Question

Use Equation 4.32 to construct ${\mathit{Y}}_{\mathbf{l}}^{\mathbf{l}}\mathbf{(}\mathit{\theta }\mathbf{,}\mathit{\varphi }\mathbf{)}$and${\mathbf{y}}_{\mathbf{3}}^{\mathbf{2}}\mathbf{(}\mathbf{\theta }\mathbf{.}\mathbf{\varphi }\mathbf{)}$ . (You can take ${\mathit{P}}_{\mathbf{3}}^{\mathbf{2}}$from Table 4.2, but you'll have to work out${\mathit{P}}_{\mathbf{l}}^{\mathbf{l}}$ from Equations 4.27 and 4.28.) Check that they satisfy the angular equation (Equation 4.18), for the appropriate values of l and m .

Short answer

The angular equation is

$$\begin{gathered} Y_l^l(\theta ,\phi )=\frac{1}{l!}\sqrt{\frac{(2l+1)!}{4\mathrm{\pi }}}{(-\frac{1}{2}{e}^{i\phi }\sin \theta )}^l \\ {Y}_l^l(\theta ,\phi )=\frac{1}{4}\sqrt{\frac{105}{2\mathrm{\pi }}}{e}^{2i\phi }{\sin }^2\theta \cos \theta \end{gathered}$$

Step-by-step solution

Define the Schrodinger equation

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.

Determine the terms in equation

Equations 4.27, 4.28 and 4.32, are given by:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{l}}^{\mathrm{m}}\left(\mathrm{x}\right)={\left(1-{\mathrm{x}}^2\right)}^{\left|\mathrm{m}\right|/2}{\left(\frac{\mathrm{d}}{\mathrm{dx}}\right)}^{\left|\mathrm{m}\right|} \\ {\mathrm{P}}_{\mathrm{l}}\left(\mathrm{x}\right){\mathrm{P}}_{\mathrm{l}}\left(\mathrm{x}\right)=\frac{1}{2^{\mathrm{l}}\mathrm{l}!}{\left(\frac{\mathrm{d}}{\mathrm{dx}}\right)}^{\mathrm{l}}{({\mathrm{x}}^2-1)}^{\mathrm{l}}{\mathrm{Y}}_{\mathrm{l}}^{\mathrm{m}}(\mathrm{\theta },\mathrm{\varphi })=\mathrm{\sigma }\sqrt{\frac{(2\mathrm{l}+1)}{4\mathrm{\pi }}\frac{(\mathrm{l}-\left|\mathrm{m}\right|)!}{(\mathrm{l}+\left|\mathrm{m}\right|)!}}{\mathrm{e}}^{\mathrm{im\varphi }}{\mathrm{P}}_{\mathrm{l}}^{\mathrm{m}}\left(\mathrm{cos\theta }\right) \end{gathered}$$

Using these equations, we need to construct $Y_l^l$and $Y_3^2$, from (3) we have:

${\mathrm{Y}}_{\mathrm{l}}^{\mathrm{l}}={(-1)}^{\mathrm{l}}\sqrt{\frac{(2\mathrm{l}+1)}{4\mathrm{\pi }}\frac{1}{\left(2\mathrm{l}\right)!}}{\mathrm{e}}^{\mathrm{il\varphi }}{\mathrm{P}}_{\mathrm{l}}^{\mathrm{l}}\left(\cos \right(\mathrm{\theta }\left)\right)$

Using (1) we can write as:

${\mathrm{P}}_{\mathrm{l}}^{\mathrm{l}}\left(\mathrm{x}\right)={(1-{\mathrm{x}}^2)}^{1/2}{\left(\frac{\mathrm{d}}{\mathrm{dx}}\right)}^{\mathrm{l}}{\mathrm{P}}_{\mathrm{l}}\left(\mathrm{x}\right)$

Substitute from (2) with $P_l$

${\mathrm{P}}_{\mathrm{l}}^{\mathrm{l}}\left(\mathrm{x}\right)=\frac{1}{2^{\mathrm{l}}\mathrm{l}!}{(1-{\mathrm{x}}^2)}^{1/2}{\left(\frac{\mathrm{d}}{\mathrm{dx}}\right)}^{2\mathrm{l}}{({\mathrm{x}}^2-1)}^{\mathrm{l}}$

but ${({\mathrm{x}}^2-1)}^{\mathrm{l}}={\mathrm{x}}^{2\mathrm{l}}+....,$, where the rest of the term has a power less than 2l, which means that when we differentiate localid="1658135583492" ${(x^2-1)}^l,2l$ times then all the terms vanishes except the first term with the power of $2l$, so:

