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 $G$ ?
 A: First of all, Lorentz symmetry can be spontaneously broken. That phenomenon is beyond the scope of the present question, though, so for this answer I'll assume that Lorentz symmetry is not spontaneously broken.
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.
