I have been introduced to the Feynman diagrams in QFT after following Wick-Dyson tedious formalism.

Two things are unclear, though, about the Feynman shortcut to compute scattering amplitudes.

What are the horizontal and vertical axes in these diagrams? Are they $x$ and $t$ respectively or $x$ and $y$?

If yes then we are drawing a particle with an exact momentun $p$.

But doesn't this violate the uncertainty principle since we are assuming exactly measured $x$ and $p$?

  • $\begingroup$ At each vertex of Feynman diagrams momentum conservation is ensured by Dirac delta function. $\endgroup$
    – Mass
    Dec 6, 2016 at 17:37
  • $\begingroup$ I mean consider an external leg in the Feynman diagram. Are both momentum and position exactly measured? $\endgroup$
    – user56963
    Dec 6, 2016 at 17:39
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    $\begingroup$ Normally, the external legs represent momentum states of the particles, so their position is undetermined. As @AccidentalFourierTransform points out in their answer, position on a Feynman diagram does not represent position or time. The graph connections are what matter - not the locations of the nodes. $\endgroup$
    – apdnu
    Dec 6, 2016 at 20:42

2 Answers 2


There are no axes in Feynman diagrams. The only important part of a diagram is what is connected to what, and not the relative orientation. You can move around the pieces of a diagram and, as long as you don't break any line, the value of the diagram remains unchanged.

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    $\begingroup$ This. If one wants a formal statement, a Feynman diagram is an isomorphism class of a certain kind of graphs, and these graphs are not to be thought to be embedded in anything. $\endgroup$
    – ACuriousMind
    Dec 6, 2016 at 17:40
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    $\begingroup$ @ACuriousMind the problem is, many introductions to QFT insist in "time flowing from left to right" in Feynman diagrams, which is of no help, not even to a beginner. I have no idea why people perpetuate this. $\endgroup$ Dec 6, 2016 at 17:45
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    $\begingroup$ @VictorVMotti see my comment above: Wikipedia, as many other informal sources (e.g., many professors), insist in the lie to the children that a Feynman diagram represents a physical process. You will have people tell you that time flows from top to bottom or statements of the sort. But you will know better: that is false. A Feynman diagram is a device to organise Wick contractions, and nothing else. $\endgroup$ Dec 6, 2016 at 17:48
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    $\begingroup$ @VictorVMotti I would call it a mnemonic rule. Nothing really moves to the past, but thinking that an antiparticle behaves as it were a particle moving backwards in time helps to remember some properties of particles and antiparticles, such as that they have opposite charge, opposite helicity, they are related by the time reversal operator, etc. $\endgroup$ Dec 6, 2016 at 18:05
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    $\begingroup$ @EmilioPisanty quite the contrary: insisting in pretending that there is a time direction in the diagram doesn't help. I guess we are gonna have to agree to disagree :-) I not only imply that the time direction of external legs has no meaning, but I reaffirm it. I hope one day people will stop using a time arrow in Feynman diagrams. There are different schools of thoughts about this, and maybe in the end your interpretation is more useful. But I am quite convinced that it is not... $\endgroup$ Dec 6, 2016 at 20:30

You have to realize that the Feynman diagram is a schematic representation of the integrals of the scattering amplitude , necessary to calculate the integral for that order. All variables will be integrated within the limits of the integral, so it is not a real space and time but an iconic one seen in a Feynman diagram. The result is a crossection, or a lifetime for that interaction, or an angular distribution. Quantum mechanical probabilities are all there.

  • $\begingroup$ Yes I see. But the diagrams looks like they are showing the scattering process in time at least. Right? $\endgroup$
    – user56963
    Dec 6, 2016 at 17:48
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    $\begingroup$ time and space on the diagram are iconal, standing in for the time and space in the integrals which are variables $\endgroup$
    – anna v
    Dec 6, 2016 at 19:01

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