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I know that a real electron has a probability (which depends on the intensity of the EM force) of emitting a photon, changing his 4-momentum. The photon should be virtual. Now, my teacher says that this photon can decay in an electron and a positron, but "obviously not all the electrons are real". I'm referring to the E1 process in the image below.

What does this mean? Does it mean that the electron and positron created are virtual? And if this is the case they should annichilate themselves to create again the virtual photon, am I right? This is because if they are virtual, being massive particles, they can't maintain their "virtuality" as long as they want (differently from the photon, which is massless and can exist as a virtual particle without time limits). But my doubts arise when the analogous process, the beta decay, occurs via weak interaction. So instead of having an electron changing his 4-momentum I have a neutron becoming a proton and emitting a $W^{-}$ virtual boson. Then the virtual boson decays into an electron and an antineutrino (real particles because they can be observed). My final question, IF what I said 'till now is right, which I'm not sure, is:

Why does the photon decays into two virtual particles and the boson into real ones?

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  • $\begingroup$ By "real electron" you seem to mean a "bare electron" (whatever that is)? $\endgroup$ – CuriousOne Jun 8 '16 at 15:21
  • $\begingroup$ I don't know what you mean by "bare electron". By real electron I mean that it's not virtual. $\endgroup$ – Luthien Jun 8 '16 at 15:32
  • $\begingroup$ A "real" electron doesn't make that process, unless there is some other matter that it can interact with. $\endgroup$ – CuriousOne Jun 8 '16 at 15:41
  • $\begingroup$ see my answer here physics.stackexchange.com/questions/261346/… $\endgroup$ – anna v Jun 8 '16 at 15:47
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Why does the photon decays into two virtual particles and the boson into real ones?

The E1 diagram does not conserve energy in the center of mass system. The incoming center of mass energy is the mass of the electron, the outgoing is at least three masses of electrons, and mind you, the electron is an elementary particle and does not decay.

Also photons interact, they do not decay. Within correct feynman diagrams they are called virtual. Similar diagrams exist for off mass shell Z boson exchange.Outgoing lines from an interaction vertex are real, not virtual. Only internal lines are virtual.

The correct pair production diagram from a photon can be seen here, where the center of mass has to take into account the nucleus also:

pair production

The virtual photon in your E1 has to interact with the field of a nucleus, by a virtual electron exchange and then a real electron positron pair can appear,

Correspondingly there will be a diagram with a Zboson exchange from your E1, but unless the energies involved are very high, the Z propagator and the weak coupling constant make the contribution to the crossection of this diagram tiny.

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  • $\begingroup$ Thank you for your answer ann v. Yes you're right, I used "decays into" meaning "destruction of .. and creation of...". I'm pretty confused. Is the E1 diagram incomplete if there is no conservation of energy? Does this mean that the $e^{-}$ and $e^{+}$ are not real particles? Thanks for your answer and for the patience, I'm new in this and I want to clarify every detail that confuses me. $\endgroup$ – Luthien Jun 8 '16 at 16:19
  • $\begingroup$ see my edit with a diagram $\endgroup$ – anna v Jun 8 '16 at 16:42
  • $\begingroup$ Thank you ann v, now it's clear to me! It's all about the conservation of 4-momentum, which tells me if a process can occur or not. The process in E1 can't occur without a nucleus because it violates the conservation of 4-momentum, whereas the beta decay can occur because there is conservation, right? Thanks again! $\endgroup$ – Luthien Jun 8 '16 at 16:52
  • $\begingroup$ Yes, one always has to conserve energy and momentum :) $\endgroup$ – anna v Jun 8 '16 at 17:11

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