One could derive the same result from QED Feynman diagrams such as in Fig. 1, see also the Supplemental Material. Since three vertices (the two XFEL photons and the external magnetic field) correspond to low energies and momenta (i.e., well below the electron mass $m$), energy-momentum conservation implies that the fourth vertex (representing the Coulomb field) does also involve low energies and momenta. In fact, for the case of forward scattering considered here, these energies and momenta are extremely low. Thus, we may use the low-energy limit of the QED Feynman diagrams, which are equivalent to the Euler-Heisenberg Lagrangian (2). Note that the situation is different for Delbrück scattering where only two vertices (the two XFEL photons) correspond to low energies and momenta such that the other two (representing the Coulomb field) can involve high momenta.

The case of smaller frequencies is discussed in the Supplemental Material [80].

Because the magnetic field ${\mathbf{B}}_{\mathrm{ext}}$ is fixed externally (and approximately constant), the $\delta \mathrm{\Psi }$ contributions from the nuclei would just add up (in the same way as their Coulomb fields ${\mathbf{E}}_{\mathrm{ext}}$ add up) and thus the total amplitude ${\mathfrak{A}}_{\delta \mathrm{\Psi }}$ is simply the sum of the amplitudes from all the nuclei separately. Note that the recoil of the heavy nuclei is negligible due to the small momentum transfer (in the eV regime) and the tight spatial confinement of the nuclear wave functions (far below 100 nm), leading to a momentum uncertainty much larger than the recoil.