# Dipole moments in QFT

We say that an electron has a dipole moment (let's arbitrarily focus on magnetic dipole moment), which we can calculate classically and also add quantum corrections. Suppose we measure the electron's spin projection. The way we interpret the setup is that we can now calculate the magnetic field generated by the dipole moment.

Does this picture still make sense in QFT? I'm used to think about these things in terms of interactions, Feynman diagrams etc. and less in terms of field configurations. Here's my concrete question:

Will the presence of some exictation in the electron field (i.e. particle), having measured its spin projection, induce a well determined photon field strength?

Somehow I'm less inclined to think that, and more inclined into thinking that you will only increase the probability of interacting with the photon field in a particular way and perhaps emit photon in particular directions.

• It will make sense to think about the magnetic dipole as an intrinsic property and the spin als a consequence. If you are new enough to physics study may be you are able to see the advantages: the explanation for the gyroscope effect, the orientations of the electrons around the nucleus, the reason for the Lorentz force. The magnetic dipole and the spin are always parallel or antiparallel. In a way they are synonyms. – HolgerFiedler Jul 18 at 4:14
• Too bad I can't downvote comments – octonion Jul 22 at 20:27

Yes, there should be a measurable magnetic field. Consider the expectation value of the magnetic field between one-electron states (with some given spin and momentum) $$\langle e^-|\vec{q}\times\vec{A}(q)|e^-\rangle$$. Through the LSZ formula this is related to the vacuum expectation value of the $$A$$ field and two electron fields.