What causes the polarization effect in optics?

[DaveC426913]

I've done this experiment several times, and once upon a time I could describe exactly what is shown - but that was a while ago. I need a refresher.

Take two polarizing lenses (let's keep it simple - transverse polarization), lens A and B, turn them 90 degrees to each other, they will block all light.

Now take a 3rd lens C and insert it between the A and B at a 45 degree angle. You will now be see to see through again.

The light passing through the lens A gets polarized, but that polarizing can be over a range of 90 degrees - 45 to the left, 45 to the right. It's enough to block any light from passing through the lens B but the insertion of lens C causes another rotation of the polarization by another 45 degrees, such that some of it can now pass through lens B.

That's a lousy description, but my question is this: is that rotation of polarization (either left or right, or both until it's measured) a quantum mechanical effect?

 

[DaveC426913]

Oh my God. Nothing like Googling for an answer.

It appears that a fine-looking (but apparently forgetful) gentleman was asking this same question 6 years ago...

How embarrassing...

 

[Ken G]

It looks like the question got answered pretty well the last time, but if you are still wondering, it might help to think of the reason that no light gets through 90 degree filters is because of destructive interference. You can correctly imagine the action of a polarizing filter is to send out a wave that is 180 degrees out of phase with, and the same magnitude of, whatever is the component of the incident wave that is perpendicular to the polarization of the filter. That "anti-wave", if you like, then behaves completely physically like a real wave would, that's just the principle of superposition. So if you have two filters at 90 degrees, the "anti-waves" of each don't interfere with each other, but they do interfere completely destructively with anything you send at them (regardless of polarization) because they have all the bases covered, if you will.

However, if you interject a third filter, the "anti-wave" it sends out, to cancel whatever perpendicular amplitude component is incident on it, will mess with the perfect destructive interference the other two polarizers are sending out. That's because the "anti-waves" also get destructively interfered with by downstream polarizers-- an anti-anti-wave if you will. That's all just the principle of superposition, so this effect is not so much classical or quantum mechanical, it applies to any theory that is linear in the amplitudes.

The key point is that when there is a principle of superposition, absorbing something is just like a source of negative that same thing, so more absorption can also be viewed as more negative emission. Negative something can actually be a thing in and of itself-- it has a positive magnitude even though it has a negative amplitude. So more negative something is not the same as less of that thing-- it might end up giving more, as with that third filter. The amplitude that survives all three filters can be traced to the effort of the third filter to cancel part of the antiwave of the second filter-- an effort that would not exist had there been no second filter.

 

[DaveC426913]

It looks like the question got answered pretty well the last time, but if you are still wondering, it might help to think of the reason that no light gets through 90 degree filters is because of destructive interference. You can correctly imagine the action of a polarizing filter is to send out a wave that is 180 degrees out of phase with, and the same magnitude of, whatever is the component of the incident wave that is perpendicular to the polarization of the filter. That "anti-wave", if you like, then behaves completely physically like a real wave would, that's just the principle of superposition. So if you have two filters at 90 degrees, the "anti-waves" of each don't interfere with each other, but they do interfere completely destructively with anything you send at them (regardless of polarization) because they have all the bases covered, if you will.

However, if you interject a third filter, the "anti-wave" it sends out, to cancel whatever perpendicular amplitude component is incident on it, will mess with the perfect destructive interference the other two polarizers are sending out. That's because the "anti-waves" also get destructively interfered with by downstream polarizers-- an anti-anti-wave if you will. That's all just the principle of superposition, so this effect is not so much classical or quantum mechanical, it applies to any theory that is linear in the amplitudes.

The key point is that when there is a principle of superposition, absorbing something is just like a source of negative that same thing, so more absorption can also be viewed as more negative emission. Negative something can actually be a thing in and of itself-- it has a positive magnitude even though it has a negative amplitude. So more negative something is not the same as less of that thing-- it might end up giving more, as with that third filter. The amplitude that survives all three filters can be traced to the effort of the third filter to cancel part of the antiwave of the second filter-- an effort that would not exist had there been no second filter.

I'm afraid I do not follow this explanation.

1] I'd like a second opinion on whether this anti-wave is a valid way of looking at polarization.

2] Assuming it is, I don't follow the aspect of interference. Seems to me, interference only works if your lenses are calibrated with the wavelength of light, otherwise the phases of each are not going to cancel out.

 

[Ken G]

1] I'd like a second opinion on whether this anti-wave is a valid way of looking at polarization.

That doesn't require a second opinion, it's a completely straightforward aspect of the superposition principle. 1-1 = 0 (absorption) is the same thing as 1 + (-1)=0 (what I mean by an "anti-wave").

2] Assuming it is, I don't follow the aspect of interference. Seems to me, interference only works if your lenses are calibrated with the wavelength of light, otherwise the phases of each are not going to cancel out.

There aren't any lenses, there are polarizing filters. The action of the filter is automatically tuned to the frequency of the light wave, there is no need to do anything to it (vibrations in the filter are stimulated by the light wave). If polarization is tricky, imagine a taut rope with small-amplitude transverse waves on it. Replace the polarizers with blocks of wood with a narrow slit cut in it, which the rope passes through. The alignment of the slit is analogous to the polarizer (a slit would reflect rather than absorb, but the net result is similar), and two perpendicular slits will block all transverse waves on a rope-- but inserting a third slit will allow some transmittance.

The system is linear, obeys a superposition principle, and can be analyzed by imagining that each slit acts as a wave source-- where the waves it produces can be negative whatever is incident. Then the action of the last slit can be conceived of as canceling the anti-waves from the previous slits too-- not just the original wave excitation. Thus, one can quite correctly state that the effect of interposing an intermediate filter is to alter the cancellation induced by the last filter, such that the last filter cancels not only a part of the original wave, but also the antiwaves (substitute "cancelling waves" or "destructively interfering waves" if you prefer) that are present to cancel the original wave! This will result in uncancelling parts of the original wave, which is just why it introduces transmittance.

 

[DaveC426913]

There aren't any lenses, there are polarizing filters.

Sorry. I was doing this is a photo lab. Lenses. Filters. I meant filter.



Is superposition not a QM behavior?

 

[Ken G]

Is superposition not a QM behavior?

No, it holds in any linear theory. Quantum mechanics has a more ontological flavor of superposition though-- as if somethng we thought had to be A or B could be a superposition of both. Music is an excellent example of a classical superposition.