A force between two particles can be described either as the action of a force field generated by one particle on the other, or in terms of the exchange of virtual force carrier particles between them.

These virtual force carrier particles are physically produced during these processes, like beta decay of neutron? Why are they called "virtual"? I know that $W$ boson was discovered in a physical "weak interaction" experiment, so it was actually produced.

  • $\begingroup$ What do you mean by "actual(ly)", "virtual", "physical(ly)", etc... Have you completely researched "virtual particles"? $\endgroup$ – Cosmas Zachos Jun 8 at 15:24

On some level, no particles are "physical"; they're all tools that we use to describe the interactions of quantum fields. This is especially true for virtual particles, which don't exist outside of the reaction they're part of.

Typically, we call the particles that enter and exit Feynman diagrams "real". They propagate "to infinity", or at least as far as their decay lifetime will let them. Because they exist for a long time, they must be "on the mass shell" if they don't decay (in other words, their energy and momentum must be consistent with their mass). If they do decay, the lifetime of the particle is related to the "width" of their mass; in other words, they can stray slightly from the mass shell (i.e. they can have an energy and momentum that is somewhat inconsistent with their mass), but not as much as if they were annihilated before decaying.

In contrast, the particles which are entirely internal to the Feynman diagram, those that are created and then annihilated within the same interaction, are typically called "virtual". Since they exist for such a short time, virtual particles can be "off the mass shell" (i.e. their energy and momentum doesn't have to correspond to the particle's actual mass). You will often hear the "virtuality" of an exchanged particle being discussed; this is a measure of how inconsistent the exchanged particle's energy and momentum are with its mass.

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  • $\begingroup$ How about the W boson in beta decay of neutron? It was revealed in an experiment, so it was off or on the mass shell? $\endgroup$ – Salmone Jun 8 at 19:19
  • $\begingroup$ @Salmone The W boson has a mass of 80 GeV. The energy involved in a beta decay reaction is on the order of an MeV. Beta decay doesn't have anywhere near enough energy to create an on-shell W boson; the W boson involved in beta decay is very far off-shell. $\endgroup$ – probably_someone Jun 8 at 19:31
  • $\begingroup$ Sorry, the experiment was with proton against antiproton. In this case, the W boson was produced? $\endgroup$ – Salmone Jun 8 at 20:10
  • $\begingroup$ @Salmone The center-of-mass energy of that experiment was 540 GeV, which is definitely enough to create on-shell W bosons. That said, the W boson lifetime is so short ($10^{-24}$ seconds!) that it doesn't make much sense to talk about a "real" W boson in the above sense. Every W boson we produce is internal to some Feynman diagram before it reaches a detector, so we observe it by looking for the peak in its production rate around its on-shell mass. See the following review: cds.cern.ch/record/2103277/files/9789814644150_0006.pdf $\endgroup$ – probably_someone Jun 8 at 20:25
  • $\begingroup$ So if the energy is enough, the W boson is produced but if energy is lower, the W boson is not poduced, but weak interaction still happens. I really can't understand. $\endgroup$ – Salmone Jun 9 at 14:08

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