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When studying the general angular momentum $\textbf{J}$, which is defined as a vector operator with its components being Hermitian operators satisfying the commutation relations

\begin{align*} \textbf{J} \times \textbf{J} = i\hbar \textbf{J} , \end{align*} and the simultaneous eigenvectors of $\textbf{J}^2$ and an abitrary component $J_z$, my book stated that the eigenvalues $j$ and $m$ are either integer (including zero) or half-odd-integer values. They came to this conclusion by using the raising and lowering operators which are defined as \begin{align*} J_{\pm} = J_x \pm iJ_y. \end{align*} More specifically, because of the inequality \begin{align*} j(j+1) \geq m^2 \end{align*} there must be an upper and lower bound $m_T , m_B$ which differ from each other an integer amount $n$.

However, what ensures that these eigenvalues are the only possible values? How can it not be that the spectrum of eigenvalues is continuous? Say for example we forget about the raising and lowering operators and just postulate that there exists a simultaneous eigenvector of both $\textbf{J}^2$ and $J_z$ with $j = 2.3$ and $m = 2.3$. Is there a way to contradict this assumption?

I have already found an answer using the representation theory of Lie algebras, but since my mathematical background is limited, I would like to have a simpler explanation. If this is the only formal explanation, then I would like to know that and move on :). The book I use for studying QM is Bransden & Joachain (chapter 6.5).

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That is because each of these operators is selfadjoint and the associated eigenvectors form a Hilbert basis in the Hilbert space. As a consequence of the spectral theorem, the set of eigenvalues coincides with the spectrum of the operator. Actually, further possible points in the spectrum could be accumulation points of the set of eigenvalues but in this case there are no such points.

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  • $\begingroup$ But how do we know that the eigenvectors that are generated through the raising and lowering operators are all the eigenvectors ? In other words, how do we know that they are complete ? $\endgroup$ – Einsteinwasmyfather Aug 13 at 17:12
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    $\begingroup$ By direct inspection, you find a family of bases of irreducible representations of $SU(2)$ and using the so called Peter-Weyl theorem for the compact group $SU(2)$. $\endgroup$ – Valter Moretti Aug 13 at 17:16

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