Feynman rules for QED

Before Feynman, computing a scattering amplitude in quantum electrodynamics meant a page-filling sum over time-orderings of virtual particles — one term for every way the intermediate states could be arranged in time, with the photon's polarizations summed by hand. Feynman's insight was that the whole calculation reduces to a picture and a dictionary. Draw the process as a diagram — lines for particles, a dot wherever a photon meets two electrons — and read off a fixed factor for each piece: a spinor for every external leg, a vertex factor $-iQe\gamma^\mu$ for every dot, a propagator $-ig_{\mu\nu}/q^2$ for every internal photon. Multiply them, with indices contracted, and the product is $-i\mathcal{M}$ , the Lorentz-invariant amplitude. This is not an approximation or a mnemonic that happens to work. Thomson stresses that the rules reproduce, term for term, what the full field-theory derivation gives: the propagator $g_{\mu\nu}/q^2$ is the summed result of all time-orderings of a massless photon, and $Qe\,\bar u\gamma^\mu u$ is the fundamental electron–photon coupling extracted from the QED Lagrangian. Every QED process — Rutherford scattering, Compton scattering, pair annihilation, the anomalous magnetic moment tested to twelve digits — is built from the same three ingredient types and the same single three-line vertex. This article is the assembly manual for the relativistic-scattering chapter: it takes the Dirac spinors from aq-pp-dirac-solutions, the bilinears from aq-pp-dirac-adjoint-bilinear-covariants, the exchange picture from aq-pp-perturbation-theory-exchange, and the photon propagator from aq-pp-photon-propagator, and forges them into the recipe every downstream cross-section calculation executes.

Key takeaways

Every QED diagram is built from a single three-line vertex coupling one photon to two spin-½ fermions; no vertex connects more than three lines, and no other vertex exists in pure QED.
The amplitude is the product of three ingredient types — external-line spinors ( $u,\bar u,v,\bar v$ and photon polarizations $\epsilon_\mu$ ), the vertex factor $-iQe\gamma^\mu$ , and the photon propagator $-ig_{\mu\nu}/q^2$ — multiplied with indices contracted to give $-i\mathcal{M}$ .
Spinor factors are written in a definite order: follow each fermion line against its arrows, and the first spinor met is always an adjoint, $\bar u$ or $\bar v$ .
Antiparticles use $v$ -spinors and are drawn with arrows running backward, but the rules are otherwise identical to the particle case.
The four-momentum $q$ in a propagator is fixed by conservation at the vertices, producing the characteristic pole $1/(q^2-m^2)$ — the massless photon's $1/q^2$ being the special case that makes electromagnetism long-ranged.
Key idea: a complicated scattering amplitude reduces to mechanical pattern-matching — draw the diagram, attach a known factor to each line and vertex, multiply, and you have $-i\mathcal{M}$ .

The core insight

A complicated scattering amplitude reduces to mechanical pattern-matching: draw the diagram, attach one known factor to each external line, each vertex, and each internal line, multiply them together with the indices contracted, and the product is $-i\mathcal{M}$ — the same answer the full time-ordered field-theory sum would give.

Where this lives today

The Feynman rules are the beating heart of every automated collider calculation. Programs such as QGRAF, FeynArts, and MadGraph enumerate the diagrams for a requested process; FeynCalc and FORM implement the vertex, propagator, and trace rules verbatim; and event generators like Pythia and Sherpa fold the results into simulated collisions compared bin-by-bin with LHC data. Nothing in this pipeline departs from the dictionary in this article — it is only executed at industrial scale, for final states with a dozen legs and thousands of interfering diagrams. The rules also underwrite the most precisely tested prediction in science: the electron's anomalous magnetic moment $a_e=(g-2)/2$ , computed from QED diagrams through five loops (12,672 diagrams at tenth order) and agreeing with experiment to better than one part in $10^{12}$ . The same three ingredient types, generalized, run the rest of the Standard Model — QCD adds three- and four-gluon vertices and colour factors, the electroweak sector adds massive $W$ and $Z$ propagators $-ig_{\mu\nu}/(q^2-M^2)$ and chiral vertices — but the QED trio of external spinor, vertex, and propagator is the template on which all of it is built.

