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Feynman diagrams show a very definite process with these particles coming in, this interaction occurring, and those particles going out. Given that all of this describes waves in a field would there be other, unimportant virtual particles created that are not shown in the diagram that just dissipate away - essentially hairs in the diagrams? Would it even be possible to predict these hairs?

Qmechanic
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foolishmuse
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  • A Feynman diagram does not describe a "very definite process", see e.g. https://physics.stackexchange.com/q/297004/50583, https://physics.stackexchange.com/q/230113/50583. – ACuriousMind Oct 04 '22 at 12:12

2 Answers2

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The probability amplitude for a process is computed$^\dagger$ by summing all possible Feynman diagrams, given the initial and final states. So a Feynman diagram does not describe a single, specific way that the process occurred. I think of it as one possible way the process could happen, and due to quantum weirdness all possible ways the process could happen contribute to the final answer.

$^\dagger$ At least, perturbatively, although there can also be non-perturbative effects not captured by Feynman diagrams.

Andrew
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  • Is it your coda sentence about "perturbatively" that I am asking about? Non-important waves that do not impact on the final state, but exist nonetheless. – foolishmuse Oct 04 '22 at 12:01
  • @foolishmuse No, I would say the "non-important waves" would correspond to complicated diagrams that do not contribute much to the final answer. For example, in electron-electron scattering, you have one diagram where two electrons come in, annihilate to form a photon, which then breaks into two electrons -- this is a "simple" process that dominates the final answer. But there are also diagrams where two electrons come in, form a photon, which breaks into two electrons, and which break into a photon and positron... and eventually recombine back into two electrons. – Andrew Oct 04 '22 at 13:09
  • The "non-perturbative" caveat is more like, "processes where the quantum field does not behave like a wave" (or more precisely, processes that can't be understood as a small perturbation of a free field, which describes waves). For example, you can have tunneling between different vacuum states. To zeroth order, you can ignore those processes, the first step to understanding is getting the idea of how many Feynman diagrams sum up to form the amplitude for a process. – Andrew Oct 04 '22 at 13:10
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A Feynman diagram represents one particular possible path between incoming and outgoing particles. But in general there will be many (often an infinite number) of possible paths between the same inputs and outputs. The trick is to find a way to add up the contribution from each of these paths to get a probability amplitude for the overall interaction. Usually the sum cannot be calculated exactly and must be approximated.

gandalf61
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