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I understand that the quanta of electromagnetic field is a photon.

Anyway, the electron field quanta is the electron. And the process of electron-positron annihilation releases two photons when the matter/anti-matter pair are re-absorbed into the electron field. This is how I understand the process.

But why don’t the photons (which are each their own anti-particles) annihilate somehow and become re-absorbed by the electromagnetic field, whose mechanism is the photon?

Here is my understanding/definition of a virtual photon as explained in Virtual and Real Photons:

an off-shell photon that is the carrier of the EM force and time-limited against detection

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  • $\begingroup$ You need to include your definition (and cite the source) of a virtual photon in your post. $\endgroup$ – user198207 Jul 7 '18 at 12:30
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    $\begingroup$ There are several confusions in your question that make it difficult to assess what you are actually asking: "I understand that the electromagnetic field is made of (virtual) photons. " No. At least, this is wrong until you explain what this is technically supposed to mean. "Anyway, the electron field force carrier is the electron. " - the electron is not a force carrier. Finally, what do you mean by photons being "reabsorbed" into the electromagnetic field? There are many interactions in quantum field theory, but none where a photon just vanishes. Why do you think this should happen? $\endgroup$ – ACuriousMind Jul 7 '18 at 12:56
  • $\begingroup$ I’m sorry I’m adding such confusion. My wording is unfortunate but I simplified the question. What is wrong with my understanding of this “re-absorption” of antiparticle pairs? Electron/position pairs come out of the electron field, right? And they go back when annihilated, no? So why don’t photons arising from the EM field act the same way? Why do they get to live on? Please be kind. $\endgroup$ – CapSix Jul 7 '18 at 13:07
  • $\begingroup$ Physical pictures are leading you (and invariably myself as well) astray, that's why the technical point is stressed in the above comment. There are lots of related posts on this site, (and you can probably find a lot of information there) and you could also read profmattstrassler.com/articles-and-posts/…. This sort of question has to phrased particularly carefully, as a correct answer will probably be math based, rather than mental pictures, unfortunately. $\endgroup$ – user198207 Jul 7 '18 at 13:19
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Here is my answer to what I think is your actual question: "why don’t the photons (which are each their own anti-particles) annihilate somehow? "

This can happen only if the photons get back into exactly the same phase and relative position as when they were created. This is practically excluded by statistics for the case of an electron positron annihilation reaction. In principle the reverse reaction can occur under the right circumstances.

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I understand that the quanta of electromagnetic field is a photon.

The electromagnetic wave is a solution of the classical maxwells equation and is out of electric and magnetic fields. It is not a field by itself.

em

The classical electromagnetic wave

In the visible frequencies it is light, but it covers a lot of frequencies in its spectrum.

There is no classical electromagnetic field. The quanta which compose light are the photons, and the classical fields emerge from zillions of photons.

Anyway, the electron field quanta is the electron.

You are misrepresenting quantum field theory here. In quantum field theory there exists an electron field all over (x,y,z,t) represented by a plane wave wave function of the solution of the Dirac equation for electrons. But it is not electrons, it is the probability amplitude of finding an electron. To get an electron you need a creation operator on the electron field, and to eliminate it an annihilation operator. By this mathematical tool the motion of an electron can be modeled as a disturbance on the electron field.

There is no absorbing by the electron field, ( or photon field or...)which is a type of coordinate system that allows calculation of cross sections and decays.

And the process of electron-positron annihilation releases two photons when the matter/anti-matter pair are re-absorbed into the electron field.

e+e-an

Again this is a misrepresentation of the model used in calculating electron positron annihilation. There is no re-absorption

Anyway, the electron field quanta is the electron.

Not in the sense you are thinking about it. In the diagram, a creation of an electron which becomes virtual and annihilates with the positron into two photons is mathematically represented.

And the process of electron-positron annihilation releases two photons

yes

when the matter/anti-matter pair are re-absorbed into the electron field.

There is no re-absorption. The electron field as well as the positron field exist whether there is an interaction or not. The interaction happens with the creation and annihilation operations.

This is how I understand the process.

You have to rethink.

But why don’t the photons (which are each their own anti-particles) annihilate somehow

The photons emerge at an angle, defined by the invariant mass of the e+e- system, and leave the interaction region with velocity c, as a disturbance on the photon field with consecutive creation and annihilation operators , if you want to visualize it, on the field. In fact if one wants to really model the leaving photons ( or specific particles) one has to use wave packets.

So they cannot meet at an interaction vertex to annihilate.

The second reason is because the all pervasive photon field is like a coordinate system, a Lorenz invariant aether, it does not contain particles, just possibilities to calculate Feynman diagrams.

and become re-absorbed by the electromagnetic field, whose mechanism is the photon?

So there is no re absorption . It is not within the quantum field theory model.

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  • $\begingroup$ Ok. Thank you very much.I just saw a lecture on QFT and I was thinking of the QFT electromagnetic force field as some type of a sea of virtual photons. @my2cts was right about my actual question which is why photons don’t become reabsorbed by what I thought of as a sea of virtual photons. So, and I’m almost afraid to ask, you’re saying the QFT elementary electromagnetic force field (The one that permeates all space forever) is pretty much separate from electromagnetic radiation, for example in the form of light from a star. (Or two photons exiting an annihilation)? $\endgroup$ – CapSix Jul 7 '18 at 21:55
  • $\begingroup$ Yes, the photon (electron, muon....) field in quantum field theory is a mathematical underlying system that allows the mathematical modeling of photon-particle interactions . It is active only with creation and annihilation operators acting on it. $\endgroup$ – anna v Jul 8 '18 at 3:17

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