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

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

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  • $\begingroup$ @ 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 $\mathbb{Z}$? $\endgroup$ – user239970 Feb 24 '20 at 12:08
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    $\begingroup$ 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? $\endgroup$ – Nikita Feb 24 '20 at 12:22
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    $\begingroup$ Rotations with $\theta = 2\pi n$ is the same as unit element $\endgroup$ – Nikita Feb 24 '20 at 12:23
  • $\begingroup$ I see, I was getting confused with SU(2) case. $\endgroup$ – user239970 Feb 24 '20 at 12:39

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