First of all, I remember <PERSON> is discussing about global U(1) in the case of superfluid. By $U(1)$ we mean that that both Hamiltonian and Lagrangian are invariant under $\phi\rightarrow e^{i\theta}\phi$ transformation.
I think he mentioned that what he is doing in that section is called "integrate out" high energy fluctuations (fields.)
The goal here is---to study excitations around (classical) ground state (with <PERSON>'s notations $\phi=\sqrt{\rho}e^{i\theta}$), $$ \phi_0=\sqrt{\rho_0}e^{i\theta_0}, $$ where $\phi_0$ and $\theta_0$ correspond to the ground state.
Comments:
1. Note that this classical ground state is asymmetric because it dose not obey the same symmetry, $U(1)$ of the Lagrangian.
2. If one write down a self-consistent bosonic theoy in a not hand-wavy way, he will find that tuning the chemical potential (which is correspond to $\rho$ in above) can obtain the transition between different phases. And for the same reason, for some range of chemical potential, one can have symmetry-breaking phase---superfluid phase in our case. Otherwise, one would get a trivial phase with symmetric ground state with no degeneracy.
Based my own experience of using AZee's book as textbook in QFT course, I prefer to derive everything in detail, independently, with my notations. 976
Thus I'll slight change notations from here.
To be more quantitative, we will apply the language of Lagrangian for non-relativistic non-interacting bosons: $$ \mathcal{L}=i \Phi(x)^\dagger\dot{\Phi}(x)-\frac{1}{2m}\nabla\Phi(x)^\dagger\nabla\Phi(x)+\mu\Phi(x)^\dagger\Phi(x), $$ where $\mu=ng$ is the chemical potential and $n$ is the particle density of the ground state. We can further include interaction term which triggers the symmetry breaking $$ \mathcal{L}=i \Phi(x)^\dagger\dot{\Phi}(x)-\frac{1}{2m}\nabla\Phi(x)^\dagger\nabla\Phi(x)+\mu\Phi(x)^\dagger\Phi(x)-\frac{g}{2}\big(\Psi(x)^\dagger\Phi(x)\big)^2\nonumber\ =i\Phi(x)^\dagger\dot{\Phi}(x)-\frac{1}{2m}\nabla\Phi(x)^\dagger\nabla\Phi(x)-\frac{g}{2}\big(n-\Phi(x)^\dagger\Phi(x)\big)^2+\text{constant}. $$ To see the Goldstone modes, we can use polar coordinates to decoupled the amplitude and phase fields, $$\Phi(x)\equiv\sqrt{\rho(x)} e^{i\theta(x)}$$. Then we have $$ \mathcal{L}=\frac{i}{2}\dot{\rho}-\rho\dot{\theta}-\frac{1}{2m}\big[\frac{1}{4\rho}(\nabla\rho)^2+\rho(\nabla\theta)^2\big]-\frac{g}{2}(n-\rho)^2, $$ where $\frac{i}{2}\dot{\rho}$ can be ignored since it's total derivative. In order to find the minimum of this Lagrangian, we can take derivative with respect to $\rho$, set it to zero and get $\rho_{min}=0$ as minimum This vacuum value has degeneracy under global $U(1)$ symmetry which will be broken if we pick out a particular vacuum---$\theta_0$. Fixing the phase of the ground state as $\theta_0$ is the same at every point in space and can take any value.
By <PERSON>'s theorem, breaking a continuous symmetry, global (or local with extra subtleties), will lead to emergence of Goldstone bosons. 804