Question regarding the three polarizer paradox

[SeanHannity]

Summary: Does the three polarizer paradox work with other materials?

Alright, so this is going to sound like a dumb question, but if you were to do the three polarizer experiment but replace the polarizers with a another material, could the different material produce a similar effect to that of the 3 polarizers? Or does the effect only work with polarizers? Thanks lol.

 

[vanhees71]

What paradox are you referring to?

A polarizer is a polarizer. It's independent of the specific material it's made of. In optics you just describe it with effective parameters. Of course there are different kind of polarizers like simple filters like a polaroid foil or polarizing beam splitters like birefringent crystals.

 

[Demystifier]

What paradox are you referring to?

https://chem.libretexts.org/Bookshe...antum_Optics/268:_The_Three-Polarizer_Paradox

 

[Dale]

Wow. It really doesn’t take much to get labeled as a “paradox” anymore.

 

[Dale]

Summary: Does the three polarizer paradox work with other materials?

if you were to do the three polarizer experiment but replace the polarizers with a another material, could the different material produce a similar effect to that of the 3 polarizers? Or does the effect only work with polarizers?

Are you asking about different kinds of polarizers, or are you thinking of a non-polarizing material? Like maybe three different color-selective filters?

 

[anorlunda]

It is a QM question, not classical.

 

[Dale]

It is a QM question, not classical.

How so? You get polarization in classical EM too.

 

[sophiecentaur]

It is a QM question, not classical.

If you want to use QM then is it not necessary to use statistics to predict an 'outcome' for a polarisation operation? In which case classical or QM give the same answer.

 

[anorlunda]

How so? You get polarization in classical EM too.

It is the case where you start with two polarizers, 90 degrees apart. No light passes. Then you insert a 3rd polarizer between the two, the 3rd angled at 45 degrees and now you get light passing through all three. In classical terms where a polarizer can only subtract light, never add, that can't happen.

It is illustrated in the link provided by @Demystifier in post #3.

 

[sophiecentaur]

A polarizer selects the field components along one axis. That is not “subtracting light” in an arithmetic sense. A second polarizer will pass any component of the light from the first that is not orthogonal. This is not paradoxical or magic. It’s just how vectors work.
A polarizer is not just ‘slots’.

 

[SeanHannity]

Are you asking about different kinds of polarizers, or are you thinking of a non-polarizing material? Like maybe three different color-selective filters?

Was thinking about the latter

 

[Dale]

In classical terms where a polarizer can only subtract light, never add, that can't happen.

No, that isn’t how it works classically. A polarizer classically passes a fraction of the light equal to $\cos^2 \theta$. So if you have $\theta=90\deg$ then it passes nothing, but if you have $\theta=45\deg$ then it passes half and the half that is passed is now polarized at $45\deg$ and will be half passed through a $90\deg$ polarizer.

 

[Dale]

Was thinking about the latter

With color selective filters it wouldn’t work. You cannot put any color selective filter between two other color selective filters and increase the total transmitted light as far as I know.

 

[Delta2]

Summary: Does the three polarizer paradox work with other materials?

Summary: Does the three polarizer paradox work with other materials?

I don't think we have the right to call it a paradox. A surprising, unexpected perhaps counter intuitive result, but not a paradox

 

[anorlunda]

The reason I said QM is because Dirac discusses the 3 polarizer experiment.

I found here https://www.informationphilosopher.com/solutions/experiments/dirac_3-polarizers/

In chapter 1 of his book The Principles of Quantum Mechanics, Paul Dirac describes our experiment. (Complete text of Chapter 1)

from section 2, The Polarization of photons, pp.5-7

It is a lengthy quote from Dirac's book, so I did not paste it here. You can see it at the link above.

Can it be explained classically? 100% of the light after the first filter is polarized at 0 degrees. 100% of the fraction of the light that passes filter 2 is polarized at 45 degrees. So the polarizer not only filtered out some of the light, but it also changed the polarization of the light passing through. It is not a question of how big the fraction was, but the polarization of the light that did pass.

 

[Dale]

Can it be explained classically? 100% of the light after the first filter is polarized at 0 degrees. 100% of the fraction of the light that passes filter 2 is polarized at 45 degrees. So the polarizer not only filtered out some of the light, but it also changed the polarization of the light passing through.

Ok, but that is what is predicted classically. There is nothing inherent in Maxwell’s equations that forbids a passive device from changing the polarization of light.

Dirac discusses the 3 polarizer experiment.

He also recognizes that the classical approach gives the same result. He said: “These values for the probabilities lead to the correct classical results for an incident beam containing a large number of photons”.

He seems to be amazed not that a new non-classical result is obtained but rather that the photon picture is compatible with the known classical result.

 

[sophiecentaur]

He seems to be amazed not that a new non-classical result is obtained but rather that the photon picture is compatible with the known classical result.

