# Calculating 2nd Order Scattering Amplitude: Feynman Diagrams

• I
• Markus Kahn

#### 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

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

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...

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?

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
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.

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?

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.

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
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