# What tree-level Feynman diagrams are added to QED if magnetic monopoles exist?

Are the added diagrams the same as for the $e-\gamma$ interaction, but with "$e$" replaced by "monopole"? If so, is the force between two magnetic monopoles described by the same virtual $\gamma$-exchange diagrams? I'm anticipating that the answer is 'no', because otherwise I don't see how one could tell magnetic monopoles and electrons apart (besides their mass and coupling strength to the photon). Obviously magnetic monopoles behave differently from electrons when placed in an $E$ or $B$ field.

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This is a very interesting question, but unfortunately, the electron is perturbative precisely when the monopole is not, since the couplings are inverse to each other. So if you do a photon description of the electron perturbation, the monopole perturbation is strong-coupling. This was a subject of several Schwinger papers in the 1960s, although I am not sure what the conclusion is. –  Ron Maimon Jun 19 '12 at 21:04

In fact, the situation for an abelian $U(1)$ gauge theory—which is the case you asked about—is a bit less clear and less well-defined than the case of a non-abelian gauge theory. Think about the running of the coupling constant, for example.

In a non-abelian theory with a Higgs field, one can have classical solutions which look like monopoles, i.e. they create magnetic flux through a sphere at infinity. Nevertheless, they are perfectly non-singular classical solutions, which almost certainly survive in the quantum theory. In a sense, they are composite, that is they are built out of fundamental fields like the gauge fields and the scalars.

From this, you can conclude that when summing up Feynman diagrams you should not include the monopoles as extra degrees of freedom. Rather, their effect should appear after resuming the entire perturbation series. If you truncate the perturbation series to any finite order, you will not capture the presence of the magnetic monopoles.

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Well you're essentially correct actually. If you didn't know the value of the coupling constant then electric and magnetic monopoles would be indistinguishable. It's actually just a matter of convention that we call the light one the electric monopole and the other one the magnetic monopole which is so heavy we don't even need to put it in our equations! This is because the probability that the vacuum could spontaneously create a magnetic monopole is enormously small the more massive it is. But the above commenter is correct you could try and draw a bunch of feynman diagrams but they would be useless once you found out g>1 since the series wouldn't converge however it would be still basically correct to think of the magnetic monopole and photon interactions in the way the diagrams describe.

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But as I pointed out in my question, if the mass and coupling are the same, they would be distinguishable: the electron responds to magnetic fields only when moving, whereas the magnetic monopole responds to magnetic fields when stationary. –  user1247 Jun 20 '12 at 9:26
Correct but you're focusing not he particular electron too much. If you decided to make a copy of the (observable) universe with 1/g (large coupling constant instead of small) then your universe would look just like this one. The only difference is if I took one of the objects you recognized as an electron back to my universe i'd be amazed to find out it's actually an electric monopole. (Ignoring the complications of traveling between hypothetical universes). You are correct about the specifics but what we're really saying is that in the big picture we're blind to the change g->1/g. –  tachyonicbrane Aug 28 '12 at 15:49