Confusion of the Proof of Goldstone Theorem in QFT

Question

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?

Answer 1

1. The pole of a generic correlation function always takes the form $$ \langle\mathcal O_1\mathcal O_2\mathcal O_3\cdots\mathcal O_n\rangle\sim\frac{1}{p^2-m^2}\langle \mathcal O_3\cdots\mathcal O_n\rangle $$ where $m^2$ is the mass of a one-particle state in your theory, see Weinberg's "polology" for the precise statement and the proof.

   Hence a pole $1/p^2$ in $\langle J\phi\rangle$ signals a massless one-particle state. The numerator $p^\mu$ is fixed by Lorentz invariance and does not matter for this argument.

   It is true that $\langle\phi\phi\rangle\sim 1/p^2$ also implies the existence of a massless state. But you cannot prove that $\langle\delta\phi\rangle\neq0\Rightarrow\langle\phi\phi\rangle\sim 1/p^2$. What you can prove, instead, is that $\langle\delta\phi\rangle\neq0\Rightarrow\langle J\phi\rangle\sim 1/p^2$, and this is enough to establish the existence of a massless state.

   Any $1/p^2$ pole, in any correlation function, demonstrates the existence of a massless state. So you work with whatever correlation function you can actually compute. In general $\langle\phi\phi\rangle$ is not computable but $\langle J\phi\rangle$ is.

   1.5. In the free scalar theory, the $U(1)$ symmetry that acts as $\phi\mapsto e^{i\alpha}\phi$ is NOT spontaneously broken, so there is no contradiction.

   That being said, the shift symmetry $\phi\mapsto\phi+\alpha$ IS spontaneously broken, now the current being $J_\mu=\partial_\mu\phi$. And here you can check that the correlation function does take the expected form.

   In general, Goldstone fields always transform linearly under the broken symmetry, so if $\phi\mapsto e^{i\alpha}\phi$, then $\phi$ cannot be a Goldstone field. Its argument can, though.

2. If $Q|\Omega\rangle=0$, then $\langle\Omega|[Q,\phi]|\Omega\rangle=0$ as you can easily check by acting with $Q$ on whatever state it is touching. Hence, $\langle\delta\phi\rangle\neq0$ implies that $Q|\Omega\rangle\neq0$, and you recover the usual definition.

References

1. Weinberg S. - Quantum theory of fields, Vol.1, §10.2.