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For example in electron, positron annihilation $e^- + e^+ \rightarrow 2\gamma$ has 2 diagrams, whereas, $e^- + e^+ \rightarrow 3\gamma$ has 6 possible diagrams. This suggests,

$\frac{\sigma_{2\gamma}}{\sigma_{3\gamma}}= \frac{2\alpha^2}{6\alpha^3}$

where the $\sigma$ is the cross section and $\alpha$ is the fine structure constant. The indices on $\alpha$ come from the different orders of the processes ($e^- + e^+ \rightarrow 2\gamma$ is order 2 so $\sigma_{2\gamma} \propto \alpha^2$ and $e^- + e^+ \rightarrow 3\gamma$ is order 3 so $\sigma_{3\gamma} \propto \alpha^3$).

Is the equation above correct? Are there caveats for other processes?

thodic
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1 Answers1

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No. Feynman diagram calculations are much more complicated; the $\alpha^n$ scaling is just one piece.

For example, consider the amplitude for $N \gamma \to (N+1) \gamma$, where $N$ photons interact to turn into $N+1$ photons. The lowest-order Feynman diagrams contain one electron loop connecting all the photons together, so the cross section, according to your heuristic, is something like $\alpha^{2N+1} (2N)!$, since there are $(2N)!$ distinct diagrams.

Except, that's not right: the cross section is zero by Furry's theorem. The amplitude for each diagram cancels with that of the diagram with the electron loop reversed.

There are many more examples where naive counting fails, but this might be one of the most dramatic ones.

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knzhou
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