91影视

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.

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

Using these three equations we have to construct

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

Combine the above we get,

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

Repeat the same procedure,

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

But

$x=\cos 胃$

So,

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

Thus,

${Y^1}_2=-\sqrt{\frac{15}{8\mathrm{蟺}}}{e}^{i蠒}\left(\sin 胃\right)\left(\cos 胃\right)$

Now we need to check the normalization

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

Finally, we need to check the orthonormality as

$鈭�Y_0^0{Y}_2^1\sin 胃d胃d蠒=-\frac{1}{\sqrt{4\mathrm{蟺}}}\sqrt{\frac{15}{8\mathrm{蟺}}}{鈭�}_0^{\mathrm{蟺}}\sin 胃\cos 胃\sin 胃d胃{鈭�}_0^{2\mathrm{蟺}}{e}^{i蠒}d蠒$

The first and second integral vanishes thus,

$鈭�Y_0^0{Y}_2^1\sin 胃d胃d蠒=0$