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The inclusive product of hadronic-current matrix elements, summed over final states with a Lorentz-invariant phase-space measure. Electromagnetic current conservation makes both indices transverse to the exchanged four-momentum.

The tree-level two-body decay width from $2m_W^{2}H{W}_{\mu }^{+}{W}^{-\mu }/v$ isabove the on-shell threshold. The charged pair is distinguishable, so no identical-particle factor divides this result. Below threshold, physical multi-particle decays through virtual W bosons require a different Lorentz-invariant phase-space measure.

When the full Lorentz-invariant phase-space measure treats $n$ identical final particles as labeled, each physical final configuration is counted $n!$ times. The unlabelled relativistic scattering cross-section therefore includes $1/n!$. For two identical scalars, this divides the full-angle result by two; equivalently, integrate over a region containing only one representative of each exchanged pair.

With relativistically normalized external states, $\mathcal{M}$ is the reduced S-matrix element after removing the four-momentum conservation distribution. A relativistic scattering cross-section combines $|\mathcal{M}{|}^2$ with the Lorentz-invariant phase-space measure and the invariant flux factor. Identical-leg factorials in Feynman vertices and the identical final-state symmetry factor in the event integral have different roles.

The invariant future mass-shell measure is $d^{3}p/E$, up to fixed normalization factors absorbed in $f$. Its relation to the Lorentz-invariant phase-space measure follows by integrating $2\theta (p^0)\delta (p_{\mu }{p}^{\mu }+m^2)d^{4}p$ over $p^0$ in a local orthonormal tetrad. A particle carries four-momentum $p^{\mu }$ and crosses a local spatial surface with velocity $p^i/p^0$. Integrating the associated momentum flux gives the displayed stress-energy tensor. For photons, $p^{\hat{a}}=E(1,\mathbf{e})$ and $|\mathbf{e}|=1$.

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.

Use an incoming spacelike momentum transfer $q$, so the physical absorption process has $p_X=P+q$, $Q^2=-q^2>0$, and $P\cdot q>0$. There is a sign inconsistency in the original PDF: its delta function has $P-p_X-q$, while its positive Bjorken x definition and target formulas use incoming $q$. With that literal delta function and $x>0$, the final-state invariant would bewhich becomes negative in the stated high-energy limit and cannot represent a physical hadronic final state. We therefore use ${\delta }^4(P+q-p_X)$, preserving the printed $x=Q^2/(2P\cdot q)$ and structure-function conventions. Equivalently, keeping an outgoing momentum variable requires reversing the signs in those definitions as well.

Assume an unpolarized target, incoherent scattering from effectively free spin-one-half partons, and the massless collinear parton approximation. In the Bjorken scaling limit, take $Q^2,P\cdot q\to \infty$ at fixed $0<x<1$, neglecting target and quark masses and intrinsic transverse momentum relative to $Q$. Write the incoming struck-quark momentum as $k=\xi P$ in the leading lightlike-reference approximation. Let $f_f(\xi )$ and ${\bar{f}}_f(\xi )$ be quark and antiquark parton distribution functions, counting all colours. For the naive scaling result also neglect radiative scaling violations; in full QCD these distributions depend logarithmically on the hard scale.

The outgoing struck quark has momentum $k^{\prime }=k+q$. Its on-shell condition givesAverage its initial spin and sum its final spin. The spin-averaged electromagnetic parton tensor has numeratorThe supplied three-gamma identity gives this directly: contract its middle index with $(k+q{)}_{\lambda }$ and trace with $/k$; the ${\gamma }_5$ trace vanishes. Thus no antisymmetric term survives the unpolarized spin average.

Define transverse tensorsOn the cut, $2k\cdot q=Q^2$, so $\tilde{k}=k+q/2=\xi \tilde{P}$. Expanding the two transverse terms givesThis also explicitly verifies electromagnetic current conservation, $q_{\mu }{S}^{\mu \nu }=q_{\nu }{S}^{\mu \nu }=0$.

The one-particle final-state Lorentz-invariant phase-space measure obeysIncluding the current's charge factor and the given $1/(4\pi )$ normalization, a single quark therefore contributesAn antiquark has the same unpolarized tensor and the same squared charge $Q_f^2$.

