No massless new goldstone boson after 1D U(1) spontanous symmetry breaking?

Question

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?

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.