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For three real scalar fields with field interaction term ${\mathcal{L}}_I=-\frac{1}{2}(g_1{\varphi }^2\Phi +g_2{\psi }^2\Phi )$ and mediator mass $M$, the connected tree-level Feynman diagram for $\varphi \psi \to \varphi \psi$ has one exchanged $\Phi$ and Mandelstam invariant $t=(p-p^{\prime }{)}^2$. Each vertex has a factor $-i{g}_j$: its two identical-field Wick contractions cancel the factor $1/2$. Thus $i\mathcal{M}=(-i{g}_1)(-i{g}_2)i/(t-M^2+i0)$. There is no $s$- or $u$-channel at this order because the interaction has no mixed $\varphi \psi \Phi$ vertex. Distinguishable external particles also require no exchange symmetrization.

The leading scattering amplitude is the sum of two tree-level Feynman diagrams, with the outgoing photons attached in either order to the fermion line. The Dirac propagator carries momentum $p-k_1$ or $p-k_2$ for incoming electron momentum $p$. Their sum obeys the two-photon fermion Ward identity, although individual diagrams need not. Crossing symmetry relates this amplitude to Compton scattering.

The four-point tree-level Feynman diagram has two cubic vertices joined by one propagator. The three ways to divide four labeled external legs into vertex pairs give the $s$, $t$, and $u$ channels. For interaction $+\lambda {\varphi }^3/3!$, the amplitude is $\mathcal{M}=-{\lambda }^2[(s-m^2{)}^{-1}+(t-m^2{)}^{-1}+(u-m^2{)}^{-1}]$, with the Feynman i-epsilon prescription understood. All channels interfere in its modulus squared.

All three fields have mass $m$. In the interaction picture the quantum fields evolve with the free Hamiltonian operator, so their equations areThe interactions evolve the states instead. For $\chi \in \{\varphi ,\psi ,\Phi \}$ the appropriate mode expansion of a free field isIndependent species have commuting creation and annihilation operators. The full set of canonical commutation relations isThese follow from the separate equal-time canonical commutation relations for the three real scalar fields. For comparison, in the Heisenberg picture the full Euler-Lagrange field equations contain interactions: $(\square +m^2){\varphi }_H=-g{\varphi }_H{\Phi }_H$, $(\square +m^2){\psi }_H=-g{\psi }_H{\Phi }_H$, and $(\square +m^2){\Phi }_H=-\frac{g}{2}({\varphi }_H^2+{\psi }_H^2)$. These are distinct from the free equations used in the interaction picture.

There are no derivative couplings, so the interaction Hamiltonian is $H_I(t)=-\int d^{3}x{\mathcal{L}}_I(x)$. The interaction picture evolution operator obeysIterating its integral equation gives the Dyson seriesThe ordered integration region is equivalent to the unrestricted region with time ordering and the factor $1/n!$. Taking the scattering limit $t_0\to -\infty$, $t\to \infty$, with the usual adiabatic prescription for perturbation theory, therefore yields the S-matrixAll fields inside this expression are interaction picture fields.

Use relativistic normalization of a one-particle state and take $|i⟩=a_{\varphi }^{†}(\mathbf{p})a_{\psi }^{†}(\mathbf{q})|0⟩$, $|f⟩=a_{\varphi }^{†}({\mathbf{p}}^{\prime })a_{\psi }^{†}({\mathbf{q}}^{\prime })|0⟩$. Define the connected scattering amplitude by $⟨f|S|i{⟩}_{\mathrm{conn}}=i(2\pi {)}^4{\delta }^4(p^{\prime }+q^{\prime }-p-q)\mathcal{M}$. The identity contribution describes no scattering; the term linear in $g$ cannot connect these four external particles. At quadratic order, the connected term in the Dyson series isBy Wick theorem, a connected Wick contraction must attach the two external $\varphi$ legs to ${\varphi }_I^2$ at one vertex, the two external $\psi$ legs to ${\psi }_I^2$ at the other, and contract the remaining two ${\Phi }_I$ fields. The two assignments of species to $x,y$ cancel the Dyson factor $1/2!$. At either vertex, the $2!$ assignments of its identical fields to the incoming and outgoing particle cancel the interaction factor $1/2$. Equivalently, the coefficient $-g^2/8$ is multiplied by $2\times 2\times 2$, giving $-g^2$. Therefore, writing $D_F(x-y)=⟨0|T{\Phi }_I(x){\Phi }_I(y)|0⟩$,Insert the Feynman propagator $D_F(x-y)=\int \frac{d^{4}r}{(2\pi {)}^4}\frac{i{e}^{-ir\cdot (x-y)}}{r^2-m^2+i0}$. The two vertex integrals impose $r=p^{\prime }-p$ and $p^{\prime }+q^{\prime }=p+q$. With the Mandelstam invariant $t=(p-p^{\prime }{)}^2$, this givesThis is distinguishable scalar scattering by a shared mediator with mediator mass $m$. In physical elastic kinematics $t\le 0$, so the denominator has no pole for $m>0$. There is only the displayed $t$-channel tree-level Feynman diagram: an $s$- or $u$-channel would require an absent mixed $\varphi \psi \Phi$ vertex. Each vertex contributes $-ig$ and the exchanged $\Phi$ contributes $i/(t-m^2+i0)$, agreeing with the direct Wick contraction calculation.

