B Photon Polarization

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Summary
1. Can a photon also have an elliptical polarization? ...
3. Is it possible to change a photon polarization state (c/l/e)?
4. In which cases/ situations do photons have which category of polarization (c/l/e)?
This text
https://en.wikipedia.org/wiki/Photon_polarization
starts with:

"Photon polarization is the quantum mechanical description of the classical polarized sinusoidal plane electromagnetic wave. An individual photon can be described as having right or left circular polarization, or a superposition of the two. Equivalently, a photon can be described as having horizontal or vertical linear polarization, or a superposition of the two."

After this it becomes very mathematical. So I can't grasp it anymore. I would like to understand this topic anyway.
(I am interested in single photons here and the consideration of them as entities. I am not considering em waves as a bunch or result of many photons.)

Questions:
1. Can a photon also have an elliptical polarization?
2. Re. the mentioned superposition: Is it just another description for a linear polarization (1st case/ superposition of two circular polarized photons) or a circular polarization (2nd case/ superposition of two linear polarized photons), or do both superposition considerations describe a general case like a circular (c), linear (l) or elliptical (e) polarized photon?
3. Is it (in specific situations) possible to change a photon polarization state (c/l/e)?
4. In which cases/ situations (production processes/ wavelength etc.) do photons have which category of polarization (c/l/e)?
Example: Light from the sun: Is the polarization of the single photons normally a linear one?
Other cases: broadcast waves, gamma ray etc.
Is there somewhere a list or overview about this which includes different cases?

I would be thankful for any sort of answer.
 

sophiecentaur

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@planet-75
That's an interesting question. As I see things, the properties of a photon are the properties of the wave of which it is part. I think some people would disagree with that - or at least say it's not enough. But detecting a photon that is 'polarised' involves detecting a wave and the apparatus to do that and to reject photons that are polarised differently must surely be based on the classical non-photon based optics. If you detect a photon (a single one) that has passed through the polarisation selector then that can only tell you that it has passed through. Only when you send a statistically large number through the experiment, can you say more. If you detect a lot of them, you can say that the wave has a polarisation that's nearly the same as the detector. If very few get through then you have 'cross polarisation'. Whether we're talking circular, elliptical or linear, I think the same argument applies.
 
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Thanks for your answer.
I think the quality of an em wave and the characteristics of the related photons can be the same or not, because of the possibility of superposition (or overlapping) of two or more categories of photons.
Examples:
1. One category of photons, all with the same sort of polarization and the same phase. Constructive Superposition.
2. Two categories of photons. All with the same kind of polarization like linear, circular etc. (If the polarization would be linear then all the photons would oscillate in the same direction or plane.) But 50% of the photons would have a phase shift of 180° to the other 50%. The em wave would completely be invisible (re. luminosity) then.
3. See Question no. 2. above.

Depending on the question if the superposition is more or less constructive or destructive the proportion between observable luminosity (or similar) and photon energy varies.
If the situation in an experiment can be classified according to example 1 above then the photon polarization can probably be determinated on the base of the polarization of the em wave.
I guess that such experiments have been done and I would be interested to know more about it.
 

sophiecentaur

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The em wave would completely be invisible (re. luminosity) then.
There is no paradox involved here. You can only get the situation where there's "invisible" light when the power is directed elsewhere, just as with any diffraction pattern. You could not, In fact (that's a challenge for you) arrange for two sources of photons (i.e. two waves) to cancel themselves out everywhere. Energy is always conserved in these situations unless you can arrange some 'lossy' medium, which is not part of this discussion.
 
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My motivation is to understand what photons are. For this I also have to know about their polarzation. Therefore questions 1 to 4.

I thought/ hoped it would be possible to say st like: In this situation … (light comming from a change of energy level of electrons, gamma ray from nuclear fusions etc.) it can be assumed that the polarizatin of the single photons is … (circular, linear etc.).

Example: Linear polarized broadcast wave from a dipol antenna. If I don’t know about the photon polarisation I have to guess:
1. Linear polarization for all photons, same directions for the oscillation, phases coordinated, constructive superposition of all photons.
2. Circular polarization for all photons, 50% with a right and 50% with a left hand orientation, coordinated phases within each of the two halfs.
This version would include some inefficiency because the superposition is not fully constructive (non optimal proportion between mesurable oscillation or frequency and photon energy).
3. All photons wuold be a superposition of two parts with a circular polarization, one time with a right and one time with a left hand orientation.
In this version there would be some inefficiency within each of the photons.

