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Author Topic: Circular polarisation of light  (Read 35763 times)
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hexa
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on: October 21, 2005, 12:23:03 pm »

Does light exist as a distinct right circularly polarized or left circularly polarized state?
If it does exist in these states, does that mean that the passage of a right circularly polarized light through a Left circular polarizer will result in no light passing through it similar to what happen when we use a linear polarizer?
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Fu-Kwun Hwang
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Reply #1 on: October 21, 2005, 04:32:47 pm » posted from:Taipei,T'ai-pei,Taiwan

circular polarization means the direction of electric and magnetic field change direction when the light is miving.
The phase determine if it is a left or right polarized.

The linear polarizer works because it only allow electric field pass through in one direction.

I do not know how to make a right/left circular poliaizer which only allow right/left circular polarized light pass through.
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hexa
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Reply #2 on: October 21, 2005, 05:22:02 pm »

Dear Prof. Hwang,

Thank you for your replies.

I am having problem with some textbooks in Quantum Mechanics where they define circularly polarized state as somewhat analogous to the linealy polarized state. They have stated that a right circularly polarized light will be cut-off by a Left circular polarizer but pass through a right circular polarizer without any loss in intensity.
Is their assertion correct?
They have mentioned that a circular polarizer is in fact a combination of a linear polarizer and a quarter-wave plate that somehow retard the propagation of light after it has been polarized. Would you please explain what essentially constitute a right circularly polarized state from a left circularly polarized state based on how the circular polarizer are made?
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Reply #3 on: October 22, 2005, 04:12:03 pm » posted from:Taipei,T'ai-pei,Taiwan

I just add a EM circular polarized wave simulaton.
The default state in the simulation is circular polarized wave, you can change the phase to zero to view a linear polarized wave.

For the linear polarized wave, the phase for the electirc field and the phase for the magnetic field are the same.
For circular polarized wave, the phase difference is 90 degree (When E field is maximum, B field is zero).

Yes. You can produce a circular polarized wave from linear polarized wave.
Let a linear polarized wave pass through a media in which the index of refractions in not uniform (i.e. the phase change of the E field in two perpendular direction are differnt). Adjust the thickness of the media so that the E field is 90 degree our of phase when the wave pass through the media.

If we let the linear polarized wave pass through the above device it become a circular polarized wave.
(Right or Left circular polarized wave depend on the phase angle is +90 or -90 degree).

If we let the circular polarized wave pass another similar device again, the phase change becomes 180 (or 0) degree which is a circular polarized wave. Then we can use a linear polarizer to detect it polarization again.

In principle, if we can have a device to determine the difference between 180 or 0 degree (may be by interference method), we can build a left/right circular polarizer.

However, I am not sure if it can be done in practice for optics and quantum wave (because the wavelength is so small). I would said that it can be done, if we can build a device with dimension similar to the wavelength of the wave.
But it can be done from theorical point of view.

Correct me if someone know such device really exists!
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hexa
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Reply #4 on: October 25, 2005, 10:48:44 am »

Thanks again Prof. Hwang.

Let me restate my question.
From our physical observation, a linear polarizer performs a definite action on a stream of unpolarized light (which we will designate as l£r> state) by realigning the orientation of the photons along its polarizing axis. Hence, a source of non-polarized photons when operated upon by a vertical or v-state polarizer (Mv) will produce photons polarized along this polarizing axis or in the lv> eigen state. Quantum Mechanics express this operation and its projection probabilities as follows: l<v l Mv l £r>l 2= 1/2. If we pass this polarized light in the lv> state through another v-state polarizer (Mv), we will get 100%: l<vl Mv lv>l 2= 1. However, if we rotate the second polarizer by 90 degree and turn it into a horizontal or h-state operator (Mh), the projection probability will be zero: l<hl Mh lv>l 2= 0.

The question I am raising is whether there is a distinct Right circularly polarized state or lR> eigen state and a Left circularly polarized state or lL> eigen state when light passes through a Right Circular Polarizer and a Left Circular polarizer respectively.

Quantum Mechanics appears to treat the lR> state and the lL> state as somewhat similar to the lv> state and the lh> state as it expresses the projection probabilities when light in the lR> state passes through a Right Circular polarizer as: l<R l MR l R>l 2= 1 and through a Left circular polarizer as: l<Ll ML lR>l 2= 0.
Is it correct for Quantum Mechanics to Make this prediction?

I thank you in anticipation of your clarification.
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Reply #5 on: October 25, 2005, 12:15:58 pm » posted from:Taipei,T'ai-pei,Taiwan

1. Yes. there is a distinct Right circularly polarized state or lR> eigen state and a Left circularly polarized state or lL> eigen state when light passes through a Right Circular Polarizer and a Left Circular polarizer respectively.

2. Quantum state is a theory. So there is no problem to make such prediction.
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hexa
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Reply #6 on: October 25, 2005, 02:58:38 pm »

Thank you Prof. Hwang.

If there is indeed a Right or Left Circularly polarized state, then we should get consistent observation similar to what is observed using linear polarizers. Unfortunately when one conduct an experiment using circular polarizers, very often we do not get the projection probabilities predicted using Quantum Mechanics. In fact, we do not get the result predicted most of the time. This is the reason that has prompted me to seek your advice.
I hope you could clarify this anomaly.
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Reply #7 on: October 25, 2005, 04:12:42 pm » posted from:Taipei,T'ai-pei,Taiwan

I only said there is a distinct Right circularly polarized state or lR> eigen state and a Left circularly polarized state or lL>.
But to conduct an experiment using circular polarizer to find out left/right state is not easy.
Measuring the intensity when light pass through is not going to give you the answer, because the intensity is the same (The only difference is phase: zero or 180 degree).
in principle: you can set up an interference experiment to check out the phase. But it is not easy to do!)
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hexa
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Reply #8 on: October 27, 2005, 03:23:44 pm »

Thanks again Prof. Hwang.

The computation of the projection probability for linearly polarized state of light using Malus Law is mathematical rather than based on any logical physical law including the wave aspect that you tried to explain. It has been shown that Local hidden variables model that tried to explain the projection probabilities of two linear polarizers does not work (see Von Neumann Impossibility proof). The claim by Quantum Mechanics is its strength in predicting an outcome based purely on mathematics-- in this case it happen to be the cosine square and not any other trigonometrical functions.

Could it be that in the case of circular polarization, Quantum Mechanics has over simplified the problem by making the assumption that there exist a distinct right or left circularly polarized state that in fact does not exist. Could this be the explanation why we do not get a consistent result most of the time?
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Life well spent is long. ..."da Vinci (1452-1519, Italian artist, sculptor, painter, architect, engineer and scientist) "
 
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