Coulomb field from QED

Question

It is well-known how to obtain Coulomb's law in perturbative QED (e.g. the answers to this question on this site). I am trying to understand if there is any reasonable way to give a meaning, within QED, to the concept of a static electric field. In the case of radiation fields, we know that they have a meaning, within QED, as coherent superpositions of photons. But what about static fields? Is there any QED observable that under proper conditions becomes the classical Coulomb field? I am interested in a formal derivation, not in hand-waving arguments.

Answer 1

When you consider QED, you need to consider the electron field as well. You cannot recover the Coulomb field exactly in QED, because the electron field will shield the EM field. However, it appears in many different ways in a quantized, pure Maxwell theory.

One way to see this is to look at the generating functional: $$ Z(J)=\int DA\exp\left(iS\right)\\ S = \int \mathcal L d^4 x\\ \mathcal L =-\frac{1}{4}F^2+JA $$ Specifying the classical source 4-current to be electrostatic: $$ J=(\rho(\vec x),0) $$ Now, the natural step is to integrate out the magnetic part. For this, you need to fix a gauge. A simple choice would be to use the analogue of the $R_\xi$ gauge for the Coulomb gauge. Essentially, this amounts to adding one final term to the Lagrangian: $$ \mathcal L = \frac{1}{2}(\partial_t A+\nabla V)^2-\frac{1}{2}(\nabla\times A)^2+\rho V+\frac{1}{2\xi}(\nabla \cdot\vec A)^2 $$ By integrating out the magnetic part, you obtain in the Fourier basis: $$ \mathcal L = \frac{\vec k{}^2}{\xi\omega^2 +\vec k{}^2}\frac{\vec k{}^2}{2}V^2 $$ and choosing the quantum analogue of the Coulomb gauge with $\xi = 0$, the propagator is given by the Coulomb potential. In fact you can check the gauge invariance since $\rho\propto \delta(\omega)$ in Fourier space so $\omega$ is effectively set to $0$ when integrating $V$ for the generating function. The result is independent of $\xi$: $$ Z = \exp\left(\frac{i}{2}\int d^3\vec xd^3\vec y \frac{\rho(\vec x)\rho(\vec y)}{4\pi|\vec x-\vec y|}\right) $$

You can therefore obtain a static field in a quantised EM theory using classical sources. To specifically get the Coulomb potential, you just need to take a single charge and the expected value of $V$ gives the Coulomb field in the Coulomb gauge. Mathematically, you set $\rho=\delta$ and take the functional, logarithmic derivative of $Z$ with respect to $\rho$ at this specific value. This is what you get from classical Maxwell theory, but now using the quantised version, you can calculate the quantum fluctuations about this mean value like higher moments (which is tractable since the theory is gaussian).

More generally, static charges give fields whose expected value are given by electrostatics in the Coulomb gauge. The true novelty is are the quantum fluctuations which are still present even without the classical charges. In particular, the Coulomb potential can be alternatively interpreted as the correlator of $V$. The link between the two interpretations is the fluctuation dissipation theorem.

In a similar spirit, Jeanbaptiste Roux outlined another standard method to get the Coulomb potential from a pure, quantised Maxwell theory using Wilson loops. Once again, this is not QED since there are no electrons. You can find the approach in Peskin and Schroeder's Introduction to Quantum Field Theory, exercise 15.3 "Colomb potential." Instead of calculating the generating functional, you calculate the Wilson loop: $$ \begin{align} W &= \left\langle \exp\left(-i\oint A dx\right)\right\rangle \\ &= \exp\left(-i\oint dx^\mu \oint dy_\mu \frac{1}{8\pi^2(x-y)^2}\right) \end{align} $$ Using the appropriate loop, you recover the Coulomb potential in the argument. Actually, you can get this from the more general approach of the generating function and choosing appropriate source localised on the chosen path.

Hope this helps.