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Studying the following scattering process at a tree-level:

$$\bar{q}^i (p_a) + q^j (p_b) \to \gamma(k_1) + \gamma(k_2)$$

Considering the “reduced” amplitude that is obtained by stripping away the polarisation vectors: $$\mathcal{M}_{\bar{q}q \to \gamma \gamma}= \mathcal{M}^{\mu_1 \mu_2}_{\bar{q}q \to \gamma \gamma}\varepsilon^*_{\mu_1} (k_1, \lambda_1)\varepsilon^*_{\mu_2} (k_2, \lambda_2)$$

Compute $\mathcal{M}^{\mu_1 \mu_2}_{\bar{q}q \to \gamma \gamma}$ using the Feynman rule for the quark-phton vertex:

$$V[\bar{\Psi}^i_q, \Psi^j_q, A_\mu ]= -ieQ_q \delta_{ij}\gamma_\mu$$

where i and j denote the colour indices of the quark legs.


I drew the Feynman diagram similar to the one in the link

https://en.wikipedia.org/wiki/Annihilation#/media/File:Mutual_Annihilation_of_a_Positron_Electron_pair.svg

and from it, I wrote the following amplitude:

$$\mathcal{M}^{\mu_1 \mu_2}_{\bar{q}q \to \gamma \gamma} = \epsilon^*_{\gamma}(k_1)(-ieQ\gamma^{k_1})(p_a)\frac{i\delta^{ij}(\require{cancel}\cancel{q}+m_{p_a})}{q^2 -m_q ^2+i\varepsilon}\epsilon^*_{\gamma}(k_2)\bar{v}(-ieQ\gamma^{k_2})(p_b)$$

This question has been set for a QCD module but I believe it only involves QED knowledge, which I have struggled with for a while now. Is my approach correct? If not, what is the correct answer and how would I obtain that answer?

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1 Answer 1

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This is close, but you need to be extra careful with the spin structure of the diagram. I find it most useful to start at the end of the fermion line, and follow the arrows backwards through the diagram (because we write left to right). I will write this out in pure QED, then after showing what the term should look like I'll show you how to add in color.

outgoing positron gets $v_{s_1}(p)$, an incoming positron is $\bar{v}_{s_1}(p)$. An outgoing electron gets $\bar{u}_{s_1}(p)$, and an incoming electron gets $u_{s_1}(p)$. Similarly incoming photons get $\epsilon_{r_1}^\mu(k)$, and an outgoing photon gets $\epsilon_{r_1}^{*\nu}(k)$. I label the helicity of the photons and spin of the electrons explicitly in the subscripts.

Start with the incoming positron line, this gets a $\bar{v}_{s_1}(p_a)$, then you get the first vertex $-ieQ\gamma^{\mu_2}\epsilon^*_{r_2\mu_2}(k_2)$.

Now you get the propagator with the four momentum of $p_a-k_2$ since energy and momentum are conserved at the vertex and the photon carried away momentum $k_2$. This looks like $i\frac{\gamma^\nu(p_{a\nu} - k_{2\nu}) + m_q}{(p_a-k_2)^2 - m_q^2}$

Next you get to the next vertex which carried $-ieQ\gamma^{\mu_1}\epsilon^*_{r_1\mu_1}(k_1)$. Finally the end of the line, the incoming electron line $u_{s_2}(p_b)$

This looks like: $$\mathcal{M}_{q\bar{q}\rightarrow{}\gamma\gamma} = \big(\bar{v}_{s_1}(p_a)\big)\big(-ieQ\gamma^{\mu_2}\epsilon^{*}_{r_2\mu_2}(k_2) \big) \bigg( i\frac{\gamma^\nu(p_{a\nu} - k_{2\nu}) + m_q}{(p_a-k_2)^2 - m_q^2}\bigg)\big(ieQ\gamma^{\mu_1}\epsilon^*_{r_1\mu_1}(k_1)\big)\big( u_{s_2}(p_b)\big)$$

Now because the polarization vectors commute with everything you can just pull them out of the term, $$\mathcal{M}_{q\bar{q}\rightarrow{}\gamma\gamma} = \mathcal{M}_{q\bar{q}\rightarrow{}\gamma\gamma}^{\mu_1\mu_2}\epsilon^*_{r_1\mu_1}(k_1)\epsilon^*_{r_2\mu_2}(k_2),$$ where the "reduced amplitude" is, $$\mathcal{M}_{q\bar{q}\rightarrow{}\gamma\gamma}^{\mu_1\mu_2} = -ie^2Q^2\left \lbrack\bar{v}_{s_1}(p_a)\gamma^{\mu_2} \frac{\gamma^\nu(p_{a\nu} - k_{1\nu}) + m_q}{(p_a-k_1)^2 - m_q^2}\gamma^{\mu_1}u_{s_2}(p_b)\right\rbrack. $$

Now you have the gamma matrices in the correct order, and they are sandwiched between the incoming and outgoing spinors forming the current. If you follow the direction of the current in this way you should be able to get all of the spin structure correct. The basic structure of the current will always be some number of gamma matrices with spinors on both sides (the left spinor will always have a bar over it, the right spinor will not).

You are correct this is a QED diagram, adding color just amounts to labeling your spinors with an $i$ and a $j$, and require it is conserved at each QED vertex, and in propagation. This is exactly what you did in your version.

Note this is only one of two diagrams at this order that contributes to this process, and the correct amplitude is the sum of the two terms. If you want to get the correct answer you need to figure out what the other diagram is, and compute it as well.

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