If a Feynman diagram only involves bosons, then can you multiply all external 4-momenta by negative one, and still get the same amplitude? Looking at the Feynman rules this seems true.(adsbygoogle = window.adsbygoogle || []).push({});

If the diagram has a fermion (as real or virtual particle), then it seems this is no longer true.

I just want to verify this. There probably should be a deeper reason for this, but just looking at the Feynman rules, it seems it ought to be true when only bosons are involved, and not true when at least one fermion is involved.

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# Changing the sign of 4-momenta in feynman diagrams

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