The physical picture

A QED diagram is a bookkeeping device, not a photograph of trajectories, and its three kinds of element map onto three kinds of factor. External legs — the real particles from the accelerator and into the detector — carry Dirac spinors $u,\bar u,v,\bar v$ or photon polarization vectors. Vertices — where the interaction happens — each carry the same factor $-iQe\gamma^\mu$ , the fundamental fermion–photon coupling. Internal lines — the virtual quanta exchanged but never observed — each carry a propagator. The first figure attaches all of these to a concrete process, $e^+e^-\to\mu^+\mu^-$ , so the "read it off and multiply" method is visible at a glance.

Figure

Anatomy of a QED amplitude for e⁺e⁻ → μ⁺μ⁻: the two vertices each contribute −iQe γ^μ, the internal photon contributes the propagator −ig_{μν}/q² with q² = s, and the four external legs contribute the spinors u, v̄, ū, v — multiplied together they give ℳ = −(e²/s)(v̄₂γ^μu₁)(ū₃γ_μv₄)

The ordering of the spinors is the one place the bookkeeping demands care: a fermion line is a chain of arrows read against the arrow direction, so the factors come out as adjoint-spinor, vertex, propagator, vertex, spinor — a matrix product closing to a number. The electron and positron of the initial state share one fermion line, so their spinors bracket the current $\bar v\,\gamma^\mu\,u$ , never $u\,\gamma^\mu\,\bar v$ (the misconception addressed below). What makes this more than accounting is that it yields hard, parameter-free numbers: the second figure turns the amplitude above into the measured cross-section $\sigma(e^+e^-\to\mu^+\mu^-)=4\pi\alpha^2/3s$ , the QED benchmark every collider normalizes against.

:::svg[The Feynman rules turned into a number: σ(e⁺e⁻ → μ⁺μ⁻) = 4πα²∕3s = 86.9 nb ∕ s[GeV²] falls as a straight line of slope −2 on log–log axes; the marked point is √s = 10 GeV giving σ = 0.87 nb, and near √s = m_Z the pure-photon prediction is about 10 pb before the Z resonance takes over]

The core equations

The master rule is the assembly law: the amplitude is the ordered product of the factors carried by every element of the diagram. The following equations are that dictionary, applied and then illustrated on $e^+e^-\to\mu^+\mu^-$ .

-i\mathcal{M}=\Big(\prod_{\text{ext}}\text{spinor}/\epsilon\Big)\times\Big(\prod_{\text{vertices}}-iQe\gamma^\mu\Big)\times\Big(\prod_{\text{int}}\text{propagator}\Big) \qquad (1)

where: the external factors are $u^s(p)$ (incoming fermion), $\bar u^s(p)$ (outgoing fermion), $\bar v^s(p)$ (incoming antifermion), $v^s(p)$ (outgoing antifermion), and $\epsilon_\mu,\epsilon^*_\mu$ for incoming/outgoing photons; spinors are ordered by reading each fermion line against its arrows. Reading: the left side is $-i$ times the invariant matrix element $\mathcal{M}$ that feeds Fermi's Golden Rule and the cross-section formulas. The vertex is the heart of QED — a single three-line junction, one photon and two fermions, appearing with the same factor everywhere. It is where the coupling constant enters, once per dot.

\text{vertex}=-iQ_f\,e\,\gamma^\mu,\qquad e=\sqrt{4\pi\alpha},\qquad \alpha=\frac{e^2}{4\pi}\approx\frac{1}{137.036} \qquad (2)

where: $e>0$ is the elementary charge (Heaviside–Lorentz natural units $\hbar=c=\varepsilon_0=1$ ); $Q_f$ is the fermion's charge in units of $e$ (electron $Q=-1$ , up-quark $Q=+\tfrac23$ ); $\gamma^\mu$ carries the Lorentz index that contracts with the photon; $\alpha$ is the fine-structure constant. Reading: the $\gamma^\mu$ makes the vertex a four-vector so that current couples to current; the overall sign of the $i$ is convention-dependent and cancels in $|\mathcal{M}|^2$ . Griffiths writes the same vertex as $ig_e\gamma^\mu$ with $g_e=\sqrt{4\pi\alpha}=e$ . An internal photon carries a propagator: the amplitude for the virtual quantum to be created at one vertex and absorbed at the other, summed over all its intermediate states. In Feynman gauge it is strikingly simple.

