Unpolarized light can be converted into polarized light. However, can polarized light be converted into unpolarized light? If so, how can this be done?

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    $\begingroup$ This question is very unclear. "Unpolarized light" can mean a lot of things. What do you mean? Randomly polarized? Mixed polarization? Circular? Less than some threshold? (ie: does 80% linear s-pol + 20% linear p-pol = "unpolarized"?). When you ask "how can this be done", are you asking for a theoretical answer or a practical one? $\endgroup$ – J... Mar 22 '18 at 13:43
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    $\begingroup$ @J... I think this is a good question precisely because it is a bit unclear. The words "unpolarized" or even "depolarized" are used a great deal, but I don't think the subtlety behind them is widely understood. $\endgroup$ – Selene Routley Mar 22 '18 at 21:38

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

  • $\begingroup$ How would scattering do this? You would need to have a changing scattering. And this adds to bandwidth. $\endgroup$ – Pieter Mar 22 '18 at 10:24
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    $\begingroup$ @Pieter, it works (as do the other techniques mentioned) because the polarization is entangled to an internal degree of freedom; for scattering the idea is that it varies across the beam profile, so if you collect the whole beam, you get a random polarization, even if you don't change the scattering. $\endgroup$ – ostrichCamel Mar 22 '18 at 16:49
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    $\begingroup$ @Pieter, if you're worried about bandwidth, note that polarizing light also sacrifices half the photons. In general, unless you entangle the polarization with external degrees of freedom, which is harder, changing the purity (increasing or decreasing) will require larger bandwidth. $\endgroup$ – ostrichCamel Mar 22 '18 at 16:57
  • $\begingroup$ @Pieter Scattering can be thought of as tiny reflections from surfaces of different angles. The output polarization from a reflection is a function of the reflection angle, hence scatter can act to depolarize a beam. $\endgroup$ – MJC Mar 23 '18 at 14:44

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 $k\,T \ll \hbar\,\omega$).

  • $\begingroup$ Cannot be done without adding to the bandwidth. $\endgroup$ – Pieter Mar 22 '18 at 10:23
  • $\begingroup$ @Pieter Of course, minimally. I don't think that bandwidth restrictions should limit a question like this. But it's actually more than that. One can put one photon states into pure superpositions of energy eigenstates, thus making them at once "broadband" whilst still being pure states. One has to achieve classical mixtures with nonpure density matrices from pure states. $\endgroup$ – Selene Routley Mar 22 '18 at 10:26
  • $\begingroup$ @Pieter Actually, I'm not even sure that one does increase the bandwidth, aside from in the sense that a time limited experiment must produce light with nonzero bandwidth. But suppose we have two pure polarization eigenstates, both with the same superposition co-efficients over their respective energy eigenstates. The power spectral density of the mixture will be the same as that of the pure state, no? $\endgroup$ – Selene Routley Mar 22 '18 at 10:32
  • $\begingroup$ @Pieter In any case, as I said, this is a really fascinating question and nontrivial topic if one digs into it. It's a great deal more subtle than people realize when they buy an off the shelf "depolarizer"! $\endgroup$ – Selene Routley Mar 22 '18 at 10:35
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    $\begingroup$ @Pieter Yes this point is irrelevant to the discussion. $\endgroup$ – MJC Mar 23 '18 at 14:45

Wikipedia discusses a bunch of methods of depolarization which include scattering:


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    $\begingroup$ +1 from the author of the first version of that article! It's been considerably improved in the ten years since I started it but I still recognise much of it. $\endgroup$ – Chris H Mar 22 '18 at 20:59
  • $\begingroup$ @ChrisH this article says that all the techniques described produce "pseudo-random polarization", but doesn't mention what's pseudo about them. In fact, I would argue that the pseudo-ness actually depends on the application (see my answer for elaboration). $\endgroup$ – ostrichCamel Mar 23 '18 at 1:36
  • $\begingroup$ @ostrichCamel that's exactly right. In one case I needed constant polarisation across the beam but it could vary over time so long as the average over a minute was unpolarised. I haven't edited it in years, and it's grown a bit so could probably handle the extra explanation now. $\endgroup$ – Chris H Mar 23 '18 at 8:34

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.

  • $\begingroup$ I didn't say one can't do it: just that it's a a great deal more subtle and involved than most people realize when they buy an off the shelf "depolarizer"! I'm not so sure about your second paragraph: does it not align with my comment about entangling the photon with the states of the birefringent crystal, or you seem to beginning with a mixed state (coherence times ...). I agree one can get a pure one photon state that has a nonzero bandwidth - that's just something with superposition weights over a range of energy eigenstates, it can still be a pure quantum state. $\endgroup$ – Selene Routley Mar 22 '18 at 10:40
  • $\begingroup$ Anyhow, I think what this question shows, whoever may turn out to be right in the arguments it is provoking, is that this question is a great deal more subtle than most people realize. Depolarized light is a concept I took a long time to get my head around. $\endgroup$ – Selene Routley Mar 22 '18 at 10:42
  • $\begingroup$ Ah, I agree that depolarizers need to be selected carefully depending on the application :) You did mention entanglement, but as I understood you were talking about entangling with external systems; this is difficult to do with photons. In my example (and actually in any simple depolarizer), you're actually entangling internal DOFs of the photon, which is much easier. $\endgroup$ – ostrichCamel Mar 22 '18 at 10:46
  • $\begingroup$ I'm starting with a pure state but a finite bandwidth. Then delaying one polarization relative to the other leads to an incoherent sum rather than a superposition of the polarizations. $\endgroup$ – ostrichCamel Mar 22 '18 at 10:48
  • $\begingroup$ I'll have to think about this. Can you give a citation of a precise experimental setup, please? I'd really like to be ably to upvote your second paragraph, because it would be a great, definitive answer to this question if correct $\endgroup$ – Selene Routley Mar 22 '18 at 10:49

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