The photon propagator & gauge choice

Every tree-level QED amplitude is a sandwich: a current where the electron couples to the electromagnetic field, a current where a second charged particle does the same, and — clamped between them — a factor for the virtual photon that carries momentum from one to the other. That middle factor is the photon propagator, and in the form used for hand calculations it is disarmingly simple: $-g_{\mu\nu}/q^2$ . The denominator $q^2$ is the massless limit of the generic exchange factor $1/(q^2-m_X^2)$ that also governs the Yukawa potential; set $m_X\to0$ and the finite-range force becomes the infinite-range Coulomb one. The numerator $-g_{\mu\nu}$ is the fingerprint of the photon's spin: it is what a sum over the exchanged photon's polarisation states leaves behind, and it is the object that stitches the electron current to the muon current index for index. But that tidy expression hides a subtlety the article's title flags: it is a choice, not the answer. The photon field has more components than physical polarisations, and gauge invariance makes the naive propagator flatly non-existent — the operator you would invert has no inverse. You fix this by choosing a gauge, which adds a free parameter $\xi$ and gives a whole family of propagators; $-g_{\mu\nu}/q^2$ is just the friendliest member, the Feynman gauge $\xi=1$ . The reassurance, and the deep point, is that nothing measurable depends on which you pick: the extra $q_\mu q_\nu$ pieces that distinguish the gauges vanish the instant they touch a conserved current. This article builds the propagator from the exchange picture, shows why gauge fixing is unavoidable, and proves the physics is gauge-blind.

Key takeaways

A QED amplitude factorises into two vertex currents and a propagator for the exchanged virtual photon; the tree-level photon propagator is $-g_{\mu\nu}/q^2$ .
The denominator is the massless limit of the exchange factor $1/(q^2-m_X^2)$ — the same $m_X\to0$ that turns the Yukawa potential into Coulomb's law.
The numerator $-g_{\mu\nu}$ comes from the polarisation sum $\sum_\lambda\varepsilon^{(\lambda)}_\mu\varepsilon^{(\lambda)*}_\nu=-g_{\mu\nu}$ ; it ties the electron current to the muon current, giving $\mathcal{M}=-Q_eQ_\mu e^2\,(j_e\cdot j_\mu)/q^2$ .
Gauge invariance makes the bare Maxwell operator singular, so the propagator does not exist until a gauge is fixed: the $R_\xi$ family is $\frac{-i}{q^2}\big[g_{\mu\nu}-(1-\xi)q_\mu q_\nu/q^2\big]$ , with Feynman gauge $\xi=1$ and Landau gauge $\xi=0$ .
Physical amplitudes are gauge-independent because current conservation $q_\mu j^\mu=0$ annihilates every $q_\mu q_\nu$ term — the choice of $\xi$ never survives to the answer.
Key idea: the photon propagator $-g_{\mu\nu}/q^2$ is the massless limit of $1/(q^2-m_X^2)$ dressed by the photon's polarisation sum $-g_{\mu\nu}$ , and it exists only after a gauge is chosen; every measurable is blind to that choice because conserved currents kill the gauge-dependent $q_\mu q_\nu$ terms.

The core insight

The photon propagator $-g_{\mu\nu}/q^2$ is the massless ( $m_X\to0$ ) limit of the generic exchange factor $1/(q^2-m_X^2)$ , with the tensor $-g_{\mu\nu}$ generated by summing over the virtual photon's polarisation states. Because gauge invariance leaves the field with unphysical components, the propagator is defined only relative to a gauge choice $\xi$ — yet current conservation $q_\mu j^\mu=0$ ensures no physical amplitude ever depends on it.

Where this lives today

The photon propagator is the single most-used internal line in particle physics, and its gauge freedom is a daily working tool. Every tree-level QED cross section computed for a collider — Bhabha scattering that calibrates luminosity at electron–positron machines, Møller scattering, $e^+e^-\to\mu^+\mu^-$ used to normalise the $R$ -ratio, deep-inelastic $ep$ scattering that maps the proton — begins by writing $-ig_{\mu\nu}/q^2$ for the exchanged photon and reading off $\mathrm d\sigma/\mathrm d\Omega\propto1/q^4$ . Practitioners deliberately switch gauges as a diagnostic: an amplitude is recomputed in Feynman and Landau gauge and the two must agree, a check built into automated tools like FeynCalc and MadGraph. The $q^2$ -dependence of the propagator, dressed by vacuum-polarisation loops, is what makes the fine-structure constant run — $\alpha(0)\approx1/137$ climbing to $\alpha(M_Z^2)\approx1/128$ , measured precisely at LEP — so the "constant" in the propagator's residue is itself a function of the momentum the propagator carries. And the same gauge-fixing machinery, generalised from the abelian photon to non-abelian gluons and weak bosons, is where Faddeev–Popov ghosts enter and where the renormalisability of the whole Standard Model was established.

The physical picture

Two charged particles never touch. An electron and a muon deflect each other by tossing a virtual photon across the gap between them: the electron's current radiates it at one vertex, the muon's current absorbs it at the other, and the photon carries the four-momentum transfer $q$ from one to the other. The amplitude is the product of what happens at each vertex with a propagator for the messenger in between. Because the photon is massless, its propagator denominator is simply $q^2$ ; because it is spin-1, a sum over its polarisation states dresses the numerator with $-g_{\mu\nu}$ , whose two indices reach out and lock onto the two currents. The first figure is that single exchange, with the amplitude it evaluates to.

Figure

The t-channel exchange for e⁻μ⁻ → e⁻μ⁻: the electron current jₑ and muon current jμ each contribute a vertex factor ieQγ, and the virtual photon carrying q = p₁ − p₃ contributes the propagator −gμν/q². The amplitude is M = −QₑQμe²(jₑ·jμ)/q²; at E = 500 MeV, θ = 60° the transfer is q² = −Q² = −0.25 GeV²

The clean expression $-g_{\mu\nu}/q^2$ is not the whole story, and the discrepancy is the subject the title names. A photon has four field components $A_\mu$ but only two physical polarisations; the redundancy is gauge freedom, the statement that $A_\mu$ and $A_\mu+\partial_\mu\chi$ describe the same physics. That redundancy has a sharp consequence: the operator whose inverse is the propagator has a zero eigenvalue and cannot be inverted at all. You buy an inverse by fixing a gauge, which introduces a parameter $\xi$ and, with it, a family of propagators. The second figure lays that family out and shows why the choice never leaks into an observable.

The core equations

The starting point is the exchange amplitude for any single-boson mediator. It factorises into a vertex at each end and a propagator denominator that carries the mediator's mass and the momentum it transfers — the structure inherited by name from the perturbation-theory-by-exchange and Yukawa articles.

\mathcal{M}=\underbrace{\langle\psi_c|V|\psi_a\rangle}_{\text{vertex 1}}\;\frac{1}{q^2-m_X^2}\;\underbrace{\langle\psi_d|V|\psi_b\rangle}_{\text{vertex 2}}\;\xrightarrow{\;m_X\to0\;}\;\frac{1}{q^2} \qquad (1)

where: $q^\mu$ — four-momentum transferred by the exchanged particle; $q^2=q_\mu q^\mu$ — its invariant square (spacelike, $q^2<0$ , for $t$ -channel scattering); $m_X$ — mediator mass. Reading it: the propagator denominator $1/(q^2-m_X^2)$ is the whole content of "particle exchange"; the photon is the $m_X\to0$ special case, which is why the electromagnetic force reaches to infinity while the massive-mediator weak force does not. The vertex is fixed by how a charged fermion couples to the electromagnetic field. Minimal substitution $\partial_\mu\to\partial_\mu+iqA_\mu$ inserts the photon field into the Dirac equation and identifies the interaction, whose matrix element between spinors is the QED vertex.

\hat V_D=q\,\gamma^0\gamma^\mu A_\mu\;\Longrightarrow\;\text{vertex factor}=\bar u(p_3)\,\big(ieQ_f\,\gamma^\mu\big)\,u(p_1) \qquad (2)

where: $Q_f$ — fermion charge in units of $e$ ( $Q_e=-1$ for the electron); $e$ — the positron charge, with $\alpha=e^2/4\pi\approx1/137$ so $e\approx0.303$ in Heaviside–Lorentz natural units; $\gamma^\mu$ — Dirac matrices; $u(p)$ — particle spinors. Reading it: each vertex contributes one power of $e$ , one factor of the charge, and a single $\gamma^\mu$ sandwiched between the incoming and outgoing spinors — the current $j^\mu=\bar u(p_3)\gamma^\mu u(p_1)$ . The photon is spin-1, so the amplitude must sum over the polarisation states of the virtual photon threaded between the vertices. That completeness sum is what promotes the bare $1/q^2$ to the full tensor propagator.

\sum_{\lambda}\varepsilon^{(\lambda)}_\mu\,\varepsilon^{(\lambda)*}_\nu=-g_{\mu\nu}\;\Longrightarrow\;\text{propagator}=\frac{-g_{\mu\nu}}{q^2} \qquad (3)

where: $\varepsilon^{(\lambda)}_\mu$ — polarisation four-vector of the exchanged photon; $\lambda$ — runs over a complete set of states; $g_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$ — the Minkowski metric. Reading it: the two indices $\mu,\nu$ are the loose ends that clamp onto the two vertex currents, and the minus sign and off-diagonal structure of $g_{\mu\nu}$ are exactly the polarisation bookkeeping of a massless spin-1 messenger. Contracting the propagator between the two currents gives the compact, manifestly Lorentz-invariant amplitude — the headline result for $e^-\mu^-\to e^-\mu^-$ .

\mathcal{M}=\big[\bar u(p_3)ieQ_e\gamma^\mu u(p_1)\big]\frac{-g_{\mu\nu}}{q^2}\big[\bar u(p_4)ieQ_\mu\gamma^\nu u(p_2)\big]=-\frac{Q_eQ_\mu e^2}{q^2}\,(j_e\cdot j_\mu) \qquad (4)

where: $j_e\cdot j_\mu=j_e^\mu\, g_{\mu\nu}\, j_\mu^\nu$ — the Lorentz-invariant contraction of the two currents; the overall sign is the $-g_{\mu\nu}$ of the propagator. Reading it: the entire spin and Lorentz structure of the process is one dot product of currents divided by $q^2$ ; everything else in a cross-section calculation is squaring and summing this object. The clean $-g_{\mu\nu}/q^2$ is a gauge choice. The general covariant ( $R_\xi$ ) propagator carries a free parameter, and only a particular value reproduces equation (3).

D_{\mu\nu}(q)=\frac{-i}{q^2+i\epsilon}\left[g_{\mu\nu}-(1-\xi)\frac{q_\mu q_\nu}{q^2}\right] \qquad (5)

where: $\xi$ — the gauge parameter; $\xi=1$ is Feynman gauge, collapsing the bracket to $g_{\mu\nu}$ ; $\xi=0$ is Landau gauge, making the propagator transverse ( $q^\mu D_{\mu\nu}=0$ ); $i\epsilon$ — the Feynman causal prescription. Reading it: there is no single "photon propagator" — there is a one-parameter family, and $-ig_{\mu\nu}/q^2$ is merely its most convenient member. What rescues physics from this ambiguity is current conservation: the electromagnetic current the photon couples to has zero four-divergence, so contracting the gauge-dependent term against it gives nothing.

q_\mu\, j^\mu=q_\mu\,\bar u(p_3)\gamma^\mu u(p_1)=0\;\Longrightarrow\;q_\mu q_\nu\text{ terms drop from }\mathcal{M} \qquad (6)

where: $q=p_1-p_3$ — the momentum transfer at the electron vertex. Reading it: the $(1-\xi)$ piece of the propagator always appears sandwiched as $q_\mu q_\nu$ between two currents, and $q_\mu j^\mu=0$ deletes it identically — the amplitude is the same in every gauge, so the parameter $\xi$ was never physical.

Derivation AFrom the exchange factor to the photon propagator via the polarisation sum

The goal is to assemble the working propagator $-g_{\mu\nu}/q^2$ and the full amplitude for $e^-\mu^-\to e^-\mu^-$ from three ingredients already owned upstream: the generic exchange factor $1/(q^2-m_X^2)$ from the perturbation-theory-by-exchange article, the QED vertex from minimal substitution, and the spin-1 polarisation sum. We take the massless limit, insert the vertices, and use the completeness relation to convert a bare denominator into the index-carrying tensor that ties the currents together — Thomson's route to the QED matrix element.

Step 1. Start from the scalar exchange amplitude, equation (1), and set $m_X=0$ : the denominator becomes $1/q^2$ . Why: the photon is massless, so the Compton wavelength that set the Yukawa range diverges — the same limit that turns $e^{-m_Xr}/r$ into the bare $1/r$ Coulomb potential, and the reason electromagnetism is a long-range force.

Step 2. Replace each scalar vertex $\langle\psi|V|\psi\rangle$ with the QED vertex of equation (2). Minimal substitution $\partial_\mu\to\partial_\mu+iqA_\mu$ turns the Dirac interaction into $\hat V_D=q\gamma^0\gamma^\mu A_\mu$ ; sandwiched between final and initial spinors it gives the current $j^\mu_e=\bar u(p_3)\gamma^\mu u(p_1)$ scaled by $Q_e e$ , and likewise $j^\nu_\mu$ at the muon end. Why: a spin-1 field couples to a vector current, not a scalar density, so each vertex now carries an open Lorentz index — the photon must supply the indices to soak them up.

Step 3. Sum over the exchanged photon's polarisations. For the physical amplitude the completeness relation collapses to $\sum_\lambda\varepsilon^{(\lambda)}_\mu\varepsilon^{(\lambda)*}_\nu=-g_{\mu\nu}$ , so the messenger contributes $-g_{\mu\nu}/q^2$ between the two currents. Why: a virtual photon is not restricted to the two transverse states of a real photon; the sum runs over a complete set, and the metric tensor is what a complete polarisation sum leaves behind once the unphysical modes are accounted for.

Step 4. Contract. The two indices of $-g_{\mu\nu}$ lower one index on each current, leaving the dot product $j_e\cdot j_\mu$ , and the charges combine to $Q_eQ_\mu e^2$ :

\mathcal{M}=-\frac{Q_eQ_\mu e^2}{q^2}\,(j_e\cdot j_\mu).

Why: this is the payoff — a manifestly Lorentz-invariant scalar built from two currents and one propagator, the template every tree-level QED cross section is squared from. *Verified: $m_X\to0$ gives $1/q^2$ ; the vertex factor $ieQ\gamma^\mu$ follows from $\hat V_D=q\gamma^0\gamma^\mu A_\mu$ ; $\sum_\lambda\varepsilon_\mu\varepsilon^*_\nu=-g_{\mu\nu}$ contracts the currents to $-Q_eQ_\mu e^2(j_e\cdot j_\mu)/q^2$ .*

Go deeper · Derivation B · 2 minWhy the naive propagator does not exist and how gauge fixing repairs it

Derivation BWhy the naive propagator does not exist and how gauge fixing repairs it

The aim is to show that the free-photon action does not, by itself, define a propagator at all, and that a gauge-fixing term is the minimal repair, yielding the $R_\xi$ family of equation (5). The obstruction is gauge invariance made concrete: the operator we must invert has a zero eigenvalue, with the null direction pointing exactly along the gauge transformations. Adding a gauge-fixing term lifts that zero and makes the operator invertible, at the price of a free parameter $\xi$ .

Step 1. From $\mathcal{L}=-\tfrac14F_{\mu\nu}F^{\mu\nu}$ with $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ , the quadratic action in momentum space is $\tfrac12 A_\mu(q)\,O^{\mu\nu}\,A_\nu(-q)$ with the operator $O^{\mu\nu}=-q^2 g^{\mu\nu}+q^\mu q^\nu$ . The propagator is the inverse of $O^{\mu\nu}$ . Why: a propagator is by definition the Green's function of the field's wave operator, so finding it is exactly the problem of inverting the quadratic form the Lagrangian defines.

Step 2. Test invertibility by acting on $q_\nu$ : $O^{\mu\nu}q_\nu=-q^2 q^\mu+q^\mu q^2=0$ . The operator annihilates $q_\nu$ , so it has a zero eigenvalue and no inverse exists. Why: $q_\nu$ is the momentum-space image of a pure gauge mode $A_\mu=\partial_\mu\chi$ ; gauge invariance guarantees the action is flat along these directions, and a flat direction is precisely a non-invertible quadratic form. The redundancy that makes the theory consistent is what blocks the propagator.

Step 3. Add a gauge-fixing term $-\tfrac{1}{2\xi}(\partial_\mu A^\mu)^2$ . It contributes $-\tfrac{1}{\xi}q^\mu q^\nu$ to the operator, giving $O^{\mu\nu}_\xi=-q^2 g^{\mu\nu}+\big(1-\tfrac1\xi\big)q^\mu q^\nu$ . Now $O^{\mu\nu}_\xi q_\nu=-\tfrac1\xi q^2 q^\mu\neq0$ : the zero mode is lifted. Why: the extra term penalises the gauge direction $\partial_\mu A^\mu$ , curving the flat direction so the quadratic form becomes non-degenerate — the smallest change that restores an inverse without disturbing the physical transverse modes.

Step 4. Invert. Writing the inverse as $D_{\nu\rho}=\frac{-i}{q^2}\big(g_{\nu\rho}-(1-\xi)\frac{q_\nu q_\rho}{q^2}\big)$ and forming $O^{\mu\nu}_\xi D_{\nu\rho}$ , the $q^\mu q_\rho$ pieces cancel because their coefficients sum to $(1-\xi)+(1-\tfrac1\xi)-(1-\tfrac1\xi)(1-\xi)=0$ , leaving $i\delta^\mu_\rho$ . This is equation (5). Why: the cancellation is the algebraic heart of gauge fixing — the $\xi$ -dependent term is engineered so the leftover is a clean delta function, and the surviving family is parametrised by the single number $\xi$ we introduced. *Verified: $O^{\mu\nu}q_\nu=0$ (bare operator singular); $O^{\mu\nu}_\xi q_\nu=-q^2q^\mu/\xi\neq0$ ; $O^{\mu\nu}_\xi D_{\nu\rho}=i\delta^\mu_\rho$ with the $q^\mu q_\rho$ coefficient $(1-\xi)+(1-1/\xi)-(1-1/\xi)(1-\xi)=0$ .*

Go deeper · Derivation C · 1 minGauge independence of the amplitude from current conservation

Derivation CGauge independence of the amplitude from current conservation

The purpose is to prove the reassuring theorem: although the propagator of equation (5) depends on $\xi$ , the physical amplitude does not. The mechanism is current conservation — the Ward identity at tree level — which says the four-divergence of the electromagnetic current vanishes, so the entire $q_\mu q_\nu$ part of the propagator contracts to zero between two external currents. Nothing measurable can carry a memory of the gauge we chose.

Step 1. Write the amplitude with the full $R_\xi$ propagator: $\mathcal{M}\propto j_e^\mu\Big[g_{\mu\nu}-(1-\xi)\tfrac{q_\mu q_\nu}{q^2}\Big]j_\mu^\nu$ . The gauge parameter lives only in the second term, multiplied by $q_\mu$ on one side and $q_\nu$ on the other. Why: isolating the $\xi$ -dependence shows the whole question reduces to a single object — the contraction $q_\mu j_e^\mu$ — so if that vanishes, gauge independence is immediate.

Step 2. Evaluate $q_\mu j_e^\mu$ with $q=p_1-p_3$ : $q_\mu\bar u(p_3)\gamma^\mu u(p_1)=\bar u(p_3)(\slashed p_1-\slashed p_3)u(p_1)$ . Why: the momentum transfer at a vertex is fixed by the external legs meeting there, so contracting $q$ into the current just replaces $\slashed q$ by the difference of the two on-shell momenta.

Step 3. Apply the Dirac equations for the external spinors: $\slashed p_1 u(p_1)=m_e u(p_1)$ and $\bar u(p_3)\slashed p_3=m_e\bar u(p_3)$ . Then $\bar u(p_3)(\slashed p_1-\slashed p_3)u(p_1)=(m_e-m_e)\,\bar u(p_3)u(p_1)=0$ . Why: both external electrons sit on the mass shell, so each $\slashed p$ collapses to the same mass $m_e$ ; the two contributions are equal and cancel, giving $q_\mu j_e^\mu=0$ — current conservation, verified on the states that actually appear.

Step 4. Substitute back: the $(1-\xi)$ term is multiplied by $q_\mu j_e^\mu=0$ , so it vanishes for every $\xi$ . The amplitude collapses to $\mathcal{M}\propto j_e^\mu g_{\mu\nu}j_\mu^\nu=(j_e\cdot j_\mu)$ , independent of the gauge. Why: this is the whole justification for using the friendly Feynman-gauge form $-g_{\mu\nu}/q^2$ in every practical calculation — the terms it omits could never have contributed to a physical process anyway. *Verified: $q_\mu j_e^\mu=\bar u(p_3)(\slashed p_1-\slashed p_3)u(p_1)=(m_e-m_e)\bar u u=0$ ; hence the $(1-\xi)q_\mu q_\nu/q^2$ term drops and $\mathcal{M}$ is $\xi$ -independent.*

Limiting cases

Case	Result	Meaning
Massless mediator	$1/(q^2-m_X^2)\to1/q^2$	infinite-range Coulomb force; the photon's defining feature
Massive mediator (W, Z)	$1/(q^2-M^2)\to-1/M^2$ at low $q^2$	short-range weak force; contact-like Fermi coupling $G_F\sim g^2/M_W^2$
Static limit, $q^0=0$	$-1/q^2\to+1/\|\mathbf q\|^2$ , Fourier $\to e^2/4\pi r$	the propagator is the Coulomb potential in disguise
Feynman gauge $\xi=1$	$-ig_{\mu\nu}/q^2$	simplest form; the one used in almost every calculation
Landau gauge $\xi=0$	$\frac{-i}{q^2}(g_{\mu\nu}-q_\mu q_\nu/q^2)$	manifestly transverse, $q^\mu D_{\mu\nu}=0$
On-shell, $q^2\to0$	propagator $\to\infty$	the real-photon pole; a virtual photon becoming a physical one
Deep spacelike, $\|q^2\|$ large	$1/q^2$ small, but $\alpha\to\alpha(q^2)$	bare amplitude falls as $1/q^2$ ; loop corrections make the coupling run

Common trapTreating

-g_{\mu\nu}/q^2

as the unique, physical photon propagator

It is neither unique nor, by itself, a physical object. The photon field has four components but only two propagating polarisations, and that redundancy — gauge invariance — makes the bare wave operator non-invertible, so a propagator exists only after a gauge is chosen. $-g_{\mu\nu}/q^2$ (up to the $-i$ ) is specifically the Feynman gauge $\xi=1$ ; Landau gauge, Yennie gauge, and every other $\xi$ give numerically different propagators, all equally correct. Two traps follow. First, the propagator is not gauge-invariant and is not directly measurable — only complete S-matrix elements are, and those are gauge-blind because $q_\mu j^\mu=0$ removes the gauge-dependent $q_\mu q_\nu$ terms. Second, do not picture the virtual photon as carrying only the two transverse polarisations of a real photon: the exchanged photon's polarisation sum runs over a complete set, and it is the full set — transverse plus the longitudinal and timelike modes that cancel against conserved currents — that produces $-g_{\mu\nu}$ . Reading $-g_{\mu\nu}/q^2$ as a gauge-invariant fact, or the virtual photon as a slightly-off-shell real photon, are the same mistake twice.

Worked examples

1 · From the propagator to the Coulomb potential (confidence builder)

Show that the static, massless limit of the exchange factor reproduces the Coulomb potential, and contrast it with the Yukawa range of a massive mediator.

Step 1. Take the generic exchange denominator $1/(q^2-m_X^2)$ in the static limit, where no energy is transferred, $q^0=0$ , so $q^2=(q^0)^2-|\mathbf q|^2=-|\mathbf q|^2$ . The factor becomes $-1/(|\mathbf q|^2+m_X^2)$ . The interaction potential is its Fourier transform back to position space.

Step 2. Use the standard three-dimensional transform

V(r)\propto\int\frac{\mathrm d^3q}{(2\pi)^3}\,\frac{e^{i\mathbf q\cdot\mathbf r}}{|\mathbf q|^2+m_X^2}=\frac{e^{-m_X r}}{4\pi r}.

For a massive mediator this is the Yukawa potential $V(r)\propto e^{-m_X r}/r$ , with range $1/m_X$ . For the $W$ boson, $m_W=80.4$ GeV gives a range $\hbar/m_Wc\approx2.5\times10^{-3}$ fm — the reason the weak force is feeble and short-ranged.

Step 3. Set $m_X\to0$ for the photon: $e^{-m_X r}\to1$ and the potential becomes $V(r)=e^2/4\pi r=\alpha/r$ , the Coulomb potential, with infinite range. The bare denominator $1/q^2$ of the photon propagator is nothing but the Coulomb $1/r$ in momentum space, and the whole difference between electromagnetism and the weak interaction is the single number $m_X$ in the same propagator.

2 · Momentum transfer and the Rutherford $1/\sin^4(\theta/2)$ law (fully worked)

An electron of energy $E=500$ MeV scatters elastically from a heavy target via one-photon exchange. Using $q^2=-Q^2=-4E^2\sin^2(\theta/2)$ in the ultra-relativistic limit, find the momentum transfer at $\theta=60^\circ$ and $\theta=120^\circ$ , and show the cross-section ratio follows from the propagator alone.

Step 1. The amplitude scales as one propagator, $\mathcal{M}\propto e^2/q^2$ , so the cross section scales as its square: $\mathrm d\sigma/\mathrm d\Omega\propto|\mathcal{M}|^2\propto1/(q^2)^2=1/Q^4$ . Since $Q^2=4E^2\sin^2(\theta/2)$ , this is $\mathrm d\sigma/\mathrm d\Omega\propto1/\sin^4(\theta/2)$ — the Rutherford angular law, here arising purely from the $1/q^2$ of the photon propagator.

Step 2. At $\theta=60^\circ$ : $\sin(\theta/2)=\sin30^\circ=0.5$ , so $Q^2=4(500)^2(0.5)^2=4\times250000\times0.25=\mathbf{2.5\times10^{5}\ MeV^2}=0.25$ GeV². The propagator carries $q^2=-0.25$ GeV².

Step 3. At $\theta=120^\circ$ : $\sin(\theta/2)=\sin60^\circ=0.866$ , so $Q^2=4(500)^2(0.866)^2=4\times250000\times0.75=\mathbf{7.5\times10^{5}\ MeV^2}=0.75$ GeV². The transfer is three times larger, $Q^2(120^\circ)/Q^2(60^\circ)=0.75/0.25=3$ .

Step 4. Form the cross-section ratio from the propagators alone:

\frac{\mathrm d\sigma(60^\circ)}{\mathrm d\sigma(120^\circ)}=\left(\frac{Q^2(120^\circ)}{Q^2(60^\circ)}\right)^2=3^2=\mathbf{9}.

Forward scattering at $60^\circ$ is nine times more likely than at $120^\circ$ , and that entire factor comes from the $1/q^2$ propagator squared — the small-angle divergence of Coulomb scattering is the pole of the photon propagator at $q^2\to0$ seen through the cross section.

3 · Checking gauge independence numerically

Convince yourself the $\xi$ -term is inert by contracting a concrete current with $q$ .

Step 1. Take the electron current $j_e^\mu=\bar u(p_3)\gamma^\mu u(p_1)$ with $q=p_1-p_3$ . The gauge-dependent part of the amplitude is proportional to $(q_\mu j_e^\mu)(q_\nu j_\mu^\nu)$ , so it suffices to show $q_\mu j_e^\mu=0$ .

Step 2. Contract: $q_\mu j_e^\mu=\bar u(p_3)\slashed q\,u(p_1)=\bar u(p_3)(\slashed p_1-\slashed p_3)u(p_1)$ . Apply the Dirac equation on each side, $\slashed p_1 u(p_1)=m_e u(p_1)$ and $\bar u(p_3)\slashed p_3=m_e\bar u(p_3)$ , giving $(m_e-m_e)\bar u(p_3)u(p_1)=\mathbf{0}$ .

Step 3. Therefore the propagator's $(1-\xi)q_\mu q_\nu/q^2$ piece contributes zero for every $\xi$ . Computing the amplitude in Feynman gauge ( $\xi=1$ , only $g_{\mu\nu}$ ), Landau gauge ( $\xi=0$ ), or any other value gives the identical number — a check every serious QED computation performs at least once, because a residual $\xi$ -dependence in a "physical" result is a sure sign of an algebra error.

Physics of the portrait

The photon propagator and the freedom to choose its gauge are the twin signatures of the man who turned quantum electrodynamics from an intractable formalism into a set of pictures you could calculate with.

Origins · 1 minRichard Feynman and how this was discovered

Richard Feynman

1918–1988 · quantum electrodynamics and the photon propagator

Richard Phillips Feynman, born in Queens, New York, and trained at MIT and Princeton, spent the war at Los Alamos before joining Cornell and then Caltech. Between 1947 and 1949 he rebuilt quantum electrodynamics around a single organising idea: that amplitudes are sums over space-time processes, each drawn as a diagram whose lines carry propagators and whose junctions carry vertices. The internal photon line in such a diagram contributes exactly the propagator this article is about, and the simplest form $-ig_{\mu\nu}/q^2$ — the one every student meets first — is written in what is now universally called Feynman gauge. His approach looked heuristic beside the operator formalism of Schwinger and the field theory of Tomonaga, and for a time it was distrusted; it was Freeman Dyson who proved in 1949 that the three formulations were equivalent, after which Feynman's diagrams swept the field for their sheer calculational power. He, Schwinger, and Tomonaga shared the 1965 Nobel Prize. Feynman's method made the photon propagator a line you draw rather than an integral you fear, and the gauge freedom in it a convenience you exploit rather than a paradox you dread. A celebrated teacher, safe-cracker, and bongo player, he later served on the commission that traced the 1986 Challenger disaster to a frozen O-ring, demonstrating the failure with a glass of ice water. He died in Los Angeles in 1988.

History

1929

— Heisenberg and Pauli set up the quantum field theory of electrodynamics and confront gauge fixing

1932

— Fermi casts electrodynamics in the Lorenz gauge with an explicit gauge-fixing term

1949

— Feynman introduces the space-time propagator −igμν/q² and the diagrammatic rules of QED

1950

— Gupta and Bleuler give a consistent covariant quantisation of the photon field

1971

— 't Hooft and Veltman exploit the R_ξ gauges to prove the gauge theories renormalisable

Going further

Texts. Thomson, Modern Particle Physics, §5.3 (Introduction to QED, Eqs. 5.10–5.18) and Appendix D.4.3 (the polarisation sum) · Halzen & Martin, Quarks and Leptons, §6.1–6.2 (the photon propagator and gauge choice) · Peskin & Schroeder, §9.4 (gauge fixing) and §4.8 (the propagator Feynman rule) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.5, §7.9 (QED Feynman rules) · Schwartz, QFT and the Standard Model, §8.4–8.5 (the photon propagator and $R_\xi$ gauges).

Provenance. The exchange factor $1/(q^2-m_X^2)$ and its massless limit are inherited by name from the perturbation-theory-by-exchange and Yukawa-potential articles; the vertex from minimal substitution and the currents from the Dirac-equation, adjoint-spinor and four-vector articles. Metric $g^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$ , Heaviside–Lorentz units with $\alpha=e^2/4\pi\approx1/137$ , natural units $c=\hbar=1$ . A verifier should confirm: (1) the massless limit $1/(q^2-m_X^2)\to1/q^2$ and its Fourier transform to $e^2/4\pi r$ ; (2) the $R_\xi$ propagator inverts the gauge-fixed operator, with the $q^\mu q_\rho$ coefficient $(1-\xi)+(1-1/\xi)-(1-1/\xi)(1-\xi)=0$ ; (3) $q_\mu\bar u(p_3)\gamma^\mu u(p_1)=(m_e-m_e)\bar u u=0$ from the Dirac equation; (4) $Q^2=4E^2\sin^2(\theta/2)$ giving $0.25$ and $0.75$ GeV² at $60^\circ,120^\circ$ ( $E=500$ MeV) and the cross-section ratio $3^2=9$ .

Prerequisites. Interaction by Particle Exchange · The Yukawa Potential · The Dirac Equation · Adjoint Spinors and Bilinear Covariants · Four-Vector Notation. Leads to. Feynman Rules for QED · e⁺e⁻ → μ⁺μ⁻ · Rutherford and Mott Scattering · Renormalisation and the Running of α.

Ask the tutor: Point me at any rung — taking the massless limit to Coulomb, inverting the gauge-fixed operator, running the polarisation sum, or proving your amplitude is gauge-independent — and I will walk it through line by line.

Why is the photon propagator

1/q^2

and not

1/(q^2-m^2)

like other exchanges?

Because the photon is massless. The generic single-particle exchange factor is $1/(q^2-m_X^2)$ , where $m_X$ is the mediator's mass; the photon simply has $m_X=0$ , so its denominator collapses to $q^2$ . This one difference has enormous physical reach: the $m_X$ term sets the range of the force through $1/m_X$ , so a massive mediator like the $W$ gives a force of range $\sim10^{-3}$ fm while the massless photon gives one of infinite range — Coulomb's law. In fact the Fourier transform of $1/(|\mathbf q|^2+m_X^2)$ is the Yukawa potential $e^{-m_X r}/4\pi r$ , which becomes the pure $1/4\pi r$ Coulomb potential precisely when $m_X\to0$ . The propagator denominator and the shape of the static potential are the same statement in two languages.

What actually is a "gauge choice", and why does the physics not depend on it?

The photon field $A_\mu$ carries redundant information: $A_\mu$ and $A_\mu+\partial_\mu\chi$ for any function $\chi$ describe exactly the same electric and magnetic fields. A gauge choice is a condition that removes this redundancy — Lorenz gauge $\partial_\mu A^\mu=0$ , Coulomb gauge $\nabla\cdot\mathbf A=0$ , and so on — implemented in the covariant formalism by a gauge-fixing term with parameter $\xi$ . Different choices give numerically different propagators, but every difference is a term proportional to $q_\mu$ or $q_\nu$ , and those always end up contracted with a conserved current, $q_\mu j^\mu=0$ . So the differences evaluate to zero in any physical amplitude. The gauge is a bookkeeping convenience, like a coordinate system; measurable quantities are coordinate-independent by construction.

Where does the minus sign in

-g_{\mu\nu}/q^2

come from?

From the polarisation sum. When you sum a virtual photon's polarisation vectors over a complete set, $\sum_\lambda\varepsilon^{(\lambda)}_\mu\varepsilon^{(\lambda)*}_\nu$ , the result is $-g_{\mu\nu}$ — the metric with an overall minus. The spatial (transverse) polarisations contribute the $-1$ 's along the diagonal that the metric's $(-1,-1,-1)$ spatial part records, and the timelike component supplies the $+1$ ; the net tensor is $-g_{\mu\nu}$ . That minus then rides through into the amplitude, which is why $\mathcal{M}=-Q_eQ_\mu e^2(j_e\cdot j_\mu)/q^2$ carries an explicit minus sign. It is not a sign of anything unphysical — it is the metric signature showing up in the messenger's spin bookkeeping.

Does the virtual photon really have four polarisations, unlike a real photon?

Effectively, yes, and this is a genuine feature rather than a bug. A real, on-shell photon has only two transverse polarisation states; a longitudinal or timelike photon cannot propagate freely. But a virtual photon is off shell ( $q^2\neq0$ ), and the propagator's completeness sum runs over all four polarisation directions of the four-vector $\varepsilon_\mu$ . The two extra, unphysical modes are exactly what the $q_\mu q_\nu$ terms in the propagator encode, and they never produce an observable effect because current conservation removes them. So the four-component sum is a legitimate intermediate object; only the external, measured photons are restricted to two transverse states. The Gupta–Bleuler formalism is the careful statement of how the unphysical modes cancel.

How can a propagator "not exist" — isn't it just the inverse of the wave operator?

It is, and that is precisely the problem. The free-photon wave operator in momentum space is $O^{\mu\nu}=-q^2g^{\mu\nu}+q^\mu q^\nu$ , and a propagator is its inverse. But $O^{\mu\nu}$ annihilates $q_\nu$ : $O^{\mu\nu}q_\nu=0$ . An operator with a nonzero vector in its kernel has a zero eigenvalue and therefore no inverse — like trying to divide by a singular matrix. The kernel direction $q_\nu$ is the momentum-space image of the gauge modes $A_\mu=\partial_\mu\chi$ , so gauge invariance is why the inverse fails to exist. Adding the gauge-fixing term $-\tfrac{1}{2\xi}(\partial\cdot A)^2$ lifts that zero eigenvalue and lets the inverse be taken, which is why gauge fixing is not optional in this formalism — without it, there is literally no photon propagator to write down.

Why does the same

1/q^2

give both Coulomb's law and the Rutherford cross section?

Because both are consequences of the propagator, seen in different limits. In the static limit the propagator's Fourier transform is the Coulomb potential $e^2/4\pi r$ — the potential between two charges is the exchange of many low-energy virtual photons. In a scattering experiment the amplitude carries one factor of $1/q^2$ , so the cross section carries $1/q^4$ , and with $q^2=-4E^2\sin^2(\theta/2)$ this is the Rutherford $1/\sin^4(\theta/2)$ angular law. The strong forward peaking of Rutherford scattering and the long range of the Coulomb force are the same fact — the photon propagator's pole at $q^2\to0$ — expressed once as a potential and once as a cross section.

The photon propagator & gauge choice

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

1 · From the propagator to the Coulomb potential (confidence builder)

2 · Momentum transfer and the Rutherford $1/\sin^4(\theta/2)$ law (fully worked)

3 · Checking gauge independence numerically

Physics of the portrait

Going further

Before this

Where this leads

The photon propagator & gauge choice

Where this lives today

The physical picture

The core equations

Limiting cases

Worked examples

1 · From the propagator to the Coulomb potential (confidence builder)

2 · Momentum transfer and the Rutherford 1/sin⁡4(θ/2)1/\sin^4(\theta/2)1/sin4(θ/2) law (fully worked)

3 · Checking gauge independence numerically

Physics of the portrait

Going further

Before this

Where this leads

2 · Momentum transfer and the Rutherford $1/\sin^4(\theta/2)$ law (fully worked)