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Let's say that we have got two light beams. Beam A is not polarized at all. Beam B is composed of half H polarized photons + half V polarized photons.

Is there any way to measure those two beams in order to distinguish one from the other ?

The same question applies for a single photon : Is it possible to distinguish a unpolarized single photon from a H or V polarized photon ?

Thanks a lot for your answers !

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Assuming the beams are white-light beams, the answer is "no", the two can't be distinguished from each other. In the original form of this answer, I wrote:

If they are coherent monochromatic beams, then they can be distinguished from each other by analyzing their polarizations. Beam A would be vertically polarized, while Beam B would be polarized circularly, elliptically, or linearly at 45 degrees.

which is not really responsive to the question. More accurately:

A coherent monochromatic beam has an unchanging polarization state: circular, elliptical, or linear. It can't really be said to be unpolarized, because by using an appropriate combination of wave plates it can be transformed to any desired polarization state. So, Beam B and Beam A are effectively the same: feed Beam B through a suitable combination of wave plates and it will be identical to Beam A. The details of how the vertical and horizontal (linear) components of B are combined, and the polarization state of A, will determine what the combination of wave plates needs to be. In other words, there is not a way to distinguish the two beams.

A beam that does not have a definite polarization state must have a range of frequency components, with different frequency components having different polarizations. Then, the net polarization in the beam at any instant is effectively random -- but then the beam is not monochromatic or coherent.

The same question applies for a single photon : Is it possible to distinguish a unpolarized single photon from a H or V polarized photon ?

This is a bit different, because a single photon can only be measured once, whereas a beam can be measured many times. For a single photon the answer is "no".

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  • $\begingroup$ Thank you for your answer but I still don't understand how you can make a difference between the two beams, even in the case of coherent monochromatic beams. How do you do to analyze their polarization ? You wrote that beam A would be vertically polarized while I wrote that it is not polarized at all. Which filters do you propose to use ? $\endgroup$ – Bizniouf Nov 8 '18 at 8:13
  • $\begingroup$ I will edit my answer to make it clearer. $\endgroup$ – S. McGrew Nov 8 '18 at 15:03
  • $\begingroup$ Thank you very much for your answer. It is clearer but I'm still in trouble because of @EmilioPisanty answer. $\endgroup$ – Bizniouf Nov 9 '18 at 10:53
  • $\begingroup$ I think @EmilioPisanti and I gave essentially the same answer, but in different words. The essence is in his statement, "it's important to note that unpolarized single photons are only possible if you allow your system to occupy mixed quantum states, which can be undesirable in several circumstances." A mixed quantum state in this sense is not coherent. It would need to span multiple wavelengths. You mention an EPR source, so do you mean that the photons are emitted in entangled pairs? If so, this changes the question substantially. $\endgroup$ – S. McGrew Nov 9 '18 at 14:19
  • $\begingroup$ "A mixed quantum state in this sense is not coherent. It would need to span multiple wavelengths" - no, that second part is completely incorrect. A mixed quantum state is indeed incoherent, but it need not have any relationship with the monochromaticity of the beam. To make a source of monochromatic photons in a mixed polarization state, simply flip a coin to determine whether to send a H or a V photon. $\endgroup$ – Emilio Pisanty Nov 10 '18 at 8:24
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This

beam B is composed of half H polarized photons + half V polarized photons

is an insufficient specification of the polarization state of beam B, because you do not specify whether the two sources are combined in a coherent or incoherent fashion.

  • If the two components are incoherent, then the combination is unpolarized, and it is indistinguishable from beam A.
  • If the two components are coherent with each other, then their superposition will be fully polarized, but the polarization state of the combination will depend on the relative phase of the superposition.

    If the two beams are in phase (in the usual convention) then the combination will be linearly polarized along an axis at 45° from both, and rotating a linear polarizer will produce a Malus-law pattern that will allow you to distinguish it from beam A.

    If the beams are 90° out of phase, on the other hand, then you will have a circular polarization, and you will need a circular analyzer to provide a suitable benchmark.

Either way, if the combination is coherent (or even partially coherent) then the thing to do is to perform a full polarization tomography measurement on the combination to measure its Stokes parameters, which encode the full information about the polarization of the beam.


As for the second half,

The same question applies for a single photon: Is it possible to distinguish a unpolarized single photon from a H or V polarized photon?

it's important to note that unpolarized single photons are only possible if you allow your system to occupy mixed quantum states, which can be undesirable in several circumstances.

The core of your question, however, is one of quantum state discrimination, and it's not completely a trivial question.

  • If you have a single copy of the photon, then it cannot be done.
  • If you have multiple copies then you can pass the photon through a polarizing beam splitter and measure the output. If you get hits on two channels, then you know it's not a single such polarization, whereas if you only get hits on a single output port, then the more hits you get the more confident you can be about the purity of the state along that polarization.
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  • $\begingroup$ Thank you very much for your answer. It is very clear and understandable but I'm still in trouble because of @S.McGrew answer. Talking about beams, photons are supposed to be emitted by an EPR source so they are coherent, both in beam A and in beam B. Talking about single photons ok. I agree (by the way, it drives me to another question about the possibility to figure out if a single photon is in a quantic superposed state or not (i.e. it is unpolarized or not). But since this is a related but different question I am going to create a new one). $\endgroup$ – Bizniouf Nov 9 '18 at 10:52
  • $\begingroup$ Beam A, which you've described as unpolarized, cannot be coherent. Coherence requires a definite polarization. $\endgroup$ – Emilio Pisanty Nov 9 '18 at 10:53
  • $\begingroup$ Constructive feedback obviously welcome. $\endgroup$ – Emilio Pisanty Nov 10 '18 at 8:25
  • $\begingroup$ *** Please excuse me for the delay of my answer. I was out for a while *** Ok. I have mixed up different things. Beam A comes from an EPR source which is supposed to produce random polarized photons. Beam B comes from a same EPR source, but half has been filtered with a H polarization filter, and half has been filtered with a V polarization filter. I understand that if the two beam B sources are phase coherent their combination is polarized. I don't know all Stokes parameters of the light provided by the EPR source. I have to dig deeper, but you helped me a lot. Thanks a lot ! $\endgroup$ – Bizniouf Nov 19 '18 at 10:13
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… we have got two light beams. Beam A is not polarized at all. Beam B is composed of half H polarized photons + half V polarized photons. Is there any way to measure those two beams in order to distinguish one from the other?

