expression $$\frac{\partial{\cal L}}{\partial F^{\mu\nu}}
 =\frac{e^2}{8\pi^2}\frac{1}{45}\frac{e^2}{m^4}\bigl[4{\cal F}F_{\mu\nu}
 +7{\cal G}{}^*F_{\mu\nu} \bigr] +{\cal O}(\varepsilon^5)\,, \label{eq:delLdelF}$$ where we counted $F^{\mu\nu}$ and ${}^*F^{\mu\nu}$ as ${\cal O}(\varepsilon)$. In our explicit calculations to be performed subsequently for an all-optical laser experiment, we will always limit ourselves to the leading order terms given explicitly in [Eq. ]{}. As the field strengths attainable in present and near future high-intensity laser facilities are small in comparison to the [*critical field strength*]{} $E_{\rm cr}\equiv\frac{m^2}{e}$ [@Heisenberg:1935qt], i.e., $\{\frac{eE}{m^2},\frac{eB}{m^2}\}\ll1$, this approximation is well justified.

The amplitude for emission of a single photon with momentum $\vec{k}$ from the vacuum subjected to the background electromagnetic field $F^{\mu\nu}(x)$ is given by $${\cal S}_{(p)}(\vec{k})\equiv\langle\gamma_{p}(\vec{k})|\int{\rm d}^4x\, f^{\mu\nu}(x)\frac{\partial{\cal L}}{\partial F^{\mu\nu}}(x)|0\rangle\,, \label{eq:S}$$ with the single photon state denoted by $|\gamma_{p}(\vec{k})\rangle\equiv a^\dag_{\vec{k},p}|0\rangle$ (cf. Fig. \[fig:Feyndiag\]). Here $p$ denotes the polarization of the emitted photons.

![Leading order contribution to the stimulated photon emission process in the weak field limit. The process is cubic in the background field represented by the wiggly lines ending at crosses; cf. Eqs.  and . It results in outgoing photons of wave vector $\vec{k}$ and polarization $p$.[]{data-label="fig:Feyndiag"}](Fig0){width="30.00000%"}

Representing the photon field in Lorentz gauge as $$a^{\mu}(x)=\sum_{p}\int\frac{{\rm d}^3{k}}{(2\pi)^3}\frac{1}{\sqrt{2k^0}}\,\Bigl(\epsilon^{\mu}_{(p)}(k)\,{\rm e}^{-ikx}\,a_{\vec{k},p}+\epsilon^{*\mu}_{(p)}(k)\,{\rm e}^{ikx}\,a^\dag_{\vec{k},p}\Bigl) ,$$ where $k^0\equiv|\vec{k}|$, and the sum is over the 