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Edit Circularly polarized photons have $$\textbf{S}\cdot\hat{\textbf{p}}=\pm \hbar\tag{1}$$ and it also satisfies $$\boldsymbol{\epsilon}\cdot\hat{\textbf{p}}=0\tag{2}$$ where $\textbf{S}$ is the spin, $\boldsymbol{\epsilon}$ is the polarization vector and $\hat{\textbf{p}}$ is the unit vector along the direction of propagation. Is it possible to derive (1) from (2) or vice-versa?

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    $\begingroup$ They are essentially the same: physics.stackexchange.com/questions/360638/… $\endgroup$ – safesphere Jan 28 at 7:38
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    $\begingroup$ Clarification request (v5): you seem to be asking whether there's a spin-based explanation why $\epsilon\cdot\hat p=0$ for circularly-polarized light. But $\epsilon\cdot\hat p$ also vanishes for linearly-polarized light, since the momentum is determined by the Poynting vector. It's hard to tell whether your question is specifically about circular polarization, or about the relationship between the circular and linear polarization bases, or about why the momentum is perpendicular to the oscillating parts of the E and B fields. Each of these is a non-trivial and interesting question. $\endgroup$ – rob Jan 28 at 14:33
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    $\begingroup$ I am trying to understand whether the fact that $\epsilon\cdot\hat{p}=0$ somehow dictates that $S.\hat{p}$ can have only two possible values because both as far as I know are related to the masslessness of photon. the latter in general has 2s+1 projections @rob $\endgroup$ – SRS Jan 28 at 22:49
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/291120/44126, physics.stackexchange.com/q/418974/44126 (neither of which address the part of your question about spin). The answer is "yes, because it's massless," but I can never keep the details straight. $\endgroup$ – rob Jan 28 at 23:08
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The gauge field $A^{\mu}(x)$ is a four-vector ($\mu=0,1,2,3$), which means that it has four internal degrees of freedom (at every point $x$ in spacetime). For the technically minded, the reason for it being a four-vector comes from the fact that it is an irreducible representation of the Poincare group, which contains the Lorentz group. These are noncompact groups. The compact part of the Lorentz group represent all the rotations, which are also associated with spin. As a result, the irreducible representations are distinguished as the different spins. The gauge field is said to be a spin-1 field.

Gauge invariance ensures that the mass of the gauge boson is zero. This (indirectly, from gauge invariance) removes one of the degrees of freedom (the temporal degree of freedom), leaving three.

Well, it turns out that the gauge invariance also remove another degree of freedom through the Ward identities. This time the degree of freedom that is removed, is the longitudinal component that would have been parallel to the direction of propagation.

Hence, we end up with only two remaining degrees of freedom for the spin of the EM field. These degrees of freedom manifest as the polarization of the EM field.

The helicity operator, which is the projection of the spin operator (the intrinsic part of the angular momentum) along the direction of propagation $$ \hat{h} = \hat{S}\cdot \mathbf{e}_{p} , $$ where $\mathbf{e}_{p}$ denotes the direction of propagation, has two eigenstates. They are the two circular polarization states, respectively with eigenvalues $\pm 1$.

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  • $\begingroup$ "The gauge field $A^\mu$ is a four-vector, which means that it has four internal degrees of freedom (spin). " 4 degrees of freedom of $A^\mu$ at a given spacetime point correspond to $\mu=0,1,2,3$. What does it have to do with spin? @flippiefanus $\endgroup$ – SRS Jan 31 at 9:10
  • $\begingroup$ @SRS: see the edited answer. $\endgroup$ – flippiefanus Feb 1 at 4:11
  • $\begingroup$ In the Hilbert space, are there operators associated with polarization? @flippiefanus $\endgroup$ – SRS Feb 10 at 13:51
  • $\begingroup$ @SRS: yes it is not too difficult to construct such an operator. That would be a different question though. $\endgroup$ – flippiefanus Feb 11 at 4:17

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