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Label the incoming momenta and spin states $(p_1,r_1),(p_2,r_2)$ and the outgoing ones $(p_3,r_3),(p_4,r_4)$, with $p_1+p_2=p_3+p_4$. Write $u_i=u^{r_i}({\mathbf{p}}_i)$ and $v_i=v^{r_i}({\mathbf{p}}_i)$. The Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. At lowest order, each connected Feynman diagram has two pseudoscalar Yukawa interaction vertices and one internal scalar propagator.

Define the scattering amplitude by $S_{fi}^{\mathrm{connected}}=i(2\pi {)}^4{\delta }^4(p_1+p_2-p_3-p_4)\mathcal{M}$. Fix the external Fock state ordering as $b_1^{†}{b}_2^{†}|0⟩$ to $b_3^{†}{b}_4^{†}|0⟩$ for the two-fermion process. The $t$ exchange connects $1$ to $3$ and $2$ to $4$; the $u$ exchange connects $1$ to $4$ and $2$ to $3$. The two Feynman vertices contribute $(-i\lambda {)}^2$, and the internal scalar propagator contributes $i$ divided by its denominator. Exchanging the two identical final fermions supplies a relative fermionic sign. ThusThis changes sign when the two outgoing labels are exchanged, as identical-fermion antisymmetry requires. There is no $s$ annihilation diagram for two incoming fermions: the interaction preserves Dirac fermion number conservation.

For the fermion-antifermion process take initial state $b_1^{†}{d}_2^{†}|0⟩$ and final state $b_3^{†}{d}_4^{†}|0⟩$. The $t$ exchange uses bilinears $({\bar{u}}_3{\gamma }^5{u}_1)({\bar{v}}_2{\gamma }^5{v}_4)$, and the $s$ annihilation uses $({\bar{v}}_2{\gamma }^5{u}_1)({\bar{u}}_3{\gamma }^5{v}_4)$. To fix the sign without a guess, expand the current $J=:\bar{\psi }{\gamma }^5\psi :$ in creation and annihilation operators. Its relevant terms have the signsThe antifermion sign of a normal-ordered bilinear makes the exchange contraction negative, whereas the product of annihilation and creation contractions is positive for the specified external ordering. Multiplying by $(-i\lambda {)}^{2}i=-i{\lambda }^2$ therefore givesA different common phase convention for an external state can reverse the overall amplitude, but never the relative sign of its two contributions. Both results are tree scattering with pseudoscalar exchange and are of order ${\lambda }^2$.

The last panel of the preceding process figure shows deep inelastic scattering through a virtual photon. The blob represents the electromagnetic hadronic tensor response, with all final hadronic states included. With charges factored into $e$ and the hadronic current as in the question, a suitable scattering amplitude, up to an irrelevant overall phase, isThe electron fermion spin sum, before averaging the two initial spin states, gives the following normalization of the leptonic tensor:The initial electron spin average is a separate factor $1/2$. Equivalently one can absorb it into a tensor $L/2$, but the prefactor in the requested formula uses the unaveraged trace $L$. The massless Dirac equations imply $q^{\mu }{L}_{\mu \nu }=q^{\nu }{L}_{\mu \nu }=0$, the relevant Ward identity.

In the target rest frame the incident flux is $4p\cdot P=4ME$. This remains valid for a massive target because the electron is massless: the general flux $4\sqrt{(p\cdot P{)}^2-p^2{P}^2}$ reduces to that expression. The outgoing electron contributes $d^3{p}^{\prime }/((2\pi {)}^{3}2E^{\prime })$, and the inclusive final-state sum is $4\pi W_H^{\mu \nu }$. Including the electron spin average therefore givesThe implicit hadron spin average in $W_H$ must not be inserted a second time.

For an unpolarized target, Lorentz invariance permits tensor structures formed from $P$, $q$ and the metric. Hermiticity relates the two index orders. The electromagnetic response is parity even, excluding the pseudotensor $i{ϵ}^{\mu \nu \rho \sigma }{P}_{\rho }{q}_{\sigma }$. A possible ordinary antisymmetric $P^{\mu }{q}^{\nu }-q^{\mu }{P}^{\nu }$ term is excluded by current conservation. The symmetric part initially has four structures, $g^{\mu \nu }$, $P^{\mu }{P}^{\nu }$, $q^{\mu }{q}^{\nu }$ and $P^{\mu }{q}^{\nu }+q^{\mu }{P}^{\nu }$. Imposing both current-conservation conditions leaves the transverse decomposition of an unpolarized hadronic tensor:For example, contracting $q$ into the first structure gives $(1+A)q^{\mu }$ in the printed parametrization, and contracting into either parenthesis gives $(1+B)\nu$ or $(1+C)\nu$. Since the two independent deep-inelastic structure functions are not generally linked by symmetry, transversality requiresThere are only two invariant variables once $P^2=M^2$ is fixed; they may be chosen as $Q^2$ and Bjorken x $x=Q^2/(2\nu )$.

