Is a polarizer a filter or a converter?

[Drakkith]

Obviously one can argue that per QM the idea of polarization of a photon being exactly X is incorrect. We have no idea what it is polarized at and most will say it doesn't have an exact polarization, only the probability of what it can be. To me, looking at the whole situation, X photons go through and less than X photons come out. Since the ones that come out don't appear to be "converted", and instead seem to have already been in the beam to begin with, I would say that a polarizer is a filter, not a converter.

 

[ibysaiyan]

interesting discussion, good points jfy and joncon, still thinking over Mr Vodka's post

just wanted to add that there is no difference between the behavior of the first polarizer versus the second because:

the first one also encounters photons with polarization at say 45%, 50% 30% etc

for 45% it again..."forces" (whatever that means) just like the second polarizer to be V or H.

for the say 30% ...it again "forces", however we know the profitability of it emerging as V or H? The probability of it emerging as H would be higher...?

we once again one of the fundamentals of QM at play...(for a given polarization angle)--you can calculate/know the probabilities but not the individual behavior

on a separate, but related, topic:
also if the photons' polarization prior to encounter with a polarizer/qwp etc is indeterminate (as per a popular QM interpretation) does it remain determinate (forever) after emerging from a polarizer?

I suppose it would return back to wave-function i.e the probability would be indefinite since it's not being observed anymore,just a guess.

 

[San K]

Here's a cute little youtube experiment that has been posted here a few times. :)

https://www.youtube.com/watch?v=ZudziPffS9E

the video says "photon polarization is indeterminate till its measured"

as per one of the popular interpretations of QM

1. is the polarization indeterminate ...before or after ...the photon goes through the 45 degree polarizer? or after it exits the third polarizer?

2. what action/setup constitutes measurement?

3. how do we convert the polarization of a photon
- from determinate to indeterminate -- does it simply happen over time, for one?
- indeterminate to determinate -- just send it through a single polarizer? or through two polarizers? with some angle between their axis so that it constitutes a measurement?

 

[San K]

Here's a cute little youtube experiment that has been posted here a few times. :)

https://www.youtube.com/watch?v=ZudziPffS9E

if we were to replace the 45 degree with 30 degree polarizer, would we still expect light to pass through however it would be a bit dimmer?

 

[agentredlum]

When he put the polarizer in the middle he should have kept rotating it so we could see if it had any effect on the polarizer closer to the camera.

The light that goes through 2 polarizers is distinctly different in color than the light that goes through all 3

With 2 polarizers the intensity changes but the color of the image does not change as he rotates the second polarizer. With 3 polarizers the image in the center looks blue but the image outside the first polarizer AND inside the third polarizer looks the same as originally with only 2 polarizers. Very curious...

 

[kith]

Okay as I seem to be ignored trying to explain it nicely, I'll be blunt: you're wrong. Say system S is a construction of two polarizers at different angle, then S is a filter because whatever comes out was already part of the original wave, there's just something left out.

No, it wasn't. This can be seen easily if you use a third polarizer at 90° to the first to analyze your wave. The first polarizer filters out all parts of the wave which could pass the third. So if the second would just filter out more parts of the wave, there would still be no transmission through the third polarizer. Actually, there is a transmission, so the second polarizer has created new parts of the wave.

How does a polarizer actually work on a QM-level? I can't figure out what it does to one photon. What is the measuring aparatus?

I'm not sure if this is a satisfying answer to your question, but like every measurement device, its a big system which can be treated classically. Similar to a Stern-Gerlach-Apparatu for measurements of the spin of the electron.

 

[kith]

But consider it as a "converter"; what is being converted!?

In the CI, the answer is "the state of the photon". It enters with initial state |i> and exits with an eigenstate of the measuring device |f> (as DrChinese has pointed out).

However, I just realized that the answer is interpretation-dependent. In the statistical interpretation, I think the situation is analogous to the classical case: the first polarisator is a simple filter, the second a converter (with limited efficiency).