This are three (very) different concepts and only one can be true.
 

sophiecentaur

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My motivation is to understand what photons are.πΠ
Photons are quanta of the energy of a wave. You can detect (and produce) them one at a time with suitable equipment but, imo, detecting just one photon does not conclusively tell you much about its source characteristics; it's just a single sample for your statistics.
You will have a problem choosing any of the three alternatives, I think.
Any thought experiment with dipole antennae will involve statistically huge numbers of photons and the wave approach tells you all you want to (or can) know. That's not a problem. Afaics, when you try to nail down a photon in terms of the wave of which it is a part, you are getting into a paradox. A single photon can only have hf of energy so how can it consist of a superposition of two photons (which would have 2hf)?

Getting a firm grip, for a start, about the wave aspect can be a help and the following is how I understand things (there are many many sources of this information).
Circular polarisation of a plane transmitted radio wave is 'easy' to describe because it can be said to consist of two linearly polarised orthogonal waves with a π/2 phase difference. Likewise, you can describe a linearly polarised wave in terms of two circularly polarised waves in quadrature and in phase.
You can translate that into your idea of photons but, whether you can really say that a single photon in a linearly polarised wave is 'really' two circularly polarised photons is something I can't consider (it means nothing to me). (Same thing goes for a single photon of circularly polarised wave).

Some people talk about photons in other ways - in terms of their polarisation states (This wiki article discusses it) and a photon can have one of two circular polarisation states (related to angular momentum) so that implies circularly polarised photons are the only ones that can be detected???? because a linearly polarised photon would have to consist of two (by the above arguments). But, again, would an an energy of 2hf be measured?
 
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Thanks for your reply.

When I said, the "luminosity" can be "invisible" in case of a destructive superposition, I should have said, the frequency or oscillation cannot be observable then.
Assumption no. 3 may be wrong, but it's a trial to refer to the Wiki quote above that says "… superposition of two"

I don't understand why the polarization of (single) photons should be meaningless.
The title of the Wiki article is "photon polarization", and it suggests that something like this should exist.

I don’t know how the polarization of an em wave or light could be manipulated or changed by polarizers, lenses etc. if there would not be specific characteristics (like photon polarization state) before the manipulation, and to understand the concepts of such manipulations it is probably helpful to know about such qualities.
Examles:
1. Light should become linearly polarized with a vertical direction. A lens should take off 50% of photons, those which have a horizontal direction.
The basis for this should probably be linear polarized photons (50% vertical and 50% horizontal) or components of the light, otherwise (with circularly polarized photons or elements at the beginning) it would not work.
2. Light should become circularly polarized with a waveplate (lambda/4).
For this concept of manipulation the same/ a similar basis is required.

Correct or not?
 

sophiecentaur

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linear polarized photons
But theory says that photons are circularly polarised. Where do your linearly polarised photons come from? This is a problem I have which can only be resolved if ideas field factors and waves are kept separate from individual particles. It's ok if you treat all optical instruments as 'statistics manipulators'.
 
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What is a "field factor", and what is the meaning of "ideas" here?
 
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Per the theory photons are circularly polarized in general (probably related to the spin quantum number of 1). But in some situation the elements or parts of light should be linear polarized (by my consideration).
If I try to conclude from this:
Is it possible that there could be something like pairs of photons (circularly polarized per the theory) in some situations which generate the impression that the parts or elements of em waves could be linearly polarized?
 

sophiecentaur

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What is a "field factor", and what is the meaning of "ideas" here?
I didn't write that bit well, sorry. I mean the 'factor' of introducing fields into the same discussion as photons. The word "field" implies a value of some quantity that varies and can be measured with position but the position of a photon is a meaningless idea because, except when and where they actually interact with matter, photons can be literally anywhere. So the approaches are pretty much mutually exclusive, afaics.
some situations which generate the impression that the parts or elements of em waves could be linearly polarized?
Well, of course a wave can be linearly polarised and interactions do not happen if the 'receiver' is set to receive cross polarised waves. What you say would seem to apply to 'parts' or samples of a wave but those parts of the wave need to be identified in some way, like a burst of monochromatic light or radio waves. You can locate such a burst / pulse in space and time because there are sufficient number of photons for the result to be statistically defined. A single photon doesn't follow those rules so how could you do it?
There is a great temptation to try to 'unify' the approaches in ones mind (that's the way our brains work) but afaik, there is still a lot about the two approaches that's mutually exclusive.
 
