Spin Angular Momentum (SAM) of plane polarized light I know that the SAM of circularly polarized light is associated with the rotating $\vec{E}$ and $\vec{B}$ fields and OAM is associated with the helical motion of the net $\vec{E}$ field. 
So how do we associate the SAM and OAM of a linearly polarized light in which there is no rotating $\vec{E}$ and $\vec{B}$ field and also the wavefronts are simply planes instead of being helical? Can we say linearly polarized light have zero OAM and SAM?
I haven't read enough about OAM and SAM, so I may be missing some subtle concept here, so please clarify it.
 A: The spin angular momentum of a light beam emerges from the spin of individual photons, but the relationship between the photons and the light beam is more subtle than you might think. A light beam isn't just a hail of photons.
In the case of linearly polarised light we describe it as built up from photons that are in a superposition of left and right handed spins so the expectation value of the photon spin is zero. That's why the corresponding light beam has a spin angular momentum of zero. At the classical level you can think of it as a sum of two beams of light with opposite circular polarisations and hence equal and opposite spin angular momenta.
Note that this is different from light made up of an equal number of left handed photons and right handed photons. A light beam built up in this way is unpolarised.
A: You are partly correct in that linearly polarized light has zero spin angular momentum (SAM). However, the orbital angular momentum (OAM) is not affected by the polarization state of the light. Instead, it is determined by the wavefront of the beam. If, for instant, the optical beam (regardless of state of polarization) has the form 
$$ \psi(r,\phi) = R(r)\exp(i\ell\phi) , $$
where $\ell$ is an integer, called the azimuthal index, then it would contain OAM that is proportional to $\ell$. There are whole sets of modes (Laguerre-Gauss; Bessel modes; etc.) that have this property the they are OAM eigenstates. Therefore each mdoe carries a well-defined quantized amount of OAM. These modes (with the exception of those with $\ell=0$) all have optical vortices (phase singularities) at the origin with topological charges given by $\ell$.
Note however that the optical vortex is not necessary to have non-zero OAM. One can also have OAM in an optical field in regions where
$$ \nabla \theta \times \nabla A \neq 0 . $$
Here $\theta$ is the phase of the optical field and $A$ is the amplitude.
It is also necessary to point out that not all optical fields with optical vortices have OAM. For instance, an optical field can have two oppositely charged optical vortices that are symmetrically displaced with respect to each other, so that the net topological charge in the field is zero, as well as the OAM.
The OAM in an optical field is computed in a way simular to how it is done in mechanics, but where we use the Poynting vector instead of the momentum vector.