 

[nonequilibrium]

No, it wasn't. This can be seen easily if you use a third polarizer at 90° to the first to analyze your wave. The first polarizer filters out all parts of the wave which could pass the third. So if the second would just filter out more parts of the wave, there would still be no transmission through the third polarizer. Actually, there is a transmission, so the second polarizer has created new parts of the wave.

Well, I still stand by my case that it's a filter, but I also understand your argument; I think are points of view aren't mutually exclusive, and this because the words "being a part of the original wave" actually seem to be pretty stupid:

Take for example a sine wave (name: X) at 45° degrees, that you can view as a sum of a sine at 0° (name: A) and at 90° (name: B). Now a polarizer at -45° won't let anything through (making a drawing helps). But if you first put a polarizer at 0° (only keeping A), and then the one at -45°, you will (not surprisingly) get something. The wave you get at the end is actually a component of A, which on its turn was a part of the original wave. But of course you might sensefully argue that that component is not a true component, because it is canceled out by a certain component of B. So in that sense, the wave you get in the end is not a part of the original wave. But as I see it, each polarizer takes a component of the wave, and to me a component of a component can still be regarded as a part of the original wave. My main point is this: it seems to be a matter of taste, wouldn't you agree?

 

[kith]

My main point is this: it seems to be a matter of taste, wouldn't you agree?

I agree that it is mainly a semantic question. I also understand your argument, which seems similar to the last paragraph of jfy4's last post: with every polarizer, you "filter out" some intensity.

Still, I don't think that calling them filters is appropriate. If I have a quantity of something and I filter out all parts with certain properties, I can't get parts with the same properties back by filtering out all parts with certain other properties.

Maybe we should just agree to disagree. As far as I see it, everything relevant has been said. ;-)

 

[nonequilibrium]

If I have a quantity of something and I filter out all parts with certain properties, I can't get parts with the same properties back by filtering out part with other properties.

Can you maybe rephrase, I don't understand the construction of your sentence (English is not my mother tongue)

 

[kith]

Can you maybe rephrase, I don't understand the construction of your sentence (English is not my mother tongue)

It isn't mine either, so maybe my sentence is more complicated than needed. ;-) I try it again with a concrete example.

We have a bowl filled with objects in different colors and different shapes. If we filter out all red objects, there is no way to get any red objects back by filtering out other objects (cubes for example). This is the behaviour I expect from filters, but it's not the way polarizers behave. If I filter out one part of the whole wave (for example the 0°-polarized), I can get it (partially) back by filtering out another part (the 45°-polarized).

 

[nonequilibrium]

Well, as you said: "partially" :p that's the key word for me there

Okay I agree to disagree ;)

 

[y33t]

It isn't mine either, so maybe my sentence is more complicated than needed. ;-) I try it again with a concrete example.

We have a bowl filled with objects in different colors and different shapes. If we filter out all red objects, there is no way to get any red objects back by filtering out other objects (cubes for example). This is the behaviour I expect from filters, but it's not the way polarizers behave. If I filter out one part of the whole wave (for example the 0°-polarized), I can get it (partially) back by filtering out another part (the 45°-polarized).

Some polarizers behave like waveguides as well depending on the polarization i/o relationships. The one posted on the video is an example of this. Internal structure of these polarizers is multiple layers of waveguides (at micro or nano level - depending on the frequency) rotating respectively from original polarization state to aimed (output) polarization state made from very low refractive index material. Linear to circular polarizer is an example of this (waveguide). Shifting from circular polarization to linear can be done by filtering and shifting from linear to circular polarization by waveguiding. There are also various polarizer types like IR polarizers which operates independent of the polarization state. 'Polarizer' is not enough to decide whether it's a waveguide or a filter, it depends on what kind of polarizer it is.

 

[Drakkith]

I agree that it is mainly a semantic question. I also understand your argument, which seems similar to the last paragraph of jfy4's last post: with every polarizer, you "filter out" some intensity.

