Accelerating charge generates electromagnetic waves and loses energy, in QFT terms it emits photons that carry it away. What of a static charge? Moving photons are usually associated with waves, which do not seem to be there in a static field, and static charge has no energy to lose. However, in the classical picture it fills the space with something, and another charge "feels" it when placed into it. So what exactly does it interact with locally?

  • $\begingroup$ You first have to recover the Coulomb potential from the tree-level QFT amplitude as in this question, and then you take usual QM for the Coulomb potential and go to the classical limit to get classical electromagnetism. (So the answer is: "Take the classical limit", as always) I'm not sure if that's your question, though. $\endgroup$ – ACuriousMind Oct 22 '15 at 20:50
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    $\begingroup$ @ACuriousMind I guess my question is what the static charge exchanges photons with when it's alone. It seems it can't just emit them away since no waves are propagating. I can see how exchange of photons between two charges creates attractive/repulsive force, but when there is only one how does the static field emerge from photons? $\endgroup$ – Conifold Oct 22 '15 at 21:16
  • $\begingroup$ Well, when the universe is empty except for the field of one lone static charge, then how do you know the electric field is there? After all, the electric field is operationally defined in terms of the force on another test charge. $\endgroup$ – ACuriousMind Oct 22 '15 at 21:20
  • $\begingroup$ @ACuriousMind People talk about creation/annihilation of virtual pairs creating vacuum fluctuations in "empty" space, so I thought there was some such picture for a static field when a lone charge is added. Also the test charge seems to "feel" the field instantly (maybe not?), whereas time is needed for it to exchange a photon with the original charge. $\endgroup$ – Conifold Oct 22 '15 at 21:30
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    $\begingroup$ The ontological status of such "virtual pairs" is highly debatable. The standard QFT formalism does not assign particle states to virtual particles, they are just lines in a Feynman diagram. If we were able to compute everything non-perturbatively, the notion of "virtual particle" would never have appeared. You should not trust any physical argument that relies on virtual particles having properties of actual particles. $\endgroup$ – ACuriousMind Oct 22 '15 at 21:35

I would have just made a comment, but I'm not allowed yet--

It sounds like your more general underlying question is: "How do we think about classical electromagnetic fields from the perspective of QED?" In particular, which configuration of quantum fields in QED do we associate with classical fields? How do we derive Maxwell's equations for these classical fields, using the QED equations of motion for the corresponding quantum fields? How do we derive the Lorentz force law describing the force on electrons?

The basic answer is that classical fields are best represented by coherent states in QED. These states are an infinite superposition of photon number states -- that is, there's nonzero probability of finding any number of photons. (In particular, probabilities for different photon numbers obey Poisson statistics.) The expected value of the electromagnetic field operators for these states corresponds to what we'd classically call the values of the electromagnetic field, and these expected values obey classical Maxwell equations.

The best notes I've found going through these questions might be: http://hitoshi.berkeley.edu/221b/photons.pdf.

You might also look at related SE questions What is the quantum state of a static electric field? and Given expectation values for E and B, can you find an associated state?.

A more direct answer to your immediate question: the classical field produced by a static charge corresponds to a quantum field (a coherent state) with high photon number. You point out that single photons are associated with traveling waves, yet somehow we have a static field. There's no contradiction because the coherent state is a superposition of many such traveling waves and thus has different behavior. Another charge placed nearby will interact with this photon field.

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  • $\begingroup$ I am glad you wrote an answer rather than a comment, it's a very nice answer. I'll wait a little for people to weigh in before accepting. $\endgroup$ – Conifold Oct 23 '15 at 0:07

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