Use $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ throughout. Vary the Maxwell action with respect to the electromagnetic four-potential. Antisymmetry of the electromagnetic field tensor gives $\delta \mathcal{L}=-F^{\mu \nu }{\partial }_{\mu }\delta A_{\nu }$. After integration by parts, the Euler-Lagrange field equation isA gauge transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$ leaves $F_{\mu \nu }$ unchanged because partial derivatives commute. It therefore leaves the Maxwell action and observable electromagnetic fields unchanged: different potentials represent the same physical field. This redundancy matters for canonical quantization: before gauge fixing, the momentum conjugate to $A_0$ vanishes and the four potential components cannot be treated as four independent physical oscillators.

Choose the Lorenz gauge $D={\partial }_{\mu }{A}^{\mu }=0$, attainable by solving $\square \chi =-D$ with suitable boundary data. The Euler-Lagrange field equation then reduces toResidual Lorenz gauge symmetry consists of transformations with $\square \chi =0$. The gauge condition and this residual redundancy ultimately leave two transverse polarizations of a free photon.

For the specified kinetic density ${\mathcal{L}}_C=-\frac{1}{2}{\partial }_{\mu }{A}_{\nu }{\partial }^{\mu }{A}^{\nu }$, one has $\partial {\mathcal{L}}_C/\partial ({\partial }_{\mu }{A}_{\nu })=-{\partial }^{\mu }{A}^{\nu }$. Its Euler-Lagrange field equation is therefore $-\square A^{\nu }=0$. This density is the Feynman-gauge Maxwell kinetic density after a boundary-term subtraction, sinceThe divergence follows by expanding both sides and commuting the second derivatives. Thus the wave equation follows from a Feynman gauge action; the additional physical-state condition will implement the Lorenz gauge. One must not impose $D=0$ inside the original action before varying it.

Take the covariant components $A_{\nu }$ as canonical coordinates. The canonical momentum computed from ${\mathcal{L}}_C$ isThe equal-time canonical Poisson brackets are $\{A_{\mu }(t,\mathbf{x}),{\Pi }^{\nu }(t,\mathbf{y})\}={\delta }_{\mu }{}^{\nu }{\delta }^3(\mathbf{x}-\mathbf{y})$, with coordinate-coordinate and momentum-momentum brackets zero. Applying canonical quantization givesIn particular $[A_{\mu },{\dot{A}}_{\nu }]=-i{g}_{\mu \nu }{\delta }^3$. The negative temporal sign is the source of the indefinite state-space norm in covariant photon quantization.

The spatial Fourier transform of the wave equation gives an oscillator of frequency $\omega =|\mathbf{k}|$ for each component. The Hermitian adjoint condition on the electromagnetic four-potential then yields the mode expansion of a free fieldThese are massless on shell modes. Consistently with the preceding equal-time canonical commutation relation,The normalization is the same as for the scalar modes, with $-g_{\mu \nu }$ replacing the species identity matrix.

Gupta-Bleuler quantization starts with the Fock vacuum annihilated by every $a_{\mu }$ and imposes only the annihilation, or positive-frequency part of a quantum field, of the Lorenz gauge condition:Imposing the entire Hermitian field divergence as a strong operator identity would be incompatible with the four-component canonical commutation relation. The weaker condition suffices: between two physical states its positive-frequency part vanishes on the ket, and its negative-frequency part vanishes on the bra. Hence $⟨{\mathrm{phys}}^{\prime }|{\partial }_{\mu }{A}^{\mu }|\mathrm{phys}⟩=0$.

