What are the real Electromagnetic fields of Circularly Polarized Light? It is my understanding that the $\vec{E}$ and $\vec{B}$ fields of a circularly polarized photon sit purely in the $(1,0)$ or $(0,1)$ representation of the (complexified) Lorentz Group, and have helicity $-1$, and $+1$ respectively. These states are described by field configurations which satisfy $$E=\pm iB$$
Clearly the Faraday tensor of these states is not real. Given that classical physics is described by a strictly real representation of the Lorentz group, how do circularly polarized fields actually appear as real, measurable fields?
 A: This formula can even be understood at a purely classical level if you take the $\vec E$ and $\vec B$ in your equation to be the positive frequency parts of the Fourier transforms $\tilde{\vec E}(\omega, \vec k)$ and $\tilde{\vec B}(\omega, \vec k)$.
The reality of $\vec E(t, \vec r)$ and $\vec B(t, \vec r)$ becomes a set of conditions like $\tilde{\vec B}(\omega)^* = \tilde{\vec B}(-\omega)$ (and so on for the other variables) in Fourier space. So you can restrict the domain of your frequency and wave vector variables to the positive frequency domain (and the negative frequency components are implied by the reality condition).
Now, if you start with a circularly polarized plane wave $\vec B(\vec r, t) = \vec e_x B \cos(\Omega t - K z) \pm \vec e_y B \sin(\Omega t - K z)$, you get the following in positive frequency Fourier space (w.l.o.g. $\vec k = k \vec e_x$)${}^1$:
\begin{align*}
 \tilde{\vec B} &= B \vec e_x  \left( \frac 1 2 \delta(\Omega - \omega) \frac 1 2 \delta(k - K) - \frac 1 {2i} \delta(\Omega - \omega) \frac 1 {2i} \delta(k - K) \right) \\
&\phantom{{}={}} \pm B\vec e_y \left( \frac 1 {2i} \delta(\Omega - \omega) \frac 1 2 \delta(k - K) - \frac 1 {2} \delta(\Omega - \omega) \frac 1 {2i} \delta(k - K) \right) \\
&= \frac 1 2 B ( \vec e_x  \pm i \vec e_y  ) \delta(\Omega - \omega) \delta(k - K)
\end{align*}
The corresponding $\tilde{\vec E} = \pm i \tilde{\vec B}$ is
\begin{align*}
 \tilde{\vec E} &= \frac 1 2  ( \pm i \vec e_x - \vec e_y ) \delta(\Omega - \omega) \delta(k - K)
\end{align*}
Transforming $\tilde{\vec E}$ back to real space (this can be done by analogy to the above) one obtains
$$ \vec E(t, \vec r) = \pm \vec e_x B \sin\left(\Omega t - K z \right) - \vec e_y B \cos\left(\Omega t - K z\right) $$
which is exactly the circularly polarized plane wave that complements $\vec B$  as a solution the vacuum Maxwell equations:
$$ \vec e_z \times \vec E = \pm \vec e_y B \sin(\Omega t - Kz) + \vec e_x B \cos(\Omega t - Kz) = \vec B$$
The computation works almost the same when including a phase offset $\phi$.
And this generalizes naturally to wave packets, etc. due to the linearity of the Maxwell equations.
The connection to QED is simply that solutions to the classical field equations describe possible modes of the quantum EM field.
${}^1$ the Fourier transform can be read off after using the addition theorems and the Fourier transforms of $\cos$ and $\sin$.