D_{\mu\nu}(q)=\frac{-ig_{\mu\nu}}{q^2}\ \ (\text{Feynman gauge}),\qquad D_{\mu\nu}(q)=\frac{-i}{q^2}\Big[g_{\mu\nu}-(1-\xi)\frac{q_\mu q_\nu}{q^2}\Big]\ \ (R_\xi) \qquad (3)

where: $q$ is the four-momentum along the internal line, fixed by conservation at the vertices; $g_{\mu\nu}$ is the Minkowski metric; $\xi$ is the gauge parameter. Reading: the pole at $q^2=0$ is the massless photon's, and it is what makes electromagnetism long-ranged; the gauge-dependent $q_\mu q_\nu$ terms never contribute, since they hit conserved currents, $q_\mu\bar u\gamma^\mu u=0$ (the Ward identity). An internal fermion carries the Dirac propagator, the momentum-space inverse of the operator $(i\slashed\partial-m)$ .

S_F(q)=\frac{i(\slashed q+m)}{q^2-m^2+i\epsilon},\qquad \slashed q\equiv\gamma^\mu q_\mu \qquad (4)

where: $m$ is the fermion mass; $+i\epsilon$ enforces causal (Feynman) boundary conditions; $\slashed q$ is the Feynman slash. Reading: the numerator $\slashed q+m$ is a $4\times4$ matrix, so a fermion propagator sits inside a spinor chain between two vertices; the pole at $q^2=m^2$ is where the virtual fermion would go on-shell. Applying the dictionary to $e^+e^-\to\mu^+\mu^-$ — one diagram, two vertices, one photon of momentum $q=p_1+p_2$ so $q^2=s$ — gives the amplitude in closed form.

\mathcal{M}=-\frac{e^2}{s}\big[\bar v(p_2)\gamma^\mu u(p_1)\big]\big[\bar u(p_3)\gamma_\mu v(p_4)\big],\qquad s=(p_1+p_2)^2 \qquad (5)

where: $p_1,p_2$ are the incoming $e^-,e^+$ momenta and $p_3,p_4$ the outgoing $\mu^-,\mu^+$ ; the two brackets are the electron and muon currents; $s$ is the Mandelstam invariant. Reading: the amplitude is a current–current interaction glued by $1/s$ — the whole diagram compressed into one line, exactly the product the first figure assembles. Squaring, spin-summing, and integrating this amplitude (jobs owned by aq-pp-matrix-element-spin-averaging and aq-pp-ee-to-mumu) yields a cross-section that is pure prediction of the rules, the QED benchmark against which every collider normalizes.

\sigma(e^+e^-\to\mu^+\mu^-)=\frac{4\pi\alpha^2}{3s}=\frac{86.9\ \text{nb}}{s\,[\text{GeV}^2]}\quad(s\gg m_\mu^2) \qquad (6)

where: the high-energy (massless-muon) limit has been taken. Reading: the $1/s$ falloff is dimensional — a squared amplitude $\propto e^4=(4\pi\alpha)^2$ over the only scale $s$ — and the coefficient is fixed once $\alpha$ is known, with no free parameters.

Derivation AWhere the vertex factor comes from — reading it off the QED Lagrangian

The goal is to show that the vertex factor $-iQe\gamma^\mu$ is not an independent postulate but is dictated by the gauge principle: it is $i$ times the interaction term of the QED Lagrangian with the fields amputated. We start from the free Dirac Lagrangian, promote the derivative to a covariant one so the theory is invariant under local phase rotations, and identify the single cubic term that appears — the electron–photon coupling whose coefficient is the vertex.

Step 1. Begin with the free Dirac Lagrangian $\mathcal{L}_0=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi$ , which is invariant under a global phase change $\psi\to e^{iQ\theta}\psi$ but not a local one $\theta\to\theta(x)$ , because $\partial_\mu$ then hits $\theta(x)$ and spoils the invariance. Why: demanding that the phase be a free choice at every spacetime point — the local gauge principle — is what forces a compensating field into existence.

Step 2. Restore it by replacing $\partial_\mu\to D_\mu=\partial_\mu+iQeA_\mu$ , where the photon field transforms as $A_\mu\to A_\mu-\frac1e\partial_\mu\theta$ . The Lagrangian becomes $\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi=\mathcal{L}_0-Qe\,\bar\psi\gamma^\mu\psi\,A_\mu$ . Why: the covariant derivative sews the photon to the fermion in the only way gauge symmetry allows, at the price of exactly one interaction term.

Step 3. Read the interaction term $\mathcal{L}_{\text{int}}=-Qe\,\bar\psi\gamma^\mu\psi\,A_\mu$ : it couples two fermion fields to one photon field — a single three-line vertex, with no four-or-more-line term anywhere in QED. Why: the number of fields in the interaction term is the number of lines at the vertex, so QED's one cubic term means one three-point vertex and nothing else.

