\frac{1}{\left(2\pi\hbar\right)^{3}} e^{-\frac{(\mathbf{p}_{f}-\mathbf{p}_{i})^{2}r_{C}^{2}}{\hbar^{2}}}.$$ The time integral in Eq. (\[eq:sdfgdr\]) is trivial, and one arrives easily at the formula: $$\begin{aligned}
 \label{eq:dasdasd}
I_{jk}^{\left(1\right)}\left(\mathbf{p}_{i},s_{i};t\right)
& = & 
\sum_{s_{f}}\int d\mathbf{p}_{f}\sqrt{\gamma_{m_{j}}\gamma_{m_{k}}}\frac{m_{j}m_{k}c^{4}}{\sqrt{E_{i}^{\left(j\right)}E_{f}^{\left(j\right)}E_{i}^{\left(k\right)}E_{f}^{\left(k\right)}}}\overline{u}(p_{f}^{\left(j\right)},s_{f})u(p_{i}^{\left(j\right)},s_{i})\overline{u}(p_{i}^{\left(k\right)},s_{i})u(p_{f}^{\left(k\right)},s_{f}) \nonumber \\
& & \cdot
\frac{1-e^{\frac{i}{\hbar}(E_{f}^{\left(k\right)}-E_{i}^{\left(k\right)}-E_{f}^{\left(j\right)}+E_{i}^{\left(j\right)})t}}{\frac{i}{\hbar}(E_{f}^{\left(j\right)}-E_{i}^{\left(j\right)}-E_{f}^{\left(k\right)}+E_{i}^{\left(k\right)})}\frac{1}{\left(2\pi\hbar\right)^{3}}e^{-\frac{(\mathbf{p}_{f}-\mathbf{p}_{i})^{2}r_{C}^{2}}{\hbar^{2}}}.\end{aligned}$$ As it is shown in the Appendix C, the integrating function (except for the Gaussian term) changes slowly within the region where the Gaussian term is appreciably different from zero. Therefore we can approximate it with the value it takes in the center of the Gaussian (where $\mathbf{p}_{f}=\mathbf{p}_{i}$) and bring it out of the integral. Taking into account that $\overline{u}\left(p,s_{i}\right)u\left(p,s_{f}\right)=\delta_{s_{i}s_{f}}$ and performing the integration of the Gaussian part, Eq. (\[eq:dasdasd\]) takes the very simple expression: $$\label{eq:syksdgar}
I_{jk}^{\left(1\right)}\left(\mathbf{p}_{i},s_{i};t\right)
=
\sqrt{\gamma_{m_{j}}\gamma_{m_{k}}}\frac{m_{j}m_{k}c^{4}}{E_{i}^{\left(j\right)}E_{i}^{\left(k\right)}}\frac{t}{\left(2\pi\right)^{3}}\frac{\pi^{3/2}}{r_{C}^{3}}.$$

We now turn our attention to the term $I_{j}^{\left(2\right)}\left(\mathbf{p}_{i},s_{i};t\right)$ in Eq. (\[eq:rtrtsfds\]). Substituting Eq. (\[eq:fdhgfg\]) in Eq. (\[eq:rtrtsfds\]), we have: $$I_{j}^{\left(2\right)}\left(\mathbf{p}_{i},s_{i};t\right)
=
-\gamma_{m_{j}}\frac{m_{j}c^{2}}{E_{i}^{\left(j\right)}}\sum_{\mathbf{p}=-\infty}^{+\infty}\frac{m_{j}c^{2}}{E_{p}^{\left(j\right)}}\frac{t}{2}\frac{1}{L^{6}}\int d\mathbf{x}_{1}\int d\mathbf{x}_{2}\frac{e^{-\left(\mathbf{x}_{1}-\mathbf{x}_{2}\right)^{2}/4r_{C}^{2}}}{\left(\sqrt{4\pi}r_{C}\right)^{3}}e^{-\frac{i}{\hbar}\left(\mathbf{p}_{i}-\mathbf{p}\right)\left(\mathbf{x}_{1}-\mathbf{x}_{2}\right)}.$$ The spatial integrals are equal to $S\left(\mathbf{p},\mathbf{p}_{i}\right)$ defined in Eq. (\[eq:zxgf\]). Performing the same type of calculation as before, and taking the limit $L \rightarrow \infty$, one arrives at the result: $$\label{eq:vcvxvxcy}
I_{j}^{\left(2\right)}\left(\mathbf{p}_{i},s_{i};t\right)
=
-\gamma_{m_{j}}\frac{m_{j}c^{2}}{E_{i}^{\left(j\right)}}\int d\mathbf{p}\frac{m_{j}c^{2}}{E_{p}^{\left(j\right)}}\frac{1}{2}t\frac{1}{\left(2\pi\hbar\right)^{3}}e^{-\frac{\left(\mathbf{p}-\mathbf{p}_{i}\right)^{2}r_{C}^{2}}{\hbar^{2}}}.$$ Once again, one can show that the integrating function (Gaussian term apart) varies slowly within the region where the Gaussian function is appreciably different from zero. Therefore one can 