I have been studying quantum mechanics, specifically angular momentum, but I have a question that concerns raising and lowering operators as a whole. For total angular momentum, you can define: $$J_\pm=J_x\pm iJ_y $$ Anyone who is familiar with angular momentum will recognize these as the raising and lowering operators, but I will continue on with the problem to better explain my question.
An analysis of this problem shows that: $$ [J_z, J_\pm]=\pm \hbar J_\pm$$ $$ [J^2, J_\pm]=0 $$ From here it is easy to see that if $J_z|\alpha\beta\rangle= \beta|\alpha\beta\rangle, $ and $J^2|\alpha\beta\rangle= \alpha|\alpha\beta\rangle$, $$ J_z(J_+|\alpha\beta\rangle)=(J_+J_z+\hbar J_+)|\alpha\beta\rangle= (J_+\beta+\hbar J_+)|\alpha\beta\rangle=(\beta +\hbar)J_+|\alpha\beta\rangle $$ And thus we can say $J_+|\alpha\beta\rangle=C|\alpha,\beta + \hbar\rangle $.
However, while this approach is very clean cut, in my mind it doesn't exactly show that the eigenvalues of $J_z$ exist only in increments of $\hbar$. For instance, if I were able to find some arbitrary set of operators $W_\pm$, such that $[J_z, W_\pm]=\pm (\hbar /4)W_\pm$, then I could easily show by the logic above that the eigenvalues of $J_z$ exist in increments of $\hbar /4$. So then, what guarantees that I cannot find such operators? More specifically, what part of the "raising and lowering operator" method guarantees that there are not more possible eigenvalues of $J_z$ (or any operator), than those found using raising and lowering operators?
Answer
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.
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