Hard to put oneself in the shoes of the early workers but the result could initially seem too good to be true when it shows the same answer for the two approaches. We have the gift of perfect hindsight.

 

[Lord Jestocost]

[PDF]I. Experimental Evidence for Quantum Mechanics - MIT

Section "How Do We Expain This Result?" might be helpful.

 

[sophiecentaur]

[PDF]I. Experimental Evidence for Quantum Mechanics - MIT

Section "How Do We Expain This Result?" might be helpful.

The second diagram has nonsense implications because it suggests a simple 'selection' between two already polarised beams. If they cannot think of a way to draw what they almost certainly mean then they should avoid even trying to write an explanatory paper. That diagram leaves out the all important Cosine factor and the use of vectors.

 

[sophiecentaur]

Ok, but that is what is predicted classically. There is nothing inherent in Maxwell’s equations that forbids a passive device from changing the polarization of light.

Exactly. There are perfect mechanical analogies of the classical polarisation of light concept and few people would talk in the same way about 'filtering out' oscillations. Why is that terminology reserved for EM polarisation? So much of written about the QM version of polarisation that almost falls over itself by insisting on talking about 'states' as if they are more than just descriptions of the probability of measuring something. A polariser can be described as a 'measuring instrument' with a result which is the probability of a photon getting through it. Probability is what it's all about.

 

[vanhees71]

https://chem.libretexts.org/Bookshe...antum_Optics/268:_The_Three-Polarizer_Paradox

Argh! That's also a paradox now. Well, I've to reformulate next week's problem set for my students to make them solve a paradox rather than some projection operator applications ;-)).

 

[sophiecentaur]

There is nothing inherent in Maxwell’s equations that forbids a passive device from changing the polarization of light.

I just read this again. A simple linear polariser has to do something with half of the energy of a non polarised beam (reflects /absorbs). Likewise, it reflects / absorbs a (sine factor) proportion of a linearly polarised beam the rest (cosine factor) is let through. The power of the resulting polarised beam will be diminished.

So far no one has introduced circular polarisation into this discussion. Perhaps there would be less of a "paradox" here because it's necessary to know a bit about what polarisation actually means so there can't be that 'comb' model lurking behind that particular mind picture. Aamof, circular polarisation should, perhaps be the best way in because it doesn't involve arbitrary angles for the H and V components - just the sense of rotation which can only have two values.

 

[Dale]

In the classical polarizer it is important to remember that there are no photons, there is only the EM field. A polarizer splits the EM field vector into a parallel component and a transverse component and then it blocks the parallel component and passes the transverse component.

I think you guys are thinking that there are classical pseudo-photons that have a definite polarization and can only be selected or not. Classically there are no pseudo-photons, just the fields. And the classical fields demonstrate this “paradox” too.

This is QM only in the sense that QM is compatible with this result, not in the sense that it is necessary for the result

 

[sophiecentaur]

@Dale . . . . . but you can't say that QM is not happening(?)
This conversation has reached the stage where all sorts of things are getting thrown up and apparently need to be dealt with. The fog level is increasing.

 

[Dale]

but you can't say that QM is not happening(?)

Certainly not! QM is always happening.

Sometimes QM is necessary for explaining the outcome of an experiment and sometimes classical theory can adequately explain the same experiment. In this case the classical theory is adequate.

Of course, QM is still happening, this is simply a case where QM agrees with the classical prediction, so it is not necessary.

 

[Vanadium 50]

Why are you guys dragging QM into this? Is the OP's difficulty that this isn't complicated enough?

 

[vanhees71]

I've also no clue, why the quantum issue came up at all. I've also no clue, why this is called a paradox, but maybe that's because I've learned classical electrodynamics and optics quite a time ago already.

As @Dale stresses, all this is completely understandable with classical electrodynamics and there's nothing mysterious nor paradoxical.

Without going into the microscopic details just consider the most simple polarizer, a polaroid foil, is a device which let's an electromagnetic wave polarized in one direction (say the $x$-direction) completely through and absorbs one polarized in the direction perpendicular to it (say the $y$ direction, and take the $z$ direction as the direction the wave travels, i.e., the direction of $\vec{k}$).

Now an arbitrary electromagnetic wave with $\vec{k}=k \vec{e}_z$ can be written as
$$\vec{E}(t,\vec{x})=(E_{0x} \vec{e}_x + E_{0y} \vec{e}_y) \exp(\mathrm{i} k z -\mathrm{i} \omega t),$$
where the physical field is of course just the real part, but the complex notation makes life much easier particularly in this polarization business.

