I How different is a calcite crystal from a polarizer?

[Frigorifico9]

TL;DR Summary

I know that polarizers block some percentage of the light, but what about a calcite crystal?

First of all, I apologize if I use incorrect terminology or I express myself poorly, I am trying my best. That said, I hope you guys are smart enough to understand me despite my shortcomings

I know that calcite has birefringence, and I know that if you take calcite crystals and cut them and join them in the right manner you can make a polarizer with them, but here's the thing

As far as I know, any polarizer made of any material will block some percentage of the light, depending on how the incoming light was polarized. Another way to say that is that the intensity of the outgoing light will always be less than the intensity of the incoming light

But a calcite crystal by itself is not quite a polarizer, so I wonder if calcite also blocks some of the incoming light or not. Maybe the intensity of the incoming and outgoing light could be the same, it seems possible to me

Of course I know that calcite has impurities and that it always reflects some light, otherwise it would be invisible, so the intensity of the outgoing light will always be less than the incoming light, but those reductions are part of other interactions, and not birefringence itself. In contrast reducing the intensity of outgoing light is indeed part polarization

I ask you to imagine an ideal situation. As far as I know an ideal polarizer would still reduce the intensity of the outgoing light, because that's part of how polarization is supposed to work. But maybe an ideal calcite crystal with ideal birefringence wouldn't, because reducing the intensity of light isn't part of how birefringence is supposed to work. I don't know but it certainly seems possible to me

Thanks for sharing some of your knowledge with me

 

[Andy Resnick]

TL;DR Summary: I know that polarizers block some percentage of the light, but what about a calcite crystal?

Thanks for sharing some of your knowledge with me

Calcite prisms (there are many designs) can be used as polarizers simply by blocking one of the refracted beams.

 

[Cutter Ketch]

I can see how it can be confusing. We use the same term “polarizer” for very different things. That’s because you can achieve the same effect in more than one way. You are used to absorptive polarizers that attenuate one polarization while letting the other pass. However, separating two polarizations is also a polarizer. Each separate beam is polarized. You are welcome to use one and dump the other as always happens in an absorptive polarizer, or you have the option of using both polarizations.

So, to answer your question, unlike an absorptive polarizer like you describe, Calcite does NOT block or even partially absorb any polarization. (Well, perfect Calcite wouldn’t. All real Calcite has defects that will lose some light, but to first order …) What you CAN do is use the birefringence to cause two polarizations to go different directions. In this way you produce two beams with orthogonal polarizations each of which can be said to be polarized.

It should be noted that birefringent polarizers are not the only non-absorptive polarizers. Polarizers are very commonly made with dielectric coating stacks. Coatings can easily be designed that reflect one polarization and transmit the other for angles of incidence well away from normal. Think of this as just enhancing the natural tendency of the Fresnel coefficients that you may recall result in Brewster’s angle. It’s actually kind of hard to make a dielectric interface that isn’t at least somewhat polarizing.

 

[vanhees71]

I'd distinguish simply between polarizers, which are in quantum-optics notation described by projection operators and polarizing beam splitters.

If you have a polarization foil oriented in horizontal direction it simply is described by the projection operator $\hat{P}_H= |H \rangle \langle H|$, maybe with some phase factor. A general pure polarization state is defined by a vector $|\psi \rangle=C_1 |H \rangle + C_2 |V \rangle$ with $|C_1|^2+|C_2|^2=1$ and the probability that a so prepared photon goes through the filter is $|C_1|^2$. The photon going through is described by the state $|H \rangle$.

Photons moving through the Calcit crystal is described by a unitary operator. Here you need to describe the full photon states, i.e., including momentum and polarization. The Unitary operator leads to an entanglement between the polarization and the momentum state (ordinary and extraordinary beam of the birefringence).