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To check if I have understood, I try to repeat this in my own words:
We have two theories (photon as particle or quantum/ em waves) which are not fully compatible, and these theories are approaches (i.e. they might be not perfect or somehow incomplete).
 

sophiecentaur

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To check if I have understood, I try to repeat this in my own words:
We have two theories (photon as particle or quantum/ em waves) which are not fully compatible, and these theories are approaches (i.e. they might be not perfect or somehow incomplete).
We do not "have two theories" there is no actual incompatibility. The only problem is that people want things to be more concrete and straightforward ( and arm waving) than they are. Quantum Field Theory began in the 1920s and reconciles classical field theory with Quantum Mechanics and Special Relativity. If you want to resolve your problems then search the Web for something that you find approachable. Wiki would be a fair start.
 

Nugatory

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To check if I have understood, I try to repeat this in my own words:
We have two theories (photon as particle or quantum/ em waves) which are not fully compatible, and these theories are approaches (i.e. they might be not perfect or somehow incomplete).
That is not a good way of thinking about it. We have one theory, quantum electrodynamics, which completely and correctly explains the behavior of photons and electromagnetic waves.

Some of the difficulty in explaining QED is that the word “particle” in quantum field theories doesn’t mean anything like what it does in ordinary English (it’s something of a historical accident that physicists ended up using that word in this context) so it’s easy to be confused when you read that”a photon is a particle”.

Several posts above you said
My motivation is to understand what photons are.
Without using math it’s hard to improve on @sophiecentaur’s answer in post #6: photons are quanta of the energy of a wave. (If you do want to get a feel for what the math looks like without confronting the whole thing, you might try http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf)
 

sophiecentaur

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Some of the difficulty in explaining QED is that the word “particle” in quantum field theories doesn’t mean anything like what it does in ordinary English (it’s something of a historical accident that physicists ended up using that word in this context) so it’s easy to be confused when you read that”a photon is a particle”.
It's easy to be wise after the event but it really was a bad idea to use such a commonly understood term as 'particle' for a fundamental entity that was (even initially) going to be very hard to deal with. The consequences of that choice are as dire - or more so - than the consequences of the chosen sign of the charge of an electron and the conflict with the direction of Conventional Current ("They got it wrong Sir!!!!").
A few years later, the West Coast Academics used all sorts of whacky terms to describe the stuff they uncovered and never really had a similar problem because no one thought of Color or Strangeness in literal terms because those terms were so 'unlikely'.
 
There is no paradox involved here. You can only get the situation where there's "invisible" light when the power is directed elsewhere, just as with any diffraction pattern. You could not, In fact (that's a challenge for you) arrange for two sources of photons (i.e. two waves) to cancel themselves out everywhere. Energy is always conserved in these situations unless you can arrange some 'lossy' medium, which is not part of this discussion.
If the linear distance between sources was zero and the phase, amplitude and frequency of the two emissions were the same would they cancel each other at all times of observation within that linear framework?
edit - if the direction of emissions between the sources were opposite along the linear dimension.
 

sophiecentaur

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If the linear distance between sources was zero and the phase, amplitude and frequency of the two emissions were the same would they cancel each other at all times of observation within that linear framework?
edit - if the direction of emissions between the sources were opposite along the linear dimension.
The region of cancellation would only be at the nodes of the standing wave. A quarter wave in either direction and the power in the beams will be four times (at antinodes), and so on, in each direction. But there is nothing magical about cancellation; it just cannot happen over all space. This is true for any two sources of equal powers and frequency. Even if you arrange for the two sources to fire directly into each other with 'zero ' spacing, the power of each beam becomes dissipated within the sources themselves - so it had to go somewhere.
 
