# Is a polarizer a filter or a converter?

1. Aug 11, 2011

### San K

Does a polarize simply filter photons
or
does it change the polarization of photons (as well)?

let's consider only one photon being sent, at a time, via the polarizer

I think the answer might be that:

for photons polarized at say 45 degree to the horizontal (or vertical) plane of the polarizer
the polarizer is a convertor, i.e. it changes the polarization of the photons into vertical or horizontal (randomly).

for photons polarized at say 0 or 90 degree to the horizontal (or vertical) plane of the polarizer
the polarizer acts as a filter because it is unable to convert the spin of such photons, the angle is too much/great.

maybe I did not understand polarization well

second question:

does a photon have a "definite" polarization prior to measurement/filtering?

Last edited: Aug 11, 2011
2. Aug 11, 2011

### y33t

Polarizer is a mechanical apparatus which only passes the light/wave depending on it's characteristics. For example a CPL (circular polarizer) will only let circularly polarized field components of the wave to pass to other side of it. Part that doesn't pass by is dissipated as heat in the polarizing filter. It's a filter not converter.

Photon it self doesn't have a polarization state, polarization is a behavior of waves. Photons has spins.

3. Aug 11, 2011

### Joncon

But photons show both wave and particle behaviour don't they? If a photon doesn't have a polarization state then what determines how it passes through a polarizer? I thought a photon followed Malus' law when encountering a polarizer.

4. Aug 11, 2011

### jfy4

The polarizer really is an interaction with the light. The light that is polarized at 45 degrees can be thought of as having the possibility of being polarized at 90 degrees, or at 0 degrees. The polarizer simply demands the photon be one or the other. Then either the photon is absorbed or passes through.
This can be written down mathematically as
$$|\psi \rangle = \sin\alpha |\psi\rangle_{||} + \cos\alpha|\psi\rangle_{\perp}$$
for 45 degrees this is
$$=\frac{1}{\sqrt{2}}(|\psi\rangle_{||}+|\psi\rangle_{\perp})$$
But from moving from a single photon to the above is about a large number of photons, not a case-by-case basis.

5. Aug 11, 2011

### Drakkith

Staff Emeritus
6. Aug 11, 2011

### Joncon

How can it not be a case by case basis? As I understand, photons don't interact with each other, so they must be acting individually.

OK, but if I hold two linear polarizers at 90 degrees to each other then no light gets through. If I then put one at 45 degrees between these two then some light *does* get through. Doesn't that suggest that the polarization of the photons is being converted by the middle polarizer?

7. Aug 11, 2011

### jfy4

The individual photons interact with the polarizer, but if you look at the number of photons on the screen behind the polarizer, it is proportional to $\sin^2\alpha$. This is only true after you let a large number of photons upon the polarizer. If you let a single photon upon it, at 45 degrees, it is equally likely to be absorbed or passed through, at random.

8. Aug 11, 2011

### Joncon

Yes but then the photon is polarized at 45 degress. So it has been converted from it's original polarization hasn't it?

9. Aug 11, 2011

### Drakkith

Staff Emeritus
10. Aug 11, 2011

### jfy4

That's an interesting point. I would put the question though, to be more precise I think: What exactly is $\alpha$ in the above?

It seems you could avoid even having to ask your question if you perhaps decided to abandon classical thinking right off the bat, i.e. what does it mean for an individual photon to be polarized at 45 degrees? You could take the math for what it's worth in experiment (a lot) and say that polarization of individual photons is defined as
$$|\psi\rangle=\sin\alpha|\psi\rangle_{||} + \cos\alpha|\psi\rangle_{\perp}$$
But then what does it mean for $\alpha=\pi /4$? Maybe you could say the preparation device for the experiment prepares a photon with the parameter $\alpha=\pi /4$, but that does seem like you are replacing polarization with just something else...

