Polarization of a single photon

[Sophrosyne]

TL;DR Summary

Does photon spin describe only circular polarization, or can a single photon have linear polarization? If so, what is the spin of a linearly polarized photon?

I was just reading on this forum (and other sources) about the relationship between photon spin and the polarization of light. From what I have gathered, photon spin corresponds to circular polarization: +1 and -1 spins correspond to right and left helical polarizations.

So I have a few questions about this:

1) How do we know helical polarization corresponds to photon “spin”?

2) how do we determine that this is spin 1? Can you use, for example, a single photon of spin 1 to turn an electron of spin +1/2 to spin -1/2 with preservation of angular momentum? Will the helicity of the photon in turn reverse to preserve angular momentum?

3) One poster had said that linear polarization corresponds to two equal populations of photons: half with right helicity and the other half with left helicity. Does that mean that you can not have a single photon with truly linear polarization?

 

[StevieTNZ]

Can you please point to URL's from this forum that suggests so, plus the other references you've used.

 

[Heidi]

you seem to forget in your questions that a photon can be annihilated in the interactions.

 

[vanhees71]

A photon is special, because it's a mass less vector boson. It has only two "spinlike" degrees of freedom. The polarization state is most naturally characterized by helicity, i.e., the projection of its total (!) angular momentum to the direction of its momentum, which can take the two values $h \in \{1,-1\}$. These refer to circular-polarized single-photon modes of the em. field. Now any superposition of these states describe also photons of the same momentum, and these superpositions describe the general polarization state of a photon (corresponding to elliptic polarizations or as special case linear polarizations in arbitrary directions).

It's highly recommended to first learn carefully about classical electromagnetic (plane) waves and only then think about field quantization!

 

[Sophrosyne]

Can you please point to URL's from this forum that suggests so, plus the other references you've used.

Sure. For example in post #11 https://www.physicsforums.com/threads/photon-spin.41358/:

“2. A bunch of photons with either -1 or +1 spin correspond to a left or right circularly polarized light (or the opposite). If you have 50% of photons in one state and 50% in the other, you have linearly polarized light. Other relative percentages corespond to the range of elliptically polarized light.

3. "Just another term" is correct in the same sense as wave-particle duality.”

So is photon polarization “just another term” for photon spin? Can you have a single linearly polarized photon, or is linear polarization only a phenomenon of multiple photons?

 

[Sophrosyne]

A photon is special, because it's a mass less vector boson. It has only two "spinlike" degrees of freedom. The polarization state is most naturally characterized by helicity, i.e., the projection of its total (!) angular momentum to the direction of its momentum, which can take the two values $h \in \{1,-1\}$. These refer to circular-polarized single-photon modes of the em. field. Now any superposition of these states describe also photons of the same momentum, and these superpositions describe the general polarization state of a photon (corresponding to elliptic polarizations or as special case linear polarizations in arbitrary directions).

It's highly recommended to first learn carefully about classical electromagnetic (plane) waves and only then think about field quantization!

So is it possible to have a single photon linearly polarized? If so, what would that correspond to in terms of spin?

 

[vanhees71]

As I said in my previous posting, photons are somewhat special, because they are massless vector bosons. They have spin 1 but only 2 polarization degrees of freedom (and not 3 as massive vector would have, because they have 3 eigenvalues of the spin-$z$ component, 1,0, and -1). This specialty is closely related to the fact that in relativistic QFT the massless vector bosons must be described as gauge bosons. The natural way to describe the 2 "spin components" of the photon is its helicity, which is the projection of the total angular momentum on the direction of its momentum, and this observable called helicity can take the two values 1 or -1. This corresponds to single-particle Fock states of left- and right-circularly polarized electromagnetic waves. You can of course have any superposition of these polarization states, and among these superpositions are also all linearly polarized photon states in any direction.

