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 $p = (E, 0, 0, p_z)$: $$\epsilon^\mu_1=(0,1,0,0)\quad\epsilon^\mu_2=(0,0,1,0)\quad\epsilon^\mu_L=(p_z/m,0,0,E/m)$$ The polarization vectors all satisfy $\epsilon^\mu_i \epsilon^{\mu *}_i=-1$.
The fact that the three polarizations mix is important, otherwise we would only have three independent degrees of freedom (like three scalar fields or something). One can see that they will mix by looking at the Little group, i.e. the lorentz transformations that leave $p^\mu$ alone.
Example: Polarization mixing by the little group
For example we could be in a frame where $$q^\mu = (m,0,0,0)$$ and $$\epsilon^\mu_1(q)=(0,1,0,0),\quad\epsilon^\mu_2(q)=(0,0,1,0),\quad\epsilon^\mu_3(q)=(0,0,0,1)$$ while the unphysical scalar degree of freedom corresponds to the polarizatoin $$\epsilon^\mu_S = (1,0,0,0).$$ In this case, the Little group beeing the subset of lorentz transformations, which preserve $q_\mu$ are the 3D rotations. Under 3D rotations, the three physical polarizations $\epsilon^\mu_i$ mix, while the fourth $\epsilon^\mu_S$ stays fixed.
The same thing will happen in a case where $\epsilon^\mu_3=\epsilon^\mu_L = (p/m,0,0,E/m)$, the $\epsilon^\mu_i$ mix under the little group SO(3) and dont mix with the new $\epsilon^\mu_S$ (which is not $(1,0,0,0)$) anymore.
# 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.
Note:
- The Ward Identity is closely related to Gauge invariance: Invariance under $A_\mu\to A_\mu +\partial_\mu \alpha$ leads to invaraiance in momentum space $\epsilon_\mu \to \epsilon_\mu + p_\mu$, which is exactly the Ward Identity.