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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?

SRS
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2 Answers2

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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 $\pm 1$.

flippiefanus
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The two statements are identical as a result of the transversality of electromagnetic waves. Transversality means that $${\bf \epsilon} \cdot {\bf p} =0 \,.$$ Photon spin is proportional to ${\bf E} \times {\bf A}$. As both are transversal the result for a plane wave can only be parallel or antiparallel to $\bf p$.

my2cts
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