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 $$ \int d^3 x \exp(i \bar p \cdot \bar x ) \sum_m \pi_m(x) \frac{d\phi_m}{d\alpha} .$$

*

*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.


*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.


*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$.
 A: 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.