Last but not least, you should forget about "wave-particle duality". This is an old-fashioned concept from the time before 1925, where the modern quantum theory has been discovered by Born, Jordan, and Heisenberg (including field quantization, which was particularly worked out by Jordan in one of the famous papers written by Born, Jordan, and Heisenberg). Shortly thereafter the same theory has also been discovered by Schrödinger in the form of wave mechanics (for non-relativistic particles) and by Dirac in its representation free form. The upshot is: The only really adequate description of photons is the quantized electromagnetic field and applying the statistical rules developed by Born as for any other "quantum system". A single photon has observable properties of both particles and waves, which was described as an engimatic "wave-particle duality" by Einstein before the discovery of the modern relativistic QFT.

Whether you observe "particle properties" or "wave properties" depends on what you decide to measure. If you have a single photon, e.g., emitted by an excited atom, and you put a detector like a photo plate or a CCD cam) around this atom, you'll register with some probability a single point somewhere on the photo plate with probabilities given by the specific properties of this photon, described by its quantum state. The same holds true for the polarization. You can measure whether a photon is in a certain linear polarization state by putting a polarization filter between the photon source and the photon detector. If the photon goes through the polarizer (with a probability given by the specific property of the photon encoded in its quantum state) you know it was polarized in the direction determined by the orientation of the polarizer.

You can also observe "wave properties" of single photons. The most famous age-old experiment to demonstrate that light is some kind of wave is Young's double-slit experiment, where you get interference effects leading to the well-known diffraction fringes. If you do such an experiment with single photons, each single photon still leaves only one spot on the detection screen, but using very many photons (all prepared in the same quantum state) you'll get the diffraction pattern as predicted for waves. In the wave picture these patterns are caused by the superposition of partial waves going through either one or the other slits (aka "Huygens's principle" used to explane these diffraction patterns). On the other hand when detecting the single photon with a screen, it always makes only a single spot not the extended diffraction pattern. So you have in some sense a particle property, when detecting the single photon but on the other hand also wave-like behavior in the explanation how the photon goes somehow through the two slits. The consistent picture of modern QFT resolving this contradictory "wave-particle duality" of the old point of view is that observables of a quantum object do not necessarily have well-determined values when prepared in some quantum state, but the quantum state only implies the probabilities for finding one of the possible values of an observable when this specific observable is measured.

 

[Sophrosyne]

As I said in my previous posting, photons are somewhat special, because they are massless vector bosons. They have spin 1 but only 2 polarization degrees of freedom (and not 3 as massive vector would have, because they have 3 eigenvalues of the spin-$z$ component, 1,0, and -1). This specialty is closely related to the fact that in relativistic QFT the massless vector bosons must be described as gauge bosons. The natural way to describe the 2 "spin components" of the photon is its helicity, which is the projection of the total angular momentum on the direction of its momentum, and this observable called helicity can take the two values 1 or -1. This corresponds to single-particle Fock states of left- and right-circularly polarized electromagnetic waves. You can of course have any superposition of these polarization states, and among these superpositions are also all linearly polarized photon states in any direction.

Last but not least, you should forget about "wave-particle duality". This is an old-fashioned concept from the time before 1925, where the modern quantum theory has been discovered by Born, Jordan, and Heisenberg (including field quantization, which was particularly worked out by Jordan in one of the famous papers written by Born, Jordan, and Heisenberg). Shortly thereafter the same theory has also been discovered by Schrödinger in the form of wave mechanics (for non-relativistic particles) and by Dirac in its representation free form. The upshot is: The only really adequate description of photons is the quantized electromagnetic field and applying the statistical rules developed by Born as for any other "quantum system". A single photon has observable properties of both particles and waves, which was described as an engimatic "wave-particle duality" by Einstein before the discovery of the modern relativistic QFT.

Whether you observe "particle properties" or "wave properties" depends on what you decide to measure. If you have a single photon, e.g., emitted by an excited atom, and you put a detector like a photo plate or a CCD cam) around this atom, you'll register with some probability a single point somewhere on the photo plate with probabilities given by the specific properties of this photon, described by its quantum state. The same holds true for the polarization. You can measure whether a photon is in a certain linear polarization state by putting a polarization filter between the photon source and the photon detector. If the photon goes through the polarizer (with a probability given by the specific property of the photon encoded in its quantum state) you know it was polarized in the direction determined by the orientation of the polarizer.