The parton convolution normalization for a hadronic tensor isThe $1/\xi$ is important: the distributions count parton numbers, whereas covariantly normalized one-particle responses carry the state normalization $2k^0$. The incoherent sum of normalized responses is $W/(2P^0)={\sum }_f\int d\xi (f_f+{\bar{f}}_f)w_f/(2k^0)$; in the collinear frame $k^0=\xi P^0$, giving the factor above. There is no extra colour multiplicity because it is already included in the distributions.

On the physical cut,Comparing the two transverse coefficients in the hadronic tensor gives, with $n_f=f_f+{\bar{f}}_f$,Therefore the requested deep-inelastic structure functions areThe notation $f_f$ here is the question's $f$ with its flavour index made explicit. The calculation proves the Callan-Gross relation, $F_2=2x{F}_1$, rather than merely citing it. Finite masses and radiative corrections modify this leading approximation; at leading QCD order the same formulas use scale-dependent parton distribution functions.

The quark number sum rule follows because a distribution is a number density. Integrating $f_f$ counts quarks and integrating ${\bar{f}}_f$ counts antiquarks; their difference is the net conserved flavour charge:Every sea quark-antiquark pair contributes zero to this difference. More formally the flavour-vector charge counts quarks minus antiquarks and has a fixed eigenvalue on the hadron state. Thus $N_f$ is the valence or net quark number, not the total number including the sea.

Use both final assumptions printed in the question: only up/down flavours and no antiquarks. The second does not follow just from restricting the flavour list. Then $\int f_u=N_u$, $\int f_d=N_d$, with squared charges $Q_u^2=4/9$ and $Q_d^2=1/9$. The proton has $(N_u,N_d)=(2,1)$ and the neutron $(1,2)$. Hence the valence-only proton and neutron structure-function moments areThe integral limits assume the scaling approximation is valid for the indicated weighted moments, for example by dominated convergence of the weighted integrands. The same net-quark sum rule alone would not fix these individual moments in the presence of antiquarks, since $F_2$ involves the sum of quark and antiquark distributions.

Use natural units and the Minkowski metric $(+---)$, and work at tree level with on-shell final particles. Write $M=m_H$ and, for a final particle of mass $m$, ${\beta }_m=\sqrt{1-4m^2/M^2}$. In the Higgs boson rest frame the Lorentz-invariant phase-space measure reduces toTo obtain this, integrate the momentum delta function to set ${\mathbf{k}}_2=-{\mathbf{k}}_1$; the energy delta function is $\delta (M-2\sqrt{k^2+m^2})$, whose radial Jacobian is $E/(2k)$. Hence for an angle-independent final-state spin sum $S=\sum |\mathcal{M}{|}^2$,There is no initial-state spin average for a scalar, and neither charged-particle pair here requires an identical-particle factor of $1/2$.

For Higgs decay to two W bosons, the Feynman vertex gives $\mathcal{M}=(2m_W^2/v){\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$, up to an overall phase. Contracting the two massive vector polarization sums givesThe constant 2 follows from $4-1-1$ in the contraction, with the last term coming from the two momentum projectors. Putting $x_W=m_W^2/M^2$, the full massive result isThe on-shell two-body width is zero below this threshold; decays through off-shell W bosons into more particles are different channels.

For Higgs decay to a fermion pair, put $y_b=m_b/v$. The amplitude is $\mathcal{M}=-y_b\bar{u}(k_1)v(k_2)$, up to an overall phase and a color Kronecker delta. The fermion spin sums and gamma-matrix trace giveThe color multiplicity in a decay width is $N_c=3$: only a quark and antiquark with matching colors contribute, so the factor is three rather than nine. ThereforeAgain the on-shell two-body width is zero below threshold. This is a partonic tree-level answer with every mass retained; hadronization is outside the specified calculation.

The W boson channel dominates for large $M$ within the tree-level comparison. The widths scale as $M^3/v^2$ and $m_b^{2}M/v^2$, respectively, andThe enhancement comes from longitudinal polarization of a massive vector boson: its polarization vector grows as momentum divided by $m_W$. Thus the longitudinal pair survives the apparently small $m_W^2$ factor in the interaction. At masses so large that the scalar sector is strongly coupled, the tree-level extrapolation itself needs corrections.