Step 4. The Feynman rule for a vertex is $i$ times the interaction coefficient with the fields stripped off: $i\times(-Qe\gamma^\mu)=-iQe\gamma^\mu$ , the $\gamma^\mu$ surviving because it sits between the two fermion fields; the coupling is $e=\sqrt{4\pi\alpha}\approx0.303$ . Why: the momentum-space rule is the functional derivative of $i\mathcal{L}_{\text{int}}$ with respect to the three fields, which returns the coupling coefficient — the vertex is the Lagrangian's fingerprint.

Go deeper · Derivation B · 1 minAssembling the amplitude for e⁺e⁻ → μ⁺μ⁻ from the rules

Derivation BAssembling the amplitude for e⁺e⁻ → μ⁺μ⁻ from the rules

The goal is to run the assembly law end to end on the archetypal QED process and recover the closed-form amplitude of equation (5), so that "read it off and multiply" becomes a concrete calculation rather than a slogan. We label the external legs, attach a spinor to each, insert the two vertices and the photon propagator in the order fixed by the fermion lines, contract indices, and collect the factors of $i$ and the charges.

Step 1. Label the momenta and identify the external factors. Incoming $e^-$ gives $u(p_1)$ ; incoming $e^+$ gives $\bar v(p_2)$ ; outgoing $\mu^-$ gives $\bar u(p_3)$ ; outgoing $\mu^+$ gives $v(p_4)$ . There is one internal photon of momentum $q=p_1+p_2$ , so $q^2=s$ . Why: fixing the external dictionary first means the rest is pure substitution — the physics of "what is coming in and going out" is entirely in this line.

Step 2. Read the $e^+e^-$ line against its arrows, adjoint first: $\bar v(p_2)\,(-iQ_e e\gamma^\mu)\,u(p_1)$ with $Q_e=-1$ . Same for the muon line: $\bar u(p_3)\,(-iQ_\mu e\gamma^\nu)\,v(p_4)$ with $Q_\mu=-1$ . Why: the against-the-arrow ordering makes each current a $1\times4$ row times $4\times4$ matrix times $4\times1$ column — a genuine number and a four-vector.

Step 3. Glue the currents with the photon propagator $-ig_{\mu\nu}/q^2$ , which contracts index $\mu$ on the electron vertex to index $\nu$ on the muon vertex:

-i\mathcal{M}=\big[\bar v_2(-iQ_e e\gamma^\mu)u_1\big]\frac{-ig_{\mu\nu}}{q^2}\big[\bar u_3(-iQ_\mu e\gamma^\nu)v_4\big].

Why: the metric $g_{\mu\nu}$ is the contraction that turns two four-vector currents into one Lorentz scalar — current coupled to current.

Step 4. Collect the scalar prefactors: $(-iQ_e e)(-i/q^2)(-iQ_\mu e)=(-i)^3Q_eQ_\mu\,e^2/q^2$ . With $(-i)^3=i$ and $Q_eQ_\mu=+1$ , $-i\mathcal{M}=\dfrac{i\,e^2}{s}\big[\bar v_2\gamma^\mu u_1\big]\big[\bar u_3\gamma_\mu v_4\big]$ , hence $\mathcal{M}=-\dfrac{e^2}{s}\big[\bar v_2\gamma^\mu u_1\big]\big[\bar u_3\gamma_\mu v_4\big]$ , matching equation (5). Why: the three factors of $-i$ and the two negative charges combine to a clean overall sign — an honest check that the bookkeeping closed.

Go deeper · Derivation C · 1 minWhy the propagator has a 1 over q² minus m² pole — the sum over time-orderings

Derivation CWhy the propagator has a 1 over q² minus m² pole — the sum over time-orderings

The goal is to show that the covariant propagator is the summed result of old-fashioned time-ordered perturbation theory, so that the single denominator $1/(q^2-m^2)$ is understood as the collapse of two separate time-orderings into one Lorentz-invariant object — the sense in which the rules "reproduce the full QFT derivation term for term." We take a virtual quantum of mass $m$ exchanged between two vertices, write the two ways it can be ordered in time, add the energy-denominator factors, and combine them.

