Use equations 4.27 4.28 and 4.32 to construct ${\mathit{Y}}_{\mathbf{0}}^{\mathbf{0}}\mathbf{,}\mathbf{}{\mathit{Y}}_{\mathbf{2}}^{\mathbf{1}}$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.

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

Using these three equations we have to construct

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

Combine the above we get,

localid="1656058765024" $Y_0^0=\frac{1}{\sqrt{4\pi }}$

Repeat the same procedure,

$$\begin{gathered} {Y^1}_2=-\sqrt{\frac{5·1}{4\mathrm{\pi }·3·2}}{e}^{i\varphi }{P}_2^1\left(\cos \theta \right) \\ {P}_2^1\left(x\right)=\sqrt{1-x^2}=\frac{d}{dx}{P}_2\left(x\right) \\ {P}_2\left(x\right)=\frac{1}{4.2}{\left(\frac{d}{dx}\right)}^2\left(x2-1\right)P_2\left(x\right)=\frac{1}{8}\frac{d}{dx}\left[2\left(x^2-1\right)+x\left(2x\right)\right] \\ {P}_2\left(x\right)=\frac{1}{2}\left[\left(x^2-1\right)+x\left(2x\right)\right] \\ {P}_2\left(x\right)=\frac{1}{2}\left[3x^2-1\right] \\ {P}_2^1\left(x\right)=\sqrt{1-x^2}\frac{d}{dx}[\frac{3}{2}{x}^2-\frac{1}{2}] \\ {P}_2^1\left(x\right)=\sqrt{1-x^2}3x \end{gathered}$$

But

$x=\cos \theta$

So,

$P_2^1\left(\cos \theta \right)=\sqrt{1-{\cos }^2\theta }3\cos \theta =3\cos \theta \sin \theta$

Thus,

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

Now we need to check the normalization

role="math" localid="1656062511219" $$\begin{gathered} \iint {\left[Y_0^0\right]}^2\sin \theta d\theta d\varphi =\frac{1}{4\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}\sin \theta d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \\ \iint {\left[Y_0^0\right]}^2\sin \theta d\theta d\varphi =\frac{1}{4\mathrm{\pi }}2\left(2\mathrm{\pi }\right) \\ \iint {\left[Y_0^0\right]}^2\sin \theta d\theta d\varphi =1 \\ \iint {\left[Y_0^0\right]}^2\sin \theta d\theta d\varphi =\frac{15}{4\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}{\sin }^2\theta {\cos }^2\theta \sin \theta d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \\ \iint {\left[Y_0^0\right]}^2\sin \theta d\theta d\varphi =\frac{15}{4\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}\left(1-{\cos }^2\theta \right){\cos }^2\theta \sin \theta d\theta {\int }_0^{2\mathrm{\pi }}d\varphi \\ y=\cos \theta ,dy=-\sin \theta \\ \iint {\left[Y_2^1\right]}^2\sin \theta d\theta d\varphi =\frac{15}{4}{\int }_1^{-1}{y}^2\left(1-y^2\right)dy \\ \iint {\left[Y_2^1\right]}^2\sin \theta d\theta d\varphi =\frac{15}{4}{{[\frac{y^3}{3}-\frac{y^5}{5}]}^1}_{-1}\iint {\left[Y_2^1\right]}^2\sin \theta d\theta d\varphi =1 \end{gathered}$$

Finally, we need to check the orthonormality as

$\iint Y_0^0{Y}_2^1\sin \theta d\theta d\varphi =-\frac{1}{\sqrt{4\mathrm{\pi }}}\sqrt{\frac{15}{8\mathrm{\pi }}}{\int }_0^{\mathrm{\pi }}\sin \theta \cos \theta \sin \theta d\theta {\int }_0^{2\mathrm{\pi }}{e}^{i\varphi }d\varphi$

The first and second integral vanishes thus,

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