an important property, because it implies that if $i\neq j$: $$\begin{aligned}
\langle \nu_{i};\mathbf{p}_{f},s_{f}|U(t)|\nu_{j};\mathbf{p}_{i},s_{i}\rangle  & = & \left\langle \Omega_{1}\left|U_{1}\left(t\right)\right|\Omega_{1}\right\rangle ...\left\langle \nu_{i};\mathbf{p}_{f},s_{f}\left|U_{i}\left(t\right)\right|\Omega_{i}\right\rangle ...\left\langle \Omega_{j}\left|U_{j}\left(t\right)\right|\nu_{j};\mathbf{p}_{i},s_{i}\right\rangle... \nonumber \\
& & ...\left\langle \Omega_{n}\left|U_{n}\left(t\right)\right|\Omega_{n}\right\rangle =0\end{aligned}$$ since $\left\langle \nu_{i};\mathbf{p}_{f},s_{f}\left|U_{i}\left(t\right)\right|\Omega_{i}\right\rangle =0$, as one can check with a direct calculation. Therefore we can write: $$\label{eq:cfhgfd}
T_{\alpha\rightarrow\beta} =\sum_{j=1}^{n}\text{U}_{\alpha j}\text{U}_{\beta j}^{*}\left[\left\langle \Omega_{1}\left|U_{1}\left(t\right)\right|\Omega_{1}\right\rangle ...\left\langle \nu_{j};\mathbf{p}_{f},s_{f}\left|U_{j}\left(t\right)\right|\nu_{j};\mathbf{p}_{i},s_{i}\right\rangle ....\left\langle \Omega_{n}\left|U_{n}\left(t\right)\right|\Omega_{n}\right\rangle \right],$$ which reduces the entire calculation to a 1-particle computation. In the next section we will focus our attention on the matrix element $\left\langle \nu_{j};\mathbf{p}_{f},s_{f}\left|U_{j}\left(t\right)\right|\nu_{j};\mathbf{p}_{i},s_{i}\right\rangle$ since, as we will show, the remaining terms contribute with an unimportant global phase factor.

The matrix elements
===================

We now focus on the main part of this work. What we need to compute the 1-particle matrix element: $$T\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)
\equiv
\left\langle \mathbf{p}_{f},s_{f}\left|U\left(t\right)\right|\mathbf{p}_{i},s_{i}\right\rangle = e^{-\frac{i}{\hbar}E_{f}t}\left\langle \mathbf{p}_{f},s_{f}\left|U_{I}\left(t\right)\right|\mathbf{p}_{i},s_{i}\right\rangle.$$ Since this part of the computation is the same for every mass eigenstate, we have dropped the label $j$. We expand the evolution operator by means of the Dyson series up to the second order: $$U_{I}\left(t,0\right)\simeq1-\frac{i}{\hbar}\int_{0}^{t}dt_{1}:N_{I}\left(t_{1}\right):-\frac{1}{\hbar^{2}}\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}:N_{I}\left(t_{1}\right)::N_{I}\left(t_{2}\right):,$$ where $N_{I}\left(t\right)$ is the interaction picture representation of Eq. (\[eq:dfgdf\]) and $: ... :$ denotes the normal ordering[^5]. Accordingly, the transition probability is the sum of three terms: $$\label{eq:uioi}
T\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)=e^{-\frac{i}{\hbar}E_{f}t}\left[T^{\left(0\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)+T^{\left(1\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)+T^{\left(2\right)}\left(\mathbf{p}_{f},s_{f};\mathbf{p}_{i},s_{i};t\right)\right],$$ corresponding to the first three terms of the Dyson 