# Calculating 2nd Order Scattering Amplitude: Feynman Diagrams

• I
• Markus Kahn
In summary, the conversation discusses the process of calculating the second order scattering amplitude in the interaction ##e^-e^- \to e^-e^-## using the relevant interaction Lagrangian and initial and final states. The probability amplitude is expanded in a series and the focus is on the second order contribution. The calculation involves using Wicks theorem and ultimately leads to the introduction of Feynman diagrams and rules. The question posed is about when and how to use these diagrams and how they relate to the initial and final states.
Markus Kahn
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.

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?

Last edited:
JD_PM
Markus Kahn said:
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.

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?
You have dropped factors of the electric charge along the way. To address your question, the final expression can be graphically represented as two Feynman diagrams, the ##t## and ##u## channels at tree level. Do you know the QED Feynman rules? One directly obtains your final expression by associating the spinor wave functions for the external states, drawing one photon exchanged with a propagator given by your ##G_{\mu \nu}## and associating to each vertex a factor ##-ie \gamma^\mu##, and multiplying the total result by ##(2 \pi)^4 \delta^4(P_{\rm in} - P_{\rm out})##.

Markus Kahn

nrqed said:
You have dropped factors of the electric charge along the way.
You're right. Sorry about that, things got a little bit messy towards the end...

nrqed said:
the final expression can be graphically represented as two Feynman diagrams, the ##t## and ##u## channels at tree level. Do you know the QED Feynman rules? One directly obtains your final expression by associating the spinor wave functions for the external states, drawing one photon exchanged with a propagator given by your ##G_{\mu \nu}## and associating to each vertex a factor ##-ie \gamma^\mu##, and multiplying the total result by ##(2 \pi)^4 \delta^4(P_{\rm in} - P_{\rm out})##.
I know the Feynman rules for QED, I also know how the diagrams look like and I also know how to "convert" Feynman diagrams using the Feynman rules into ##M_{\alpha\beta\gamma\delta}##.

The thing is, once I have the expression
\begin{align*} M_{\alpha\beta\gamma\delta} &\propto 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)\\ &-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)\\ \end{align*}
the Feynman diagrams are hardly useful anymore... I need them before I start doing this tedious calculation, i.e. probably at this point in the above calculation:
\begin{align*} F^{(2)} &= \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 \end{align*}
But the problem is that I don't know how to "read off" Feynman diagrams from this expression. How do you get the Feynman diagrams for the ##t## and ##u## channel just from looking at this expression?

Markus Kahn said:
Thanks for the answer!You're right. Sorry about that, things got a little bit messy towards the end...I know the Feynman rules for QED, I also know how the diagrams look like and I also know how to "convert" Feynman diagrams using the Feynman rules into ##M_{\alpha\beta\gamma\delta}##.

The thing is, once I have the expression
\begin{align*} M_{\alpha\beta\gamma\delta} &\propto 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)\\ &-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)\\ \end{align*}
the Feynman diagrams are hardly useful anymore... I need them before I start doing this tedious calculation, i.e. probably at this point in the above calculation:
\begin{align*} F^{(2)} &= \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 \end{align*}
But the problem is that I don't know how to "read off" Feynman diagrams from this expression. How do you get the Feynman diagrams for the ##t## and ##u## channel just from looking at this expression?
I am not completely sure I understand your point of view. If one uses Feynman diagrams, one does not start from the expression for ##F^{(2)}##. One does not write down any of the steps you went through, one draws the diagrams and one gets the final answer directly. So I am not sure if your question is about deriving the Feynman rules or about how to use them.

vanhees71
nrqed said:
So I am not sure if your question is about deriving the Feynman rules or about how to use them.
It's not about deriving them. You would do that by going through the calculations that I did for the "elemental building blocks" of the Feynman diagrams.

nrqed said:
I am not completely sure I understand your point of view. If one uses Feynman diagrams, one does not start from the expression for ##F^{(2)}##. One does not write down any of the steps you went through, one draws the diagrams and one gets the final answer directly.
Maybe I can rephrase my question to make it clearer. Where do you get the Feynman diagrams from? The information you have is ##e^-e^- \to e^-e^-## scattering and the interaction Lagragian ##\mathcal{L}_{\rm int}##. So you basically know that two electrons go in, two electrons come out and that's it. How does one find diagrams from this information?

Markus Kahn said:
The information you have is e−e−→e−e−e−e−→e−e−e^-e^- \to e^-e^- scattering and the interaction Lagragian LintLint\mathcal{L}_{\rm int}.

If you already have the Feynman rules, you don't need ##\mathcal{L}_\mathrm{int}## again, this information is now in the rules for the vertex, in this case the one in which two fermion lines and a photon line meet.

Markus Kahn said:
So you basically know that two electrons go in, two electrons come out and that's it. How does one find diagrams from this information?

Draw a line (with corresponding arrow) for each incoming electron, one for each outgoing electron, and connect them in all possible ways that are consistently using the building blocks of the Feynman rules (the order at which you are calculating basically limits how many vertices you get to use, and you usually only want connected diagrams). The first non-trivial way is by using two vertices, with their photon lines connected and one fermion line of each vertex connected to an incoming electron and the other one to one of the outgoing ones. Of course, with more complicated examples it might need some experience to be confident that you are done.

Markus Kahn and vanhees71
Dr.AbeNikIanEdL said:
Draw a line (with corresponding arrow) for each incoming electron, one for each outgoing electron, and connect them in all possible ways that are consistently using the building blocks of the Feynman rules
Perfect, this is what I was looking for. So once the Feynman rules are know I can just draw "all" permissible diagrams and use the rules to compute them instead of going through the detailed computations, i.e. computing the ##F^{(n)}## individually.

Thanks for the help!

vanhees71

## 1. What is the purpose of calculating 2nd order scattering amplitude using Feynman diagrams?

The purpose of calculating 2nd order scattering amplitude using Feynman diagrams is to understand and predict the interactions between particles at a quantum level. Feynman diagrams provide a visual representation of the mathematical calculations involved in calculating the scattering amplitude, making it easier to interpret and analyze the results.

## 2. How do Feynman diagrams help in calculating 2nd order scattering amplitude?

Feynman diagrams use graphical representations to simplify the complex mathematical calculations involved in calculating the 2nd order scattering amplitude. They show the exchange of virtual particles between the interacting particles, making it easier to calculate the amplitude and interpret the results.

## 3. What are the key components of a Feynman diagram for calculating 2nd order scattering amplitude?

The key components of a Feynman diagram for calculating 2nd order scattering amplitude are the external particles (incoming and outgoing), the vertices (where particles interact), and the internal lines (representing virtual particles). These components are used to determine the mathematical expression for the scattering amplitude.

## 4. How does the number of Feynman diagrams affect the calculation of 2nd order scattering amplitude?

The number of Feynman diagrams involved in calculating 2nd order scattering amplitude depends on the number of interacting particles. The more particles involved, the more diagrams need to be considered. However, some diagrams may cancel each other out, simplifying the calculation process.

## 5. What is the significance of calculating 2nd order scattering amplitude in particle physics?

Calculating 2nd order scattering amplitude is essential in particle physics as it helps us understand the fundamental interactions between particles at a quantum level. This information is crucial in developing theories and models to explain the behavior of particles and the universe as a whole.

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