Since Newton's law of gravitation can be gotten out of Einstein's field equatons as an approximation, I was wondering whether the same applies for the electromagnetic force being the exchange of photons. Is there an equation governing the force from the exchange of photons? Are there any links which would show how the Coulomb force comes out of the equations for photon exchange? I know that my question is somewhat similar to the one posted here The exchange of photons gives rise to the electromagnetic force, but it doesn't really have an answer to my question specifically.

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    $\begingroup$ For Rutherford scattering in QM & QED, see e.g. M.D. Schwartz, QFT & the standard model, 2014; section 13.4. $\endgroup$
    – Qmechanic
    Sep 14, 2021 at 7:37

2 Answers 2


The classical Coulomb potential can be recovered in the non-relativistic limit of the tree-level Feynman diagram between two charged particles.

Applying the Born approximation to QM scattering, we find that the scattering amplitude for a process with interaction potential $V(x)$ is

$$\mathcal{A}(\lvert p \rangle \to \lvert p'\rangle) - 1 = 2\pi \delta(E_p - E_{p'})(-\mathrm{i})\int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r$$

This is to be compared to the amplitude obtained from the Feynman diagram:

$$ \int \mathrm{e}^{\mathrm{i}k r_0}\langle p',k \rvert S \lvert p,k \rangle \frac{\mathrm{d}^3k}{(2\pi)^3}$$

where we look at the (connected) S-matrix entry for two electrons scattering off each other, treating one with "fixed" momentum as the source of the potential, and the other scattering off that potential. Using the Feynman rules to compute the S-matrix element, we obtain in the non-relativistic limit with $m_0 \gg \lvert \vec p \rvert$

$$ \langle p',k \rvert S \lvert p,k \rangle \rvert_{conn} = -\mathrm{i}\frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}(2m)^2\delta(E_{p,k} - E_{p',k})(2\pi)^4\delta(\vec p - \vec p')$$

Comparing with the QM scattering, we have to discard the $(2m)^2$ as they arise due to differing normalizations of momentum eigenstate in QFT compared to QM and obtain:

$$ \int V(\vec r)\mathrm{e}^{-\mathrm{i}(\vec p - \vec p')\vec r}\mathrm{d}^3r = \frac{e^2}{\lvert \vec p -\vec p'\rvert^2 - \mathrm{i}\epsilon}$$

where Fourier transforming both sides, solving the integral and taking $\epsilon \to 0$ at the end will yield

$$ V(r) = \frac{e^2}{4\pi r}$$

as the Coulomb potential.

  • $\begingroup$ Could you explain whether or not this method yields the overall sign, so that we can see that opposite charges attract whereas like charges repel? $\endgroup$ Aug 11, 2022 at 21:58
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    $\begingroup$ @AndrewSteane It does, but in order to demonstrate how I'd have to explain in much more detail how the S-matrix element is computed from the Feynman diagram. $\endgroup$
    – ACuriousMind
    Aug 12, 2022 at 11:35

There is the (non-genetal) relation between the free energy of interacting of two currents $J^{a}, J^{b}$ and the propagator: $$ U = -\frac{1}{2} \int d^{4}xd^{4}y J^{a}(x) D_{ab}(x - y)J^{b}(y). $$ It's not general, but it realizes the simple example which can help you to understand how to get the expression for force.

The structure of field which causes structure of propagator helps us to get the expression for the force. For example, for intercation via scalar field (by setting $D_{ab}(x - y) = \frac{1}{p^{2} - m^{2}}$) after simple transformations for "point-like" currents $J(x) = \delta (\mathbf x - \mathbf x_{0})$ we can get $$ U = -\frac{1}{4 \pi |\mathbf r|}e^{-mr}. $$ For case $m = 0$ we get the Coulomb-like law of interaction.

Absolutely the same thing you may do with the case of EM field (in Feynman gauge $D_{\mu \nu} = -\frac{g_{\mu \nu}}{p^{2}}$).

If you need some explicit derivation I'll give it later.

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    $\begingroup$ Hi, I have a question – do you know how the first formula in this answer generalizes to non-abelian gauge theories? I presume the integral becomes a Wilson loop somehow, because as it stands it is not gauge-invariant. Is that true? $\endgroup$ Jul 4, 2018 at 17:03
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    $\begingroup$ @SolenodonParadoxus that's pretty much right. In QED calculating the effective action for a static source is essentially the expectation value of a Wilson loop. If you run the same derivation through for a nonabelian gauge theory, you discover the equivalent of Coulombs law, which is basically to the old formula only with a factor of the quadratic casimir for the rep of the gauge group $\endgroup$
    – user213887
    Nov 30, 2018 at 17:28

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