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.

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

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,

localid="1656058765024" $Y_0^0=\frac{1}{\sqrt{4\mathrm{Ï€}}}$

Repeat the same procedure,

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

But

$x=\cos \mathrm{θ}$

So,

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

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="1656062511219" $$\begin{gathered} âˆ\neg {\left[Y_0^0\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{1}{4\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}\sin \mathrm{θ}d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}d\mathrm{Ï•} \\ âˆ\neg {\left[Y_0^0\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{1}{4\mathrm{Ï€}}2\left(2\mathrm{Ï€}\right) \\ âˆ\neg {\left[Y_0^0\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=1 \\ âˆ\neg {\left[Y_0^0\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}{\sin }^2\mathrm{θ}{\cos }^2\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}d\mathrm{Ï•} \\ âˆ\neg {\left[Y_0^0\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}\left(1-{\cos }^2\mathrm{θ}\right){\cos }^2\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}d\mathrm{Ï•} \\ y=\cos \mathrm{θ},dy=-\sin \mathrm{θ} \\ âˆ\neg {\left[Y_2^1\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4}{∫}_1^{-1}{y}^2\left(1-y^2\right)dy \\ âˆ\neg {\left[Y_2^1\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=\frac{15}{4}{{[\frac{y^3}{3}-\frac{y^5}{5}]}^1}_{-1}âˆ\neg {\left[Y_2^1\right]}^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=1 \end{gathered}$$

Finally, we need to check the orthonormality as

$âˆ\neg Y_0^0{Y}_2^1\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}=-\frac{1}{\sqrt{4\mathrm{Ï€}}}\sqrt{\frac{15}{8\mathrm{Ï€}}}{∫}_0^{\mathrm{Ï€}}\sin \mathrm{θ}\cos \mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}{e}^{i\mathrm{Ï•}}d\mathrm{Ï•}$

The first and second integral vanishes thus,

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