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Feynman rules translate each diagram into momentum-space factors for propagators, vertices, external states, momentum conservation and loop integration.

Feynman rules translate a Feynman diagram into a contribution to a scattering amplitude or correlation function in perturbation theory. They specify factors for external states, interaction vertices, internal propagators, momentum-conservation constraints and loop integrals, with statistics-dependent signs and symmetry factors. For quantum electrodynamics, a photon-fermion vertex contains the coupling $e$, while a photon propagator supplies an inverse squared four-momentum. The complete amplitude also contains currents, so propagator powers alone do not determine the energy dependence of a scattering cross-section.

Gerard 't Hooft showed how theories with massive gauge fields described by Yang-Mills theory generated through spontaneous symmetry breaking could support consistent renormalization. His 1971 paper constructs models, derives Feynman rules and checks the associated Ward identities. Controlling these identities while removing divergent intermediate expressions is what permits finite higher-order predictions without losing the theory's gauge consistency. The work made spontaneously broken gauge theories usable for precision particle calculations.^[ref]

A physicist working on theoretical physics at Utrecht.^[ref]

The neutral vector mass term is $\frac{1}{2}{m}_Z^2{Z}_{\mu }{Z}^{\mu }(1+H/v{)}^2$. Expanding it gives the cubic $HZZ$ and quartic $HHZZ$ interactions in unitary gauge. Differentiating the action with respect to identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$. A single $Z$ does not couple to two copies of the real neutral radial Higgs boson through an elementary vertex.

The tree-level Feynman diagrams are the two ordered quark-exchange graphs and the $s$-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary QCD Lagrangian density. Let $q=k_1+k_2=p_1+p_2$, $r_t=p_1-k_1$, $r_u=p_1-k_2$, and $s=q^2$, $t=r_t^2$, $u=r_u^2$. Define $i\mathcal{M}$ as the product of the stated Feynman rules.

Suppressing external colour indices, the quark graphs areThe massless Dirac equations give $\bar{u}/p_1=0$ and $/p_{2}v=0$. ConsequentlyThe last equality uses $\bar{u}/qv=0$. The quark-exchange sum is therefore $-g^2\bar{u}/k_1[T^a,T^b]v$.

For the gluon graph, use all incoming momenta $(k_1,k_2,-q)$ in the three-gluon tensor:Its double contraction is $k_{1\mu }{k}_{2\nu }{V}^{\mu \nu \rho }=s(k_2-k_1{)}^{\rho }/2$. The vertex and gluon propagator giveCombining the graphs yieldsThus the requested spinor proportionality factor is $-g^2\bar{u}(p_1)/k_{1}v(p_2)$ when the colour matrix is displayed separately; retaining it inside the spinor bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself $\mathcal{M}$, rather than $i\mathcal{M}$, the entire result gains one factor of $i$.

Route a directed ghost loop so that its internal four-momenta are $l$ and $l+k$. At the external gluon with indices $(\mu ,a)$, let the incoming ghost carry colour $c$ and the outgoing ghost carry colour $d$; the other Feynman vertex has the reverse order. The stated ghost-gluon Feynman rules supply $g_D^2{f}^{acd}{f}^{bdc}(l+k{)}^{\mu }{l}^{\nu }$. Their colour factor isThe two ghost propagators supply $i^2=-1$, while the closed ghost loop supplies another $-1$. Thus, with the insertion convention stated above,This establishes every prefactor before integration. To isolate its ultraviolet pole, take generic off shell $k$, initially $k^2<0$, so the massless integral has no overlapping infrared divergence. Introduce a Feynman parameter, shift $r=l+xk$, and set $\Delta =-x(1-x)k^2>0$. Odd powers of $r$ vanish, givingThe needed dimensional regularization integrals follow from the supplied gamma function formula. In $D=4-ϵ$,Here $\Gamma (ϵ/2)=2/ϵ+O(1)$, rather than $1/ϵ$. Using $1/D=1/4+O(ϵ)$ and ${\int }_0^{1}x(1-x)dx=1/6$, obtainTherefore the ghost-loop pole of the gluon self-energy isThe factor ${\mu }^{ϵ}$ in $g_D^2$ affects only finite terms at this order. The isolated ghost loop is not transverse: contracting this result gives $g^2{C}_A{\delta }^{ab}{k}^2{k}^{\nu }/(32{\pi }^2ϵ)$. Transversality is a condition on the complete gluon self-energy, including the other loop graphs.

