Timeline for When can a global symmetry be gauged?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2014 at 18:11 | comment | added | Qmechanic♦ | Related preprint by OP: arxiv.org/abs/1001.5212 page 6. | |
| Apr 20, 2013 at 1:22 | comment | added | Diego Mazón | Let me point out that the reason why my argument does not work is because one would need a real, massless Klein-Gordon field. Real and Klein-Gordon are no problem. The problem is "massless" since then one cannot take the non-relativistic limit. | |
| Apr 18, 2013 at 8:28 | vote | accept | Tomáš Brauner | ||
| Apr 18, 2013 at 8:23 | comment | added | Tomáš Brauner | @drake: The nonrelativistic limit has a two-parametric shift symmetry: $i\int(\psi-\psi^\dagger)$ generates real shifts of $\psi$ while $\int(\psi+\psi^\dagger)$ imaginary shifts. The commutator of these two generators is a constant, i.e., a central charge, which represents the obstruction to gauging this symmetry. This comes from the fact that $\psi$ and $\psi^\dagger$ are canonically conjugate. In the case of a complex Klein-Gordon field, $\psi$ and $\psi^\dagger$ are independent dynamical variables; classically, the symmetry is the same, but now there is no central charge in the commutator. | |
| Apr 18, 2013 at 3:58 | comment | added | Diego Mazón | Thanks and sorry, but I do not understand that. 1)The non-relativistic limit of a real Klein-Gordon field is a (complex) Schrödinger field. The former obeys a 2nd order EOM, while the latter a 1st EOM. So one physical degree of freedom in both cases. 2) The transformation is uniparametric so that there is only one generator ($\int \, \theta \, \psi^{\dagger} + h.c.$). Sorry about the brevity, I'm in a rush. @DavidBarMoshe | |
| Apr 17, 2013 at 14:52 | comment | added | Tomáš Brauner | @DavidBarMoshe: Thanks for clarifying this point! Yes, I chose the Schrödinger Lagrangian precisely because the commutator of the two shift symmetries has a central charge, unlike e.g. the Klein-Gordon Lagrangian for a massless complex scalar field. | |
| Apr 17, 2013 at 14:34 | comment | added | David Bar Moshe | Tomáš Brauner: Regarding the remark by drake, Please see section 7.5 of the lecture notes by: Riccardo Rattazzi: itp.epfl.ch/files/content/sites/itp/files/groups/ITP-unit/…. @drake: The scalar Stueckelberg field is real and so is its shifting transformation. The extension to complex values, together with the fact that the Schrodinger Lagrangian is linear in the time derivatives make the shifting transformation anomalous. | |
| Apr 17, 2013 at 12:39 | answer | added | David Bar Moshe | timeline score: 16 | |
| Apr 17, 2013 at 10:20 | answer | added | user10001 | timeline score: 4 | |
| Apr 17, 2013 at 7:30 | comment | added | Tomáš Brauner | @drake: Sorry, but I don't see the connection of my example to Proca's theory. I am talking about a classical nonrelativistic scalar field. If you think the global symmetry in this example can be gauged, can you write down the gauge-invariant Lagrangian? Anyway, thanks for the comment! | |
| Apr 17, 2013 at 6:19 | comment | added | Diego Mazón | I think that one can gauge this symmetry, which is the responsible of the renormalizability of Proca theory physics.stackexchange.com/q/16931 Have you read the opposite? Ref or link? | |
| Apr 16, 2013 at 23:41 | answer | added | Qmechanic♦ | timeline score: 10 | |
| Apr 16, 2013 at 10:55 | history | tweeted | twitter.com/#!/StackPhysics/status/324113704490786816 | ||
| Apr 16, 2013 at 9:38 | comment | added | Tomáš Brauner | Replacing ordinary derivatives with covariant ones is indeed what one usually does. But that doesn't work in this trivial example. The covariant derivative for the shift symmetry will be something like $D_\mu\psi=\partial_\mu\psi-A_\mu$, and the term with the time derivative will not be gauge invariant. (There are theories whose action is gauge invariant yet the gauge field does not appear solely in covariant derivatives. An example is the low-energy effective theory for a ferromagnet. Generally this happens when the Lagrangian changes under the gauge transformation by a surface term.) | |
| Apr 16, 2013 at 9:19 | comment | added | joshphysics | Ok well in that case, I'm not sure Lubos' answer is very satisfying. I have to think a bit more about this, but I think the answer, if phrased in the manner of my first comment above, is that given any action of a Lie algebra on the fields that is either (i) an internal symmetry or (ii) a spacetime symmetry or both, making the symmetry parameter local, introducing a gauge field, introducing the corresponding covariant derivative, and replacing partial derivatives with covariant derivatives will do the trick. | |
| Apr 16, 2013 at 9:07 | comment | added | Tomáš Brauner | Yes, this is exactly what I have in mind! I just wanted to avoid being too formal. But sometimes it is better if one wants to be precise :) | |
| Apr 16, 2013 at 9:02 | comment | added | joshphysics | Would it be accurate to say that one could rephrase your question as follows: "Under what conditions can one take a Lagrangian leading to an action which possesses a global Lie-algebraic symmetry and find a new Lagrangian depending on a gauge field that leads to an action with local Lie-algebraic symmetry such that the new Lagrangian agrees with the old Lagrangian when the gauge field is set to zero and the symmetry parameter is constant on spacetime?" | |
| Apr 16, 2013 at 8:39 | answer | added | Luboš Motl | timeline score: 7 | |
| Apr 16, 2013 at 8:07 | history | asked | Tomáš Brauner | CC BY-SA 3.0 |