Still, I don't think that calling them filters is appropriate. If I have a quantity of something and I filter out all parts with certain properties, I can't get parts with the same properties back by filtering out all parts with certain other properties.

Maybe we should just agree to disagree. As far as I see it, everything relevant has been said. ;-)

Why does that not make it a filter?

 

[harrylin]

Why does that not make it a filter?

Again: a pure filter only removes, it doesn't modify.
Perhaps a counter example is useful: An attenuator is a pure filter, as it doesn't modify the polarization.

 

[Drakkith]

Again: a pure filter only removes, it doesn't modify.
Perhaps a counter example is useful: An attenuator is a pure filter, as it doesn't modify the polarization.

I have yet to see where a polarizer modifies the polarization of individual photons.

 

[Joncon]

I have yet to see where a polarizer modifies the polarization of individual photons.

So if a photon polarized at 45° to the vertical passes through a polarizer set at 0° to the vertical, would you say that photon is still polarized at 45°?

 

[Drakkith]

So if a photon polarized at 45° to the vertical passes through a polarizer set at 0° to the vertical, would you say that photon is still polarized at 45°?

Yes. To my knowledge, at that polarization the photon still has a chance of passing through the polarizer. Another photon polarized also at 45 degrees might not make it through.

 

[Joncon]

Yes. To my knowledge, at that polarization the photon still has a chance of passing through the polarizer. Another photon polarized also at 45 degrees might not make it through.

OK, so if I hold two identical polarizers together, at the same angle, are you saying the light passing through is dimmer than if I only used one polarizer? i.e. a photon at 45° has a 50/50 chance of passing the first and then a 50/50 chance of passing the second.

 

[DrChinese]

Joncon: So if a photon polarized at 45° to the vertical passes through a polarizer set at 0° to the vertical, would you say that photon is still polarized at 45°?

Yes. To my knowledge, at that polarization the photon still has a chance of passing through the polarizer. Another photon polarized also at 45 degrees might not make it through.

There is absolutely no evidence this is a true assessment. After any photon passes through a 0 degree polarizer, it will continue to pass similarly oriented polarizers - no problem. However, its orientation at 45 degrees is completely indeterminate. So I would NOT say it is oriented at 45 degrees. For if it were, it would pass 45 degree polarizers.

 

[Joncon]

There is absolutely no evidence this is a true assessment. After any photon passes through a 0 degree polarizer, it will continue to pass similarly oriented polarizers - no problem. However, its orientation at 45 degrees is completely indeterminate. So I would NOT say it is oriented at 45 degrees. For if it were, it would pass 45 degree polarizers.

Yes, that's what I was thinking.

Is it correct then to say that, while a polarizer acts like a filter (either letting photons pass or not), it also "rotates" the polarization of photons which pass successfully?

 

[DrChinese]

Is it correct then to say that, while a polarizer acts like a filter (either letting photons pass or not), it also "rotates" the polarization of photons which pass successfully?

It does appear to have that effect. Any quantum observation has this appearance (following the HUP). It is a bit difficult to know exactly what is happening during the process of collapse, and there are varying interpretations.

 

[ibysaiyan]

It does appear to have that effect. Any quantum observation has this appearance (following the HUP). It is a bit difficult to know exactly what is happening during the process of collapse, and there are varying interpretations.

Are there any links/articles which go a little in depth about what happens during the collapse of the wavefunction ? I will be grateful if you know of any.

P.S: Sorry for having hijacked this thread, slightly.


-ibysaiyan

 

[harrylin]

I have yet to see where a polarizer modifies the polarization of individual photons.

The point of the demonstration with three polarizers - as also Joncon and others tried to explain - is that evidently photons that come out of the middle polarizer have a different polarization than the photons that enter it. If the photons coming out had the same polarization as the ones that enter it, then no light would come out. It's a basic characteristic of polarizers that they attenuate by changing the polarization, in contrast with pure attenuators that do not change the polarization.

