Timeline for Vacuum Degeneracy for Massless Free Scalar Field
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2021 at 4:28 | vote | accept | anon123456789 | ||
| Oct 4, 2021 at 14:12 | history | edited | Chiral Anomaly | CC BY-SA 4.0 |
Enumerated the options and added a third option
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| Oct 4, 2021 at 0:51 | history | edited | Chiral Anomaly | CC BY-SA 4.0 |
Highlighted the cousin of the other highlighted statement
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| Oct 4, 2021 at 0:48 | comment | added | Chiral Anomaly | @bittermania It's because of option (2) in your comment. Taking the zero-mode factor of the state to be an infinite plane wave (that is, independent of the zero-mode variable) makes the state non-normalizable, so the $\alpha$ term in your second-to-last equation might as well be zero because the other terms are "infinite". Loosely speaking, the equations $\delta(0)=\delta(0)+\alpha$ and $\delta(0)=\delta(0)$ are perfectly consistent with each other, because $\delta(0)=\infty$. | |
| Oct 3, 2021 at 20:13 | comment | added | anon123456789 | One last point of clarification: I did try and explicitly account for the zero mode in my question, essentially by claiming it to be an infinite wavelength "plane wave" in field space. However, it did not help the matter. I can see two potential problems with this: (1) These "plane wave" modes don't have a particle interpretation (2) These "plane waves" are only normalizable to $\delta$-functions. Is this approach to incorporating the zero mode wrong because of (1), (2), both, or something else? | |
| Oct 3, 2021 at 16:40 | comment | added | Chiral Anomaly | @bittermania The "vacuum degeneracy" phenomenon is still physically meaningful, because we can still take the low-energy limit of expectation values like I explained in the answer. | |
| Oct 3, 2021 at 16:40 | comment | added | Chiral Anomaly | @bittermania Regarding the linear sigma model: I have not thought carefully enough about how to take the limit in which only the goldstone modes remain. (I assume that's the decoupling limit you're talking about.) However, generally speaking, theories with massless particles don't have vacuum states in the strict sense of a (normalizable) state-vector in the Hilbert space unless we exclude the zero-momentum operator(s) from the set of observables, like we normally do for the quantum electromagnetic field. | |
| Oct 3, 2021 at 16:37 | history | edited | Chiral Anomaly | CC BY-SA 4.0 |
Replaced another statement
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| Oct 3, 2021 at 16:15 | comment | added | anon123456789 | Yes, that is what I mean | |
| Oct 3, 2021 at 16:12 | history | edited | Chiral Anomaly | CC BY-SA 4.0 |
Replaced an incorrect statement (doesn't affect the conclusion)
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| Oct 3, 2021 at 15:11 | comment | added | Chiral Anomaly | @bittermania Before I try to respond, I want to make sure I understand your comments: By $U(1)$ linear sigma model, do you mean the model whose lagrangian is a polynomial in two real scalar fields, with an internal $O(2)$ symmetry that mixes the two scalar fields, with a potential term chosen to break that symmetry spontaneously, and no gauge fields? | |
| Oct 3, 2021 at 5:09 | comment | added | anon123456789 | Perhaps it is not sensible to ignore the goldstone zero mode though, if this would require us to view the broken $U(1)$ as gauge symmetry. | |
| Oct 3, 2021 at 4:59 | comment | added | anon123456789 | Thanks for the answer. My example was motivated with SSB in mind, and this model is the decoupling limit of the standard $U(1)$ linear sigma model, which has degenerate vacua. I would have also expected the linear sigma model to possess a zero mode, i.e. a very soft goldstone. I may be misunderstanding, but should I take your answer to imply that none of the vacua are normalizable? Moreover, if I do not insist that the goldstone zero mode is observable, does the vacuum become unique? | |
| Oct 3, 2021 at 4:10 | history | answered | Chiral Anomaly | CC BY-SA 4.0 |