Complex Scattered polarization vector? (Conceptual)

[PhDeezNutz]

Homework Statement

This is more so a conceptual question rather than a specific homework problem. I always thought that the polarization vector (i.e. direction of the $\vec{E}$ - field denoted as $\vec{\varepsilon}$) in an EM wave was a Euclidean three vector but apparently I am wrong because in one equation in my book they mention the complex conjugate $\vec{\varepsilon}$. I'm using Zangwill's "Modern Electrodynamics" btw.

The specific part I'm confused about is

Call this equation 1

$\frac{d \sigma_{scatt}}{d \Omega} = \frac{\langle \frac{dP}{d \Omega} \rangle}{\frac{1}{2} \varepsilon_0 c E_0^2} = r^2 \frac{\vec{E}_{rad}^2}{\vec{E}_{0}^2} = \left| \vec{f} \left( \vec{k} \right) \right|^2$

"If our interest is a scattered electric field with a particular polarization $\vec{\varepsilon}$, the differential scattering cross section (above) generalizes to"

Call this equation 2


$\left. \frac{d \sigma_{scatt}}{d \Omega} \right|_{\vec{\varepsilon}} = r^2 \frac{\left|\vec{\varepsilon}^* \cdot \vec{E}_{rad} \right|^2}{\left| \vec{E}_0\right|^2} = \left| \vec{\varepsilon}^* \cdot \vec{f}\left( \vec{k} \right) \right|^2$

Again I always thought $\vec{\varepsilon}$ was a 3D euclidean vector and I'm trying to reconcile what it would mean for it to be complex.

Relevant Equations

The differential scattering cross section is defined as

$\frac{d \sigma_{scatt}}{d \Omega} = \frac{\langle \vec{S}_{rad} \rangle \cdot r^2 \hat{r}}{\left| \langle \vec{S}_{inc} \rangle \right|}$

In the far field

$\vec{E} = \vec{E}_{inc} + \vec{E}_{rad} = E_0 \left[ \hat{e}_0 e^{i \vec{k}_0 \cdot \vec{r}} + \frac{e^{ikr}}{r} \vec{f} \left( \vec{k} \right)\right]e^{-i \omega t}$

I guess I will show my work for substantiating equation 1 and hopefully by doing so someone will be able to point out where I could generalize.

$\langle \vec{S}_{rad} \rangle = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{rad} \times \vec{B}^*_{rad}\right) = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{rad} \times \left( \hat{r} \times \vec{E}_{rad}^*\right) \sqrt{\mu \varepsilon} \right) = \sqrt{\frac{\varepsilon}{\mu}} \mathfrak{R} \left(\vec{E}_{rad} \times \left( \hat{r} \times \vec{E}_{rad}^* \right) \right) = \sqrt{\frac{\varepsilon}{mu}} \mathfrak{R} \left( \hat{r} \left(\vec{E}_{rad} \cdot \vec{E}_{rad}^* \right) - \vec{E}_{rad}^* \left( \hat{r} \cdot \vec{E}_{rad} \right) \right) = \sqrt{\frac{\varepsilon}{\mu}} \frac{E_0^2 \left|\vec{f} \left(\vec{k} \right) \right|^2}{r^2} \hat{r}$

$\langle \vec{S}_{inc} \rangle = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{inc} \times \vec{B}_{inc}^*\right) = \sqrt{\frac{\varepsilon}{\mu}} \mathfrak{R} \left( \hat{r} \left( \vec{E}_{inc} \cdot \vec{E}_{inc}^*\right) \right) = \sqrt{\frac{\varepsilon}{\mu}} E_0^2$

Therefore

$\frac{d \sigma_{scatt}}{d \Omega} = \left| \vec{f} \left( \vec{k} \right) \right|^2$

 

[PhDeezNutz]

I think I get it. The polarization vector contains information about phase as well as the direction of the electric field.

$\vec{E} = \vec{E}_0 e^{i \vec{k} \cdot \vec{r} - i \omega t}$

$e^{i \vec{k} \cdot \vec{r} - i \omega t}$ indicates only that the wave is transverse and provides information about frequency. It says nothing about phase and amplitude.

I think the convention is to wrap up information about phase and amplitude into $\vec{E}_0$ and because $\vec{E}_0$ contains information about phase it must be complex (Euler's formula).