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There's interesting statement about amplitude of Goldstone bosons:

I wanna to understand simple argument for statement of vanishing of amplitudes in the single scalar soft limit. In principle, it is obvious that this have direct relation to symmetry of model. But technically I can't find argument for this.

So, how to demonstrate, that amplitudes for Goldstone bosons vanish in the single scalar soft limit?

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    $\begingroup$ looks like the amplitude has to be proportional to $p_\mu$, which vanishes in the limit $p_\mu \rightarrow 0$ $\endgroup$ Nov 22, 2020 at 8:28

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This statement is called in the literature Adler zero. Nambu-Goldstone boson couples to the associated Noether current with a strength parametrized by the decay constant $f$ : $$ \langle 0 | J_\mu (x)| \phi(p) \rangle = -i p_\mu f e^{-i p x} $$ The matrix element between physical states has a pole for $p^2 \rightarrow 0$ and the residue corresponds to the scattering amplitude for NGB emission. For the $S$-matrix element between states $\alpha$ and $\beta$ : $$ \langle \alpha | J_\mu (x)| \beta \rangle = f \frac{p_\mu}{p^2} A_n (\alpha + \phi(p), \beta ) + R_{\mu}(p) $$ where $A_n$ is on-shell amplitude for emission of the $\phi$ with $p = P_\alpha - P_\beta$, and $R_\mu (p)$ is regular in limit $p^2 \rightarrow 0$. Due to the conservation of the current $p_\mu \langle \alpha | J^\mu (x)| \beta \rangle = 0$, so: $$ A_n (\alpha + \phi(p), \beta ) = -\frac{p_\mu R^\mu (p)}{f} $$ Further assuming that $R_\mu (p)$ is regular in the limit $p \rightarrow 0$ (it is an additional assumption), one has the vanishing of the amplitude in the soft limit: $$ \lim_{p \rightarrow 0}A_n (\alpha + \phi(p), \beta ) = 0 $$

I've taken this extract from the paper- https://www.researchgate.net/publication/340008599_New_Soft_Theorems_for_Goldstone-Boson_Amplitudes.

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