# Angular momentum: why must descending and ascending chains terminate?

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?

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