# Electron-positron annihilation diagram

1. Jan 31, 2016

### CAF123

Just a quick question regarding the tree level Feynman diagram(s) contributing to this process - I am wondering if I wanted to compute the unpolarised transition amplitude for the annihilation $e^+ e^- \rightarrow \gamma \gamma$, are there two tree level diagrams that contribute or just one?

I am thinking of an electron and positron as initial state, an electron/positron being the virtual particle and the two photons as the external state. (e.g photon a) tagged at vertex with the electron b) and another photon c) tagged at vertex with positron d)) That's one diagram. But I also thought, to take into account the bose symmetry, I would also need to consider an another diagram where photon a) is tagged at vertex with positron d) and another where photon c) is tagged at vertex with electron b)?

Is it correct?

2. Feb 1, 2016

### Simon Bridge

... what was the question?

3. Feb 1, 2016

### CAF123

Hi SimonBridge,

I have drawn what I think are the two contributing tree level processes for $e^+ e^- \rightarrow 2\gamma$. I want to understand why we don't consider the $t-$ channel diagram on the left to be different from the $u$ channel diagram on the right.

Thanks!

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4. Feb 1, 2016

### Dr.AbeNikIanEdL

Who is doing this? I think you indeed have to consider both diagrams. But this is not exclusive to bosons, you would also get two diagrams of this kind with outgoing (indistinguishable) fermions.

5. Feb 1, 2016

### vanhees71

Of course, in the leading (tree-level) order there are the two diagrams you draw, and that's indeed important to fulfill Bose symmetry for the outgoing photons. The diagrams are $t$- and $u$-channel diagrams.

6. Feb 1, 2016

### CAF123

Ok, many thanks for the confirmation - it is as I thought. So, if I was computing the (unpolarised) transition amplitude for this process I would sum the amplitudes for each of the t and u channel processes? Then a probability would be the square of this:

$$P(e^+ e^- \rightarrow 2 \gamma) = \frac{1}{4} \sum_{\text{spins,polarisations}} |\mathcal M_1+ \mathcal M_2|^2$$

7. Feb 1, 2016

### vanhees71

That's right.

8. Feb 1, 2016

### CAF123

Ok thanks, I am aware of a crossing symmetry that exists between the unpolarised transition probability for compton scattering $(e^- + \gamma \rightarrow e^- + \gamma)$ and that for the case of diphoton production. We did compton scattering in the lecture and I put the notation we used for the process in an attachment, together with the case at hand in another notation.

It seems that by putting $k \rightarrow k_1, \,\,\ q \rightarrow p_2\,\,\,\ q' \rightarrow p_1$ I can get agreement between the two formulas for the unpolarised transition probability. The middle condition here I don't understand. Even though the photon with momentum q is in the initial state in compton scattering is it simply ok to put it to a photon of momentum $p_2$ in the final state of diphoton production?
Thanks!

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