Do conservation laws still hold for spontaneously broken symmetries?

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?

• Why do you think they'd still hold? Also, question of whether Ward identities stilll hold after ssb very relevant here Jan 8 '17 at 12:16
• @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.
– jak
Jan 8 '17 at 12:23
• 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. Jan 8 '17 at 15:52
• 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. Jan 9 '17 at 21:00

• 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.)