series. We now give a pictorial representation of each term by means of Feynman diagram and we compute each of them. The first term corresponds to the free propagation: $$\begin{array}{cclcclccl}
\displaystyle  T^{\left(0\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right) & = & \Diagram{\vertexlabel^{i}fsfAfs\vertexlabel^{f}}
\end{array}$$ where the solid line represent the particle. This term is trivial: $$\label{eq:cvhjcg}
T^{\left(0\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)
\equiv
\left\langle \mathbf{p}_{f},s_{f}|\mathbf{p}_{i},s_{i}\right\rangle =\delta_{s_{f}s_{i}}\delta_{\mathbf{p}_{f},\mathbf{p}_{i}}.$$ The second term correspond to the diagram: $$\begin{array}{cclcclccl}
\displaystyle T^{\left(1\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right) & = & \Diagram{\vertexlabel^ifA\vertexlabel_1 hu \\ f0 fdA\vertexlabel^{\;\;f}} 
\end{array}$$ where the dotted line represents the noise field. This term is: $$\label{eq:T1}
T^{\left(1\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)
\equiv
i\sqrt{\gamma_{m}}\int_{0}^{t}dt_{1}\int d\mathbf{x}_{1}w\left(x_{1}\right)\langle \mathbf{p}_{f},s_{f}|:\bar{\psi}_{I}\left(x_{1}\right)\psi_{I}\left(x_{1}\right):|\mathbf{p}_{i},s_{i}\rangle.$$ In order to compute the matrix element in Eq. (\[eq:T1\]), we use the series expansion of the fields as given in Eq. (\[eq:wave\]). The non-null terms are those containing two $b$ and two $b^{\dagger}$ operators. After some calculations, one finds that: $$\begin{aligned}
\label{eq:T1me}
\lefteqn{\langle \Omega|b\left(p_{f},s_{f}\right):\bar{\psi}_{I}\left(x_{1}\right)\psi_{I}\left(x_{1}\right):b^{\dagger}\left(p_{i},s_{i}\right)|\Omega\rangle =} \nonumber  \\
& = & \sum_{s,s'=1}^{2}\sum_{\mathbf{p},\mathbf{p'}=-\infty}^{+\infty}\frac{1}{L^{3}}\frac{mc^{2}}{\sqrt{E_{p}E_{p'}}}e^{\frac{i}{\hbar}\left(p'^{\mu}-p^{\mu}\right)x_{1\mu}}\overline{u}\left(p',s'\right)u\left(p,s\right)\langle \Omega|b\left(p_{f},s_{f}\right)b^{\dagger}\left(p',s'\right)b\left(p,s\right)b^{\dagger}\left(p_{i},s_{i}\right)|\Omega\rangle \nonumber \\
& = & \frac{1}{L^{3}}\frac{mc^{2}}{\sqrt{E_{i}E_{f}}}e^{\frac{i}{\hbar}\left(p_{f}^{\mu}-p_{i}^{\mu}\right)x_{1\mu}}\overline{u}\left(p_{f},s_{f}\right)u\left(p_{i},s_{i}\right). \end{aligned}$$ Here we introduced the four momentum $p^{\mu}=\left(E_{p}/c,\mathbf{p}\right)$. If we substitute Eq. (\[eq:T1me\]) in the definition of $T^{\left(1\right)}$, we get: $$\label{eq:T1again}
T^{\left(1\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)=i\sqrt{\gamma_{m}}\frac{mc^{2}}{\sqrt{E_{i}E_{f}}}\overline{u}\left(p_{f},s_{f}\right)u\left(p_{i},s_{i}\right)\frac{1}{L^{3}}\int_{0}^{t}dt_{1}\int d\mathbf{x}_{1}w\left(x_{1}\right)e^{\frac{i}{\hbar}\left(p_{f}^{\mu}-p_{i}^{\mu}\right)x_{1\mu}}.$$

The last term in Eq. (\[eq:uioi\]) is the more complicated to compute, since it involves the product of four fields. It gives the following contribution: $$\begin{aligned}
\lefteqn{T^{\left(2\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right) 
\equiv
