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.
