I'm studying theory of angular momentum, but I haven't understood a step.

$M_1, M_2, M_3$ are the components of angular momentum $\vec{M}$.

Let's consider ladder operators $M_+=M_1+iM_2$ and $M_-=M_1-iM_2$: $$\begin{align} M_+\lvert l, m\rangle &= \lvert l, m+1\rangle & M_-\lvert l, m\rangle &= \lvert l, m-1\rangle. \end{align}$$ Then $$M_3^2=M^2-M_1^2-M_2^2 \tag{$\ast$}$$

If we take the average of the relation $(\ast)$ on the state $\lvert l, m\rangle$ we obtain $$m^2\leq l^2.\tag{$\ast\!\ast$}$$

Well, my textbook says:

The ascending chain (that we obtain applying the operator $L_+$) and the descending chain (that we obtain applying the operator $L_-$) have to be interrupted because the relation $(\ast\ast)$ must be valid.

Well, I understand why the ascending chain has to be interrupted, but why does the relation $(\ast\ast)$ imply the interruption of the descending chain?

  • $\begingroup$ Uh, because $m^2\leq l^2$ means that $\lvert m \rvert \leq l$, so $m$ can't be lower than $-l$? $\endgroup$ – ACuriousMind Oct 27 '15 at 15:19
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    $\begingroup$ @ACuriousMind No need to use a condescending tone. $\endgroup$ – childofsaturn Oct 27 '15 at 15:32
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    $\begingroup$ @ACuriousMind Care to turn that into an answer? $\endgroup$ – Emilio Pisanty Oct 27 '15 at 15:34

$m^2\leq l^2$ implies that $\lvert m \rvert \leq l$ (for positive $l$, which it is in this case).

Therefore, both the ascending and the descending chain have to terminate at $m=l$ and $m=-l$, respectively.

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