 

[Drakkith]

There is absolutely no evidence this is a true assessment. After any photon passes through a 0 degree polarizer, it will continue to pass similarly oriented polarizers - no problem. However, its orientation at 45 degrees is completely indeterminate. So I would NOT say it is oriented at 45 degrees. For if it were, it would pass 45 degree polarizers.

Ah, ok. I had not seen anything like that before.

I found this page that explains the effect: http://demonstrations.wolfram.com/LightBeamsThroughMultiplePolarizers/

The point of the demonstration with three polarizers - as also Joncon and others tried to explain - is that evidently photons that come out of the middle polarizer have a different polarization than the photons that enter it. If the photons coming out had the same polarization as the ones that enter it, then no light would come out. It's a basic characteristic of polarizers that they attenuate by changing the polarization, in contrast with pure attenuators that do not change the polarization.

Yes, I see now. It would appear that a polarizer is both a filter and a converter then.

 

[sgb27]

We have a bowl filled with objects in different colors and different shapes. If we filter out all red objects, there is no way to get any red objects back by filtering out other objects (cubes for example). This is the behaviour I expect from filters, but it's not the way polarizers behave. If I filter out one part of the whole wave (for example the 0°-polarized), I can get it (partially) back by filtering out another part (the 45°-polarized).

Consider this. The 1st polariser filters out red balls, the 2nd yellow balls, and the 3rd green balls. Now consider that red balls are equal to (yellow-green) balls, and green balls are equal to (yellow-red) balls. You'll see that the 2nd filter will "turn" green balls into (negative) red balls, but it's not adding anything, only subtracting the yellow component.

Maybe the bit that is confusing is that a light wave polarised in the negative "red" direction is just as visible as one polarised in the positive "red" direction. With real objects you can't see a "negative red" ball.

 

[harrylin]

Consider this. The 1st polariser filters out red balls, the 2nd yellow balls, and the 3rd green balls. Now consider that red balls are equal to (yellow-green) balls, and green balls are equal to (yellow-red) balls. You'll see that the 2nd filter will "turn" green balls into (negative) red balls, but it's not adding anything, only subtracting the yellow component. [..]

Your illustration doesn't work. For it to work, no ball at all must pass with two filters, and some balls must pass with an additional filter added between them. That won't be easy.

Harald

 

[sgb27]

Your illustration doesn't work. For it to work, no ball at all must pass with two filters, and some balls must pass with an additional filter added between them. That won't be easy.

Harald

Assuming you just start with red and green balls, the red will be filtered out by the 1st polariser and leave the green unchanged, and the 3rd polariser (at 90 degrees) will filter out the green leaving no balls passing. The key point is that each polariser considers the "colours" relative to it's optical axis, the 1st and 3rd use red/green as their axes, and the 2nd uses another set of colours. I think that's about as good as you can get with this analogy.

 

[harrylin]

Assuming you just start with red and green balls, the red will be filtered out by the 1st polariser and leave the green unchanged, and the 3rd polariser (at 90 degrees) will filter out the green leaving no balls passing. The key point is that each polariser considers the "colours" relative to it's optical axis, the 1st and 3rd use red/green as their axes, and the 2nd uses another set of colours. I think that's about as good as you can get with this analogy.

Indeed the analogy only serves to show that a mere filtering doesn't work: either a filter filters out a full colour (which isn't the case in your example, the second filter distinguishes and filters a component), or it only filters out a component (which isn't the case in your example, the third filter doesn't let the yellow component pass).

 

[sgb27]

In terms of polarisers there is no such thing as a "full colour", ie an absolute or reference axis for polarisation. You must consider every case with reference to the optical axis of the material being considered.

Even in the analogy, who said that red was a "full colour" and yellow wasn't? Red can be made up of yellow and negative green, and also yellow can be made up of red and positive green. It all depends on the point of view of the specific filter/polariser as to what is the "full" colour. In terms of the 2nd polariser, yellow is the full colour, it no more or less full than the red the first polariser sees.