0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @ACuriousMind No need to use a condescending tone. $\endgroup$ – childofsaturn Oct 27 '15 at 15:32
  • 1
    $\begingroup$ @ACuriousMind Care to turn that into an answer? $\endgroup$ – Emilio Pisanty Oct 27 '15 at 15:34
1
$\begingroup$

$m^2\leq l^2$ 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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.