I am going to give you, not a rigourous explanation, but a feeling.

The discretization of angular momentum is a consequence of the quantum nature of particles.

Imagine a classical rotating particle, the variation of action could be written :

$\Delta S = J \Delta \theta$

where $J$ is the angular momentum, and $\theta$ an angle.

At this point, there is no reason why $J$ needs to have discrete values.

Now, turn to quantum mechanics.
The main point is that, in quantum mechanics, $ \large \frac{S}{\hbar}$ is a phase. For instance, if we want to calculate the transition amplitude (propagator), we have :

$$<x't'|xt> = \int D \Phi ~e^{i\large \frac{\Delta S(\Phi)}{\hbar}}$$
where the sum is done for all the paths $\Phi$, with $\Phi(t) =x$ and $\Phi(t') =x'$

It is very clear, in this expression, that $ \large \frac{S}{\hbar}$ is really a phase.

Now, we may write, formally, $e^{i\large \frac{\Delta S}{\hbar}}$ as, for your question, as $e^{i\large \frac{ J \Delta \theta}{\hbar}}$

But, because $\theta$ is an angle, it is identified to $\theta + 2 \pi$, and if we want that $e^{i\large \frac{ J \Delta \theta}{\hbar}}$ does not change, we need :

$$\frac{J (2 \pi)}{\hbar} = 2 \pi n$$
That is :

$$J = n \hbar$$

Now, if you want to be more rigourous, one may say, that, in quantum mechanics, we are going to search unitary representations of groups. For instance, we may look at the group of rotations $SO(3)$. This is a compact group (because of the identification $\theta$ with $\theta + 2 \pi$), so we can find finite dimensional representations. To find all representations, we have to consider in fact $SU(2)$. The representations of $SU(2)$ are labelled by a semi-integer or an integer : $0,1/2,1,3/2,2$, etc...
In a representation $s$, you have $2s+1$ states, labelled by $-s, -s+1,....,s-1,s$.
For the representation $s= 1/2$, you have 2 states labelled $-1/2, +1/2$

Everything depends on how your fields (vectors and spinors are fields in the classical theory, and when you quantize in QFT, they become operator-valued fields) transform when you make a Lorentz transform:

An **scalar** is a field that doesn't change at all: $\phi'(x') = \phi(x)$. Examples are the Higgs and pions.

A **vector field** is a field that transform like a relativistic four-vector $A'_\mu (x') = {\Lambda^\nu}_\mu A_\nu(x)$, where $\Lambda$ is a Lorentz transformation. Examples are the electromagnetic field (photons) and gluons.

An **spinor field** transform using a different set of matrices $$\psi'(x') = \exp\left[(-i \vec{\theta} \pm\vec{\eta}) \cdot \frac{\vec{\sigma}}{2}\right] \psi(x)$$ where $\vec{\theta}$ are the angles of rotation along the axes, $\vec{\eta}$ the rapidity and $\vec{\sigma}$ the Pauli matrices. As you can see, Pauli matrices are 2x2 matrices, so this transforms acts on objects with two components, the **Weyl spinors**. I have written two signs, $\pm$, because there are two types of transformations that act on two types of spinors: $-$ for *left-handed* spinors $\psi_L$ and $+$ for *right-handed* spinors $\psi_R$. But Weyl spinors have two problems: when you make a parity transformation ($\vec{r} \to -\vec{r}$), spinors change their *handeness*, and we know that QED and QCD are invariant under parity. And the other, as you say, is that Weyl fields must be massless.

The **Dirac spinor** solves both problems. It is just (in the chiral representation) a left-handed and a right-handed Weyl spinors side-by-side $$\psi = \begin{pmatrix} \psi_L\\ \psi_R \end{pmatrix}$$ The Dirac spinor can have mass (although massless Dirac spinors are fine). Electrons, muons, taus, neutrinos and quarks are described Dirac fields.

The **Majorana spinor** is a special Dirac spinor, where the left-handed and the right-handed parts are not independent. This relationship means that a Majorana particle is equal to its antiparticle. Therefore, the Majorana field has no electric charge. Although you only need one Weyl spinor to determine a Majorana spinor, Majorana fields still can have mass. It is conjectured that neutrinos might be Majorana particles (there are several experiments researching this).

So, where is spin? Angular momentum is a conserved quantity related to rotations. Whe you apply NĂ¶ther's theorem to a field, you get two terms: one depends on the movement of the particles (the orbital angular moment) and the other not (the **spin**). The spin part is related to the type of Lorentz transform that the field uses: in scalar fields there is no spin term (they have spin 0), in spinor fields it is a representation of rotations of dimension 2 (spin 1/2), and in vector fields a representation of rotations of dimension 3 (spin 1).

## Best Answer

The spin of a fundamental fermionic particle always has the absolute value $\frac{\hbar}{2}$. This does not relate to its orbital angular momentum, it's just what the spin of a fundamental fermion is.

It now turns out that, for massless particles, there is the notion of

helicity(see also What is polarisation, spin, helicity, chirality and parity?), which is the projection (i.e. the relative direction) of spin and momentum.In your picture, someone decided to orient the axis along which one component of the spin is measured parallel to the particles momentum, so that $+1/2$ spin corresponds to the spin being parallel to the linear momentum, and $-1/2$ to them being anti-parallel. In situations where "neutrinos are massless" is a good approximation, this picture is Lorentz-invariant and indeed a good description for the otherwise rather technical notion of

chirality.The electron cannot correspond to an anti-neutrino because the latter has zero electrical charge. Spin is only one of many quantum numbers.