Why is the state equation of a circularly polarised photon as such? $\newcommand{\bra}[1]{\left< #1 \right|}
\newcommand{\ket}[1]{\left| #1 \right>}\renewcommand{\vec}{\boldsymbol}$This question can be regarded as a follow up of this question.
In this lecture video (jump to 12:20) of Prof. Binney, he writes out the electric field equations of circularly polarized waves.
\begin{align}
\text{Right hand polarised: } \vec{E_+} = \frac{E_0}{\sqrt{2}}\operatorname{Re}\left[(\hat{\vec{e}}_x + \mathrm{i}\hat{\vec{e}}_y) (\mathrm{e}^{\mathrm{i}\omega t})\right]\tag{1}\label{1}\\
\text{Left hand polarised: } \vec{E_-} = \frac{E_0}{\sqrt{2}}\operatorname{Re}\left[(\hat{\vec{e}}_x - \mathrm{i}\hat{\vec{e}}_y)(\mathrm{e}^{\mathrm{i}\omega t})\right]\tag{2}\label{2}
\end{align}
I understand them. What I don't understand are the equations for the quantum mechanical state (jump to 15:35). If $\ket{+}$ and $\ket{-}$ denote the states of a right and left circularly polarised photon respectively, then 
\begin{align}
\text{Right hand polarised: } \ket{+} = \frac{1}{\sqrt{2}}\left(\ket{\rightarrow} + \mathrm{i} \ket{\uparrow}\right)\tag{3}\label{3}\\
\text{Left hand polarised: } \ket{-} = \frac{1}{\sqrt{2}}\left(\ket{\rightarrow} - \mathrm{i} \ket{\uparrow}\right)\tag{4}\label{4}
\end{align}
where $\ket{\rightarrow}$ denotes the state of a photon which is polarised in the horizontal plane and $\ket{\uparrow}$ in the vertical plane. In the linked question, the state equations were obatined by parametrisation. How can we achieve \eqref{3} and \eqref{4} in this case? Or are they the definitions of state of a circularly polarised photon? How do we get or why do we need $\mathrm{i}$ in \eqref{3} and \eqref{4}?
Ref: Binney, James; Skinner, David The Physics of Quantum Mechanics, Oxford University Press, 2014, pp. 20-22. Google books link
 A: The definitions of right and left circularly polarized light may be motivated from the requirement that these states must be eigenstates of the rotation operator ; Physically, the most reasonable definition of circular polarization is that the state of polarization is left invariant by an arbitrary rotation.
In terms of the notation you have used, the action of Rotation by angle $\theta$ is described as $R(\theta) = cos(\theta)(\left|\uparrow \right>\left<\uparrow\right| + \left|\to \right>\left<\to\right|) + sin(\theta)(\left|\uparrow \right>\left<\to\right| - \left|\to \right>\left<\uparrow\right|)$
We can check that $\left|\pm  \right> = \frac{1}{\sqrt{2}}(\left|\uparrow \right> \pm i\left|\to \right>)$ are eigenstates of the rotation operator for arbitrary $\theta$ :
$R(\theta)\left|\pm  \right> = \frac{1}{\sqrt{2}}[(cos(\theta)\left|\uparrow \right>-sin(\theta)\left|\to \right>) \pm i(cos(\theta)\left|\to \right>+ sin(\theta)\left|\uparrow \right>)] =  e^{\pm i \theta}\left|\pm  \right>$
In addition, we may check that either of the circularly polarized states gives a physically reasonable expectation value for the projection operator corresponding to any direction of (linear) polarization: 
Defining $\left|\alpha \right> \equiv cos (\alpha) \left|\uparrow \right> + sin(\alpha)\left|\to \right>$, we may check that $\left<+|\alpha \right>\left<\alpha|+\right> = \frac{1}{2} $, independent of $\alpha$. This makes physical sense, as passage of circularly polarized light through a linear polarizer oriented along any direction halves the intensity. Imposing this condition may also be used to derive the form of $\left|\pm  \right>$
