Quantization and Ward identity
In the following we quantize the fields for the Spin 0 and massive- and massless Spin 1 case. To do so we only have to use a seperate set of creation an annihilation operators for every d.o.f.
# Complex scalar
For a complex scalar $\phi = \phi_1 + i\phi_2$ (2 d.o.f.) we just quantize two Spin-0 (scalar) fields $$\phi_1(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}}(a_{p,1}e^{-ipx} + a^\dagger_{p,1}e^{ipx})$$ $$\phi_2(x) = \int \frac{d^3p}{(2\pi)^3 }\frac{1}{\sqrt{2\omega_p}} (a_{p,2}e^{-ipx} + a^\dagger_{p,2}e^{ipx})$$
This can be understood as a real doublet $\phi = \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix}$, which would correspond to a vector representation with the polarization vectors $$\epsilon_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad \epsilon_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$$
# Massive spin 1
For the massive spin 1 we have $$A_\mu(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}} \sum^3_{j=1} \left[\epsilon^j_\mu(p) a_{p,j} e^{-ipx} + \epsilon^{j*}\mu(p) a^\dagger{p,j}e^{ipx}\right]$$ where we sum over the basis of polarization vectors of the representations and give each representation an own creation and annihilation operator. Since the operators $a^\dagger_{p,j}$ now have indices for the polarization $\epsilon_j$ we also need both to specify states: $$a^\dagger_{p,j}\ket{0} = \frac{1}{\sqrt{2\omega_p}}\ket{p,\epsilon^j}$$ and $$\braket{0|A_\mu(x)|p,\epsilon^j} = \epsilon^j_\mu e^{-ipx}$$ The correct basis of polarization vectors is three dimensional, representing the three d.o.f. for the Spin 1. Under lorentz transformations these three polarizations dont mix with the fourth Spin 0 d.o.f., instead they only mix among another, as described by the little group, which is an $SO(3)$ for a fixed $p$. The space of polarization vectors $\epsilon^\mu(p) = c_j\epsilon^\mu_j(p)$ is closed under lorentz transformation. For $p = (E, 0, 0, p_z)$: $$\epsilon^\mu_1=(0,1,0,0)\quad\epsilon^\mu_2=(0,0,1,0)\quad\epsilon^\mu_L=(p_z/m,0,0,E/m)$$
# Massless spin 1
The quantization for massless spin 1 is the same with only 2 polarization vectors: $$A_\mu(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}} \sum^2_{j=1} \left[\epsilon^j_\mu(p) a_{p,j} e^{-ipx} + \epsilon^{j*}\mu(p) a^\dagger{p,j}e^{ipx}\right]$$ However like in the massive spin 1 case the two polarizations will mix with the third under lorentz transformation. Again WOLOG for $p = (E, 0, 0, p_z)$ our basis vectors are $$\epsilon^\mu_1=(0,1,0,0)\quad\epsilon^\mu_2=(0,0,1,0),$$ however they will mix with the third which is proportional to $p^\mu$ in the limit. $$\lim_{m\to 0}\epsilon^\mu_L(p)\propto p^\mu,$$ so that for a general porlarization under lorentz transformation $\epsilon^\mu \to c_1 \epsilon^\mu_1 + c_2 \epsilon^\mu_2 + c_3 p^\mu$, which is not physical because of the $p^\mu$-term. For an amplitude $$\mathcal{M} = \epsilon^\mu M_\mu,$$ to be lorentz invariant on the restriced space of physical polarizations (spanned by $\epsilon^\mu_1$ and $\epsilon^\mu_2$) , we need $$\mathcal{M}\to(a_1\epsilon^\mu_1 + a_2\epsilon^\mu_2+a_3p^\mu)M^\prime_\mu = (c_1\epsilon^\mu_1 + c_2\epsilon^\mu_2)M^\prime_\mu,$$ which is only possible if $p^\mu M_\mu=0$
Ward Identity
Far matrix elements $\mathcal{M}=\epsilon^\mu M_\mu$ with the polarization $\epsilon^\mu$ of a massless spin 1 particle the Ward Identity_ holds:
$$p^\mu M_\mu=0$$ where $p^\mu$ is the momentum of the particle.
This follows directly form lorentz invariance and the fact that the representaiton of a massless spin 1 only has two polarizations.