Now make the leading massless collinear parton approximation. In the hard parton calculation, use a lightlike collinear reference for $P$, neglecting target-mass effects of relative order $M^2/Q^2$ as well as quark masses. Literally combining a massive $P$ with an exactly massless constituent momentum $\xi P$ would be inconsistent. The exact rest-frame flux derived above does not require this approximation; it is the following partonic response that does.

Let $r=\xi P$ and $k=r+q$. Strip the quark charge out of the current, since its square is already the factor $Q_f^2$ in the parton convolution. Averaging the initial quark spin and summing the final spin gives the spin-averaged electromagnetic parton tensorThe spatial momentum delta fixes $\mathbf{k}=\mathbf{q}+\xi \mathbf{P}$. The remaining one-particle Lorentz-invariant phase space, together with $1/(4\pi \xi )$ in the definition, suppliesSince $r^2=k^2=0$ and $r\cdot k=\xi P\cdot q$, the full parton response isTerms with a free $q$ index vanish when contracted with the conserved leptonic tensor. Thus the part of $\tilde{W}$ contributing to the cross-section iswhere $\doteq$ denotes equality in that contraction, rather than an equality of the full tensors. The omitted terms are necessary for the full hadronic tensor to be transverse; the wording about contributing terms in the PDF is important here.

To prove the delta identity with the correct Jacobian, set $\omega =E_q+\xi E_P$. On physical support $E_k>0$ and $\omega >0$. Factoring $E_k^2-{\omega }^2=(E_k-\omega )(E_k+\omega )$ gives the positive-energy on-shell delta function identityThis is restricted to the positive-energy branch; the other root must not be included. Using the spatial momentum constraint and $P^2=0$ in the leading hard approximation,The Dirac delta function change of variable therefore gives the parton on-shell delta identityHere $\nu >0$. This is the actual printed identity: the converted TeX's final $\delta (x)/E_k$ is a transcription error. If target mass is formally retained in the delta argument, it becomes $Q^2-2\xi \nu -{\xi }^2{M}^2$, so the stated simple root and Jacobian no longer apply.

Let $D(x)={\sum }_f{Q}_f^2(q_f(x)+{\overline{q}}_f(x))$. The parton integral collapses at $\xi =x$, givingThe transverse decomposition of an unpolarized hadronic tensor, in the same contraction, is $-g^{\mu \nu }{F}_1+P^{\mu }{P}^{\nu }{F}_2/\nu$. Comparing the two independent structures yieldsThis derives the Callan-Gross relation from the spin-one-half trace and on-shell constraint. Antiquarks give the same symmetric response and therefore the sum $q_f+{\overline{q}}_f$. The result applies at leading order in the massless parton model; target masses, intrinsic transverse momentum and Quantum chromodynamics radiative effects can generate a nonzero longitudinal deep-inelastic structure function and modify the relation.

The interaction contains no time derivatives, so $\pi =\dot{\varphi }$. The Legendre transform in mechanics givesThusHere $H_0=\int {\mathcal{H}}_0$ is the free Hamiltonian operator of mass $m$, and $H_I=\int {\mathcal{H}}_I$ is the interaction Hamiltonian.

In the interaction picture, operators evolve with $H_0$ and the states with $H_I(t)=e^{i{H}_{0}t}{H}_I{e}^{-i{H}_{0}t}$. With $U_I(t_0,t_0)=I$, the state evolution satisfies $i{\partial }_t{U}_I(t,t_0)=H_I(t)U_I(t,t_0)$. Integrating once and iterating gives the Dyson seriesEquivalently it is the time ordering of the exponential. Taking the infinite-time scattering limit, with adiabatic switching understood, gives the scattering matrixThe field in the interaction is the free interaction-picture field, which is essential for applying the free-field contractions.

Use relativistic normalization of a one-particle state, $|\mathbf{p}⟩=\sqrt{2E_{\mathbf{p}}}{a}^{†}(\mathbf{p})|0⟩$, so each external scalar contraction contributes its plane-wave phase without an extra $1/\sqrt{2E}$ factor. For the connected on-shell two-to-two scattering amplitude, the first Dyson series term is $-i\lambda \int d^{4}x{\varphi }_I(x{)}^4/4!$. The Wick theorem gives $4!$ ways to attach the two incoming and two outgoing external particles to the four fields, canceling the denominator. ConsequentlywhereThe delta function expresses four-momentum conservation. The connected scattering convention removes spectator, vacuum and external self-energy contributions; equivalently at this order one can normal-order the interaction and keep the physical external mass fixed. These contributions should not be mistaken for additional connected four-point diagrams.

