# Vacuum Manifold of $U(1)$ theory and Goldstone theorem

I want to know if my understanding of the Goldstone theorem is correct.

What I know is that the number of Goldstone is equal to the rank of $$G/H$$ where $$G$$ is the symmetry of the Lagrangian before symmetry breaking and $$H$$ is the symmetry of the vacuum manifold

$$\rho (h)\phi_0 = \phi_0 , \forall h\in H.$$

Now as an example consider a potential $$V=(|\phi|^2-2r^2)^2 ~,~~~ \sqrt{2}\phi=\varphi+i\chi$$ Where $$\phi$$ is in the fundamental representation of $$U(1)$$. Now solving $$\delta V(\phi_0)=0$$ gives

$$\varphi^2_0 + \chi^2_0 = r^2 \implies (\varphi_0,\chi_0)\in S^1 \cong SO(2)$$ So in this case do I have

$$\color{red}{H=\mathbb{Z}~~?}$$

Because $$e^{i\theta}\phi_0 = \phi_0 ~,~~~\forall \theta = 2n\pi ~~~?$$

• Did you mean that you solve $V^\prime (\phi_0) = 0$? – Nikita Feb 24 '20 at 11:35
• @Nikita Yes. Sorry I didn't mention this but $\phi_0$ is found by solving $\delta V(\phi_0)=0$ – user239970 Feb 24 '20 at 11:37

1) You found minimum of potential: $$V^\prime (\phi_0)= 0 \Rightarrow |\phi_0|^2 = 2r^2$$

2) You choose minimum, for example: $$\phi_0 = \sqrt{2} r$$

3) This minimum is not invaritant under action of any nontrivial subgroup of initial $$U(1)$$ group. So $$H = {1}$$ is group with one trivial element.

4) So $$dim(G/H) = dim (U(1)) = 1$$ and we have one Goldstone boson.

5) Actually, one can easily check it, if consider $$\phi$$ in following form:

$$\phi = \varphi e^{i\chi}$$

It is trivial to see that potential doesn't depend on field $$\chi$$, so it can't generate mass for $$\chi$$ after expansion over minimum of potential.

• @ Nikita Thank you for your answer. When you said "trivial element" do you mean $H=\{\mathbb{1}\}$? The reason why I thought $H=\mathbb{Z}$ is because $U(1)$ is a double cover of $SO(2)$ so we have $U/\mathbb{Z}=SO(2)=S^1$? in which case we need $H$ to be $\mathbb{Z}$? – user239970 Feb 24 '20 at 12:08
• About trivial element yes. U(1) is isomorphic to SO(2) check for example math.stackexchange.com/questions/308418/… It is different for example in case of $SO(3) = SU(2)/Z_2$. Did I correctly understand you? – Nikita Feb 24 '20 at 12:22
• Rotations with $\theta = 2\pi n$ is the same as unit element – Nikita Feb 24 '20 at 12:23
• I see, I was getting confused with SU(2) case. – user239970 Feb 24 '20 at 12:39