Label the four-momenta as $e^{-}(p_1){\mu }^{-}(p_2)\to e^{-}(p_3){\mu }^{-}(p_4)$. There is one tree-level Feynman diagram, the $t$-channel photon exchange shown in the lower-left panel. The two lepton species are distinct, so there is no exchanged-final-lepton diagram. With $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$, the Feynman rules giveThe massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$. Averaging the two initial spins of each lepton supplies $1/4$, soThe gamma matrix trace identities imply, for arbitrary null external four-momenta,To see the contraction, each trace is $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$; the terms proportional to $(a\cdot b)(c\cdot d)$ cancel in four dimensions, leaving the displayed two products. Here $p_1\cdot p_2=p_3\cdot p_4=s/2$ and $p_1\cdot p_4=p_2\cdot p_3=-u/2$. ThereforeThe requested proportionality constant is $2e^4=32{\pi }^2{\alpha }_{\mathrm{em}}^2$, where ${\alpha }_{\mathrm{em}}$ is the fine-structure constant. Both numerator terms are required by the spin sum; no final-state statistical factor enters a squared scattering amplitude for distinct leptons.

The proper quark-gluon vertex colour factors come from two one-particle-irreducible Feynman diagrams, shown in the upper-middle and upper-right panels. One has a virtual gluon joining the open quark line on opposite sides of the external gluon insertion. The other has two virtual gluons meeting the external gluon at a three-gluon Feynman vertex, with their other ends joined by an internal quark. A Faddeev-Popov ghost field has no direct quark coupling, so it supplies no additional one-loop Feynman vertex graph here.

Let $T^a$ be the Hermitian generators of the fundamental representation of $SU(N)$, with $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, $[T^a,T^b]=i{f}^{abc}{T}^c$, ${\sum }_b{T}^b{T}^b=C_{F}1$, and $f^{abc}{f}^{dbc}=C_A{\delta }^{ad}$. Here $C_F=(N^2-1)/(2N)$ and $C_A=N$ are quadratic Casimir operators evaluated in the fundamental representation and the Adjoint representation.

For the first graph, the ordered product is ${\sum }_b{T}^b{T}^a{T}^b$. Expand the double commutator:Thus its colour factor isFor the second graph choose the orientation of the three-gluon Feynman vertex so that its group tensor is $f^{abc}$ and the ordered product along the quark line is $T^b{T}^c$. Antisymmetry removes the anticommutator part:The corresponding real group coefficient is $C_A/2=3/2$ after the conventional factor of $i$ is incorporated into the remaining Feynman rules. If the vertex is written with $i{f}^{abc}$, its raw matrix product is instead $-C_A{T}^a/2$. Reversing the ordered product reverses its sign together with the orientation-dependent three-gluon Feynman rule; these conventions describe the same scattering amplitude. External-leg corrections and local counterterms are separate from these two proper one-loop graphs.

Put $Q_u=2/3$, $s=q_s^2$, $u=q_u^2$, and suppress the common colour factor ${\delta }_{ji}$. Define $i\mathcal{M}$ by the product of Feynman rules. With massless external Dirac spinors the two graphs giveThe massless Dirac equation gives $/p_{1}u(p_1)=0$ and $\bar{u}(p_3)/p_3=0$. ThusBy four-momentum conservation, $p_2=p_3-q_u$, soThe two terms cancel:This is the Ward identity for the incoming photon; replacing its photon polarization vector by its four-momentum produces zero. The outgoing photon obeys the analogous identity. Each exchange graph separately fails this test, so the cancellation requires their sum. Consequently the physical scattering amplitude is invariant under ${\epsilon }^{\mu }(p_2)\mapsto {\epsilon }^{\mu }(p_2)+c{p}_2^{\mu }$.

