e^{i(qx)}\hat{e}^*_l(q)\frac{1}{\hat{\Pi}(\phi)+\hat{k}-m+i0}\gamma_{\lambda}U_s(p,x)\\
&=-e^3\int d^4x\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{k^2-\kappa^2+i0}\\
&\quad\times\bar{U}_{s'}(p',x)\gamma^{\lambda}[\hat{\Pi}(\phi)+\hat{k}+m]\frac{1}{[\hat{\Pi}(\phi)+\hat{k}]^2-m^2+i0} e^{i(qx)}\hat{e}^*_l(q)\\
&\qquad\times\frac{1}{[\hat{\Pi}(\phi)+\hat{k}]^2-m^2+i0}[\hat{\Pi}(\phi)+\hat{k}+m]\gamma_{\lambda}U_s(p,x),
\end{split}$$ where $\Pi^{\mu}(\phi)=i\partial^{\mu}-eA^{\mu}(\phi)$. By using the fact that $[\hat{\Pi}(\phi)-m]U_s(p,x)=[\hat{\Pi}(\phi)-m]U_{s'}(p',x)=0$, we obtain $$\begin{split}
-ie\Gamma_{s,s',l}(p,p',q)&=-e^3\int d^4x\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{k^2-\kappa^2+i0}\\
&\quad\times\bar{U}_{s'}(p',x)[2\Pi^{\lambda}(\phi)+\gamma^{\lambda}\hat{k}]\frac{1}{[\hat{\Pi}(\phi)+\hat{k}]^2-m^2+i0} e^{i(qx)}\hat{e}^*_l(q)\\
&\qquad\times\frac{1}{[\hat{\Pi}(\phi)+\hat{k}]^2-m^2+i0}[2\Pi_{\lambda}(\phi)+\hat{k}\gamma_{\lambda}]U_s(p,x).
\end{split}$$ Now, we notice that \[see Eq. (\[E\_p\])\] $$\Pi^{\lambda}(\phi)U_s(p,x)=\left[\pi^{\lambda}_p(\phi)+i\frac{e\hat{n}\hat{A}'(\phi)}{2p_-}n^{\lambda}\right]U_s(p,x),$$ where $$\pi^{\lambda}_p(\phi)=p^{\lambda}-eA^{\lambda}(\phi)+\frac{e(pA(\phi))}{p_-}n^{\lambda}-\frac{e^2A^2(\phi)}{2p_-}n^{\lambda}$$ is the classical kinetic four-momentum of an electron in the plane wave $A^{\mu}(\phi)$, with $\lim_{\phi\to\pm\infty}\pi^{\lambda}_p(\phi)=p^{\lambda}$. The kinetic four-momentum $\pi^{\lambda}_p(\phi)$ is clearly a gauge-invariant four-vector and, by using the tensor $\mathscr{F}^{\mu\nu}(\phi)$ (see Sec. \[Notation\]), it can be written in the manifestly gauge-invariant form as $$\label{pi}
\pi^{\lambda}_p(\phi)=p^{\lambda}-\frac{ep_{\mu}\mathscr{F}^{\mu\lambda}(\phi)}{p_-}+\frac{e^2p_{\mu}\mathscr{F}^{\mu\rho}(\phi)\mathscr{F}_{\rho\nu}(\phi)p^{\nu}}{2p^3_-}n^{\lambda}.$$ In this way, the quantity $-ie\Gamma_{s,s',l}(p,p',q)$ can be written as $$\begin{split}
-ie\Gamma_{s,s',l}(p,p',q)&=-e^3\int d^4x\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{k^2-\kappa^2+i0}\\
&\quad\times\bar{U}_{s'}(p',x)\left[2\pi_{p'}^{\lambda}(\phi)+i\frac{e\hat{n}\hat{A}'(\phi)}{p'_-}n^{\lambda}+\gamma^{\lambda}\hat{k}\right]\frac{1}{[\hat{\Pi}(\phi)+\hat{k}]^2-m^2+i0} e^{i(qx)}\hat{e}^*_l(q)\\
&\qquad\times\frac{1}{[\hat{\Pi}(\phi)+\hat{k}]^2-m^2+i0}\left[2\pi_{p,\lambda}(\phi)+i\frac{e\hat{n}\hat{A}'(\phi)}{p_-}n_{\lambda}+\hat{k}\gamma_{\lambda}\right]U_s(p,x).
