# Goldstone theorem in Schwartz

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

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

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

• What do you mean by your first question? It is a product of the two functional derivatives evaluated from (28.13,14). Do you wish to edit it out? Commented Mar 11, 2020 at 14:13
• Have you heeded (28.8)? Have you commuted the annihilator to the right? Commented Mar 11, 2020 at 14:21
• @CosmasZachos I know how to do the functional derivative with $L$, but to do the other, I don’t know how a symmetry transformation leads to a derivative. If it’s simply $\pi(x)=F_\pi \theta$, then $\frac{\delta \pi}{\delta \theta}=F_\pi$. Commented Mar 11, 2020 at 14:25
• Ughhhh! The symmetry transformation means, informally, $\pi \mapsto \pi + \delta\pi = \pi + F \theta.$ Often there are higher terms in $\theta$, but one keeps the linear term to lowest order. Your instructor has failed to detail this? Commented Mar 11, 2020 at 14:31
• @CosmasZachos -i consider (28.8) merely as a constructed state defined as Goldstone bosons, and (28.9) gives a way to identify it. It’s not necessarily the same state mentioned in the following example,(so we are now trying to identity it). -I haven’t. I just let the terms with creation operator to the bra and anihilation operator to the ket vanish. Commented Mar 11, 2020 at 14:38

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}