91Ó°ÊÓ

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.

$$\begin{gathered} {{P_l}^m}^{\text{'}}\left(x\right)=(1-x^2{)}^{\frac{m}{2}}(\frac{d}{dx}{)}^m{P}_l\left(x\right) \\ {P}_l\left(x\right)=\frac{1}{2^l\overline{)!}}\left(\frac{d}{dx}\right){\left(x^2-1\right)}^l \\ {{Y^m}_l}_l(\mathrm{θ},\mathrm{Ï•})=\mathrm{Î\mu }\sqrt{\frac{(2l+1)(l-m)!}{4\mathrm{Ï€}(l+m)!}}{e}^{im\mathrm{Ï•}}{P}_l^m\left(cos\mathrm{θ}\right) \end{gathered}$$

Using these three equations we have to construct

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

Combine the above we get,

$Y_0^0=\frac{1}{\sqrt{4\mathrm{Ï€}}}$

Repeat the same procedure,

_{$$\begin{gathered} {Y^1}_2=-\sqrt{\frac{5â‹\dots 1}{4\mathrm{Ï€}â‹\dots 3â‹\dots 2}}{e}^{i\mathrm{Ï•}}{P}_2^1\left(cos\mathrm{θ}\right) \\ {P}_2^1\left(x\right)=\sqrt{1-x^2}\frac{d}{dx}{P}_2\left(x\right) \\ {P}_2\left(x\right)=\frac{1}{4â‹\dots 2}{\left(\frac{d}{dx}\right)}^2(x^2-1) \\ {P}_2\left(x\right)=\frac{1}{8}\frac{d}{dx}\left[2\right(x^2-1\left)2x\right] \\ {P}_2\left(x\right)=\frac{1}{2}\left[(x^2-1)+X\left(2X\right)\right] \\ {P}_2\left(x\right)=\frac{1}{2}\left[3X^2-1\right] \\ {P}_2\left(x\right)=\sqrt{1-X^2}\frac{d}{dx}[\frac{3}{2}{x}^2-\frac{1}{2}] \\ {P}_2\left(x\right)=\sqrt{1-X^2}3x \end{gathered}$$}

But

_{$x=\cos \mathrm{θ}$}

So,

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

Thus,

${Y^1}_2=-\sqrt{\frac{15}{8\mathrm{π}}}{e}^i\mathrm{ϕ}\left(sin\mathrm{θ}\right)\left(cos\mathrm{θ}\right)$

Now we need to check the normalization

_{role="math" localid="1658466244130" $$\begin{gathered} âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{1}{4\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}sin\mathrm{θ}d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}d\mathrm{Ï•} \\ âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{1}{4\mathrm{Ï€}}2\left(2\mathrm{Ï€}\right) \\ âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=1 \\ âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}si{n}^2\mathrm{θ}c{o}^2\mathrm{θ}sin\mathrm{θ}d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}d\mathrm{Ï•} \\ âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}(1-co{s}^2\mathrm{θ})co{s}^2\mathrm{θ}sin\mathrm{θ}d\mathrm{θ}{∫}_0^2\mathrm{Ï€}d\mathrm{Ï•} \\ y=cos\mathrm{θ},dy=-sin\mathrm{θ} \\ âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4\mathrm{Ï€}}{∫}_1^{-1}{y}^2(1-y^2)dy \\ âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4\mathrm{Ï€}}{\left[\frac{y^3}{3}{\frac{y}{5}}^5\right]}^1-1 \\ âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=1 \end{gathered}$$}

Finally, we need to check the orthonormality as

_{role="math" localid="1658466365776" $âˆ\neg {\left[Y_0^0\right]}^{2}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{1}{4\mathrm{Ï€}}\sqrt{\frac{15}{8\mathrm{Ï€}}}{∫}_0^{\mathrm{Ï€}}sin\mathrm{θ}cos\mathrm{θ}sin\mathrm{θ}d\mathrm{θ}{∫}_0^2\mathrm{Ï€}{e}^i\mathrm{Ï•}d\mathrm{Ï•}$}

The first and second integral vanishes thus,

_{$âˆ\neg Y_0^0{y}_2^{1}sin\mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=0$}