For a given gauge symmetry $G$, we get via Noether's theorem conservation laws
$$ \partial_\mu j^\mu = 0 . $$
Do these conservation laws still hold, when $G$ gets broken spontaneously through a non-zero vacuum expecation value of some scalar field?
Answer
The takeaway, since this question keeps being asked and answered, in several forms, e.g. 33917,150644, and is at the heart of the Goldstone theorem:
- In SSB, the current is conserved, $ \partial_\mu j^\mu = 0$, as you wrote, and does crucial yeomanly work--that is why one is interested in SSB theories, to start with. The current starts out as linear as opposed to the normal bilinear, in the fields, so goes like $j_\mu = -v^2 \partial_\mu \phi ~$ for U(1), for instance. This means the goldstons shift under the related symmetry transformation, and so can push into and out of the nontrivial vacuum.
- But the corresponding Noether charge is not well-defined, so its time derivative is also ill defined, and not vanishing, as per the Fabry-Picasso theorem--an infrared phenomenon. So ${d\over dt} Q = {d\over dt} \int_x J^0(x) \neq 0 $ because Q itself is not meaningful. (So using it in the C-M theorem is definitely unwarranted.)
- You may live dangerously and utilize conservation of the unbroken charge in the lagrangian, as, e.g. in 149324, but you must ensure neither the Higgs that effects SSB nor its consequences (in fermion masses) are present in the terms involved. It is highly dangerous and contraindicated to novices.
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