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- TL;DR Summary
- Starting from an interaction Lagragian ##\mathcal{L}_{\rm int}## I'd like to calculate the second order scattering amplitude and understand specific steps in this calculation better, especially when and where Feynman diagrams and rules appear and are supposed to be helpful.

In the following I will try to deduce the scattering amplitude for a specific interaction. My question is at the bottom, the entire rest is my reasoning to explain how I came to the results I present.

Let's assume I would like to calculate the second order scattering amplitude in ## e^- e^- \to e^-e^-##. The relevant interaction Lagragian is then given by

$$\mathcal{L}_{\rm int}(x) := -i e A_\mu(x)\bar{\psi}(x)\gamma^\mu\psi(x).$$

We prepare an initial and final state at times ##t_{\rm in}## and ##t_{\rm out}## and assume that they are independent at these times

$$\vert i \rangle := a_\alpha^\dagger(\vec{p}_1)a_\beta^\dagger(\vec{p}_2)\vert 0\rangle \quad\text{and}\quad \langle f\vert := \langle 0\vert a_\delta(\vec{q}_2) a_\gamma(\vec{q}_1).$$

The probability amplitude of interest is then

$$F = \lim_{t_{\rm in},t_{\rm out} \to \infty}\langle f\vert U_{\rm int}(t_{\rm out}, t_{\rm in})\vert i\rangle \quad \text{where} \quad U_{\rm int} = T \exp\left( i \int d^4x \mathcal{L}_{\rm int} \right).$$

We can now formally expand ##F## in a series of the form

$$F= \sum_{i\in\mathbb{N}_0} F^{(i)}$$

where each ##F^{(i)}## just corresponds to a higher order contribution from the Dyson series. A quick calculation reveals that ##F^{(0)}= \langle f\vert i\rangle## and ##F^{(1)}=0##, so the interesting part starts at ##F^{(2)}##, where

$$\begin{align*}

F^{(2)}

&= \frac{1}{2}i^2 \int d^4x d^4y \langle f \vert T\{\mathcal{L}_{\rm int}(x)\mathcal{L}_{\rm int}(y)\}\vert i\rangle\\

&= \frac{1}{2} e^2 \int d^4x d^4y\,\langle f\vert T\{A_\mu(x) \bar{\psi}(x)\gamma^\mu\psi(x)A_\nu(y)\bar{\psi}(y)\gamma^\nu\psi(y) \}\vert i\rangle\\

&= -\frac{i}{2} e^2 \int d^4x d^4y\, G_{\mu\nu}^F (x-y) \langle f\vert T\{ \bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y) \}\vert i\rangle,

\end{align*}$$

where ##G_{\mu\nu}^F(x-y)## is the photon propagator and we introduced it since I know that my final state will not have any asymptotic photons. One can now apply Wicks theorem to get

$$T\{ \bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y) \} = N\{\bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y) + \text{contractions}\}.$$

It turns out that all the contractions will give loop contributions or vacuum bubbles, so we want to focus on the first term, namely:

$$\begin{align*}

\langle f\vert N\{&\bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y)\}\vert i \rangle\\

&=i q^{2} \int d^{4} x d^{4} y G_{\mu \nu}^{\mathrm{F}}(x-y) e^{i p_{1} \cdot x+i p_{2} \cdot y-i q_{1} \cdot x-i q_{2} \cdot y} \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\gamma}\left(\vec{q}_{1}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\delta}\left(\vec{q}_{2}\right)\\

&- i q^{2} \int d^{4} x d^{4} y G_{\mu \nu}^{\mathrm{F}}(x-y) e^{i p_{1} \cdot x+i p_{2} \cdot y-i q_{2} \cdot x-i q_{1} \cdot y} \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\delta}\left(\vec{q}_{2}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\gamma}\left(\vec{q}_{1}\right)\\

& = (2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) G_{\mu\nu}^F(q_1-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\gamma}\left(\vec{q}_{1}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\delta}\left(\vec{q}_{2}\right)\\

& -(2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) G_{\mu\nu}^F(q_2-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\delta}\left(\vec{q}_{2}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\gamma}\left(\vec{q}_{1}\right)\\

&= (2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) \left\{G_{\mu\nu}^F(q_1-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\gamma}\left(\vec{q}_{1}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\delta}\left(\vec{q}_{2}\right)\right.\\

& \left.\hspace{4.75cm}-G_{\mu\nu}^F(q_2-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\delta}\left(\vec{q}_{2}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\gamma}\left(\vec{q}_{1}\right)\right\}\\

&=: (2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) M_{\alpha\beta\gamma\delta}

\end{align*}$$

Now this last calculation was a nightmare to do and as far as I understand here is where Feynman diagrams and Feynman rules are supposed to be useful. What I don't understand is

