Given $L_+|l,m\rangle \propto |l,m+1\rangle$
and $L_-|l,m\rangle \propto |l,m-1\rangle$
Why isn't it the case that $\langle l,m|L_+L_-|l,m\rangle = 1$? Perhaps naively, but I assumed
$\langle l,m|L_+L_-|l,m\rangle = \langle l,m|L_+|l,m-1\rangle = \langle l,m|l,m\rangle = 1.$
(Where the L's are operators.)
I ask this as I'm trying to calculate $(\Delta L_1)^2$ and the above is involved as I'm using $L_1=\frac{1}{2}(L_++L_-)$.
No, you really should use $\propto$ instead of $\approx$ to describe the result of applying $L_{+}, L_{-}$ to the eigenstate $\mid l,m \rangle$ of angular momentum operators.
In fact, $$\begin{cases} L_{+} L_{-} - L_{-} L_{+} &=& [ L_{+}, L_{-} ] = [ L_x + i L_y, L_x - i L_y ] = -2 i [L_x,L_y] = 2 L_z\\ L_{+} L_{-} + L_{-} L_{+} &=& 2 ( L_x^2 + L_y^2 ) = 2 ( L^2 - L_z^2 ) \end{cases}\\ \implies \begin{cases} L_{-}L_{+} &=& L^2 - L_z^2 - L_z\\ L_{+}L_{-} &=& L^2 - L_z^2 + L_z\\ \end{cases}$$ Notice $$ L^2 \mid l, m \rangle = l(l+1) \mid l, m\rangle \quad\text{ and }\quad L_z \mid l, m \rangle = m \mid l, m\rangle $$ One has $$\begin{cases} \langle l, m \mid L_{-} L_{+} \mid l, m \rangle &=& l(l+1) - m(m+1) = (l-m)(l+m+1)\\ \langle l, m \mid L_{+} L_{-} \mid l, m \rangle &=& l(l+1) - m(m-1) = (l+m)(l-m+1) \end{cases}$$
The convention is fixing the phases among $\mid l, m \rangle$ such that
$$ L_{\pm} \mid l, m \rangle = \sqrt{l(l+1) -m(m \pm 1)} \mid l, m\pm 1 \rangle $$
Please note that $L_{+} \mid l, l\rangle = L_{-} \mid l, -l\rangle = 0$. A consequence for this is for each $l$, there are only $2l + 1$ admissible values of $m$ (i.e $-l, -l+1, \ldots, l-1, l$).