11. Aug 11, 2011

### nonequilibrium

I understand it intuitively seems to be converting, but upon breaking it down step by step it's clear it's just a filter: (I'll write it down for classical waves)

---
In the first scenario (the one filter), the filter asks "does the polarization have a component parallel to my polarization axis? If "yes", let that part through; if "no", stop everything; answer in this case: "no""

In the second scenario, the first filter asks the same question, sees a parallel component, and lets that component through; the second filter asks the same question, again sees a component parallel and lets that pass.

You see that at each step all a polarizer does is either let a part go through or stop everything; it is obviously a filter, not a converter. The part where your intuition is led astray is because the notion of "having a component parallel to a direction" can change when stopping/deleting certain parts of your wave (as the first filter in the 2nd scenario does). To understand this even more clearly: imagine two sines, in phase, oscillating around the same line, but each in a different plane, and the planes are perpendicular to each other. Call one plane A, the other B. Rotate the whole in your mind such that you're looking at a cross (i.e. the direction of propagation of the waves is pointing away from you). Draw a new plane C that cuts the X you see with a horizonal line through the midpoint. The two sines (i.e. the net sum) does not have a component parallel to C. But if a polarizer deletes the sine in plane A, then the wave suddenly does have a component parallel to C, yet all you did was filter.
---

However, you might say "yes classically I agree it's a filter, but it seems when I think about it quantum mechanically, it looks like it's converting, because once a photon that is polarized by 45° goes through a 90° filter, its polarization has shifted, yet a photon is one particle, so the explanation "the component parallel blabla" doesn't work." (not trying to put words into your mouth, I just think it would be an intuitively acceptable thought)

The thing is --and I might be wrong because my knowledge of quantum mechanics is shabby; someone please correct me if I'm wrong-- the photon isn't really polarized by 45° degrees to begin with. A polarization of 45° degrees simply demonstrates your ignorance about its actual polarization (or in the more popular interpretation of QM: it demonstrates that nature hasn't decided yet what the polarization is); in either interpretation it comes down to: a "45° polarization" means a superposition (in equal amount) of a 0° polarization and a 90° polarization; when you do an experiment to determine its polarization (with a 90° polarizer), there's a 50% chance it'll have a 90° polarization (and thus pass through), and it has a 50% chance it'll have a 0° polarization (and thus be stopped). You see the only "converting" that has been done, is going from ignorance/undeterminateness to knowledge/determinacy (but that can be said about any QM-measurement).[*]

Now the connection between the classical explanation and the QM explanation: say all the photons are polarized at an angle alpha w.r.t. the axis of polarization, this means that the polarization of one photon is the superposition of $\cos \alpha \{ \textrm{axis of polarization} \} + \sin \alpha \{ \textrm{perpendicular direction} \}$, and thus the chance of a photon having the polarization in the "right" direction is (Born's rule) $|\psi|^2 = (\cos \alpha)^2$; now in the classical limit we have a huge number of photons, so the law of large numbers says the fraction of photons that pass through is cos(alpha)², and the numer of photons is proportionate to the classical intensity of a wave (such that the relative intensity of the part of the wave passing through is cos(alpha)²), and as the intensity is the square of the electric field, this means that a fraction cos(alpha) of the original classical wave passed through, which is indeed equivalent to "the component parallel to the axis of polarization".

[*] this reasoning does imply that a photon only can be polarized in either the direction of the polarizer or perpendicular to it; the intuitive strangeness is avoided in the usual interpretation because then there is no polarization beforehand (nature is not yet determined); I don't know how deterministic QM interprets this, but this may be off-topic...

EDIT: I started writing my post before jfy4 posted, hence the fact that I'm not referring to his post although I seem to be discussing similar things

Last edited: Aug 11, 2011
12. Aug 11, 2011

### ThomasT

If you put a polarizer in the path of, say, sunlight, then the intensity of the light transmitted by the polarizer is half that of the original sunlight. If you then put a polarizer in the path of the light transmitted by the first polarizer, then the intensity of the light transmitted by the second polarizer varies according to the law of Malus. Both the first and second polarizers are clearly filters. And the second polarizer seems to be a converter as well. But what about the first polarizer?

