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In QED, photon number is not conserved and photons can produce electron-positron pairs in the vacuum.

But if we take away electromagnetism and have a pure electron theory. Does this still produce electron-positron pairs in the vacuum?

i.e. starting with a single electron, in the free theory is there zero chance of detecting two electrons and a positron at a later time? There is no Feynman diagram that can do this....

On the other hand I hear a lot about electron-poistron pairs being created out of the vacuum. I can understand this would happen near a black hole if a high energy graviton decayed into an electron-posistron pair.

In the free theory even if electron-posistron pairs were created and destroyed would they have any impact if they can't interact with ordinary electrons via electromagnetism?

Any ideas?

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    $\begingroup$ In free theories there is no creation nor annihilation of anything, precisely because they are free. I don't understand the question. In pure electron theories, the leading interactions are given by non-renormalisable terms a la Fermi. Is this what you had in mind? $\endgroup$ – AccidentalFourierTransform Aug 22 '17 at 16:48
  • $\begingroup$ Why did you not put this as an answer if this is the answer? $\endgroup$ – zooby Aug 22 '17 at 16:54
  • $\begingroup$ I did not put that as an answer because it is not an answer: I'm asking for clarification, mostly because I don't really understand the question. $\endgroup$ – AccidentalFourierTransform Aug 22 '17 at 16:55
  • $\begingroup$ If I understood what I was saying I wouldn't need an answer.... $\endgroup$ – zooby Aug 22 '17 at 17:17
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If you forget about the $U(1)$ gauge group of QED you can certainly take a theory of fermions (e.g. electrons) and add all sorts of coupling terms (so long as these are sensible terms respecting Poincaré and global $U(1)$) that will give you vacuum graphs and pair creation-annihilation. For example, you could have a quartic coupling of the form $(\bar{\psi}\psi)^2$ in your theory of electrons and positrons. In a free theory, there would be no interactions (by definition), and no pair-creation or annihilation.

EDIT: By the way, I think you are misunderstanding what "free" means. A QFT is free if it is at most quadratic in its fields (such that equations of motion are linear), with no two different fields coupled together. Such theory only has kinetic and mass terms. You can have a theory without photons which is still not a free theory (e.g. fermions with a quartic vertex like the one above).

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