Use equations 4.27 4.28 and 4.32 to construct${\mathit{y}}_{\mathbf{0}}^{\mathbf{0}}\mathbf{}\mathbf{,}\mathbf{}{\mathit{y}}_{\mathbf{2}}^{\mathbf{1}}$Check that they are normalized and orthogonal

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

$$\begin{gathered} {{P_l}^m}^{\text{'}}\left(x\right)=(1-x^2{)}^{\frac{m}{2}}(\frac{d}{dx}{)}^m{P}_l\left(x\right) \\ {P}_l\left(x\right)=\frac{1}{2^l\overline{)!}}\left(\frac{d}{dx}\right){\left(x^2-1\right)}^l \\ {{Y^m}_l}_l(\theta ,\varphi )=\epsilon \sqrt{\frac{(2l+1)(l-m)!}{4\pi (l+m)!}}{e}^{im\varphi }{P}_l^m\left(cos\theta \right) \end{gathered}$$

Using these three equations we have to construct

_{$$\begin{gathered} Y_0^0=\frac{1}{\sqrt{4\pi }}{P}_0^0\left(cos\theta \right) \\ {P}_0^0\left(x\right)=P_0\left(x\right) \\ {P}_0\left(x\right)=1 \end{gathered}$$}

Combine the above we get,

$Y_0^0=\frac{1}{\sqrt{4\pi }}$

Repeat the same procedure,

_{$$\begin{gathered} {Y^1}_2=-\sqrt{\frac{5\cdot 1}{4\pi \cdot 3\cdot 2}}{e}^{i\varphi }{P}_2^1\left(cos\theta \right) \\ {P}_2^1\left(x\right)=\sqrt{1-x^2}\frac{d}{dx}{P}_2\left(x\right) \\ {P}_2\left(x\right)=\frac{1}{4\cdot 2}{\left(\frac{d}{dx}\right)}^2(x^2-1) \\ {P}_2\left(x\right)=\frac{1}{8}\frac{d}{dx}\left[2\right(x^2-1\left)2x\right] \\ {P}_2\left(x\right)=\frac{1}{2}\left[(x^2-1)+X\left(2X\right)\right] \\ {P}_2\left(x\right)=\frac{1}{2}\left[3X^2-1\right] \\ {P}_2\left(x\right)=\sqrt{1-X^2}\frac{d}{dx}[\frac{3}{2}{x}^2-\frac{1}{2}] \\ {P}_2\left(x\right)=\sqrt{1-X^2}3x \end{gathered}$$}

But

_{$x=\cos \theta$}

So,

_{$$\begin{gathered} p_2^1\left(\cos \theta \right)=\sqrt{1-{\cos }^2\theta 3\cos \theta } \\ =3\cos \theta \sin \theta \end{gathered}$$}

Thus,

${Y^1}_2=-\sqrt{\frac{15}{8\pi }}{e}^i\varphi \left(sin\theta \right)\left(cos\theta \right)$

Now we need to check the normalization

_{role="math" localid="1658466244130" $$\begin{gathered} \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =\frac{1}{4\pi }{\int }_0^{\pi }sin\theta d\theta {\int }_0^{2\pi }d\varphi \\ \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =\frac{1}{4\pi }2\left(2\pi \right) \\ \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =1 \\ \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =\frac{15}{4\pi }{\int }_0^{\pi }si{n}^2\theta c{o}^2\theta sin\theta d\theta {\int }_0^{2\pi }d\varphi \\ \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =\frac{15}{4\pi }{\int }_0^{\pi }(1-co{s}^2\theta )co{s}^2\theta sin\theta d\theta {\int }_0^2\pi d\varphi \\ y=cos\theta ,dy=-sin\theta \\ \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =\frac{15}{4\pi }{\int }_1^{-1}{y}^2(1-y^2)dy \\ \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =\frac{15}{4\pi }{\left[\frac{y^3}{3}{\frac{y}{5}}^5\right]}^1-1 \\ \iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =1 \end{gathered}$$}

Finally, we need to check the orthonormality as

_{role="math" localid="1658466365776" $\iint {\left[Y_0^0\right]}^{2}sin\theta d\theta d\varphi =\frac{1}{4\pi }\sqrt{\frac{15}{8\pi }}{\int }_0^{\pi }sin\theta cos\theta sin\theta d\theta {\int }_0^2\pi e^i\varphi d\varphi$}

The first and second integral vanishes thus,

_{$\iint Y_0^0{y}_2^{1}sin\theta d\theta d\varphi =0$}