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In this Quora post: https://qr.ae/pv5tac, it states that the electron quantum field "wants" to reduce the energy it has, so when a particle and an anti-particle interact and the charges cancel out (so conservation of charge not violated) the electron field uses the opportunity to get rid of the energy, which is turned into the photon field.

My understanding of quantum fields is that they are sort of entities where if you input enough energy you get a particle.

I understand that these are probably oversimplified explanations, but I was wondering, wouldn't the photon field not 'want' that energy? Is it that the photon field is somehow more accepting of energy than electron field? If so, why?

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When we are calculating the scattering probability for particles we can do this using Feynman diagrams. For example the tree level diagram for electron positron scattering to two photons is:

electron positron annihilation

(diagram from the question Electron Positron annihilation Feynman Diagram)

The scattering probability is related to the number of nodes in the diagram. For more on this see David Z's answer to Why does electron-positron annihilation prefer to emit photons? In this case the probability is proportional to g2em where gem is the electromagnetic coupling constant.

But suppose we reverse the direction of the time axis to change the diagram to:

photon photon scattering

Now the diagram shows two photons scattering into an electron-positron pair. But it's the same diagram so the scattering probability is the same i.e. proportional to g2em. The scattering probability is the same in both directions, so to say that:

the electron quantum field "wants" to reduce the energy it has, so when a particle and an anti-particle interact and the charges cancel out (so conservation of charge not violated) the electron field uses the opportunity to get rid of the energy, which is turned into the photon field.

is not a statement that makes much sense.

It is certainly true that in everyday life we frequently see electrons and positron annihilating to photons, but we rarely see two photons annihilating to an electron and positron (NB ignore pair production as this is a different mechanism). However this is not due to any fundamental asymmetry. It's simply because the initial conditions are diferent and in particular photons move at the speed of light while electrons (typically) don't. In fact if we accelerate electrons to speeds near c we find the scattering probability decreases with increasing kinetic energy.

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Be careful of metaphorical statements like the electron field "wanting" something. The electron-photon system described by QED has an interaction where either an electron can radiate a photon or an electron & positron can turn into a photon, or the time reversal of either of these. (Neither of these interactions as described conserve energy/momentum and angular momentum, so physical interactions involve 2 or more photons.) So, an electron & positron that are in the same place at the same time can interact - annihilate - to form 2 photons. Likewise, 2 photons aimed just right can "annihilate" to form an e- and e+.

Many phenomena that look like a system "wanting" to lower its energy are really about initial conditions and entropy. For example, people will say that an atom in an excited electronic state "wants" to lower its energy by radiating a photon. But really what's going on is that there's a time-reversal invariant interaction that allows the decay event, an initial condition where the system is prepared in an excited state, and a final state involving a lower energy atom plus escaping photon that is overwhelmingly more likely (in phase space volume) than the initial one with the excited atom and no photon.

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