For a one-photon state $|\mathbf{k},ϵ⟩={ϵ}^{\nu }{a}_{\nu }^{†}(\mathbf{k})|0⟩$, commuting the constraint through the creator givesThus the allowed photon polarization vectors satisfy $\boxed{k\cdot ϵ=0}$. Their state inner product, with the common momentum-normalization factor suppressed, is $⟨{ϵ}^{\prime }|ϵ⟩=-{ϵ}^{\prime \ast }\cdot ϵ$. Momentum eigenstates are delta-normalized; the same statements hold with finite norms for normalizable wave packets.

Rotate the spatial axes so that $k=\omega (1,0,0,1)$. The constraint is ${ϵ}^0={ϵ}^3$, so the complete allowed family isThe first vector is the allowed longitudinal gauge polarization, proportional to the null four-momentum $k$. Its state has zero norm because $k^2=0$; it is also orthogonal to every constrained state because $k\cdot ϵ=0$. A purely spatial vector $(0,0,0,1)$ alone is not an allowed physical polarization: its time component must accompany it to satisfy the constraint. The other two vectors are transverse polarizations, perpendicular to $\mathbf{k}$, with positive unit norms. For the displayed family,The timelike negative-norm excitation has been excluded by the constraint, and the remaining form is positive semidefinite. The Gupta-Bleuler null-state quotient identifies $ϵ$ and $ϵ+\lambda k$, removing the null direction. It leaves exactly two transverse physical photon polarizations with a positive inner product; their bosonic Fock space is the physical photon state space.

Finally use Feynman slash notation, $/a={\gamma }^{\mu }{a}_{\mu }$, and impose four-momentum conservation $p+q=k+k^{\prime }$. The external Dirac spinors obeySet $r=p-k$ and $r^{\prime }=p-k^{\prime }$. Replacing $ϵ$ by $k$ in the first term of the electron-positron annihilation into two photons amplitude, the rightmost contracted vertex obeysSince the Clifford algebra implies $(/r+m)(/r-m)=(r^2-m^2)I_4$, that term becomesFor the second term, four-momentum conservation gives $k=q+r^{\prime }$. The contracted vertex is now adjacent to the outgoing positron spinor on the left, soConsequentlyThe two terms cancel, proving the two-photon fermion Ward identityExchanging the two photon labels proves the same result for ${ϵ}^{\prime }=k^{\prime }$. By linearity the amplitude is invariant under $ϵ\mapsto ϵ+\lambda k$ on either external photon leg. Therefore the null longitudinal gauge polarization has zero production amplitude: physical scattering produces only the two transverse photon polarizations. This cancellation also shows why both photon orderings in the tree-level Feynman diagrams are essential; either term alone fails the Ward identity.

Fix the phi-fourth theory convention ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4/4!$, with mostly-minus Minkowski spacetime and $\hslash =c=1$. Define the scattering amplitude by stripping the overall momentum-conserving delta function from $S-1=i\mathcal{M}$. If instead the coupling is defined by ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4$, every vertex coupling below must be replaced by $24\lambda$; the question does not specify this factorial convention.

The momentum-space Feynman rules are a Feynman propagator $i/(k^2-m^2+i0)$ for each internal scalar line, a Feynman vertex $-i\lambda$ with four scalar legs, and four-momentum conservation at each vertex. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. For an amputated scattering amplitude there are no external propagators; external states have the usual relativistic normalization. Sum the connected diagrams and all inequivalent assignments of the labelled external momenta. The overall delta function is $(2\pi {)}^4{\delta }^4(\sum p_{\mathrm{in}}-\sum p_{\mathrm{out}})$.

These Feynman rules follow by expanding the Dyson series $T\exp (i\int d^{4}x{\mathcal{L}}_{\mathrm{int}})$ and applying the Wick theorem to the free-field correlation functions. A Wick contraction gives the free Feynman propagator; the $4!$ ways of contracting a vertex cancel its factorial denominator. The expansion's $1/V!$ cancels permutations of identical interaction insertions, while any remaining automorphisms give the Feynman-diagram symmetry factor. Fourier transformation produces momentum conservation, and the LSZ reduction formula amputates external propagators to obtain the scattering amplitude.

