# Circularly polarized waves

Consider a time-harmonic E field, given as:

$\mathbf{E}(\mathbf{r},t) = \text{Re}\Big[\mathbf{E}(\mathbf{r})\text{e}^{j\omega t}\Big]$

For a plane wave, the $$\mathbf{E}$$ field phasor is

$\mathbf{E}(\mathbf{r}) = \hat{n}~E_{0}~\text{e}^{-j~\mathbf{k}\cdot \mathbf{r}}$

Now, if we consider a plane wave with $$\mathbf{E}$$ field of amplitude $$E_{0}$$, pointing in a direction $$45^{circ}$$ with respect to the $$x$$-axis and traveling in the $$+z$$ direction, the $$\mathbf{E}$$ field phasor is

$\mathbf{E}(\mathbf{r}) = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{e}^{-j~k~z}$

The $$\mathbf{E}$$ field in this wave always points along the line defined by the diagonal $$\hat{\mathbf{y}} + \hat{\mathbf{z}}$$. It turns out that we still have a linearly polarized wave, rotated by $$45^{\circ}$$ and with an amplitude divided by $$\sqrt{2}$$.

$\mathbf{E}(\mathbf{r}) = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{e}^{-j~k~z} \; \Rightarrow \; \mathbf{E}(\mathbf{r},t) = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{cos}~(\omega t - kz)$

If an observer sits at $$z=0$$, along the direction of propagation, he sees an oscillating $$\mathbf{E}$$ field given by

$\mathbf{E}(\mathbf{r},t)\Big|_{z=0} = \Bigg(\frac{\hat{x}+\hat{y}}{\sqrt{2}}\Bigg)~E_{0}~\text{cos}~(\omega t)$

Now, if we consider a plane wave given by the phasor

$\mathbf{E}(\mathbf{r}) = (\hat{x} - j~\hat{y})~E_{0}~\text{e}^{-j~k~z}$

we notice that it has a complex polarization vector.

Finding the time-dependent $$\mathbf{E}$$ field:

$\mathbf{E}(\mathbf{r},t) = \text{Re}\Big[\mathbf{E}(\mathbf{r})\text{e}^{j\omega t}\Big] = E_{0}~\big[\hat{x}~\text{cos}(\omega t - kz) + \hat{y}~\text{sin}(\omega t -kz)\big]$

The important thing to notice here is that the $$x$$- and $$y$$-components of the $$\mathbf{E}$$ field have the same amplitude, but are $$\pi/2$$ out of phase.

Now, an observer sitting at $$z=0$$, will see the $$\mathbf{E}$$ field vector rotating in a circle

$\mathbf{E}(\mathbf{r}) = (\hat{x}-j~\hat{y})~E_{0}~\text{e}^{-j~k~z} \; \Rightarrow \; \mathbf{E}(\mathbf{r},t) = E_{0} \big[\hat{x}~\text{cos}(\omega t - kz) + \hat{y}~\text{sin}(\omega t - kz)\big]$

$\mathbf{E}(\mathbf{r},t)\Big|_{z=0} = E_{0}~\big[\hat{x}~\text{cos}(\omega t) + \hat{y}~\text{sin}(\omega t)\big]$