So we can describe the polarizer oriented such that it let's $x$-polarized waves through and absorbs $y$-polarized waves completely by the linear (projection) operator.
$$\vec{E}'(t,\vec{x})=\hat{P}_x \vec{E}(t,\vec{x})=E_{0x} \vec{e}_x \exp(\ldots).$$
The intensity $I$ is given (up to an uninteresting constant) by
$$I=\langle \vec{E}^* \cdot \vec{E} \rangle_{T}=\vec{E}_0^{*} \vec{E}_0.$$
Now if you have linearly polarized waves we have $E_{0x},E_{0y} \in \mathbb{R}$ and you can write
$$\vec{E}_0=E_0 (\cos \varphi \vec{e}_x + \sin \varphi \vec{e}_y).$$
The intensity (before the polaroid foil) thus is
$$I=E_0^2=E_{0x}^2+E_{0y}^2.$$
and after
$$I'=E_{0x}^2=E_0^2 \cos^2 \varphi,$$
which is Malus's Law for linearly polarized light.

Now it's easy to show that for a polaroid with it's direction oriented with an angle $\vartheta$ to the $x$ axis acts as the projection operator
$$\hat{P}_{\vartheta}= \vec{e}_{\vartheta} \vec{e}_{\vartheta}^{\text{T}}=\begin{pmatrix} \cos \vartheta \\ \sin \vartheta \end{pmatrix} (\cos \vartheta,\sin \vartheta),$$
i.e., for the amplitude after the polaroid you have
$$\vec{E}_0'=\vec{e}_{\vartheta} (\vec{e}_{\vartheta} \cdot \vec{E}_0).$$
Now it's clear that with two filters in perpendicular relative orientation (say one in $x$ and one in $y$ direction) you get
$$\vec{E}_0'=\hat{P}_{0} \hat{P}_{\pi/2} \hat{E}_0 = \vec{e}_x (\vec{e}_x \cdot \vec{e}_y) \vec{e}_y \cdot \vec{E}_0=0.$$
If you now have first one in $x$ then one in $\pi/4$ and then one in $\pi/2$ orientation you have
$$\vec{E}_0' = \hat{P}_{\pi/2} \hat{P}_{\pi/4} \hat{P}_{0} = \vec{e}_y (\vec{e}_y \cdot \vec{e}_{\pi/4}) (\vec{e}_{\pi/4} \cdot \vec{e}_x) E_{0x}=\frac{1}{2} \vec{e}_y E_{0x}.$$
The intensity after the filter thus is
$$I'=\frac{1}{4} |E_{0x}|^2.$$
Where is here a paradox? Where do I need photons?

 

[Ibix]

I've also no clue, why this is called a paradox

If you think as a wave as something either passing through the polaroid or not, it's slightly surprising. I've certainly seen polaroids illustrated as a gate with a plastic sinusoid slid through it (which will only go one way, and can't be rotated by a polaroid). That's a wrong picture, of course.

I think a better picture is to think of the polaroid as an array of antennae that are driven by the incident wave to emit a new wave. Then it's easier to believe that the intermediate polariser could re-emit radiation with components that wouldn't be completely absorbed by the third polariser.

 

[sophiecentaur]

@Dale I knew you knew that!

 

[anorlunda]

I've also no clue, why the quantum issue came up at all.

My bad. I thought it was a QM-only thing.

 

[sophiecentaur]

My bad. I thought it was a QM-only thing.

You don't use QM to describe what happens in your VP VHF boat antenna. RF electromagnetism make so much more sense when people want some 'understanding'. You can 'feel' the currents flowing and the result of tilting an antenna.

 

[Dale]

I think a better picture is to think of the polaroid as an array of antennae that are driven by the incident wave to emit a new wave.

I had never heard that before. Very Huygens-like.

 

[Ibix]

I had never heard that before. Very Huygens-like.

Really? I thought it was a fairly standard explanation of how wire grid (and similar) polarisers work, an anisotropic relative of the usual explanation of why EM waves don't penetrate metals. The charges can't move very far across the grid.

 

[vanhees71]

Usually the treatment is using "macroscopic electrodynamics", i.e., Maxwell's equations for $\vec{E}$, $\vec{B}$ and $\vec{D}$ and $\vec{H}$ with consitutive equations motivated from linear-response theory (in very advanced special treatments the latter derived from quantum many-body theory). Then of course the microscopic picture that the electromagnetic field is due to the superposition of the incoming (external) em. field and the em. field from the motion of the charges in the material.

For a treatment in the latter spirit see the very nice textbook by Schwartz:

M. Schwartz, Principles of Electrodynamics, Dover Publications (1972)

It's a bit like a book with the didactical intentions of Purcell in the Berkeley physics book done right, though there are still some rough edges in there (e.g., the use of the concept of "relativistic mass").

 

[Demystifier]

Argh! That's also a paradox now. Well, I've to reformulate next week's problem set for my students to make them solve a paradox rather than some projection operator applications ;-)).

Well, it's always more fun to solve a straightforward problem when it is presented as a "paradox". At least to me.