I understand the interpretation given in topics like this one that gauge symmetries are "fake" in the sense that they do not represent an actual difference in physical states. I also know that gauge symmetries are those that are local. But how would I see the relation between the two? Or, to put it another way: given a Lagrangian, how can I see directly whether the "symmetry" is a real symmetry or not?

For concreteness, consider the following Lagrangians:

$$\mathcal{L}_S = \frac12 \partial_\mu \vec\phi \cdot \partial^\mu \vec\phi - \frac12 m^2 \vec\phi \cdot \vec\phi $$

$$\mathcal{L}_M = -\frac14 F_{\mu\nu}F^{\mu\nu}$$

In the first case we have that $\vec \phi \to R \vec\phi$ is a symmetry when $R \in SO(3)$, in the second case we have the usual gauge symmetry of pure electromagnetism. Why is the latter a gauge symmetry and not the former? If we label states with the value of the fields, why do $|A_\mu\rangle$ and $|A_\mu + \partial_\mu \lambda\rangle$ represent the same physical state but $|\vec\phi\rangle$ and $|R\vec\phi\rangle$ do not?

I had a feeling this might have to do with the Noether currents, so I calculated them: in the first case we have three currents given by $J_a^\mu = i \partial^\mu \phi_a (T_a)_{bc} \phi_b$ with $T_a$ the generators of $SO(3)$, in the second case I get a current of the form $J^\mu = -F^{\mu\nu}\partial_\nu \lambda$. Does it have to do with the fact that there are an infinite number of currents, one for each $\lambda$?

  • $\begingroup$ hint: a synonym for gauge symmetry is "local symmetry". [also, google "first and second Noether theorems"] $\endgroup$ – AccidentalFourierTransform Jul 5 '17 at 18:39
  • $\begingroup$ Gauge transformation = transformation that depends on space-time coordinates in some way $\endgroup$ – JamalS Jul 5 '17 at 18:40
  • $\begingroup$ Those are good points, let me edit the question to make my point clearer. $\endgroup$ – Javier Jul 5 '17 at 18:40

You can't tell just by looking at the Lagrangians. In fact, gauge "symmetry" is something you put in by hand. For instance, you could quantize the $U(1)$ theory you denote as $\mathcal{L}_M$ without gauging, taking the $U(1)$ to be global (you can quantize either using harmonic oscillators or in the path integral picture). This would give you a completely well-defined quantum theory, but one that has too many modes to describe the physical Maxwell theory, and you'd get some negative-norm states. Likewise, you could gauge the $SO(3)$ theory $\mathcal{L}_S$ (you'd have to replace the $\partial_\mu$'s by covariant derivatives $D_\mu = \partial_\mu + A_\mu$ where the $A_\mu$'s transform in the adjoint of $SO(3)$).

By the way, the formula for your $U(1)$ Noether current is certainly not right, it shouldn't depend on $\lambda$. By Lorentz invariance and dimensional analysis it should be something like $J_\mu = a_1 \, A_\mu \partial_\nu A^\nu + \ldots$.

  • 1
    $\begingroup$ Downvoter: could you explain why this answer is wrong? I don't have any particular opinion, but I'd like to know. $\endgroup$ – Javier Jul 5 '17 at 19:32

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