Radiative corrections to Coulomb’s law and Euler-Heisenberg theory

Question

Maxwell's electrodynamics is the classical limit of QED (quantum electrodynamics). Using Maxwell's equations, the electrostatic (Coulomb) potential of a point charge is obtained as $\Phi \propto \frac{1}{r}$. But, in QED, the Coulomb potential is modified due to the radiative corrections. The QED result for the electrostatic potential of a point charge-like electron (i.e., radiative corrections to Coulomb's law) is found to be as $^1$

$$\Phi (r) = \frac{e}{r}\left( {1 + \frac{\alpha }{{4\sqrt \pi }}\frac{{{e^{ - 2m{ _e}r}}}}{{{{(m{_e}r)}^{3/2}}}}} \right),\qquad({m_e}r \gg 1) \tag 1$$

and

$$\Phi (r) = \frac{e}{r}\left[ {1 + \frac{{2\alpha }}{{3\pi }}\left( {\ln \left( {\frac{1}{{m{ _e}r}}} \right) - \gamma - \frac{5}{6}} \right)} \right],\qquad({m_e}r \ll 1) \tag 2$$

where $e$ and $m_e$ are the electric charge and the mass of the electron respectively, $\alpha$ is the fine-structure constant and $\gamma$ is the Euler–Mascheroni constant. The second term is responsible for the electric polarization of the vacuum with the order-$\alpha$ vacuum polarization diagram as

On the other hand, Euler and Heisenberg constructed an effective Lagrangian in their famous paper$^2$ for taking into account vacuum polarization to one loop. In the weak field limit, the Euler-Heisenberg (EH) Lagrangian reduces to

$$\mathcal{L}_{\mathrm{eff}} = -\frac14 F_{\mu\nu} F^{\mu\nu}+\frac{\alpha^2}{90 m_e^4}\left[\left(F_{\mu\nu} F^{\mu\nu}\right)^2 +\frac{7}{4}\left(F_{\mu\nu} \tilde{F}^{\mu\nu}\right)^2\right] +\mathcal{O}(m_e^{-6}), \tag 3$$ or equivalently $$\mathcal{L}_{\mathrm{eff}} = \frac12 \left(\mathbf{E}^2 - \mathbf{B}^2\right)+\frac{2\alpha^2}{45 m_e^4}\left[\left(\mathbf{E}^2 - \mathbf{B}^2\right)^2 +7\left(\mathbf{E} \cdot \mathbf{B}\right)^2\right]+\mathcal{O}(m_e^{-6}). \tag 4$$

In this effective field theory, the electrostatic potential is found to be as $$\Phi (r) = \frac{e}{{{r}}} - \frac{{8{\alpha ^2}}}{{225m_e^4}}\frac{{{e^3}}}{{{r^5}}} + O({\alpha ^3}). \tag 5$$

However, I have difficulty interpreting these results.

1. Correction in the electric potential formula (5) in EH theory is higher order in the fine structure constant ($\alpha$) than the correction in the electric potential formulas (1) and (2) in QED. What does this mean? Do they not describe the same physics? (I know that Maxwell equations in EH theory are no longer linear.)

2. As far as I know, EH theory as nonlinear electrodynamics gives us the QED $β$-function, the prediction of light-by-light scattering and its cross-section, Schwinger pair creation, etc. Can we claim that the EH theory could describe some results of QED (e.g., quantum mechanical nonlinearity of the fields) without incorporating any quantum mechanical (QED) treatment? To what extent is this claim true?

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_{$^1$ Berestetskii, V. B.; Lifshitz, E. M.; Pitaevskii, L., Quantum Electrodynamics, Section 114 (1980)}

_{$^2$ Heisenberg, W., and H. Euler. "Consequences of Dirac theory of the positron." arXiv preprint physics/0605038 (2006)}

Answer 1

EH Lagrangian is derived in the approximation of constant (or slowly-varying) electromagnetic fields. It is an effective Lagrangian only containing E and B fields, but not their derivatives.

Even if you consider external point charges (and not the electrons themselves, which you have integrated out,) it would be wrong to derive the corrected Coulomb potential from this Lagrangian.

One would have to include all terms in the effective action containing derivatives of E and B, to obtain the fully corrected Coulomb potential. This is unrealistically hard, but of course to the lowest order, only a few additional higher-derivative terms would be enough.

What you observed is that the EH Lagrangian by itself is not enough. In other words, the answer (1), (2) is more correct. It is derived using different techniques, for example, by computing the loop corrections to the photon two-point function, as you noted. Such a computation usually bypasses explicitly writing the effective action. However, if you were to do that, you would discover that the leading $O(\alpha)$ effect is not captured by the EH Lagrangian, which as you noticed starts contributing only at the sub-leading order.