Timeline for Why Goldstone Bosons? (A Question about VEVs)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2013 at 20:31 | vote | accept | Edward Hughes | ||
| Oct 23, 2013 at 13:30 | comment | added | Adam | @Dan: as I said, it's mainly a matter of taste. But I don't think you need the S-matrix and LSZ theorem to describe what is a Goldstone mode, or even what is a particle. You just need to look at the Lehmann representation (spectral function) and see if there is a peak at a given energy/mass. | |
| Oct 23, 2013 at 9:19 | comment | added | Edward Hughes | Ah - I've reread and have a different interpretation. Is your answer essentially that zero VEV is a mathematical requirement due to the LSZ theorem? If so that makes me very happy (since I'm really a mathematician at heart, so I like mathematical reasoning)! | |
| Oct 23, 2013 at 9:16 | comment | added | Edward Hughes | @Dan - thanks. I still don't really understand why you have to identify the Goldstone bosons as the on-shell particles though. Why can't the original fields $\phi^n$ give rise to these on-shell species? Assuming that the Goldstone bosons represent the particle content of the theory, I can prove in several ways that they must be massless. But I have no idea how to justify this assumption! | |
| Oct 23, 2013 at 5:01 | comment | added | Dan | The author referenced the Wightman axioms, and wanted to know why zero VEV was part of the definition of a particle, so I assumed he wanted the answer for relativistic QFT. In RQFT neither fields nor their expectation values are physical, and the definition of a particle has been settled since Wigner. This much is made clear by the on-shell formulations for scattering amplitudes. I think statmech is less clear. You might disagree, but I find it hard to call an order parameter a particle, and goldstones in stat-mech are really just long-wavelength modes in a classical field, not a particle. | |
| Oct 23, 2013 at 2:47 | comment | added | Adam | I don't agree with all you said. Fields are not physical, but their expectation values are. In the sense that they are measurable. You have a HEP point of view, which is no wrong it itself, of course, but that renders everything more convoluted (LSZ formula, etc.) to make sense out of them. A stat-phys approach makes everything easier. It's simpler to grasp what is the mean value of a spin (the magnetisation), than that of the Higgs field. Same with the Goldstone mode (that are never there in HEP anyway). | |
| Oct 23, 2013 at 0:36 | history | answered | Dan | CC BY-SA 3.0 |