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?
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$\begingroup$ Why do you think they'd still hold? Also, question of whether Ward identities stilll hold after ssb very relevant here $\endgroup$– innisfreeCommented Jan 8, 2017 at 12:16
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$\begingroup$ @innisfree I'm currently trying to understand the famous no-go theorems (Coleman-Mandula's, Haag–Lopuszanski–Sohnius'). Their argument is that when we put the internal symmetry group and the Poincare group in some larger group, we get additional conservation laws. These addtional conservation laws overconstrain the S-matrix in such a way that no interactions are possible, i.e. the S-matrix is trivial in such theories. I'm trying to figure out to what extend this argument holds for broken symmetries. $\endgroup$– jakCommented Jan 8, 2017 at 12:23
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1$\begingroup$ Of course they do, cf Goldstone's argument. However, the conventional step in parlaying current conservation to time-invariant charges fails by Picasso's thm (in that WP article) on account of infrared troubles. It is this very persistence of current conservation that endows SSB with its miraculous properties! A SB symmetry is just the same symmetry realized in the nonlinear Nambu-Goldstone mode, but it is magnificently still present, albeit hidden. $\endgroup$– Cosmas ZachosCommented Jan 8, 2017 at 15:52
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1$\begingroup$ As far as I understand the proof of Coleman-Mandula, it doesn't hold for spontaneously broken symmetries. See the comment here en.wikipedia.org/wiki/… . Coming to you question: of course the conservation equation for the current holds; the variation of a field under an infinitesimal transformation, i.e. the commutator with the conserved charge $Q$ exists too. But $Q$ isn't well defined on the Hilbert space unless $Q|0\rangle=0$, which isn't the case by assumption. $\endgroup$– TwoBsCommented Jan 9, 2017 at 21:00
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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.
answered Jan 8, 2017 at 18:22