introduced in [Eq. ]{} above. The resulting electric and magnetic fields amount to standing waves and read $$\begin{gathered}
 \vec{E}(x)=\vec{E}_+(x)+\vec{E}_-(x)=2{\cal E}\,\vec{e}_{\rm x}\,{\rm e}^{-\frac{4({\rm x}^2+{\rm y}^2)}{w^2}}{\rm e}^{-\frac{{\rm z}^2}{\zeta^2}}\sin(\Omega t)\sin(\Omega {\rm z})\,, \nonumber\\
 \vec{B}(x)=\vec{B}_+(x)+\vec{B}_-(x)=2{\cal E}\,\vec{e}_{\rm y}\,{\rm e}^{-\frac{4({\rm x}^2+{\rm y}^2)}{w^2}}{\rm e}^{-\frac{{\rm z}^2}{\zeta^2}}\cos(\Omega t)\cos(\Omega {\rm z})\,. \label{eq:resfields}\end{gathered}$$ Figure \[fig:cartoon\] is a cartoon of the experimental situation we have in mind.

![Two counter-propagating Gaussian laser beams with wave vectors $\vec{k}_{\pm}=\pm\vec{e}_{\rm z}$ (electric and magnetic field vectors $\vec{E}_\pm$ and $\vec{B}_\pm$) are superimposed to form a standing electromagnetic wave about the beam focus with nonvanishing field invariant $\cal F$. The idea is to look for induced photons emitted from this strong-field region, to be detected in the field free region. For later reference, we also depict the beam divergence $\theta$.[]{data-label="fig:cartoon"}](Fig1){width="20.00000%"}

For the particular electromagnetic field configuration , the invariants  are $${\cal F}(x)={\cal E}^2\,{\rm e}^{-\frac{8({\rm x}^2+{\rm y}^2)}{w^2}}{\rm e}^{-\frac{2{\rm z}^2}{\zeta^2}}\bigl[\cos(2\Omega t)+\cos(2\Omega z)\bigr]\,,\quad {\cal G}=0\,,$$ and all components of the field strength tensor $F_{\mu\nu}$ apart from $$F_{10}(x)=-F_{01}(x)=\vec{e}_{\rm x}\cdot\vec{E}(x)\,,\quad F_{31}(x)=-F_{13}(x)=\vec{e}_{\rm y}\cdot\vec{B}(x)\,,$$ vanish. Thus, the emission amplitude  can be expressed concisely as $${\cal S}_{(p)}(\vec{k})
=\frac{2i}{\sqrt{2k^0}}\biggl(\hat f^{10}_{(p)}(k)\,\frac{\partial{\cal L}}{\partial F^{10}}(-k)
 +\hat f^{31}_{(p)}(k)\,\frac{\partial{\cal L}}{\partial F^{31}}(-k)\biggr)\,, \label{eq:Sp1031}$$ with $$\frac{\partial{\cal L}}{\partial F^{\mu\nu}}(-k)
 =\frac{e^2}{\pi^2}\frac{1}{90}\frac{e^2}{m^4}\int{\rm d}^4x\,{\rm e}^{ikx} {\cal F}(x)F_{\mu\nu}(x) + {\cal O}(\varepsilon^5) \,. \label{eq:delLdelFexpl}$$ The Fourier integrals in [Eq. ]{} 