Position quantization in vacuum is forbidden by rotational, translational, and boost invariance. There is no rotationally invariant grid. On the other hand, if you have electrons in a periodic potential, the result in any one band is mathematically the theory of an electron on a discrete lattice. In this case, the position is quantized, so that the momentum is periodic with period p.
Fourier duality
The quasimomentum p in a crystal is defined as i times the log of the eigenvalue of the crystal translation operator acting eigenvector. Here I give the definition of crystal position operators and momentum operators, which are relevant in the tight-binding bands, and to describe the analog of the canonical commutation relations which these operators obey. These are the discrete space canonical commutation relations.
In 1d, consider a periodic potential of period 1, the translation by one unit on an energy eigenstate commutes with H, so it gives a phase, which you write as:
$$ e^{ip}$$
and for p in a Brillouin zone $-\pi<0<\pi$ this gives a unique phase. The p direction has become periodic with period $2\pi$. This means that any superposition of p waves is a periodic function in p-space.
The Fourier transform is a duality, and a periodic spatial coordinate leads to discrete p. In this case, the duality takes a periodic p to a discrete x. Define the dual position operator using eigenstates of position. The position
eigenstate is defined as follows on an infinite lattice:
$$ |x=0\rangle = \int_0^{2\pi} |p\rangle $$
Where the sum is over the Brillouin zone, and the sum is over one band only. This state comes with a whole family of others, which are translated by the lattice symmetry:
$$ |x=n\rangle = e^{iPn} |x=0\rangle = \int_0^{2\pi} e^{inp} |p\rangle $$
These are the only superpositions which are periodic on P space. This allows one to define the X operator as;
$$ X = \sum_n n |x=n\rangle\langle x=n| $$
The X operator has discrete eigenvalues, it tells you which atom you are bound to. It only takes you inside one band, it doesn't have matrix elements from band to band.
The commutation relations for the quasiposition X and quasimomentum P is derived from the fact that integer translation of X is accomplished by P:
$$ X+ n = e^{-inP} X e^{inP}$$
This is the lattice analog of the canonical commutation relation. It isn't infinitesimal. If you make the translation increment infinitesimal, the lattice goes away and it becomes Heisenberg's relation.
If you start with a free particle, any free $|p\rangle$ state is also a quasimomentum p state, but for any given quasimomentum p, all the states
$$ |p + 2\pi k\rangle $$
have the same quasimomentum for any integer k. If you add a small periodic potential and do perturbation theory, these different k-states at a fixed quasimomentum mix with each other to produce the bands, and the energy eigenstates $|p,n\rangle$ are labelled by the quasimomentum and the band number n:
you define the discrete position states as above for each band
$$ |x,n\rangle = \int_0^{2\pi} e^{inp} |p,n\rangle $$
These give you the discrete position operator and the discrete band number operator.
$$ N |x,n\rangle = n |x,n\rangle $$
if you further make the crystal of finite size, by imposing periodic boundaries in x, the discrete X become periodic and p becomes the Fourier dual lattice, so that the number of lattice points in x and in p are equal, but the increments are reciprocal.
This is what finite volume discrete space QM looks like, and it does not allow canonical commutators, since these only emerge at small lattice spacing.
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 $-1/2, +1/2$
Best Answer
Quantisation does not imply discreteness. If a system has been quantised, we just mean we have taken the set of states, and replaced it by a vector space of states. In other words, one can add states in quantum mechanics, allowing a system to be in two states "at once". Observable quantities become certain operators acting on this vector space of states.
As you can see, this doesn't have anything, on the face of it, to do with discretisation. It turns out, however, that a lot of the operators we are interested in have discrete eigenvalues, and this implies that the corresponding physical values are discrete. Position, however, has a continuous spectrum, as do many other quantum observables.
There are plenty of sources explaining exactly how one goes from classical sets of states and numerical observables to quantum states (vector spaces - Hilbert spaces in particular) and quantum observables (operators); I won't cover that. All I shall say is that quantisation is a big mathematical process replacing a load of classical things with quantum things, and this sometimes leads to certain physical quantities being discretised.