$P_l^l\left(x\right)=\frac{1}{2^{l}l!}{(1-x^2)}^{1/2}{\left(\frac{d}{dx}\right)}^{2l}{x}^{2l}$

but,

localid="1658135800919" ${\left(\frac{\mathrm{d}}{\mathrm{dx}}\right)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}-\mathrm{n}!$

thus:

${\mathrm{P}}_{\mathrm{l}}^{\mathrm{l}}=\frac{\left(2\mathrm{l}\right)!}{2^{\mathrm{l}}\mathrm{l}!}{(1-{\mathrm{x}}^2)}^{1/2}$

for $x=\cos \left(\theta \right),1-x^2={\sin }^2\left(\theta \right)$we get:

${\mathrm{P}}_{\mathrm{l}}^{\mathrm{l}}=\frac{\left(2\mathrm{l}\right)!}{2^{\mathrm{l}}\mathrm{l}!}{\sin }^{\mathrm{l}}\left(\mathrm{\theta }\right)$

Substitute into (4) with $P_l^l$, so we get:

$$\begin{gathered} Y_l^l={(-1)}^l\sqrt{\frac{(2l+1)}{4\mathrm{\pi }\left(2\mathrm{l}\right)!}}{e}^{il\varphi }\frac{\left(2l\right)}{2^{l}l!}{\sin }^l\left(0\right) \\ ={(-1)}^l\sqrt{\frac{\left(2l\right)!(2l+1)}{4\mathrm{\pi }}}{e}^{il\varphi }\frac{\left(2l\right)}{2^{l}l!}{\sin }^l\left(\theta \right) \\ =\frac{1}{l!}\sqrt{\frac{(2l+1)}{4\mathrm{\pi }}}{\left(-\frac{1}{2}{e}^{i\varphi }\sin \theta \right)}^l \\ {Y}_l^l=\frac{1}{l!}\sqrt{\frac{(2l+1)!}{4\mathrm{\pi }}}{\left(-\frac{1}{2}{e}^{i\varphi }\sin \theta \right)}^l \end{gathered}$$

Determine the terms in Schrodinger equation

Now we need to find$y_3^2$, following the same method, from equations (1), (2) and (3) we have:

$$\begin{gathered} Y_3^2=\sqrt{\frac{7}{4\mathrm{\pi }}.\frac{1}{5!}}{e}^{2i\varphi }{p}_3^2\left(\cos \right(\theta \left)\right) \\ {P}_3^2\left(x\right)=(1-x^2){\left(\frac{d}{dx}\right)}^2{p}_3\left(x\right) \\ {p}_3\left(x\right)=\frac{1}{8.3!}{\left(\frac{d}{dx}\right)}^3{(x^2-1)}^3 \end{gathered}$$

First we do the differentiation in the last equation as:

$$\begin{gathered} P_3=\frac{1}{8.3.2}{\left(\frac{d}{dx}\right)}^2\left[6x{(x^2-1)}^2\right] \\ =\frac{1}{8}\frac{d}{dx}\left[{(x^2-1)}^2+4x^2(x^2-1)\right] \\ =\frac{1}{8}\left[4x(x^2-1)+8x(x^2-1)+4x^2.2x\right] \\ =\frac{1}{2}(x^3-x+2x^3-2x+2x^3) \\ =\frac{1}{2}(5x^3-3x) \end{gathered}$$

Then we substitute into the second one, and also do the differentiation as:

$$\begin{gathered} P_3^2\left(x\right)=\frac{1}{2}(1-x^2){\left(\frac{d}{dx}\right)}^2(5x^3-3x) \\ =\frac{1}{2}(1-x^2)\frac{d}{dx}(15x^2-3) \\ =\frac{1}{2}(1-x^2)30x \\ =15x(1-x^2) \end{gathered}$$

for$x=\cos \left(\theta \right),1-x^2={\sin }^2\left(\theta \right)$we get:

$P_3^2\left(\cos \right(\theta \left)\right)={\sin }^2\left(\theta \right)\cos \left(\theta \right)$

Now substitute into the first one, so we get:

$$\begin{gathered} Y_3^2=\sqrt{\frac{7}{4\mathrm{\pi }}\frac{1}{5!}}15e^{2i\varphi }\cos \left(\theta \right){\sin }^2\left(\theta \right) \\ =\frac{1}{4}\sqrt{\frac{105}{2\mathrm{\pi }}}{e}^{2i\varphi }{\sin }^2\left(\theta \right)\cos \left(\theta \right) \\ {Y}_3^2=\frac{1}{4}\sqrt{\frac{105}{2\mathrm{\pi }}}{e}^{2i\varphi }{\sin }^2\left(\theta \right)\cos \left(\theta \right) \end{gathered}$$