Why do $sl(2,\mathbb{C})$ raising and lowering operators $J_{\pm}$ guarantee quantized eigenvalues?
The formal answer lies in representation theory, in this case, the representation theory of the Lie algebra $\mathfrak{su}(2)$, which is spanned by the three operators $J_z,J_+,J_-$. That there are no more eigenvalues of $J_z$ than those found by the ladder operator method follows from two facts:
Every representation of $\mathfrak{su}(2)$ is completely decomposable, i.e. the direct sum of irreducible representations.
The irreducible representations of $\mathfrak{su}(2)$ are precisely the "spin representations" of physics, labeled by the half-integer largest eigenvalue ("highest weight") $s$ of $J_z$, which have dimension $2s+1$, consisting of the states with eigenvalues $-s,-s+1,\dots,s-1,s$.
$s$ has to be half-integer because one can directly show that if $s$ is the highest weight, then the lowest eigenvalue is $-s$, and if the difference between the highest and the lowest weight were not an integer, we would be able to reach an even lower weight by appling the lowering operator to the highest weight state.
- There is no combination of angular momentum operator that satisfy a condition like $[J_z,W_{\pm}]=\pm (\hbar /4)W_\pm$. The only possible ladder operators constructed out of $J_x$ and $J_y$ are $J_\pm$, and their commutation relations are $[J_z,J_{\pm}]=\pm \hbar J_\pm$, which does imply that neighbouring $m$ values differ by $1$. (Since we only have $J_x,J_y$ and $J_z$ to play with, it's not hard to show that $[J_z,J_{\pm}]=\pm \hbar J_\pm$: just start with a generic $J_+=a L_x+bL_y$ and you will find that $b=\pm i a$. The actual value of $a$ is irrelevant for computing the shift in $m$.)
- It is possible for an operator $\hat A$ to satisfy (for instance) $[J_z,\hat A]=2 \hbar \hat A$. An example is any operator proportional to $(x+iy)^2$. The action of this operator changes $m$ by $+2\hbar$ but $\hat A$ is NOT an angular momentum operator.
- Angular momentum operators have a Lie algebraic structure, and from the representation theory of Lie algebras we know that the set $\{\vert jm\rangle\}$ must contain $2j+1$ elements and must contain $m=j$ and $m=-j$. Thus the ladder by angular momentum ladder operators can only change $m$ by one unit of $\hbar$.