\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