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Maxwell's equations are famously linear and are the classical limit of QED. The thing is QED even without charged particles is pretty non-linear with photon-photon interaction terms. Can these photon-photon interaction terms have a "classical" limit as well? Can we just perturbatively expand the QED Lagrangian and find the classical limit of certain non-linear terms?

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    $\begingroup$ arxiv.org/abs/1009.2313 (Usually people say that there is an extra factor of hbar for each loop and so these go away in the classical limit... but per the linked paper, there can be complications...) $\endgroup$
    – hft
    Commented Jun 18 at 3:10
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    $\begingroup$ The other common (but not super rigorous) argument is that the path integral has a factor of $e^{iS/\hbar}$ and so only the classical paths contribute in the $\hbar\to 0$ limit... $\endgroup$
    – hft
    Commented Jun 18 at 3:12
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    $\begingroup$ Photon-photon interaction involves virtual charged particles . $\endgroup$
    – my2cts
    Commented Jun 18 at 3:31
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    $\begingroup$ Integrate out the electron, and you get terms like $F^4$ in the action, which gives an effective photon-photon interaction at energy scales below the electron mass. $\endgroup$
    – Prahar
    Commented Jun 18 at 7:37
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    $\begingroup$ This monograph covers the subject quite thoroughly. archive.org/details/springer_10.1007-3-540-45585-X/mode/2up $\endgroup$
    – Buzz
    Commented Jun 19 at 23:18

3 Answers 3

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In Quantum Electrodynamics by A. I. Akhiezer; V. B. Berestetskii, photon-photon scattering is considered in paragraph 54, and the effective classical Lagrangian for nonlinear vacuum electrodynamics is derived in 54.3 (see Equation 54.14 and further). The nonlinear correction is written as $$ L_1 = a\left(E^2 - H^2\right)^2 + b(\vec{E}\vec{H})^2. $$

These terms arise from 4-point electron loop diagrams. To my knowledge, they can be found in two steps:

  1. The perturbative expansion of the effective action for EM field interacting with charged fermions (or whatever) in one-loop order,
  2. Gradient expansion in order to make the action local.
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    $\begingroup$ Akhiezer, not for the faint of heart ... $\endgroup$
    – my2cts
    Commented Jun 18 at 8:29
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    $\begingroup$ On some level the corrections have to be of this form, since they're the only possible terms that are frame-independent and parity-even. The real question is what the values of $a$ and $b$ are. $\endgroup$ Commented Jun 18 at 12:41
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In addition to what the other answer already mentions, the non-linear extension required to include QED effects classically can also be found in [Jackson].

Eq. 1.5

But the claim the question makes, that "even without charged particles" the theory is non-linear is debatable. The nonlinear terms always come from interaction with the electron, as in Jackson's [Fig. 1.3].

The question also asks specifically for a second order nonlinear addition, but the nonlinearities we find are in fact third order. They can be expressed (as Jackson does) as 2nd order additions to the permittivity and permeability of vacuum, but in the Maxwell equations those quantities appear multiplied with the fields so they become third order.

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    $\begingroup$ I'd say it is debatable, because OP's formulation can be understood as "in solutions without charged particles" even though the theory is of a kind that can have charged particles. (OK, a word game I guess...) $\endgroup$ Commented Jun 18 at 7:45
  • $\begingroup$ Why is that crazy book taught to first year graduate students, I don't understand. I've never ever concieved of there being a nonlinear QED term to Maxwell's equations, and yet of course it is in Jackson, the standard textbook for "classical electrodynamics". $\endgroup$ Commented Jul 9 at 13:53
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Good answers. I just want to add that

  1. as I commented electromagnetism is strictly linear without coupling to charged

  2. if there ís coupling to charges or spins then upon integrating out the virtual charge or spin degrees of freedom you should quite generally end up with effectively nonlinear electromagnetism expressible in the invariants $F_{\mu\nu} F^{\mu\nu}$ and $\epsilon^{\mu\nu\rho\sigma} F_{\mu\nu} F_{\rho\sigma}$.

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    $\begingroup$ Thanks! I slightly updated my answer to emphasize that EM field should be coupled to some charged field. $\endgroup$
    – E. Anikin
    Commented Jun 18 at 10:08

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