Baby Astronaut Posted November 29, 2009 Share Posted November 29, 2009 If two particles were in the act of exchanging bosons, and in the meanwhile one/each particle disappeared, would the boson headed towards it "miss" the vanished particle and now exist on its own? Example 1: virtual particles disappear before the exchange is complete. Example 2: with normal particles, but one's velocity was suddenly altered before the exchange got completed. Link to comment Share on other sites More sharing options...
ajb Posted November 29, 2009 Share Posted November 29, 2009 My answer might sound like a deflection, but I think your question comes from taking the interpretation of virtual particle and exchange bosons too literally. I do not think your question is very well-formulated. Maybe a better answer is that the S-matrix (in which the virtual particle is really a term in an expansion of) involves an integration over all positions (or momenta) of the exchange boson. (This is in part the origin of infinities and renormalisation) Link to comment Share on other sites More sharing options...
Baby Astronaut Posted December 16, 2009 Author Share Posted December 16, 2009 It's been over a month, I just don't get your answer in the least. Perhaps the real question is simply can a boson exist by itself? Link to comment Share on other sites More sharing options...
ajb Posted December 16, 2009 Share Posted December 16, 2009 Perhaps the real question is simply can a boson exist by itself? Photons exist for example. Composite bosons exist, like the Helium-4. Your question about exchange particles is really a question of perturbation theory in quantum field theory. The internal legs of a Feynman diagram correspond to what we call propagators (or sometimes Green's functions). These propagators do not correspond to incoming or outgoing particles. We call them virtual particles. These virtual particles need not obey the classical equations of motion for the corresponding observable particles. Indeed, the propagator diverges when the equations of motion hold. More importantly for your question is that in the S-matrix or correlation function propagators come with an integration over all position or momentum, depending on how you formulate things. It is inherently an non-local object. I simply think your original question is ill posed. Link to comment Share on other sites More sharing options...
swansont Posted December 16, 2009 Share Posted December 16, 2009 I simply think your original question is ill posed. IOW, your question is framed in a classical context, in which the boson has a trajectory and can miss its target. That's not consistent with the QM formulation. Link to comment Share on other sites More sharing options...
Bob_for_short Posted December 16, 2009 Share Posted December 16, 2009 (edited) If two particles were in the act of exchanging bosons, and in the meanwhile one/each particle disappeared, would the boson headed towards it "miss" the vanished particle and now exist on its own? Yes, if the boson is real, not virtual. For example, if you go out of the room, photons miss you and hit the wall. Example 1: virtual particles disappear before the exchange is complete. Example 2: with normal particles, but one's velocity was suddenly altered before the exchange got completed. A virtual photon is to a great extent a usual Coulomb interaction potential 1/|r1 - r2|. Whatever motion of particles happens, it remains a function of their distance. If one particle decays into another charge and a neutral piece, the Coulomb force will act between the new and the other charges. Edited December 16, 2009 by Bob_for_short Link to comment Share on other sites More sharing options...
ajb Posted December 16, 2009 Share Posted December 16, 2009 IOW, your question is framed in a classical context, in which the boson has a trajectory and can miss its target. That's not consistent with the QM formulation. Yes, thank you. That is what I was trying to say, in a round about way Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now