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This question already has an answer here:

From what I understand, a global symmetry is a symmetry that holds at every point in space-time. On the other hand, a gauge symmetry is a redundancy in the description of a physical state.

Now, when localizing a global symmetry, we usually impose $\partial_\mu \rightarrow D_\mu = \partial_\mu + A_\mu$. For example, for scalar field theory we might have

$$\mathcal{L} = |\partial_\mu \phi|^2 \rightarrow |D_\mu \phi|^2$$

Now, this Lagrangian still enjoys the global symmetry $\phi \rightarrow e^{i\theta}\phi$ for constant $\theta$ after localizing the global symmetry as the global symmetry is a subset of the local symmetry. However, in performing this localization my understanding is that you end up gauging the symmetry.

  1. How can I see that the global symmetry was originally not a gauge symmetry, but after localization is a gauge symmetry? Naively I expected $\phi \rightarrow e^{i\theta} \phi$ to correspond to physically equivalent states similar to how quantum states are only defined up to an overall phase factor.
  2. Are there general principles in which one can easily see that some symmetry is in fact a gauge symmetry?
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marked as duplicate by Qmechanic Jan 27 '17 at 2:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The global symmetry is not a gauge symmetry after localization! Only the non-constant $U(1)$-valued functions are gauge transformations. The global U(1) acts non-trivially. It has, too, because it's the charge operator. You don't want to set all the charges to zero! $\endgroup$ – user1504 Jan 27 '17 at 1:39
  • $\begingroup$ Note that the physics term global transformation means an $x$-independent transformation, not a globally defined transformation. Possible duplicates: physics.stackexchange.com/q/48188/2451, physics.stackexchange.com/q/122726/2451 and links therein. $\endgroup$ – Qmechanic Jan 27 '17 at 2:08