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On page 34 of A. Zee's book QFT in a Nutshell, he states:

I expect you to remember the concept of polarization from your course on electromagnetism. A massive spin 1 particle has three degrees of polarization for the obvious reason that in its rest frame its spin vector can point in three different directions.

  1. I understand that the spin of a spin-1 particle is a 3-vector, because it is the fundamental representation of $SO(3)$.
  2. I also know from classical ED that the polarization of light can be quantitatively described using a 3-vector.

But why is this the same? Why is the second implication obvious:

$$ \text{Spin}\,1 \implies \text{Spin} \in \mathbb{R}^3 \implies \text{Polarization} \in \mathbb{R}^3 \, ? $$

DanielSank
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Bass
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    Polarization is a 3-vector, but can't point in three directions. – Ryan Unger Oct 21 '15 at 17:03
  • That was both a correction of something you said and a hint. The spin vector of a photon can't point in three directions either. – Ryan Unger Oct 21 '15 at 17:57
  • @0celo7: They're both 3-vectors. What's their connection? Are they the same QM observable? – Bass Oct 21 '15 at 17:59
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    The polarization states are the eigenstates of the S_z operator (for a massive particle, in its rest frame). – Andrew Oct 21 '15 at 18:28
  • Thanks! Could you please elaborate this in more detail, or provide me a reference where it is explained? Does that belong to standard non-relativistic QM or to QFT? – Bass Oct 22 '15 at 10:47
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    @BastianTreichler It applies to both QM and to (the single particle states of) QFT. From the QFT side the hardcore reference is Weinberg QFT vol 1 chapter 2, but that might be a bit much. I don't know a good reference off the top of my head. You can sort of see how it comes about from the Stern-Gerlach experiment. The different spots you see on the screen are the different polarizations, and they correspond to the different eigenstates of $S_z$. – Andrew Oct 22 '15 at 18:08
  • @Andrew: I always thought the different spots in the Stern-Gerlach experiment are the different eigenstates of the spin, measured in the z axis. So you're saying that for the photon, the z-axis spin is the "QM version" of the light polarization in classical ED? – Bass Oct 22 '15 at 18:20
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    @BastianTreichler Essentially yes that's what I'm saying. For a photon it's slightly trickier because the photon is massless and has no rest frame. The analogue of the z-component of spin in that case is known as helicity, it is the projection of the spin vector along the direction of motion of the particle. – Andrew Oct 22 '15 at 19:50
  • @Andrew Not sure if I understand this. Isn't the helicity of a photon more or less trivial, since the spin axis is always in the direction of the momentum? So the photon's spin has just one degree of freedom, but its polarization has two of them. How are they related exactly? – Bass Jan 24 '16 at 14:20
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    @Bass I'm not 100% sure what you're asking but let's see if this helps. The number of polarization states depends on the spin and the mass. For a massive spin 1 particle, there are 3 polarizations. This is because in the rest frame of the particle you can apply the ordinary undergrad quantum arguments (S=1, S_z=-1,0,1). For a massless spin 1 particle there are 2 polarizations. The undergrad quantum arguments don't apply because you can't go into the rest frame of a photon. We characterize the polarization states by helicity (roughly S.P where P is momentum), and there are 2 helicity states. – Andrew Jan 24 '16 at 14:33
  • @Andrew If I understand it correctly, the polarization vector of a photon is always orthogonal to the momentum. So the polarization plane is spanned by two vectors $\epsilon^{(1)}$ and $\epsilon^{(2)}$. Are these two polarization states somehow related to the helicity? If yes, how exactly? – Bass Jan 24 '16 at 18:30
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    @Bass What you are saying is exactly correct for massless spin-1 particles. There are two polarization states, and you can associate those with the two basis vectors in the plane perpendicular to the momentum direction. Those two basis vectors / helictity eigenstates are a basis for the helicity (to be more correct the real helicity eigenstates are the things that transform as $e^{\pm i \theta}$ if you rotate by $\theta$ around the momentum direction which arent quite $\epsilon^{(1,2)}$ but morally speaking they're the same). For particles with mass or different spin it's a little different. – Andrew Jan 24 '16 at 18:40
  • @Andrew OK. So if we have a photon with momentum in the positive $z$ axis, its polarization must be in the $x$-$y$-plane. Now if momentum and spin are aligned (positive helicity, spin parallel to momentum), then what polarization does it have? (Sorry if I'm formulating a bit blurry or heuristically, I hope it's clear what I mean). – Bass Jan 24 '16 at 19:15
  • @Bass yeah that's exactly point, for a massless spin 1 particle the helicity never points in the direction of motion. In other words there is 0 probability to measure the polarization direction along the direction of propagation. For a massive spin 1 there is a probability to measure the polarization along the direction of motion. This is associated with their being an extra polarization state for a massive spin 1 and this extra polarization can point along the direction of motion. – Andrew Jan 24 '16 at 20:38
  • @Andrew Wiki says that helicity is the projection of the angular momentum onto the direction of motion. You're saying for a massless spin 1 particle helicity never points in the direction of motion. How does that fit? – Bass Jan 24 '16 at 21:01
  • Let us continue this discussion in chat. – Andrew Jan 25 '16 at 03:42

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