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An electron can emit a real photon, which can be absorbed by another electron. Thus two electrons can exchange a real photon. However, when two electrons repulse, the math of their interactions is described by a virtual photon. Thus it seems that two electrons can exchange either a real or virtual photon (at least conceptually). What conditions define whether the photon that two electrons exchanged was real or virtual?

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An electron can emit a real photon,

Real photon means it can be measured/detected. For example in this single photon at a time experiment, the dots are measured photons, and the photons are real.

which can be absorbed by another electron.

Electrons do not absorb photons. They interact with photons. For example in Compton scattering:

comptonscat

A photon interacts with the electron, the electron becomes off shell because it absorbs part of the four momentum of the incoming photon and a lower energy photon leaves.

Thus two electrons can exchange a real photon.

Wrong . If the second photon hits another electron, the process will be repeated. It is not an exchange but a scattering.

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    $\begingroup$ Then the question becomes, how do we tell a short range scattering from exchange? Example: two hydrogen atoms repel off each other. (1) Their electrons repell by exchanging a virtual photon. (2) One atom emitted a real photon (the electron changed the energy level) absorbed by the other atom. The momentum conservation would make the atoms repel as well. So how do we know it was 1 or 2? $\endgroup$ – safesphere Sep 23 '17 at 7:57
  • $\begingroup$ In X-Ray interaction with matter, photoelectric absorption means that the whole phonon energy is transferred to an electron - therefore, your statement that electrons do not absorb photons confuses me. $\endgroup$ – Sanya Sep 23 '17 at 8:00
  • $\begingroup$ @Sanya Atoms are compoosite particles with bound states and can absorb energy by raising the energy levels . The electron is an elementary point like particle with a fixed mass. In the center of mass of the final electron, the mass of the electron is fixed, the energy balance from "before hitting the electron" to "after", when there would be only an electron would violate energy conservation. $\endgroup$ – anna v Sep 23 '17 at 12:38
  • $\begingroup$ @VictorStorm see my answer to Sanya. Atoms are a different story. $\endgroup$ – anna v Sep 23 '17 at 12:40
  • $\begingroup$ @annav so what you wrote applies to free electrons? $\endgroup$ – Sanya Sep 24 '17 at 13:07
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The process involving real particles:

Real particles

cannot occur because it cannot simultaneously conserve both energy and momentum. In effect the left vertex would be inverse Compton scattering with an incoming photon energy of zero and the right vertex would be Compton scattering with an outgoing photon energy of zero. In both cases the probability of the process is zero. Electrons can certainly emit (real) photons, in bremsstrahlung for example, but only if some other particle is present to balance the energy and momentum changes.

This process occurs in Feynman diagrams because a Feynman diagram does not show something that is actually happening i.e. the particles shown in a Feynman do not exist. This is discussed in Do virtual particles actually physically exist? and probably many other questions on this site. The Feynmann diagram is a graphical representation of an integral called a propagator not a real process that actually happens.

Feynman diagrams are only used in QFT calculations so there is never any ambiguity about whether the represent a real process (they don't). In a Feynman diagram the real particles, i.e. the ones observable in an experiment, are the ones that come in from infinity and go out to infinity. Any particle that appears and disappears in the diagram is virtual.

A perhaps more concrete description is that all real particles obey the relativistic energy equation:

$$ E^2 = p^2c^2 + m^2c^4 $$

and virtual particles do not.

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An electron can emit a real photon, which can be absorbed by another electron.

In a closer view the following happens:

  1. The electron emits photons if it gets accelerated. For example an electron - crossing a magnetic field - gets deflected, moves in a spiral path (this is an acceleration) and loses its kinetic energy in the form of photon emission.
  2. Or the electron is braked by another particle and some of its kinetic energy over goes to this particle. This phenomenon is accompanied by the loss of some energy in the form of photon emission. The greater the speed difference and the greater the moment of inertia of the hitted particle (in relation to the electron) the more energy lose.

However, when two electrons repulse, the math of their interactions is described by a virtual photon.

What you describe now is an elastic collision. As long as no photons observable the kinetic energy over goes from one to the other electron without energy loses. Colliding two electrons on high velocities one observe photon emission and it seems to be obvious that such collisions could have an lower limit only if one suppose the elasticity of the electrons electric field.

Thus it seems that two electrons can exchange either a real or virtual photon (at least conceptually).

In Feynman diagrams the interaction of two electrons is described in an idealized way (conceptually). I got used to use a model of electric field which makes it easier to predict or comprehend what happens with charged particles:

  1. In analogy to rigid bodies the electric field of subatomic particles claims place around himself. Where a body/field is, there can’t be a second.
  2. This field may have an elasticity in low energy interactions (no photon emission like in the related Feynman diagram). In higher energy interaction the energy exchange is accompanied by energy losses / photon emission.

It would be interesting if you could found a counterexample where this simple model could not be applied to physical processes.

Additional you may want to read about examples of the emission of photons .

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