13. Aug 11, 2011

### nonequilibrium

ThomasT - I think I explained your question "And the second polarizer seems to be a converter as well. But what about the first polarizer?" in my previous post, between the --- symbols (its conclusion is that nothing converts). Unless there is still a reason to call something a converter, but I wouldn't see what.

14. Aug 11, 2011

### Drakkith

Staff Emeritus
Neither polarizer converts anything.

15. Aug 11, 2011

### ThomasT

Thanks Drakkith and mr. vodka (I must read your explanation more carefully and think about it).
The problem I have in understanding this is that the first polarizer has a particular orientation. So, all of the light transmitted by it can be said to be polarized wrt that orientation. Then the orientation of the second polarizer can be varied and if some light is being transmitted then we can say that that light is polarized wrt the second polarizer's orientation. And if the orientation of the second polarizer is offset from that of the first, then how can it be said that the polarization of the light hasn't been converted by the second polarizer?

16. Aug 11, 2011

### San K

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?

Last edited: Aug 11, 2011
17. Aug 11, 2011

### nonequilibrium

@Thomas: Hm, I understand your problem: you're thinking "okay at first I have a collection of sines (i.e. all kinds of polarizations) coming in into the first polarizer, then the polarizer takes a part of it and lets only one sine wave go through, more specifically: a sine with polarization matching the (first) polarizer. Then seeing this "pure" sine wave encountering the second polarizer, I see a new sine wave exiting the 2nd polarizer, rotated with a certain angle (and also a different amplitude, but that's not of the essence here). So basically your problem is: you see one sine wave coming in, and one sine wave coming out, this gives the illusion of conversion, cause what is there to select out of one sine wave, since it is indeed one wave?

Well, the thing is: the concept of "one wave" is faulty. Take a random sine wave, and in a plane perpendicular to the direction of propagation, take two perpendicular axes x and y (make a drawing). Is it not true that this "one" sine wave can be written as the sum of two other (sine) waves? In particular it's the sum of the projections of the original wave onto the two planes formed by the axes and the direction of propagation (again, easier to follow if you made a simple drawing)

My point: the situation at the 2nd polarizer is exactly the same as at the 1st polarizer: what intuitively seems as "one wave" is in fact a sum of waves with different polarizations; it's just a matter of perspective

I'm convinced what I'm trying to say is very simple and convincing, I just find it hard to state in clear terms, but let me know if something is getting through

18. Aug 11, 2011

### dlgoff

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

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

19. Aug 12, 2011

### ThomasT

Your terms are clear enough. And if I have to say that presently I might not fully understand it, I expect that by tomorrow I will. I should say that my intuitive inclination was to say that polarizers act solely as filters.

I'll either come up with another (what I, imho, think is a good) question tomorrow, or agree with you.

20. Aug 12, 2011

### jfy4

I'm going to have to answer this like this:

The state vector, which seems dependent on $\alpha$, just gives the statistical properties of the photon upon interaction. So when an alpha is chosen, the state vector is chosen, but this can't mean a single result is guaranteed from an interaction. I think it simply means that a statistical distribution of possible results has been set. So specifying the "polarization" of a photon doesn't mean the same thing as it intuitively seems to mean... So a photon with polarization $\alpha=\pi /4$ doesn't mean what it means classically. Instead it would mean that for various classes of interactions with the photon, a particular distribution of results can be expected. That particular distribution of results is what I would call the state vector which is dependent on the parameter we call polarization.

So say we have $\alpha=\pi /4$ which says that for a particular experiment there are two possible results: $\alpha=0$ and $\alpha=\pi$. After an interaction $\alpha=\pi$, this decides a whole other possible distribution of results for the photon for all different classes of interactions.

So did the polarization get converted, sure. But thinking about it this way seems to make it unimportant...

Any thoughts?

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