91Ó°ÊÓ

A normalizing constant guarantees that the probability of a probability density function is 1. The constant can appear in many forms, including scalar values, equations, and even functions.

As a result, there is not a "one size fits all" constant; instead, any probability distribution that does not sum to 1 will have its own normalization constant.

The operator $L_{+}$ is the raising operator that is acting on the function $Y_I^m$. It gives a result proportional to $Y_I^{m+1}$but the maximum value of m is I . So, $Y_I^I$is the top function.

Thus, the value of $L_{+}{Y}_I^I$ is 0.

Consider $L_z{Y}_I^I=hI{Y}_I^I$and use the definition of $L_z$to solve for $Y_I^I$.

localid="1658204848834" $$\begin{gathered} L_z{Y}_{\mathit{1}}^I\mathit{(}\mathit{θ}\mathit{,}\mathit{ϕ}\mathit{)}\mathit{=}\sout{h}I{Y}_{\mathit{1}}^I\mathit{(}\mathrm{θ}\mathit{,}\mathrm{ϕ}\mathit{)}\mathit{-}i\sout{h}\frac{\mathit{∂}{Y}_{\mathit{1}}^I\mathit{(}\mathrm{θ}\mathit{,}\mathrm{ϕ}\mathit{)}}{\mathit{∂}\mathrm{ϕ}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\sout{h}I{Y}_{\mathit{1}}^I\mathit{(}\mathit{θ}\mathit{,}\mathit{ϕ}\mathit{)}\mathit{}\frac{\mathit{∂}{Y}_{\mathit{1}}^I}{Y_{\mathit{1}}^I}\mathit{} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}iI\mathit{∂}\mathrm{ϕ}{Y}_{\mathit{1}}^I\mathit{(}\mathrm{θ}\mathit{,}\mathrm{ϕ}\mathit{)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}f\left(\mathrm{θ}\right)e^{\mathit{i}\mathit{I}\mathit{ϕ}} \end{gathered}$$

It can be observed that from part (a) localid="1658204942389" $L_{\mathit{+}}{Y}_I^I\mathit{=}\mathit{0}$. So, substitute 0 for $L_{\mathit{+}}{Y}_{\mathrm{I}}^I$in the above expression.

$L_{\mathit{+}}f\left(\mathrm{θ}\right)e^{iI\mathrm{ϕ}}=0$

Unravele the definition.

$$\begin{gathered} \sout{h}e\left[\frac{∂f\left(\mathrm{θ}\right)}{∂\mathrm{θ}}.e^{iI\mathrm{ϕ}}+icot\mathrm{θ}f\left(\mathrm{θ}\right)\frac{∂e^{iI\mathrm{ϕ}}}{∂\mathrm{ϕ}}\right]=0 \\ \frac{∂f\left(\mathrm{θ}\right)}{∂\mathrm{θ}}.e^{iI\mathrm{ϕ}}+icot\mathrm{θ}f\left(\mathrm{θ}\right)\left(il{e}^{il\mathrm{ϕ}}\right)=0 \\ \frac{∂f\left(\mathrm{θ}\right)}{∂\mathrm{θ}}=lcot\mathrm{θ}f\left(\mathrm{θ}\right) \end{gathered}$$

Integrate both sides of the above equation.

$$\begin{gathered} ∫\frac{∂f\left(\mathrm{θ}\right)}{f\left(\mathrm{θ}\right)}d\mathrm{θ}=∫Icot\mathrm{θ}d\mathrm{θ} \\ Inf\left(\mathrm{θ}\right)=lIn\sin \mathrm{θ}+k \\ Inf\left(\mathrm{θ}\right)=lIn\left(A{\sin }^I\mathrm{θ}\right) \\ f\left(\mathrm{θ}\right)=A{\sin }^I\left(\mathrm{θ}\right) \\ {Y}_I^I\left(\mathrm{θ},\mathrm{ϕ}\right)=A{\sin }^I\mathrm{θ}{e}^{il\mathrm{ϕ}} \end{gathered}$$

Here, A isanormalization constant.

Thus, the value of $Y_I^I\left(\mathrm{θ},\mathrm{ϕ}\right)$is A ${\left[\sin \mathrm{θ}{e}^{i\mathrm{ϕ}}\right]}^l$.

To normalize the function $1=∫{\left|Y_I^I\left(\mathrm{θ},\mathrm{ϕ}\right)\right|}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ}$is required.

Substitute in $A{\left[\sin \mathrm{θ}{e}^{i\mathrm{ϕ}}\right]}^I$for $Y_I^I$in the expression.

$$\begin{gathered} 1=A^2{∫}_0^{2\mathrm{π}}{∫}_0^{\mathrm{π}}{\sin }^{2I}\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}d\mathrm{ϕ} \\ =2\mathrm{π}{A}^2{∫}_0^{\mathrm{π}}{\sin }^{2I}\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ} \\ =2\mathrm{π}{A}^2{∫}_0^{\mathrm{π}}{\sin }^{2I+1}\mathrm{θ}d\mathrm{θ} \end{gathered}$$

Use integral table for solving the expression.

$$\begin{gathered} =4\mathrm{Ï€}{A}^2\frac{2.4.6...\left(2l\right)}{1.3.5...\left(2l+1\right)} \\ =4\mathrm{Ï€}{A}^2\frac{{\left(2.4.6...\left(2l\right)\right)}^2}{1.2.3.5...\left(2l\right)\left(2l+1\right)} \\ =4\mathrm{Ï€}{A}^2\frac{{\left(2ll\right)}^2}{\left(2l+1\right)\mathit{!}} \\ A=\frac{1}{2^{I+1}l\mathit{!}}\sqrt{\frac{\left(2l+1\right)\mathit{!}}{\mathrm{Ï€}}} \end{gathered}$$

Thus, the normalization constant is$\frac{1}{2^{I+1}l\mathit{!}}\sqrt{\frac{\left(2l+1\right)\mathit{!}}{\mathrm{Ï€}}}$.$\frac{1}{2^{I+1}l\mathit{!}}\sqrt{\frac{\left(2l+1\right)\mathit{!}}{\mathrm{Ï€}}}$.