For six-point amplitudes in phi-fourth theory, an original pair of diagrams is:

In the tree-level Feynman diagram, let the internal four-momentum be $r=p_1+p_2-q_1=q_2+q_3+q_4$. The labelled diagram has symmetry factor one, so its contribution isThis is one channel. The full tree scattering amplitude sums the ten unordered partitions of the six labelled external legs into two groups of three, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$.

For the one-loop triangle Feynman diagram, set $P=p_1+p_2$, $Q_{12}=q_1+q_2$, $Q_{34}=q_3+q_4$, so $P=Q_{12}+Q_{34}$. A consistent routing has denominatorsWith the external labels held fixed, no nontrivial graph automorphism remains, so again the symmetry factor is one. The contribution isThere are other labelled assignments and other one-loop topologies in the full scattering amplitude; the displayed expression belongs to the particular drawn diagram.

One can also express this scalar triangle Feynman integral using Feynman parameters. With $x+y+z=1$, completing the square in $x{D}_0+y{D}_1+z{D}_2$ givesThe identity $1/(D_0{D}_1{D}_2)=2{\int }_{x,y,z\ge 0}dxdydz\delta (1-x-y-z)/(x{D}_0+y{D}_1+z{D}_2{)}^3$ and the shifted momentum integral yieldThus ${\mathcal{M}}_{△}=-{\lambda }^3/(16{\pi }^2)$ times the displayed parameter integral. It is ultraviolet finite in four dimensions, although physical threshold singularities must retain the Feynman i-epsilon prescription.

For the fixed-target production threshold, the incoming four-momenta are $(E,\mathbf{p})$ and $(m,0)$, with $E=\sqrt{m^2+|\mathbf{p}{|}^2}$. Their invariant mass obeysIn the centre-of-momentum frame, four final particles have total energy at least $4m$, attained when all four are at rest in that frame. Therefore $s\ge 16m^2$, equivalently $E\ge 7m$, andThis is the kinematic threshold for $m>0$. Exactly at equality the final phase space has zero volume; a nonzero production rate requires a strict inequality.

Use the mostly-plus Minkowski metric, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the source term $+\int J\varphi$. Define the vacuum-normalized generating functional, connected generating functional, and quantum effective action byThe Legendre transform is taken after expressing $J$ in terms of $\phi$, locally where this relation is invertible. Varying it gives $\delta \Gamma /\delta \phi =J$ and ${\Gamma }^{(2)}=(W^{(2)}{)}^{-1}$. The logarithm generates connected graphs; the Legendre transform removes graphs joined by a single propagator, leaving one-particle-irreducible vertices. This sign convention is important: its classical limit is $\Gamma =-S$, consistent with the positive vertices in this question.

For the free real scalar field, integration by parts puts the action into the form $S=-\frac{1}{2}\varphi A\varphi$, where $A=-\square +m^2$ has Fourier symbol $p^2+m^2$. Include the Feynman i-epsilon prescription $A\mapsto A-i0$. Completing the square yieldsand consequentlyThe physical scalar propagator is $-i{A}^{-1}$, whereas the inverse quadratic kernel appearing in $W_0$ is $A^{-1}$. Differentiating this quadratic functional and Fourier transforming givesThe infinitesimal prescription specifies propagators and is dropped when quoting the local vertex kernel.

At the order represented by tree-level Feynman diagrams, stationary phase gives $W_{mtree}=S[\phi ]+\int J\phi$, with the classical field equation $\delta S/\delta \phi +J=0$. The Legendre transform cancels the source term, soThusOther connected tree graphs with several quartic vertices are one-particle reducible and belong to $W$, not to the proper vertices of $\Gamma$. This explains why no extra tree correction is hidden in these formulas.

The one-loop proper two-point graph is the tadpole diagram. The one-loop proper four-point graphs are the three bubble diagrams, with pairings $(12|34)$, $(13|24)$ and $(14|23)$ of the external legs. Each has Feynman-diagram symmetry factor $1/2$. The figure shows the topologies and every four-point channel; external momenta are incoming.

The tadpole factor can also be counted directly: there are $4\cdot 3=12$ assignments of two labeled external legs to one quartic vertex, while its $4!$ normalization leaves $12/24=1/2$. For a fixed four-point partition, the interchange of the two internal lines gives the same factor $1/2$; there are three distinct partitions. Graphs with an external self-energy insertion are one-particle reducible and are not additional contributions to ${\tau }_4$.

