Unpolarized light can be converted into polarized light. However, can polarized light be converted into unpolarized light? If so, how can this be done?
Yes you can convert polarised light to unpolarised light. Scattering will do this.
You can also buy optics that depolarise light, https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=870
The answer is yes.
There is a certain, practical sense wherein you can readily convert polarized to unpolarized light, as in Matt Cliffe's answer. Russell A. Chipman and his PhD student Hannah Noble (Hannah Dustan Noble, "Mueller Matrix Roots', PhD Dissertation, U. Arizona 2011) were active in this field some years ago. But see my comment at the end of this answer.
However, to depolarize light "properly" in a theoretical sense is much harder, although the answer is still, theoretically, yes.
Essentially what one needs to do here is to convert a pure quantum state into a mixed state. Conceptually, one way to imagine this happenning is if one could "industrialize" the Wigner's Friend experiment: we'd take the role of the Friend and impart some observable to each photon state so that each photon is forced into an eigenstate of the observable. Naturally, we'd need to find an observable whose eigenstates are two orthogonal polarization eigenstates. The eigenstates of course are chosen at "random", where "random" is intended in the sense of what happens at quantum measurement. Finding nondestructive observables for photons in practice is hard; most observations tend to impart an observable and absorb the light in the process, but Non demolition measurement is possible, if tricky. Such measurement would have to be imparted to every photon in the whole light beam.
Another way to do this would be to entangle the pure photon states with an outside system so that each becomes mixed. Again this is certainly can be theoretically done, but is much harder in practice.
This is a fascinating question when one really digs into it. The answers saying yes it can be done are certainly correct in a practical sense, but I'd argue that one-photon states scattered from irregular surfaces are still pure one photon states: if the surface is cold and unthermalized, the same surface always scatters each one photon state in the same way, so that the scattered state the same for each photon and the light is thus still in a pure and thus has a polarization, albeit a very complicated one. In other words, one can still ascribe definite, constant electromagnetic field directions to each spatial point in the scattered field, even though they vary wildly all over the place. It's simply that its not a plane wave (pure momentum state) anymore, but it's still pure. For optical photons, a "cold" surface in the sense of this argument is one of a few hundred kelvin or less: the argument breaks down when we don't have $k\,T \ll \hbar\,\omega$).
Wikipedia discusses a bunch of methods of depolarization which include scattering:
The answers above are probably already more than you were looking for. The standard techniques are adequate for most applications, though, as WetSavannaAnimal aka Rod Vance pointed out, they don't produce a truly mixed state of the incident photon.
However, I disagree with him that they cannot produce a truly mixed state of polarization. In fact, it is quite easy to entangle the internal degrees of freedom of a photon. For example, one can entangle arrival time and polarization by sending a photon through a sufficiently thick slab of birefringent material (the slab must be thick enough that the two polarization states end up in completely distinguishable time bins, i.e., one polarization should be delayed relative to the other by more than the coherence time of the light source). Certainly the photon is still in a pure state, but if frequency is traced over (i.e., ignored), its polarization is completely mixed.
Similarly, one can produce a spatially-varying pure polarization: here, the polarization is entangled with the spatial mode, and if the spatial mode is traced over, the output is completely unpolarized.
For any such technique to work in a specific application, the key point is that the extra degree of freedom needs to be traced over. For example, if we filter out a very narrow frequency bandwidth before measuring our photon's polarization, the first depolarizer I described will cease to work. Similarly, the second scheme won't work if you only collect light from a small portion of the beam where the polarization doesn't vary much.
protected by Qmechanic♦ Mar 22 '18 at 11:32
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