(a) What is${\mathbf{L}}_{\mathbf{+}}{\mathbf{Y}}_{\mathbf{1}}^{\mathbf{I}}$? (No calculation allowed!)

(b) Use the result of (a), together with Equation 4.130 and the fact that${\mathbf{L}}_z{\mathbf{Y}}_{\mathbf{1}}^{\mathbf{I}}\mathbf{=}\sout{\mathit{h}}\mathit{I}{\mathit{Y}}_{\mathit{I}}^{\mathit{I}}$ to determine${\mathit{Y}}_{\mathit{I}}^{\mathit{I}}\left(\theta ,\varphi \right)$ , up to a normalization constant.

(c) Determine the normalization constant by direct integration. Compare your final answer to what you got in Problem 4.5.

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" $$\begin{gathered} L_z{Y}_{\mathit{1}}^I\mathit{(}\mathit{\theta }\mathit{,}\mathit{\varphi }\mathit{)}\mathit{=}\sout{h}I{Y}_{\mathit{1}}^I\mathit{(}\theta \mathit{,}\varphi \mathit{)}\mathit{-}i\sout{h}\frac{\mathit{\partial }{Y}_{\mathit{1}}^I\mathit{(}\theta \mathit{,}\varphi \mathit{)}}{\mathit{\partial }\varphi } \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\sout{h}I{Y}_{\mathit{1}}^I\mathit{(}\mathit{\theta }\mathit{,}\mathit{\varphi }\mathit{)}\mathit{}\frac{\mathit{\partial }{Y}_{\mathit{1}}^I}{Y_{\mathit{1}}^I}\mathit{} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}iI\mathit{\partial }\varphi Y_{\mathit{1}}^I\mathit{(}\theta \mathit{,}\varphi \mathit{)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}f\left(\theta \right)e^{\mathit{i}\mathit{I}\mathit{\varphi }} \end{gathered}$$

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.

$L_{\mathit{+}}f\left(\theta \right)e^{iI\varphi }=0$

Unravele the definition.

$$\begin{gathered} \sout{h}e\left[\frac{\partial f\left(\theta \right)}{\partial \theta }.e^{iI\varphi }+icot\theta f\left(\theta \right)\frac{\partial e^{iI\varphi }}{\partial \varphi }\right]=0 \\ \frac{\partial f\left(\theta \right)}{\partial \theta }.e^{iI\varphi }+icot\theta f\left(\theta \right)\left(il{e}^{il\varphi }\right)=0 \\ \frac{\partial f\left(\theta \right)}{\partial \theta }=lcot\theta f\left(\theta \right) \end{gathered}$$

Integrate both sides of the above equation.

$$\begin{gathered} \int \frac{\partial f\left(\theta \right)}{f\left(\theta \right)}d\theta =\int Icot\theta d\theta \\ Inf\left(\theta \right)=lIn\sin \theta +k \\ Inf\left(\theta \right)=lIn\left(A{\sin }^I\theta \right) \\ f\left(\theta \right)=A{\sin }^I\left(\theta \right) \\ {Y}_I^I\left(\theta ,\varphi \right)=A{\sin }^I\theta e^{il\varphi } \end{gathered}$$

Here, A isanormalization constant.

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

To normalize the function $1=\int {\left|Y_I^I\left(\theta ,\varphi \right)\right|}^2\sin \theta d\theta d\varphi$is required.

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

$$\begin{gathered} 1=A^2{\int }_0^{2\pi }{\int }_0^{\pi }{\sin }^{2I}\theta \sin \theta d\theta d\varphi \\ =2\pi A^2{\int }_0^{\pi }{\sin }^{2I}\theta \sin \theta d\theta \\ =2\pi A^2{\int }_0^{\pi }{\sin }^{2I+1}\theta d\theta \end{gathered}$$

Use integral table for solving the expression.

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

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