On this page, polarization incoherence and methods to simulate it are discussed.
Problems with direct simulation
In a similar manner to temporal phase incoherence, the polarization of a beam or the orientation of a dipole source depends on time. In the case of a polarized beam we have
$$ \vec{E}(t)=\vec{u}(t) E_{0} \cos (\omega t) $$
where the unit vector \(u(t)\) defines the beam polarization and varies on a time scale \( \tau_c \gt\gt T \).
In the case of a dipole source, we have
$$ \vec{p}(t)=\vec{u}(t) p_{0} \cos (\omega t) $$
where the unit vector \(u(t)\) defines the dipole source polarization and varies on a time scale \( \tau_c \gt\gt T \).
In both cases, the time scale for the variation of the polarization is much larger than the optical cycle, making it unpractical to simulate the statistics of temporal polarization incoherence with FDTD.
Derivation
To calculate the response of a system to an unpolarized beam, we need to average over all possible input polarizations:
$$ \left\langle \left\vert E \right\vert^2 \right\rangle = \frac{1}{2\pi} \int^{2\pi}_0 \left\vert \vec{E}\left(\theta\right) \right\vert^2 d\theta $$
Due to the linearity of Maxwell’s equations, we can represent the electric field of an arbitrarily polarized incoming beam as a sum of two orthogonal polarizations:
$$ \vec{E} \left( \theta \right) = \vec{E}_s sin(\theta) + \vec{E}_p cos(\theta) $$
Therefore, the integral can be computed as follows:
\begin{align} \left\langle \left\vert E \right\vert^2 \right\rangle & = \frac{1}{2\pi} \int^{2\pi}_0 \left\vert \vec{E}\left(\theta\right) \right\vert^2 d\theta \\
& = \frac{1}{2\pi} \int^{2\pi}_0 \left\vert \vec{E}_s sin(\theta) + \vec{E}_p cos(\theta) \right\vert^2 d\theta \\
& = \frac{1}{2\pi} \int^{2\pi}_0 \left( \left\vert \vec{E}_s \right\vert^2 sin^2(\theta) + \left\vert \vec{E}_p \right\vert^2 cos^2(\theta) + 2 \left\vert \vec{E}_s \vec{E}_p \right\vert sin(\theta)cos(\theta) \right) d\theta \\
& = \frac{\left\vert \vec{E}_s \right\vert ^2}{2\pi} \int^{2\pi}_0 sin^2(\theta)d\theta + \frac{\left\vert \vec{E}_p \right\vert ^2}{2\pi} \int^{2\pi}_0 cos^2(\theta)d\theta + \frac{2 \left\vert \vec{E}_s\vec{E}_p \right\vert}{2\pi} \int^{2\pi}_0 sin(\theta)cos(\theta)d\theta \\
& = \frac{1}{2} \left\vert \vec{E}_s \right\vert^2 + \frac{1}{2} \left\vert \vec{E}_p \right\vert^2 \end{align}
The following identities are required to simplify the above integral:
\begin{align} \int^{2\pi}_0 sin^2(\theta)d\theta & = \pi \\
\int^{2\pi}_0 cos^2(\theta)d\theta & = \pi \\
\int^{2\pi}_0 cos(\theta)sin(\theta)d\theta & = 0 \\ \end{align}
Recommended simulation method
FDTD simulations have well defined polarization. For a beam, unpolarized results are obtained by adding the results of 2 orthogonal polarization simulations incoherently using the equation
$$ \left\langleE^{2}\right\rangle=\frac{1}{2}\left\vec{E}_{s}\right^{2}+\frac{1}{2}\left\vec{E}_{p}\right^{2} $$
In the case of a dipole, the results of the three orthogonal polarizations can be added incoherently using the equation
$$ \left\langleE^{2}\right\rangle=\frac{1}{3}\vec{E} p x^{2}+\frac{1}{3}\vec{E} p y^{2}+\frac{1}{3}\vec{E} p z^{2} $$
In practice, this means that we simulate a beam with a “polarization angle” of 0 and then a beam with a “polarization angle of 90”, as shown below. The quantity \( \left\langleE^{2}\right\rangle \) refers to the time averaged electric field intensity of an unpolarized beam source.
$$ \left\langleE^{2}\right\rangle=\frac{1}{2}\vec{E}_ 0^{2}+\frac{1}{2}\vec{E} _{90}^{2} $$
Polarization incoherence examples
The files usr_unpolarized_beam.fsp and usr_unpolarized_beam.lsf can be used to reproduce the following results.
A focused beam propagating at an angle of 60 degrees. The beam is incident upon a surface with an index of 2.
The angle of incidence is very close to Brewster’s angle for this structure. Therefore, we expect quite different behavior for the two polarizations.

P polarized light will have higher transmission (low reflection)

S polarized light will have lower transmission (high reflection)
Field profile from a P polarized beam
Notice that the transmitted fields have an amplitude of about 0.3 (highlighted with a green circle).
Field profile from a S polarized beam
Notice that the transmitted fields have an amplitude of about 0.2 (highlighted with a green circle).
Field profile from an unpolarized beam
The field profile is simply the average of the P and S polarized beams (highlighted with a green circle).
Fraction of power transmitted into the substrate
Transmission (P polarization): 98%
Transmission (S polarization): 72%
Transmission (unpolarized): 85%
As expected, the P polarized beam has a higher transmission. Once again, the unpolarized transmission is simply the average of the transmissions from the P and S simulations.