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Why does the circular polarization of light determine the photon's spin state? I understand the difference between what spin and polarization are, but I don't understand why one would affect the other, as I was of the understanding that spin is an intrinsic property of a particle that is unrelated to any other property of the particle. This seems to not be the case for the photon - where does this relationship come from?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Oct 28 '17 at 15:48
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To understand the spin you should appreciate the fact that all the particles we know are actually excitations of certain fields. Those fields may have certain non-trivial transformation properties under rotations, like electromagnetic field is a 4-vector whereas e.g. electrons are described by spinor fields.

The conserved currents $\partial_\mu j^{\mu}=0$ originate from symmetries and can be obtained using Noether theorem. Integrating their temporal component over space $Q=\int d^3x j^0$ give you some conserved quantities. E.g. symmetries under translations give you stress-energy tensor $T^{\mu\nu}$ that corresponds to the conserved energy and momentum, \begin{equation} E=\int d^3x T^{00},\quad P^k=\int d^3x T^{0k} \end{equation}

What about angular momentum $M^{\mu\nu}$? It originates from the symmetry under rotations and Lorentz boosts and is associated with the angular momentum current $\mathcal{M}^{\alpha\mu\nu},\,\partial_\alpha\mathcal{M}^{\alpha\mu\nu}$. Now knowing that in mechanics $M^{\mu\nu}=x^\mu P^\nu-x^\nu P^\mu$ you may expect that this current takes the form, \begin{equation} \mathcal{M}_0^{\alpha\mu\nu}=x^\mu T^{\alpha\nu}-x^\nu T^{\alpha\mu} \end{equation} and in case of the scalar field you would be right. However because electromagnetic field is a 4-vector it transforms non-trivially under rotations. That means that the rotation of the field itself gives the contribution into the Noether current and it is actually, \begin{equation} \mathcal{M}^{\alpha\mu\nu}=\mathcal{M}_0^{\alpha\mu\nu}+\mathcal{S}^{\alpha\mu\nu} \end{equation} where $\mathcal{S}^{\alpha\mu\nu}$ is known as the spin momentum and it originates already in the classical field theory. If you consider left and right circular polarizations you will discover that they correspond to different signs of the spin momentum.

So the spin is the angular momentum associated with rotation of the field as in the circular polarization and its analogs for tensors and spinors.

All other properties of the spin come from the quantum properties of the angular momentum (its quantization and noncommutativity of the components) We usually select the basis of $S_z$ eigenstates and for photon those happen to correspond to the field excitations in the left and right circular polarization.

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  • $\begingroup$ I must admit I don't know much of field theory - but I believe I understand the meaning of your answer. Would that mean that the spin of an electron also causes it to have some kind of rotation that we cannot detect because the electric field from an electron is radially symmetric? $\endgroup$ – Billy Kalfus Oct 27 '17 at 18:49
  • $\begingroup$ @BillyKalfus No, the reason is that electrons are fermions whereas photons are bosons. The bosonic field may be in the so-called coherent state that behaves very like some classical field configuration. That's why we get nice limit of classical electrodynamics - the classical fields are actually nice and smooth coherent states of the quantum electromagnetic field. Unlike that fermionic fields behave in the significantly quantum way and we can't directly measure them in the same way as we do with EM field. But in relativistic regime we have to describe them with spinor fields and... $\endgroup$ – OON Oct 27 '17 at 19:05
  • $\begingroup$ @BillyKalfus and spin states correspond to certain polarizations of those spinor fields. Please also note that it's not a rotation of the electron as some small ball (this picture is broken long before you encounter spin). It is the rotation of the electron field. $\endgroup$ – OON Oct 27 '17 at 19:08
  • $\begingroup$ Ah, I believe I understand - so the spin corresponds to the state of the excitation in the quantum electromagnetic field, and polarization corresponds to states in the classical electromagnetic field, and because the classical fields are coherent states of the quantum field, when we apply a rotation in the quantum field (represented by the particle having some nonzero spin angular momentum) a corresponding rotation is cohered in the classical field, meaning a rotation of polarization? $\endgroup$ – Billy Kalfus Oct 27 '17 at 19:23
  • $\begingroup$ Do I understand your comment correctly that the Bose-Einstein distribution allows the coherent state of the quantum field and this allows the classical field to emerge? And that in contrast fermionic fields cancel each other due to the Fermi-Dirac distribution and this results in no observable classical field? Makes sense to me, just wanted to confirm if this is in the ballpark of accurate. $\endgroup$ – safesphere Oct 27 '17 at 23:28
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... I was of the understanding that spin is an intrinsic property of a particle that is unrelated to any other property of the particle. This seems to not be the case for the photon...

That really is the point. There is a property of photons called the intrinsic spin. And as “intrinsic” suggested, it is a property of the photon which is independent from any circumstances. That intrinsic spin is not the spin which is related to circular or elliptical polarized photons. These circular and elliptical polarizations beside the directions lefthanded and righthanded have an additional value of how fast the rotation happens. This value is not limited by two numbers. I regret that this is no longer taught.

So why it was introduced the intrinsic spin for photons? As you see in this excellent animation

and in these sketches

enter image description here... enter image description here

that the two field components of the photon could have two possible sequences of following one another, lefthanded or righthanded. This is right even for this sketch

enter image description here

Now about the relation of photons intrinsic spin to the intrinsic spin of the subatomic particles. Have you ever thought about the fact, that accelerating electrons in the antenna rod one will get photon emission with always the same sequence of their electric and magnetic field components? Otherwise the magnetic field components would cancel each other out. Really think about this statement for a while (and give me a counterexample if this statement seems to be wrong for you).

Making - imaging you can do it - radio waves with positrons one will get the second possible sequence of the electric and magnetic field components. The relation between the spin of the electron/positron and the intrinsic spin of the emitted photons is uniquely assigned.

BTW, a circular “polarization” of the photon is achievable not only by an artfully designed polarization foil. An revolving electron which emits a photon gives some amount of its momentum to the photon and his electric and magnetic field components are rotating. But as I stated above the value of this rotation is not limited to two numbers like for the intrinsic spin.

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