# Electron-positron annihilation

1. Jan 18, 2016

### sandy stone

I am presently slogging through an introductory text on QFT (a Christmas present - does that officially make me a nerd?) and I have gotten as far as the Dyson expansion of the S operator and Wick's Theorem. The author dismisses all first-order terms of the expansion as being represented by single-vertex Feynman diagrams, which are unphysical. As an example, he uses the interaction where an electron and positron enter and a photon leaves, which is not possible because a real photon cannot carry away the 4-momentum the massive particles entered with. However, I am under the impression that an electron and positron annihilating to produce a high-energy photon is an experimentally observed interaction. Where am I going wrong?

2. Jan 18, 2016

### Staff: Mentor

Annihilation produces TWO high-energy photons.

3. Jan 18, 2016

### A. Neumaier

Hence the corresponding tree diagram has two vertices joined by a fermion line, each with two external legs (a fermion and a photon).

4. Jan 18, 2016

### sandy stone

Ohhh... OK, that wasn't too hard. Thanks. On a somewhat related note, I understand that QFT (QED) has made the most precise experimental predictions in history, but isn't it somewhat unsettling that after a thorough mathematical analysis you have to just ignore all your first-order expansion terms because they don't apply to reality?

5. Jan 18, 2016

### A. Neumaier

This is just the simplest instance of a complicated process called renormalization that in fact does much more - it also shrinks all subdiagrams that are connected to a bigger diagram by exactly two (external or internal) lines of the same kind to a single line of this kind, no matter how complicated the mess in between. The result is that one is left with 1PI (1-particle-irreducible) diagrams with more than one vertex - these contain the real physics. Everything else is only scaffolding needed to set up the perturbative formalism in a consistent way.

Last edited: Jan 18, 2016
6. Jan 18, 2016

### A. Neumaier

In QED, their contribution to scattering processes is in fact exactly zero: External legs must correspond to on-shell particles with the correct mass. But a straightforward calculation shows (that you should do yourself - it is really instructive. Doing the same for the above tree diagram will reveal the difference!) that there is no way to assign momenta to the legs such that total momentum is conserved. Therefore the delta function in the corresponding integral forces the value of the integral to zero.

So only zeros are thrown away, which doesn't change any result but simplifies the analysis.

Full renormalization, on the other hand, achieves a miracle since it makes all infinities disappear. If you are interested in understanding why the miracle (stated everywhere, but hardly ever explained in an understandable way) is not just a happy accident, read my tutorial paper on renormalization!

Last edited: Jan 18, 2016