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I am having some trouble understanding this Feynman diagram, it seems to indicate that the electron produces the positron, as the arrow of the positron is pointing from the electron.

Additionally the arrow is in directed downwards, implying that the particle is going backwards in time? Is this diagram wrong, or does the arrow mean something else, or does the positron go back in time?

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    $\begingroup$ Ahrrgggg! Goofy cosmology convention for time and space! Real physicists know that time increases to the right, damnit! $\endgroup$ Commented Nov 27, 2011 at 19:37
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    $\begingroup$ don't listen to dmckee. Everyone knows that time is the queen of the dimensions and deserves the vertical slot (and the plus sign in the signature) ! $\endgroup$
    – twistor59
    Commented Nov 27, 2011 at 19:43
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    $\begingroup$ Oh by the way, related question $\endgroup$
    – twistor59
    Commented Nov 27, 2011 at 19:44
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    $\begingroup$ @Jonathan: I tend to just look to see if the arrow on the line is pointing in the opposite direction to time flow, if yes, I assume it's a positron. $\endgroup$
    – twistor59
    Commented Nov 27, 2011 at 20:53
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    $\begingroup$ @dmckee and twistor59 are both wrong! Time must go from right to left! That is how it is done in a great book by Itzykson and Zuber! $\endgroup$
    – Kostya
    Commented Nov 29, 2011 at 9:56

5 Answers 5

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The first thing to notice, as pointed out in the comments, is that time increases going up. So if you are more familiar with viewing Feynman diagrams where time increases to the right, this problem is easily solved: just rotate the diagram by 90 degrees when you are interpreting it.

If the problem is that you're not all that familiar with matter lines in Feynman diagrams in general, here's the rule:

  • If the arrow points in the direction of increasing time, it's a matter particle (in this case, an electron).
  • If the arrow points in the direction of decreasing time, it's an antimatter particle (in this case, a positron).

That's it. The arrow on the line doesn't indicate the direction in which anything is actually moving.* It's simply a convention to show whether the particle is matter or antimatter.

This rule is convenient because, if the diagram is correctly drawn, you can trace along a lepton (or quark) line from one endpoint to another, always going with the arrows.

Note that sometimes you'll encounter an arrow which points in the spatial direction, perpendicular to the time axis, such as the horizontal line segment in the middle of this diagram. Don't take that segment too literally. It doesn't mean that the electron actually moves in a spacelike direction. It's just "code" for whatever (leading order) processes actually do occur in which the electron and positron annihilate with each other while emitting two photons in the process.

--

*Actually, it does indicate the orientation of the current of the corresponding field, but for purposes of interpreting the diagram, I'd suggest you just consider that a technical detail of the QFT calculations and not worry about it.

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It is more than curious that to preserve O(3,1) symmetry in the Maxwell equations both charge conjugation and time reversal must both occur. In that sense the Feynman diagram is not a mere visual aid, but accurately reflects the physics. Under normal conditions that implies that the positron is going backwards in time, BUT we must always remember that spacetime cannot be subjected to anthropomorphic visualisation in which man's universal time plays the chief role. There are 4 dimensions, but what is time and what is space are human conventions; Nature is ambivalent over what we call timelike and spacelike intervals (See Misner, Thorne, Wheeler, p. 409, Gravitation). Coordinates are the work of man, not Nature.

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The positron is the electron going back in time. This is what the arrow means. But you have to remember that in physics, cause and effect are not by the direction of the arrows, but by the direction of entropy increase, which is up in the diagram regardless of which way the particles go relative to their own proper time.

The idea that positrons are electrons going back in time is due to Ernest Stueckelberg, but it was misattributed to Feynman, who got it indirectly from Stueckelberg through Wheeler. The idea is correct, and it solves the problem of creating a relativistic particle formalism. But it requires that the particle trajectories are not causal, they either produce an S-matrix, or correlation functions. The notion of causality is only for a higher level description.

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    $\begingroup$ If the positron were going back in time, it wouldn't be a positron :-P $\endgroup$
    – David Z
    Commented Nov 28, 2011 at 3:05
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The above diagram is wrong. The correct diagram is not a traditional qed diagram. First this is the most profound inelastic event. Two fermions in, generally two out with 2 photon lines out and connected by photon exchange, not an electron propagator. You'll never get an amplitude or cross section from that diagram. Remember, this is inelastic "scattering." The outgoing Fock space must have 2 photons and 2 fermions, the incident Fock space has 2 vacuum states for photons and 2 incident Fermion momenta. You should find the amplitude of annihilation is infinite when there are no outgoing fermions.

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    $\begingroup$ The question is about QED, not about some other, possibly unknown, theory. $\endgroup$
    – my2cts
    Commented Aug 11 at 15:01
  • $\begingroup$ You will not make sense of annihilation with that diagram. The correct diagram looks much like Klein-Nishina ... it is all qed tho .. $\endgroup$
    – LiveProton
    Commented Aug 12 at 17:59
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Think of this as Bahbha scattering with two photons on the outgoing fermion lines. It is a radiative prosess

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