Do the apperance in the atomic nucleus of virtual matter-antimatter particle pairs play a role in the random nature of radioactive decay?
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The random character of the process itself doesn't depend on any specific virtual particles - instead, the random character of all microscopic processes in the world is a basic consequence of quantum mechanics. The existence of virtual particles is a consequence of quantum mechanics, too. Obviously, virtual particle pairs are important - in a particular computational scheme - to calculate the actual decay rate of a particular unstable object.
So the virtual pairs are not the cause of the randomness itself - they're a consequence of the randomness - but in any calculation, the virtual pairs are a part of the cause of the actual numerical value of the decay rate.
answered May 4, 2011 at 6:10
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$\begingroup$ A virtual particle then can't "knock" a neutron or proton out of the nucleus? $\endgroup$– Andersi2Commented May 4, 2011 at 7:53
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$\begingroup$ @Anders I Nuclear interactions and the energies available for them are of the order of MeVs and keVs . The nuclear "force" can be well approximated by a potential well, as in the shell model which describes the behavior of nuclei quite well. This force is a manifestation of a spill over from the fundamental interactions with quarks and gluons that hold the protons and neutrons together, a bit like the Van der Waals forces for electromagetism in atomic physics. The virtual particles are way off their mass shell because of the small energies available and can play role only in decays $\endgroup$– anna vCommented May 4, 2011 at 18:24
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$\begingroup$ continued: that are allowed energetically. No, in general they cannot knock off a proton or neutron. $\endgroup$– anna vCommented May 4, 2011 at 18:26
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$\begingroup$ Thanks, but does the hypothetical "homerun" depend on the spatial proximity to a given proton or neutron of the manifestation of the virtual particle? $\endgroup$– Andersi2Commented May 4, 2011 at 23:15
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$\begingroup$ @Anders I: I don't think it makes sense in this case to talk about positions or spatial proximity. The virtual particle is not a billiard ball that knocks out something from the nucleus. In the case of beta decay, if you look at the Feynman diagram (en.wikipedia.org/wiki/…), a neutron turns into a proton, while "emitting" a virtual W boson (that turns into an antineutrino and an electron). The decay probability depends on the probability of the neutron to emit the W, among other factors. $\endgroup$– jdmCommented May 5, 2011 at 7:10