For the two-point divergence, perform a Wick rotation. The Euclidean tadpole contribution to the proper inverse kernel isAssume $m^2>0$ to separate the ultraviolet calculation from infrared issues. A Schwinger parameterization gives $1/(k_E^2+m^2)={\int }_0^{\infty }ds{e}^{-s(k_E^2+m^2)}$. Using the supplied Gaussian integral, initially in the convergence region, givesThe last expression provides the meromorphic dimensional regularization continuation. Set $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,If the coupling is written as ${\mu }^{ϵ}\lambda$, that factor multiplies the intermediate integral but does not change the displayed pole residue. The negative pole is a property of analytic continuation; it must not be inferred from the positive cutoff integrand without carrying out the continuation.

The added mass counterterm contributes $+B$ to ${\tau }_2$, because ${\Gamma }_{mtree}=-S$ and its Lagrangian term is $-B{\varphi }^2/2$. Hence cancellation requiresup to an arbitrary finite subtraction specifying the renormalization scheme. No external momentum occurs in the tadpole, so no one-loop wave-function renormalization is needed. At $m=0$ the unregulated tadpole is scaleless and vanishes in dimensional regularization; the massive calculation identifies the mass-dependent ultraviolet pole without conflating it with a massless infrared limit.

For completeness, the signs of the four-point graphs in this effective-action convention can be read from the Euclidean one-loop functional $\frac{1}{2}\mathrm{Tr}\log (A_E+\lambda {\phi }^2/2)$. Its expansion has terms $\lambda \mathrm{Tr}(A_E^{-1}{\phi }^2)/4$ and $-{\lambda }^2\mathrm{Tr}(A_E^{-1}{\phi }^2{A}_E^{-1}{\phi }^2)/16$. Differentiating four times givesThis is the Euclidean form, analytically continued to the physical external momenta. It confirms the three channel diagrams and their symmetry factors; evaluating its divergence is not needed for the requested mass counterterm.

Use metric signature $(+---)$ and average over the four initial spin states, summing over final spins and the $N_c=3$ quark color charge states. The Electrons carry no color charge, so their initial color charge average is trivial. A final color charge average would not give the inclusive relativistic cross-section requested.

Photon exchange gives the tree amplitude, up to an irrelevant overall phase,Massless spin sums give $\sum u\bar{u}=/p$ and $\sum v\bar{v}=/p$. ThereforeThe color charge multiplicity is ${\sum }_{a,b}|{\delta }_{ab}{|}^2=N_c$, not $N_c^2$. Each trace equals $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$. Contracting them, the terms proportional to $(p_1\cdot p_2)(q_1\cdot q_2)$ cancel, leaving $32[(p_1\cdot q_1)(p_2\cdot q_2)+(p_1\cdot q_2)(p_2\cdot q_1)]$. Thus

In the centre-of-momentum frame every energy is $\sqrt{s}/2$. Let $\theta$ be the angle between $p_1$ and $q_2$. ThenConsequently $\overline{|\mathcal{M}{|}^2}=N_c{e}^4{Q}_q^2(1+{\cos }^2\theta )$.

The supplied two-body Lorentz-invariant phase space formula has a factor-of-two error. Deriving the normalization explicitly,so the massless two-body scattering flux normalization isFor example, integrating the spatial Dirac delta in $d{\Phi }_2$ leaves $[4(2\pi {)}^2{]}^{-1}dkd\Omega \delta (\sqrt{s}-2k)$; its radial integral supplies the factor $1/2$. The printed denominator $64{\pi }^2{p}_1\cdot p_2$ would double the relativistic cross-section and cannot reproduce the requested result with the standard scattering amplitude.

The correctly normalized differential and total results for electron-positron annihilation into a quark pair areand hence

At energies where photon exchange and the parton description apply, sum over kinematically accessible quark flavors. The produced quarks undergo hadronization, so the leading inclusive hadronic rate isThis is the hadronic R ratio. Perturbative QCD adds a leading relative correction ${\alpha }_s(s)/\pi$; thresholds, narrow resonances and, near the Z boson scale, electroweak exchange require additional treatment. The calculation is a photon-exchange, massless result from a tree-level Feynman diagram.

Take incoming electron momentum $p$ and positron momentum $q$, with outgoing photon momenta $k_1,k_2$, so $p+q=k_1+k_2$. The leading electron-positron annihilation into two photons has two tree-level Feynman diagrams, differing by the order in which the two outgoing photons attach to the charged-fermion line.

