Timeline for Proper definition of spontaneously broken symmetry
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 8, 2023 at 23:50 | vote | accept | Truth and Beauty and Hatred | ||
| Dec 8, 2023 at 23:36 | answer | added | Nandagopal Manoj | timeline score: 1 | |
| Dec 8, 2023 at 13:23 | vote | accept | Truth and Beauty and Hatred | ||
| Dec 8, 2023 at 13:23 | comment | converted from answer | Truth and Beauty and Hatred | The answer is found in the comments to this post. Thanks again to Nandagopal for his excellent explanation. | |
| Dec 7, 2023 at 15:26 | comment | added | Nandagopal Manoj | Yes, that sounds correct! | |
| Dec 7, 2023 at 10:35 | comment | added | Truth and Beauty and Hatred | Thank you a lot for your answer! I think I understand now. Just to make sure the point is that the expectation value $\langle W[C_p] \rangle$ can be regarded (under an appropriate deformation) as the two point function of charged loops. Then having $\langle W[C_p] \rangle \neq 0$ is reminiscent of the fact that in the finite volume case (before thermodynamic limit) we had an ordered phase with perimeter law, thus we have SSB. Is my logic correct? | |
| Dec 6, 2023 at 18:24 | comment | added | Nandagopal Manoj | We use operators like $\langle O^\dagger(x) O(y)\rangle $ where $x$ and $y$ are well seperated to diagnose spontaneous symmetry breaking (this is commonly referred to as long range order). Are you familiar with this? I agree that the higher form case is more subtle, but this answer of mine may help. physics.stackexchange.com/questions/743794/… | |
| Dec 6, 2023 at 10:16 | comment | added | Truth and Beauty and Hatred | Something of the form $\langle O^{\dagger}(x) O(y) \rangle$ would be a good example since the full object in the expectation value is neutral under the symmetry. In particular in the new version of the paper this is done at the end of page 12, where they evaluate the expectation value of the operator $W^{\dagger}[C_p] which, from my understanding, is neutral under the symmetry they are considering. | |
| Dec 6, 2023 at 10:03 | comment | added | Truth and Beauty and Hatred | I apologize. Yes indeed i was referring to an older version of that paper. In the new version i am not able to locate the exact statement i found previously but nonetheless the claim is the same. They used an operator with no charge to establish SSB. | |
| Dec 5, 2023 at 21:24 | comment | added | Nandagopal Manoj | You say "... operator O has 0 charge under the symmetry." Are you referring to the expectation value of the form $\langle O^\dagger(x) O(y)\rangle$, where $O$ is charged under the symmetry? | |
| Dec 5, 2023 at 21:22 | comment | added | Nandagopal Manoj | I think you are referring to an earlier version of the paper, it would be better to mention that. I could not find those statements in the new version. | |
| Dec 3, 2023 at 15:41 | history | edited | Truth and Beauty and Hatred |
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| Dec 2, 2023 at 16:58 | history | asked | Truth and Beauty and Hatred | CC BY-SA 4.0 |