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.
1) "When a quantum wave function is in a potential well, what causes the
quantization? The finiteness of the well, or only the term with ℏ in
Schrödinger's equation?"
For the quantum finite potential well, the discrete possible values for $E_n \sim \hbar ^2 v_n$ where the $v_n$ are discrete solutions to non-trivial equations due to the boudary conditions (see the details in the Wikipedia reference above). You may see directly in the formula, that both the Schrodinger equation (so quantum mechanics and $\hbar$), and the boudary conditions are necessary to have discrete values for $E_n$
2) Is there an analogy between these two approaches? Is the Schrödinger
equation fundamentally due to a sort of boundary condition, which
gives its value to the Planck constant ℏ?
No, this is not due to boudary conditions.
The basis of quantum mechanics is that position and momentum are no more commutative quantities, but are linear operators (infinite matrices), such that,at same time, $[X^i,P_j]= \delta^i_j ~\hbar$.
Now, you may have different representations for these operators.
In the Schrodinger representation, we consider that these linear operators apply on vectors $|\psi(t)\rangle$ (called states). The probability amplitude $\psi(x,t)$ is the coordinate of the vector $|\psi(t)\rangle$ in the basis $|x\rangle$. In this representation, you have $X^i\psi(x,t) = x^i\psi(x,t), P_i\psi(x,t) = -i\hbar \frac{\partial}{\partial x^i}\psi(x,t)$ . This extends to energy too, with $E\psi(x,t) = i\hbar \frac{\partial}{\partial t}\psi(x,t)$. This last equality is coherent with the momentum operator definition if we look at the de Broglie waves
3) One can obtain an analog of Schrödinger's equation if space was
discrete. Is it possible to derive Schrödinger equation from such a
description of space and time?
In the reference you gave, there is no discrete space, and there is no discrete time, the $\psi_i(t)$ are only the coordinates of the vector $|\psi(t)\rangle$ in some basis $|i\rangle$
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Best Answer
First and second quantization
Quantization is a misleading term, since it implies discreteness (e.g., of the energy levels), which is not always the case. In practice (first) quantization refers to describing particles as waves, which in principle allows for discrete spectra, when boundary conditions are present.
The electromagnetic waves behave in a similar fashion, exhibiting discrete spectra in resonators. Thus, technically, quantization of the electromagnetic field corresponds to second quantization of particles.
Second quantization arises when dealing with many-particle systems, when the focus is not anymore on the wave nature of the states, but on the number of particles in each state. The discreteness (of particles) is inherent in this approach. For the electromagnetic field this corresponds to the first quantization, and the filling particles, whose number is counted, are referred to as photons. Thus, photon is not really a particle, but an elementary excitation of electromagnetic field. Associating a photon with a wave packet is misleading, although it appeals to intuition. (One could argue however that physically observed photons are always wave packets, since to have truly well-defined energy they would have to exist for infinite time, which is not possible.)
This logic of quantization is applied to other wave-like fields, such as wave excitations in crystals: phonons (sound), magnons, etc. One speaks sometimes even about diffusons - quantized excitation of a field described by the duffusion equation.
Uncertainty relation
An alternative way to look at quantization is from the point of view of the Heisenberg uncertainty relation. One switches from classical to quantum theory by demanding that canonically conjugate variables cannot be measured simultaneously (e.g., position and momentum, $x,p$ can be simultaneously measured in classical mechanics, but not in quantum mechanics). Mathematically this means that the corresponding operators do not commute: $$ [\hat{x}, \hat{p}_x]_- = \imath\hbar \Rightarrow \Delta x\Delta p_x \geq \frac{\hbar}{2}.$$ The discreteness of spectra then shows up as discrete eigenvalues of the operators.
This procedure can be applied to anything - particles or fields - as long as we can formulate it in terms of Hamiltonian mechanics and identify effective position and momenta, on which we then impose the non-commutativity. E.g., for electromagnetic field, one demands the non-commutativity of the electric and the magnetic fields at a given point.