Let us consider a scalar field $\phi$ with the potential $V(\phi)=\frac{1}{2}\mu^2\phi^2+\frac{1}{4}\lambda\phi^4$ for $\lambda>0$ and $\mu^2<0$. The Lagrange density then yields

$$ \mathcal{L}(\phi)=\frac{1}{2}(\partial_u\phi)(\partial^u\phi)-V(\phi) $$ with a non-zero vacuum expectation value $v=\sqrt{-\frac{\mu^2}{\lambda}}$. One can show that the Lagrangian is invariant under global U(1) gauge phase space transformation. If we expand the Lagrangian about the vacuum expectation value by writing $\phi(x)=v + \eta(x)$, where $\eta(x)$ is a small perturbation one easily can show that $$ \mathcal{L}(\eta)=\frac{1}{2}(\partial_u\eta)(\partial^u\eta)-\lambda v^2\eta^2 - \lambda v \eta^3 - \frac{1}{4}\lambda\eta^4 + \frac{1}{4}\lambda v^4 $$ That is, the global U(1) symmetry has been broken and a new massive gauge boson $\eta$ with mass $m_{\eta}=\sqrt{2v^2\lambda}$ introduced. Now, the goldstone boson theorem states, that for each spontanous symmetry breaking at least one new $\textbf{massless}$ boson emerges. How is this compatible with what I have shown?

  • 2
    $\begingroup$ It is real simple. For a real φ there is no global continuous symmetry: your U(1) is a bluff. Exhibit it. You say it is global, but you called it gauge a few paragraphs before. Surely, you mean discrete $Z_2$. Goldstone's theorem has little to do with what you display. $\endgroup$ Commented Jul 20, 2017 at 19:05
  • $\begingroup$ Ok, so basically this is because there is only a discrete number of vacuum expectation values, namely +v and -v? $\endgroup$
    – Mark
    Commented Jul 20, 2017 at 19:11
  • $\begingroup$ Yes, of course. Your teacher did not do this first, before going on to continuous groups?? $\endgroup$ Commented Jul 20, 2017 at 19:12
  • $\begingroup$ It is more a course on Higgs phenomenology, therefore the focus was not so much on the theoretical side ;) $\endgroup$
    – Mark
    Commented Jul 20, 2017 at 19:14
  • $\begingroup$ Look at p 162 here... $\endgroup$ Commented Jul 20, 2017 at 19:52

1 Answer 1


Your Lagrangian is not invariant under U(1) transformations $\phi\to e^{i\theta}\phi$. Hence there is no contradiction. The Lagrangian you wrote is that of a real scalar field $\phi$ and it is only invariant under $\phi\to \pm \phi$ i.e., under the discrete symmetry $\mathbb{Z}_2$. The Goldstone theorem applies for continuous global symmetries.

  • $\begingroup$ I understood some of these words. $\endgroup$ Commented Aug 7, 2017 at 14:56
  • $\begingroup$ @LegitimateWorkUser Which part you did not understand? $\endgroup$
    – SRS
    Commented Aug 7, 2017 at 15:10

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