Goldstone theorem in Schwartz

Question

On page 566, Schwartz’s QFT book, to see the $\pi$ is the Goldstone boson, it reads: $$J^\mu=\frac{\partial L}{\partial(\partial_\mu \pi)} \frac{\delta \pi}{\delta \theta}=F_\pi \partial_\mu \pi \tag{28.15}$$ $$\langle\Omega|J^\mu(x)|\pi(p)\rangle=ip^\mu F_\pi e^{-ipx} \tag{28.16}$$ My question is:

• in the first equation, how is $\frac{\delta \pi}{\delta \theta}=F_\pi$ derived from the symmetry translation $\pi(x) \rightarrow \pi(x)+F_\pi \theta$ ?
• how to derive the second equation?

My attempt to the second equation: $$\langle \Omega|J^\mu(x)|\pi(p)\rangle= F_\pi \langle \Omega|\partial_\mu\pi \pi|\Omega\rangle$$ Substitute $\pi=\int \frac{d^3 p}{(2\pi)^3\sqrt{2\omega_p}}[a_p e^{-ipx}+a_p^\dagger e^{ipx}]$ into it, I get

$$F_\pi \langle \Omega| \int \frac{d^3 p}{(2\pi)^3\sqrt{2\omega_p}}[a_p (-ip^\mu)e^{-ipx}+a_p^\dagger (ip^\mu)e^{ipx}] \int \frac{d^3 k}{(2\pi)^3\sqrt{2\omega_k}}[a_k e^{-ikx}+a_k^\dagger e^{ikx}] |\Omega\rangle$$ $$=F_\pi \langle \Omega| \int \frac{d^3 p}{(2\pi)^3\sqrt{2\omega_p}}[a_p (-ip^\mu)e^{-ipx}] \int \frac{d^3 k}{(2\pi)^3\sqrt{2\omega_k}}[a_k^\dagger e^{ikx}] |\Omega\rangle$$ $$=F_\pi \int \frac{d^3 p}{(2\pi)^3\sqrt{2\omega_p}}\int \frac{d^3 k}{(2\pi)^3\sqrt{2\omega_k}}e^{-i(p-k)x}\langle \Omega|a_p(-ip^\mu)a_k^\dagger |\Omega\rangle$$ $$=F_\pi( -ip^\mu) \int \frac{d^3 p}{(2\pi)^3\sqrt{2\omega_p}}\int \frac{d^3 k}{(2\pi)^3\sqrt{2\omega_k}}e^{-i(p-k)x}(2\pi)^3\delta^3(p-k)$$

$$=F_\pi( -ip^\mu) \int \frac{d^3 p}{(2\pi)^3 2\omega_p}$$

Answer 1

For the first equation, consider an infinitesimal transformation, $\pi(x) \rightarrow \pi(x)+F_\pi \delta \theta$. We have $\delta \pi(x) = F_\pi \delta \theta$, so $\frac{\delta \pi(x)}{\delta \theta}= F_\pi$.

For the second equation, your first mistake is on equating $|\pi(p)\rangle$ with $\pi|\Omega\rangle$. $\pi$ is a field, not a single creation operator.

To derive that result, you just need to show that since $|\pi(p)\rangle$ is defined to be the state created by the $\pi$ field, $\langle \Omega |\pi(x)|\pi(p)\rangle=e^{-ipx}$. Then: $$ \begin{align} \langle \Omega |J^\mu(x)|\pi(p)\rangle= & F_\pi\langle \Omega |\partial^\mu\pi(x)|\pi(p)\rangle \\ =&F_\pi \partial^\mu e^{-ipx} \\ =&-ip^\mu F_\pi e^{-ipx}. \end{align} $$