\end{split}$$ At this point, it is convenient to use the representations in Eq. (\[G\_1\]) and in Eq. (\[G\_2\]) for the second and the first square Volkov propagator $1/\{[\hat{\Pi}(\phi)+\hat{k}]^2-m^2+i0\}$, respectively: $$\begin{split}
&-ie\Gamma_{s,s',l}(p,p',q)=e^3\int d^4x\int\frac{d^4k}{(2\pi)^4}\int_0^{\infty}ds\int_0^{\infty}du\,\frac{e^{i(qx)}}{k^2-\kappa^2+i0}\\
&\quad\times\bar{U}_{s'}(p',x)\left[2\pi_{p'}^{\lambda}(\phi)+i\frac{e\hat{n}\hat{A}'(\phi)}{p'_-}n^{\lambda}+\gamma^{\lambda}\hat{k}\right]e^{-im^2s}e^{-2is(p_--q_-+k_-)(P_{\phi}-k_++q_+)}\\
&\quad\times e^{-i\int_0^sds'[\bm{p}_{\perp}-\bm{q}_{\perp}+\bm{k}_{\perp}-e\bm{A}_{\perp}(\phi+2s'(p_--q_-+k_-))]^2}\left\{1+\frac{e\hat{n}[\hat{A}(\phi+2s(p_--q_-+k_-))-\hat{A}(\phi)]}{2(p_--q_-+k_-)}\right\}\\
&\quad\times \hat{e}^*_l(q)e^{-im^2u}\left\{1-\frac{e\hat{n}[\hat{A}(\phi-2u(p_-+k_-))-\hat{A}(\phi)]}{2(p_-+k_-)}\right\}e^{-i\int_0^udu'[\bm{p}_{\perp}+\bm{k}_{\perp}-e\bm{A}_{\perp}(\phi-2u'(p_-+k_-))]^2}\\
&\quad\times e^{-2iu(p_-+k_-)(P_{\phi}-k_+)}\left[2\pi_{p,\lambda}(\phi)+i\frac{e\hat{n}\hat{A}'(\phi)}{p_-}n_{\lambda}+\hat{k}\gamma_{\lambda}\right]U_s(p,x),
\end{split}$$ where we have exploited the fact that Volkov states are eigenstates of the operators $P_T$ and $\bm{P}_{\perp}$. Indeed, the only operator remaining in this equation is $P_{\phi}$. Now, we use the translation property in Eq. (\[Trans\_P\]) and, analogously to the vacuum case, we write the amplitude $-ie\Gamma_{s,s',l}(p,p',q)$ in the form $$\label{Gamma^mu}
-ie\Gamma_{s,s',l}(p,p',q)=-ie\int d^4x\,e^{i(qx)}\bar{U}_{s'}(p',x)\Gamma^{\mu}(p,p',q;\phi)U_s(p,x)e^*_{l,\mu}(q),$$ where $$\label{Gamma_3}
\begin{split}
-ie\Gamma^{\mu}(p,p',q;\phi)&=e^3\int\frac{d^4k}{(2\pi)^4}\int_0^{\infty}ds\int_0^{\infty}du\,\frac{e^{-im^2(s+u)}}{k^2-\kappa^2+i0}\\
&\quad\times e^{2ik_+[s(p'_-+k_-)+u(p_-+k_-)]}e^{i\left\{p'_+(\phi_s-\phi)+\int_{\phi}^{\phi_s}d\phi'\left[-\frac{e\bm{p}'_{\perp}\cdot\bm{A}_{\perp}(\phi')}{p'_-}+\frac{e^2\bm{A}^2_{\perp}(\phi')}{2p'_-}\right]\right\}}\\
&\quad\times e^{-i\int_0^sds'[\bm{p}'_{\perp}+\bm{k}_{\perp}-e\bm{A}_{\perp}(\phi_{s'})]^2}e^{-i\int_0^udu'[\bm{p}_{\perp}+\bm{k}_{\perp}-e\bm{A}_{\perp}(\phi_{u'})]^2}\\
&\quad\times e^{i\left\{-p_+(\phi_u-\phi)-\int_{\phi}^{\phi_u}d\phi'\left[-\frac{e\bm{p}_{\perp}\cdot\bm{A}_{\perp}(\phi')}{p_-}+\frac{e^2\bm{A}^2_{\perp}(\phi')}{2p_-}\right]\right\}}M^{\mu}(\phi,k,s,u).
\end{split}$$ Here, 