Timeline for When can a global symmetry be gauged?
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Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
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May 25, 2016 at 10:11 | history | edited | AccidentalFourierTransform | CC BY-SA 3.0 |
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Apr 18, 2013 at 8:28 | vote | accept | Tomáš Brauner | ||
Apr 18, 2013 at 7:46 | comment | added | Tomáš Brauner | I think I have understood your point, thanks once again! | |
Apr 17, 2013 at 15:56 | comment | added | David Bar Moshe | @Tomáš Brauner: Yes, the obstruction to gauging can be understood intuitively by means of the Dirac's constraint theory. If the theory can be gauged, the currents couple to the gauge field. Since the time component of the gauge field is non-dynamical, the charge densities of the symmetry currents will become constraints, i.e., must become zero on the gauge surface which can be thought as a reformulation of the theory with gauge invariant fields. But, then how can the bracket of two vanishing quantities give a nonzero constant. | |
Apr 17, 2013 at 15:03 | comment | added | Tomáš Brauner | I read your previous answer and got the point with the classical anomaly, thanks! Before I get into the mathematical details, is there an intuitive way to understand why an obstruction like a central charge in the Poisson brackets of the symmetry generators prevents gauging the symmetry on the classical level? Also, is there an elementary way to decide whether the symmetry can be gauged which does not use the canonical structure? (I think of low-energy effective Lagrangians which can contain an arbitrary number of derivatives.) | |
Apr 17, 2013 at 13:10 | comment | added | Tomáš Brauner | I see, so my guess that the obstruction to gauging is related to the central charge (in the commutator of the generators of real and imaginary shifts of $\psi$) was in the right direction. Thanks for a clear and general statement! I will study your previous answer carefully. Apparently, there is a lot they didn't tell me at school :) | |
Apr 17, 2013 at 12:39 | history | answered | David Bar Moshe | CC BY-SA 3.0 |