Typically in QFT one derives the Feynman rules by differentiating certain terms in the Lagrangian w.r.t the relevant fields. So for instance if our term is $\mathscr{L} =\phi_1\phi_2\phi_{\mathrm{Goldstone}}$ we can find the corresponding Feynman rule by performing a Fourier Transformation and calculating $\frac{i\delta \mathscr{L}}{\delta \phi_1(p_1) \delta \phi_2(p_2) \delta \phi_{\mathrm{Goldstone}}(p_3)}$. Such a calculation will give us a Feynman diagram with 3 external legs: \begin{equation}\begin{split} ----&\cdot----\\ &|\\ &|\\ &|\\ &| \end{split}\end{equation}

The Problem

Now consider that the theory mentioned above has a Noether current that can create or destroy Goldstone bosons. By create/destroy I mean that the matrix element $$<0|J^{\mu}|Goldstone> \neq 0.$$ I'm wondering if there is a neat way to append the Noether current to one of the three external legs. That is I want to know if there is some clever way I can use to determine the Feynman rule of a diagram with 3 legs were we attach the Noether current to only one of them. Schematically it would look something like \begin{equation}\begin{split} ----&\cdot----\cdot J^{\mu}\\ &|\\ &|\\ &|\\ &| \end{split}\end{equation} For concreteness let us say that we have an explicit expression for the Noether current that looks like $J^{\mu} = p^{\mu}\phi_{\mathrm{Goldstone}}$, where $p^{\mu}$ is the four-momentum of the Goldstone field.

Initial idea

I suspect that a way to go about solving this problem is to view the Feynman diagram as a scattering process $\phi_1\rightarrow \phi_{\mathrm{Goldstone}}+\phi_2$. In that way we can define an initial state $|i>$ containing $\phi_1$, and a final state $|f>$ containg the Goldstone boson, that was created by the current $J^{\mu}$, and the field $\phi_2$. If we do that I guess the Feynman diagram would correspond to the scalar $<f|J^{\mu}|i>$. To proceed it seems natural to write $J^{\mu} = p^{\mu}\phi_{\mathrm{Goldstone}}$ in terms of creation and annhilation operators: $$\phi_{\mathrm{Goldstone}} = \int \frac{d^3 k}{(2\pi)^3}\frac{1}{\sqrt{\mathrm{Energy}}}\left(a_{\mathbf{p}}e^{-i\mathbf{p}\mathbf{x}}+ a_{\mathbf{p}}^{\dagger}e^{i\mathbf{p}\mathbf{x}}\right).$$ To obtain the Feynman rule I suspect we then have to do Wick contractions. However, I'm unsure how to do this for this case.

  • $\begingroup$ just a comment: your initial lagrangian $\phi_1\phi_2 \phi_{Goldstone}$ actually breaks the shift symmetry making the would be Goldstone not a Goldstone in the end. What you could look instead at is something like $(\phi_1 \partial_\mu \phi_2-\partial_\mu\phi_1 \phi_2)\partial_\mu\phi_{goldstone}$ $\endgroup$ – TwoBs Apr 30 '17 at 7:10

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