Step 1. In time-ordered perturbation theory a virtual particle of three-momentum $\mathbf q$ and energy $E_q=\sqrt{\mathbf q^2+m^2}$ can be emitted at the earlier vertex and absorbed at the later, or vice versa — two distinct orderings, each on-shell but not conserving energy. Why: on-shell intermediate states with energy unconserved is the hallmark of the old formalism; each ordering carries a factor $1/(E-E_{\text{intermediate}})$ .

Step 2. The two orderings contribute $\dfrac{1}{2E_q}\dfrac{1}{E_0-E_q}$ and $\dfrac{1}{2E_q}\dfrac{1}{-E_0-E_q}$ , where $E_0$ is the transferred energy. Why: the $1/2E_q$ is the relativistic state normalization, and the two denominators are the two orders in which the intermediate state can appear.

Step 3. Add them: $\dfrac{1}{2E_q}\Big[\dfrac{1}{E_0-E_q}-\dfrac{1}{E_0+E_q}\Big]=\dfrac{1}{E_0^2-E_q^2}=\dfrac{1}{E_0^2-\mathbf q^2-m^2}=\dfrac{1}{q^2-m^2}$ , since $q^2=E_0^2-\mathbf q^2$ . Why: the two frame-dependent orderings sum to a single Lorentz-invariant denominator — the covariant propagator hides the time-ordering entirely.

Step 4. For the photon, $m=0$ gives the denominator $1/q^2$ , and summing over the physical polarizations supplies the numerator $-g_{\mu\nu}$ (Feynman gauge), so $D_{\mu\nu}=-ig_{\mu\nu}/q^2$ . Why: the massless pole is why electromagnetism has infinite range, and the $g_{\mu\nu}$ is the polarization completeness relation — both from the same summed exchange.

Limiting cases

Case	Result	Meaning
Tree level (no loops)	$\mathcal{M}$ finite, no $\int d^4q$	leading order in $\alpha$ ; every internal momentum is fixed by conservation
Static / low-energy	$\dfrac{e^2}{q^2}\to$ Coulomb $\dfrac{Q_1Q_2\,e^2}{4\pi r}$	Fourier transform of the $1/\mathbf q^2$ propagator is the $1/r$ potential
Massless fermion ( $m\to0$ )	$S_F\to i\slashed q/q^2$	chirality conserved along the line; numerator loses its mass term
Real (external) photon	use $\epsilon_\mu(p)$ , not a propagator	on-shell $q^2=0$ ; only virtual photons carry $-ig_{\mu\nu}/q^2$
Higher order (loops)	each extra pair of vertices $\times\,\alpha$ , plus $\int\dfrac{d^4q}{(2\pi)^4}$	perturbation series in $\alpha\approx1/137$ ; loops need renormalization
Heavy mediator (contrast)	$\dfrac{-ig_{\mu\nu}}{q^2-M^2}\xrightarrow{q^2\ll M^2}\dfrac{ig_{\mu\nu}}{M^2}$	massive $W,Z$ give short-range, contact (Fermi) interactions

Common trapReading the spinors left-to-right off the page

The single most common blunder is to write the spinor factors in the order they appear across the page — left to right — instead of against the arrows of the fermion line. The rule is exact: pick a fermion line, start at the head (where the arrow points out of the diagram), and walk backward against every arrow, writing each factor as you meet it. The first factor is therefore always an adjoint spinor — $\bar u$ for an outgoing particle, $\bar v$ for an incoming antiparticle — and the last is a plain $u$ or $v$ . For $e^+e^-\to\mu^+\mu^-$ this makes the electron current $\bar v(p_2)\gamma^\mu u(p_1)$ , the positron's adjoint on the left even though the electron entered "first". Reverse it and you compute $u\gamma^\mu\bar v$ , a $4\times4$ matrix rather than the Lorentz scalar an amplitude must be; the indices will not contract. A related trap: the momentum $q$ inside a propagator is a virtual, off-shell quantum ( $q^2\ne m^2$ ), not a real particle you could detect — the propagator is an amplitude summed over intermediate states, not a snapshot of a travelling particle.

Worked examples

1 · Assembling e⁻μ⁻ → e⁻μ⁻ from scratch (confidence builder)

Write down $-i\mathcal{M}$ for electron–muon scattering, the t-channel workhorse of QED, using only the rules. Take incoming $e^-(p_1)$ , $\mu^-(p_2)$ and outgoing $e^-(p_3)$ , $\mu^-(p_4)$ .