Expand the gauge fixing square:The second term is a derivative cubic photon interaction; the third is a quartic photon contact interaction. Both arise from nonlinear Abelian gauge fixing, rather than an additional physical electromagnetic force.

For $\xi =1$, differentiating the quartic term with respect to four identical gauge fields gives eight copies for each pairing, henceWith the Fourier convention $A(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{ipx}A(p)$ and all four-momenta incoming, the differentiated cubic term givesmatching the supplied Feynman rules. Reversing the Fourier convention reverses this cubic rule's sign; each exchange graph contains two such Feynman vertices, so its contribution is unchanged.

Let $g$ and $g^{\prime }$ be the two gauge couplings. In unitary gauge, $\Phi =(0,0,v+h{)}^T$ and the weak hypercharge term in the gauge covariant derivative vanishes. Using $(T^a{)}_{bc}=-i{ϵ}_{abc}$, the kinetic term becomesThus, in the basis $(W^1,W^2,W^3,B)$, the gauge-boson mass matrix and its eigenvalues areHere $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$. The two massless neutral gauge bosons may be rotated to any orthonormal basis; no neutral mass term selects a unique mixing angle.

Expanding the same kinetic term to first order in $h$ givesFunctional differentiation gives the Feynman rulesand, in the real basis, $h{W}_{\mu }^a{W}_{\nu }^b:2i{g}^{2}v{\delta }^{ab}{g}_{\mu \nu }$ for $a,b\in \{1,2\}$. All such three-point Feynman vertices involving $W^3$ or $B$ vanish. The factor two in the real basis comes from the two identical gauge boson fields; in the charged basis it is already explicit in ${\mathcal{L}}_{hVV}$.

Use the Feynman rules and Feynman gauge convention from part a. Put $s=q^2$, $t=r_t^2$, $u=r_u^2$, the Mandelstam variables, and define the three-boson tensor with incoming four-momenta $(q,-k_1,-k_2)$:The contributions, with external Dirac spinors included and both external gauge boson propagators amputated, areFor example, $i{\mathcal{M}}_t$ contains $(+ig{)}^{2}i=-i{g}^2$. In the annihilation channel, the current vertex, propagator, and three-boson vertex give $(+ig)(-i)g=g^2$, so division by the defining overall $i$ gives $-i{g}^2$. These relative signs are essential for the cancellation in part d. The antisymmetric colour factor makes the last expression symmetric under simultaneously exchanging $(a,\mu ,k_1)$ and $(b,\nu ,k_2)$, as required for identical outgoing gauge bosons. The on shell external Dirac spinors obey $(/p_1-m)u(p_1)=0$, $\bar{v}(p_2)(/p_2+m)=0$.

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/a15272h1corrected.pdf

Solutions to all seventeen lettered parts. The source is the version publicly labelled corrected; no claim about the contents of an earlier version is needed. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$, and $\square ={\partial }_{\mu }{\partial }^{\mu }$ for the d'Alembert operator. In question 2, every denominator has its Feynman i-epsilon prescription $+i0$, and ${\int }_k={\mu }_R^{2ϵ}\int d^{4-2ϵ}k/(2\pi {)}^{4-2ϵ}$. Here $ϵ$ is the dimensional regularization parameter; the small parameter in 1(f) is denoted $\eta$ to distinguish it. Signs follow the insertion Feynman rules printed on source page 4.

The first panel is the one-particle-irreducible Feynman diagram for photon vacuum polarization. The two internal Dirac propagators carry $k$ and $k+q$, and the external photons carry $q$. The other panels fix the routings used in parts c and d.

