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there are a couple of similar posts here on StackExchange about bound particles being off shell in QFT. From posts on other sites, I've been told that the idea of off shell particles doesn't apply to bound states - but I still see this language in many places.

My question is then, what is the context for the use of the term off shell when referring to bound states? Is there a reason why bound particles can be considered 'off shell'. I'd always heard of off shell particles are being considered to be unphysical (as in virtual particles in perturbative calculations) but clearly this is not the case.

Also, states involving two or more free particles are also off shell - but I'm not grasping the definition correctly (since, e.g., QFT has creation operators for on shell particles only).

Qmechanic
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asimo
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The meanings of the words "off-shell" and "on-shell" just describe the contribution of the states to the propagators. There is known the so-called spectral representation of the propagator, which reads $$ D_{ab}(p) = \int \limits_{0}^{\infty}dm^{2}\rho(m^{2})D_{ab}(p,m) $$ Here, $\rho(m^{2})$ is the so-called state spectral density depending on details of interacting theory, while $D_{ab(p,m^{2})}$ is the free propagator depending on the squared mass $m^{2}$.

Being integrated over the momentum, the propagator can be approximated by the contributions of its poles. There is well-known that one-particle and bound states are approximated by the poles, while the free many-particle states just give branch cuts. We talk that for the particular process the particle ot the bound state is on-shell if the momentum is close to the corresponding pole, while it is off-shell when the momentum squared is far from the pole.

If, for example, the CM frame momenta of electron and proton in the electron-proton collision are significantly larger than $e^2m_{e}$, the bound state is off-shell; however, near this energy the bound state becomes to contribute into the process since it becomes to be on-shell.

Name YYY
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  • Sorry - this might be a bit over my head. If we integrate over all the propagators, to we obtain a physical result? – asimo Sep 23 '17 at 14:10
  • @asimo : are you asking adout the spectral representation? If yes, then the answer on your question is indeed yes. – Name YYY Sep 23 '17 at 18:19
  • Thanks @Name YYY - ok, so here's a scenario: I have a system I wish to measure, which is sitting in a bunch of external fields (unavoidable), and so all particles are off shell ... now I have my measurement apparatus, which is in exactly the same situation ...

    I make my measurement - can that measurement be exactly, precisely on-shell? I would say no - and I think your answer above implies that we may be closer to of farther from, but never exactly obtain an on shell measurement.

     – asimo Sep 26 '17 at 12:07
  • And, actually, does it matter - if the results are still physical? I'd understood we had to measure on-shell values ... but (and I'm not sure I'm correct here), can we use techniques like the LSZ reduction formula to calculate scattering amplitudes even assuming off-shell conditions? – asimo Sep 26 '17 at 12:19