The arrows indicate fermion flow. An incoming positron has the opposite flow direction to its physical momentum; the lower row uses an outgoing electron instead.

The required QED Feynman rules, with interaction $-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$, are a vertex $-ie{\gamma }^{\mu }$, a Dirac propagator $i(/r+m)/(r^2-m^2+i0)$, an incoming electron spinor $u(p)$, an incoming positron adjoint $\bar{v}(q)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. Here Feynman slash notation means $/a={\gamma }^{\mu }{a}_{\mu }$, and the external Dirac equations are $(/p-m)u(p)=0$ and $\bar{v}(q)(/q+m)=0$.

Put $\mathcal{S}(r)=(/r+m)/(r^2-m^2+i0)$, omitting only the propagator's overall factor $i$. Multiplying the two vertices and the propagator givesThe diagrammatic expression is $i\mathcal{M}$. The relative sign is plus because the two photons are bosons; exchanging their attachments does not exchange external fermions. The amplitude is of order $e^2$.

A physical photon polarization vector obeys $k^2=0$ and $k\cdot ϵ=0$, with the gauge equivalence $ϵ\sim ϵ+\alpha k$. One can choose a transverse representative with ${ϵ}^0=0$ and normalize ${ϵ}^{\ast }\cdot ϵ=-1$. There are two physical transverse polarizations.

To prove the requested two-photon fermion Ward identity, replace ${ϵ}_1^{\ast }$ by $k_1$. Away from an internal pole, $\mathcal{S}(r)$ is the inverse of $/r-m$, with the boundary prescription understood. The first term simplifies becauseFor the second term, four-momentum conservation gives $k_1=(p-k_2)+q$, soThe two contributions are therefore $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Interchanging the photon labels proves the other contraction vanishes. By linearity, adding a multiple of either photon momentum to its polarization leaves the summed amplitude unchanged. This is gauge invariance: longitudinal gauge representatives do not describe an additional physical photon state. An individual diagram does not generally have this property.

For Compton scattering, let $p,k$ be incoming electron and photon momenta and $p^{\prime },k^{\prime }$ the outgoing ones. The diagrams in the lower row have intermediate electron momenta $p+k$ and $p-k^{\prime }$ respectively. The same QED Feynman rules giveTheir denominators are $(p+k{)}^2-m^2+i0$ and $(p-k^{\prime }{)}^2-m^2+i0$, the $s$- and $u$-channel denominators. In annihilation the two exchanged-electron channels instead carry $p-k_1$ and $p-k_2$. Crossing symmetry relates the two expressions by moving the positron to an outgoing electron and one outgoing photon to an incoming photon, with the corresponding momentum and polarization replacements. There is no two-photon contact vertex for the minimally coupled Dirac electron. Both processes require the sum of the two photon orders to satisfy the Ward identity.

Variation with respect to the Dirac adjoint, the Dirac field, and the real scalar field respectively givesThe derivative in the adjoint equation acts to the left; its sign follows by integrating $i\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\delta \psi$ by parts. The Yukawa interaction supplies a spacetime-dependent effective fermion mass and a scalar source.

Use $/p={\gamma }^{\mu }{p}_{\mu }$ and the convention $S_{fi}^{\mathrm{conn}}=i(2\pi {)}^4{\delta }^4(p_f-p_i)\mathcal{M}$. The momentum-space Feynman rules are: a scalar internal line contributes $i/(p^2-m^2+i0)$; an oriented fermion line contributes $i(/p+m)/(p^2-m^2+i0)$; and each scalar-fermion vertex contributes $-ig$ times the identity in spinor space. Impose four-momentum conservation at each vertex. Incoming and outgoing fermions supply $u$ and $\bar{u}$, while incoming and outgoing antifermions supply $\bar{v}$ and $v$; scalar external legs supply one. Loop momenta are integrated with $d^{4}p/(2\pi {)}^4$, a closed fermion loop contributes a minus sign, and graph symmetry factors are included. Relative signs between distinct contractions of identical external fermions follow from their anticommutation relations. These specify the Feynman rules also beyond the tree approximation.

Label incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, with all external particles on shell. Write the Mandelstam variables as $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, and abbreviate $D_r=r-m^2+i0$. Spin indices on $u_i,v_i$ are implicit. The six required tree-level Feynman diagrams are: