From Relativistic QM to QED

An external Coulomb potential can be recovered as the leading static interaction produced by dynamical photon exchange. The calculation shows both the connection and the extra information in QED: the target is a quantum participant with a current, recoil, and its own external-state normalization. Here two distinguishable massive charged fermions scatter at tree level; their low-momentum amplitude matches the signed potential $V(r)=q_1q_2/(4\pi r)$ in rationalized natural units. This is a leading matching calculation, not a complete infrared treatment or a unique exact relativistic potential.

Required background. Fermion Fields fixes external spinors; Gauge Theory adds electromagnetic dynamics; From Scattering to LSZ fixes the amplitude workflow; First Born Approximation supplies the potential-to-amplitude map.

Helpful background. Invariant Phase Space sets the two-body flux convention; NRQED extends matching to a controlled low-energy operator expansion.

One-photon exchange between two distinct species

Use $\hbar=c=1$, metric $(+---)$, and two different Dirac species with masses $m_1,m_2>0$ and signed charges $q_1,q_2$. The interaction is

$$\mathcal L_{\rm int} =-\sum_{a=1}^2q_a\bar\psi_a\gamma^\mu\psi_a A_\mu.$$

Consider elastic particle scattering $1+2\to1+2$ with future-directed external momenta $p_a,p'_a$ and $p_a^2=(p'_a)^2=m_a^2$. Different species avoid identical-particle exchange and same-species particle–antiparticle annihilation diagrams. Electron–positron scattering, for example, is not a process with only the diagram retained here.

Use covariant states and spinors, $u_s^\dagger(p)u_r(p)=2E_p\delta_{sr}$. Suppress spin labels temporarily and define

$$\begin{aligned} j_a^\mu&=\bar u_a(p'_a)\gamma^\mu u_a(p_a),\\ k&=p'_1-p_1=p_2-p'_2. \end{aligned}$$

With $\langle f|S-I|i\rangle =i(2\pi)^4\delta^{(4)}(P_f-P_i)\mathcal M$, the vertex is $-iq_a\gamma^\mu$ and the Feynman-gauge photon kernel is $-i\eta_{\mu\nu}/(k^2+i0)$. Their product gives

$$\begin{aligned} i\mathcal M &=(-iq_1)(-iq_2) \frac{-i\,j_1^\mu\eta_{\mu\nu}j_2^\nu}{k^2+i0},\\ \mathcal M &=\frac{q_1q_2\,j_1\cdot j_2}{k^2+i0}. \end{aligned}$$

The photon line is an internal quantum-field contraction. It is not an additional observed real photon with an independently imposed on-shell momentum.

Current conservation checks the gauge dependence:

$$\begin{aligned} k_\mu j_1^\mu &=\bar u_1(p'_1) (p'_1\!\!\!/-p_1\!\!\!/)u_1(p_1)\\ &=(m_1-m_1)\bar u_1(p'_1)u_1(p_1)=0. \end{aligned}$$

The second line obeys the same identity with the opposite momentum-transfer orientation. Consequently the longitudinal $k_\mu k_\nu$ part of a covariant-gauge photon propagator does not change this on-shell tree amplitude. This check uses the external Dirac equations; it is not a claim that every off-shell Green function is gauge independent.

The static limit retains the charge sign

In the center-of-momentum frame, elastic scattering preserves each particle’s energy, so $k^0=0$ and $k^2=-|\mathbf K|^2$, where $\mathbf K=\mathbf p'_1-\mathbf p_1$. At low momentum, $|\mathbf p_a|/m_a\ll1$, the normalized spinors give

$$j_a^0=2m_a\,\chi_a'{}^\dagger\chi_a +O(|\mathbf p_a|^2/m_a),$$

while the spatial current is of order $|\mathbf p_a|$. The leading interaction is therefore diagonal in the chosen Pauli-spin basis. For spin-preserving matrix elements,

$$\mathcal M_{\rm static} =-\frac{4m_1m_2q_1q_2}{|\mathbf K|^2}.$$

The minus sign comes from the spacelike photon denominator. It has not changed the sign of either physical charge. Spatial-current, recoil, and spin-dependent corrections enter beyond this leading static reduction.

Match the amplitude to a potential

Define the potential transform by

$$\widetilde V(\mathbf K) =\int d^3r\,e^{-i\mathbf K\cdot\mathbf r}V(\mathbf r).$$

For reduced mass $\mu=m_1m_2/(m_1+m_2)$, the accepted Born convention is

$$f_{\rm Born}(\mathbf K) =-\frac{\mu}{2\pi}\widetilde V(\mathbf K).$$

In the same scattering-state phase convention, the elastic center-of-momentum amplitude is $f_{\rm CM}=\mathcal M/(8\pi\sqrt s)$. Its modulus reproduces $d\sigma/d\Omega=|\mathcal M|^2/(64\pi^2s)$ for distinguishable elastic final particles with fixed spin states. The phase convention matters: a cross section alone would not determine the sign of an interaction.

At leading nonrelativistic order $\sqrt s=m_1+m_2+O(\mathbf p^2/m_a)$. Equating the two amplitudes gives

$$\widetilde V(\mathbf K) =-\frac{\mathcal M_{\rm static}}{4m_1m_2} =\frac{q_1q_2}{|\mathbf K|^2}.$$

The factors $2m_a$ in the external currents have been removed in passing to the nonrelativistic normalization. With the inverse Fourier convention $V(\mathbf r)=\int d^3K\,(2\pi)^{-3} e^{i\mathbf K\cdot\mathbf r}\widetilde V(\mathbf K)$, this is

$$V(r)=\frac{q_1q_2}{4\pi r}.$$

Like charges repel; opposite charges attract. Restoring SI gives $q_1q_2/(4\pi\epsilon_0r)$ when the $q_a$ are expressed in coulombs. Tong (2006, section 6.6.1) gives the photon-exchange origin of this static interaction.

The Coulomb transform and forward singularity need a distributional or screened limiting prescription. For example, $|\mathbf K|^{-2}$ can be obtained as the limit of $(|\mathbf K|^2+\lambda^2)^{-1}$ with $\lambda>0$ and then $\lambda\to0^+$. This device checks the static Fourier matching; it is not a gauge-invariant photon-mass completion of QED. The unscreened Coulomb problem has long-range asymptotic phases, so the displayed Born matching should not be mistaken for a proof of ordinary short-range scattering limits.

Recovering a prescribed source is another limit

The starting amplitude contains both particles’ external currents and exact four-momentum conservation. If particle 2 is much heavier and transfers small energy, its recoil can be neglected to a stated accuracy. Its current and normalization must also be converted into the selected source density or external potential.

The unreduced invariant $\mathcal M$ grows with the heavy mass through the target spinor normalization. That factor is not an infinite physical potential; it cancels in the normalization conversion above. Simply deleting the target leg while keeping the same numerical amplitude would miss this cancellation.

The Mott page starts directly from a prescribed static potential and keeps the projectile relativistic. It therefore answers a different intermediate problem from the low-momentum two-body matching here. Comparing their predictions requires the same source profile, charge, recoil approximation, and flux convention.

Beyond the leading potential, field theory organizes additional operators and channels. The NRQED matching page shows how on-shell form factors determine selected low-energy coefficients. The Dirac-to-field boundary explains which photon and number sectors are absent from an external-field wave equation. Neither a very accurate Dirac eigenvalue nor this tree graph includes radiative loops or a complete bound-state QED calculation.

At higher orders, charged scattering also requires an infrared prescription for unresolved radiation or dressed states. The LSZ qualification remains in force. Recovering the leading Coulomb interaction does not establish an exact finite-photon exclusive QED $S$ matrix.

Exercises

A repulsive amplitude with a minus sign. Take $q_1=q_2=e>0$. Explain why a negative $\mathcal M_{\rm static}$ gives a repulsive, positive potential.

Solution

The conversion is $\widetilde V=-\mathcal M_{\rm static}/(4m_1m_2)$, so $\widetilde V=e^2/|\mathbf K|^2$ and $V=e^2/(4\pi r)>0$. The negative invariant amplitude reflects the propagator and scattering-phase conventions, not an attractive force.

Screen the Fourier check. Replace $|\mathbf K|^{-2}$ by $(|\mathbf K|^2+\lambda^2)^{-1}$. What static potential results, and which limit recovers Coulomb?

Solution

The inverse is $V_\lambda(r)=q_1q_2e^{-\lambda r}/(4\pi r)$. It follows equivalently from the decaying Green function of $-\nabla^2+\lambda^2$. At fixed $r>0$, taking $\lambda\to0^+$ recovers Coulomb. Forward scattering and infinite-distance limits require separate care. The relative error $1-e^{-\lambda r}$ is not uniformly small over unbounded distances, although the absolute potential error is bounded by $|q_1q_2|\lambda/(4\pi)$ uniformly for $r>0$.

Remove the heavy normalization. At fixed small transfer and fixed projectile mass, let $m_2\to\infty$. Track $\mathcal M_{\rm static}$, $\mu$, and $\widetilde V$.

Solution

$\mathcal M_{\rm static}$ is proportional to $m_2$ because the target covariant current is approximately $2m_2$. Meanwhile $\mu\to m_1$, and division by $4m_1m_2$ leaves $\widetilde V=q_1q_2/|\mathbf K|^2$ finite. The corresponding Born amplitude tends to $-m_1q_1q_2/(2\pi|\mathbf K|^2)$.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press (2014). doi:10.1017/9781139540940. QED amplitudes, nonrelativistic matching, and infrared qualifications.
• Tong, David. Quantum Field Theory. University of Cambridge lecture notes (2006), sections 6.3–6.6. Quantum electrodynamics. Charged currents, photon exchange, and the Coulomb potential.