Step 1. Identify the external factors: incoming $e^-\to u(p_1)$ , outgoing $e^-\to\bar u(p_3)$ , incoming $\mu^-\to u(p_2)$ , outgoing $\mu^-\to\bar u(p_4)$ . The exchanged photon carries the transferred momentum $q=p_1-p_3$ , so $q^2=t$ (Mandelstam $t$ ), spacelike, $t<0$ .

Step 2. Build the two currents against the arrows. The electron line runs $p_1\to p_3$ , read backward as $\bar u(p_3)\,(-iQ_e e\gamma^\mu)\,u(p_1)$ ; the muon line gives $\bar u(p_4)\,(-iQ_\mu e\gamma^\nu)\,u(p_2)$ . With $Q_e=Q_\mu=-1$ both vertices are $+ie\gamma$ .

Step 3. Glue with the photon propagator $-ig_{\mu\nu}/t$ and collect: $(-iQ_e e)(-i/t)(-iQ_\mu e)=i\,Q_eQ_\mu e^2/t=ie^2/t$ . Hence

-i\mathcal{M}=\frac{ie^2}{t}\big[\bar u_3\gamma^\mu u_1\big]\big[\bar u_4\gamma_\mu u_2\big]\ \Rightarrow\ \mathcal{M}=-\frac{e^2}{t}\big[\bar u(p_3)\gamma^\mu u(p_1)\big]\big[\bar u(p_4)\gamma_\mu u(p_2)\big].

Step 4. Sanity checks. The structure is identical to $e^+e^-\to\mu^+\mu^-$ but with $s\to t$ and particle spinors replacing antiparticle ones — exactly what "crossing" predicts. In the non-relativistic, low- $|t|$ limit the currents reduce to charge densities and $\mathcal{M}\to e^2/|\mathbf q|^2$ , whose Fourier transform is the Coulomb potential — the amplitude reproduces Rutherford scattering. $\mathcal{M}=-\dfrac{e^2}{t}(\bar u_3\gamma^\mu u_1)(\bar u_4\gamma_\mu u_2)$ , the t-channel one-photon-exchange amplitude.

2 · The QED benchmark cross-section, with numbers

Use the amplitude for $e^+e^-\to\mu^+\mu^-$ to evaluate the total cross-section at $\sqrt s=10$ GeV (well above threshold, below the $Z$ ), and state the coupling strength that entered.

Step 1. The vertex strength. From $\alpha=1/137.036$ , the coupling at each dot is $e=\sqrt{4\pi\alpha}=\mathbf{0.303}$ . The amplitude carries two vertices, so $\mathcal{M}\propto e^2$ and $|\mathcal{M}|^2\propto e^4=(4\pi\alpha)^2\propto\alpha^2$ — one-photon exchange is order $\alpha^2$ in the rate.

Step 2. The spin-averaged, integrated result (derived in aq-pp-ee-to-mumu) is $\sigma=\dfrac{4\pi\alpha^2}{3s}$ . Insert the coefficient: $\dfrac{4\pi\alpha^2}{3}=\dfrac{4\pi(5.325\times10^{-5})}{3}=2.231\times10^{-4}\ \text{GeV}^{-2}\!\cdot\!\text{GeV}^2$ , so $\sigma=\dfrac{2.231\times10^{-4}}{s}\ \text{GeV}^{-2}$ .

Step 3. Convert to cross-section units with $1\ \text{GeV}^{-2}=0.3894\ \text{mb}=3.894\times10^5\ \text{nb}$ : $\sigma\,s=2.231\times10^{-4}\times3.894\times10^{5}=\mathbf{86.9\ nb\cdot GeV^2}$ , the famous "point cross-section". At $\sqrt s=10$ GeV, $s=100\ \text{GeV}^2$ , so

\sigma=\frac{86.9\ \text{nb}}{100}=\mathbf{0.87\ nb}.

Step 4. Interpret. The $1/s$ falloff drops the cross-section a hundredfold from $\sqrt s=1$ to $\sqrt s=10$ GeV — pure dimensional analysis of a fixed-coupling amplitude. Measured $e^+e^-\to\mu^+\mu^-$ rates at PETRA, PEP, and LEP track this curve across three decades of energy, and the R-ratio (hadrons over this muon benchmark) is how the number of quark colours was counted. $\sigma(e^+e^-\to\mu^+\mu^-)=86.9\ \text{nb}/s[\text{GeV}^2]=0.87$ nb at $\sqrt s=10$ GeV.