Use the printed interaction, so the Dyson series contributes $-ig$ per vertex before its spinor matrix. The momentum-space Feynman rules areEach interaction vertex has one pseudoscalar line and one incoming and one outgoing fermion-number line. Conserve four-momentum at every vertex, integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, and divide by any Feynman-diagram symmetry factor. A closed fermion loop contributes an extra minus sign. The external wavefunctions are $u$ for incoming fermions, $\bar{u}$ for outgoing fermions, $\bar{v}$ for incoming antifermions and $v$ for outgoing antifermions; spinor matrices are multiplied in order along the fermion line. An external pseudoscalar has unit wavefunction factor in standard relativistic normalization. Relative fermionic signs also occur when the external contractions are permuted.

The leading scattering amplitudes are of order $g^2$. Label the incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, all with positive energy and $p_i^2=m^2$. Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The tree scattering with pseudoscalar exchange is shown below; dashed edges are pseudoscalar propagators, and arrows indicate fermion-number flow, opposite to the motion of an antifermion.

Define the scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^4(p_3+p_4-p_1-p_2)\mathcal{M}$. Use standard relativistic normalization for external states and spinors. To fix even the overall signs, suppress the positive state-normalization factors and write the creator order of the two-fermion states as $|i⟩=b_1^{†}{b}_2^{†}|0⟩$, $|f⟩=b_3^{†}{b}_4^{†}|0⟩$, and the fermion-antifermion states as $|i⟩=b_1^{†}{d}_2^{†}|0⟩$, $|f⟩=b_3^{†}{d}_4^{†}|0⟩$. Spin labels on the $u_i,v_i$ are implicit.

For two fermions, the direct $t$ graph connects $1\to 3$ and $2\to 4$, while the exchanged $u$ graph connects $1\to 4$ and $2\to 3$. The canonical anticommutation relations give signs $+$ and $-$ for these contractions in the chosen state order. Multiplication of the two vertices and the scalar propagator gives $(-ig{)}^{2}i=-i{g}^2$. HenceInterchanging the labels of either pair of identical external fermions reverses this amplitude, as required for the antisymmetric states of fermionic Fock space.

For a fermion and an antifermion, the graphs are $t$ exchange and $s$ annihilation. Their signs can be derived without guessing an analogy with the first process. The normal-ordered product of the pseudoscalar bilinear contains terms of the formsThe minus in the second term is the antifermion sign of a normal-ordered bilinear. Between the chosen external states, exchange has sign $-$ and pair annihilation followed by pair creation has sign $+$. ThereforeChanging an external-state phase can change the common overall sign; it cannot change the relative minus between the two channels. Both amplitudes use the coefficient exactly as printed. For the Hermitian convention $g=i{g}_P$ discussed in part (a), substitute $g^2=-g_P^2$ throughout, or equivalently use vertex $g_P{\gamma }^5$ from the outset.

The asymptotic scattering wavefunction is $\Psi \sim e^{ikz}+f(\theta )e^{ikr}/r$. For a real central potential, the scattering phase shift ${\delta }_l$ multiplies the outgoing component of each partial wave relative to the incoming one by $e^{2i{\delta }_l}$. Matching the incident-wave normalization givesSubtract the given incident plane wave expansion. Since $e^{2i{\delta }_l}-1=2i{e}^{i{\delta }_l}\sin {\delta }_l$, the scattering amplitude isOrthogonality ${\int }_{-1}^1{P}_l{P}_{j}dx=2{\delta }_{lj}/(2l+1)$ eliminates cross terms in $\sigma =\int |f{|}^{2}d\Omega$, giving $\boxed{\sigma =(4\pi /k^2)\sum _l(2l+1){\sin }^2{\delta }_l}$.

For the spherical square well, the regular reduced radial $l=0$ wave is $A\sin (\kappa r)$ inside and $B\sin (kr+{\delta }_0)$ outside, with ${\kappa }^2=k^2+{\gamma }^2$. Continuity of the wave and its derivative gives the logarithmic-derivative matching relation $\kappa \cot (\kappa a)=k\cot (ka+{\delta }_0)$, orAt zeros of the denominators the original matching relation supplies the limiting interpretation.

