-\frac{1}{\hbar^{2}}\langle \mathbf{p}_{f},s_{f}|\left[\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}:N_{I}\left(t_{1}\right)::N_{I}\left(t_{2}\right):\right]|\mathbf{p}_{i},s_{i}\rangle =}  \\
& = & -\frac{\gamma_{m}}{2}\int_{0}^{t}\!dt_{1}dt_{2}\int\! d\mathbf{x}_{1} d\mathbf{x}_{2}\,w\left(x_{1}\right)w\left(x_{2}\right) \langle \mathbf{p}_{f},s_{f}|T\left[:\bar{\psi}_{I}\left(x_{1}\right)\psi_{I}\left(x_{1}\right)::\bar{\psi}_{I}\left(x_{2}\right)\psi_{I}\left(x_{2}\right):\right]|\mathbf{p}_{i},s_{i}\rangle, \nonumber\end{aligned}$$ here “$T$” is the time-ordering product. Using Wick’s theorem, and discarding all tadpole terms, which involve a contraction between two fields at the same spacetime point, we have: $$\begin{aligned}
 \label{eq:dsfdfbv}
T\left[\bar{\psi}_{1a}\psi_{1a}\bar{\psi}_{2b}\psi_{2b}\right]
& = & :\bar{\psi}_{1a}\psi_{1a}\bar{\psi}_{2b}\psi_{2b}:-S_{ab}\left(x_{1}-x_{2}\right)S_{ba}\left(x_{2}-x_{1}\right) \nonumber \\
& & +iS_{ab}\left(x_{1}-x_{2}\right):\bar{\psi}_{1a}\psi_{2b}:-iS_{ba}\left(x_{2}-x_{1}\right):\psi_{1a}\bar{\psi}_{2b}:,\end{aligned}$$ where $a$ and $b$ label the spinor components and the Dirac propagator is: $ iS_{ab}\left(x_{1}-x_{2}\right)\equiv\langle \Omega|T\left[\psi_{1a}\bar{\psi}_{2b}\right]|\Omega\rangle$. Here, Einstein’s summation convention is used for the spinor indices. We momentarily drop the pedex $I$ related to the interaction picture and we write the dependence on $x_{1}$ and $x_{2}$ simply as $_{1}$ and $ _{2}$. The diagramatic representation of the different terms in Eq. (\[eq:dsfdfbv\]) is: $$\begin{array}{cclcclccl}
\Diagram{\vertexlabel^ifA\vertexlabel_1 hu \\ f0 fdA\vertexlabel^{\;\;f}}\;\;\;\;\Diagram{\vertexlabel^ifA\vertexlabel_2 hu \\ f0 fdA\vertexlabel^{\;\;f}} & = & \displaystyle :\bar{\psi}_{1a}\psi_{1a}\bar{\psi}_{2b}\psi_{2b}:\;, \qquad &
\feyn{h\vertexlabel_1 f0 {flA}{}
{fluV}{} f0 \vertexlabel_2h}  & = & \displaystyle -S_{ab}\left(x_{1}-x_{2}\right)S_{ba}\left(x_{2}-x_{1}\right)\;,
\end{array}$$ $$\begin{array}{cclcclccl}
\Diagram{\vertexlabel_i hd\vertexlabel_{2}  fA \vertexlabel_{1}
hu\\
fuA f0 fdA \vertexlabel^{\;\;f}}  & = & \displaystyle iS_{ab}\left(x_{1}-x_{2}\right):\bar{\psi}_{1a}\psi_{2b}:\;,\qquad\qquad\qquad
\Diagram{\vertexlabel_i hd\vertexlabel_{1}  fA \vertexlabel_{2}
hu\\
fuA f0 fdA \vertexlabel^{\;\;f}}  & = & \displaystyle -iS_{ba}\left(x_{2}-x_{1}\right):\psi_{1a}\bar{\psi}_{2b}:\;.
\end{array}$$ We can easily see that the first term is zero, since we are studying the case with only one particle in the initial and final 