Saturday, July 6, 2019

symmetry - Noether charge of local symmetries


If our Lagrangian is invariant under a local symmetry, then, by simply restricting our local symmetry to the case in which the transformation is constant over space-time, we obtain a global symmetry, and hence a corresponding Noether charge.


Because, however, this Noether charge didn't come from just any old symmetry, but in fact, a local symmetry, we might be able to say something special about it. In particular, I believe that the Noether charge should vanish, but I don't know why. If this is indeed the case, how do we prove it?


(Note that I don't want to make the assumption that local=gauge (i.e. non-physical).)




Answer



In its most comprehensible derivation, Noether's procedure derives the current by considering the global symmetry transformation whose parameters $\epsilon$ are made to depend on the spacetime coordinates. Because $\delta S$ has to vanish if $\epsilon$ is constant, the actual variation $\delta S$ in the generalized case has to be proportional to the integral of spacetime derivatives $\partial_\mu \epsilon$ multiplied by some coefficients $J^\mu$, the currents. By integrating by parts, one may then show that the current obeys the continuity equation if the equations of motion are satisfied.


Now, when the symmetry is actually local, the "generalization" of the global transformation isn't a real generalization: it's a symmetry by itself. So because the action is locally symmetric, $\delta S$ vanishes for any configuration $\epsilon(x^\alpha)$, including a non-constant one, which means that all the coefficients $J^\mu$ actually vanish themselves, as you said. These conditions (constraints plus equations of motion) can be equivalently obtained from the variation of fields like $A_\mu$.


Because the currents are classically vanishing – or, using a more general description in quantum mechanics with an extended Hilbert space, they have to annihilate the physical states in quantum mechanics – it really means that the local symmetry is gauge. You don't need to assume this fact; we have just derived it. So you can't avoid it, either.


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