Firstly, the OP is forgetting that the classic microwave polariser experiment is done with EM radiation in a pure state, not a mixture. We simply have polarised light from, say, a Gunn diode and this pure quantum superposition is forced into a polarisation eigenstate by the polariser. So we begin with near to zero entropy light, absorb some of it (adding entropy to the polariser as it heats), and the leftover light is also of near zero entropy. No problem. So this easily explains how the OP claims to remember microwave polarisation experiments shown to him when he was 17 and that they work perfectly at 300K.
But what about depolarised microwaves? In this case, the second law of thermodynamics puts a limit on how well a polariser can work at a given temperature.
Let's imagine a finite-length, depolarised beam of microwaves moving through deep space. All quantum state variables - direction, spin, frequency - aside from polarisation set to known states so the light is fully describable by a $2\times 2$ density matrix written with respect to the polarisation eigenstate basis. The photons don't need to be in frequency (energy) eigenstates and direction (momentum) eigenstates, but instead can be a quantum superposition of such eigenstates (not a mixture), so that they form a pulse can be finite in space and time. The photon ensemble is moving through deep space separated from its source so that we can think of the whole initial system as:
- The microwave photon ensemble, described by a classical mixture of two pure states (quantum superposition of energy and momentum eigenstates) that differ only by being in orthogonal polarisation states;
- The freespace quantum light field, in thermodynamic equilibrium at the CMBR temperature of $T_{CMBR}=2.7K$, so it is a mixture of thermalised photons described by the Planckian BB spectrum at $T_{CMBR}$;
- The polariser itself, also initially at thermodynamic equilibrium at $T_{CMBR}$.
It should be pointed out that depolarised microwaves whose other quantum properties are otherwise perfectly known are exotic creatures and I cannot think of how one might produce such things experimentally, contrasting with the situation for light where such mixtures are much more plausible. Nonetheless, there doesn't seem to be any in principle reason why such mixtures can't exist for microwaves if they do for visible light.
We consider the pulse sundered from its source, so we have already accounted for the entropy increase in the production of the light, which, as Wolphram jonny's Answer points out, will produce a great deal of entropy. There is no problem with the second law here. But we now wish to know what the entropy changes are for the system that begins as incoming microwave energy, polariser and CMBR.
As By Symmetry's Answer discusses, initially the polariser is not at equilibrium and heating is irreversible. It will begin to heat up. But it must eventually reach a steady state: it may have temperature gradients in it with a "hot spot" where the beam is absorbed, but it will eventually be described as a temperature distribution $T(\vec{x})$. This steady state is reached when the sum of the incoming power from the beam together with the heat absorbed by the polariser from the CMBR equals the heat re-radiated to the CMBR radiation field in unit time. I sketch these ideas below.
At steady state, the polariser's macrostate is unchanging with time, therefore its total entropy must be steady. We can therefore think conceptually of the conversion happening as the light is absorbed by the polariser and, later, the equal nett power is radiated to the CMBR field as drawn in my sketch below:
At steady state, therefore, we need to account for:
- The entropy lost from the beam through any polarisation: the entropy of the partially polarised beam output less that of the input beam;
- The entropy increase of the radiation field as the energy input to the polariser from the beam is re-radiated.
Now the maximum entropy that the radiation field can "soak up" in absorbing power $\mathrm{d}\,P$ is $\frac{\mathrm{d}P}{T_{CMBR}}$; this is because:
- The proportional error between the total state information content of a system of particles (here thermalised EM field quantum oscillators) and the information calculated assuming that the system is in its maximum entropy, thermodynamic equilibrium state approaches nought as the number of particles increases without bound, as discussed further in my answer here to the question "Why are the laws of thermodynamics “supreme among the laws of Nature”?" and
- The definition of thermodynamic temperature is $\frac{1}{T} = \frac{\partial\,S}{\partial\,U}$ where $U$ is the internal energy of a system and $S$ its entropy.
Therefore, the entropy increase of the radiation field must exceed he entropy lost from the beam through any polarisation in all cases, and so:
$$(1-\alpha)\,\frac{\mathrm{d}P}{T_{CMBR}} \geq \frac{\mathrm{d}P}{h\,\nu}\,k_B\,\left(-\mathrm{tr}\left(\rho_{in}\,\log\,\rho_{in}\right)+\alpha\,\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right)\right)$$
and this is the statement of the second law of thermodynamics manifesting itself as a limit on how much a polariser can actually polarise light, where $\alpha$ is the fraction of the input beam's that is transmitted by the polariser, $\rho_{in}\approx\left(\begin{array}{cc}\frac{1}{2}&0\\0&\frac{1}{2}\end{array}\right)$ is the density matrix describing the mixed state of each photon input to the polariser and $\rho_{out}$ is the density matrix describing the mixed state of each photon directly transmitted by the polariser. Therefore, the second law limits the quality of polarisation possible and:
$$-\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right) \geq \frac{\log\,2}{\alpha}-\frac{1-\alpha}{\alpha}\,\frac{h\,\nu}{k_B\,T}$$
defines the maximum "quality" of polarisation allowed by the second law of thermodynamics, where the quantity $-\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right)$ on the left hand side is positive, gets smaller with increasing quality of polarisation and has a value of nought nats when the polarisation is perfect. Here $T$ is the effective temperature of the ambient radiation field, and it must be greater than or equal to $T_{CMBR}$. For a given quality of polarisation to happen, a necessary condition is then:
$$T \leq \frac{(1-\alpha)\,h\,\nu}{k_B\,\left(-\mathrm{tr}\left(\rho_{in}\,\log\,\rho_{in}\right)+\alpha\,\mathrm{tr}\left(\rho_{out}\,\log\,\rho_{out}\right)\right)}$$
which reduces to the OPs formula in the case of fully depolarised input light and perfectly polarised output light.
