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When a symmetry is spontaneously broken what happens to the corresponding Noether's charge? Since the action never breaks the symmetry only the vacuum does do we expect the charge to remain conserved?

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    $\begingroup$ The charge operator becomes a ill-defined operator (the one you define as space integral of the time-component of the conserved current). However, its commutator is still well defined and that's the only ingredient you need to generate the variation of the fields, namely $[Q,\phi] = \delta \phi$ is still a good definition of variation. $\endgroup$ – apt45 Dec 29 '17 at 17:53
  • $\begingroup$ Look at this post physics.stackexchange.com/a/228616/47373 $\endgroup$ – apt45 Dec 29 '17 at 18:32
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The conservation of Noether's charge relies on the assumption that the 3-current vanish at the boundary of spacetime which in turn requires the field $\phi(x)$ itself to vanish at spatial infinity. In case of spontaneously broken symmetry, although the action does not break the symmetry, the field at infinity does not vanish but goes to a constant value corresponding to one of the minima of the classical potential. This is the same reason why odd correlation functions can be nonzero in a spontaneously broken $\phi^4$-theory even though the action itself respects the $\phi\to-\phi$ symmetry.

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