Inequalities of orbital angular momentum eigenvalues (show that is bounded)

Question

If we apply the rising/lowering operators for angular momentum to a state $|{l,m}\rangle$, we get:

\begin{align} L_+|{l,m}\rangle = C_+(l,m)|{l,m+1}\rangle \\ L_-|{l,m}\rangle = C_-(l,m)|{l,m-1}\rangle \end{align}

it can be shown that: \begin{align} C_+(l,m) = \hbar\sqrt{(l-m)(l+m+1)}\\ C_-(l,m) = \hbar\sqrt{(l+m)(l-m+1)} \end{align}

thus:

\begin{align} L_+|{l,m}\rangle = \hbar\sqrt{(l-m)(l+m+1)}|{l,m+1}\rangle \\ L_-|{l,m}\rangle = \hbar\sqrt{(l+m)(l-m+1)}|{l,m-1}\rangle \end{align}

On the other hand:

\begin{align} \langle{L_\pm(l,m)|L_\pm(l,m)}\rangle \geq 0 \end{align}

And it follows from (7) that:

\begin{align} \langle{L_\pm(l,m)|L_\pm(l,m)}\rangle &= \langle{l,m|L_\mp L_\pm|l,m}\rangle\\ & = \langle{l,m|L^{2} - L_{z}^{2} \pm\hbar L_z|l,m}\rangle\\ & = \hbar^2 [l(l+1) - m(m\mp 1)] \geq 0 \end{align}

This implies that both:

\begin{align} l(l+1) - m(m- 1) \geq 0\\ l(l+1)- m(m+ 1) \geq 0 \end{align}

Now here comes my doubt, gasiorowicz's book (Quantum Physics, Thrid Edition, Chapter 7) claims: Since $l\geq 0$ It follows form the above that:

\begin{align} -l\leq m \leq l \end{align}

But i cant see how the above inequalities imples the last one. Any help would be appriciate, thanks.

Answer 1

If $0\leq l<m$, then $l+1<m+1$, so multiplying these inequalities yields \begin{align} l(l+1)<m(m+1) \end{align} (we can do this since everything is non-negative), so this violates the second inequality you wrote down.

Next, suppose $m<-l$, which means $0\leq l<-m$, and thus by the same reasoning as above, we have \begin{align} l(l+1)&< (-m)[(-m)+1]\\ &=m(m-1) \end{align}

which violates your first inequality.

Therefore, we must have $-l\leq m\leq l$.