How are the two independent states of polarization of photon related to the two helicity states? (1) In the canonical quantization of the free electromagnetic field, the Coulomb gauge condition $$A^0=0,~~ \nabla\cdot\textbf{A}=0\tag{1}$$ implies that the polarization vector $\epsilon^\mu$ satisfies $$\epsilon^0=0,~~\boldsymbol{\epsilon}\cdot\hat{\textbf{p}}=0\tag{2}$$ which says that the electromagnetic field has two independent transverse states of polarization.
(2) From the representation theory of Poincare group, it is known that for photons $\textbf{S}\cdot\hat{\textbf{p}}$ has eigenvalues $h=\pm 1$ where $\textbf{S}$ denotes the spin operator.
Both the descriptions above make it clear that the electromagnetic field has two independent degrees of freedom. The description (1) says the electromagnetic field has two independent states of polarization and the description (2) says that it has two independent helicity states.

Question
$\bullet$ Does it mean states of polarization are identical to states of helicity? 
$\bullet$ Is there is a unique one-to-one correspondence between the states of helicity and the independent states of polarization? In that case, $h=+1$ corresponds to which polarization and $h=-1$ corresponds to which? How can such a correspondence, if exists, be understood?
A similar question was asked here.
 A: States of definite helicity are states of definite spin measured along a particular axis. You can, in principle, use any axis to define your spin eigen-basis, it's just not commonly done because the result isn't Lorentz invariant and you have to be careful about what spins are forbidden by the lack of a helicity 0 state. It can be shown that there is a unique one to one mapping between the choice of spin basis and the polarization state basis. The polarization vectors, $\boldsymbol{\epsilon}_\lambda$, handle that mapping (the $\lambda$ indices exist in spin/polarization space, and the spatial index exists in physical space).
A: In the radiation gauge, the 3-vector potential has the most general Fourier mode expansion given by $$\textbf{A}(\textbf{x})=\int\frac{d^3\textbf{p}}{(2\pi)^3\sqrt{2E_{\textbf{p}}}}\sum\limits_{r=1}^{2}[\boldsymbol{\epsilon}_r(\textbf{p})a_{\textbf{p},r}e^{i\textbf{p}\cdot\textbf{x}}+\boldsymbol{\epsilon}^*_r(\textbf{p})a^\dagger_{\textbf{p},r}e^{-i\textbf{p}\cdot\textbf{x}}].\tag{1}$$ Using the definition of spin operator $$S^{ij}=\int d^3\textbf{x}:(A^i\partial_0 A^j-A^j\partial_0 A^i):$$ and Eq.(1), one obtains after performing the integral over space
$$S^{ij}=i\int\frac{d^3\textbf{p}}{(2\pi)^3}\sum\limits_{r,s}[\epsilon_r^i(\textbf{p})\epsilon_s^{j*}(\textbf{p})-\epsilon_s^{i*}(\textbf{p})\epsilon_r^j(\textbf{p})]a^\dagger_{\textbf{p},r}a_{\textbf{p},s}.$$
The action of the spin operator $S^{ij}$, as obtained in Eq. (2), on a one-particle state $a^\dagger_{\textbf{k},m}|0\rangle$, one finds, $$S^{ij}a^\dagger_{\textbf{k},m}|0\rangle=i\sum\limits_{s=1}^{2}\Bigg[\epsilon_m^i(\textbf{p})\epsilon_s^{j*}(\textbf{k})-\epsilon_s^{i*}(\textbf{k})\epsilon_m^j(\textbf{k})\Bigg]a^\dagger_{\textbf{k},s}|0\rangle.$$
Let us choose $\textbf{k}=(0,0,k)$ so that the helicity is measured by $\textbf{S}\cdot\hat{\textbf{k}}=S^3=S^{12}$. We choose, $\boldsymbol{\epsilon}_{1}(\textbf{k})=1/\sqrt{2}(1,i,0)$ and $\boldsymbol{\epsilon}_{2}(\textbf{k})=1/\sqrt{2}(1,-i,0)$. Therefore, $$S^3a^\dagger_{\textbf{k},1}|0\rangle=(+1)a^\dagger_{\textbf{k},1}|0\rangle,\\
S^3a^\dagger_{\textbf{k},2}|0\rangle=(-1)a^\dagger_{\textbf{k},2}|0\rangle.$$ 
Conclusion A one-particle state with right circular polarization corresponds to helicity $+1$, and the one-particle state with left-circular polarization corresponds to a state with helicity $-1$.

Update
The electric field is given by $$\textbf{E}(\textbf{x})=-\frac{\partial\textbf{A}}{\partial t}=(-i)\int\frac{d^3\textbf{p}}{(2\pi)^3}\sqrt{\frac{E_\textbf{p}}{2}}\sum\limits_{r=1}^{2}[\boldsymbol{\epsilon}_r(\textbf{p})a_{\textbf{p},r}e^{i\textbf{p}\cdot\textbf{x}}-\boldsymbol{\epsilon}^*_r(\textbf{p})a^\dagger_{\textbf{p},r}e^{-i\textbf{p}\cdot\textbf{x}}].\tag{2}$$ and
$$\textbf{B}(\textbf{x})=\nabla\times\textbf{A}=(i)\int\frac{d^3\textbf{p}}{(2\pi)^3}\sqrt{\frac{E_\textbf{p}}{2}}\sum\limits_{r=1}^{2}[\hat{\textbf{p}}\times\boldsymbol{\epsilon}_r(\textbf{p})a_{\textbf{p},r}e^{i\textbf{p}\cdot\textbf{x}}-\hat{\textbf{p}}\times\boldsymbol{\epsilon}^*_r(\textbf{p})a^\dagger_{\textbf{p},r}e^{-i\textbf{p}\cdot\textbf{x}}].\tag{3}$$ where I used the fact that $E_{\textbf{p}}=|\textbf{p}|$. Using $\hat{\textbf{p}}=(0,0,1)$, it should be easily checked that the operators $\textbf{E}\pm i\textbf{B}$ acting on the vacuum $|0\rangle$ respectively creates one-particle states of photon with right-circular and left-circular states of polarization. I'm lazy to work out the algebra.
Reference A Modern Introduction to Quantum Field Theory-Michele Maggiore.
