Are electron fields and photon fields part of the same field in QED?

Question

I know in classical field theory we have the electromagnetic field. And Maxwell's equations show how electromagnetic radiation can propagate through empty space.

I also have been reading about QED and I gather the electric repulsion between two electrons is mediated by a virtual photon.

Also, as I understand it, in quantum field theory we speak of particles as manifestation of an underling field. For instance, a photon is a manifestation of a photon field.

Two Questions:

1. Are quantum fields like electron fields or photon fields one big field (like we assume gravity to be one field) or are there separate ones? Meaning, can I have several electron fields?

2. I often here the term electromagnetism and people say they are the same force. Are electron fields and photons fields part of the same underlying field or are they separate fields that just interact?

Answer 1

In our modern understanding, every electron is thought to be a localized excitation of the electron (or Dirac) (spinor) field $\Psi(x^\mu)$, while every photon is considered to be an excitation of the photon (vector) field $A^\nu(x^\mu)$, which is the quantum field-theoretic counterpart of the classical four-potential.

Thus, the answer to your questions are:

1. All particles of the same type (e.g. photons or electrons) is understood to be 'coming from' one all-permeating quantum field. It should be noted that these fields also give rise to the corresponding anti-particles, so the positron field is the same as the electron field.

2. The different particle types are truly separated in quantum field theory: Each type is represented by one field, and the fields interact. These interactions are quantified by the Lagrangian (density), which essentially determines everything about the theory. In pure electrodynamics, the quantum field-theoretic Lagrangian density is (using 'mostly minus' sign convention for the metric)

$$\mathcal{L}_{\text{QED}}= \bar\Psi(i\gamma^\mu D_\mu-m)\Psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} =\bar\Psi(i\gamma^\mu (\partial_\mu+ieA_\mu)-m)\Psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$ where $F_{\mu\nu}\equiv \partial_\mu A_\nu-\partial_\nu A_\mu$ is the electromagnetic field strength tensor. The 'covariant derivative' $D_\mu\equiv \partial_\mu+ie A_\mu$ encodes the interaction between the two fields $A_\mu$ and $\Psi$, and the 'strength' of the interaction is given by $e$, the charge of the electron.

Answer 2

For what it's worth, I showed in my recent article http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf (published in European Phys. J. C) that one can eliminate the Dirac field from the Dirac-Maxwell electrodynamics after introduction of a complex electromagnetic 4-potential (producing the same electromagnetic field as the real 4-potential), so modified Maxwell equations can describe both electrons and photons.