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Obviously the machinery of QFT allows us to calculate processes, such as QED diagrams, to great precision, and whilst it is effective, it seems there are many processes that make calculations (say by hand) significantly slow.

Are there any recent developments in our machinery to compute Feynman diagrams that makes it faster to analytically compute matrix elements, widths and cross sections?

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There are a number of computer algebra systems for evaluating Feynman diagrams and doing other particle physics calculations, such as FeynCalc, FORM, GiNAC, Package-X, etc.

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  • $\begingroup$ Sorry for the misunderstanding, those packages I'm aware of, I meant by hand. Are the feyman rules really the fastest way we have to compute diagrams? $\endgroup$ – MKF Feb 9 at 20:40
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    $\begingroup$ As far as I know, the Feynman rules are the fastest way to compute amplitudes for a general QFT. In some specific supersymmetric theories, new methods like the amplituhedron are being explored. I don’t understand the relationship between that method and the Feynman diagram method. $\endgroup$ – G. Smith Feb 9 at 22:04
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Nowadays, almost nobody uses Feynman diagrams for precision calculations (that is, anything beyond a tree level 4 or 5 point amplitude). There is a whole field dedicated to finding better methods of calculating scattering amplitudes (using recursion and unitarity for example). In fact this is how most calculations for the LHC are done in practice.

A nice review of some of the techniques is https://arxiv.org/abs/1308.1697. Most of the work is motivated by the observation that scattering amplitudes are much simpler than one could have thought by just staring at the Feynman rules. An example of this is the famous Parke-Taylor amplitude https://en.wikipedia.org/wiki/MHV_amplitudes. Take a look and compare to the QCD Feynman rules.

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  • $\begingroup$ Please don't link to the PDF on arxiv, link to the abstract page (consider mobile users). $\endgroup$ – Kyle Kanos Feb 10 at 3:24
  • $\begingroup$ The first sentence of this answer is simply false. $\endgroup$ – QuantumDot Sep 6 at 17:43

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