**My working**Let's assume I would like to calculate the second order scattering amplitude in ## e^- e^- \to e^-e^-##. The relevant interaction Lagragian is then given by

$$\mathcal{L}_{\rm int}(x) := -i e A_\mu(x)\bar{\psi}(x)\gamma^\mu\psi(x).$$

We prepare an initial and final state at times ##t_{\rm in}## and ##t_{\rm out}## and assume that they are independent at these times

$$\vert i \rangle := a_\alpha^\dagger(\vec{p}_1)a_\beta^\dagger(\vec{p}_2)\vert 0\rangle \quad\text{and}\quad \langle f\vert := \langle 0\vert a_\delta(\vec{q}_2) a_\gamma(\vec{q}_1).$$

The probability amplitude of interest is then

$$F = \lim_{t_{\rm in},t_{\rm out} \to \infty}\langle f\vert U_{\rm int}(t_{\rm out}, t_{\rm in})\vert i\rangle \quad \text{where} \quad U_{\rm int} = T \exp\left( i \int d^4x \mathcal{L}_{\rm int} \right).$$

We can now formally expand ##F## in a series of the form

$$F= \sum_{i\in\mathbb{N}_0} F^{(i)}$$

where each ##F^{(i)}## just corresponds to a higher order contribution from the Dyson series. A quick calculation reveals that ##F^{(0)}= \langle f\vert i\rangle## and ##F^{(1)}=0##, so the interesting part starts at ##F^{(2)}##, where

$$\begin{align*}

F^{(2)}

&= \frac{1}{2}i^2 \int d^4x d^4y \langle f \vert T\{\mathcal{L}_{\rm int}(x)\mathcal{L}_{\rm int}(y)\}\vert i\rangle\\

&= \frac{1}{2} e^2 \int d^4x d^4y\,\langle f\vert T\{A_\mu(x) \bar{\psi}(x)\gamma^\mu\psi(x)A_\nu(y)\bar{\psi}(y)\gamma^\nu\psi(y) \}\vert i\rangle\\

&= -\frac{i}{2} e^2 \int d^4x d^4y\, G_{\mu\nu}^F (x-y) \langle f\vert T\{ \bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y) \}\vert i\rangle,

\end{align*}$$

where ##G_{\mu\nu}^F(x-y)## is the photon propagator and we introduced it since I know that my final state will not have any asymptotic photons. One can now apply Wicks theorem to get

$$T\{ \bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y) \} = N\{\bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y) + \text{contractions}\}.$$

It turns out that all the contractions will give loop contributions or vacuum bubbles, so we want to focus on the first term, namely:

$$\begin{align*}

\langle f\vert N\{&\bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y)\}\vert i \rangle\\

&=i q^{2} \int d^{4} x d^{4} y G_{\mu \nu}^{\mathrm{F}}(x-y) e^{i p_{1} \cdot x+i p_{2} \cdot y-i q_{1} \cdot x-i q_{2} \cdot y} \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\gamma}\left(\vec{q}_{1}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\delta}\left(\vec{q}_{2}\right)\\

&- i q^{2} \int d^{4} x d^{4} y G_{\mu \nu}^{\mathrm{F}}(x-y) e^{i p_{1} \cdot x+i p_{2} \cdot y-i q_{2} \cdot x-i q_{1} \cdot y} \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\delta}\left(\vec{q}_{2}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\gamma}\left(\vec{q}_{1}\right)\\

& = (2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) G_{\mu\nu}^F(q_1-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\gamma}\left(\vec{q}_{1}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\delta}\left(\vec{q}_{2}\right)\\

& -(2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) G_{\mu\nu}^F(q_2-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\delta}\left(\vec{q}_{2}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\gamma}\left(\vec{q}_{1}\right)\\

&= (2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) \left\{G_{\mu\nu}^F(q_1-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\gamma}\left(\vec{q}_{1}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\delta}\left(\vec{q}_{2}\right)\right.\\

& \left.\hspace{4.75cm}-G_{\mu\nu}^F(q_2-p_1) \bar{v}_{\alpha}\left(\vec{p}_{1}\right) \gamma^{\mu} v_{\delta}\left(\vec{q}_{2}\right) \bar{v}_{\beta}\left(\vec{p}_{2}\right) \gamma^{\nu} v_{\gamma}\left(\vec{q}_{1}\right)\right\}\\

&=: (2\pi)^4 \delta(P_{\rm in} - P_{\rm out}) M_{\alpha\beta\gamma\delta}

\end{align*}$$

**Question**Now this last calculation was a nightmare to do and as far as I understand here is where Feynman diagrams and Feynman rules are supposed to be useful. What I don't understand is

**where exactly**am I supposed to introduce them and from where can I read off the Feynman diagrams? Am I supposed to guess the diagrams from ##N\{\bar{\psi}(x)\gamma^\mu\psi(x)\bar{\psi}(y)\gamma^\nu\psi(y)\}##? What about the ##\vert i\rangle## and ##\vert f\rangle##, how do they play into the diagrams?
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