# A How does a polarising filter work?

#### cmb

Being an old school Newtonian engineering type, I have a model of a polarising filter that generally works for day to day stuff like taking photos, explaining sun glasses etc..

But it is clearly wrong.

Given that opening remark, the 'A' index, and the forum subject area, I trust you will understand I want a bit more of a fundamental discussion on this. Please can anyone give it a go? Personally I think it is going to get tricky, but maybe I am seeing problems where there are simple answers?

... or maybe polarising filters really are too complicated for us to explain at the moment?

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#### hutchphd

Being an old school Newtonian engineering type, I have a model of a polarising filter that generally works for day to day stuff like taking photos, explaining sun glasses etc..

But it is clearly wrong.

Given that opening remark, the 'A' index, and the forum subject area, I trust you will understand I want a bit more of a fundamental discussion on this. Please can anyone give it a go? Personally I think it is going to get tricky, but maybe I am seeing problems where there are simple answers?

... or maybe polarising filters really are too complicated for us to explain at the moment?
Here is a start: There are several fundamentally different kinds of polarizers. In addition their exact construction depends upon the wavelength (or frequency if you prefer) of the radiation in question.
1. Sometimes the radiation is polarized by orienting the emitters. For instance a dipole antenna will emit linearly polarized radiation.
2. The most common "filters" look like "combs" that absorb preferentially because they conduct and dissipate along the teeth of the comb. These could be parallel aluminum rods or metal deposited on glass or long molecules oriented in a matrix again depending upon wavelength
3. Reflection from a surface will also polarize
This is a very big and useful subject and you can find lots of published sources. I will not attempt to reproduce them. If you have specific questions from your researches they should asked here

#### cmb

OK, well let's start with what process determines if a photon passes through the polarising filter or not? Or do no photons pass through and all are absorbed in which case which ones are re-radiated?

(I am assuming an 'optical' polarising filter, like sunglasses are, whatever they are)

#### Cryo

Gold Member
I think all of the stuff relating to polarizers is covered by Maxwell's Equations. As already mentioned by hutchphd there are many different kinds, but if you are comfortable with Maxwell's Equations then all of it is simple.

Pick a filter you like, e.g. https://www.thorlabs.com/navigation.cfm?guide_id=8 and then try to understand it. It will be much more easier to help with specific questions.

#### Cryo

Gold Member
Oh, you want photons. Well in this case have a look in Loudons "The Quantum Theory of Light". Basically you quantize the electromagnetic field, and get the electric and magnetic field operators. After that a lot of the treatment of filters is like in classical electrodynamics. It's the detection that is tricky

#### Cryo

Gold Member
In the simplest case you have electromagnetic field, quantized, before the filter and after the filter. So you may have a creation operator for the photon before the filter $\hat{a}_{in,h}^\dagger$, $\hat{a}_{in,v}^\dagger$ and the boundary condition (matrix) that represents the filter which will tell you how these operators translate into the output field $\hat{a}_{out,h}^\dagger$, $\hat{a}_{out,v}^\dagger$. If the filter is lossy you will probably have to work with the density matrix rather than the pure states, though there are other ways, e.g. https://journals.aps.org/pra/abstract/10.1103/PhysRevA.57.2134 .

Filters do not let or stop photons from passing. You have a quantum state of light which is affected by the filter, but nothing is determined until you get to the detection, at least in the standard interpretation.

#### hutchphd

OK, well let's start with what process determines if a photon passes through the polarising filter or not? Or do no photons pass through and all are absorbed in which case which ones are re-radiated?
I should have said absorbed or scattered to be correct. Typically the scattering (or reradiation) is isotropic and hence lost from the beam direction.

#### cmb

In the simplest case you have electromagnetic field, quantized, before the filter and after the filter. So you may have a creation operator for the photon before the filter $\hat{a}_{in,h}^\dagger$, $\hat{a}_{in,v}^\dagger$ and the boundary condition (matrix) that represents the filter which will tell you how these operators translate into the output field $\hat{a}_{out,h}^\dagger$, $\hat{a}_{out,v}^\dagger$. If the filter is lossy you will probably have to work with the density matrix rather than the pure states, though there are other ways, e.g. https://journals.aps.org/pra/abstract/10.1103/PhysRevA.57.2134 .

Filters do not let or stop photons from passing. You have a quantum state of light which is affected by the filter, but nothing is determined until you get to the detection, at least in the standard interpretation.
This is in the right direction but far too complicated and also inspecific at the same time. I don't really know what that means, and don't think I need to either.

Say I buy a polarising filter from the camera shop. What is the ratio of light that will pass through it?