The closed fermion loop sign contributes a minus sign; there is no extra symmetry factor. Applying the given Feynman rules, with the external propagators amputated,Equivalently, the required amplitude isThe Feynman slash notation is $/k={\gamma }^{\rho }{k}_{\rho }$ and $\mathrm{tr}1=4$ in the chosen dimensional regularization convention.

For ${\mu }^2<0$, minimize the Higgs potential to obtain $v^2=-{\mu }^2/\lambda$. In unitary gauge, the physical scalar fluctuation is real andComparing with $m_W^2{W}_j^{\mu }{W}_{j\mu }/2$ and $-m_{\sigma }^2{\sigma }^2/2$ givesAll three gauge generators are broken by this complex doublet; this pure SU(2) model has no surviving massless photon. Three scalar angular degrees of freedom supply the longitudinal vector polarizations, leaving one real Higgs boson. The count is $3\times 2+4=3\times 3+1=10$. The corresponding fundamental-doublet example also appears in Gauge Theory by David Tong.^[ref]

The remaining interactions are $g^{2}v\sigma W_j{W}_j/4$, $g^2{\sigma }^2{W}_j{W}_j/8$, $-\lambda v{\sigma }^3$ and $-\lambda {\sigma }^4/4$. They give vertices with two equal-species vector legs and one or two scalar legs, or three or four scalar legs, as drawn above. The coefficients in the Lagrangian must be differentiated with respect to identical fields when obtaining fully normalized Feynman rules; the written coefficients alone are not all the vertex factors.

For $m^2>0$ and initially $\Re n>d/2$, Schwinger parameterization and the supplied Gaussian integral giveSubstitute $u=\alpha m^2$ to obtain the Euclidean massive loop integralFor a positive integer $n$, $\Gamma (n)=(n-1)!$. Outside the convergence region this expression defines its meromorphic continuation, used in dimensional regularization.

Combine the two denominators with a Feynman parameter and shift $k$ by $(1-x)p$. This givesThe gamma function has $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the remaining factors tend to $(4\pi {)}^{-2}$ and one. Hence the massive scalar bubble pole in four dimensions isThe pole is independent of the external momentum, as required for a local quartic counterterm.

Use the mostly-plus Minkowski metric and expand $e^{iS}$. The momentum-space Feynman rules are: an internal scalar propagator $i/(-p^2-m^2+i0)$; a four-scalar interaction vertex $-i{\mu }^{ϵ}\lambda$; a momentum conservation delta function at each vertex; and $\int d^{d}k/(2\pi {)}^d$ per independent loop momentum. Divide by the Feynman-diagram symmetry factor. All momenta can be taken incoming. External propagators are omitted when computing an amputated one-particle-irreducible vertex.

The connected one-loop one-particle-irreducible Feynman diagrams are the two-point tadpole diagram and the three four-point bubble diagrams, in the $s,t,u$ channels. Each has symmetry factor $1/2$. A tadpole attached to an external leg through a single bridge would be reducible, so is excluded here.

The Euclidean massive loop integral at $n=1$ has $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$, so $I_{d,1}=-m^2/(8{\pi }^2ϵ)+O(1)$. Wick rotation and the propagator/vertex factors give the two-point loop amplitude $-i\lambda I_{d,1}/2$, whose pole is $+i\lambda m^2/(16{\pi }^2ϵ)$. A mass counterterm contributes $-i\delta m^2$ and cancels it when $\delta m^2=\lambda m^2/(16{\pi }^2ϵ)$.

Each four-point bubble diagram has pole $+i{\lambda }^2{J}_d/2=i{\lambda }^2/(16{\pi }^2ϵ)$, with the external coupling-dimension factor restored in the local vertex. Summing its three channels gives the quartic counterterm $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. Therefore the one-loop massive phi-fourth counterterms are exactlyThere is no one-loop wave-function renormalization, since the tadpole is independent of external momentum. Vacuum graphs would additionally require a vacuum-energy subtraction.

