will need only the case $\mathbf{p}_{f}=\mathbf{p}_{i}$ and $s_{f}=s_{i}$. In this case, we have: $$\begin{aligned}
T^{\left(2\right)}\left(\mathbf{p}_{i},s_{i};\mathbf{p}_{i},s_{i};t\right)
& = & 
-\gamma_{m}\frac{1}{L^{3}}\frac{m c^{2}}{E_{i}}\sum_{s=1}^{2}\sum_{\mathbf{p}=-\infty}^{+\infty}\frac{1}{L^{3}}\frac{m c^{2}}{E_{p}} \nonumber \\
& & 
\cdot\left\{ \int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}\int d\mathbf{x}_{1}\int d\mathbf{x}_{2}\, w(x_{1})w(x_{2})e^{\frac{i}{\hbar}(p_{i}^{\mu}-p^{\mu})\left(x_{1\mu}-x_{2\mu}\right)}\right. \nonumber \\
& & \cdot \; \overline{u}_{a}(p_{i},s_{i})u_{a}(p^,s)\overline{u}_{b}(p^,s) u_{b}(p_{i},s_{i}) \nonumber \\
& &
-\int_{0}^{t}dt_{2}\int_{0}^{t_{2}}dt_{1}\int d\mathbf{x}_{1}\int d\mathbf{x}_{2}w\left(x_{1}\right)w\left(x_{2}\right)e^{\frac{i}{\hbar}(p_{i}^{\mu}-p^{\mu})\left(x_{1\mu}-x_{2\mu}\right)} \nonumber \\
& & \cdot \left.
\overline{u}_{a}(p_{i},s_{i})v_{a}(p,s)\overline{v}_{b}(p,s)u_{b}(p_{i},s_{i})\phantom{\frac{1}{2}}\!\!\!\!\right\}.\end{aligned}$$ Using the standard relations [@Greiner]: $$\sum_{s=1}^{2}u_{a}(p,s)\overline{u}_{b}(p,s)=\left(\frac{p^{\mu}\gamma_{\mu}+mc}{2mc}\right)_{ab},\;\;\;\;\;\;\sum_{s=1}^{2}v_{a}(p,s)\overline{v}_{b}(p,s)=\left(\frac{p^{\mu}\gamma_{\mu}-mc}{2mc}\right)_{ab},$$ we can see that the terms containing $p^{\mu}\gamma_{\mu}$ cancel each other, while those containing the mass give a $\delta_{ab}$. Thus, if we also use: $\overline{u}(p_{i},s_{f})u(p_{i},s_{i})=\delta_{s_{f},s_{i}}$, we obtain: $$\begin{aligned}
 \label{eq:fdhgfg}
T^{\left(2\right)}\left(\mathbf{p}_{i},s_{i};\mathbf{p}_{i},s_{i};t\right)
& = &
-\gamma_{m}\frac{1}{L^{3}}\frac{mc^{2}}{E_{i}}\sum_{\mathbf{p}=-\infty}^{+\infty}\frac{1}{L^{3}}\frac{mc^{2}}{E_{p}}\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}\int d\mathbf{x}_{1}\int d\mathbf{x}_{2}w\left(x_{1}\right)w\left(x_{2}\right) \nonumber \\
& & \times e^{\frac{i}{\hbar}(p_{i}^{\mu}-p^{\mu})\left(x_{1\mu}-x_{2\mu}\right)}.\end{aligned}$$ Now we have all the elements we need, in order to compute the transition probability. We will do this in the next section.

The transition probability
==========================

The physical quantity we are interested in, is the transition probability, which corresponds to $\left|T_{\alpha\rightarrow\beta}\right|^{2}$, averaged over the noise, and integrated over the final momentum and polarization states: $$\label{eq:sfdso}
P_{\alpha\rightarrow\beta}
\equiv
\sum_{s_{f}}\sum_{\mathbf{p}_{f}=-\infty}^{+\infty}\mathbb{E}\left|T_{\alpha\rightarrow\beta}\right|^{2}
=
\sum_{k=1}^{n}\sum_{j=1}^{n}\text{U}_{\alpha k}^{*}\text{U}_{\beta k}\text{U}_{\alpha j}\text{U}_{\beta j}^{*}P_{kj}\left(\mathbf{p}_{i},s_{i};t\right),$$ where: $$P_{kj}\left(\mathbf{p}_{i},s_{i};t\right)\equiv\sum_{s_{f}}\sum_{\mathbf{p}_{f}=-\infty}^{+\infty}e^{\frac{i}{\hbar}(E_{f}^{\left(k\right)}-E_{f}^{\left(j\right)})t}\,\mathbb{E}\left[T_{k}^{*}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right) T_{j}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)\right],$$ and $T_{j}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)$ is given by Eq. (\[eq:uioi\]), where now we have explicitly indicated the label $j$ associated to the mass eigenstate $m_{j}$ and $E_{f}^{\left(j\right)}=\sqrt{\mathbf{p}_f^{2}c^{2}+m_j^{2}c^{4}}$. When averaging, one has to remember that only 