Context

I found many proof of the symmetry breaking is the following:

1. Let $\langle \delta \phi(x) \rangle$ be the corresponding symmetry transformation w.r.t. the charge $Q = \int d^Dx\; J^0(x)$.

2. $\langle \delta\phi(p=0) \rangle = \int d^Dx\; \langle \delta \phi(x) \rangle \neq 0$ indicates the symmetry is spontaneous breaking.

3. If the symmetry is spontaneous breaking, one can prove: $$\langle J^{\mu}(p)\phi(-p) \rangle \sim \frac{p^\mu}{p^2}$$ This pole structure indicates massless state excitations.

Such proof can be found at page 10 section II.C. in 2303.01817. Similar proof can be found in S.Weinberg well known QFT book(vol. 2) at page 169.

Question

1. Why the pole structure in step 3 indicates massless state excitations? It seem more reasonable to me that one say the pole structure of $\langle \phi\phi \rangle \sim \frac{1}{p^2}$ indicates massless state. Further, even in the most trivial example of massless complex scalar field $\mathcal{L} = \frac{1}{2}|\partial\phi|^2$, such pole structure doesn't exist. Namely, the current in the theory is $J^{\mu} = i(\phi\partial\phi^* - \phi^*\partial\phi)$ and $\langle J^{\mu}\phi \rangle = 0 $ since there are odd number of $\phi$ in the v.e.v. There is no reason for me to believe that step 3. is correct.

2. From my understanding, one say a symmetry is spotaneous breaking if $Q|\Omega\rangle \neq 0$, in which $|\Omega\rangle$ is the ground state. Is it equivalence to the definition in the step 2?