How scalars appear in Goldstone's theorem

Let $$G$$ be a lie group and $$G_i$$ its generators. Suppose for a set of fields $$\chi_{\alpha}(x)$$ we have $$\left[G_{i}, \chi_{\alpha}(x)\right]=-\left(g_{i}\right)_{\alpha \beta} \chi_{\beta}(x)$$

If the Lagragian $$\mathcal{L}$$ is invariant under the $$G$$ then we have the conserved charges $$G_{i} \equiv Q_{i}=\int d^{3} x J_{i}^{o}=\int d^{3} x\left[\frac{\partial \mathcal{L}}{\partial \partial_{o} \chi_{\alpha}} \frac{1}{i}\left(g_{i}\right)_{\alpha \beta} \chi_{\beta}\right]$$ Suppose that $$\left\langle 0\left|\left[G_{i}, \chi_{\alpha}(x)\right]\right| 0\right\rangle=-\left(g_{i}\right)_{\alpha \beta}\left\langle 0\left|\chi_{\beta}(x)\right| 0\right\rangle \neq0.$$ then we have \begin{aligned} \left(g_{i}\right)_{\alpha \beta}\left\langle 0\left|\chi_{\beta}(x)\right| 0\right\rangle=& \sum_{n} \int d^{3} y\left\{\left\langle 0\left|e^{-i P y} J_{i}^{o}(0) e^{i P y}\right| n\right\rangle\left\langle n\left|\chi_{\alpha}(x)\right| 0\right\rangle\right.\\ &\left.-\left\langle 0\left|\chi_{\alpha}(x)\right| n\right\rangle\left\langle n\left|e^{-i P y} J_{i}^{o}(0) e^{i P y}\right| 0\right\rangle\right\} \\ =& \sum_{n} \int d^{3} y e^{i P_{n} y}\left\langle 0\left|J_{i}^{o}(0)\right| n\right\rangle\left\langle n\left|\chi_{\alpha}(x)\right| 0\right\rangle \\ &-\sum_{n} \int d^{3} y e^{-i P_{n} y}\left\langle 0\left|\chi_{\alpha}(x)\right| n\right\rangle\left\langle n\left|J_{i}^{o}(0)\right| 0\right\rangle \\ =& \sum_{n}(2 \pi)^{3} \delta^{3}\left(\vec{p}_{n}\right)\left\{e^{-i P_{n}^{o} y^{o}}\left\langle 0\left|J_{i}^{o}(0)\right| n\right\rangle\left\langle n\left|\chi_{\alpha}(x)\right| 0\right\rangle\right.\\ &\left.-e^{+i P_{n}^{o} y^{o}}\left\langle 0\left|\chi_{\alpha}(x)\right| n\right\rangle\left\langle n\left|J_{i}^{o}(0)\right| 0\right\rangle\right\} \end{aligned} By assumption this expression does not vanish and, furthermore, since the LHS is independent of $$y^{o}$$ it must also be independent of $$y^{o} .$$ Clearly this can only happen if in the theory there exist some massless one-particle states $$|n\rangle$$ and only these states contribute to the sum.

This is the proof that I found of Goldstone theorem .

But the proof only shows that particles are massless . How do we show that these particles are scalars and how the fields $$\phi_i$$ associated with this $$|p,i\rangle$$ particles transform under the group $$G$$ ?

In this case, the vacuum state is Lorentz invariant: if $$U$$ is the unitary operator implementing a Lorentz transform, then $$U|0\rangle=|0\rangle$$. The condition implies $$\langle 0| \chi' |0\rangle = \langle 0|U^\dagger \chi U|0\rangle = \langle 0| \chi |0\rangle$$ where $$\chi'\equiv U^\dagger \chi U$$ is the Lorentz transform of $$\chi$$. This says that the vacuum expectation value of $$\chi$$ is invariant under Lorentz transformations, so either $$\chi$$ is a scalar or its vacuum expectation value is zero. The assumption of SSB is that its vacuum expectation value is nonzero, so $$\chi$$ must be a scalar field. Then the only single-particle states that $$\chi$$ can create from the vacuum are scalars, so if $$|n\rangle$$ is a one-particle state, then the factor $$\langle 0|\chi|n\rangle$$ cannot be nonzero unless that particle is a scalar.