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 metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the usual Dirac basis, with $({\gamma }^0{)}^{†}={\gamma }^0$ and $({\gamma }^i{)}^{†}=-{\gamma }^i$. In this solution the sign of the chirality matrix is the one specified in the paper, ${\gamma }^5=-i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$. Define $t_0=1$ and $t_i=-1$, with no summation in componentwise transformation formulas.

Complex conjugation changes the explicit $-i$ to $+i$. Three spatial gamma matrices each contribute another minus sign, soAlso $({\gamma }^5{)}^2=1$ and $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ by the Clifford algebra.

The quantum time-reversal operator is antiunitary: it conjugates numerical coefficients, including the Dirac spinors and plane-wave exponentials in the mode expansion of a Dirac field. To fix the spin phase explicitly, put ${ϵ}_s=(-1{)}^{1/2-s}$ and useThis convention gives ${\hat{T}}^2=-1$ on a one-fermion state, since ${ϵ}_s{ϵ}_{-s}=-1$. In the transformed expansion, change variables from $(p,s)$ to $(p_T,-s)$. The Lorentz-invariant phase-space measure is unchanged, and $p_T\cdot x=-p\cdot x_T$. The coefficient of $b^s(p)e^{-ip\cdot x_T}$ is thereforeThe same calculation applies to the antiparticle coefficient. Thus, with this explicitly fixed spin convention,The spin phase in the Dirac time-reversal matrix matters here: a common change of the one-particle time-reversal phase replaces $B$ by $-B$; the phase-independent relation is $B\propto {\gamma }^{5}C$. It changes neither the defining conjugation property nor any bilinear result below. Normalize $B$ to be a unitary matrix. In the Dirac basis, ${\gamma }^0$ is real and commutes with $B$, so transforming the Dirac adjoint givesHere complex conjugation of the matrix defining the Dirac adjoint is essential. An explicit realization consistent with the sign of ${\gamma }^5$ used here is $C=i{\gamma }^2{\gamma }^0$ and $B={\gamma }^1{\gamma }^3$; it has $B^{-1}=-B$, so conjugation by $B$ and by $B^{-1}$ coincides. The usual changes of Dirac spinor basis by unitary matrices carry the adjoint and time-reversal matrix with them.

For the charge-conjugation matrix, the gamma matrix adjoint and transpose identities give ${\gamma }^{\mu T}=t_{\mu }{\gamma }^{\mu \ast }$. Consequently $B{\gamma }^{\mu T}{B}^{-1}={\gamma }^{\mu }$. Since $C={\gamma }^{5}B$ in the chosen phase convention,This proof uses a temporal Hermitian matrix and spatial skew-Hermitian matrices explicitly; the transpose identity is preserved when $C$ is transformed appropriately with the gamma matrices.

For the Fermi interaction, let $V_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }n$ and $A_{pn}^{\mu }=\bar{p}{\gamma }^{\mu }{\gamma }^{5}n$. In the displayed Dirac basis, the inverse version of the conjugation relation also holds. Antiunitarity, $B^{-1}{\gamma }^{\mu \ast }B=t_{\mu }{\gamma }^{\mu }$ and $B^{-1}{\gamma }^{5\ast }B={\gamma }^5$ giveThe leptonic weak charged current has the same component signs. In the contraction of leptonic and hadronic currents, the two $t_{\mu }$ factors cancel. Its two independent operators thus keep their form while their coefficients become $g_V^{\ast }$ and $g_A^{\ast }$. The Hermitian-conjugate term transforms separately; its presence does not remove a relative complex phase between the vector and axial couplings.

With all intrinsic phases fixed to one and a real positive Fermi constant, invariance requires real coefficients in that convention. A common phase of $g_V$ and $g_A$ can instead be absorbed into a rephasing of the nucleon fields, and hence into their intrinsic time-reversal phases. The convention-independent condition, for $g_V\ne 0$, isThis is the relative weak phase condition for time reversal. If one coefficient vanishes, there is no relative phase to constrain; the remaining common phase can be removed. A nonreal ratio violates time-reversal symmetry even though the Lagrangian density includes its Hermitian conjugate.