If I buy two of those filters and set them at an angle to each other, what is the ratio of light that passes then?

Gold Member

#### A. Neumaier

Given that opening remark, the 'A' index, and the forum subject area, I trust you will understand I want a bit more of a fundamental discussion on this. Please can anyone give it a go? Personally I think it is going to get tricky, but maybe I am seeing problems where there are simple answers?
Look at my Insight article for a simple introduction to some of the theory, and follow up with Wikipedia on the items menitioned there.

Filters do not let or stop photons from passing. You have a quantum state of light which is affected by the filter, but nothing is determined until you get to the detection, at least in the standard interpretation.
That's not true. Filters reduce the intensity of light, which means only part of the light passes, the remainder heats up the filter and can in principle be measured. Thus (unlike a Stern-Gerlach magnet, which is unitary), a filter is always measuring.
I think all of the stuff relating to polarizers is covered by Maxwell's Equations.
No. Maxwell's equations cover only fully polarized light. Unpolarized light and how to polarize it is not described by it.

#### Cryo

Gold Member
A. Neumaier, I would be very glad to learn from you. Can you please clarify a couple of things?

That's not true. Filters reduce the intensity of light, which means only part of the light passes, the remainder heats up the filter and can in principle be measured. Thus (unlike a Stern-Gerlach magnet, which is unitary), a filter is always measuring.
In the reference I gave above, (Barnett, Jeffers, Gatti, Loudon, "Quantum optics of lossy beam splitters", Phys. Rev. A 57, 2134 (1998) ) they treat a lossy beam splitter. The loss in the beam splitter is treated by introducing a bath of quantum oscillators that can be used to dump the energy of the light into. Would the same formalism not account for response of a lossy polarizer? I understand that you would not get a pure state out of such a thing, but a mixed state may be ok too.

No. Maxwell's equations cover only fully polarized light. Unpolarized light and how to polarize it is not described by it.
What do you mean by unpolarized light? Classically unpolarized light is light with polarization that is changing randomly at the time-scale faster than your detector. Treatment of propagation of such light is the same as polarized light, but with arbitrary polarization. Extra care is needed when you detect this light, sure, but as far as I can see classical unpolarized light = Maxwell's equations + random variables describing polarization.

In a non-classical case, again, propagation of light, i.e. the bit that Maxwell's equation are needed for, is the same for any state - you simply specify the state (or the density matrix) and time-evolve the relevant operators.

Is there something I am missing?

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#### vanhees71

Gold Member
No. Maxwell's equations cover only fully polarized light. Unpolarized light and how to polarize it is not described by it.
But you can describe it with some statistics. Incoherent light can, e.g., be described by taking the intensity of coherent light and randomize phase differences. The same holds for polarization. So it's well possible to describe the optics of "natural light" within classical Maxwell theory.

#### A. Neumaier

In the reference I gave above, (Barnett, Jeffers, Gatti, Loudon, "Quantum optics of lossy beam splitters", Phys. Rev. A 57, 2134 (1998) ) they treat a lossy beam splitter. The loss in the beam splitter is treated by introducing a bath of quantum oscillators that can be used to dump the energy of the light into. Would the same formalism not account for response of a lossy polarizer? I understand that you would not get a pure state out of such a thing, but a mixed state may be ok too.
Any matter light goes through, acts as a filter and is slightly lossy, though this can be often neglected in simplified discussions. A polarizer is necessarily lossy (and not only slightly) when the input light is not of the polarization of the output, whereas it lets correctly polarized light through almost without loss.
What do you mean by unpolarized light? Classically unpolarized light is light with polarization that is changing randomly at the time-scale faster than your detector. Treatment of propagation of such light is the same as polarized light, but with arbitrary polarization. Extra care is needed when you detect this light, sure, but as far as I can see classical unpolarized light = Maxwell's equations + random variables describing polarization.
So it is described by a stochastic version of the Maxwell equation, not by Maxwell's equation itself.
Is there something I am missing?
If your usage of Maxwell's equations includes the stochastic case then nothing is missing.
But you can describe it with some statistics. Incoherent light can, e.g., be described by taking the intensity of coherent light and randomize phase differences. The same holds for polarization. So it's well possible to describe the optics of "natural light" within classical Maxwell theory.
Stochastic Maxwell equations need much more complex mathematical machinery (of stochastic fields) than the ordinary Maxwell equations; hence I treat them as not the same.

#### hutchphd

You can make overly complicated formulations of anything. The OP asked about how sunglasses work. I am reminded of my favorite "peanuts" cartoon

"How does a polarising filter work?"

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