Under the stated deep-well approximation, put $x=ka$ and $B_0=\tan (\gamma a)/(\gamma a)$. The condition ${\delta }_0\in \pi \mathbb{Z}$ is $\tan x/x=B_0$. If $B_0<0$, its first positive solution lies in $(\pi /2,\pi )$; if $B_0=0$, it is $\pi$. For $0<B_0\le 1$, a solution lies in $(\pi ,3\pi /2)$ because there $\tan x/x$ rises continuously from zero to infinity. If $B_0>1$, the first positive solution lies in $(0,\pi /2)$, where $\tan x/x$ rises from one to infinity. The limiting pole case gives $x=\pi /2$ by the cotangent relation. Thus $\boxed{0<k_0<3\pi /(2a)}$, without calculating the integer phase index. The use of $\kappa \simeq \gamma$ presumes the located wave numbers remain small compared with $\gamma$.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

The conjugate complex scalar field has the opposite charge. Thus the derivative on ${\varphi }^{\ast }$ in the kinetic term must mean $(D_{\mu }\varphi {)}^{\ast }=({\partial }_{\mu }-ie{A}_{\mu }){\varphi }^{\ast }$, while the derivative on $\varphi$ is $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. This is the conjugate gauge covariant derivative. Applying the same plus-charge differential operator to both fields would not give the stated gauge-invariant theory. For example, take $A=0$, $\varphi =x^0$ and $\alpha =c{x}^0$; the literal plus-charge kinetic product changes from $1$ to $1+2iec{x}^0$. The opposite-charge interpretation removes this term.

For an arbitrary real function $\alpha (x)$, takeThen $D_{\mu }^{\prime }{\varphi }^{\prime }=e^{-ie\alpha }{D}_{\mu }\varphi$ and the conjugate derivative transforms oppositely. Also $F_{\mu \nu }^{\prime }=F_{\mu \nu }$ because mixed derivatives commute. The kinetic contraction and $|\varphi {|}^2$ are unchanged, proving local U(1) gauge symmetry.

Expanding the covariant kinetic term identifies the interactions:The first term gives the scalar electrodynamics three-point vertex, and the second gives the seagull vertex. For scalar charge flow from an incoming particle of momentum $p$ to an outgoing one of momentum $p^{\prime }$, the factors areThe factor two in the second rule comes from the two identical photon fields. Equivalently, with all momenta incoming, the three-point rule is $ie(r-q{)}^{\mu }$ for a $\varphi$ leg of momentum $q$ and a ${\varphi }^{\ast }$ leg of momentum $r$. An antiparticle line has the opposite charge-flow sign. The vertices are shown with dashed scalar lines and wavy photon lines:

For particle-antiparticle scattering, use incoming $p_1,p_2$ and outgoing $p_3,p_4$, with $p_1,p_3$ particle momenta. At order $e^2$, there are exactly two tree-level Feynman diagrams: $t$-channel photon exchange and $s$-channel annihilation. The seagull vertex has only two scalar legs and cannot alone provide four external scalar legs.

Set $q_t=p_1-p_3=p_4-p_2$ and $q_s=p_1+p_2=p_3+p_4$. The external currents areAll are transverse to the corresponding exchanged momentum: for instance $q_t\cdot J_t=p_1^2-p_3^2=0$ and $q_s\cdot J_s=p_1^2-p_2^2=0$. This is the on-shell scalar quantum electrodynamics Ward identity. With the vertex convention above,The relative sign comes from the opposite particle/antiparticle charge in the exchange diagram and the two equal annihilation-vertex signs.

To see how the Coulomb gauge expression becomes covariant, let $J,K$ be either pair, $q=(q^0,\mathbf{q})$, and initially take $|\mathbf{q}|\ne 0$. The displayed photon propagator givesCurrent conservation gives $\mathbf{J}\cdot \mathbf{q}=q^0{J}^0$ and the same identity for $K$. Since $q^2=(q^0{)}^2-|\mathbf{q}{|}^2$, the temporal coefficient simplifies asThereforeThe common Feynman pole prescription is understood in these formulas. This Coulomb-gauge propagator between conserved currents identity proves equality of the physical amplitudes despite the noncovariant individual propagator components. It should be formed before taking limits such as $\mathbf{q}\to 0$: individual instantaneous and transverse terms can be undefined separately in that limit while their sum has a finite covariant limit away from a physical pole.

Substitution gives the final scattering amplitudeHere $s,t,u$ are the Mandelstam variables for the equal-mass external scalars. This is scalar particle-antiparticle tree scattering; overall external-state phases do not affect its relative channel sign.

Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol ${\rho }_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere $d{\Pi }_X$ is the product of final-particle measures $d^{3}k/[(2\pi {)}^{3}2E]$, including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention ${\rho }_h$ is the hadronic photon spectral density, with mass dimension $-2$, rather than a density of a stationary time series.

The electron annihilation amplitude can be written $e\bar{v}(p_{+}){\gamma }_{\mu }u(p_{-})⟨X|A^{\mu }|0⟩$, up to an overall phase. The leptonic tensor after the initial spin average isIt obeys $q^{\mu }{L}_{\mu \nu }=0$ and $g^{\mu \nu }{L}_{\mu \nu }=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt{s}/2$. Contracting the tensors gives the inclusive relativistic cross-sectionHere $\alpha$ is the fine-structure constant. The $1/4$ in the initial spin average is essential because the hint's particles were spinless.