A version of the Noether's theorem applies to local gauge symmetries. What is the Noether's charge associated with a non-abelian gauge symmetry such as the color $SU(3)$ and how is that derived? I want an expression for the color charge operator like we have an expression for the electric charge operator. Please see Eq. (9) and (11) of the answer here.


1 Answer 1


The $\mathrm{SU}(3)$ gauge symmetry is a local symmetry, and therefore it is not Noether's first, but Noether's second theorem that applies to it, which does not yield conserved quantities.

For $\mathrm{U}(1)$ gauge symmetries like the electromagnetic symmetry, there is also a global $\mathrm{U}(1)$ symmetry, and hence a conserved quantity. But the global symmetry associated to a non-Abelian gauge symmetry is just the center of the gauge group, which is discrete for $\mathrm{SU}(3)$, and hence there is no conserved quantity associated to it. This center symmetry has physical significance e.g. in models of confinement, see this question and its answer.

  • $\begingroup$ Could you please elaborate on the global symmetry associated to a gauge symmetry? (e.g. what it is, how to find it, does it always exist..) $\endgroup$
    – ersbygre1
    Oct 19, 2018 at 3:11
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    $\begingroup$ Or why it‘s the center of SU(3) and not just a global SU(3)? $\endgroup$
    – ersbygre1
    Oct 20, 2018 at 3:45
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    $\begingroup$ @Stephan Note that a non-Abelian gauge transformation acts as $A\mapsto gAg^{-1} + \mathrm{d}g$, so constant transformations that are not in the center still change the gauge field, and are therefore still gauge transformations between physically identical states. In this answer, by "global symmetry" I mean a symmetry that does not change the gauge field and therefore transforms between physically distinct states, since these are the symmetries to which one can meaningfully apply Noether's theorem. $\endgroup$
    – ACuriousMind
    Oct 20, 2018 at 10:58
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    $\begingroup$ That makes sense, thank you! So for the case of U(1), the center of the group coincides with the group itself, so that's why we have the additional U(1)? $\endgroup$
    – ersbygre1
    Oct 20, 2018 at 12:36

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