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?

  • 1
    $\begingroup$ Not quite. Sometimes they cancel out. $\endgroup$ – knzhou May 3 '16 at 18:36
  • 1
    $\begingroup$ 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. $\endgroup$ – AccidentalFourierTransform May 3 '16 at 18:40
  • $\begingroup$ @knzhou Could you elaborate? Do they cancel out in electron/positron annihilation? $\endgroup$ – thodic May 3 '16 at 18:40
  • $\begingroup$ @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 $k$? $\endgroup$ – thodic May 3 '16 at 18:52
  • $\begingroup$ @knzhou That should be an answer $\endgroup$ – 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.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.