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

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

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

  • $\begingroup$ 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? $\endgroup$ Mar 11, 2020 at 14:13
  • $\begingroup$ Have you heeded (28.8)? Have you commuted the annihilator to the right? $\endgroup$ Mar 11, 2020 at 14:21
  • $\begingroup$ @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 $\frac{\delta \pi}{\delta \theta}=F_\pi$. $\endgroup$
    – RicknJerry
    Mar 11, 2020 at 14:25
  • 1
    $\begingroup$ 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? $\endgroup$ Mar 11, 2020 at 14:31
  • $\begingroup$ @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. $\endgroup$
    – RicknJerry
    Mar 11, 2020 at 14:38

1 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: $$ \begin{align} \langle \Omega |J^\mu(x)|\pi(p)\rangle= & F_\pi\langle \Omega |\partial^\mu\pi(x)|\pi(p)\rangle \\ =&F_\pi \partial^\mu e^{-ipx} \\ =&-ip^\mu F_\pi e^{-ipx}. \end{align} $$


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