The strong-interaction matrix element between two spin-zero pseudoscalar mesons has only $p^{\mu }$ and $k^{\mu }$ available. The product of the two intrinsic parities is positive. An axial current would require a pseudovector constructed from these momenta, but an expression involving the Levi-Civita symbol needs three independent four-vectors and therefore vanishes. This is a consequence of parity conservation in the hadronic matrix element, not of parity conservation in the weak interaction. The vector current can have the two independent structures $p+k$ and $p-k$. Its coefficients are Lorentz scalars; with $p^2=m_K^2$ and $k^2=m_{\pi }^2$ fixed, their only varying invariant is $s=q^2$. HenceThese are the pseudoscalar-to-pseudoscalar form factors. With relativistically normalized states they are dimensionless. The vanishing axial matrix element and this decomposition explain the two equalities separately.

Write $q_1$ and $q_2$ for the outgoing electron and antineutrino momenta. From the Fermi interaction, an invariant scattering amplitude, up to an irrelevant overall sign or phase, isThe CKM matrix element $V_{us}$ multiplies the quark current in the convention of this interaction. Let ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$. For massless leptons, the massless Dirac equation and chirality matrix anticommutation giveIn the second term move $/q_2$ through the chiral projector before applying $/q_{2}v=0$. Since $p+k=2p-q$, this transverse massless leptonic current givesThe disappearance of $f_{-}$ uses the massless approximation; for a massive charged lepton its contraction is proportional to the lepton mass.

Use the fermion spin sum and the supplied gamma matrix trace identities. The symmetric part of the leptonic tensor isThe Levi-Civita symbol term is antisymmetric and drops out when contracted with $p^{\mu }{p}^{\nu }$. ThusThere is no initial-spin average because the kaon is spinless. For the integrated massless leptonic tensor, keep every factor of $2\pi$ explicit and define the unnormalized two-lepton Lorentz-invariant phase spaceThe leptons are massless, so $q_i^0=|{\mathbf{q}}_i|$. Its trace is $g^{\mu \nu }{I}_{\mu \nu }=\pi s$. ThereforeThe three Lorentz-invariant phase-space measures and their momentum delta function contribute $1/[8(2\pi {)}^5]$, in addition to $1/(2m_K)$ in the decay rate. Combining them with the spin sum givesThis massless semileptonic pseudoscalar decay rate uses a two-lepton integral over future-timelike $q$ and the pion integral is restricted to the physically allowed region. The null endpoint follows by continuity. The coefficient has mass dimension $-5$, so the complete expression has mass dimension one, as a decay rate must in natural units.

In the kaon centre-of-momentum frame, put $E=k^0$ and $\kappa =|\mathbf{k}|$. ThenThe dimensionally consistent Källén function isThe pion-only mass term must have fourth power: the second power printed in the PDF is dimensionally inconsistent. This repair also follows directly from squaring $E=(m_K^2+m_{\pi }^2-s)/(2m_K)$. Angular integration and the change of variable givewhere the negative sign reverses the endpoints. Combining this with the bracket $\lambda /4$ yieldsThus $\boxed{B=G_F^2|V_{us}{|}^2/(192{\pi }^3{m}_K^3),a=0,b=(m_K-m_{\pi }{)}^2}$. The lower limit is the minimum invariant mass of two massless leptons; at the upper limit the pion is at rest. The coefficient has mass dimension $-7$, while $ds{\lambda }^{3/2}$ has dimension eight. The Källén function also shows why the differential decay rate vanishes at zero pion momentum.

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.

Assume a complete relativistically normalized basis of scalar momentum eigenstates transforms asThis states both reversal of spatial momentum and preservation of the normalization of basis quantum states. Antilinearity alone would not suffice: multiplying a conjugation operator by two gives an antilinear operator that multiplies squared norms by four.

Expand $|\varphi ⟩=\int d{\Pi }_p\varphi (\mathbf{p})|\mathbf{p}⟩$ and similarly for $|\psi ⟩$, with $d{\Pi }_p=d^{3}p/((2\pi {)}^{3}2E_p)$. This Lorentz-invariant phase-space measure is unchanged under $\mathbf{p}\mapsto -\mathbf{p}$. The antilinearity of the quantum time-reversal operator gives conjugated expansion coefficients. Orthogonality of the momentum eigenstates and cancellation of their unit-modulus phases yieldSince momentum reversal is a bijection of the complete basis, $\hat{T}$ is also onto. ThusThis proves antiunitarity, with the normalization hypothesis explicitly included. For scalar multiparticle quantum states the same argument uses the complete occupation-state basis and reverses all momenta; it is not restricted to a single-particle wave packet.