That’s easy. Take polarization filter (designed for the range of the wavelengths you use) and measure the light behind the filter. The trick is to rotate the filter. In the case of the unpolarized light, the intensity of the light beam is not changing. In the case of half V- and H-polarized light the intensity is swelling four times for a 360°-rotation.

The same question applies for a single photon : Is it possible to distinguish a unpolarized single photon from a H or V polarized photon?

No and yes. For a single photon it is impossible, see the answer from S. McGrew. For a stream of single emitted photons a statistics with measurements under different angles of rotation of the filter will show a swelling amount of measured behind the filter photons (for the V- or H-polarized photons) or a equally distributed statistic for nonpolarized photons.

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  • $\begingroup$ Thank you for your answer but I still doubt. Since Sin(angle)^2+Cos(angle)^2=1 the light intensity shouldn't swell while you rotate the polarization filter behind beam B, should it ??? I was under the impression that some assymetric polarizer would be required, so that the 90° angle between the two half of beam B could change, but I don't know if such an assymetric polarizer exists… $\endgroup$ – Bizniouf Nov 8 '18 at 8:22
  • $\begingroup$ Good point. Use a trick. Take a filter near the used wavelengths, which than should be transparent not for 90° rotation but only for 30° . I’m curious, please stay in contact with the results. Or is it only a theoretical question? $\endgroup$ – HolgerFiedler Nov 8 '18 at 10:35
  • $\begingroup$ At this point this is still a theoretical question. I am thinking about a device for which I need to make the difference. I have got other tracks but this one would be the simplest. I am afraid that using a 30° filter wouldn't change anything because whatever the angle, Sin(angle)^2+Cos(angle)^2 still equals 1. I may be wrong, of course... $\endgroup$ – Bizniouf Nov 8 '18 at 15:17
  • $\begingroup$ "In the case of half V- and H-polarized light the intensity is swelling four times for a 360°-rotation" - that's completely incorrect. There is absolutely no polarization state of a monochromatic beam that will produce this behaviour. If you're thinking that you'll get two lobes per polarization and that those will add up to produce four lobes, then you're plain wrong: Malus's law tells you that each set of two lobes will behave as $\sin^2(\theta)$ and $\cos^2(\theta)$ respectively. If you add them together you very much don't get a four-lobe pattern. $\endgroup$ – Emilio Pisanty Nov 8 '18 at 15:36
  • $\begingroup$ And as for the "trick" that you mentioned - the filter you've described is physically impossible. @Bizniouf's observation is right on the dot, the method just won't work. $\endgroup$ – Emilio Pisanty Nov 8 '18 at 15:57
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  • If the 2 parts (H and V)of beam B are in phase (they come from the same source along the same distance), then H+V polarization = diagonal linear polarization. That can be easily seen with a linear polarization filter.

  • If the 2 parts (H and V)of beam B come from the same source but are dephased you get an elliptic polarization $\to$ can be seen with a circular polarized filter.

The same question applies for a single photon : Is it possible to distinguish a unpolarized single photon from a H or V polarized photon ?

A single photon must always be polarized

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  • $\begingroup$ Your last statement is incorrect - it applies only to single-photon pure states. It is perfectly possible to produce single photons in mixed states that are basically fully de-polarized. $\endgroup$ – Emilio Pisanty Nov 8 '18 at 15:46
  • $\begingroup$ what you mean @EmilioPisanty is that the polarization can be undertermined until it's measured right ? If so there is still a polarization. $\endgroup$ – Manu de Hanoi Nov 8 '18 at 15:48
  • $\begingroup$ No. In such a configuration, saying that "there is still a polarization" is a counter-factual statement which is not within the statements that can be made in quantum mechanics. $\endgroup$ – Emilio Pisanty Nov 8 '18 at 15:54
  • $\begingroup$ Please elaborate because I dont know how you can talk about a (EM) wave that has no polarization. That would mean that the EM field have no direction, meaning they are 0, meaning it's a constant(not a wave). In QM you are not supposed to talk about both a particle and a classical wave in the same time. But if you break the rule and allow yourself to give a wave property like polarization to a particle like a photon, then be consistent. EDIT: in other words, you can either say a photon always has a polarisation or never. But not "sometimes". $\endgroup$ – Manu de Hanoi Nov 8 '18 at 16:06
  • $\begingroup$ @ Manu I would recommend looking up "unpolarized light" in your favourite optics textbook or web resource; if you want a specific pointer, the Wikipedia page is here and a good in-site link is here. It is a perfectly standard concept and (since it is necessary to explain most light sources that you'll encounter in everyday light) it is explained in depth in any introductory resource to polarization in optics. $\endgroup$ – Emilio Pisanty Nov 8 '18 at 19:31

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