Okay thanks, that makes sense. I'm really interested in the discreet observation of phenomena within a 3 physical dimension framework, and how that translates in a space-time context. It seems to me that wave-partical duality arises from different moments of observation. Any suggestions on information or research along those lines? Thanks.
 

Nugatory

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It seems to me that wave-partical duality arises from different moments of observation.
Wave-particle duality is pretty much an obsolete concept. It came into vogue before the modern theory of quantum mechanics was hammered out in the few years after 1925; you won’t find it any modern textbook except as a historical aside. Thus, if your goal is to understand QM, your best bet is to try to forget that you ever heard the term.

However, no quantum physics, wave-particle duality, or photons are involved in the question you asked above and in @sophiecentaur’s answer. That’s just straight classical electromagnetism.
 
Thanks for the reply. I'll work on my terminology.
 

Andy Resnick

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Summary: 1. Can a photon also have an elliptical polarization? ...
3. Is it possible to change a photon polarization state (c/l/e)?
4. In which cases/ situations do photons have which category of polarization (c/l/e)?
There are several separate (and complex) concepts mixed together here. Let's try and untangle them a bit.

First: a single photon- by which I mean something specific (an elementary excitation of a plane-wave electromagnetic field, equivalently a Fock state with occupancy number 1, or |1>)- has a polarization state described by a Jones vector and is formally identical to the quantum-mechanical notion of 'spin'. Single photons emitted from an atom may be circularly polarized ('σ-polarized', corresponding to the selection rule ΔL = ±1, dipole transition) or 'π-polarized' (linearly polarized from a quadrupole transition). So for question #1, the answer is 'no' for emitted photons.

For question #3 (no #2?), the polarization state can be changed by, for example, passing the emitted photon through a retarder. If you start with a photon of known polarization (for example, passing the photon through a linear polarizer), you can choose any final polarization state. Note, however, that passing a single σ-polarized photon through a linear polarizer means that the photon may not pass through the polarizer; we could instead use polarizing beamsplitters and 'which way' arguments. Again, the mathematics are formally identical to that of spin-polarized atoms moving through magnetic fields.

Which brings up the notion of polarization in terms of many photons: this requires the notion of 'coherence' or, in terms of quantum mechanics, a 'density matrix'. Most of the time we simply assume that the photons are either 100% mutually coherent (a perfect laser) or 0% mutually coherent (blackbody radiation). Mandel and Wolf's "Optical Coherence and Quantum Optics" is the essential reference for partially coherent light.

If you have many correlated photons, you can construct the entire range of (macroscopic) electromagnetic field polarizations. Combining many photons to generate arbitrary polarization states is an active area of research: radial and azimuthal polarization states, for example. But again, the underlying assumption is that the photons are all 100% correlated: decreasing the coherence degrades the polarization state in a way that cannot be described by a Jones vector, but instead a Stokes vector (and Mueller matrices) are required which retain statistical information the same way that density matrices do.

Does that help?
 

sophiecentaur

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Does that help?
Yes, thanks, I think so. That explanation allows a photon approach to all the 'wave' phenomena. I just have a problem with what goes on in changing the polarisation of a single photon with a conventional polariser. It's almost as if you need to introduce another photon to allow the quantisation of the Energy on the way through and that extra photon disappears as the new photon comes out the other end. (I'm assuming that vectors are involved during the process.)
 

Andy Resnick

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Yes, thanks, I think so. That explanation allows a photon approach to all the 'wave' phenomena. I just have a problem with what goes on in changing the polarisation of a single photon with a conventional polariser. It's almost as if you need to introduce another photon to allow the quantisation of the Energy on the way through and that extra photon disappears as the new photon comes out the other end. (I'm assuming that vectors are involved during the process.)
Some of that conceptual difficulty comes from the fact that photon number is not conserved (absorption and emission). Photon number (in the Fock basis) is not correlated with Electric field amplitude.
 

sophiecentaur

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Some of that conceptual difficulty comes from the fact that photon number is not conserved (absorption and emission). Photon number (in the Fock basis) is not correlated with Electric field amplitude.
I'll need to sit and think about that one. But I guess we should always expect some weirdness in these matters.
 

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