# For processes with multiple Feynman diagrams is the cross section increased?

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?

• Not quite. Sometimes they cancel out. – knzhou May 3 '16 at 18:36
• Was it Dyson who observed that the $n$-loop amplitude of a process scales roughly like $n!$, indicating the fact that the series are always asymptotic? or was it Schwinger? anyway, that equation is in general wrong. It would look better if you wrote $\sim$ instead of $=$, but still, in general is wrong. – AccidentalFourierTransform May 3 '16 at 18:40
• @knzhou Could you elaborate? Do they cancel out in electron/positron annihilation? – thodic May 3 '16 at 18:40
• @AccidentalFourierTransform Are you discussing how the number of possible diagrams scale with the number of indices? Because in the above example the $2\gamma$ annihilation has 2 indices and $2! = 2$ diagrams and the $3\gamma$ annihilation has 3 indices and $3! = 6$ diagrams. My understanding is that for an $n$-order EM process the cross section is proportional to $\alpha^n$, so if a process has $k$ possible $n$-order diagrams does the cross section not scale proportionally with $k$? – thodic May 3 '16 at 18:52
• @knzhou That should be an answer – David Z May 3 '16 at 18:53

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.