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I came across this question here which was asking my exact initial question (I even came from Griffiths problem 2.4 too!). But the given answer gave me a lot more questions which I think constitute too much content to fit in a comment, so I hope this warrants its own question here.

To summarize, we have Bhabha scattering (with time flowing left to right):

Bhabha scattering.

As the accepted answer puts it, we have energy and momentum conservation at each vertex, as well as overall. Thus for each vertex we have $E_{e,\text{in}} + E_\gamma = E_{e,\text{out}}$, and since $E_{e,\text{in}} = E_{e,\text{out}} = m_e$ (in natural units), we have $E_\gamma = 0$.

My primary question is: how do you determine the direction of the photon (alternatively, which vertex "happens first")? The answer on the linked question seems to imply they happen simultaneously; but in that case, how do you know which side of the energy conservation equation the $E_\gamma$ goes on? In this case, it of course doesn't matter either way since $E_\gamma$ is 0, but I could imagine interactions where this does matter.

Also, were it not for total energy conservation (say we had the identical diagram for the invalid "scattering" $\mu^-\mu^+\to e^-e^+$), I assume the photon in fact could take on energy. But if one vertex happens first, could the the photon not take on a variable amount of the excess energy $m_\mu - m_e$, with momentum conservation then being ensured by the second vertex? I have only ever seen variable energy to be integrated over in diagrams with loops, so it feels like this intuition is wrong.

Hopefully this isn't too many small little questions; help is appreciated.

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Feynman diagrams are a diagrammatic representation of the terms needed to calculate interactions of elementary particles. What is really measurable in experiments are the four vectors inputed, and the four vectors going out. The drawing and labeling the internal line as a photon is only done for a mnemonic representation of the quantum numbers exchanged during the interaction. The four vector of the line is "virtual". Example

virtual

This means that the four vector of the "virtual particle" does not have its mass, it is off mass shell and within the complexities of the diagram can have any value that is allowed by the conservation laws of energy and momentum. The four vectors of real particles have as length the mass of the particle.

how do you determine the direction of the photon (alternatively, which vertex "happens first")?

In Feynman diagrams the axis of the variables are defined. In the Bhabha case the horizontal axis implies the same time for the two vertices. Considering that experimentally we do not see first one and then the other real particle with a time delay, the time axis in the diagram is just a coordinate confirming this.

But if one vertex happens first, could the the photon not take on a variable amount of the excess energy mμ−me, with momentum conservation then being ensured by the second vertex?

As I state above, all experiments show simultaneous time for the exit particles. Energy can be carried by the virtual particle, example:

In Compton scattering the virtual particle is a virtual electron, ( see diagram here )its four vector will be off mass shell, but there is energy transferred in the interaction.

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  • $\begingroup$ Thanks! I have a follow-up/clarification question: when I asked which vertex "happens" first, I understood that none of this stuff is truly "happening" at distinct times (at least not measurably so), but it seems to me that there might be quantitative differences depending on whether the photon is drawn to be "emitted" by the electron and "absorbed" by the positron or vice versa. Namely, I could imagine another interaction where the photon does not have zero energy, in which case it matters in that energy conservation eq. which side the photon energy is on. Would this be 2 different diagrams? $\endgroup$
    – Vedvart1
    May 23, 2022 at 21:57
  • $\begingroup$ This is the state of the mainstream Feynman model QFT calculational theory. There could be other mathematical theories that might fit the specific BhaBha data that in their mathematical representation have no virtual particles, or behave as you suggest. These models would fail to fit the electromagnetic data that the mainstream QFT fits, which only has interactions and not absorptions and emissions, because no time factor is seen in the data, the output particles. $\endgroup$
    – anna v
    May 24, 2022 at 4:04

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