Timeline for Dirac operator on 4-dimensional rectangle with the periodic boundary conditions is self-adjoint? What is its spectrum?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 21, 2022 at 0:24 | comment | added | Michael Engelhardt | I'm not sure there would be a reference for these things - they're not that complicated. You just solve the problem for zero gauge potential and then treat the gauge potential as a perturbation. For large $n$, that's justified since the gauge potential is suppressed by a factor $1/n$ compared to the unperturbed operator. | |
| Apr 20, 2022 at 16:17 | comment | added | Isaac | Oh, sorry for not clarifying my intention. In which references can I find the results you mentioned in your comments? In particular, the spectrum becoming asymptotically $2\pi n$ in the leading order. | |
| Apr 20, 2022 at 2:16 | comment | added | Michael Engelhardt | If discretized versions are of interest to you, there is a large field of research called Lattice QCD, with yearly symposia including reviews, and textbooks that you can find by googling ... | |
| Apr 19, 2022 at 7:02 | comment | added | Isaac | Thank you for your detailed comments. Perhaps, could you recommend me any relavant reference as well? | |
| Apr 19, 2022 at 1:04 | comment | added | Michael Engelhardt | There is not much to say - say you have found an eigenfunction. If you multiply it by the fixed external gauge potential, the result is bounded, as opposed to the contribution from the derivative part of the Dirac operator, which diverges as $n$. | |
| Apr 18, 2022 at 19:47 | comment | added | Isaac | I see. Why does the gauge potential only give corrections of order one? Could you clarify more? | |
| Apr 18, 2022 at 18:10 | comment | added | Michael Engelhardt | My comment had nothing to do with the asymptotic freedom of QCD. I merely meant that the spectrum does become $2\pi n$ for large $n$ to leading order, since the gauge potential will only give corrections of order one. | |
| Apr 18, 2022 at 17:15 | comment | added | Isaac | The notion of asymptotic freedom pertains to the strength of interaction, or the coupling constant, as far as I know. | |
| Apr 18, 2022 at 17:12 | comment | added | Isaac | I am considering any kind of gauge theories, in which the gauge field is the classical background field and the Dirac field is the one to be quantized. Could you explain what you mean by "become free"? The spectrum is essentially the collection of energy levels, and what you mean by it becoming free? | |
| Apr 18, 2022 at 15:51 | comment | added | Jon | This is a physicist's point of view. It is not clear to me what kind of gauge potential you are considering. Is it Abelian or non-Abelian? For the former, there are exact solutions you can use to understand the spectrum (e.g. a sinusoidal potential). For the latter, you can have confinement and the Dirac spectrum can change dramatically. If we can assume that the Dirac field is not a source for the gauge field (external background), you can find in literature some examples also for the non-Abelian case of some exact solutions but I cannot say if they can be used for your compact space. | |
| Apr 18, 2022 at 15:14 | comment | added | Michael Engelhardt | Asymptotically, the spectrum does become free, since you can treat the gauge potential as a perturbation when the wavenumber becomes large. | |
| Apr 18, 2022 at 14:28 | comment | added | Isaac | I see.. In fact, what I am interested in is the asymptotic decay growth rate of the spectrum. If the gauge potential is a compactly supported continuous function in $[-1,1]^4$, can we estimate the asymptotic growth rate of the spectrum? For example, would it be polynomial growth of degree $4$? | |
| Apr 18, 2022 at 13:55 | comment | added | Michael Engelhardt | The statement on $S^1 $ does not generalize to higher dimensions. You can only use a gauge transformation to get rid of one of the components of the gauge potential (modulo a zero mode). | |
| Apr 18, 2022 at 11:22 | comment | added | Isaac | I think the periodic boundary condition is equivalent to $(S^1)^4$. So, couldn't we just generalize the result in the link to 4D? | |
| Apr 18, 2022 at 11:21 | comment | added | Isaac | But, I heard that on $S^1$, the Dirac operator may be rescaled to the momentum operator and have the spectrum $2 \pi n$ for $n \in \mathbb{Z}$: as in mathoverflow.net/questions/389920/… | |
| Apr 17, 2022 at 15:32 | comment | added | Michael Engelhardt | Sure, the spectrum depends on the gauge potential. And of course, along with it being difficult to find explicit eigenfunctions for general gauge potential, it's concomitantly difficult to find the corresponding eigenvalues, and therefore, even more so, the complete spectrum. | |
| Apr 17, 2022 at 7:44 | comment | added | Isaac | @MichaelEngelhardt I am more interested in the spectrum. Is the spectrum also dependent on the gauge potential? | |
| Apr 17, 2022 at 3:05 | comment | added | Michael Engelhardt | For general gauge potential, you'll have a hard time finding explicit eigenfunctions. | |
| Apr 16, 2022 at 22:43 | comment | added | Ryan Budney | Does "periodic boundary conditions" mean the domain is $(S^1)^4$? | |
| Apr 16, 2022 at 21:45 | history | asked | Isaac | CC BY-SA 4.0 |