# I QFT Interpretation of Electron and Attached EM Field

1. Dec 7, 2018

### CSnowden

This is an elementary question on visualizing the interaction of an electron with the surrounding EM field in QFT. I believe in QFT the electron is viewed as an excitation of the Electron matter field with an associated coupling constant between the electron field and EM field (say q) - q 'couples' the two fields and allows the electron to feel a force from the local value of the extended (free) EM field. But in classical physics many electrons together would be said to generate a higher surrounding EM field value, presumably by the combination of the attached EM field each generates. In QFT does the q coupling from each electron also raise the local EM field value, in addition to allowing the the electron (electron field excitation) to 'feel' the force from the EM field.

2. Dec 8, 2018

### A. Neumaier

Yes. Each electron is accompanied by an electromagnetic field, which is additive and approximately of Coulomb form.

3. Dec 8, 2018

### CSnowden

That's very helpful, thanks. Just to clarify, in the use of the term 'attached EM field of an electron' does this refer to a elevation of the energy level of the local EM field neighboring the electron, ie. is there truly just one pervasive EM field and not separate 'attached' fields in addition. Are the electron and EM quantum field equations asymmetrical in some respect so that electron field quantum (local field excitation) couples to raise the local EM field energy level but the reverse does not happen (ie. an EM quantum like a photon does not raise the energy level of the surrounding electron field)? Or perhaps this is intrinsic to the nature of the fields themselves, with one a field of spinor values and the other a field of vector values?

4. Dec 8, 2018

### king vitamin

If I understand your question correctly, the answer is that there is symmetry - just as the interaction causes a change in the local properties of electrons, the interaction also causes a change in the local property of photons. For example, if $g=0$, photons will never interact - they simply pass through each other. In contrast, the simple fact that $g\neq 0$ means that actually two photons will scatter off of each other, even if there is no electron (excitation of the electron field) in the universe.

In an interacting QFT, the general rule is that everything affects everything; it is a true many-body problem. For example, there is no coupling constant between the electron and the muon, but since they both couple to the photon, the existence of one has an effect on the other. (Since the electron is lighter, is has a larger effect on the muon than vice-versa.)

5. Dec 9, 2018

### A. Neumaier

No, just the opposite. Since the electron is lighter, is has a smaller effect on the muon than vice-versa.
There is just one electromagnetic field, just as in classical mechanics there is only one Hamiltonian. But the electromagnetic field can be decomposed into a sum of pieces coming from different sources and the remainder, just as a Hamiltonian can be decomposed into a sum of energy contributions from each particle in the noninteracting case, and particle pairs, triples, etc., in case of interactions.

6. Dec 9, 2018

### CSnowden

Thanks to all, this has been very helpful!

7. Dec 9, 2018

### king vitamin

It would be a horrible thing if physics worked this way. We would need to know about all of the most massive particles (up to the Planck scale and beyond) in order to calculate the properties of the electron!

But it's not true. Schwinger only needed to consider the electron to compute g-2 to leading order, and contributions due to muon/tauon loops are much smaller (though still needed to match the extraordinary experimental precision we have). Similarly, since the electron strongly dominates over (for example) the leading correction to the photon field polarization, it dominates over certain higher-order contributions of g-2 of the muon compared to similar loops with virtual muons and tauons. (What I mean is that removing the electron field from the theory would have a bigger effect on the muon's g-2 than the corresponding thought experiment of the effect of removing the muon to the electron's g-2.) This is because propagators come with factors of the mass of the particle in the denominator, so heavier particles always contribute less in perturbation theory (and I think one can use RG arguments to make this statement nonperturbative, though this isn't needed for QED).

In fact, this is why the discrepancies in the muon's g-2 (compared to the electron) are considered predictors for more massive particles somewhere beyond the Standard Model. Corrections to this physical value due to some unobserved more massive particle with mass $m_X$ scale like $m_{\mu}/m_{X}$, so it makes sense to see the discrepancy in the more massive muon rather than the electron where this correction is smaller. (For a reference, see for example the first section of this review article: https://arxiv.org/abs/hep-ph/0703125v3).

8. Dec 10, 2018

### A. Neumaier

Ah, your statements are about virtual processes involving the electron and the muon, where indeed the most massive particles have the least influence.
But mine was about real electrons or muons, and a real muon affects the dynamics much more than a real electron.