You can also observe "wave properties" of single photons. The most famous age-old experiment to demonstrate that light is some kind of wave is Young's double-slit experiment, where you get interference effects leading to the well-known diffraction fringes. If you do such an experiment with single photons, each single photon still leaves only one spot on the detection screen, but using very many photons (all prepared in the same quantum state) you'll get the diffraction pattern as predicted for waves. In the wave picture these patterns are caused by the superposition of partial waves going through either one or the other slits (aka "Huygens's principle" used to explane these diffraction patterns). On the other hand when detecting the single photon with a screen, it always makes only a single spot not the extended diffraction pattern. So you have in some sense a particle property, when detecting the single photon but on the other hand also wave-like behavior in the explanation how the photon goes somehow through the two slits. The consistent picture of modern QFT resolving this contradictory "wave-particle duality" of the old point of view is that observables of a quantum object do not necessarily have well-determined values when prepared in some quantum state, but the quantum state only implies the probabilities for finding one of the possible values of an observable when this specific observable is measured.

I see. Thank you for the response. It’s making some sense to me, but I have to admit I am still struggling to understand.

So in trying to understand the correspondence of the quantum states to the observables, let me ask a few questions and see how it would be answered.

How would you complete the blanks in this sentence (if it even makes sense to talk of photon spin in this way):

1) Spin +1 and -1 photon spin states of a single photon correspond respectively to right and left-handed helical polarizations of a single photon, and —— and —— photon spin states correspond to a linearly polarized photon (whose wave function has been already measured and collapsed).

2) How does one measure (ie, collapse the wave function) of the spin of a particular photon? Is there some way to do it other than measuring its polarization?

 

[A. Neumaier]

Summary:: Does photon spin describe only circular polarization, or can a single photon have linear polarization? If so, what is the spin of a linearly polarized photon?

Yes. The spin is in both cases $+1$. The spin of a single photon is always $+1$, since it belongs to a spin 1 representation of angular momentum. See also this post.

Can you use, for example, a single photon of spin 1 to turn an electron of spin +1/2 to spin -1/2

No. This contradicts conservation of energy/momentum.

How would you complete the blanks in this sentence (if it even makes sense to talk of photon spin in this way):
1) Spin +1 and -1 photon spin states of a single photon correspond respectively to right and left circular polarizations of a single photon, and —— and —— photon states correspond to a linearly polarized photon

Linearly polarized photons are superpositions of helicity $\pm1$ (i.e., chiral, left and right circularly polarized) photons. This has nothing to do with measurement or collapse.

A bunch of photons with either -1 or +1 spin correspond to a left or right circularly polarized light (or the opposite). If you have 50% of photons in one state and 50% in the other, you have linearly polarized light. Other relative percentages correspond to the range of elliptically polarized light.

This is a very inaccurate statement. The percentage is sloppy language for the detection probability. If you take the 50% at face value you do not get linearly polarized light but unpolarized light.

 

[Sophrosyne]

Linearly polarized photons are superpositions of helicity $\pm1$ (i.e., chiral, left and right circularly polarized) photons. This has nothing to do with measurement or collapse.

But light measured to have polarization at, say, 90 degrees, is an observed phenomenon, not a superposition state. You can’t have a single photon polarized at 90 degrees?

 

[A. Neumaier]

But light measured to have polarization at, say, 90 degrees, is an observed phenomenon, not a superposition state. You can’t have a single photon polarized at 90 degrees?

"Polarization at 90 degrees" is not a meaningful physical statement. One can linearly polarize orthogonal to any given line perpendicular to the light beam. After passing a corresponding polarization filter, every photon is linearly polarized in this way.

You should read more about polarization and then ask more informed questions.