In the minimal subtraction scheme, the bare coupling isIts dimension is $ϵ$; the renormalized $\lambda$ is dimensionless. Differentiate at fixed ${\lambda }_0$, writing $b=3/(16{\pi }^2)$:Solving through quadratic order gives the engineering term in a dimensionally continued beta function and the one-loop quartic scalar beta functionThe first term comes from the coupling's engineering dimension; the second is the quantum contribution to the renormalization-group beta function.

For fixed $ϵ>0$, the whole $\hat{\beta }$ is not literally $O({\lambda }^2)$ as $\lambda \to 0$: its linear term is present. Interpret the final instruction as keeping terms through quadratic order. Set ${\lambda }_{\ast }=ϵ/b=16{\pi }^2ϵ/3$. Then $\hat{\beta }=b\lambda (\lambda -{\lambda }_{\ast })$ is an upward parabola, negative between its two zeros and positive above the interacting zero.

For positive initial coupling ${\lambda }_r=\lambda ({\mu }_r)$, the quartic running coupling below four dimensions follows by solving the linear equation for $1/\lambda$:If $0<{\lambda }_r<{\lambda }_{\ast }$, then $\lambda (\mu )\to {\lambda }_{\ast }$ as $\mu \to 0$ and $\lambda (\mu )\to 0$ as $\mu \to \infty$. If ${\lambda }_r={\lambda }_{\ast }$ it stays fixed; the zero solution also stays fixed. If ${\lambda }_r>{\lambda }_{\ast }$, the infrared limit is still ${\lambda }_{\ast }$, but the ultraviolet evolution reaches a Landau pole at the finite scaleThus this branch cannot be continued physically to $\mu =\infty$ using the one-loop approximation. These are statements about the quartic coupling direction. The Wilson-Fisher fixed point of the full scalar theory also has a thermal relevant direction at the Wilson-Fisher fixed point; approaching the critical theory requires mass tuning. For negative initial coupling, the quartic potential is unstable: the same formula tends to zero from below in the ultraviolet and reaches a pole at finite scale toward the infrared.

Continuum amplitudes. The leading quantum electrodynamics diagram is electron-positron annihilation into a virtual photon, followed by conversion to a muon pair. Replacing the outgoing muons with a quark-antiquark pair gives the quark-level hadronic diagram. The first two panels below show these processes.

The Feynman rules give, up to signs and factors of $i$,The two vertices supply $e^2$, and the photon propagator supplies $1/s$. Each external current scales as $\sqrt{s}$, so the complete amplitude is of order $e^2$, and its squared magnitude is of order $e^4\propto {\alpha }^2$. The two-body relativistic scattering cross-section supplies the remaining $1/s$ from flux and phase space. Integrating the angular distribution yields ${\sigma }_{\mu \mu }=4\pi {\alpha }^2/(3s)$. Keeping only the squared propagator while ignoring the external currents would incorrectly predict $1/s^2$.

A quark of charge $Q_{f}e$ contributes $N_c{Q}_f^2$ times the massless muon rate. With $u,d,s,c,b$ open and $N_c=3$, the hadronic R ratio isThis is the requested massless leading-QED continuum approximation. Strong corrections, physical hadron thresholds, resonance contributions and Z exchange are separate effects; the pure photon formula is not a complete high-energy electroweak cross section throughout the entire stated range.

Efficiency normalization. Read the original PDF plot as a continuum of about $0.8\mathrm{nb}$ and a maximum around $1.7\mathrm{nb}$ at $E\simeq 10.58\mathrm{GeV}$. The supplied conversion constant givesHence near the peak,If $C$ multiplies the displayed cross sections, the required scale isHere $ϵ$ is the implied detection efficiency under the common-scale assumption. Using continuum readings $0.75$–$0.85\mathrm{nb}$ gives approximately $C=3.4$–$3.8$, illustrating the graph-reading precision. The nearby open-bottom threshold also limits the precision of the ideal continuum approximation.