Physics of the portrait

The rules bear one man's name because he changed how the calculation is thought, not just how it is done. Feynman replaced an abstract operator formalism with a picture a physicist could draw on a napkin and a dictionary that turned it into an integral — and the pictures turned out to be more than a mnemonic.

Origins · 1 minRichard Feynman and how this was discovered

Richard Feynman

1918–1988 · quantum electrodynamics and the diagrammatic method

Richard Phillips Feynman was, by his own account, more interested in doing physics than in formalizing it, and the diagrams are the purest expression of that instinct. Working at Cornell in the late 1940s on the problem of QED's infinities, he developed a "space-time approach" in which particles are drawn as lines and interactions as vertices, with antiparticles represented as particles moving backward in time — the Stückelberg–Feynman interpretation that lets a single $v$ -spinor stand in for a positron. His 1949 papers, The Theory of Positrons and Space-Time Approach to Quantum Electrodynamics, laid out the rules essentially as they are used today: a factor for each line, a factor for each vertex, momentum conservation at every junction, and an integral over each undetermined loop momentum. What made the method revolutionary was not just its economy but that Freeman Dyson, in 1949, proved it equivalent to the operator-based renormalization theories that Julian Schwinger and Sin-Itiro Tomonaga had built independently — three routes to the same finite answers. Feynman shared the 1965 Nobel Prize with Schwinger and Tomonaga, and in 1986 famously traced the Challenger disaster to a failed O-ring by dropping a seal into iced water on live television. The diagrams that carry his name are now the visual language of every field theory, from QED to the Standard Model to string perturbation theory.

History

1927

— Dirac quantizes the electromagnetic field, the first quantum-electrodynamic calculation

1929

— Heisenberg and Pauli build the general quantum field theory of interacting fields

1948

— Feynman, Schwinger and Tomonaga complete renormalized QED

1949

— Feynman publishes the space-time diagrams and the rules for QED

1949

— Dyson proves the Feynman, Schwinger and Tomonaga formulations equivalent

1965

— Feynman, Schwinger and Tomonaga share the Nobel Prize for QED

Going further

Texts. Thomson, Modern Particle Physics, §5.4 (Feynman rules for QED) and §5.4.1 (treatment of antiparticles) · Peskin & Schroeder, An Introduction to Quantum Field Theory, §4.7–§4.8 and the inside-cover rules · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.5 and Appendix D (Feynman rules, tree level) · Halzen & Martin, Quarks and Leptons, §6.1–§6.9 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 7 and Ch. 13.

Provenance. The external spinors $u,\bar u,v,\bar v$ are recalled by name from aq-pp-dirac-solutions; the current structure $\bar u\gamma^\mu u$ from aq-pp-dirac-adjoint-bilinear-covariants; the exchange picture and time-ordered origin of the propagator from aq-pp-perturbation-theory-exchange; and the explicit form $-ig_{\mu\nu}/q^2$ from aq-pp-photon-propagator. A verifier should confirm: (1) the vertex $-iQe\gamma^\mu$ follows as $i$ times the interaction term of $\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi$ with $D_\mu=\partial_\mu+iQeA_\mu$ ; (2) the assembled amplitude $\mathcal{M}=-(e^2/s)(\bar v_2\gamma^\mu u_1)(\bar u_3\gamma_\mu v_4)$ with the sign from $(-i)^3Q_eQ_\mu=i$ ; (3) the two-time-ordering sum yields $1/(q^2-m^2)$ ; (4) $e=\sqrt{4\pi\alpha}=0.303$ and $\sigma=4\pi\alpha^2/3s=86.9\ \text{nb}/s[\text{GeV}^2]=0.87$ nb at $\sqrt s=10$ GeV.

Prerequisites. Dirac Spinor Solutions · The Adjoint Spinor and Bilinear Covariants · Interaction by Particle Exchange · The Photon Propagator. Leads to. e⁺e⁻ → μ⁺μ⁻ · Matrix Elements and Spin Averaging · Compton Scattering · Trace Techniques.

Ask the tutor: Name any tree-level QED process and I will assemble its amplitude from the rules step by step — spinors, vertices, propagator, and the final $-i\mathcal{M}$ .

Why is there a factor of i in every single rule?

