It is known that the Coulomb potential can be obtained by Fourier transform of the propagator from E&M. Is this because one of Maxwell's equations have the form $\nabla \cdot \mathbf{E}=\rho$?


To obtain the classical (in the sense of non quantum mechanical) potential due to a pair of charged particles, says, two electrons, you may calculate the scattering amplitude and compare it to

$$\langle p'|iT|p \rangle=-i2\pi\tilde{V}( \textbf{q} )\delta^{3}(E_{p'}-E_{p}), \hspace{2cm} \bf{q}=\bf{p'}-\bf{p'} $$

this is equation $(4.123)$ in Peskin's & Schroeder - An Introduction to Quantum Field Theory, and you in the first chapter of this lecture notes (link) that expression is justified. In the book this is first done to the Yukawa potential (scalar-fermion theory) and then generalised to Quantum Electrodynamics (vector-fermion theory).

The form of the Hamiltonian interaction for QED, which is the relevant piece of information to do the scattering amplitude computation (in particular, the lowest order contribution) has nothing to do with the classical form of Maxwell's equations but with the imposition of the $U(1)$ symmetry of the theory. Pretty much that underlying symmetry dictates the form of the quantum amplitudes and classical field theory equations.

In particular, Gauss law is pretty much the same thing (contains the same information) as the Coulomb equation, the latter can be derived from the former and vice versa.

You may find the following question interesting Alternative methods to derive the static potential in the NR limit of QED

  • $\begingroup$ Thank you for your answer and the lecture note. The scattering amplitude in the equation has information of interaction. But the derivation from the Fourier transform needs only propagator, which does not have interaction information. This is the point which is not so clear to me. $\endgroup$ – Traveler Jul 3 '13 at 5:56
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    $\begingroup$ Hi Traveler. Where did you see that the Fourier transform of the photon propagator gives the Coulomb interaction? The Fourier transform of the propagator in the position space gives you the propagator in the momentum space. The Coulomb potential is the classical potential for point-like charged particles, and a photon has no charge at all! In order to derive the Coulomb potential you need somewhere charged particles, which is where the charged fermions in the scattering amplitude enter. Hope that helps. $\endgroup$ – Jorge Lavín Jul 3 '13 at 9:00
  • $\begingroup$ You can see en.wikipedia.org/wiki/Yukawa_potential $\endgroup$ – Traveler Jul 3 '13 at 14:41
  • $\begingroup$ I see, as a useful reference en.wikipedia.org/wiki/…. I can't answer your question then, my apologies. $\endgroup$ – Jorge Lavín Jul 3 '13 at 18:52
  • $\begingroup$ Thank you for helping me understand the problem, Nivalth. I really appreciate it. If I have a question next time, let us discuss. $\endgroup$ – Traveler Jul 4 '13 at 6:26

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