Resonance quantum numbers and widths. Direct formation through a single photon requiresThese are the quantum numbers of the electromagnetic current. This concerns single-photon formation; associated production or two-photon reactions need not obey this restriction.

For a Breit-Wigner resonance cross section, subtract the continuum before finding the full width at half maximum. The plotted excess is about $0.9\mathrm{nb}$, so its half-height occurs at a total displayed cross section near $1.25\mathrm{nb}$, not $0.85\mathrm{nb}$. The crossings are roughly $10.57$ and $10.59\mathrm{GeV}$, with a separation around $0.02$–$0.025\mathrm{GeV}$. ThusFor spin-one formation from unpolarized electrons and positrons, $g=3/4$ and $p\simeq M/2$. With ${\Gamma }_{\mathrm{had}}\simeq \Gamma$, the peak resonance cross section and branching fractions relation becomesThe corrected excess is $C(1.7-0.8)\simeq 3.2\mathrm{nb}$. Taking $\Gamma \simeq 24\mathrm{MeV}$ as a representative graph reading givesThus the plot supportsThe width and electronic partial width are estimates from the printed graph under the prescribed Breit-Wigner and efficiency assumptions; these are not precision world-average values. A scale correction changes the peak height and inferred partial width, but not the peak position or width.

Why the lower Upsilon states are narrow. In electromagnetic production and decay of vector quarkonium, every $nS$ state can annihilate as $b\bar{b}\to {\gamma }^{\ast }\to e^{+}{e}^{-}$. The electronic width probes the wavefunction at the origin, ${\Gamma }_{ee}\propto {\alpha }^2{Q}_b^2|{\psi }_{nS}(0){|}^2/M_n^2$. The radial overlaps differ, but the same short-distance annihilation mechanism operates, so electronic widths remain of comparable order.

The hadronic mechanism changes sharply at $2m_B\simeq 10.560\mathrm{GeV}$. The first three Upsilon mesons lie below it and cannot produce an on-shell B-meson pair. Their leading purely strong annihilation channel is $b\bar{b}\to ggg$, followed by hadronization: a color-singlet pair cannot annihilate to one gluon, and two gluons cannot match its negative charge-conjugation parity. Three-gluon annihilation carries three powers of ${\alpha }_s$ in the rate and requires quark-antiquark overlap. Electromagnetic annihilation to light quarks and, for excited states, transitions to lower bottomonium plus pions or photons also contribute.

The fourth state lies above the pair threshold and can instead undergo ${\rm Y}(4S)\to B\bar{B}$ through light-pair creation, keeping the b and anti-b in separate outgoing mesons. The final panel depicts this quark-line topology; it represents a nonperturbative strong decay, not a free on-shell b quark radiating a real gluon. For two pseudoscalar B mesons the pair is in a P wave to give $1^{-}$. Although close to threshold, this open strong channel has a much larger rate than the annihilation channels. Consequently a total width of tens of MeV can coexist with an electronic width below a keV; below threshold the total widths are smaller by hundreds or more, while their electronic widths remain of similar order.

Hyperfine estimate. For two spin-half constituents,The meson hyperfine mass splitting therefore raises the vector mass above the pseudoscalar mass. To infer its coefficient from only one measured mass, adopt the intended simple constituent ansatz that the spin-independent baseline is $m_b+m_q=5.31\mathrm{GeV}$. ThenFor the spin-one partner,The estimate is the same for $\bar{b}u$ and $\bar{b}d$ with the supplied equal light-quark masses. The splitting is below $m_{\pi }$, so the strong decay $B^{\ast }\to B\pi$ is closed. An electromagnetic transition is open and dominates over weak decay:It preserves electric charge, giving the corresponding charged or neutral B meson; the photon energy is approximately $40\mathrm{MeV}$, slightly reduced by recoil. Strictly, the statement that the mass merely includes a spin-spin term does not fix $A$: an independent spin-independent binding energy would leave the baseline unknown. The numerical prediction above explicitly uses the additive constituent-mass approximation.