Relationship of Photon Spin States to Beam Polarization What is the relationship between the individual spin states of photons in a light beam and the polarization of the light beam, if any? What about the spin states determines the degree of polarization, if anything?
 A: There is a wiki article on this:


Left and right circular polarization and their associate angular momenta

As seen in the graphic the orientation of the photon spins defines the left and right circular polarization.
Light emerges from the superposition of the wavefunctions of zillions of photons, an illustration is given in the graphic above. The full complex wavefunction of all those photons contains the E and B fields that will appear in the classical electromagnetic wave equation. The total probability for the superposed photons will be given by the $ψ*ψ$ of this sum. Thus even if an individual photon in that beam when measured has a +1 or -1 spin in its direction of motion, according to initial conditions,this builds up a polarized electromagnetic field.
A: As in AnnaV's answer, left and right handed polarized states are but one possible vector basis for the possible polarization states of light. Light is perfectly polarized when all of its photons are in the same state, and we write the polarization down as a vector which defines that state. We can equally well use orthogonal linear polarization states as our basis and get a different expression for the same vector. This is simply like a change of co-ordinates. 
The degree of polarization has nothing to do directly with spin. It refers to the general case where the photons are in different states and no one polarization state defines the light. It measures how much the "dominant" polarization state predominates.
In this case, we think of the light as being a mixture of pure polarization states; there is a fraction $p_j$ of photons in pure polarization state $\psi_j$. Here $\psi_j$ is a $2\times1$ vector of complex weights with respect to the basis we have chosen, which could, for example, be the left and right handed spin eigenstates of AnnaV's answer. It can be shown that, in this $2\times 1$ case, that any mixture can be thought of as equivalent to a mixture of only two orthogonal polarization basis states, the left and right handed spin states, for example. If the probability of a given photon's being in either state is equal, then the light is totally depolarized. 
We work out the density matrix for the mixture (I say more about density matrices for light here and more about the general concept here):
$$\rho = \sum\limits_j p_j\,\psi_j\,\psi_j^\dagger= s_0\,\mathrm{id}+s_x\,\sigma_x+s_y\,\sigma_y+s_z\,\sigma_z$$
where the $\sigma_j$ are the Pauli spin matrices. The entity $\psi_j\,\psi_j^\dagger$ is a $2\times 2$ Hermitian matrix, so that $\rho$ is also a Hermitian matrix and can thus be uniquely resolved into a sum of the Pauli matrices plus a leftover multiple $s_0$ of the identity matrix $\mathrm{id}$ (which there must be if the density matrix has a nonzero trace). The four Stokes parameters for the light are $s_0,\,s_x,\,s_y,\,s_z$ and a common measure of the degree of polarization is:
$$DOP = \frac{\sqrt{s_x^2+s_y^2 +s_z^2}}{s_0}$$
and this equals unity if and only if $s_0=0$, $\rho$ is traceless and the light is in a pure polarization state. 
The degree of polarization can also be measured by the von Neumann Entropy:
$$S = -\mathrm{tr}(\rho\,\log\rho) = -\sum\limits_jp_j\,\log p_j$$
where $p_j$ are the probabilities to find a photon in the each of the pure state constituents of the mixture. $S$ is nought if and only if the light is perfectly polarized. As the degree of polarization shrinks, $S$ rises and becomes equal to its maximum possible value of precisely one bit per photon when the light is totally depolarized. $S$s the Shannon entropy of the source when we consider each photon pure state as a letter in the alphabet transmitted by the source.
