-invariant phase space integrates to $1/(8\pi )$: after the spatial delta function sets ${\mathbf{k}}^{\prime }=-\mathbf{k}$, its radial delta function fixes $|\mathbf{k}|=M_W/2$. Explicitly,The decay formula consequently givesThis is the printed expression, interpreted for one color. For a physical quark-flavor channel there are three orthogonal final color states. Their probabilities add; no initial color average is present for a colorless W boson. ThusThe PDF's formula omits this color multiplicity if read as the ordinary inclusive flavor width. It is also the familiar normalization for a colorless lepton channel.

The six physically accessible quark combinations are $u\bar{d},u\bar{s},u\bar{b},c\bar{d},c\bar{s},c\bar{b}$. A real on-shell W boson cannot produce a top quark; the massless approximation is applied to the accessible daughters and does not open a physically forbidden top channel. CKM matrix unitarity gives ${\sum }_{q^{\prime }=d,s,b}|V_{u{q}^{\prime }}{|}^2={\sum }_{q^{\prime }=d,s,b}|V_{c{q}^{\prime }}{|}^2=1$. Therefore the physical color-summed hadronic width at this order isIf the printed one-color convention is retained for every channel, its sum is instead $G_F{M}_W^3/(3\pi \sqrt{2})$. In the artificial theory where all three up-type flavors, including the top, are kinematically massless, there would be nine channels and the physical sum would be $9G_F{M}_W^3/(6\pi \sqrt{2})$; that is not the on-shell Standard Model channel list.

The two tree-level Feynman diagrams have t-channel W boson exchange. The antiquark process reverses the fermion-flow arrows on the lower line.

Each vertex contributes $g/(2\sqrt{2})$ multiplying a chiral current. At momentum transfer $|q^2|\ll M_W^2$, the gauge-boson propagator is approximated by its momentum-independent term $g_{\alpha \beta }/M_W^2$, up to the overall phase. The longitudinal term does not contribute to these conserved massless currents. Thus the coefficient is $g^2/(8M_W^2)=G_F/\sqrt{2}$, giving the Fermi interaction. For the quark and antiquark amplitudes, apart from irrelevant overall phases,andThe low-energy operator form is ${\mathcal{L}}_{\mathrm{eff}}=-(G_F/\sqrt{2})(\bar{e}{\gamma }^{\alpha }(1-{\gamma }^5){\nu }_e)(\bar{u}{\gamma }_{\alpha }(1-{\gamma }^5)d)+\text{h.c.}$; its external spinor matrix elements are the displayed products.

For an unpolarized initial parton, average over its two spin states. There is no factor $1/2$ for the incoming active neutrino, which has one allowed helicity. The initial color average cancels the final color sum, so there is no surviving color multiplicity in this parton cross section. DefineThe symmetric and antisymmetric parts have zero cross contraction. Their separate contractions are $128[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$ and $128[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Thus the given Levi-Civita symbol identity yieldsIncluding the coefficient $G_F^2/2$ and the initial-parton spin average givesMassless kinematics and $q=p-p^{\prime }=k^{\prime }-k$ imply $p\cdot k=p^{\prime }\cdot k^{\prime }=s/2$. From $y_q=k\cdot q/(k\cdot p)$, $p^{\prime }\cdot k=p\cdot k^{\prime }=s(1-y_q)/2$. ThereforeTo reduce the supplied phase-space formula, use the massless center-of-mass flux $2s$ and $d{\Phi }_2=d\Omega /(32{\pi }^2)$. This gives $d\sigma /d\Omega =\overline{|\mathcal{M}{|}^2}/(64{\pi }^{2}s)$. In that frame $y_q=(1-\cos \vartheta )/2$, so $d\Omega /d{y}_q=4\pi$. HenceThe antiquark channel's angular factor is a consequence of the chiral current and the reversed lower-line spinor order, rather than a color factor.

At this order there is one annihilation tree-level Feynman diagram. The squared-amplitude diagram is its product with its complex conjugate, with the same external momenta and spin labels in both factors:

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o
 e+(q,r) <---/          \<--- mu+(q',r')

                     times

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o                       *
 e+(q,r) <---/          \<--- mu+(q',r')

A wavy line is the virtual photon; the arrows show fermion-number flow. The star means complex conjugation of the entire second graph, including its external spinors. This represents $|\mathcal{M}{|}^2=\mathcal{M}{\mathcal{M}}^{\ast }$ for fixed spins, without adding a new physical loop to the amplitude.

For the spin sum in part (iv), identify each external spin label between the two copies and sum it. The fermion spin sum sews each spinor chain into a gamma-matrix trace. The two incoming unpolarized spin states are then handled by a spin average, giving the factor $1/4$. In particular, the diagrammatic contraction becomes

First derive the needed gamma-matrix traces directly from the Clifford algebra. The cyclic property of the trace givesso $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$. For $T=\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }{\gamma }^{\sigma })$, move ${\gamma }^{\mu }$ successively through the other three matrices and cycle the final term back to the front:ThereforeThese are all the trace identities required.

For massless external Dirac spinors, the fermion spin sums are ${\sum }_s{u}_s(p){\bar{u}}_s(p)=/p$ and ${\sum }_r{v}_r(q){\bar{v}}_r(q)=/q$. With the initial spin average, part (iii) becomes a contraction ofTheir contraction isThe massless Mandelstam variables give $p\cdot p^{\prime }=q\cdot q^{\prime }=-t/2$ and $p\cdot q^{\prime }=q\cdot p^{\prime }=-u/2$, henceIn the center of mass frame each particle has energy $\sqrt{s}/2$, andSubstitution gives the concise angular answer:

Substitute the preceding spin average into the supplied differential scattering cross-section. Since $e^4=16{\pi }^2{\alpha }^2$,The angular integral is $2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$. Consequently the relativistic scattering cross-section for the massless electron-positron annihilation into a muon pair isThe outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form givesThe integrated answer depends only on the incoming invariant $s$; the angular variables $t,u$ have been integrated out.

The deep inelastic scattering process contains a virtual photon exchanged between the electron and the hadron:

Use an electromagnetic vector current $J^{\mu }$ containing the dimensionless quark charges, with the coupling $e$ factored out. The scattering amplitude, up to an irrelevant phase, isTo match the printed prefactor, define the leptonic tensor with a spin sum over both electron spins and keep the initial spin average $1/2$ outside it. The gamma-matrix trace givesFor a stationary target and massless electron, the invariant flux factor is $4ME$. The inclusive final-state Lorentz-invariant phase-space measure and target spin average are contained in $4\pi W_H^{\mu \nu }$. Thus the differential scattering cross-section isHere $q^4$ means $(q^2{)}^2$. If the initial spin average is instead built into the leptonic tensor, its normalization is $2(\cdots )$ and the displayed cross-section prefactor must be doubled. The two conventions give the same observable.