Does an electron beam have a polarity, in the way that light has a polarity? I'll concede I didn't do any heavy research on the topic, but it dawned on me that if electrons have polarity, then polarity could be manipulated for storage - i.e., some range of polarities corresponds to some state of memory.
This would be useful, because electrons can be contained, unlike photons.
So again, my question is, does an electron beam, or a single electron, have a polarity?
 A: Yes it does. The physical basis for polarity at the particle level is its spin, so the gross or macroscopic polarization is the net spin of the collection of particles. Generating such a group of polarized electrons is relatively easy. There is an atomic analog as well; applying an electric field to an atom will "adjust" the electron's orbitals that result in them being polarized as well.
Polarized electrons are commonly used in modern variations of the Stern-Gerlach Experiment.
A: Electrons possess a purely quantum mechanical property known as spin, which can take two values, conventionally called "up" and "down".
Congratulations, you've just discovered Spintronics! You might find this article interesting.
A: Electrons are spin $s=1/2$, so they have 2 states: $|\uparrow>, |\downarrow>$, which are so-called spin-up and spin-down states with angular momentum projection
$$ J_z = \pm\hbar/2$$
onto an arbitrary axis (usually, the $z$-axis).
Note: the total angular momentum is
$$J = \hbar\sqrt{s(s+1)} = \hbar\frac{\sqrt 3}2$$
so that the $J_x$ and $J_y$ projections cannot be simultaneously known.
The two orthogonal combinations of the states:
$$ |\uparrow'> = \cos{\frac{\theta}2}e^{-i\phi/2}|\uparrow>+\sin{\frac{\theta}2}e^{+i\phi/2}|\downarrow>$$
$$ |\downarrow'> = \sin{\frac{\theta}2}e^{-i\phi/2}|\uparrow>+\cos{\frac{\theta}2}e^{+i\phi/2}|\downarrow>$$
correspond to the two projections along a different axis with respect to the z-axis. ($\theta$ is the polar angle and $\phi$ is the azimuthal angle). Thus, an electron in a pure state has an $\hbar/2$ projection on some axis.
Photons are spin $s=1$, so they have angular momentum
$$J = \hbar\sqrt{s(s+1)} = \hbar{\sqrt 2}$$
The general spin-one particle has three eigenstates of $J_z$: $|\pm\rangle, |0\rangle$ with $J_z = \pm\hbar$, and $J_z=0$ projection on an axis.
Because photons are massless, the $J_z=0$ state is forbidden, and the projection of spin is taken along the direction of propagation (they have no rest frame, after all). These states are referred to helicity states:
$$ \vec S \cdot \hat k = \pm \hbar $$
which have the spin projection parallel and antiparallel with the direction of propagation. They correspond to the 2 states of left and right circular polarization.
And arbitrary combination of these states:
$$ |\psi\rangle = \alpha |+\rangle + \beta|-\rangle \ \ \ \ (||\alpha||^2+||\beta||^2 = 1)$$
does point not the spin in a new direction, rather, it corresponds to states of elliptical and linear polarization, e.g.:
$$ |\psi\rangle = \frac 1{\sqrt 2}|+\rangle + \frac 1{\sqrt 2}|-\rangle$$
corresponds to a tensor alignment of the electric field along an axis that is perpendicular to the wave vector $\vec k$.