Because the amplitude the rules build is $-i\mathcal{M}$ , and that $-i$ is threaded through the whole perturbation expansion. Each term in the Dyson series carries a factor of $i$ from expanding $\exp(i\!\int\!\mathcal{L}_{\text{int}})$ , so a vertex carries $i\times(\text{coupling})$ and a propagator $i\times(\text{inverse kinetic operator})$ . The bookkeeping is arranged so that multiplying all the factors and stripping the overall $-i$ leaves $\mathcal{M}$ , whose square is a real, positive rate. The individual $i$ 's are not physical on their own; only $|\mathcal{M}|^2$ is measurable, and every convention-dependent phase (including the sign of the vertex) cancels there. If your $i$ 's do not combine to the expected factor, you have dropped a rule.

Why must the adjoint spinor be written first?

Because a fermion current has to be a Lorentz scalar or four-vector — one number per index — and only the ordering $\bar u\,\Gamma\,u$ produces that. In matrix terms $\bar u$ is a $1\times4$ row, $\Gamma$ (a product of gammas) is $4\times4$ , and $u$ is a $4\times1$ column, so $\bar u\Gamma u$ multiplies down to a $1\times1$ number. Reverse it to $u\,\Gamma\,\bar u$ and you multiply a column by a row, getting a $4\times4$ matrix — not an amplitude. "Follow the line against the arrows, adjoint first" is just the rule that guarantees the row-matrix-column pattern for every fermion line; interior propagators $(\slashed q+m)$ slot in as further $4\times4$ matrices, and the whole chain still closes to a number.

Are the particles on internal lines real?

No — internal lines are virtual, off-shell: their four-momentum does not satisfy $q^2=m^2$ . A real photon has $q^2=0$ , but the photon exchanged in $e^+e^-\to\mu^+\mu^-$ has $q^2=s>0$ , which no real photon can have. The propagator is not a particle travelling from A to B; it is the amplitude, summed over intermediate states, for the field to carry momentum $q$ between two vertices. This is why it can be large (near its pole, $q^2\to m^2$ ) or small (far off-shell) with no "real particle" present — and why a resonance appears when $s$ approaches $M^2$ for a massive mediator.

Why is the propagator −ig_{μν}/q² and not something gauge-dependent?

It is gauge-dependent in general — the full $R_\xi$ propagator is $\frac{-i}{q^2}[g_{\mu\nu}-(1-\xi)q_\mu q_\nu/q^2]$ , and $-ig_{\mu\nu}/q^2$ is just the Feynman-gauge choice $\xi=1$ . What saves you is that the extra $q_\mu q_\nu$ piece never contributes to a physical amplitude: each propagator index contracts into a vertex on a conserved fermion current, and $q_\mu\bar u(p_3)\gamma^\mu u(p_1)=0$ by the Dirac equation (the Ward identity). So the $q_\mu q_\nu$ term hits a zero, and every gauge gives the same answer. Feynman gauge is chosen simply because it is the least cluttered; picking a different $\xi$ in a loop calculation is a standard cross-check, since the cross-section must not depend on it.

How can a positron be an electron going backward in time?

It is a bookkeeping identity, not a claim about time travel. In the field expansion, the operator that creates an outgoing positron is the same one that annihilates an incoming negative-energy electron state, so the mathematics is identical whether you call the line "a positron forward" or "an electron backward"; the diagram encodes this by drawing the antiparticle's arrow against the momentum flow. Practically, you need no separate rules for antiparticles: an incoming positron uses $\bar v(p)$ and an outgoing positron uses $v(p)$ , the $v$ -spinors being the reinterpreted negative-energy Dirac solutions. Every observable comes out with the correct positive-energy, opposite-charge values, so nothing acausal happens — the "backward in time" language just keeps one set of rules covering particles and antiparticles alike.

Do the QED rules extend to QCD and the weak interaction?

Yes — the three ingredient types are universal, only the specific factors change. QCD keeps the fermion structure but replaces the photon with eight gluons; the quark–gluon vertex is $-ig_s\gamma^\mu T^a$ (with a colour matrix $T^a$ ), and because gluons carry colour there are also three- and four-gluon vertices with no QED analogue. The weak interaction uses massive $W,Z$ propagators $-ig_{\mu\nu}/(q^2-M^2)$ that collapse to a contact Fermi interaction at low energy, and its vertices are chiral, containing $\gamma^\mu(1-\gamma^5)$ so they couple only to left-handed fermions — parity violation. But the assembly law, the " $i$ times the coupling" vertex rule, the against-the-arrow ordering, the loop integral $\int d^4q/(2\pi)^4$ , and the $(-1)$ per closed fermion loop are inherited unchanged from QED. Learn the QED rules and you have learned the grammar of the Standard Model.