1
$\begingroup$

If virtual particles cancel out after being created from spacetime flunctuations because they come in matter-antimatter pairs, how do virtual photons cancel out, due to a photon being its own antiparticle?

$\endgroup$
13
  • 2
    $\begingroup$ Virtual particles are more of a calculational tool than a particle that actually exists and can be observed. This is why they don't have to conform to some of the rules that real, observable particles do. $\endgroup$ – probably_someone Nov 19 '18 at 12:09
  • $\begingroup$ virtual photons can be interpreted as external legs that die off if they don't encounter an asymptotic on-shell particle to merge with, when there is an encounter their energy exchange is consistent with the asymptotic behavior of the on-shell components $\endgroup$ – lurscher Nov 19 '18 at 12:13
  • $\begingroup$ Also Energy conservation of Virtual Particles - Quantum Fluctuation? $\endgroup$ – John Rennie Nov 19 '18 at 12:16
  • 5
    $\begingroup$ Re the proposed duplicates: the point made in all three questions is that virtual particles are not physically real - they are a computational device and they do not actually exist. Let me emphasise this: virtual particles do not exist. So the question of how they appear and disappear is meaningless since they don't appear and disappear. $\endgroup$ – John Rennie Nov 19 '18 at 12:56
  • 2
    $\begingroup$ @AbdulMoizQureshi There's at least one other example of a computational device in physics that doesn't actually exist: the ideal "wires" that are drawn in circuit diagrams. They do not correspond to any actual physical object; rather, they are there to demonstrate the way in which the other (real) components are connected together. This is why they can have properties that would be strange for any real object to have (for example, having no voltage drop across them, and no resistance, but also carrying finite current). $\endgroup$ – probably_someone Nov 19 '18 at 21:00

Browse other questions tagged or ask your own question.