Why is a perfect blackbody radiation completely unpolarized? Given a collection of photons obeying the Planck distribution at some temperature $T$, is it possible to compute the average electric field they produce and argue that a perfect blackbody radiation is completely unpolarized? I am looking for a mathematical understanding of the nature of polarization of perfect thermal radiation.
 A: Yes, you surely can compute things about the electric field of black body radiation.
As stated in the question, the electromagnetic field obeys the Planck law.
The Planck law usually refers to the spectral radiance of a black body.
To simplify things, we can instead refer to Bose-Einstein statistics, which tell us the mean occupation of each mode of the electromagnetic field.
The Bose-Einstein law is
$$n_i(E_i) = \frac{1}{\exp(E_i/k_b T) - 1}$$
where $E_i$ is the energy of the $i^\text{th}$ mode and $k_b$ is the Boltzmann constant.
Noting that $n_i$ depends only on energy, and that two polarizations of the electromagnetic field (in vacuum) have the same energy, we can see that there's is no preferred occupation of any particular polarization.
A: The simplest way to see this is by thermodynamics. The radiation an ideal blackbody can emit is completely determined by what it can absorb. Since blackbodies can absorb all polarizations of radiation, they must be able to emit all polarizations of radiation. Hence they emit a statistical mixture of all of them. 
Of course, if the surface of the "blackbody" was a reflective polarization filter, which reflected horizontally polarized light, then it would not emit horizontally polarized light either.
