For a single photon, the only similar physically meaningful question is whether the circular polarization is left-handed or right-handed. Quantum mechanics may predict the probabilities of these two answers. An experiment, a measurement of L/R, produces one of these answers, too. After the measurement, the photon is either left-handed or right-handed circularly polarized.

If a photon is prepared in a general state, it has nonzero probabilities both for L and R. In such a "superposition", we may perhaps say that the single photon has no circular polarization. This statement means that we are uncertain which of the polarizations will be measured if it is measured. But when the circular polarization is measured, one always gets an answer, according to the result of the measurement.

Linear polarizations are the simplest nontrivial superpositions of L and R. The absolute value of both coefficients, $c_L$ and $c_R$, is the same while the relative phase encodes the axis on which the photon is polarized.

The paper quoted in the question is completely wrong. An example of a very wrong statement is that the linearly polarized photon moving in the $z^+$ direction carries $J_z=0\cdot\hbar$. In reality, a linearly polarized photon or any photon is certain not to have $J_z=0\cdot\hbar$. A linearly polarized photon has the 50% probability to be $J_z=+1\cdot\hbar$ and 50% to have $J_z=-1\cdot\hbar$. The expectation value $\langle J_z\rangle = 0$ but it's still true that the value $J_z=0\cdot\hbar$ is forbidden.

A different question is the polarization of an electromagnetic wave. For a wave, e.g. light, one may distinguish left-right and right-handed and $x$-linearly and $y$-linearly and elliptic polarizations of all kinds one may think of. In terms of photons, a macroscopic electromagnetic wave is the tensor product of many photons. If all these tensor factors are linearly (or circularly) polarized, then the wave may be said to be linearly (or circularly) polarized. Because the polarization of the whole wave requires some correlation in the state of individual photons, a wave may be measured not to be circularly polarized in either direction. But an individual photon is *always* circularly polarized in one of the directions when the answer to this question is measured.

The paper may present a proposed experiments which may be done but what is completely invalid is the author's interpretation of this experiment – even "possible interpretations" before the experiment is actually performed. The correct description by quantum mechanics isn't included among their candidate theories with which they want to describe the experiment.

There are two kinds of circular polarizers: right (clockwise) and left (counterclockwise). The first will transmit right circularly polarized light and absorb left circularly polarized light; the second does the opposite.

Turn the glasses over so the light goes the other way, and you will see that the effect is different.

A quarter wave plate, as @Jasper pointed out, is always wavelength (and light propagation angle) dependent. Most polarizers work less well with shorter wavelengths.

If light is randomly polarized, then passes through a linear polarizer, it becomes linearly polarized. Then when the light subsequently passes through the quarter-wave plate, the linear polarization converts to circular polarization. A quarter wave plate has two axes: a "slow" axis and a "fast" axis. Light propagates a little bit slower if it is polarized in the "slow" direction. The thickness of the quarter wave plate is precisely enough to slow light polarized in the "slow" direction, by a quarter of a wave (relative to light polarized in the "fast" direction. If you draw the waves, you'll see that the QW plate does nothing to light that is polarized precisely along either axis; but if light entering the QW plate is polarized at 45 degrees it comes out circularly polarized in, say, the *clockwise* direction. Rotate the QW plate by 90 degrees and the light will come out circularly polarized in the *counterclockwise* direction instead. If light entering So, the orientation of the QW plate relative to the linear polarizer is crucial, but when light enters from the linear polarizer side everything works right because the linear polarizer absorbs the portion of light that's polarized "wrong".

On the other hand, when randomly polarized light enters from the QW plate side, essentially nothing happens to it because it is still randomly polarized when it exits the QW plate. When it subsequently enters the linear polarizer, the light simply comes out polarized in the direction determined by the orientation of the linear polarizer.

Now consider what happens if *circularly* polarized light passes through from either side. If it enters from the QW plate side, *and if the handedness (clockwise or counterclockwise) is right, the QW plate converts it to linearly polarized light that passes almost 100% through the linear polarizer. If the handedness is wrong, the QW plate converts it to light that is linearly polarized light but in the wrong direction so it gets nearly 100% absorbed by the polarizer. Rotating the polarizer makes no difference to the amount of light that passes through, but it affects the polarization angle of the exiting light.

However, if circularly polarized light enters from the linear polarizer side, the polarizer absorbs the component that's not aligned with the polarizer, then converts the remainder to circularly polarized light with a certain handedness. Rotating the circular polarizer in this case makes no difference.

The third case to consider is that of linearly polarized light entering the filter from the QW side or linear polarizer side. If linearly polarized light enters from the QW side, the QW plate converts it to elliptically polarized light: circularly polarized if the linearly polarized incident light is at 45 degrees to the axes of the QW plate. The linear polarizer then transmits the portion of the elliptically polarized light that has the proper linear polarization. If linearly polarized light passes first through the linear polarizer, then through the QW plate, the linear polarizer passes only the fraction of the light having the proper linear polarization (note: linear polarization can be treated as a vector, so light polarized at 45 degrees from the vertical is equivalent to two vector components polarized vertically and horizontally, each with amplitude $1/sqrt(2)$). The QW plate then converts the passed light into right- or left-circularly polarized light.

There is a really good Harvard lecture here about polarization. Lots of good graphics in the Wikipedia article on circular polarization.

You will get color effects whenever different wavelengths are affected differently. In your experiment, you found that rotating the 3D glasses 90 degrees to the screen's vertical axis changed the color of light passing through the glasses. That means that the light emitted by the screen is polarized differently for different wavelength ranges. You don't know without further experimentation, in what way the polarization differs by wavelength. However, if you turn your 3D glasses over, they will act effectively as linear polarizers. You can explore the linear polarization of different wavelength ranges by putting colored cellophane between the polarizer and your eye as you rotate the polarizer while looking at the screen.

To see some really beautiful effects, look at some wrinkled cellophane tape between two crossed linear filters.

## Best Answer

With two circular and one linear polarisers you can do it. Use an unpolarised light source. First show that half the light passes the two circular polarisers when aligned and none passes when they are anti-aligned. So there is polarisation. Then replace the second filter by the linear filter and show its orientation does not matter. So the polarisation is not linear.

If that does not help, repeat the experiment by Richard Beth in 1935. Or get your colleague fired.