Question

The raising and lowering operators change the value of m by one unit:

$\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{L}}_{\mathbf{\pm }}{\mathit{f}}_{\mathbf{l}}^{\mathbf{m}}\mathbf{}\mathbf{=}\left(A_{\mathcal{l}}^m\right){\mathit{f}}_{\mathbf{l}}^{\mathbf{m}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{,}$ (4.120).

Where ${\mathbf{A}}_{\mathbf{l}}^{\mathbf{m}}\mathbf{}$are constant. Question: What is ${\mathbf{A}}_{\mathbf{l}}^{\mathbf{m}}\mathbf{}$, if the Eigen functions are to be normalized? Hint: First show that${\mathit{L}}_{\mathbf{\pm }}$is the Hermitian conjugate of ${\mathit{L}}_{\mathbf{\pm }}$(Since ${\mathit{L}}_x\mathbf{}\mathbf{and}\mathbf{}{\mathbf{L}}_{\mathbf{y}}$are observables, you may assume they are Hermitian…but prove it if you like); then use Equation 4.112.

Short answer

If the Eigen functions are to be normalized, they are upper and lower signs.

Step-by-step solution

Given.

The raising and lowering operators are:

$$\begin{gathered} {\mathbf{L}}_{\mathbf{+}}{\mathbf{f}}_{\mathbf{l}}^{\mathbf{m}}\mathbf{}\mathbf{=}\left(A_l^m\right){\mathbf{f}}_{\mathbf{l}}^{\mathbf{m}\mathbf{+}\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{-}}{\mathbf{f}}_{\mathbf{l}}^{\mathbf{m}}\mathbf{}\mathbf{=}\mathbf{\left(}{\mathbf{B}}_{\mathbf{l}}^{\mathbf{m}}\mathbf{\right)}{\mathbf{f}}_{\mathbf{l}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\mathbf{,} \end{gathered}$$

Eigen functions to be normalized

We start solving the problem by taking the inner product between a hydro genic set acted upon $L_{\pm }$and $L_{\mp }$as following:

$$\begin{gathered} {\mathrm{A}}_{\mathcal{l}}^{\mathrm{m}}=\mathrm{ħ}\sqrt{\mathcal{l}(\mathcal{l}+1)-\mathrm{m}(\mathrm{m}+1)} \\ =\mathrm{ħ}\sqrt{\mathcal{l}(\mathcal{l}+1)-\mathrm{m}(\mathrm{m}+1)}, \\ {\mathrm{B}}_{\mathcal{l}}^{\mathrm{m}}=\mathrm{ħ}\sqrt{\mathcal{l}(\mathcal{l}+1)-\mathrm{m}(\mathrm{m}+1)} \\ =\mathrm{ħ}\sqrt{\mathcal{l}(\mathcal{l}+1)-\mathrm{m}(\mathrm{m}+1)}, \\ {\mathrm{L}}^2={\mathrm{L}}_{\pm }{\mathrm{L}}_{\mp }+{\mathrm{L}}_{\mathrm{z}}^2\pm {\mathrm{ħL}}_{\mathrm{z}}(4.112). \end{gathered}$$

Note what happens at the top and bottom of the ladder (i.e. when you apply $$\begin{gathered} {\mathrm{L}}_{+}\mathrm{L}+\mathrm{to}{\mathrm{f}}_{\mathcal{l}}^{\mathcal{l}}\mathrm{or}{\mathrm{L}}_{-}\mathrm{to}{\mathrm{f}}_{\mathcal{l}}^{-\mathcal{l}} \\ \mathrm{Now},\mathrm{using}\mathrm{Eq}.4.112,\mathrm{in}\mathrm{the}\mathrm{from}{\mathrm{L}}_{\pm }{\mathrm{L}}_{\mp }={\mathrm{L}}^2-{\mathrm{L}}_{\mathrm{z}}^2\mp {\mathrm{ħL}}_{\mathrm{z}}\mathrm{L} \\ {\mathrm{L}}^2={\mathrm{L}}_{\pm }{\mathrm{L}}_{\mp }+{\mathrm{L}}_{\mathrm{z}}^2\mp {\mathrm{ħL}}_{\mathrm{z}}\mathrm{L}2=\mathrm{L}\left(4.112\right). \end{gathered}$$

At the top of the ladder$\left(\mathrm{m}=\mathcal{l}\right)$we get $A_{\mathcal{l}}^{\mathcal{l}}=0$$\mathrm{we}\mathrm{get}{\mathrm{A}}_{\mathrm{l}}^{\mathrm{l}}=0$, so there is no higher rung; at the bottom of the ladder $\left(\mathrm{m}=\mathcal{l}\right)$ we get$B_{\mathcal{l}}^{-\mathcal{l}}=0$ , so there is no lower rung.