All magnetic quantum numbers are separated by integer steps The most common algebraic derivation of the properties of angular momentum in quantum mechanics follows the usual line of thought.

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*We start with an (unknown) Hilbert space $\mathcal H$ on which there is a triple $\mathbf J=(J_1,J_2,J_3)$ of self-adjoint operators which satisfy $$ [J_i,J_j]=i\epsilon_{ijk}J_k $$(for simplicity, let $\hbar=1$).

*The operators $J^2=\sum_iJ_i^2$ and $J_\pm=J_1\pm iJ_2$ are defined. We have $[J^2,J_i]=0$, $[J^2,J_\pm]=0$ and $[J_3,J_\pm]=\pm J_\pm$.

*Since $J_3$ and $J^2$ commute, they have a common eigenbasis written as $$ J^2\varphi_{\lambda,m}=\lambda^2\varphi_{\lambda,m} \\ J_z\varphi_{\lambda,m}=m\varphi_{\lambda,m}, $$with $-\lambda\le m\le \lambda$ thus we can write the decomposition $$ \mathcal H=\bigoplus_\lambda S_\lambda,\quad S_\lambda=\bigoplus_m S_{\lambda,m}, $$where $S_\lambda$ is the eigensubspace of $J^2$ corresponding to the eigenvalue $\lambda$ and $S_{\lambda,m}$ is the common eigensubspace of $J^2$ and $J_3$ corresponding to $(\lambda,m)$. Although this is not true some of the time, we may suppose that the eigenspaces $S_{\lambda,m}$ are one dimensional and thus spanned by $\varphi_{\lambda, m}$ (take quotients otherwise??? but I think the general argument should work with eigenspaces instead of states).

*It follows from the restriction $-\lambda\le m\le \lambda$ that for any fixed $\lambda$ we must have a maximum $m_+$ and a minimum $m_-$ eigenvalue, and it follows from the properties of $J_\pm$ that $S_\lambda$ is an invariant subspace of $J_\pm$ and $J_\pm:S_{\lambda,m}\rightarrow S_{\lambda,m\pm1}$. It also follows from some algebraic relations that $m_-=-m_+\equiv j$ and we have $\lambda^2=j(j+1)$

*The usual argument to show that $j$ is an integer multiple of $1/2$ is that if we take the state $\varphi_{j,m_-}$ belonging to the minimum value $m_-$, then there is a nonnegative integer $k$ such that $J^k_+\varphi_{j,m_-}=C\varphi_{j,m_-+k}\neq 0$ but $J_+^{k+1}\varphi_{j,m_-}=0$. One then infers that $m_+=m_-+k$, thus $2j=k$ and therefore $j$ is an integer multiple of $1/2$.
It is this inference that has been highlighted by the use of bold letters that I have a problem with. I see no logical reason why is it that $m_-+k=m_+$. I do not see why it is not possible that there is a number $m^\prime<m_+$ such that $m^\prime$ is an eigenvalue of $J_3$ but $J_+\varphi_{j,m^\prime}=0$.
Alternatively, it is clear that starting from $\varphi_{j,m_-}$, there is a tower of eigenstates $\varphi_{j,m_-+1},\dots,\varphi_{j,m_-+k},\dots$ separated by integer values and terminating after a finite number of elements, but I see no reason why the index of termination should be $m_+$ and that it is not possible that there is a number $\mu$ with $m_-<\mu\neq m_-+k$ ($k$ is an integer) such that there is also a second tower $\varphi_{j,\mu},\varphi_{j,\mu+1},\dots$ "in between" the elements of the first tower, terminating at a different element whose value $\mu_+$ is less than the maximum $m_+$.
If $m_+$ is indeed the maximum then it only follows that there exists one tower which terminates at $m_+$, not that all towers terminate there.
The question is then how to see that once we have the minimum and maximum eigenvalues $m_-$ and $m_+$, the only allowed ones are $m_-+1,m_-+2,\dots,m_+-1,m_+$ (and with $m_+-m_-$ being an integer) instead of having several towers running concurrently?
Remark: A similar proof is given when one classifies the irreducible representations of $\mathfrak{sl}(2,\mathbb C)$, however there one explicitly makes the assumptions that 1) each representation space (corresponding to $S_\lambda$ here) is finite dimensional and 2) that each representation is irreducible.
I do not see on what basis would one make these assumptions when discussing angular momentum from a physical point of view, therefore I have not made these assumptions. Of course making the above assumptions would resolve my problem but as I have said I want the eigenvalue problem of angular momentum, not to classify the irreducible representations of $\mathfrak{sl}(2,\mathbb C)$. The two problems might turn out to be equivalent, but they are not so a priori.
 A: To illustrate the importance of the hermiticity condition, consider the map
\begin{align}
L_+\mapsto \Gamma(L_+)&= e^{-i \phi}\left(j+i \frac{d}{d\phi}\right)\, ,\\
L_-\mapsto \Gamma(L_-)&= e^{+i \phi}\left(j-i \frac{d}{d\phi}\right)\, ,\\
L_z\mapsto \Gamma(L_z)&= i\frac{d}{d\phi}\, .
\end{align}
You can verify that this is a realization of the algebra in the sense that
$$
[L_i,L_j]\mapsto [\Gamma(L_i),\Gamma(L_j)]
$$
The realization $\Gamma$ acts naturally on functions of the form $\Phi_k=e^{-i(j+k)\phi}$ and in fact
\begin{align}
\Gamma(L_-)\Phi_0&=0\, ,\\
\Gamma(L_-)\Phi_k&=-k\Phi_{k-1}\, ,\\
\Gamma(L_+)\Phi_k&=(2j+k)\Phi_{k+1}
\end{align}
for any j.  There is no restriction here on the possible $j$'s or the possible $m$'s, and the resulting representation is "legitimate" but fails the hermiticity test as clearly $\Gamma^\dagger(L_+)\ne \Gamma(L_-)$: this representation does not exponentiate to a unitary representation
Thus, hermiticity is essential to the argument that the $m$'s are bounded by $\pm j$.
A: Don't forget what you're trying to do here: The angular momentum algebra is the infinitesimal version of the rotations $\mathrm{SO}(3)$. You're looking for unitary linear representations of the group $\mathrm{SU}(2)$ as its linear representations constitute the projective representations of $\mathrm{SO}(3)$.
All unitary representations are completely reducible and since $\mathrm{SU}(2)$ is compact all of its irreducible representations are finite-dimensional. Thus your additional "assumptions" are not actually assumptions, but just consequences of the compactness of $\mathrm{SU}(2)$.
