# Goldstone modes/momentum generation from the vacuum

I'm quite confused by two short paragraphs in Schwarz 28. He proves that $$Q = \int d^3 x J_0(x) = \int d^3 x \sum_m \frac{\partial L}{\partial \dot \phi_m} \frac{\delta \phi_m}{\delta \alpha} \tag{28.5}$$ (where $$J_\mu$$ is a conserved current) is a generator for the symmetry transformation, which is fine, and that $$Q | \Omega\rangle$$ is degenerate with the ground state due to $$H$$ commuting with $$Q$$ due to charge conservation.

Fine, the issue is when he states $$| \pi(\bar p ) \rangle = \frac{-2i}{F} \int d^3 x \exp(i \bar p \cdot \bar x ) J_0(x) | \Omega \rangle .\tag{28.8}$$ If I insert one of the definitions above into the below, we end up with something like $$\int d^3 x \exp(i \bar p \cdot \bar x ) \sum_m \pi_m(x) \frac{d\phi_m}{d\alpha} .$$

1. Is this just a definition, or is there a deeper mathematical reason? Intuitively, it makes sense since we expect $$p = 0$$ to produce $$| \pi(0 ) \rangle$$ so it's just like we're moving it in momentum space.

2. Furthermore, he then states they have energy $$E(p) + E_0$$, which also confuses me: when applying $$H$$ to the above, assuming we can pass it into the integral, I'd expect to see some kind of obvious separation like $$H( \int d^3 x J_0 (x) | \Omega\rangle ) + H (something else)$$ to correspond to the form above, but this is also unclear to me on how to proceed.

3. I'm also confused on his final assertion, that since $$E(p) \rightarrow 0$$ as $$p \rightarrow 0$$, (which makes sense), that there is a massless dispersion relation. this means that $$E(k)$$ is not dependent on mass. this isn't obvious to me, since if it is of the form $$m k$$ or something silly of this nature, then it will still drop out as $$k \rightarrow 0$$.

The formula for $$|\pi(p)\rangle$$ is a definition. It is just some state in the theory, but we know that it has 3-momentum $$p$$ because by acting with a momentum operator $$P$$
$$P_k |\pi(p)\rangle = \frac{-2i}{F}\int d^3 x e^{ip\cdot x} P_k J_0(x)|\Omega\rangle =\frac{-2i}{F}\int d^3 x e^{ip\cdot x} [P_k, J_0(x)]\,|\Omega\rangle = \frac{-2i}{F}\int d^3 x e^{ip\cdot x} i\partial_k J_0(x)\,|\Omega\rangle = p_k |\pi(p)\rangle$$ where the last step follows from integration by parts. (Depending on your sign conventions you might get $$-p_k$$ instead. If so just define your state with the opposite sign of exponential.)
When he says the energy is $$E(p)+E_0$$, that is also something like a definition. The energy of all states is shifted by the energy of the vacuum and the dispersion relation $$E(p)$$ is defined with reference to that. The rest mass is defined to just be $$E(0)$$ (e.g. the standard relativistic dispersion relation is $$E(p)=\sqrt{p^2+m^2}$$). The argument that the energy of $$|\pi(0)\rangle$$ is $$E_0$$ shows that the rest mass is zero. (Maybe everything would be clearer if we just set $$E_0=0$$ throughout.)
You can have a dispersion relation like $$E(k) = mk$$, but the quantity "$$m$$" is dimensionless (in units $$c=1$$) and is interpreted as a velocity not a mass.