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Why leptons annihilation is not 4-particles, and it is represented in the form of two 3-particles? If the pair of an electron and a positron has formed a positronium particle, then what is virtual state before the formation of two photons? What is located on the Feynman diagram between the photons emission points by the electron and the positron? enter image description here

Why is the formation of a virtual lepton with zero mass not shown for each photon emission at above picture from https://en.wikipedia.org/wiki/Annihilation? enter image description here

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    $\begingroup$ I don't think that your attempt at drawing the Feynman diagram as a process in the spacetime is very meaningful $\endgroup$
    – OON
    Commented Feb 9, 2022 at 15:58
  • $\begingroup$ The first diagram is from en.wikipedia.org/wiki/Annihilation. What should be for more meaningful? $\endgroup$
    – Imyaf
    Commented Feb 9, 2022 at 16:08
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    $\begingroup$ well, if it is in wikipedia it does not mean it's correct. The diagram is fine. The space and time axes are not $\endgroup$
    – OON
    Commented Feb 9, 2022 at 16:38
  • $\begingroup$ At the beginning there are 2 leptons at the bottom. Then there are 2 photons above. Therefore, time from bottom to top. Why are axes wrong? $\endgroup$
    – Imyaf
    Commented Feb 9, 2022 at 16:58
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    $\begingroup$ See also physics.stackexchange.com/q/297004/50583, physics.stackexchange.com/q/279315/50583 for more discussion of the meaning of internal lines "constant in time". $\endgroup$
    – ACuriousMind ♦
    Commented Feb 10, 2022 at 13:42

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In fact, the first Feynman diagram in your question is the sum of two time-ordered diagrams. One in which the electron sends out a real photon and later in time meets the positron at the vertex to annihilate with it and emit another photon.
The other time ordered diagram presents a positron that first emits a photon, travels on to meet the electron, and annihilates with it at the vertex to give the other photon.
The Feynman diagram comprises both diagrams at once, in a symmetric way.

So, in fact, there are 4 particles involved. Two in each time-ordered diagram. But the process itself of course involves two particles only.

By the way, space and time can have both the axes. If space were vertical and time horizontal, you had photon scattering!

Time ordered diagram:

enter image description here

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  • $\begingroup$ Electron and positron sends out real photons simultaneously and turn into virtual electron and positron with zero mass? $\endgroup$
    – Imyaf
    Commented Feb 10, 2022 at 5:50
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    $\begingroup$ While a popular misconception, this isn't how Feynman diagrams work on a technical level. There is no "time-order" in them (they're mathematically just graphs where the external lines are label as "input" or "output"), the diagrams represent terms in a perturbation series, not the actual sequence of a process (so drawing space and time axes doesn't make a lot of sense), and the term that the diagram in the question represents is not "the sum" of the two you claim. $\endgroup$
    – ACuriousMind ♦
    Commented Feb 10, 2022 at 13:37
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    $\begingroup$ @ACuriousMind Then what are the space and time axes for? I mean, the two time-ordered diagrams are both comprised by the Feynman diagram. $\endgroup$
    – MatterGauge
    Commented Feb 10, 2022 at 13:57
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    $\begingroup$ The space and time axes are nonsense, cf physics.stackexchange.com/a/297006/50583. $\endgroup$
    – ACuriousMind ♦
    Commented Feb 10, 2022 at 14:35

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