Matrix element of the currents associated with the broken generators between the vacuum and Goldstone's bosons

Question

Let $G$ be a Lie group and $L^i$ the generators of this group. Suppose we have $L^{j}|0\rangle \neq 0$ where $|0\rangle \neq 0$ denotes the vacuum. If $G$ is associated with a symmetry of the Lagrangian $L$ according to Noether's theorem we have a set of conserved currents $J_i^ \mu$.

According to Goldstone's theorem $L^{j}|0\rangle \neq 0$ implies that we have $j$th massless boson $|p,j\rangle $.

Now in this article Exact and Broken Symmetries in Particle Physics by Peccei they are claiming that we should have $$ \left\langle 0\left|J_{i}^{\mu}(0)\right| p , j\right\rangle=i f_{j} \delta_{i j} p^{\mu}\tag{49} $$ where $f_{j}$ are some non-vanishing constants.

Why this last expression is true?

Answer 1

Define $$ \left\langle 0\left|J_{i}^{\mu}(0)\right| p , j\right\rangle=:F^\mu_{ij}(p) $$

The only object with a $\mu$ index is $p$ itself, so by Lorentz invariance $F^\mu_{ij}(p)=p^\mu F_{ij}(p^2)$ for some scalar function $F$. But $p^2=m^2$ is just a constant, so $F_{ij}(p^2)=f_{ij}$ for some numbers $f_{ij}$: $$ \left\langle 0\left|J_{i}^{\mu}(0)\right| p , j\right\rangle=p^\mu f_{ij} $$

Finally, diagonalize $f_{ij}$ to get the expression in the OP.

See also §19 in ref.1.

References.

1. Weinberg S., Quantum theory of fields, Vol.2. Modern Applications.